PROBLEM 10.1
KNOWN: Water at 1 atm with Ts – Tsat = 10°C.
FIND: Show that the Jakob number is much less than unity; what is the physical significance of the
result; does result apply to other fluids?
ASSUMPTIONS: (1) Boiling situation, Ts > Tsat .
PROPERTIES: Table A-5 and Table A-6, (1 atm):
Water
Ethylene glycol
Mercury
R-12
hfg (kJ/kg)
cp,v (J/kg⋅K)
Tsat (K)
2257
812
301
165
2029
2742*
135.5*
1015*
373
470
630
243
* Estimated based upon value at highest temperature cited in Table A-5.
ANALYSIS: The Jakob number is the ratio of the maximum sensible energy absorbed by the vapor to
the latent energy absorbed by the vapor during boiling. That is,
(
)
Ja = c p ∆T / h fg = cp,v ∆Te / h fg
v
For water with an excess temperature ∆Ts = Te - T∞ = 10°C, find
Ja = ( 2029 J / k g ⋅ K ×10K ) /2257 ×103 J/kg
Ja = 0.0090.
Since Ja << 1, the implication is that the sensible energy absorbed by the vapor is much less than the
latent energy absorbed during the boiling phase change. Using the appropriate thermophysical
properties for three other fluids, the Jakob numbers are:
Ethylene glycol:
Ja = ( 2742J/kg ⋅ K ×10K ) /812 ×103 J / k g = 0.0338
<
Mercury:
Ja = (135.5J/kg ⋅ K ×10K ) /301 ×103 J / k g = 0.0045
<
Refrigerant, R-12:
Ja = (1015J/kg ⋅ K ×10K ) /165 ×103 J / k g = 0.0615
<
For ethylene glycol and R-12, the Jakob number is larger than the value for water, but still much less
than unity. Based upon these example fluids, we conclude that generally we’d expect Ja to be much
less than unity.
COMMENTS: We would expect the same low value of Ja for the condensation process since cp,g
and cp,f are of the same order of magnitude.
PROBLEM 10.2
KNOWN: Horizontal 20 mm diameter cylinder with ∆Te = Ts – Tsat = 5°C in saturated water, 1 atm.
FIND: Heat flux based upon free convection correlation; compare with boiling curve. Estimate
maximum value of the heat transfer coefficient from the boiling curve.
SCHEMATIC:
ASSUMPTIONS: (1) Horizontal cylinder, (2) Free convection, no bubble information.
3
PROPERTIES: Table A-6, Water (Saturated liquid, Tf = (Tsat + Ts)/2 = 102.5 °C ≈ 375K): ρ l = 956.9 kg/m , cp,l =
-6
4220 J/kg⋅K, µl = 274 × 10
2
-6
N⋅s/m , k l = 0.681 W/m⋅K, Pr = 1.70, β = 761 × 10
-1
K .
ANALYSIS: To estimate the free convection heat transfer coefficient, use the Churchill-Chu
correlation,
Nu D
hD
0.387Ra1D/ 6
=
= 0.60 +
8/27
k
1 + ( 0.559/Pr )9/16
2
.
Substituting numerical values, with ∆T = ∆Te = 5°C, find
Ra D =
g β ∆T D3
να
=
9.8m/s2 × 761 ×10− 6 K− 1 × 5° C( 0.020m) 3
274 ×10
−6
2
N ⋅ s / m /956.9 k g / m
3
× 1.686 ×10
3
−7
2
= 6.178 × 10 6
m /s
-7
2
where α = k/ρ cp = (0.681 W/m⋅K/956.9 kg/m × 4220 J/kg⋅K) = 1.686 × 10 m /s. Note that RaD is
within the prescribed limits of the correlation. Hence,
(
)
2
1/6
0.387 6.178 × 106
Nu D = 0.60 +
= 27.22
8/27
1 + ( 0.559/1.70) 9/16
k 27.22 × 0.681W/m ⋅ K
hfc = NuD =
= 928W/m2 ⋅ K.
D
0.020m
<
Hence, q ′′s = h fc∆ Te = 4640 W / m 2
From the typical boiling curve for water at 1 atm, Fig. 10.4, find at ∆Te = 5°C that
q ′′s ≈ 8.5 ×10 3 W / m2
<
The free convection correlation underpredicts (by 1.8) the boiling curve. The maximum value of hbc
can be estimated as
h max ≈ q ′′max / ∆Te =1.2 ×106 M W / m 2 /30 °C = 40,000W/m 2 ⋅ K.
COMMENTS: (1) Note the large increase in h with a slight change in ∆Te.
<
(2) The maximum value of h occurs at point P on the boiling curve.
PROBLEM 10.3
KNOWN: Spherical bubble of pure saturated vapor in mechanical and thermal equilibrium with its
liquid.
FIND: (a) Expression for the bubble radius, (b) Bubble vapor and liquid states on a p-v diagram; how
changes in these conditions cause bubble to collapse or grow, and (c) Bubble size for specified
conditions.
SCHEMATIC:
ASSUMPTIONS: (1) Liquid-vapor medium, (2) Thermal and mechanical equilibrium.
PROPERTIES: Table A-6, Water (Tsat = 101°C = 374.15K): psat = 1.0502 bar; (Tsat = 100°C =
-3
373.15K): psat = 1.0133 bar, σ = 58.9 × 10 N/m.
ANALYSIS: (a) For mechanical equilibrium, the difference in pressure between the vapor inside the
bubble and the liquid outside the bubble will be offset by the surface tension of the liquid-vapor
interface. The force balance follows from the free-body diagram shown above (right),
( )
( )
Fst = π rb2 ∆p = (p i − p o ) π rb2
( 2π rb )σ
( )
= π rb2 (p i − p o )
rb = 2σ / ( p i − p o )
(1)
Thermal equilibrium requires that the temperatures of the vapor and liquid be equal. Since the vapor
inside the bubble is saturated, pi = psat,v (T). Since po < pi, it follows that the liquid outside the bubble
must be superheated; hence, po = pl (T), the pressure of superheated liquid at T. Hence, we can
write,
(
rb = 2σ / p sat,v − pl
)
(2)
<
(b) The vapor [1] and liquid [2] states are represented on the following p-v diagram. Thermal
equilibrium requires both the vapor and liquid to be at the same temperature [3]. But mechanical
equilibrium requires that the outside liquid pressure be less than the inside vapor pressure [4]. Hence
the liquid must be in a superheated state. That is, its saturation temperature, Tsat (po) [5] is less than
Tsat (pi); Tl = Tsat (po) and po = p l .
Continued …..
PROBLEM 10.3 (Cont.)
<
The equilibrium condition for the bubble is unstable. Consider situations for which the pressure of the
surrounding liquid is greater or less than the equilibrium value. These situations are presented on
portions of the p-v diagram
When p ′o < p o, Tl′ < Tsat,v and
heat must be transferred out of
the bubble and vapor condenses.
Hence, the bubble collapses.
A similar argument for the condition p ′o > p o leads to Tl′ > Tsat,v and heat is transferred into the
bubble causing evaporation with the formation of vapor. Hence, the bubble begins to grow.
(c) Consider the specific conditions
Tsat,v = 101°C
and
Tl = Tsat ( po ) = 100°C
and calculate the radius of the bubble using the appropriate properties in Eq. (2).
rb = 2 × 58.9 ×10 −3
rb = 0.032mm.
N
N
/ (1.0502 −1.0133 ) bar × 10 5
/bar
m
m2
<
Note the small bubble size. This implies that nucleation sites of the same magnitude formed by pits and
crevices are important in promoting the boiling process.
PROBLEM 10.4
KNOWN: Long wire, 1 mm diameter, reaches a surface temperature of 126°C in water at 1 atm
while dissipating 3150 W/m.
FIND: (a) Boiling heat transfer coefficient and (b) Correlation coefficient, Cs,f, if nucleate boiling
occurs.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Nucleate boiling.
3
PROPERTIES: Table A-6, Water (saturated, 1 atm): Ts = 100°C, ρ l = 1/vf = 957.9 kg/m , ρ f =
3
-6
2
1/vg = 0.5955 kg/m , cp,l = 4217 J/kg⋅K, µl = 279 × 10 N⋅s/m , Prl = 1.76, hfg = 2257 kJ/kg, σ =
-3
58.9× 10 N/m.
ANALYSIS: (a) For the boiling process, the rate equation can be rewritten as
h = q′′s / ( Ts − Tsat ) =
h=
q ′s
/ ( Ts − Tsat )
πD
3150W/m
W
/ (126 − 100 ) °C = 1.00 × 106
/26°C = 38,600W/m2 ⋅ K.
2
π × 0.001m
m
<
Note the heat flux is very close to q ′′max , and nucleate boiling does exist.
(b) For nucleate boiling, the Rohsenow correlation may be solved for Cs,f to give
1/3
1/6
g ( ρl − ρ v ) c p,l ∆Te
µ l h fg
.
Cs,f =
n
′′
q
σ
s
h fg Prl
Assuming the liquid-surface combination is such that n = 1 and substituting numerical values with ∆Te =
Ts –Tsat , find
1/6
1/3 9.8 m ( 957.9 − 0.5955 ) kg
279 × 10−6 N ⋅ s / m 2 × 2257 ×103 J/kg
s2
m3
Cs,f =
×
1.00 × 106 W / m 2
58.9 × 10− 3 N / m
4217J/kg ⋅ K × 26K
2257 ×103 J / k g ×1.76
Cs,f = 0.017.
<
COMMENTS: By comparison with the values of Cs,f for other water-surface combinations of Table
10.1, the Cs,f value for the wire is large, suggesting that its surface must be highly polished. Note that
the value of the boiling heat transfer coefficient is much larger than values common to single-phase
convection.
PROBLEM 10.5
KNOWN: Nucleate pool boiling on a 10 mm-diameter tube maintained at ∆Te = 10°C in water at 1
atm; tube is platinum-plated.
FIND: Heat transfer coefficient.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Nucleate pool boiling.
3
PROPERTIES: Table A-6, Water (saturated, 1 atm): Ts = 100°C, ρ l = 1/vf = 957.9 kg/m , ρ v =
3
-6
2
1/vg = 0.5955 kg/m , cp,l = 4217 J/kg⋅K, µl = 279 × 10 N⋅s/m , Prl = 1.76, hfg = 2257 kJ/kg, σ =
-3
58.9 × 10 N/m.
ANALYSIS: The heat transfer coefficient can be estimated using the Rohsenow nucleate-boiling
correlation and the rate equation
1/2
µl hfg g ( ρ l − ρ v )
q′′
h= s =
∆ Te
∆ Te
σ
3
.
Cs,f h fg Pr n
l
cp,l ∆ Te
From Table 10.1, find Cs,f = 0.013 and n = 1 for the water-platinum surface combination. Substituting
numerical values,
1/2
2
3
279 ×10−6 N ⋅ s / m 2 × 2257 ×103 J/kg 9.8m/s ( 957.9 − 0.5955 ) k g / m
h=
−
3
10K
58.9 × 10 N / m
3
4217J/kg ⋅ K ×10K
0.013 × 2257 ×103 J / k g × 1.76
h = 13,690 W / m 2 ⋅ K.
×
<
COMMENTS: For this liquid-surface combination, q ′′s = 0.137MW/m2 , which is in general
agreement with the typical boiling curve of Fig. 10.4. To a first approximation, the effect of the tube
diameter is negligible.
PROBLEM 10.6
KNOWN: Water boiling on a mechanically polished stainless steel surface maintained at an excess
temperature of 15°C; water is at 1 atm.
FIND: Boiling heat transfer coefficient.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Nucleate pool boiling occurs.
3
PROPERTIES: Table A-6, Saturated water (1 atm): Tsat = 100°C, ρ l = 957.9 kg/m , ρ v = 0.596
3
-6
2
-3
kg/m , cp,l = 4217 J/kg⋅K, µl = 279 × 10 N⋅s/m , Prl = 1.76, σ = 58.9 × 10 N/m, hfg = 2257
kJ/kg.
ANALYSIS: The heat transfer coefficient can be expressed as
h = q ′′s / ∆ Te
where the nucleate pool boiling heat flux can be estimated using the Rohsenow correlation.
1/2
c ∆ Te
p,l
g ( ρl − ρ v )
q′′s = µ l hfg
σ
3
.
Cs,f h fg Pr n
l
From Table 10.1, find for this liquid-surface combination, Cs,f = 0.013 and n = 1, and substituting
numerical values,
1/2
9.8m/s2 ( 957.9 − 0.596) k g / m3
q′′s = 279 ×10−6 N ⋅ s / m 2 × 2257 ×103 J/kg
58.9 ×10−3 N / m
3
4217 J / k g ⋅ K× 15°C
0.013 × 2257kJ/kg ×1.76
×
q′′s = 461.9kW/m 2 .
Hence, the heat transfer coefficient is
h = 461.9 ×103 W / m 2 /15 °C = 30,790 W / m 2 ⋅ K.
<
COMMENTS: Note that this value of q ′′s for ∆Te = 15°C is consistent with the typical boiling curve,
Fig. 10.4.
PROBLEM 10.7
KNOWN: Simple expression to account for the effect of pressure on the nucleate boiling convection
coefficient in water.
FIND: Compare predictions of this expression with the Rohsenow correlation for specified ∆ Te and
pressures (2 and 5 bar) applied to a horizontal plate.
ASSUMPTIONS: (1) Steady-state conditions, (2) Nucleate pool boiling, (3) Cs,f = 0.013, n = 1.
3
PROPERTIES: Table A-6, Saturated water (2 bar): ρ l = 942.7 kg/m , cp,l = 4244.3 J/kg⋅K, µl =
-6
2
-3
3
230.7 × 10 N⋅s/m , Prl = 1.43, hfg = 2203 kJ/kg, σ = 54.97 × 10 N/m, ρ v = 1.1082 kg/m ; Saturated
3
-6
2
water (5 bar): ρ l = 914.7 kg/m , cp,l = 4316 J/kg⋅K, µl = 179 × 10 N⋅s/m , Prl = 1.13, hfg =
-3
3
2107.8 kJ/kg, σ = 48.4 × 10 N/m, ρ v = 2.629 kg/m .
ANALYSIS: The simple expression by Jakob [51] accounting for pressure effects is
n
0.4
h = C( ∆ Te ) ( p / pa )
(1)
where p and pa are the system and standard atmospheric pressures. For a horizontal plate, C = 5.56
and n = 3 for the range 15 < q ′′s < 2 3 5 k W / m2 . For ∆Te = 10°C,
p = 2 bar
h = 5.56 (10 )
3
( 2bar/1.0133bar )0.4 = 7,298W/m 2 ⋅ K,
q ′′s = 7 3 k W / m2
p = 5 bar
h = 5.56 (10 )
3
( 5bar/1.0133bar )0.4 = 10,529W/m 2 ⋅ K,
q ′′s = 105kW/m2
<
<
where q ′′s = h∆T e . The Rohsenow correlation, Eq. 10.5, with Cs,f = 0.013 and n = 1, is of the form
3
1/2
cp, l ∆Te
g ( ρl − ρv )
.
q′′s = µ l h fg
(2)
n
σ
C
h
Pr
s,f
fg
l
1/2
p = 2 bar:
q ′s = 230.7 × 10
−6
N⋅s
m
3
× 2203 ×10
2
9.8 m ( 942.7 − 1.1082) kg
2
3
J
s
m
−3
kg
54.97 × 10 N / m
4244.3J/kg ⋅ K × 10K
×
J
0.013 × 2203 × 10 3 × 1.431
kg
q′′s = 232 k W / m 2
p = 4 bar:
3
<
<
q′′s = 439 k W / m 2 .
COMMENTS: For ease of comparison, the results with pa = 1.0133 bar are:
(
Correlation/p (bar)
Simple
Rohsenow
q′′s k W / m2
)
1
2
56
73
135 232
4
105
439
Note that the range of q ′′s is within the limits of the Simple correlation. The comparison is poor and
therefore the correlation is not to be recommended. By manipulation of the Rohsenow results, find that
m
the (p/po) dependence provides m ≈ 0.75, compared to the exponent of 0.4 in the Simple correlation.
PROBLEM 10.8
KNOWN: Diameter of copper pan. Initial temperature of water and saturation temperature of
boiling water. Range of heat rates (1 ≤ q ≤ 100 kW).
FIND: (a) Variation of pan temperature with heat rate for boiling water, (b) Pan temperature shortly
after start of heating with q = 8 kW.
SCHEMATIC:
ASSUMPTIONS: (1) Conditions of part (a) correspond to steady nucleate boiling, (2) Surface of
pan corresponds to polished copper, (3) Conditions of part (b) correspond to natural convection from
a heated plate to an infinite quiescent medium, (4) Negligible heat loss to surroundings.
PROPERTIES: Table A-6, saturated water (Tsat = 100°C): ρ " = 957.9 kg / m3 , ρν = 0.60 kg / m3 ,
c p," = 4217 J / kg ⋅ K, µ" = 279 × 10
−6
6
2
N ⋅ s / m , Pr" = 1.76, h fg = 2.257 × 10 J / kg, σ = 0.0589 N / M.
3
-6
Table A-6, saturated water (assume Ts = 100°C, Tf = 60°C = 333 K): ρ = 983 kg/m , µ = 467 × 10
2
-6 -1
-6 2
N⋅s/m , k = 0.654 W/m⋅K, Pr = 2.99, β = 523 × 10 K . Hence, ν = 0.475 × 10 m /s, α = 0.159 ×
-6 2
10 m /s.
ANALYSIS: (a) From Eq. (10.5),
∆ Te = Ts − Tsat =
Cs,f h fg
Pr n
"
cp,"
1/ 3
q /µ h A
s " fg s
×
1/ 2
g ( ρ" − ρν ) / σ
2
2
For n = 1.0, Cs,f = 0.013 and As = πD /4 = 0.0707 m , the following variation of Ts with qs is
obtained.
S u tfa c e te m p e ra tu re (C )
125
120
115
110
<
105
100
0
20000
40000
60000
80000
100000
H e a t ra te (W )
As indicated by the correlation, the surface temperature increases as the cube root of the heat rate,
permitting large increases in q for modest changes in Ts. For q = 1 kW, Ts = 104.7°C, which is barely
sufficient to sustain boiling.
7
11
(b) Assuming 10 < RaL < 10 , the convection coefficient may be obtained from Eq. (9.31). Hence,
with L = As/P = D/4 = 0.075m,
Continued …..
PROBLEM 10.8 (Cont.)
1/ 3
9.8 m / s 2 × 523 × 10−6 K −1 ( T − T )( 0.075m )3
k
3 0.654 W / m ⋅ K
s
i
=
h = 0.15 Ra1/
0.15
L
−
12
4
0.075m
L
0.475 × 0.159 × 10
m / s2
(
= 1.308 2.86 × 107
2
)
1/ 3
(Ts − Ti )1/ 3 = 400 (Ts − Ti )1/ 3
2
With As = πD /4 = 0.0707 m , the heat rate is then
(
)
q = hAs ( Ts − Ti ) = 400 W / m 2 ⋅ K 4 / 3 0.0707 m 2 ( Ts − Ti )
4/3
With q = 8000 W,
<
Ts = Ti + 69°C = 89°C
9
COMMENTS: (1) With (Ts – Ti) = 69°C, RaL = 1.97 × 10 , which is within the assumed Rayleigh
number range. (2) The surface temperature increases as the temperature of the water increases, and
bubbles may nucleate when it exceeds 100°C. However, while the water temperature remains below
the saturation temperature, the bubbles will collapse in the subcooled liquid.
PROBLEM 10.9
KNOWN: Fluids at 1 atm: mercury, ethanol, R-12.
FIND: Critical heat flux; compare with value for water also at 1 atm.
ASSUMPTIONS: (1) Steady-state conditions, (2) Nucleate pool boiling.
PROPERTIES: Table A-5 and Table A-6 at 1 atm,
ρv
hfg
3
(kJ/kg)
Mercury
Ethanol
R-12
Water
3
ρl
(kg/m )
301
846
165
2257
3.90
1.44
6.32
0.596
12,740
757
1,488
957.9
σ × 10
Tsat
(N/m)
(K)
417
17.7
15.8
58.9
630
351
243
373
ANALYSIS: The critical heat flux can be estimated by the modified Zuber-Kutateladze correlation,
Eq. 10.7,
1/4
σ g ( ρl − ρ v )
q′′max = 0.149 h fg ρv
ρ 2v
.
To illustrate the calculation procedure, consider numerical values for mercury.
q′′max = 0.149 × 301 ×103 J / k g × 3.90kg/m3 ×
1/4
417 ×10−3 N / m × 9.8m/s 2 (12,740 − 3.90 ) k g / m3
2
3
3.90kg/m
(
)
q′′max = 1.34 M W / m 2 .
For the other fluids, the results are tabulated along with the ratio of the critical heat fluxes to that for
water.
Flluid
(
q′′max M W / m2
)
q′′max / q′′max,water
Mercury
Ethanol
1.34
0.512
1.06
0.41
R-12
Water
0.241
1.26
0.19
1.00
<
COMMENTS: Note that, despite the large difference between mercury and water properties, their
critical heat fluxes are similar.
PROBLEM 10.10
KNOWN: Copper pan, 150 mm diameter and filled with water at 1 atm, is maintained at 115°C.
FIND: Power required to boil water and the evaporation rate; ratio of heat flux to critical heat flux;
pan temperature required to achieve critical heat flux.
SCHEMATIC:
ASSUMPTIONS: (1) Nucleate pool boiling, (2) Copper pan is polished surface.
3
3
PROPERTIES: Table A-6, Water (1 atm): Tsat = 100°C, ρ l = 957.9 kg/m , ρ v = 0.5955 kg/m ,
-6
2
-3
cp,l = 4217 J/kg⋅K, µl = 279 × 10 N⋅s/m , Prl = 1.76, hfg = 2257 kJ/kg, σ = 58.9 × 10 N/m.
ANALYSIS: The power requirement for boiling and the evaporation rate can be expressed as
follows,
q boil = q′′s ⋅ As
& = q boil / h fg.
m
The heat flux for nucleate pool boiling can be estimated using the Rohsenow correlation.
3
1/2
g ( ρ l − ρ v ) c p,l ∆Te
q′′s = µ l h fg
.
σ
Cs,f h fg Prln
Selecting Cs,f = 0.013 and n = 1 from Table 10.1 for the polished copper finish, find
q ′′s = 279 × 10
−6
9.8 m ( 957.9 − 0.5955 ) kg
s2
3
N ⋅s
3 J
m
× 2257 × 10
2
−3
kg
m
589 × 10 N / m
1/2
J
4217
× 15 °C
kg ⋅ K
0.013 × 2257 × 103 J × 1.76
kg
3
q′′s = 4.619 ×105 W / m 2 .
The power and evaporation rate are
q boil = 4.619 ×105 W / m 2 ×
π
( 0.150m )2 = 8.16kW
4
<
& boil = 8.16kW/2257 ×103 J / k g = 3.62 × 10−3 k g / s = 13kg/h.
m
<
The maximum or critical heat flux was found in Example 10.1 as
q′′max = 1.26MW/m 2 .
Hence, the ratio of the operating to maximum heat flux is
q′′s
= 4.619 × 105 W / m 2 /1.26MW/m 2 = 0.367.
′′
qmax
From the boiling curve, Fig. 10.4, ∆Te ≈ 30°C will provide the maximum heat flux.
<
<
PROBLEM 10.11
KNOWN: Nickel-coated heater element exposed to saturated water at atmospheric pressure;
thermocouple attached to the insulated, backside surface indicates a temperature To = 266.4°C when
7
3
the electrical power dissipation in the heater element is 6.950 × 10 W/m .
FIND: (a) From the foregoing data, calculate the surface temperature, Ts, and the heat flux at the
exposed surface, and (b) Using an appropriate boiling correlation, estimate the surface temperature
based upon the surface heat flux determined in part (a).
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Water exposed to standard atmospheric pressure
and uniform temperature, Tsat, and (3) Nucleate pool boiling occurs on exposed surface, (4) Uniform
volumetric generation in element, and (5) Backside of heater is perfectly insulated.
PROPERTIES: Table A-6, Saturated water, liquid (100°C): ρ" = 1/ vf = 957.9 kg / m3 ,
c p," = c p,f = 4.217 kJ / kg ⋅ K, µ" = µf = 279 × 10−6 N ⋅ s / m 2 , Pr" = Prf = 1.76, hfg = 2257 kJ/kg,
σ = 58.9 × 10
−3
3
N/m; Saturated water, vapor (100°C): ρv = 1/vg = 0.5955 kg/m .
ANALYSIS: (a) From Eq. 3.43, the temperature at the exposed surface, Ts, is
6.95 × 107 W / m3 ( 0.015 m )
2
qL
Ts = To −
= 266.4°C −
2k
2 × 50 W / m ⋅ K
2
<
Ts = 110.0°C
The heat flux at the exposed surface is
q′′s = q / L = 6.95 × 107 W / m3 / 0.015 m = 4.63 × 109 W / m 2
<
(b) Since ∆Te = Ts – Tsat = (110 – 100)°C = 10°C, nucleate pool boiling occurs and the Rohsenow
correlation, Eq. 10.5, with q′′s from part (a) can be used to estimate the surface temperature, Ts,c,
1/ 2
g ( ρ" − ρ v )
q′′s = µ" h fg
σ
3
c ∆T
p," e,c
C h Pr n
s,f fg "
From Table 10.1, for the water-nickel surface-fluid combination, Cs,f = 0.006 and n = 1.0.
Substituting numerical values, find ∆Te,c and Ts,c.
Continued …..
PROBLEM 10.11 (Cont.)
4.63 × 109 W / m 2 = 279 × 10−6 N ⋅ s / m 2 × 2257 × 103 J / kg
1/ 2
9.8 m / s 2 (957.9 − 0.5955 ) kg / m3
×
3
−
58.9 × 10 N / m
3
4.217 × 103 J / kg ⋅ K × ∆T
e,c
×
0.006 × 2257 ×103 J / kg ×1.76
∆Te,c = Ts,c − Tsat = 9.1°C
Ts,c = 109.1°C
<
COMMENTS: From the experimental data, part (a), the surface temperature is determined from the
conduction analyses as Ts = 110.0°C. Using the traditional nucleate boiling correlation with the
experiential value for the heat flux, the surface temperature is estimated as Ts,c = 109.1°C. The two
approaches provide excess temperatures that are 10.0 vs. 9.1°C, which amounts to nearly a 10%
difference.
PROBLEM 10.12
KNOWN: Chips on a ceramic substrate operating at power levels corresponding to 50% of the critical
heat flux.
FIND: (a) Chip power level and temperature rise of the chip surface, and (b) Compute and plot the chip
temperature Ts as a function of heat flux for the range 0.25 ≤ q′′s q′′max ≤ 0.90 .
SCHEMATIC:
ASSUMPTIONS: (1) Nucleate boiling, (2) Fluid-surface with Cs,f = 0.004, n = 1.7 for Rohsenow
correlation, (3) Backside of substrate insulated.
PROPERTIES: Table A-5, Refrigerant R-113 (1 atm): Tsat = 321 K = 48°C, ρ " = 1511 kg/m3 , ρv =
7.38 kg/m3 , hfg = 147 kJ/kg, σ = 15.9 × 10-3 N/m; R-113, sat. liquid (given, 321 K): cp," = 983.8
J/kg⋅K, µ " = 5.147 × 10-4 N⋅s/m2 , Pr " = 7.183.
ANALYSIS: (a) The operating power level (flux) is 0.50 q′′max , where the critical heat flux is
estimated from Eq. l0.7 for nucleate pool boiling,
1/ 4
q′′max = 0.149h fg ρ v σ g ( ρ" − ρ v ) ρ v2
1/ 4
2
J
kg
N
m
kg
kg
3
3
−
q′′max = 0.149 × 147 × 10
15.9 × 10
/ 7.38
× 7.38
× 9.8 (1511 − 7.38 )
kg
m
m3
s2
m3
m3
q′′max = 233kW m 2 .
Hence, the heat flux on a chip is 0.5 × 233 kW/m2 = 116 kW/m2 and the power level is
2
qchip = q′′s × As = 116 × 103 W m 2 × 25 mm2 10−3 m mm = 2.9 W.
)
(
<
To determine the chip surface temperature for this condition, use the Rohsenow equation to find ∆Te = Ts
- Tsat with qs′′ = 116 × 103 W/m2 . The correlation, Eq. 10.5, solved for ∆Te is
Cs,f h fg Pr"n q′′s
∆Te =
µ" h fg
cp,"
1/ 3
1/ 6
σ
g ( ρ" − ρ v )
1/ 3
116 ×103 W m 2 ⋅
5.147 × 10−4 N ⋅ s ×147 × 103 J
kg
m2
0.004 × 147 × 103 J kg ( 7.18 )
1.7
=
983.8 J kg ⋅ K
×
1/ 6
15.9 × 10−3 N m
9.8 m (1511 − 7.38 ) kg
s2
m3
= 19.9$ C.
Hence, the chip surface temperature is
Continued...
PROBLEM 10.12 (Cont.)
Ts = Tsat + ∆Te = 48$ C + 19.9$ C ≈ 68$ C.
<
(b) Using the IHT Correlations Tools, Boiling, Nucleate Pool Boiling -- Heat flux and Maximum heat
flux, the chip surface temperature, Ts, was calculated as a function of the ratio q′′s q′′max . The required
thermophysical properties as provided in the problem statement were entered directly into the IHT
workspace. The results are plotted below.
Chip temperature, Ts (C)
75
70
65
60
0.2
0.4
0.6
0.8
1
Ratio q''s/q''max
COMMENTS: (1) Refrigerant R-113 is attractive for electronic cooling since its boiling point is
slightly above room temperature. The reliability of electronic devices is highly dependent upon
operating temperature.
(2) A copy of the IHT Workspace model used to generate the above plot is shown below.
// Correlations Tool - Boiling, Nucleate pool boiling, Critical heat flux
q''max = qmax_dprime_NPB(rhol,rhov,hfg,sigma,g)
// Eq 10.7
g = 9.8
// Gravitational constant, m/s^2
/* Correlation description: Critical (maximum) heat flux for nucleate pool boiling (NPB). Eq 10.7, Table
10.1 . See boiling curve, Fig 10.4 . */
// Correlations Tool - Boiling, Nucleate pool boiling, Heat flux
qs'' = qs_dprime_NPB(Csf,n,rhol,rhov,hfg,cpl,mul,Prl,sigma,deltaTe,g) // Eq 10.5
//g = 9.8
// Gravitational constant, m/s^2
deltaTe = Ts - Tsat
// Excess temperature, K
Ts = Ts_C + 273
// Surface temperature, K
Ts_C = 68
// Surface temperature, C
//Tsat =
// Saturation temperature, K
/* Evaluate liquid(l) and vapor(v) properties at Tsat. From Table 10.1. */
// Fluid-surface combination:
Csf = 0.004
// Given
n = 1.7
// Given
/* Correlation description: Heat flux for nucleate pool boiling (NPB), water-surface combination (Cf,n), Eq
10.5, Table 10.1 . See boiling curve, Fig 10.4 . */
// Heat rates:
qsqm = qs'' / q''max
// Ratio, heat flux over critical heat flux
qsqm = 0.5
// Thermophysical Properties (Given):
Tsat = 321
// Saturation temperature, K
Tsat_C = Tsat - 273
// Saturation temperature, C
rhol = 1511
// Density, liquid, kg/m^3
rhov = 7.38
// Density, vapor, kg/m^3
hfg = 147000
// Heat of vaporization, J/kg
sigma = 15.9e-3
// Surface tension/ N/m
cpl = 983.3
// Specific heat, saturated liquid, J/kg.K
mul = 5.147e-4
// Viscosity, saturated liquid, N.s/m^2
Prl = 7.183
// Prandtl number, saturated liquid
PROBLEM 10.13
KNOWN: Saturated ethylene glycol at 1 atm heated by a chromium-plated heater of 200 mm
diameter and maintained at 480K.
FIND: Heater power, rate of evaporation, and ratio of required power to maximum power for critical
heat flux.
SCHEMATIC:
ASSUMPTIONS: (1) Nucleate pool boiling, (2) Fluid-surface, Cs,f = 0.010 and n = 1.
PROPERTIES: Table A-5, Saturated ethylene glycol (1atm): Tsat = 470K, hfg = 812 kJ/kg, ρ f =
3
-3
3
1111 kg/m , σ = 32.7 × 10 N/m; Saturated ethylene glycol (given, 470K): ρ v = 1.66kg/m , µl = 0.38
-3
2
× 10 N⋅s/m , cp,l = 3280 J/kg⋅K, Prl = 8.7, k l = 0.15 W/m⋅K.
ANALYSIS: The power requirement for boiling and the evaporation rate are qboil = q ′′s ⋅ A s and
& = q boil / h fg . Using the Rohsenow correlation,
m
1/2
g ( ρl − ρv )
q ′′s = µl h fg
σ
−3
q ′′s = 0.38 ×10
3
Cs,f h fg Prln
cp, l ∆Te
9 . 8 m / s2 ( 1111 − 1.66 ) k g / m
× 812 × 10
2
−3
kg
m
32.7 × 10 N / m
N ⋅s
q′′s = 1.78 ×104 W / m 2
3
3 1 / 2
3 J
3280J/kg ⋅ K ( 480 − 470 ) K
0.01 × 812 × 103 J ( 8.7 )1
kg
q boil = 1.78 ×10 4 W / m 2 ×π / 4 ( 0.200m ) = 559 W
2
<
<
& = 559 W / 8 1 2 ×103 J / k g = 6.89 ×10−4 kg/s.
m
For this fluid, the critical heat flux is estimated from Eq. 10.7,
1/4
q ′′max = 0.149 h fg ρv σ g ( ρl − ρ v ) / ρv2
1/4
−3
2
3
32.7
×
10
N
/
m
×
9.8m/s
1111
−
1.66
k
g
/
m
J
kg
(
)
q ′′max = 0.149 × 812 ×10 3 × 1.66
3
2
kg
3
m
1.66kg/m
(
)
q ′′max = 6.77 × 105 W / m2 .
Hence, the ratio of the operating heat flux to the critical heat flux is,
q′′s
1.78 × 10 4 W / m2
=
≈ 0.026
q′′max 6.77 ×105 W / m2
or
2.6%.
<
COMMENTS: Recognize that the results are crude approximations since the values for Cs,f and n
are just estimates. This fluid is not normally used for boiling processes since it decomposes at higher
temperatures.
PROBLEM 10.14
KNOWN: Copper tubes, 25 mm diameter × 0.75 m long, used to boil saturated water at 1 atm operating
at 75% of the critical heat flux.
FIND: (a) Number of tubes, N, required to evaporate at a rate of 750 kg/h; tube surface temperature, T s,
for these conditions, and (b) Compute and plot Ts and N required to provide the prescribed vapor
production as a function of the heat flux ratio, 0.25 ≤ q′′s q′′max ≤ 0.90 .
SCHEMATIC:
ASSUMPTIONS: (1) Nucleate pool boiling, (2) Polished copper tube surfaces.
PROPERTIES: Table A-6, Saturated water (100°C): ρ" = 957.9 kg/m3, cp," = 4217 J/kg⋅K, µ" =
279 × 10-6 N⋅s/m2, Pr" = 1.76, hfg = 2257 kJ/kg, σ = 58.9 × 10-3 N/m, ρv = 0.5955 kg/m3.
ANALYSIS: (a) The total number of tubes, N, can be evaluated from the rate equations
q = q′′s As = qs′′ Nπ DL
fg q′′s π DL .
N = mh
fg
q = mh
(1,2,3)
The tubes are operated at 75% of the critical flux (1.26 MW/m2, see Example 10.1). Hence, the heat flux
is
q′′s = 0.75 q′′max = 0.75 × 1.26 M W m 2 = 9.45 × 105 W m 2 .
Substituting numerical values into Eq. (3), find
N = 750 kg h (1h 3600s ) × 2257 × 103 J kg 9.45 × 105 W m 2 × π × 0.025 m × 0.75 m = 8.5 ≈ 9.
)
(
<
To determine the tube surface temperature, use the Rohsenow correlation,
Cs,f h fg Pr"n q′′s
∆Te =
µ" h fg
cp,"
1/ 3
1/ 6
σ
g ( ρ" − ρ v )
.
From Table 10.1 for the polished copper-water combination, Cs,f = 0.013 and n = 1.0.
1
3
5
2
∆Te =
0.013 × 2257 × 10 J kg (1.76 )
4217 J kg ⋅ K
1/ 3
279 × 10−6 N ⋅ s m 2 × 2257 × 103 J kg
9.45 × 10 W m
×
1/ 6
58.9 × 10 −3 N m
9.8 m s 2 (957.9 − 0.5955 ) kg m3
= 19.0$ C.
Hence,
$
Ts = Tsat + ∆Te = (100 + 19 ) C = 119$ C.
<
Continued...
PROBLEM 10.14 (Cont.)
(b) Using the IHT Correlations Tool, Boiling, Nucleate Pool Boiling, Heat flux and the Properties Tool
for Water, combined with Eqs. (1,2,3) above, the surface temperature Ts and N can be computed as a
function of q′′s q′′max . The results are plotted below.
Ts (C) and N
300
200
100
0
0.2
0.4
0.6
0.8
1
Ratio q''s/q''max
Tube surface temperature, Ts (C)
Number of tubes, N*10
Note that the tube surface temperature increases only slightly (113 to 120°C) as the ratio q′′s q′′max
increases. The number of tubes required to provide the prescribed evaporation rate decreases markedly
as q′′s q′′max increases.
COMMENTS: (1) The critical heat flux, q′′max =1.26 MW/m2, for saturated water at 1 atm is
calculated in Example 10.1 using the Zuber-Kutateladze relation, Eq. 10.7. The IHT Correlation Tool,
Boiling, Nucleate pool boiling, Maximum heat flux, with the Properties Tool for Water could also be
used to determine q′′max .
PROBLEM 10.15
KNOWN: Diameter and length of tube submerged in pressurized water. Flowrate and inlet
temperature of gas flow through the tube.
FIND: Tube wall and gas outlet temperatures.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) Uniform tube wall temperature, (3) Nucleate boiling at outer
surface of tube, (4) Fully developed flow in tube, (5) Negligible flow work and potential and kinetic
energy changes for tube flow, (7) Constant properties.
6
PROPERTIES: Table A-6, saturated water (psat = 4.37 bars): Tsat = 420 K, hfg = 2.123 × 10 J/kg,
3
3
ρ " = 919 kg / m , ρν = 2.4 kg/m , µ" = 185 × 10
−6
2
N ⋅ s / m , c p," = 4302 J / kg ⋅ K, Pr" = 2.123 ×
10 J / kg ⋅ K, σ = 0.0494 N/m. Table A-4, air ( p = 1atm, Tm ≈ 700 K ) : cp = 1075 J/kg⋅K, µ = 339 × 10
6
7
-
2
N⋅s/m , k = 0.0524 W/m⋅K, Pr = 0.695.
ANALYSIS: From an energy balance performed for a control surface that bounds the tube, we know
that the rate of heat transfer by convection from the gas to the inner surface must equal the rate of heat
transfer due to boiling at the outer surface. Hence, from Eqs. (8.44), (8.45) and (10.5), the energy
balance for a single tube is of the form
3
1/ 2
cp," ∆Te
∆To − ∆Ti
g ( ρ" − ρν )
hAs
(1)
= As µ" h fg
n
σ
ln ( ∆To / ∆Ti )
C
h
Pr
s,f
fg
"
where U = h
and
Cs,f = 0.013 and n = 1.0 from Table 10.1. The corresponding unknowns are the wall
temperature Ts and gas outlet temperature, Tm,o, which are also related through Eq. (8.43).
π DL
h
= exp −
m
cp
Ts − Tm,i
Ts − Tm,o
(2)
/ π Dµ = 119, 600, the flow is turbulent, and with n = 0.3, Eq. (8.60) yields,
For Re D = 4m
k 0.023 Re 4 / 5 Pr 0.3 = 0.0524 W / m ⋅ K 0.023 (119, 600 )4 / 5 ( 0.695 )0.3 = 502 W / m 2 ⋅ K
D
0.025m
D
h = h fd =
Solving Eqs. (1) and (2), we obtain
Ts = 152.6°C,
Tm,o = 166.7°C
<
c (Tm,i – Tm,o) = 45,930 W, and the total heat
COMMENTS: (1) The heat rate per tube is q = m
p
rate is Nq = 229,600 W, in which case the rate of steam production is m steam = q / h fg = 0.108 kg / s.
(2) It would not be possible to maintain isothermal tube walls without a large wall thickness, and Ts,
as well as the intensity of boiling, would decrease with increasing distance from the tube entrance.
However the foregoing analysis suffices as a first approximation.
PROBLEM 10.16
KNOWN: Nickel wire passing current while submerged in water at atmospheric pressure.
FIND: Current at which wire burns out.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Pool boiling.
ANALYSIS: The burnout condition
will occur when electrical power
dissipation creates a surface heat flux
exceeding the critical heat flux, q ′′max .
This burn out condition is illustrated on
the boiling curve to the right and in
Figure 10.6.
The criterion for burnout can be
Expressed as
q′elec = I 2 R′e .
q′′max ⋅ π D = q′elec
(1,2)
That is,
I = [q ′′maxπ D / R ′e ]
1/2
.
(3)
For pool boiling of water at 1 atm, we found in Example 10.1 that
q′′max = 1.26MW/m 2 .
Substituting numerical values into Eq. (3), find
1/2
I = 1.26 × 106 W / m 2 ( π × 0.001m ) /0.129Ω / m
= 175A.
<
COMMENTS: The magnitude of the current required to burn out the 1 mm diameter wire is very
large. What current would burn out the wire in air?
PROBLEM 10.17
KNOWN: Saturated water boiling on a brass plate maintained at 115°C.
2
FIND: Power required (W/m ) for pressures of 1 and 10 atm; fraction of critical heat flux at which plate is operating.
SCHEMATIC:
ASSUMPTIONS: (1) Nucleate pool boiling, (2) ∆Te = 15°C for both pressure levels.
3
PROPERTIES: Table A-6, Saturated water, liquid (1 atm, Tsat = 100°C): ρ l = 957.9 kg/m , cp,l = 4217 J/kg⋅K, µl
-6
=279 × 10
3
2
-3
N/m; Table A-6, Saturated water, vapor (1 atm): ρ v =
N⋅s/m , Pr = 1.76, h fg = 2257 kJ/kg, σ = 58.9 × 10
3
0.596 kg/m ; Table A-6, Saturated water, liquid (10 atm = 10.133 bar, Tsat = 453.4 K = 180.4°C): ρ l = 886.7 kg/m ,
-6
2
-3
cp,l = 4410 J/kg⋅K, µl = 149 × 10 N⋅s/m , Prl = 0.98, h fg = 2012 kJ/kg, σ = 42.2 × 10 N/m; Table A-6, Water,
3
vapor (10.133 bar): ρ v = 5.155 kg/m .
ANALYSIS: With ∆Te = 15°C, we expect nucleate pool boiling. The Rohsenow correlation with Cs ,f = 0.006 and n =
1.0 for the brass-water combination gives
1/2
g ( ρl − ρv )
q ′′s = µ l h fg
σ
3
C h Pr n
s,f fg l
c p,l ∆Te
1/2
1 atm: q ′′s = 279 × 10 −6 N ⋅ s / m 2 × 2257 × 10 3 J / k g
9 . 8 m / s2 ( 957.9 − 0.596 ) k g / m3
−3
58.9 × 10 N / m
4217 J / k g ⋅ K × 15K
0.006 × 2257 ×10 3J / kg ×1.761
10 atm:
×
3
2
= 4.70MW/m
q ′′s = 23.8MW/m2
From Example 10.1, q ′′max (1atm ) = 1 . 2 6 M W / m 2 . To find the critical heat flux at 10 atm, use the
correlation of Eq. 10.7,
1/4
q ′′max = 0.149 h fg ρ v σ g ( ρl − ρ v ) / ρ v2
.
q ′′max (1 0 a t m ) = 0.149 × 2012 × 10 J / k g × 5.155kg/m ×
3
3
1/4
42.2 × 10−3 N / m × 9 . 8 m / s 2 ( 886.7 − 5.16 ) k g / m3
2
3
5.155kg/m
(
)
2
= 2.97MW/m .
For both conditions, the Rohsenow correlation predicts a heat flux that exceeds the maximum heat flux, q ′′max . We
conclude that the boiling condition with ∆Te = 15°C for the brass-water combination is beyond the inflection point
3
(P, see Fig. 10.4) where the boiling heat flux is no longer proportional to ∆ Te .
′′ (1atm ) ≤ 1.26MW/m2
q ′′s ≈ q max
′′ (10 atm) ≤ 2.97MW/m2 .
q s′′ ≈ q max
<
PROBLEM 10.18
KNOWN: Zuber-Kutateladze correlation for critical heat flux, q ′′max .
FIND: Pressure dependence of q ′′max for water; demonstrate maximum value occurs at
approximately 1/3 pcrit ; suggest coordinates for a universal curve to represent other fluids.
ASSUMPTIONS: Nucleate pool boiling conditions.
PROPERTIES: Table A-6, Water, saturated at various pressures; see below.
ANALYSIS: The Z-K correlation for estimating the critical heat flux, has the form
1/4
g σ ( ρl − ρv )
q′′max = 0.149 ρv h fg
ρv2
where the properties for saturation conditions are a function of pressure. The properties (Table A-6)
and the values for q ′′max are as follows:
p
p/pc
ρl
3
(bar)
2
(MW/m )
1.01
11.71
26.40
44.58
61.19
82.16
123.5
169.1
221.2
ρv
(kg/m )
0.0045
0.053
0.120
0.202
0.277
0.372
0.557
0.765
1.000
957.9
879.5
831.3
788.1
755.9
718.4
648.9
562.4
315.5
0.5955
5.988
13.05
22.47
31.55
43.86
72.99
117.6
315.5
3
h
σ×10
(kJ/kg)
(N/m)
2257
1989
1825
1679
1564
1429
1176
858
0
58.9
40.7
31.6
24.5
19.7
15.0
8.4
3.5
0
q ′′max
1.258
3.138
3.935
4.398
4.549
4.520
4.047
2.905
0
The q ′′max values are plotted as a function of p/pc , where pc is the critical pressure. Note the rapid
decrease of hfg and σ with increasing pressure. The universal curve coordinates would be
q ′′max / q′′max (1 / 3 p crit ) vs. p / pc .
PROBLEM 10.19
KNOWN: Kutateladze’s dimensional analysis and the bubble diameter parameter.
FIND: (a) Verify the dimensional consistency of the critical heat flux expression, and (b) Estimate
heater size with water at 1 atm required such that the Bond number will exceed 3, i.e., Bo ≥ 3.
ASSUMPTIONS: Nucleate pool boiling.
ANALYSIS: (a) Kutateladze postulated that the critical flux was dependent upon four parameters,
(
q′′max = q′′max h fg , ρ v , σ , Db
)
where Db is the bubble diameter parameter having the form
1/2
Db = σ / g ( ρl − ρv )
.
(1)
The form of the critical heat flux expression was presumed to be
−1 / 2 σ 1 / 2
q′′max = C h fg ρ1/2
v Db
(2)
where C is a constant. It is not possible to derive this equation from a dimensional (Pi) analysis. We
can only determine that the equation is dimensionally consistent. Using SI units, check Eq. (1) for Db ,
(
)(
)(
)
)(
)(N
1/2
Db => Nm −1 m −1s 2 kg −1m 3
1/2
s2
=> N
m 3
2
kg ⋅ m
=> [m ]
and in Eq. (2) for q ′′max ,
(
)(
q ′′max => Jkg −1 kg1 / 2 m −3 / 2 m −1 / 2
1/2
m
−1 / 2
)
2 1 / 2
=> J ⋅ N ⋅ s m −2 => W .
s kg ⋅ m
m2
Hence, the equations are dimensionally consistent.
(b) The Bond number, Bo, is defined as the ratio of the characteristic length L (width or diameter) of
the heater surface to the bubble diameter parameter, Db . That is, Bo ≡ L/DB. The number squared is
also indicative of the ratio of the buoyant to capillary forces. For water at 1 atm (see Example 10.1 for
properties listing), Eq. (1) yields
1/2
N
m
kg
Db = 58.9 ×10−3 /9.8 ( 957.9 − 0.5955 )
m
s2
m3
= 0.0025m = 2.5mm.
Eq. 10.7 for the critical heat flux is appropriate for an “infinite” heater (Bo ≥ 3). To meet this
requirement, the heater dimension must be
L ≥ Bo ⋅D b = 3 × 2.5mm = 7.5mm.
<
COMMENTS: As the heater size decreases (Bo decreasing), the boiling curve no longer exhibits the
characteristic q ′′max and q ′′min features. The very small heater, such as a wire, is enveloped with
vapor at small ∆Te and film boiling occurs.
PROBLEM 10.20
KNOWN: Lienhard-Dhir critical heat flux correlation for small horizontal cylinders.
FIND: Critical heat flux for 1 mm and 3 mm diameter horizontal cylinders in water at 1 atm.
SCHEMATIC:
ASSUMPTIONS: Nucleate pool boiling.
3
3
-3
PROPERTIES: Table A-6, Water (1 atm): ρ l = 957.9 kg/m , ρ v = 0.5955 kg/m , σ = 58.9 × 10
N/m.
ANALYSIS: The Lienhard-Dhir correlation for small horizontal cylinders is
q′′max = q′′max,Z 0.94 ( Bo ) −1/4
0.15 ≤ Bo ≤ 1.2
(1)
where q ′′max,Z is the critical heat flux predicted by the Zuber-Kutateladze correlation for the infinite
heater (Eq. 10.6) and the Bond number is
r
1/2
Bo =
= r / σ / g ( ρ l − ρ v ) .
(2)
Db
Note the characteristic length is the cylinder radius. From Example 10.1, using Eq. 10.6,
q′′max,Z = 1.11MW/m2
and substituting property values for water at 1 atm into Eq. (2),
1/2
N
m
kg
Db = 58.9 ×10−3 /9.8 ( 957.9 − 0.5955 )
m
s2
m3
= 2.51mm.
Substituting appropriate values into Eqs. (1) and (2),
1 mm dia cylinder
Bo = 0.5 mm/2.51 mm = 0.20
q′′max = 1.11MW/m 2 0.94 ( 0.20 )−1/4 = 1.56MW/m 2 .
3 mm dia cylinder
<
Bo = 1.5 mm/2.51 mm = 0.60
q′′max = 1.11MW/m 2 0.94 ( 0.60 )−1/4 = 1.19 M W / m 2 .
<
Note that for the 3 mm diameter cylinder, the critical heat flux is 1.19/1.11 = 1.07 times larger than the
value for a very large horizontal cylinder.
COMMENTS: For practical purposes a horizontal cylinder of diameter greater than 3 mm can be
considered as a very large one. The critical heat flux for a 1 mm diameter cylinder is 40% larger than
that for the large cylinder.
PROBLEM 10.21
KNOWN: Boiling water at 1 atm pressure on moon where the gravitational field is 1/6 that of the
earth.
FIND: Critical heat flux.
ASSUMPTIONS: Nucleate pool boiling conditions.
3
3
PROPERTIES: Table A-6, Water (1 atm): Tsat = 100°C, ρ l = 957.9 kg/m , ρ v = 0.5955 kg/m , hfg
-3
= 2257 kJ/kg, σ = 58.9 × 10 N/m.
ANALYSIS: The modified Zuber-Kutateladze correlation for the critical heat flux is Eq. 10.7.
1/4
q′′max = 0.149 ρ1/2
v hfg σ g ( ρl − ρv )
.
The relation predicts the critical flux dependency on the gravitational acceleration as
q′′max ~ g1/4.
2
It follows that if gmoon = (1/6) gearth and recognizing q ′′max,e = 1.26 MW/m for earth acceleration
(see Example 10.1),
1/4
q′′max,moon = q′′max,earth ( g moon / g earth )
q′′max,moon = 1.26
MW 1 1/4
= 0.81MW/m 2 .
2
m 6
<
COMMENTS: Note from the discussion in Section 10.4.5 that the g1/4 dependence on the critical
heat flux has been experimentally confirmed. In the nucleate pool boiling regime, the heat flux is nearly
independent of the gravitational field.
PROBLEM 10.22
KNOWN: Concentric stainless steel tubes packed with dense boron nitride powder. Inner tube has
3
heat generation while outer tube surface is exposed to boiling heat flux, q ′′s = C ( Ts − Tsat ) .
Saturation temperature of boiling liquid and temperature of the outer tube surface.
FIND: Expressions for the maximum temperature in the stainless steel (ss) tubes and in the boron
nitride (bn).
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional (cylindrical) steady-state heat
transfer in tubes and boron nitride.
ANALYSIS: Construct the thermal circuit shown above where R ′23 and R ′34 represent the
resistances due to the boron nitride between r2 and r3 and to the outer stainless steel tube, respectively.
From an overall energy balance,
q′gen = q′boil ,
(
)
q& π r22 − r12 = ( 2π r4 ) C ( Ts − Tsat )3 .
With prescribed values for Tsat , Ts and C, the required volumetric heating of the inner stainless steel
tube is
q& =
2r4
(r22 − r12 )
C ( Ts − Tsat )3 .
Using the thermal circuit, we can write expressions for the maximum temperature of the stainless steel
(ss) and boron nitride (bn).
Stainless steel: Tss,max occurs at r1. Using the results of Section 3.4.2, the temperature distribution in
a radial tube of inner and outer radii r1 and r2 is
T (r ) = −
q& 2
r + C1 ln r + C 2
2k ss
for which the boundary conditions are
BC#1:
r = r1
dT
=0
dr
2
&
qr
q&
C
0=−
2r1 + 1 + 0 → C1 = + 1
2k ss
r1
kss
Continued …..
PROBLEM 10.22 (Cont.)
T ( r2 ) = T2
r = r2
BC#2:
2
&
q& 2 qr
T2 = −
r2 + 1 lnr2 + C 2
2k ss
kss
& 12
q& 2 qr
C2 = T2 +
r −
l n r2
2k ss 2 k ss
Hence,
(
)
& 12
qr
q&
2
2
T (r ) = −
r − r2 +
ln ( r / r2 ) + T2.
2k ss
k ss
Using the thermal circuit, find T2 in terms of known parameters Ts, Tsat and C.
T2 − Ts
= ( 2π r4 ) C ( Ts − Tsat )3 .
R′23 + R′34
Hence, the maximum temperature in the inner stainless steel tube (r = r1) is
(
)
& 12
qr
q&
2
2
Tss,max = T ( r1 ) = −
r −r +
ln ( r1 / r2 ) + Ts
2k ss 1 2
k ss
+ ( R′23 + R′34 )( 2π r4 ) C ( Ts − Tsat )
3
<
where from Eq. 3.27
R ′23 =
ln ( r3 / r2 )
2π k bn
R ′34 =
ln ( r4 / r3 )
2π kss
.
Boron nitride: Tbn,max occurs at r1. Hence
Tbn,max = T ( r1 )
as derived above for the inner stainless steel tube.
<
PROBLEM 10.23
KNOWN: Thickness and thermal conductivity of a silicon chip. Properties of saturated fluorocarbon
liquid.
FIND: (a) Temperature at bottom surface of chip for a prescribed heat flux and 90% of CHF, (b) Effect
of heat flux on chip surface temperatures; maximum allowable heat flux.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Uniform heat flux and adiabatic sides, hence onedimensional conduction in chip, (3) Constant properties, (4) Nucleate boiling in liquid.
PROPERTIES: Saturated fluorocarbon (given): cp," = 1100 J/kg⋅K, hfg = 84,400 J/kg, ρ" = 1619.2
kg/m3, ρv = 13.4 kg/m3, σ = 8.1 × 10-3 kg/s2, µ" = 440 × 10-6 kg/m⋅s, Pr" = 9.01.
ANALYSIS: (a) Energy balances at the top and bottom surfaces yield q′′o = q′′cond = k s ( To − Ts ) L =
q′′s ; where Ts and q′′s are related by the Rohsenow correlation,
Cs,f h fg Pr"n q′′s
Ts − Tsat =
µ" h fg
cp,"
1/ 3
1/ 6
σ
g ( ρ" − ρ v )
Hence, for q′′s = 5 × 104 W/m2,
1/ 3
0.005 (84, 400 J kg ) 9.011.7
5 × 104 W m 2
Ts − Tsat =
440 ×10−6 kg m ⋅ s × 84, 400 J kg
1100 J kg ⋅ K
1/ 6
8.1×10−3 kg s 2
×
9.807 m s 2 (1619.2 − 13.4 ) kg m3
= 15.9$ C
$
Ts = (15.9 + 57 ) C = 72.9$ C .
From the rate equation,
q′′o L
5 × 104 W m 2× 0.0025 m
$
To = Ts +
= 72.9 C +
= 73.8$ C
ks
135 W m⋅ K
<
For a heat flux which is 90% of the critical heat flux (C1 = 0.9), it follows that
1/ 4
σ g ( ρ" − ρ v )
q′′o = 0.9q′′max = 0.9 × 0.149h fg ρ v
ρ v2
= 0.9 × 0.149 × 84, 400 J kg × 13.4 kg m3
Continued...
PROBLEM 10.23 (Cont.)
1/ 4
8.1×10−3 kg s 2 × 9.807 m s 2 (1619.2 − 13.4 ) kg m3
×
2
3
13.4 kg m
)
(
q′′o = 0.9 × 15.5 × 104 W m 2 = 13.9 × 104 W m 2
From the results of the previous calculation and the Rohsenow correlation, it follows that
(
∆Te = 15.9$ C q′′o 5 × 104 W m 2
)
1/ 3
= 15.9$ C (13.9 5 )
1/ 3
= 22.4$ C
Hence, Ts = 79.4°C and
To = 79.4$ C +
13.9 ×104 W m 2× 0.0025 m
= 82$ C
135 W m⋅ K
<
(b) Using the energy balance equations with the Correlations Toolpad of IHT to perform the parametric
calculations for 0.2 ≤ C1 ≤ 0.9, the following results are obtained.
82
Temperature (C)
80
78
76
74
72
70
30000
60000
90000
120000
150000
Chip heat flux, qo''(W/m^2)
To
Ts
The chip surface temperatures, as well as the difference between temperatures, increase with increasing
heat flux. The maximum chip temperature is associated with the bottom surface, and To = 80°C
corresponds to
q′′o,max = 11.3 × 104 W m 2
<
which is 73% of CHF ( q′′max = 15.5 × 104 W/m2).
COMMENTS: Many of today’s VLSI chip designs involve heat fluxes well in excess of 15 W/cm2, in
which case pool boiling in a fluorocarbon would not be an appropriate means of heat dissipation.
PROBLEM 10.24
KNOWN: Operating conditions of apparatus used to determine surface boiling characteristics.
FIND: (a) Nucleate boiling coefficient for special coating, (b) Surface temperature as a function of heat
flux; apparatus temperatures for a prescribed heat flux; applicability of nucleate boiling correlation for a
specified heat flux.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional, steady-state conduction in the bar, (2) Water is saturated at 1
atm, (3) Applicability of Rohsenow correlation with n = 1.
PROPERTIES: Table A.6, saturated water (100°C): ρ" = 957.9 kg/m3, cp," = 4217 J/kg⋅K, µ" =
279 × 10-6 N⋅s/m2, Pr" = 1.76, hfg = 2.257 × 106 J/kg, σ = 0.0589 N/m, ρ v = 0.5955 kg/m3.
ANALYSIS: (a) The coefficient Cs,f associated with Eq. 10.5 may be determined if qs′′ and Ts are
known. Applying Fourier’s law between x1 and x2,
$
(158.6 − 133.7 ) C = 6.64 × 105 W m2
T −T
q′′s = q′′cond = k 2 1 = 400 W m⋅ K ×
x 2 − x1
0.015 m
Since the temperature distribution in the bar is linear, Ts = T1 - (dT/dx)x1 = T1 - [(T2 - T1)/(x2 - x1)]x1.
Hence,
Ts = 133.7$ C − 24.9$ C 0.015 m 0.01m = 117.1$ C
From Eq. 10.5, with n = 1,
1/ 3
cp," ∆Te µ" h fg
Cs,f =
h fg Pr" q′′s
Cs,f =
(
$
4217 J kg ⋅ K 17.1 C
) 279 ×10
2.257 × 10 J kg (1.76 )
Cs,f = 0.0131
6
g ( ρ" − ρ v )
σ
1/ 6
−6
1/ 3
kg s ⋅ m × 2.257 × 10 J kg
6
5
6.64 × 10 W m
2
1/ 6
9.8 m s 2 × 957.3 kg m 3
2
0.0589 kg s
<
(b) Using the appropriate IHT Correlations and Properties Toolpads, the following portion of the
nucleate boiling regime was computed.
Continued...
PROBLEM 10.24 (Cont.)
1E6
800000
Heat flux, qs''(W/m^2)
600000
400000
200000
100000
8
10
12
14
16
18
20
Excess temperature, DeltaTe(C)
For qs′′ = 106 W/m2 = q′′cond , Ts = 119.6°C and
T1 = 144.6°C
and
<
T2 = 182.1°C
With the critical heat flux given by Eq. 10.7,
1/ 4
σ g ( ρ" − ρ v )
q′′max = 0.149h fg ρ v
ρ v2
1/ 4
2 × 9.8 m s 2 × 957.3kg m3
0.0589
kg
s
q′′max = 0.149 2.257 × 106 J kg 0.5955 kg m3
2
3
0.5955 kg m
(
)
(
)
q′′max = 1.25 × 106 W m 2
Since q′′s = 1.5 × 106 W/m2 > q′′max , the heat flux exceeds that associated with nucleate boiling and the
foregoing results can not be used.
COMMENTS: For q′′s > q′′max , conditions correspond to film boiling, for which Ts may exceed
acceptable operating conditions.
PROBLEM 10.25
KNOWN: Small copper sphere, initially at a uniform temperature, Ti, greater than that corresponding to
the Leidenfrost point, TD, suddenly immersed in a large fluid bath maintained at Tsat.
FIND: (a) Sketch the temperature-time history, T(t), during the quenching process; indicate temperature
corresponding to Ti , TD, and Tsat , identify regimes of film, transition and nucleate boiling and the singlephase convection regime; identify key features; and (b) Identify times(s) in this quenching process when
you expect the surface temperature of the sphere to deviate most from its center temperature.
SCHEMATIC:
ANALYSIS: (a) In the right-hand schematic above, the quench process is shown on the “boiling curve”
similar to Figure 10.4. Beginning at an initial temperature, Ti > TD , the process proceeds as indicated by
the arrows: film regime from i to D, transition regime from D to C, nucleate regime from C to A, and
single-phase (free convection) from A to the condition when ∆Te = Ts - Tsat = 0. The quench process is
shown on the temperature-time plot below and the boiling regimes and key temperatures are labeled..
The highest temperature-time change should occur in the nucleate pool boiling regime, especially near
the critical flux condition, Tc . The lowest temperature-time change will occur in the single-phase, free
convection regime.
(b) The difference between the center and surface temperature will occur when Bi = hro /3k ≥ 0.1. This
could occur in regimes with the highest convection coefficients. For example, h = 10,000 W/m2 ⋅K
which might be the case for water in the nucleate boiling regime, C-A, Bi ≈ 10,000 W/m2
(0.010m)/3×400 W/m⋅K = 0.08. For a sphere of larger dimension, in the nucleate and film pool boiling
regimes, we could expect temperature differences between the center and surface temperatures since Bi
might be greater than 0.1.
PROBLEM 10.26
KNOWN: A sphere (aluminum alloy 2024) with a uniform temperature of 500°C and emissivity of
0.25 is suddenly immersed in a saturated water bath maintained at atmospheric pressure.
FIND: (a) The total heat transfer coefficient for the initial condition; fraction of the total coefficient
contributed by radiation; and (b) Estimate the temperature of the sphere 30 s after it has been
immersed in the bath.
SCHEMATIC:
ASSUMPTIONS: (1) Water exposed to standard atmospheric pressure and uniform temperature,
Tsat, and (2) Lumped capacitance method is valid.
PROPERTIES: See Comment 2; properties obtained with IHT code.
ANALYSIS: (a) For the initial condition with Ts = 500°C, film boiling will occur and the
coefficients due to convection and radiation are estimated using Eqs. 10.9 and 10.11, respectively,
1/ 4
g ( ρ − ρ ) h ′ D3
h conv D
"
v
fg
Nu =
(1)
= C
D
h rad =
ηv k v (Ts − Tsat )
kv
(
4
εσ Ts4 − Tsat
Ts − Tsat
)
(2)
-8
2
4
where C = 0.67 for spheres and σ = 5.67 × 10 W/m ⋅K . The corrected latent heat is
h ′fg = h fg + 0.8 cp,v ( Ts − Tsat )
(3)
The total heat transfer coefficient is given by Eq. 10.10a as
4/3 + h
1/ 3
h 4 / 3 = h conv
(4)
rad ⋅ h
The vapor properties are evaluated at the film temperature,
Tf = ( Ts + Tsat ) / 2
(5)
while the liquid properties are evaluated at the saturation temperature. Using the foregoing relations
in IHT (see Comments), the following results are obtained.
Nu D
(
hcnv W / m 2 ⋅ K
)
(
h rad W / m 2 ⋅ K
)
226
867
12.0
The radiation process contribution is 1.4% that of the total heat rate.
(
h W / m2 ⋅ K
)
876
<
(b) For the lumped-capacitance method, from Section 5.3, the energy balance is
dT
(6)
− hAs ( Ts − Tsat ) = ρs Vcs s
dt
where ρs and cs are properties of the sphere. To determine Ts(t), it is necessary to evaluate h as a
function of Ts. Using the foregoing relations in IHT (see Comments), the sphere temperature after
30s is
Ts (30s ) = 333°C.
<
Continued …..
PROBLEM 10.26 (Cont.)
COMMENTS: (1) The Biot number associated with the aluminum alloy sphere cooling process for
the initial condition is Bi = 0.09. Hence, the lumped-capacitance method is valid.
(2) The IHT code to solve this application uses the film-boiling correlation function, the water
properties function, and the lumped capacitance energy balance, Eq. (6). The results for part (a),
including the properties required of the correlation, are shown at the outset of the code.
/* Results, Part (a): Initial condition, Ts = 500C
NuDbar
hbar
hcvbar hradbar
F
226
875.5
866.5
11.97
0.01367
*/
/* Properties: Initial condition, Ts = 500 C, Tf = 573 K
Pr
cpv
h'fg
hfg
kv
nuv
rhol
1.617 5889
3.291E6 1.406E6 0.0767 4.33E-7 712.1
rhov
45.98 */
/* Results: with initial condition, Ts = 500 C; after 30 s
Bi
F
Ts_C
hbar
t
0.09414
0.01367 500
875.5
0
0.04767
0.01587 333.2
443.3
30
// LCM analysis, energy balance
- hbar * As * (Ts - Tsat) = rhos * Vol * cps * der(Ts,t)
As = pi * D^2 / 4
Vol = pi * D^3 / 6
/* Correlation description: coefficients for film pool boiling (FPB). Eqs. 10.9, 10.10 and 10.11 .
See boiling curve, Fig 10.4 . */
NuDbar = NuD_bar_FPB(C,rhol,rhov,h'fg,nuv,kv,deltaTe,D,g)
// Eq 10.9
NuDbar = hcvbar * D / kv
g = 9.8
// gravitational constant, m/s^2
deltaTe = Ts - Tsat
// excess temperature, K
// Ts_C = 500
// surface temperature, K
Tsat = 373
// saturation temperature, K
// The vapor properties are evaluated at the film temperature,Tf,
Tf = Tfluid_avg(Ts,Tsat)
// The correlation constant is 0.62 or 0.67 for cylinders or spheres,
C = 0.67
// The corrected latent heat is
h'fg = hfg + 0.80*cpv*(Ts - Tsat)
// The radiation coefficient is
hradbar = eps * sigma * (Ts^4 - Tsat^4) / (Ts - Tsat) // Eq 10.11
sigma = 5.67E-8
// Stefan-Boltzmann constant, W/m^2·K^4
eps = 0.25
// surface emissivity
// The total heat transfer coefficient is
hbar^(4/3) = hcvbar^(4/3) + hradbar * hbar^(1/3) // Eq 10.10a
F = hradbar / hbar
// fraction contribution of radiation
// Input variables
D = 0.020
rhos = 2702
cps = 875
ks = 186
Bi = hbar * D / ks
// Sphere properties, aluminum alloy 2024
// Biot number
// Water property functions :T dependence, From Table A.6
// Units: T(K), p(bars);
x=1
// Quality (0=sat liquid or 1=sat vapor)
xx = 0
rhov = rho_Tx("Water",Tf,x)
// Density, kg/m^3
rhol = rho_Tx("Water",Tf,xx)
// Density, kg/m^3
hfg = hfg_T("Water",Tf)
// Heat of vaporization, J/kg
cpv = cp_Tx("Water",Tf,x)
// Specific heat, J/kg·K
nuv = nu_Tx("Water",Tf,x)
// Kinematic viscosity, m^2/s
kv = k_Tx("Water",Tf,x)
// Thermal conductivity, W/m·K
Pr = Pr_Tx("Water",Tf,x)
// Prandtl number
// Conversions
Ts_C = Ts - 273
PROBLEM 10.27
KNOWN: Steel bar upon removal from a furnace immersed in water bath.
FIND: Initial heat transfer rate from bar.
SCHEMATIC:
ASSUMPTIONS: (1) Uniform bar surface temperature, (2) Film pool boiling conditions.
3
PROPERTIES: Table A-6, Water, liquid (1 atm, Tsat = 100°C): ρ l = 957.9 kg/m , hfg = 2257 kJ/kg;
3
Table A-6, Water, vapor (Tf = (Ts + Tsat )/2 = 550K): ρ v = 31.55 kg/m , cp,v = 4640 J/kg⋅K, µv = 18.6
-6
2
× 10 N⋅s/m , kv = 0.0583 W/m⋅K.
ANALYSIS: The total heat transfer rate from the bar at the instant of time it is removed from the
furnace and immersed in the water is
qs = h As ( Ts − Tsat ) = h As ∆ Te
(1)
where ∆Te = 455 – 100 = 355K. According to the boiling curve of Figure 10.4, with such a high ∆Te,
film pool boiling will occur. From Eq. 10.10,
3
4/3 + h
1/3
h 4 / 3 = hconv
or
h = hconv + h rad ( if h conv > h rad ) .
(2)
rad ⋅ h
4
To estimate the convection coefficient, use Eq. 10.9,
1/4
g ( ρ − ρ ) h′ D3
h conv D
l
v fg
Nu D =
= C
kv
ν v k v ∆Te
(3)
where C = 0.62 for the horizontal cylinder and h ′fg = h fg + 0.8 c p,v ( Ts − Tsat ) . Find
hconv =
0.0583W/m ⋅ K
0.020 m
9 . 8 m / s 2 ( 957.9 − 31.55 ) k g / m3 2257 × 103 + 0.8 × 4640 × 355 J / k g ( 0.020m) 3
0.62
−
6
2
18.6 × 10 /31.55 m / s × 0.0583W/m ⋅ K × 355K
(
1/4
)
hconv = 690 W / m 2 ⋅ K.
To estimate the radiation coefficient, use Eq. 10.11,
4
ε σ Ts4 − Tsat
0.9 × 5.67 ×10 −8 W / m 2 ⋅ K 4 728 4 − 3734 K 4
hrad =
=
= 37.6W/m 2 ⋅ K.
Ts − Tsat
355K
Substituting numerical values into the simpler form of Eq. (2), find
h = 690 + 3 / 4 37.6 W / m2 ⋅ K = 718W/m2 ⋅ K.
(
(
)
(
(
)
)
)
Using Eq. (1), the heat rate, with As = π D L, is
q s = 718 W / m 2 ⋅ K (π × 0.020m × 0.200m ) × 355K = 3.20kW.
COMMENTS: For these conditions, the convection process dominates.
<
PROBLEM 10.28
KNOWN: Electrical conductor with prescribed surface temperature immersed in water.
FIND: (a) Power dissipation per unit length, q′s and (b) Compute and plot q′s as a function of surface
temperature 250 ≤ Ts ≤ 650°C for conductor diameters of 1.5, 2.0, and 2.5 mm; separately plot the
percentage contribution of radiation as a function of Ts .
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Water saturated at l atm, (3) Film pool boiling.
PROPERTIES: Table A-6, Water, liquid (1 atm, Tsat = 100°C): ρ " = 957.9 kg/m3 , hfg = 2257 kJ/kg;
Table A-6, Water, vapor (Tf = (Ts + Tsat ) / 2 = 600 K): ρv = 72.99 kg/m3, cp,v = 8750 J/kg⋅K, µv = 22.7
× 10-6 N⋅s/m2 , kv = 0.0929 W/m⋅K.
ANALYSIS: (a) The heat rate per unit length due to electrical power dissipation is
q
A
q′s = s = h s ( Ts − Tsat ) = hπ D∆Te
where ∆Te = (555 - 100)°C = 455°C. According to the boiling curve of Figure 10.4, with such a high
∆Te, film pool boiling will occur. From Eq 10.10,
4/3 + h
1/ 3
h 4 / 3 = h conv
rad ⋅ h
3
h = h conv + h rad
4
or
(if
h conv > h rad ).
To estimate the convection coefficient, use Eq. 10.9,
1/ 4
g ( ρ − ρ ) h ′ D3
h conv D
"
v fg
Nu D =
= C
ν v k v ∆Te
kv
where C = 0.62 for the horizontal cylinder and h′fg = hfg + 0.8cp,v (Ts - Tsat). Find
hconv =
9.8 m
× 0.62
0.002 m
h conv = 2108 W m 2 ⋅ K .
0.0929 W m ⋅ K
s
2
(957.9 − 72.99 ) kg
(
22.7 × 10
−6
m
3
2257 × 103 + 0.8 × 8750 × 455 J
)
72.99 m
2
s × 0.0929 W m ⋅ K × 455 K
kg ( 0.002m )
3
1/ 4
To estimate the radiation coefficient, use Eq. 10.11.
h rad =
(
4
εσ Ts4 − Tsat
) = 0.5 × 5.67 ×10−8 W m2 ⋅ K4 (8284 − 3734 ) K 4 = 28 W m2⋅ K.
Ts − Tsat
455 K
Since hconv > hrad , the simpler form of Eq. 10.10b is appropriate. Find,
h = ( 2108 + (3 4 ) × 28 ) W m 2 ⋅ K = 2129 W m 2 ⋅ K
Continued...
PROBLEM 10.28 (Cont.)
The heat rate is
q′ = 2129 W m 2 ⋅ K × π ( 0.002m ) × 455K = 6.09 kW m.
<
(b) Using the IHT Correlations Tool, Boiling, Film Pool Boiling, combined with the Properties Tool for
Water, the heat rate, q′ , was calculated as a function of the surface temperature, Ts , for conductor
diameters of 1.5, 2.0 and 2.5 mm. Also, plotted below is the ratio (%) of q′rad q′ as a function of
surface temperature.
2
20000
12000
q'rad/q' (%)
Heat rate, q' (W/m)
16000
8000
1
4000
0
0
250
350
450
550
Surface temperature, Ts (C)
D = 1.5 mm
D = 2.0 mm
D = 2.5 mm
650
250
350
450
550
650
Surface temperature, Ts (C)
D = 1.5 mm
D = 2.0 mm
D = 2.5 mm
From the q′ vs. Ts plot, note that the heat rate increases markedly with increasing surface temperatures,
and, as expected the heat rate increases with increasing diameter. The discontinuity near Ts = 650°C is
caused by the significant changes in the thermophysical properties as the film temperature, Tf ,
approaches the critical temperature, 647.3 K. From the q′rad q′ vs. Ts plot, the maximum contribution
by radiation is 2% and surprisingly doesn’t occur at the maximum surface temperature. By examining a
plot of q′rad vs. Ts , we’d see that indeed q′rad increases markedly with increasing Ts; but q′conv
increases even more markedly so the relative contribution of the radiation mode actually decreases with
increasing temperature for Ts > 350°C.
PROBLEM 10.29
KNOWN: Horizontal, stainless steel bar submerged in water at 25°C.
FIND: Heat rate per unit length of the bar.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Film pool boiling, (3) Water at 1 atm.
PROPERTIES: Table A-6, Water, liquid (1 atm, Tsat = 100°C): ρ l = 957.9 kg/m3, hfg = 2257
3
kJ/kg; Table A-6, Water, vapor, (Tf = (Ts + Tsat )/2 ≈ 450K): ρ v = 4.81 kg/m , cp,v = 2560 J/kg⋅K, µv
-6
2
= 14.85 × 10 N⋅s/m , kv = 0.0331 W/m⋅K.
ANALYSIS: The heat rate per unit length is
q′s = q s / l = q′′π D = hπ D ( Ts − Tsat ) = h π D ∆Te
where ∆Te = (250-100)°C = 150°C. Note from the boiling curve of Figure 10.4, that film boiling will
occur. From Eq. 10.10,
3
4 / 3 + h h1/3
h 4 / 3 = hconv
or
h = hconv + h rad
if hconv > h rad .
rad
4
(
)
To estimate the convection coefficient, use Eq. 10.9,
1/4
g ( ρ − ρ ) h′ D3
h conv D
l
v fg
Nu D =
= C
kv
ν v k v ∆Te
where C = 0.62 for the horizontal cylinder and h ′fg = h fg + 0.8cp,v ( Ts − Tsat ) . Find
1/4
h conv =
0.0331W/m ⋅ K
0.050m
9.8m/s 2 ( 957.9 − 4.81 ) k g / m 3 2257 × 10 3 + 0.8 × 2560J/kg ⋅ K × 150 K ( 0.050m)3
0.62
−6
2
14.85 × 10 /4.81 m / s × 0.0331 W / m ⋅ K × 150K
(
)
hconv = 273 W / m 2 ⋅ K.
To estimate the radiation coefficient, use Eq. 10.11,
4
ε σ Ts4 − Tsat
0.50 × 5.67 ×10 −8 W / m 2 ⋅ K 4 5234 − 3734 K 4
hrad =
=
= 1 1 W / m 2 ⋅ K.
Ts − Tsat
150K
Since hconv > hrad, the simpler form of Eq. 10.10 is appropriate. Find,
h = 273 + 3 / 4 ×11 W / m2 ⋅ K = 2 8 1 W / m2 ⋅ K.
(
)
(
(
)
)
Using the rate equation, find
q′s = 281W/m 2 ⋅ K × π × ( 0.050m ) ×150K = 6.62kW/m.
COMMENTS: The effect of the water being subcooled (T = 25°C < Tsat ) is considered to be
negligible.
<
PROBLEM 10.30
KNOWN: Heater element of 5-mm diameter maintained at a surface temperature of 350°C when
immersed in water under atmospheric pressure; element sheath is stainless steel with a mechanically
polished finish having an emissivity of 0.25.
FIND: (a) The electrical power dissipation and the rate of evaporation per unit length; (b) If the
heater element were operated at the same power dissipation rate in the nucleate boiling regime, what
temperature would the surface achieve? Calculate the rate of evaporation per unit length for this
operating condition; and (c) Make a sketch of the boiling curve and represent the two operating
conditions of parts (a) and (b). Compare the results of your analysis. If the heater element is operated
in the power-controlled mode, explain how you would achieve these two operating conditions
beginning with a cold element.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, and (2) Water exposed to standard atmospheric
pressure and uniform temperature, Tsat.
PROPERTIES: Table A-6, Saturated water, liquid (100°C): ρ" = 957.9 kg / m3 , c p," = 4217
J/kg⋅K, µ" = 279 × 10 −6 N ⋅ s / m 2 , Pr" = 1.76, hfg = 2257 kJ/kg, h ′fg = h fg + 0.80 c p,v ( Ts − Tsat ) =
−3
3
2905 kJ/kg, σ = 58.9 × 10 N/m; Saturated water, vapor (100°C): ρv = 0.5955 kg/m ; Water vapor
-6
(Tf = 498 K): ρv = 1/vv = 12.54 kg/m3, cp,v = 3236 J/kg⋅K, kv = 0.04186 W/m⋅K, ηv = 1.317 × 10
2
m /s.
ANALYSIS: (a) Since ∆Te > 120°C, the element is operating in the film-boiling (FB) regime. The
electrical power dissipation per unit length is
q′s = h (π D )( Ts − Tsat )
(1)
where the total heat transfer coefficient is
4/3 + h
1/ 3
h 4 / 3 = h conv
rad h
The convection coefficient is given by the correlation, Eq. 10.9, with C = 0.62,
(2)
1/ 4
g ( ρ − ρ ) h′ D3
h conv D
"
v fg
= C
kv
ηv k v ( Ts − Tsat )
(3)
1/ 4
9.8 m / s 2 (833.9 − 12.54 ) kg / m3 × 2.905 × 106 J / kg ⋅ K (0.005 m )3
h conv = 0.62
6
2
−
1.31× 10 m / s × 0.04186 W / m ⋅ K (350 − 100 ) K
h conv = 626 W / m 2 ⋅ K
−8
The radiation coefficient, Eq. (10.11), with σ = 5.67 × 10
2
4
W/m ⋅K , is
Continued …..
h rad =
h rad =
(
4
εσ Ts4 − Tsat
(Ts − Tsat )
PROBLEM 10.30 (Cont.)
)
(
)
0.25σ 6234 − 3734 K 4
(350 − 100 ) K
= 4.5 W / m 2 ⋅ K
Substituting numerical values into Eq. (2) for h, and into Eq. (1) for q′s , find
h = 630 W / m 2 ⋅ K
q′s = 630 W / m 2 ⋅ K (π × 0.005 m )(350 − 100 ) K = 2473 W / m
<
q′′s = q′s / π D = 0.157 MW / m 2
The evaporation rate per unit length is
′b = q′s / h fg = 3.94 kg / h ⋅ m
m
<
(b) For the same heat flux, q′′s = 0.157 MW / m 2 , using the Rohsenow correlation for the nucleate
boiling (NB) regime, find ∆Te, and hence Ts.
1/ 2
g ( ρ" − ρ v )
q′′s = µ" h fg
σ
3
c ∆T
p," e
C h Pr n
s,f fg "
where, from Table 10.1, for stainless steel mechanically polished finish with water, Cs,f = 0.013 and n
= 1.0.
0.157 × 106 W / m 2 = 279 × 10−6 N ⋅ s / m 2 × 2.257 × 106 J / kg
1/ 2
9.8 m / s 2 (957.9 − 0.5955 ) kg / m3
×
−3 N / m
58.9
10
×
4217 J / kg ⋅ K × ∆Te
×
0.013 × 2.257 ×106 J / kg ×1.76
∆Te = Ts − Tsat = 10.5 K
3
<
Ts = 110.5°C
The evaporation rate per unit length is
′b = q′′s (π D ) h fg = 3.94 kg / h ⋅ m
m
<
Continued …..
PROBLEM 10.30 (Cont.)
(c) The two operating conditions are shown on the boiling curve, which is fashioned after Figure 10.4.
For FB the surface temperature is Ts = 350°C (∆Te = 250°C). The element can be operated at NB
2
with the same heat flux, qs′′ = 0.157 MW/m , with a surface temperature of Ts = 110°C (∆Te = 10°C).
Since the heat fluxes are the same for both conditions, the evaporation rates are the same.
If the element is cold, and operated in a power-controlled mode, the element would be brought to the
NB condition following the arrow shown next to the boiling curve near ∆Te = 0. If the power is
increased beyond that for the NB point, the element will approach the critical heat flux (CHF)
condition. If q′′s is increased beyond q′′max , the temperature of the element will increase abruptly,
and the burnout condition will likely occur. If burnout does not occur, reducing the heat flux would
allow the element to reach the FB point.
PROBLEM 10.31
KNOWN: Inner and outer diameters, outer surface temperature and thermal conductivity of a tube.
Saturation pressure of surrounding water and convection coefficient associated with gas flow through
the tube.
FIND: (a) Heat rate per unit tube length, (b) Mean temperature of gas flow through tube.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) Uniform surface temperature, (3) Water is at saturation
temperature, (4) Tube is horizontal.
PROPERTIES: Table A-6, saturated water, liquid (p = 1.523 bars): Tsat = 385 K, ρ " = 950 kg / m3 ,
3
hfg = 2225 kJ/kg. Table A-6, saturated water, vapor (Tf = 480 K): ρv = 9.01 kg/m , cp,v = 2940
J/kg⋅K. µv = 15.9 × 10 N⋅s/m , kv = 0.0381 W/m⋅K, ν v = 1.77 × 10−6 m 2 / s.
-6
2
ANALYSIS: (a) The heat rate per unit length is q ′ = h oπ Do ( Ts,o − Tsat ) , where ho includes
contributions due to convection and radiation in film boiling. With C = 0.62 and h ′fg = h fg + 0.80
(
)
6
c p,v Ts,o − Tsat = 2.67 × 10 J / kg, Eq. 10.9 yields
1/ 4
hconv,o
9.8 m / s 2 ( 950 − 9 ) kg / m3 × 2.67 × 106 J / kg ( 0.03m )3
0.0381 W / m ⋅ K
=
0.62
−6 2
0.030m
1.77 × 10 m / s × 0.0381 W / m ⋅ K × 190 K
From Eq. 10.11, the radiation coefficient is
h rad,o =
0.30 × 5.67 × 10
−8
2
W/m ⋅K
4
(575
4
− 385
4
(575 − 385 ) K
)K
2
= 376 W / m ⋅ K
4
2
= 7.8 W / m ⋅ K
From Eq. 10.10b, it follows that
2
ho = hconv,o + 0.75 hrad,o = 382 W / m ⋅ K
and the heat rate is
(
)
q ′ = hoπ D o Ts,o − Tsat = 382 W / m ⋅ K (π × 0.03m )190 K = 6840 W
<
2
(
(b) From the thermal circuit, with R ′conv,i = ( h iπ Di )−1 = 500 W / m 2 ⋅ K × π × 0.026m
)
−1
= 0.0245 m ⋅ K / W
and R ′cond = "n ( Do / Di ) / 2π k s = "n ( 0.030 / 0.026 ) / 2π (15 W / m ⋅ K ) = 0.00152 m ⋅ K / W,
q′ =
Tm − Ts,o
Tm − 575 K
=
R conv,i + R ′cond ( 0.0245 + 0.00152 ) m ⋅ K / W
Tm = 575 K + 6840 W / m ( 0.0260 m ⋅ K / W ) = 753 K
<
COMMENTS: Despite the large temperature of the gas, the rate of heat transfer is limited by the
large thermal resistances associated with convection from the gas and film boiling. The resistance
due to film boiling is R ′fb = (π D o ho )−1 = 0.0278 m ⋅ K / W.
PROBLEM 10.32
KNOWN: Cylinder of 120 mm diameter at 1000K quenched in saturated water at 1 atm
FIND: Describe the quenching process and estimate the maximum heat removal rate per unit length
during cooling.
SCHEMATIC:
ASSUMPTIONS: Water exposed to 1 atm pressure, Tsat = 100°C.
ANALYSIS: At the start of the quenching process, the surface temperature is Ts(0) = 1000K.
Hence, ∆Te = Ts – Tsat = 1000K – 373K = 627K, and from the typical boiling curve of Figure 10.4,
film boiling occurs.
As the cylinder temperature decreases, ∆Te decreases, and the cooling process follows the boiling
curve sketched above. The cylinder boiling process passes through the Leidenfrost point D, into the
transition or unstable boiling regime (D → C).
2
At point C, the boiling heat flux has reached a maximum, q ′′max = 1.26 MW/m (see Example 10.1).
Hence, the heat rate per unit length of the cylinder is
q′s = q′max = q′′max (π D ) = 1.26MW/m 2 π ( 0.120m ) = 0.475MW/m.
<
As the cylinder cools further, nucleate boiling occurs (C → A) and the heat rate drops rapidly. Finally,
at point A, boiling no longer is present and the cylinder is cooled by free convection.
COMMENTS: Why doesn’t the quenching process follow the cooling curve of Figure 10.3?
PROBLEM 10.33
KNOWN: Horizontal platinum wire of diameter of 1 mm, emissivity of 0.25, and surface temperature of
800 K in saturated water at 1 atm pressure.
FIND: (a) Surface heat flux, q′′s , when the surface temperature is Ts = 800 K and (b) Compute and plot
on log-log coordinates the heat flux as a function of the excess temperature, ∆Te = Ts - Tsat, for the range
150 ≤ ∆Te ≤ 550 K for emissivities of 0.1, 0.25, and 0.95; separately plot the percentage contribution of
radiation as a function of ∆Te.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Film pool boiling.
PROPERTIES: Table A.6, Saturated water, liquid (Tsat = 100°C, 1 atm): ρ" = 957.9 kg/m3, hfg = 2257
kJ/kg; Table A.6, Water, vapor (Tf = (Ts + Tsat)/2 = (800 + 373)K/2 = 587 K): ρv = 58.14 kg/m3, cp,v =
7065 J/kg⋅K, µv = 21.1 × 10-6 N⋅s/m2, kv = 81.9 × 10-3 W/m⋅K.
ANALYSIS: (a) The heat flux is
q′′s = h ( Ts − Tsat ) = h∆Te
where ∆Te = (800 - 373)K = 427 indicative of film boiling. From 10.10,
4 / 3 + h h −1/ 3
h 4 / 3 = h conv
or
h = h conv + (3 4 ) h rad
rad
if hrad < hconv. Use Eq. 10.9 with C = 0.62 for a horizontal cylinder,
3 1/ 4
′
ρ
ρ
g
h
D
−
(
)
h
D
"
v fg
Nu D = conv = C
kv
ν v k v ( Ts − Tsat )
1/ 4
3
9.8 m s 2 (957.9 − 58.14 ) kg m3 × 4670 k J kg ( 0.001m )
hconv × 0.001m
= 0.62
−3
6
2
3
−
81.9 × 10 W m ⋅ K
21.1 × 10 N ⋅ s m 58.14 kg m × 0.0819 W m ⋅ K (800 − 373 ) K
)
(
h conv = 2155 W m 2 ⋅ K
where h ′fg = h fg + 0.8c p,v ( Ts − Tsat ) = 2257 kJ kg + 0.8 × 7065 J kg ⋅ K (800 − 373 ) K = 4670 kJ kg. To
estimate the radiation coefficient, use Eq. 10.11,
4
εσ Ts4 − Tsat
0.25σ 8004 − 3734 K 4
h rad =
=
= 13.0 W m 2 ⋅ K .
Ts − Tsat
800
373
K
−
(
)
Since h rad < h conv , use the simpler expression,
(
)
(
)
h = 2155 W m 2⋅ K + (3 4 )13.0 W m 2⋅ K = 2165 W m 2⋅ K.
Using the rate equation, find
Continued...
PROBLEM 10.33 (Cont.)
q′′s = 2165 W m 2 ⋅ K (800 − 373) K = 0.924 MW m 2 .
<
(b) Using the IHT Correlations Tool, Boiling, Film Pool Boiling, combined with the Properties Tool for
Water, the heat flux, q′′s , was calculated as a function of the excess temperature, ∆Te for emissivities of
0.1, 0.25 and 0.95. Also plotted is the ratio (%) of q′′rad q′′ as a function of ∆Te .
3
8E6
1E6
600000
2
q''rad/q''s (%)
Heat flux, q'' (W/m^2)
4E6
200000
80000
40000
1
10000
100
400
Temperature excess, deltaTe (K)
eps = 0.1
eps = 0.25
eps = 0.95
Critical heat flux, q''max (W/m^2)
Minimum heat flux, q''min (W/m^2)
800
0
100
200
300
400
500
600
Temperature excess, deltaTe (K)
eps = 0.1
eps = 0.25
eps = 0.95
From the q′′s vs. ∆Te plot, note that the heat rate increases markedly with increasing excess temperature.
On the plot scale, the curves for the three emissivity values, 0.1, 0.25 and 0.95, overlap indicating that the
overall effect of emissivity change on the total heat flux is slight. Also shown on the plot are the critical
heat flux, q′′max = 1.26 MW/m2, and the minimum heat flux, q′′min = 18.9kW/m2, at the Leidenfrost
point. These values are computed in Example 10.1. Note that only for the extreme value of ∆Te is the
heat flux in film pool boiling in excess of the critical heat flux. The relative contribution of the radiation
mode is evident from the q′′rad qs′′ vs. ∆Te plot. The maximum contribution by radiation is less than
3% and surprisingly doesn’t occur at the maximum excess temperature. By examining a plot of q′′rad vs.
∆Te , we’d see that indeed q′′rad increases markedly with increasing ∆Te ; however, q′′conv increases
even more markedly so that the relative contribution of the radiation mode actually decreases with
increasing temperature for ∆Te > 250 K. Note that, as expected, the radiation heat flux, q′′rad , is
proportional to the emissivity.
COMMENTS: Since q′′s < q′′max = 1.26 MW m 2 , the prescribed condition can only be achieved in
power-controlled heating by first exceeding q′′max and then decreasing the flux to 0.924 MW/m2.
PROBLEM 10.34
KNOWN: Surface temperature and emissivity of strip steel.
FIND: Heat flux across vapor blanket.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Vapor/jet interface is at Tsat for p = 1 atm, (3)
Negligible effect of jet and strip motion.
3
PROPERTIES: Table A-6, Saturated water (100°C): ρ l = 957.9 kg/m , hfg = 2257 kJ/kg;
3
-6
2
Saturated water vapor (Tf = 640K): ρv = 175.4 kg/m , cp,v = 42 kJ/kg⋅K, µv = 32 × 10 N⋅s/m ,
-6 2
k = 0.155 W/m⋅K, ν v = 0.182 × 10 m /s.
ANALYSIS: The heat flux is
q′′s = h∆Te
where
∆Te = 907 K − 373 K = 534 K
h = h conv + ( 3 / 4 ) h rad .
and
4/3
h 4 / 3 = hconv
+ h rad h1/3
With
h ′fg = h fg + 0.80cp,v ( Ts − Tsat ) = 2.02 ×107 J/kg
or
Equation 10.9 yields
(
)
1/4
9.8m/s 2 ( 957.9 −175.4 ) k g / m 3 2.02 ×10 7 J / k g (1 m )3
Nu D = 0.62
−
6
2
0.182 ×10 m / s ( 0.155 W / m⋅ K )( 907 − 373) K
= 6243.
Hence,
hconv = Nu D k v / D = 6243W/m 2 ⋅ K ( 0.155 W / m ⋅K / 1 m ) = 968 W / m 2 ⋅ K
hrad =
(
4
ε σ Ts4 − Tsat
Ts − Tsat
) = 0.35 × 5.67 ×10−8 W / m 2 ⋅ K 4 (9074 − 3734 ) K 4
( 907 − 373) K
hrad = 24 W / m 2 ⋅ K
(
)
Hence,
h = 968 W / m2 ⋅ K + ( 3 / 4) 24 W / m 2 ⋅ K = 986 W / m 2 ⋅ K
And
q′′s = 986 W / m 2 ⋅ K (907 − 373 ) K = 5.265 ×105 W / m 2 .
<
COMMENTS: The foregoing analysis is a very rough approximation to a complex problem. A more
rigorous treatment is provided by Zumbrunnen et al. In ASME Paper 87-WA/HT-5.
PROBLEM 10.35
KNOWN: Copper sphere, 10 mm diameter, initially at a prescribed elevated temperature is quenched in
a saturated (1 atm) water bath.
FIND: The time for the sphere to cool (a) from Ti = 130 to 110°C and (b) from Ti = 550°C to 220°C .
SCHEMATIC:
ASSUMPTIONS: (1) Sphere approximates lumped capacitance, (2) Water saturated at 1 atm.
PROPERTIES: Table A-1, Copper: ρ = 8933 kg/m3; Table A.11, Copper (polished) : ε = 0.04, typical
value; Table A.4, Water: as required for the pool boiling correlations.
ANALYSIS: Treating the sphere as a lumped capacitance and performing an energy balance, see Eq.
5.14,
E in − E out = E st
− qs′′ ⋅ As = ρ c∀
dT
dt
(1,2)
For the sphere, V = πD3 / 6 and As = πD2 . Using the IHT Lumped Capacitance Model to solve this
differential equation, we need to specify (1) the specific heat of the copper sphere as a function sphere
temperature; use IHT Properties Tool, Copper; and (2) the heat flux, qs′′ , associated with the pool
boiling processes; use IHT Correlations Tool, Boiling:
(a) Cooling from Ti =130° to 110°: Nucleate pool boiling, Rohsenhow correlation, Eq. 10.5,
(b) Cooling from Ti = 550 to 220°C : Film Pool Boiling, Eq. 10.9 with C = 0.67 (sphere).
The thermophysical properties for water required of the correlations are provided by the IHT Tool,
Properties-Water. The specific heat of copper as a function of sphere temperature is provided by the IHT
Tool, Properties-Copper. The temperature-time histories for each of the cooling processes are plotted
below.
Nucleate pool boiling quench process
Film pool boiling quench process
130
500
Sphere temperature, T (s)
Sphere temperature, Ts (C)
400
120
110
100
300
200
100
0
1
2
3
Elapsed time, t (s)
4
5
0
10
20
30
Elapsed time, t (s)
Continued...
PROBLEM 10.35 (Cont.)
Using the Explore feature in the IHT Plot Window, the elapsed times for the quench process were found
as:
Quench process
Ti - Tf (°C)
∆t(s)
Nucleate pool boiling
130-110
0.76
Film pool boiling
550-220
13.5
COMMENTS: (1) Comparing the elapsed times for the two processes, the nucleate pool boiling process
cools 20°C in 0.76s (26.3°C/s) vs. 330°C in 13.5s (24.4°C/s) for the film pool boiling process.
(2) The IHT Workspace used to generate the temperature-time history for the nucleate pool boiling
process is shown below.
// Correlations Tool - Boiling, Nucleate Pool Boiling, Heat flux
qs'' = qs_dprime_NPB(Csf,n,rhol,rhov,hfg,cpl,mul,Prl,sigma,deltaTe,g) // Eq 10.5
g = 9.8
// Gravitational constant, m/s^2
deltaTe = Ts - Tsat
// Excess temperature, K
Ts = Ts_C + 273
// Surface temperature, K
//Ts_C = 130
Tsat = 100 + 273
// Saturation temperature, K
/* Evaluate liquid(l) and vapor(v) properties at Tsat. From Table 10.1 (Fill in as required), */
// fluid-surface combination:
Csf = 0.013
// Polished copper-water combination, Table 10.1
n = 1.0
/* Correlation description: Heat flux for nucleate pool boiling (NPB), water-surface combination (Cf,n), Eq 10.5,
Table 10.1 . See boiling curve, Fig 10.4 . */
// Properties Tool- Water:
// Water property functions :T dependence, From Table A.6
// Units: T(K), p(bars);
xv = 1
// Quality (0=sat liquid or 1=sat vapor)
rhov = rho_Tx("Water",Tsat,xv)
// Density, kg/m^3
hfg = hfg_T("Water",Tsat)
// Heat of vaporization, J/kg
sigma = sigma_T("Water",Tsat)
// Surface tension, N/m (liquid-vapor)
// Water property functions :T dependence, From Table A.6
// Units: T(K), p(bars);
xl = 0
// Quality (0=sat liquid or 1=sat vapor)
rhol = rho_Tx("Water",Tsat,xl)
// Density, kg/m^3
cpl = cp_Tx("Water",Tsat,xl)
// Specific heat, J/kg·K
mul = mu_Tx("Water",Tsat,xl)
// Viscosity, N·s/m^2
Prl = Pr_Tx("Water",Tsat,xl)
// Prandtl number
// Lumped Capacitance Model:
/* Conservation of energy requirement on the control volume, CV. */
Edotin - Edotout = Edotst
Edotin = 0
Edotout = As * ( + qs'' )
Edotst = rho * vol * cp * Der(Ts,t)
/* The independent variables for this system and their assigned numerical values are */
As = pi * D^2 / 4
// surface area, m^2
vol = pi * D^3 / 6
// volume, m^3
D = 0.01
rho = 8933
// density, kg/m^3
// Properties Tool - Copper
// Copper (pure) property functions : From Table A.1
// Units: T(K)
cp = cp_T("Copper",Ts)
// Specific heat, J/kg·K
PROBLEM 10.36
KNOWN: Saturated water at 1 atm is heated in cross flow with velocities 0 – 2 m/s over a 2 mmdiameter tube.
FIND: Plot the critical heat flux as a function of water velocity; identify the pool boiling and transition
regions between the low and high velocity ranges.
SCHEMATIC:
ASSUMPTIONS: Nucleate boiling in the presence of external forced convection.
3
3
PROPERTIES: Table A-6, Water (1 atm): Tsat = 100°C, ρ l = 957.9 kg/m , ρ v = 0.5955 kg/m , hfg
-3
= 2257 kJ/kg, σ = 58.9 × 10 N/m.
ANALYSIS: The Lienhard-Eichhorn correlations for forced convection boiling with cross flow over a
cylinder are appropriate for estimating q ′′max , Eqs. 10.12 and 10.13.
Low Velocity
1/3
ρ v h fg 4σ
q′′max =
1 + ρ V 2D V
π
v
1/3
1
kg
J 4 × 58.9 ×10 −3 N / m
3
q′′max = 0.5955
× 2257 ×10
1+
V
π
kg 0.5955kg/m3 V2 0.002m
m3
q′′max = 4.2782 ×105 V + 2.921×10 6 V1/3.
High Velocity
1/3
1/2
ρ v h fg 1 ρl 3 / 4
1 ρl σ
q′′max =
+
V
π 169 ρ v
19.2 ρ v ρv V2 D
q′′max =
1
kg
J 1 957.9 3 / 4
0.5955
× 2257 ×103
+
0.5955
3
π
kg
169
m
1/3
1 957.9 1/2
58.9 ×10−3 N / m
19.2 0.5955 0.5955kg/m3 V2 0.002m
V
q′′max = 6.4299 ×105 V + 3.280 ×10 6 V1/3
Continued …..
PROBLEM 10.36 (Cont.)
The transition between the low and high velocity regions occurs when
1/2
0.275 ρl
′′
q max = ρ v h fg V
+ 1
π ρv
q′′max = 0.5955
J 0.275 957.9 1/2
3
× 2257 ×10
V
+ 1 = 6.0627 ×106 V.
3
kg
m
kg
π
0.5955
(3)
For pool boiling conditions when the velocity is zero, the critical heat flux must be estimated according
to the correlation for the small horizontal cylinder as introduced in Problem 10.22. If the cylinder were
2
“large,” the critical heat flux would be 1.26 MW/m as given by the Zuber-Kutateladze correlation, Eq.
10.7. Following the analysis of Problem 10.22, find Bo = 0.40 and the critical heat flux for the “small”
2 mm cylinder is
q′′max ) pool = 1.18 ×1.26 M W / m 2 = 1.49 W / m 2.
The graph below identifies four regions: pool boiling where q ′′max = 1.49 M W / m2 from V = 0 to 0.15
m/s and the low velocity, transition and high velocity regimes.
PROBLEM 10.37
KNOWN: Saturated water at 1 atm and velocity 2 m/s in cross flow over a heater element of 5 mm
diameter.
FIND: Maximum heating rate, q ′[ W / m] .
SCHEMATIC:
ASSUMPTIONS: Nucleate boiling in the presence of external forced convection.
3
3
PROPERTIES: Table A-6, Water (1 atm): Tsat = 100°C, ρ l = 957.9 kg/m , ρ v = 0.5955 kg/m , hfg
-3
= 2257 kJ/kg, σ = 58.9 × 10 N/m.
ANALYSIS: The Lienhard-Eichhorn correlation for forced convection with cross flow over a cylinder
is appropriate for estimating q ′′max . Assuming high-velocity region flow, Eq. 10.13 with Eq. 10.14 can
be written as
1/3
1/2
ρ v h fg V 1 ρl 3 / 4
1 ρl σ
q′′max =
+
.
169 ρ
π
19.2 ρ v ρ v V 2D
v
Substituting numerical values, find
q′′max =
1 957.9 3 / 4
1
0.5955kg/m3 × 2257 ×103 J / k g × 2 m / s
+
π
169 0.5955
1/2
1 957.9
19.2 0.5955
1/3
0.5955kg/m 3 ( 2 m / s ) 2 0.005m
58.9 × 10− 3 N / m
q′′max = 4.331MW/m 2 .
The high-velocity region assumption is satisfied if
1/2
q′′max ? 0.275 ρl
+1
ρ v h fg V < π
ρv
4.331×106 W / m2
0.5955kg/m 3 × 2257 ×10 3 J / k g × 2 m / s
? 0.275 957.9 1/2
= 1.61<
π
0.5955
The inequality is satisfied. Using the q ′′max estimate, the maximum heating rate is
q′max = q′′max ⋅π D = 4.331MW/m 2 ×π ( 0.005m ) = 68.0kW/m.
+ 1 = 4.51.
<
COMMENTS: Note that the effect of the forced convection is to increase the critical heat flux by
4.33/1.26 = 3.4 over the pool boiling case.
PROBLEM 10.38
KNOWN: Correlation for forced-convection local boiling inside a vertical tube.
FIND: Boiling heat transfer rate per unit length of the tube.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Local boiling occurs when tube wall is 15°C above
the saturation temperature.
ANALYSIS: From experimental results, the heat transfer coefficient can be estimated by the
correlation
p
h = 2.54 ( ∆Te )3 exp
15.3
W / m2 ⋅ K
where ∆Te is the excess temperature, Ts − Tsat [ K ] , and p is the pressure [bar]. The heat transfer
rate per unit length is
q′ = π D h ∆Te .
Evaluating the heat transfer coefficient, find
h = 2.54 (15K ) exp (4 bar/15.3) = 11,134 W / m 2 ⋅ K.
3
The heat rate is then.
q′ = π ( 0.050m ) ×11,134W/m 2 ⋅ K ×15K = 26.2 kW/m.
COMMENTS: The saturation temperature at 4 bar is Tsat = 406.5K according to Table A-6.
<
PROBLEM 10.39
KNOWN: Forced convection and boiling processes occur in a smooth tube with prescribed water
velocity and surface temperature.
FIND: Heat transfer rate per unit length of the tube.
SCHEMATIC:
ASSUMPTIONS: (1) Fully-developed flow, (2) Nucleate boiling conditions occur on inner wall of
tube, (3) Forced convection and boiling effects can be separately estimated.
3
PROPERTIES: Table A-6, Water (Tm = 95°C = 368K): ρ l = 1/vf = 962 kg/m , ρ v = 1/vg = 0.500
3
-6
2
kg/m , hfg = 2270 kJ/kg, cp,l = 4212 J/kg⋅K, µl = 296 × 10 N⋅s/m , k l = 0.678 W/m⋅K, Prl = 1.86,
-3
-7
2
σ = 60 × 10 N/m, ν l = 3.08 × 10 m /s.
ANALYSIS: Experimentation has indicated that the heat transfer rate can be estimated as the sum of
the separate effects due to forced convection and boiling. On a per unit length basis,
q′ = q′fc + q ′boil.
For forced convection, ReD = u m D / ν l = 1.5m/s × 0.015m/3.08 × 10 −7 m2 / s = 73,052. Since Re
> 2300, flow is turbulent and since fully developed, use the Dittus-Boelter correlation but with the 0.023
coefficient replaced by 0.019 and n = 0.4,
NuD = h D / k = 0.019Re 4D/ 5 Pr n
k
0.678W/m ⋅ K
h = Nu D =
× 0.019 ( 73,052 ) 4 / 5 (1.86 )0.4 = 8563W/m 2 ⋅ K.
D
0.015m
q′fc = h π D ( Ts − Tm ) = 8,563W/m 2 ⋅ K ⋅ π ( 0.015m )(110 − 95 ) ° C = 6052W/m.
For boiling, ∆Te = (110 – 100)°C = 10°C and hence nucleate boiling occurs. From the Rohsenow
equation, with Csf = 0.006 and n = 1.0,
3
1/2
c p,l ∆Te
g ( ρl − ρ v )
q′′boil = µl h fg
σ
Csf hfg Pr n
l
q ′boil = 296 × 10
−6 N ⋅ s
3 J
× 2270 × 10
2
kg
m
q ′′boil = 1.22 ×106 W / m2
1/2
9 . 8 m / s2 ( 962 − 0.5) k g / m3
−3
60 × 10 N / m
4212J/kg ⋅ K × 10K
J
0.006 × 2270 × 103 × ( 1.86 )1.0
kg
3
′ = qboil
′′ ( π D ) = 1.22 × 106 W / m2 (π × 0.015m ) = 57,670W/m.
q boil
The total heat rate for both processes is
q′ = ( 6052 + 57,670 ) W / m = 6.37 ×104 W/m.
<
COMMENTS: Recognize that this method provides only an estimate since the processes are surely
coupled.
PROBLEM 10.40
KNOWN: Saturated steam condensing on the outside of a brass tube and water flowing on the inside
of the tube; convection coefficients are prescribed.
FIND: Steam condensation rate per unit length of the tube.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions.
3
PROPERTIES: Table A-6, Water, vapor (0.1 bar): Tsat ≈ 320K, hfg = 2390 × 10 J/kg; Table A-1,
Brass ( T = ( Tm + Tsat ) / 2 ≈ 300K ) : k = 110 W / m ⋅ K
ANALYSIS: The condensation rate per unit length follows from Eq. 10.33 written as
& ′ = q ′/ h ′fg
m
(1)
where the heat rate follows from Eq. 10.32 using an overall heat transfer coefficient
q′ = Uo ⋅π Do ( Tsat − Tm )
and from Eq. 3.31,
1 D /2
D
D 1
Uo = + o ln o + o
k
Di Di h i
ho
(2)
−1
(3)
1
0.0095m
19
19
1
Uo =
+
ln
+
6800W/m 2 ⋅ K 1 1 0 W / m ⋅ K 16.5 16.5 5200W/m 2 ⋅ K
Uo = 147.1× 10−6 + 12.18× 10−6 + 192.3× 10−6
−1
−1
W / m 2 ⋅ K = 2627W/m 2 ⋅ K.
Combining Eqs. (1) and (2) and substituting numerical values (see below for h ′fg ), find
& ′ = U oπD o ( Tsat − Tm ) / h ′fg
m
& ′ = 2627W/m 2 ⋅ Kπ ( 0.019m )( 320 − 303 ) K/2410 ×103 J / k g = 1.11×10 −3 kg/s.
m
<
COMMENTS: (1) Note from evaluation of Eq. (3) that the thermal resistance of the brass tube is
not negligible. (2) From Eq. 10.26, with Ja = c p, l ( Tsat − T s ) / h fg , h ′fg = h fg [1 + 0.68Ja ] . Note from
expression for Uo, that the internal resistance is the largest. Hence, estimate Ts,o ≈ To – (Ro/ΣR) (To
– Tm) ≈ 313K. Hence,
h ′fg ≈ 2390 ×103 J / k g 1 + 0.68 × 4179J/kg ⋅ K (320 − 313 )K / 2390 ×103J /kg
h ′fg = 2410kJ/kg
where c p, l for water (liquid) is evaluated at Tf = (Ts,o + To)/2 ≈ 317K.
PROBLEM 10.41
KNOWN: Insulated container having cold bottom surface and exposed to saturated vapor.
FIND: Expression for growth rate of liquid layer, δ(t); thickness formed for prescribed conditions;
compare with vertical plate condensate for same conditions.
SCHEMATIC:
ASSUMPTIONS: (1) Side wall effects are negligible and, (2) Vapor-liquid interface is at Tsat , (3)
Temperature distribution in liquid is linear, (4) Constant properties.
3
PROPERTIES: Table A-6, Saturated vapor (p = 1.0133 bar): Tsat = 100°C, ρ v = 0.596 kg/m , hfg =
3
-6
2257 kJ/kg; Table A-6, Saturated liquid (Tf = 90°C = 363K): ρl = 1000 kg/m , µl = 313 × 10
2
N⋅s/m , kl = 0.676 W/m⋅K, c p, l = 4207 J/kg⋅K.
ANALYSIS: Perform a surface energy balance on the interface (see above) recognizing that
& l / A = ρl dδ /dt from an overall mass rate balance on the liquid to obtain
m
&
m
T −T
dδ
T −T
E& ′′in − E& ′′out = q′′conds − q′′cond = h fg − k l sat s = ρl
h fg − k l sat s = 0 (1)
A
δ
dt
δ
where q ′′conds is the condensation heat flux and q ′′cond is the conduction heat flux into the liquid layer
of thickness δ with linear temperature distribution. Eq. (1) can be rewritten as
ρl h fg
dδ
T −T
= k l sat s .
dt
δ
Separate variables and integrate with limits shown to obtain the liquid layer growth rate,
δ
t kl ( Tsat − Ts )
δ dδ =
dt
0
0
ρl hfg
∫
∫
1/2
or
2k ( T − T )
δ = l sat s t
ρl hfg
.
(2)
<
For the prescribed conditions, the liquid layer thickness and condensate formed in one hour are
1/2
W
kg
J
δ (1hr ) = 2 × 0.676
(100 − 80 ) °C × 3600s/1000 3 × 2257 × 103
m⋅K
kg
m
= 6.57mm
<
M (1hr ) = ρl Aδ = 1000kg/m 3 × 200 ×10 −6 m 2 × 6.57 ×10−3m = 1.314 ×10 −3 kg.
<
Continued …..
PROBLEM 10.41 (Cont.)
The condensate formed on a vertical plate with the same conditions follows from Eq. 10.33,
& ⋅ t = h L A ( Tsat − Ts ) ⋅ t / h ′fg
M vp = m
where h ′fg and h L follow from Eqs. 10.26 and 10.30, respectively.
(
h ′fg = h fg (1 + 0.68Ja ) = h fg 1 + 0.68cp,l ∆T / hfg
)
J
h ′fg = 2257 ×103 J / k g 1+ 0.68 × 4207
(100 − 80 ) °C/2257 ×103 J / k g = 2314kJ/kg
kg ⋅ K
1/4
hL = 0.943 g ρl ( ρl − ρv ) k3l h′fg / µl ( Tsat − Ts ) L
hL = 0.943 9.8m/s2 × 1000kg/m3 (1000 − 0.596) k g / m3 ( 0.676W/m ⋅ K )3
1/4
×2314 ×103 J/kg/313 ×10 −6 N ⋅ s / m 2 (100 − 80 ) °C × 0.2m
hL = 8155 W / m 2 ⋅ K.
Hence,
M vp = 8155W/m 2 ⋅ K × 200 ×10 −6 m 2 (100 − 80 ) °C × 3600s/2314 × 103 J/kg
M vp = 5.08 × 10−2 kg.
<
COMMENTS: (1) Note that the condensate formed by the vertical plate is an order of magnitude
larger. For the vertical plate the rate of condensate formation is constant. For the container bottom
surface, the rate decreases with increasing time since the conduction resistance increases as the liquid
layer thickness increases.
(2) For the vertical plate, assumed to be square 14.1 × 14.1 mm, the Reynolds number, Eq. 10.35 and
10.33, is
Reδ =
Reδ =
&
4m
4 hL A ( Tsat − Ts )
=
µl b µl b
h′fg
(
)
4
8155W/m 2 ⋅ K 200 × 10−6 m2 (100 − 80) ° C
313 ×10 −6 N ⋅ s / m 2 ×14.1×10 −3 m
2314 kJ/kg
Reδ = 12.8.
Hence, using Eq. 10.30 to estimate h L is correct since, in fact, the film is laminar.
PROBLEM 10.42
KNOWN: Vertical tube experiencing condensation of steam on its outer surface.
FIND: Heat transfer and condensation rates.
SCHEMATIC:
ASSUMPTIONS: (1) Film condensation, (2) Negligible non-condensibles, (3) D/2 >> δ, vertical plate
behavior.
3
PROPERTIES: Table A-6, Water, vapor (1.0133 bar): Tsat = 100°C, ρ v = 0.596 kg/m , hfg = 2257
3
-6
2
kJ/kg; Table A-6, Water, liquid (Tf = 97°C): ρl = 960.6 kg/m , µl = 289 × 10 N⋅s/m , c p, l = 4214
J/kg⋅K, kl = 0.679 W/m⋅K.
ANALYSIS: The heat transfer and condensation rates are
q = h L (π D L )( Tsat − Ts )
& = q / h ′fg
m
where h ′fg = h fg (1 + 0.68Ja ) and Ja = c p,l ( Tsat − Ts) / h fg. Hence Ja = 4214 J/kg⋅K (100 3
94)K/2257 × 10 J/kg = 0.0112 and h ′fg = 2274 kJ/kg. Assume laminar film condensation and use Eq.
10.31 to estimate h L ,
1/4
Nu L =
hL =
hLL
kl
0.679W/m ⋅ K
Hence,
1.0m
ρ l g ( ρ l − ρ v ) h ′fg L3
= 0.943
µ l k l ( Tsat − Ts )
1/4
960.6kg/m 3 × 9 . 8 m / s 2 ( 960.6 − 0.596 ) k g / m 3 × 2274 × 103J / kg × (1m ) 3
× 0.943
−6
2
289 × 10 N ⋅ s / m × 0.679W/m ⋅ K ( 100 − 94 ) K
q = 7360W/m 2 ⋅ K ( π × 0.100m × 1m )(100 − 94 ) K = 13.87kW.
& = 13.9 ×103 W/2274 ×10 3 J / k g = 0.00610kg/s.
m
-6
2
Check the laminar film assumption: Reδ = 4 m& / µ lb = 4 × 0.00610 kg/s/289 × 10 N⋅s/m × (π ×
0.100m) = 269. Since 30 < Reδ < 1800, the flow is wavy, not laminar. By combining Eqs. 10.33 and
10.35 with 10.38 (see Example 10.3), find Reδ,
Reδ µ l b h ′fg
4 A s ( Tsat − Ts )
=
Reδ
1.22
1.08Reδ
− 5.2
⋅
kl
(ν / g )
2
l
1/3
Continued …..
2
= 7 3 6 0 W / m ⋅ K.
PROBLEM 10.42 (Cont.)
−6
289 × 10
( π 0.10m ) × 2274 × 103J / kg
1
=
⋅
1.22
4 ( π × 0.10m × 1m ) (100 − 94 ) K
1.08Reδ − 5.2
N⋅s / m
2
(
0.679W/m ⋅ K
−6
289 × 10
/960.6
)
2
4
2
1/3
2
m / s /9.8m/s
Solving, we obtain Reδ = 311. Using Eq. 10.38, find
(
2
hL ν l / g
)
1/3
=
kl
Reδ
1.22
1.08Reδ − 5.2
2
hL = 8 5 0 7 W / m ⋅ K
q = 8 5 0 7 W / m ⋅ K ( π × 0.10m × 1m )(100 − 94 ) K = 16.0kW
<
& = 16.0 ×10 3 W / 2 2 7 4 ×10 3 J / k g = 7.05 ×10 −3 kg/s.
m
<
2
COMMENTS: To determine whether the assumption D/2 >> δ is satisfied, use Eq. 10.25 to estimate
δ(L) ≈ 0.12mm. Despite the laminar film assumption, clearly the assumption is justified and the vertical
plate correlation is applicable.
PROBLEM 10.43
KNOWN: Vertical tube experiencing condensation of steam on its outer surface.
FIND: Heat transfer and condensation rates.
SCHEMATIC:
ASSUMPTIONS: (1) Film condensation, (2) Negligible condensibles in steam, (3) D/2 >> δ, vertical
plate behavior.
3
PROPERTIES: Table A-6, Water, vapor (1.5 bar): Tsat ≈ 385K, ρ v = 0.876 kg/m , hfg = 2225
3
-6
kJ/kg; Table A-6, Water, (liquid Tf = 376K): ρl = 956.2 kg/m , c p, l = 4220 J/kg⋅K, µl = 271 × 10
2
N⋅s/m , kl = 0.681 W/m⋅K.
ANALYSIS: The heat transfer and condensation rates are
q = h L (π D L )( Tsat − Ts )
& = q / h ′fg
m
where h ′fg = h fg (1 + 0.68Ja ) and Ja = cp, l ( Tsat − Ts ) / h fg . Hence, Ja = 4220 J/kg⋅K (385 –
3
367)K/2225 × 10 J/kg = 0.0171 and h ′fg = 2277 kJ/kg. Assume the flow is wavy. Combine Eqs.
10.33 and 10.35 with 10.38, find Reδ.
Reδ µ l b h ′fg
Reδ
=
⋅
4 As ( Tsat − Ts ) 1.08Re1.22 − 5.2
δ
(
kl
ν2 /g
)
1/3
271× 10−6 N ⋅ s / m 2 (π × 0.10m ) × 2277 × 103 J/kg
4 × ( π × 0.10m ×1m )( 385 − 367 ) K
=
1
⋅
0.681W/m ⋅ K
1/3
1.08Re1.22
δ − 5.2 271×10 −6 /956.2 2 m 4 / s 2 /9.8m/s2
Reδ = 832.
Using Eq. 10.38, find
(
h L ν l2 / g
kl
)
(
)
1/3
=
Reδ
1.08Re1.22
δ − 5.2
h L = 7,127W/m 2 ⋅ K.
q = 7127W/m 2 ⋅ K (π × 0.1m ×1m )( 385 − 367 ) K = 40.3kW
<
<
& = 40.3 ×103 W/2277 ×103 J / k g = 0.0177kg/s.
m
Continued …..
PROBLEM 10.43 (Cont.)
COMMENTS: Since 30 < Reδ < 1800, the wavy flow film assumption is justified. By comparing
these results with those of Problem 10.42, the effect of increased pressure on condensation can be
seen.
p (bar)
Tsat(K)
Tsat-Ts(K)
373
385
6
18
(
2
hL W / m ⋅ K
)
q (kW)
& ⋅ 10 3 ( k g / s )
m
1.01
1.5
8507
7127
16.0
40.3
7.05
17.7
The effect of increasing the pressure from 1.01 to 1.5 bar is to increase the excess temperature threefold, to decrease hL by 16%, and to increase the rates by a factor of 2.5.
PROBLEM 10.44
KNOWN: Saturated steam at one atmosphere condenses on the outer surface of a vertical tube;
water flow within tube experiences 4°C temperature rise.
FIND: Required flow rate to maintain tube wall at 94°C.
SCHEMATIC:
ASSUMPTIONS: (1) Laminar wavy film condensation on a vertical surface, (2) Negligible
concentration of non-condensible gases in the stream, (3) Thermal resistance of tube wall is negligible,
(4) Water flow is fully developed, (5) Tube wall surface is at uniform temperature Ts.
-6
2
PROPERTIES: Table A-6, Water (Assume Tm ≈ 300K): cp = 4179 J/kg⋅K, µ = 855 × 10 N⋅s/m ,
k = 0.613 W/m⋅K, Pr = 5.83.
ANALYSIS: From the results of Problem 10.42, the heat rate for laminar wavy condensation on the
outside surface of the tube was found to be q = 16.0 kW. From an energy balance on the water
flowing within the tube, the flow rate is
& = q / c p Tm,o − Tm,i = 16.0 ×103 W/4179J/kg ⋅ K × 4K = 0.957kg/s.
m
(1)
To determine the inlet temperature of the water, the rate equation is required.
(
)
q = U A ∆Tlm
<
U=
1
1 / hi + 1 / ho
∆Tlm =
∆T1 − ∆ T2
.
ln ( ∆T1 / ∆T2 )
(2,3,4)
2
From Problem 10.42, ho = 8507 W/m ⋅K. Evaluate Re for the water flow using Eq. 8.6
& π Dµ = 4 × 0.957kg/s/ π × 0.092m × 855 ×10 −6 N ⋅ s / m 2 = 15,493.
Re = 4m/
The flow is turbulent and since fully-developed, Eq. 8.60 is an appropriate correlation.
Nu = h i D i / k = 0.023Re4D/ 5 Pr 0.4 = 0.023 (15,493)
4/5
( 5.83 )0.4 = 104.7
h i = Nu ⋅ k / Di = 104.7 × 0.613W/m⋅ K/0.092m = 6 9 8 W / m 2 ⋅ K.
2
Hence, U = 1/[1/698 + 1/8507] = 645 W/m ⋅K. Substituting numerical values into the rate equation,
Eq. (2), with A = π Di L, find
∆Tlm = q / U A = 16.0 × 103 W / 6 4 5 W / m 2 ⋅ K × (π 0.092m × 1m ) = 85.6K.
Recalling now Eq. (4), note that ∆T1 - ∆T2 = 4K and that Tm,o – Tm,i = 4K, hence,
85.6K = 4K/ln
94 − Tm,i
(
94 − Tm,i + 4
)
giving
Tm,i = 6.3°C.
COMMENTS: Note that the Tm = 300K assumption is not reasonable and an iteration should be
made. Also, it is likely that the thermal resistance of the tube wall is not negligible.
PROBLEM 10.45
KNOWN: Cooled vertical plate 500-mm high and 200-mm wide condensing saturated steam at 1 atm.
= 25 kg/h, (b)
FIND: (a) Surface temperature, Ts, required to achieve a condensation rate of m
≤ 50 kg h , and (c)
Compute and plot Ts as a function of the condensation rate for the range 15 ≤ m
, but if the plate is 200 mm high and 500 mm wide (vs. 500
Compute and plot Ts for the same range of m
mm high and 200 mm wide for parts (a) and (b)).
SCHEMATIC:
ASSUMPTIONS: (1) Film condensation, (2) Negligible non-condensables in steam.
PROPERTIES: Table A-6, Water, vapor (1.0133 bar): Tsat = 100°C, ρv = 0.5963 kg m3 , hfg = 2257
kJ/kg; Table A-6, Water, liquid (Tf ≈ (74 + 100)°C/2 ≈ 360 K): ρ" = 967.1 kg/m3 , cp," = 4203
J/kg⋅K, µ" = 324 × 10-6 N⋅s/m2 , k " = 0.674 W/m⋅K.
ANALYSIS: (a) The surface temperature can be determined from the rate equation, Eq. 10.32, written as
′fg h L As
Ts = Tsat − q h L As = Tsat − mh
where h ′fg = hfg (1 + 0.68 Ja) and Ja = c p," (Tsat − Ts )/hfg . To evaluate Ts, we need values of hL and
h ′fg , both of which require knowledge of Ts . Hence, we need to assume a value of Ts and iterate the
solution until good agreement with calculated Ts value is achieved. Assume Ts = 74°C and evaluate h ′fg
and Reδ .
(
)
h ′fg = 2257 kJ kg 1 + 0.68 4203J kg ⋅ K (100 − 74 ) K 2257 × 103 J kg = 2331kJ kg
−6
µ b = 4 × ( 25 3600 ) kg s 324 × 10
Reδ = 4m
N ⋅ s m 2 × 0.2m = 429 .
"
Since 30 < Reδ < 1800, the flow is wavy-laminar and Eq. 10.38 is appropriate,
Reδ
k"
hL =
⋅
1/ 3
1.08 Re1.22
δ − 5.2 ν 2 g
( )
"
hL =
429
108 ( 429 )
1.22
⋅
− 5.2
0.674 W m ⋅ K
(
)
1/ 3
= 7312 W m 2 ⋅ K
2
4 2
2
−6
324 × 10 967.1 m s 9.8 m s
kg
J
W
7312
Hence, Ts = 100$ C − ( 25 3600 ) × 2331 × 103
× ( 0.2 × 0.5 ) m 2 = 78$ C .
2
kg
s
m ⋅K
<
This value is to be compared to the assumed value of 74°C. See comment 1.
(b,c) Using the IHT Correlations Tool, Film Condensation, Vertical Plate for laminar, wavy-laminar
and turbulent regions, combined with the Properties Tool for Water, the surface temperature Ts was
Continued...
PROBLEM 10.45 (Cont.)
, considering the two plate configurations as
calculated as a function of the condensation rate, m
indicated in the plot below.
Plate temperature, Ts (C)
100
80
60
40
15
25
35
45
Condensation rate, mdot (kg/h)
500 mm high x 200 mm wide
200 mm high x 500 mm wide
As expected the condensation rate increases with decreasing surface temperature. The plate with the
shorter height (L = 200 mm vs 500 mm) will have the thinner boundary layer and, hence, the higher
average convection coefficient. Since both plate configurations have the same total surface area, the 200mm height plate will have the larger heat transfer and condensation rates. For the range of conditions
examined, the condensate flow is in the wavy-laminar region.
COMMENTS: (1) With the IHT model developed for parts (b) and (c), the result for the part (a)
2
= 25 kg/h is Ts = 78.2°C (Reδ = 439 and h L = 7403 W/m ⋅ K) . Hence, the assumed
conditions with m
value (Ts = 74°C) required to initiate the analysis was a good one.
(2) A copy of the IHT Workspace model used to generate the above plot is shown below.
/* Correlations Tool
- Film Condensation, Vertical Plate, Laminar, wavy-laminar and turbulent regions: */
NuLbar = NuL_bar_FCO_VP(Redelta,Prl) // Eq 10.37, 38, 39
NuLbar = hLbar * (nul^2 / g)^(1/3) / kl
g = 9.8
// Gravitational constant, m/s^2
Ts = Ts_C + 273
// Surface temperature, K
Ts_C = 78
// Initial guess value used to solve the model
Tsat = 100 + 273
// Saturation temperature, K
// The liquid properties are evaluated at the film temperature, Tf,
Tf = Tfluid_avg(Ts,Tsat)
// The condensation and heat rates are
q = hLbar * As * (Tsat - Ts)
// Eq 10.32
As = L * b
// Surface Area, m^2
mdot = q / h'fg
// Eq 10.33
h'fg = hfg + 0.68 * cpl * (Tsat - Ts)
// Eq 10.26
// The Reynolds number based upon film thickness is
Redelta = 4 * mdot / (mul * b)
// Eq 10.35
// Assigned Variables:
L = 0.5
// Vertical height, m
b = 0.2
// Width, m
mdot_h = mdot * 3600
// Condensation rate, kg/h
//mdot_h = 25
// Design value, part (a)
// Properties Tool - Water:
// Water property functions :T dependence, From Table A.6
// Units: T(K), p(bars);
xl = 0
// Quality (0=sat liquid or 1=sat vapor)
rhol = rho_Tx("Water",Tf,xl) // Density, kg/m^3
hfg = hfg_T("Water",Tsat)
// Heat of vaporization, J/kg
cpl = cp_Tx("Water",Tf,xl)
// Specific heat, J/kg·K
mul = mu_Tx("Water",Tf,xl) // Viscosity, N·s/m^2
nul = nu_Tx("Water",Tf,xl)
// Kinematic viscosity, m^2/s
kl = k_Tx("Water",Tf,xl)
// Thermal conductivity, W/m·K
Prl = Pr_Tx("Water",Tf,xl)
// Prandtl number
PROBLEM 10.46
KNOWN: Plate dimensions, temperature and inclination. Pressure of saturated steam.
FIND: (a) Heat transfer and condensation rates for vertical plate, (b) Heat transfer and condensation
rates for inclined plate.
SCHEMATIC:
ASSUMPTIONS: (1) Conditions correspond to the turbulent film region, (2) Constant properties.
3
PROPERTIES: Table A-6, saturated vapor (p=1.0133 bars): Tsat = 100°C, ρv = 0.596 kg/m , hfg =
2257 kJ/kg. Table A-6, saturated liquid (Tf = 75°C): ρ " = 975 kg / m3 , µ" = 375 × 10−6 N ⋅ s / m 2 ,
k " = 0.668 W / m ⋅ K, c p," = 4193 J / kg ⋅ K.
ANALYSIS: (a) Expressing hL in terms of Reδ by combining Eqs. (10.33) and (10.35) and
substituting into Eq. (10.38), it follows that
Reδ µ" h′fg
4 L (Tsat − Ts )
=
Reδ
⋅
(
k"
)
1/ 3
1.08 Re1.22
δ − 5.2 ν "2 / g
(1)
where, with Ja = c p," ( Tsat − Ts ) / h fg = 0.0929, h ′fg = h fg (1 + 0.68 Ja ) = 2400 kJ / kg. From an iterative
solution to Eq. (1), we obtain Reδ = 2370, and the assumption of a turbulent film is justified. From
Eqs. (10.35) and (10.33) the condensation and heat rates are then
=
m
µ" b Reδ
= 0.444 kg / s
4
<
h′fg = 0.444 kg / s × 2.4 × 106 J / kg = 1.065 × 106 W
q=m
<
From Eq. (10.32), we also obtain hL = q / [( bL )( Tsat − Ts )] = 5325 W / m 2 ⋅ K.
(b) With hL(incl ) ≈ ( cos θ )1/ 4 hL , we obtain hL(incl ) ≈ 0.917 × 5325 W / m 2 ⋅ K = 4880 W / m 2 ⋅ K. If
the inclination reduces hL by 8.73%, the heat and condensation rates are reduced by equivalent
amounts. Hence,
= 0.407 kg / s,
m
q = 0.977 × 106 W
<
COMMENTS: The initial guess of a turbulent film region was motivated by the value of L = 2m,
which was believed to be large enough for transition to turbulence. Note that the solution could also
have been obtained by accessing the Film Condensation correlations of IHT, implementation of which
does not require an assumption of flow conditions.
PROBLEM 10.47
KNOWN: Saturated ethylene glycol (1 atm) condensing on a vertical plate at 420K.
FIND: Heat transfer rate to the plate and condensation rate.
SCHEMATIC:
ASSUMPTIONS: (1) Film condensation, (2) Negligible non-condensible gases in vapor.
3
PROPERTIES: Table A-5, Ethylene glycol vapor (1 atm): Tsat = 470K, ρ v ≈ 0 kg/m , h fg = 812 kJ/kg; Table
A-5, Ethylene glycol, liquid (Tf = (Ts + Tsat )/2 ≈ 445K; use properties at upper limit of table 373K): ρl = 1058.5
3
-2
kg/m , c p, l = 2742 J/kg⋅K, µl = 0.215 × 10
2
N⋅s/m , kl = 0.263, W/m⋅K.
ANALYSIS: The heat transfer and condensation rates are given by Eqs. 10.32 and 10.33.
q = h L A s ( Tsat − Ts )
& = q / h ′fg ,
m
where h ′fg = h fg (1 + 0.68 Ja) and Ja = c p, l ( Tsat − Ts ) / h fg . Substituting property values at Tf = (Ts +
3
Tsat )/2, find h ′fg = 812 kJ/kg (1 + 0.68 [2742 J/kg⋅K (470 – 420)K/812 × 10 J/kg]) = 905 kJ/kg.
Assuming the flow is laminar, use Eq. 10.30 to evaluate hL .
hL
g ρ l ( ρl − ρv ) k 3l h ′fg
= 0.943
µ l ( Tsat − Ts ) L
1/4
1/4
9 . 8 m / s 2 × 1 0 5 8 . 5 k g / m 3 ( 1058.5 − 0 ) k g / m 3 ( 0 . 2 6 3 W / m ⋅ K )3 × 905 × 10 3J / k g
−2
2
0.215 × 10 N ⋅ s / m ( 470 − 420 ) K × 0.3m
2
find hL = 1451 W/m ⋅K. Using the rate equations, find
q = 1451 W / m ⋅ K ( 0.3 × 0.1 ) m
2
2
( 470 − 420 ) K = 2.18kW
& = 2.18 × 10 3 W / 9 0 5 ×10 3J / k g = 0.002405kg/s = 8.66kg/h.
m
Determine whether the flow is indeed laminar:
& / µlb = 4 × 0.002405
Re δ = 4 m
-2
2
kg/s/0.215 × 10 N⋅s/m ×
0.1m = 44.7. Since 30 < Reδ < 1800, the flow is in the wavy-laminar region. Hence, the correlation of
Eq. 10.38 is more appropriate. Combining Eq. 10.33 and 10.35 with 10.38 (see Example 10.3),
Reδ µ l b h ′fg
4 A s ( Tsat − Ts )
−2
0.215 × 10
=
Reδ
1.22
1.08Reδ − 5.2
2
3
N ⋅ s / m × 0.1m × 905 × 10 J / k g
4 × ( 0.3 × 0.1 ) m
2
( 470 − 420 ) K
⋅
kl
(ν / g )
2
l
1/3
0.263W/m ⋅ K
1
=
1.22
1.08Reδ
− 5.2
(
−2
0.215 × 10
)
4
2
2
/1058.5 m / s / 9 . 8 m / s
1/3
find Reδ = 45 and using Eq. 10.38, find
hL =
Hence,
Reδ
1.22
1.08Reδ − 5.2
q = 2.21kW
⋅
kl
(ν / g )
2
l
1/3
2
=1 4 7 0 W / m ⋅ K.
& = 2.44 × 10 −3 kg/s.
m
COMMENTS: Note the wavy-laminar value of Reδ is within 1.3% of the laminar value.
<
PROBLEM 10.48
KNOWN: Vertical plate 2.5 m high at a surface temperature Ts = 54°C exposed to steam at atmospheric
pressure.
FIND: (a) Condensation and heat transfer rates, (b) Whether turbulent flow would still exist if the height
were halved, and (c) Compute and plot the condensation rates for the two plate heights (2.5 m and 1.25
m) as a function of surface temperature for the range, 54 ≤ Ts ≤ 90°C.
SCHEMATIC:
ASSUMPTIONS: (1) Film condensation, (2) Negligible non-condensables in steam.
PROPERTIES: Table A-6, Water, vapor (1 atm): Tsat = 100°C, ρv = 0.596 kg/m3, hfg = 2257 kJ/kg;
Table A-6, Water, liquid (Tf = (100 + 54)°C/2 = 350 K): ρ" = 973.7 kg/m3, k " = 0.668 W/m⋅K, µ" = 365
× 10-6 N⋅s/m2 , cp," = 4195 J/kg⋅K, Pr" = 2.29.
ANALYSIS: (a) The heat transfer and condensation rates are given by Eqs. 10.32 and 10.33,
q′ = h L L ( Tsat − Ts )
′ = q′ h ′fg
m
(1,2)
where, from Eq. 10.26, with Ja = c p," (Tsat − Ts)/hfg ,
(
h′fg = h fg 1 + 0.68 c p," (Tsat − Ts ) h fg
h′fg = 2257
)
4195 J kg ⋅ K (100 − 54 ) K
kJ
1 + 0.68
= 2388 kJ kg .
3 J kg
kg
2257
10
×
Assuming turbulent flow conditions, Eq. 10.39 is the appropriate correlation,
1/ 3
2
(
hL ν " g
)
k"
=
Reδ
(
8750 + 58 Pr −0.5 Reδ0.75 − 253
Reδ > 1800
)
(3)
Not knowing Reδ or h L , another relation is required. Combine Eq. 10.33 and 10.35,
h ′fg
Re µ b
= δ "
.
A ( Tsat − T )
4
A ( Tsat − T )
Substitute Eq. (4) for h L into Eq. (3), with A = bL,
Reδ µ" bh′fg
Reδ
k"
.
=
⋅
1/ 3
4 ( bL )(Tsat − T ) 8750 + 58 Pr −0.5 Re0.75 − 253
2
ν" g
"
δ
hL =
′fg
mh
(
)(
)
(4)
(5)
Using appropriate properties with L = 2.5 m, find
Continued...
PROBLEM 10.48 (Cont.)
365 × 10−6 N ⋅ s m 2 × 2388 × 103 J kg
(6)
4 × 2.5m (100 − 54 ) K
1
0.668 W m ⋅ K
=
⋅
1/ 3
−0.5
2 4 2
8750 + 58 ( 2.29 )
Reδ0.75 − 253
6
2
−
365 × 10 973.7 m s 9.8m s
Reδ = 2979 .
) (
(
)
Note that Reδ > 1800, so indeed the flow is turbulent, and using Eq. (4) or (3), find
h L = 5645 W m 2 ⋅ K .
From the rate equations (1) and (2), the heat transfer and condensation rates are
q′ = 5645 W m 2 ⋅ K × 2.5m 100 − 54 K = 649k W m
(
)
<
<
′ = 649 ×103 W m 2388 ×103 J kg = 0.272 kg s ⋅ m .
m
(b) If the height of the plate were halved, L = 1.25 m, Eq. (6) would only need to be modified for this
new value. Using the calculated values for the LHS and the last term on the RHS, Eq. (6) becomes,
3.78960 =
1
−0.5
8750 + 58 ( 2.29 )
(
Reδ0.75 − 253
)
× 27, 493
(7)
and after some manipulation , find
Reδ = 1280 .
Since 1800 > Reδ , the flow is not turbulent, but wavy-laminar. Now the procedure follows that of
Example 10.3. For L = 1.25 m with wavy-laminar flow, Eq. 10.38 is the appropriate correlation. The
calculations yield these results:
Reδ = 1372
h L = 5199 W m 2 ⋅ K
q′ = 299 kW m
<
′ = 0.125 kg s ⋅ m .
m
Note that the height was decreased by a factor of 2 while the rates decreased by a factor of 2.2! Would
you have expected this result?
(c) Using the IHT Correlation Tool, Film Condensation, Vertical Plate for laminar, wavy-laminar, and
turbulent regions, combined with the Properties Tool for Water, the condensation rates were calculated
as a function of the surface temperature considering the two plate heights indicated.
Condensation rate, mdot' (kg/s.m)
0.5
0.375
0.25
0.125
0
50
60
70
80
90
Plate temperature, Ts (C)
2.5 m high plate
1.25 m high plate
Continued...
PROBLEM 10.48 (Cont.)
The condensation rate decreases nearly linearly with increasing surface temperature. The inflection in
the upper curve (L = 2.5 m) corresponds to the flow transition at Reδ = 1800 between wavy-laminar and
turbulent. For surface temperature lower than 76°C, the flow is turbulent over the 2.5 m plate. The flow
over the 1.25 m plate is always in the wavy-laminar region. The fact that the 2.5 m plate experiences
turbulent flow explains the height-rate relationship mentioned in the closing sentences of part (b).
COMMENTS: A copy of the IHT model used to generate the above plot is shown below.
/* Correlations Tool
- Film Condensation, Vertical Plate, Laminar, wavy-laminar and turbulent regions: */
NuLbar = NuL_bar_FCO_VP(Redelta,Prl) // Eq 10.37, 38, 39
NuLbar = hLbar * (nul^2 / g)^(1/3) / kl
g = 9.8
// Gravitational constant, m/s^2
Ts = Ts_C + 273
// Surface temperature, K
Ts_C = 54
// Part (a) design condition
Tsat = 100 + 273
// Saturation temperature, K
// The liquid properties are evaluated at the film temperature, Tf,
Tf = Tfluid_avg(Ts,Tsat)
// The condensation and heat rates are
q = hLbar * As * (Tsat - Ts)
// Eq 10.32
As = L * b
// Surface Area, m^2
mdot = q / h'fg
// Eq 10.33
h'fg = hfg + 0.68 * cpl * (Tsat - Ts)
// Eq 10.26
// The Reynolds number based upon film thickness is
Redelta = 4 * mdot / (mul * b)
// Eq 10.35
// Assigned Variables:
L = 1.25
// Height, m
b=1
// Width, m
// Properties Tool - Water:
// Water property functions :T dependence, From Table A.6
// Units: T(K), p(bars);
xl = 0
// Quality (0=sat liquid or 1=sat vapor)
rhol = rho_Tx("Water",Tf,xl) // Density, kg/m^3
hfg = hfg_T("Water",Tsat)
// Heat of vaporization, J/kg
cpl = cp_Tx("Water",Tf,xl)
// Specific heat, J/kg·K
mul = mu_Tx("Water",Tf,xl) // Viscosity, N·s/m^2
nul = nu_Tx("Water",Tf,xl)
// Kinematic viscosity, m^2/s
kl = k_Tx("Water",Tf,xl)
// Thermal conductivity, W/m·K
Prl = Pr_Tx("Water",Tf,xl)
// Prandtl number
PROBLEM 10.49
KNOWN: Two vertical plate configurations maintained at 90°C for condensing saturated steam at 1
atm: single plate L × w and two plates each L/2 × w where L and w are the vertical and horizontal
dimensions, respectively.
FIND: Which case will provide the larger heat transfer or condensation rate.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible concentration of non-condensible gases in the steam.
3
PROPERTIES: Table A-6, Saturated water vapor (1 atm): Tsat = 100°C, ρ v = (1/vg) = 0.596 kg/m ,
hfg = 2257 kJ/kg; Saturated water (Tf = (Ts + Tsat )/2 = (90 + 100)°C/2 = 95°C = 368K): ρ l = (1/vf) =
3
-6
2
962 kg/m , µl = 296 × 10 N⋅s/m , k l = 0.678 W/m⋅K, cp,l = 4212 J/kg⋅K.
ANALYSIS: The heat transfer and condensation rates are
q = h LA s ( Tsat − Ts )
& = q / h ′fg
m
where, for the two cases,
hL,1As,1 = hL,1 ( L) [ L × w]
hL,2 As,2 = hL,2 ( L / 2) 2 ( L / 2× w )
and the average convection coefficients are evaluated at L and L/2, respectively. Hence,
h L,1 ( L ) [ L × w ]
h L,1 ( L )
&
q1 m
= 1 =
=
.
& 2 hL,2 ( L / 2) 2 ( L / 2 × w ) hL,2 ( L / 2)
q2 m
-1/4
For laminar film condensation on both plates, using the correlation of Eq. 10.30, with hL α L
q1 / q 2 = ( L / [ L / 2 ])
−1/4
= 0.84.
<
Hence, case 2 is preferred and provides 16% more heat transfer.
When Reδ = 30 for case 1 with the given conditions, find from Eq. 10.37
(
h L ν l2 / g
kl
)
1/3
1/3
2
−
6
2
3
2
h L 296 × 10 N ⋅ s / m /962kg/m
/9.8m/s
=
0.678W/m ⋅ K
(
,
)
Continued …..
PROBLEM 10.49 (Cont.)
(
h L ν l2 / g
kl
)
1/3
−1/3
= 1.47Reδ−1/3 = 1.47 ( 30 )
hL = 15,061W/m 2 ⋅ K
and then from Eq. 10.30,
1/4
g ρ ( ρ − ρ )k 3 h ′
l l
v l fg
hL = 0.943
µ l ( Tsat − Ts ) L
where
h ′fg = h fg + 0.68c p,l ( Tsat − Ts )
h ′fg = 2257kJ/kg + 0.68 × 4212J/kg ⋅ K (100 − 90 ) K = 2286kJ/kg,
15,061W/m2 ⋅ K =
1/4
9.8m/s2 × 962kg/m 3 ( 962 − 0.596) k g / m3 ( 0.678W/m ⋅ K )3
0.943
2286kJ/kg
296 ×10−6 N ⋅ s / m 2 (100 − 90 ) K L
L = 34 mm.
We can anticipate for other, larger values of L that the comparison of hL values cannot be so easily
made. However, according to Figure 10.15, we expect the same behavior of hL in the wavy region
and anticipate that indeed case 2 will provide the greater condensation rate. Note that in the turbulent
region with the increase in hL with Reδ, we cannot conclude with certainty which case is preferred.
COMMENTS: In dealing with single-phase, forced or free convection, we associate thin thermal
boundary layers with higher heat transfer rates. For vertical plates, we would expect the shorter plate
to have the higher convection heat transfer coefficient. The results of this problem suggest the same is
true for condensation on the vertical plate.
PROBLEM 10.50
KNOWN: Number, diameter and wall temperature of condenser tubes in a square array. Pressure of
saturated steam around tubes.
FIND: Rates of heat transfer and condensation per unit length of the array.
SCHEMATIC:
ASSUMPTIONS: (1) Laminar film condensation on tubes, (2) Negligible concentration of
noncondensable gases in steam.
PROPERTIES: Table A-6, saturated vapor (psat = 0.105 bar): Tsat = 320 K = 47°C, ρv = 0.0715
3
kg/m , hfg = 2390 kJ/kg. Table A-6, saturated liquid (Tf = 32°C = 305 K): ρ " = 995 kg / m3 ,
µ" = 769 × 10
−6
2
N ⋅ s / m , k " = 0.620 W / m ⋅ K, c p," = 4178 J / kg ⋅ K.
ANALYSIS: The average heat rate per unit length for a single tube is q1′ = hD,N (π D )( Tsat − Ts ) ,
where hD,N is obtained from Eq. 10.41. With Ja = c p," ( Tsat − Ts ) / h fg = 0.052 and h ′fg = h fg (l +
6
6
0.68 Ja) = 1.04 (2.390 × 10 J/kg) = 2.48 × 10 J/kg,
1/ 4
h D,N
g ρ " ( ρ " − ρ v ) k 3" h ′fg
= 0.729
N µ" ( Tsat − Ts ) D
1/ 4
hD,N
9.8 m / s 2 × 995 kg / m3 (995 − 0.0715 ) kg / m3 ( 0.62 W / m ⋅ K )3 2.48 × 106 J / kg
= 0.729
2
−6
25 × 769 × 10 N ⋅ s / m ( 30°C ) 0.025m
2
= 3260 W / m ⋅ K
The heat rate per unit length of the array is q ′ = N 2q ′l . Hence,
q′ = N 2 h D,N (π D )( Tsat − Ts ) = 625 × 3260 W / m 2 ⋅ K (π × 0.025m ) 30°C = 4.79 × 106 W / m
<
The corresponding condensation rate is
′=
m
q′
4.79 × 106 W / m
=
= 1.93kg / s ⋅ m
h ′fg 2.48 × 106 J / kg
<
COMMENTS: Because of turbulence generation due to splashing from one tube to another in a
vertical column, the foregoing value of hD,N is expected to underestimate the actual value of hD,N
and hence to underpredict the heat and condensation rates.
PROBLEM 10.51
KNOWN: Tube wall diameters and thermal conductivity. Mean temperature and flow rate of water
flow through tube. Pressure of saturated steam around tube.
FIND: (a) Rates of heat transfer and condensation per unit length, (b) Effect of flow rate on heat
transfer.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible concentration of noncondensible gases in the steam, (2) Uniform
tube surface temperatures, (3) Laminar film condensation, (4) Fully-developed internal flow, (5)
Constant properties.
2
PROPERTIES: Table A-6, water (Tm = 290 K): µ = 0.00108 N⋅s/m , k = 0.598 W/m⋅K, Pr = 7.56.
3
Table A-6, saturated vapor (p = 0.135 bar): Tsat = 325 K = 52°C, ρv = 0.0904 kg/m , hfg = 2378
kJ/kg. Table A-6, saturated liquid (Tf ≈ Tsat): ρ " = 987 kg / m3 , c p," = 4182 J / kg ⋅ K,
µ" = 528 × 10
−6
2
N ⋅ s / m , k " = 0.645 W / m ⋅ K.
ANALYSIS: (a) From the thermal circuit, the heat rate may be expressed as
q′ =
where,
Tsat − Tm
R ′fc + R ′cond + R ′conv
(1)
R ′cond = n ( Do / Di ) / 2π k s = 0.00152 m ⋅ K / W
/ π D µ = 11, 336, the flow is
The convection resistance is R ′conv = (π Di h i )−1 . With Re D = 4m
i
turbulent and the Dittus-Boelter correlation yields
k
4/5
0.4
4 / 5 0.4 0.598 W / m ⋅ K
2
hi =
0.023 (11, 336 ) ( 7.56 ) = 2082 W / m ⋅ K
0.023 Re D Pr =
0.026m
Di
The convection resistance is then
−1
−1
R ′conv = (π Di hi ) = π × 0.026m × 2082 W / m 2 ⋅ K
= 0.00588 m ⋅ K / W
)
(
The resistance associated with the condensate film is R ′fc = (π D o ho ) , where ho is given by Eq.
10.40. With C = 0.729,
1/ 4
ho
1/ 4
gρ " ( ρ " − ρ v ) k 3" h fg
′
=C
µ" ( Tsat − Ts,o ) Do
9.8 m / s 2 × 987 (987 − 0.09 ) kg 2 / m 6 ( 0.645 W / m ⋅ K )3 h ′fg
= 0.729
2
−6
528 × 10 N ⋅ s / m (325 − Ts,o ) × 0.030m
1/ 4
W3 ⋅ kg
h o = 462
m8 ⋅ K3 ⋅ s
(
1/ 4
h′fg
325
T
−
s,o
)
(
)
where h ′fg = h fg + 0.68c p," Tsat − Ts,o = 2.38 × 106 J / kg + 2844 J / kg ⋅ K 325 − Ts,o
The unknown surface temperature may be determined from an additional rate equation, such as
Continued …..
q′ =
PROBLEM 10.51 (Cont.)
Ts,o − Tm
(2)
R ′cond + R ′conv
Substituting the thermal resistances into Eqs. (1) and (2), an iterative solution yields
q′ = 4270 W / m
Ts,o = 321.6 K = 48.6°C
<
The condensation rate is then
′cond =
m
q′
4270 W / m
=
= 0.00179 kg / s ⋅ m
h ′fg 2.39 ×106 J / kg
<
The corresponding values of the condensate convection coefficient and resistance are
h o = 13,380 W / m 2 ⋅ K
R ′fc = 0.000793m ⋅ K / W
and
Because R ′conv is much larger than R ′cond and R ′fc , attention should be paid to reducing the
convection resistance in order to increase the heat rate. The resistance to heat flow by convection is
the limiting factor.
(b) The effects of varying the flow rate are shown below
50
10000
Su rfa ce te m p e ra ture , Ts ,o (C )
H e a t ra te (W /m )
9000
8000
7000
6000
5000
48
46
44
42
40
4000
0
0 .2 0 .4 0 .6 0 .8
1
1 .2 1 .4 1 .6 1 .8
Tu b e flo w ra te (kg /s )
2
0
0 .2 0 .4 0 .6 0 .8
1
1 .2 1 .4 1 .6 1 .8
Tu be flo w ra te (kg /s )
The effect of increasing m on q ′ is significant and is accompanied by a reduction in Ts,o.
COMMENTS: (1) Use of the IHT convection and condensation correlations, as well as its
temperature-dependent properties of water facilitated the numerical solution. (2) Evaluation of the
film properties at Tsat is reasonable for part (a), since Tf = (Ts,o + Tsat)/2 = 50.3°C ≈ Tsat. However,
with increasing m and hence decreasing Ts,o, the approximation would become inappropriate.
2
PROBLEM 10.52
KNOWN: Inner surface of a vertical thin-walled container of length L and diameter D experiences
condensation of a saturated vapor. Container wall maintained at a uniform surface temperature by
flowing cold water across its outer surface.
FIND: Expression for the time, tf , required to fill the container with condensate assuming the condensate
film is laminar. Express your result in terms of D, L, (Tsat − Ts), g and appropriate fluid properties.
SCHEMATIC:
ASSUMPTIONS: (1) Laminar film condensation on a vertical surface, (2) Uniform temperature
container wall surface, and (3) Mass of liquid condensate in the laminar film negligible compared to
liquid mass on bottom of container.
ANALYSIS: From an instantaneous mass balance on the container,
m(t)
=
dM
dt
(1)
(t) is the condensate rate and the liquid mass in the container, M, is
Where m
M = ρ π D2 4 L − x
"
(
)(
)
(2)
The condensate rate from Eq. 10.33 can be expressed as
(t ) =
m
h A ( T − Ts )
q
= s s sat
h′fg
h′fg
where the average film coefficient over the height 0 to x from Eq. 10.30 is,
1/ 4
gρ ( ρ − ρ ) k 3h ′
"
"
v
fg
"
hs = 0.943
µ" (Tsat − Ts ) x
(3)
(4)
and the surface area over which condensation occurs is
As = π Dx
(5)
Continued...
PROBLEM 10.52 (Cont.)
Substituting Eqs (2-5) into Eq. (1),
1/ 4
gρ ( ρ − ρ ) k 3h ′
"
"
ν " fg
0.943
µ" ( Tsat − Ts ) L
L1/ 4
x1/ 4
(π Dx )
(
h ′fg = − ρ" π D2 4
) dxdt
(6)
Separate variables and identify the limits of integration,
1/ 4
gρ ( ρ − ρ ) k 3h ′
t
0
"
"
fg
ν
1/
4
2
"
L (π D ) h′fg ρ " π D 4 ∫ f dt = − ∫
x −3/ 4dx
0.943
x =L
0
µ" ( Tsat − Ts ) L
(
)
(7)
The RHS integrates to
0
− x1/ 4 (1 4 ) = 4L1/ 4
L
(8)
and solving for tf,
2
ρ" π D 4 Lh ′fg
tf = 4
1/
4
gρ ( ρ − ρ ) k 3h ′
"
"
v " fg
0.943
π
DL
T
T
−
(
)( sat s )
µ" ( Tsat − Ts ) L
(
)
<
COMMENTS: The numerator and denominator in the bracketed expression are of special significance.
The numerator is product of the mass in the filled container and the latent heat of vaporization; that is,
the total energy removed by the cold water. What is physical significance of the denominator? Can you
interpret the time-to-fill, tf , expression in light of these terms?
PROBLEM 10.53
KNOWN: Tube of Problem 10.42 in horizontal position experiences condensation on its outer surface.
FIND: Heat transfer and condensation rates.
SCHEMATIC:
ASSUMPTIONS: (1) Laminar film condensation, (2) End effects negligible, (3) Negligible
concentration of non-condensible gases in steam.
3
PROPERTIES: Table A-6, Water, vapor (1 atm): Tsat = 100°C, ρ v = 0.596 kg/m , hfg = 2257 kJ/kg;
3
Table A-6, Water, liquid (Tf = (Ts + Tsat )/2 = 370K): ρl = 960.6 kg/m , c p, l = 4214 J/kg⋅K, µl =
-6
2
289 × 10 N⋅s/m , kl =0.679 W/m⋅K.
ANALYSIS: From Eq. 10.32 with A = π D L and Eq. 10.33, the heat transfer and condensation rates
are
& = q / h ′fg
q = h L (π D L ) ( Tsat − Ts )
m
where from Eq. 10.26 with Ja = c p, l ( Tsat − Ts ) / h fg, find
h ′fg = h fg [1 + 0.68Ja ] = 2257kJ/kg 1 + 0.68 4214J/kg ⋅ K (100 − 94 ) K / 2 257 × 10 J / k g = 2274
3
kJ
.
kg
For laminar film condensation, Eq. 10.40 is the appropriate correlation for a cylinder with C = 0.729,
1/4
g ρ ( ρ − ρ ) k3 h′
l
l
v
fg
l
.
hD = 0.729
µ l ( Tsat − Ts ) D
1/4
9 . 8 m / s 2 × 960.6kg/m 3 ( 960.6 − 0.596 ) k g / m 3 ( 0.679W/m ⋅ K )3 ×2274 ×10 3 J / k g
h D = 0.729
−6
2
289 × 10 N ⋅ s / m ( 100 − 94 ) K × 0.1m
hD = 10,120 W / m 2 ⋅ K.
Hence, the heat transfer and condensation rates are
2
q = 10,120 W / m ⋅ K (π × 0.1m ×1m )(100 − 94 ) K = 19.1kW
<
<
& = 19.1 ×103 W/2274 ×10 3 J / k g = 8.39 ×10−3 kg/s.
m
COMMENTS: A comparison of the above results for the horizontal tube with those for a vertical
tube (Problem 10.42) follows:
Position
Vertical
Horizontal
(
2
h W / m ⋅K
)
8,507
10,120
The rates are higher for the horizontal case. Why?
& ⋅ 10 3 ( k g / s )
m
q(kW)
16.0
19.1
7.05
8.39
PROBLEM 10.54
KNOWN: Horizontal pipe passing through an air space with prescribed temperature and relative
humidity.
FIND: Water condensation rate per unit length of the pipe.
SCHEMATIC:
ASSUMPTIONS: (1) Film condensation occurs on horizontal tube.
PROPERTIES: Table A-6, Water, vapor (T∞ = 37°C = 310K): pA,sat = 0.06221 bar; Table A-6,
3
Water, vapor (pA = φ ⋅ pA,sat = 0.04666 bar): TA,sat ≈ 305K, ρ v = 0.04 kg/m , hfg = 2426 kJ/kg; Table
3
A-6, Water, liquid (Tf = (Ts + TA,sat )/2 = 297K): ρl = 997.2 kg/m , c p, l = 4180 J/kg⋅K, µl = 917 ×
-6
2
10 N⋅s/m , kl = 0.609 W/m⋅K.
ANALYSIS: From Eq. 10.33, the condensate rate per unit length is
&′=
m
h (π D) ( Tsat − Ts )
q′
= L
h ′fg
h ′fg
where, from Eq. 10.26, with Ja = cp, l ( Tsat − Ts ) / h fg ,
h ′fg = h fg 1 + 0.68c p, l ( Tsat − T s ) / h fg = 2426
1 + 0.68 × 4180 J 305 − 288 K /2 4 2 6 × 10 3 J
(
)
kg
kg ⋅ K
kg
kJ
h ′fg = 2474 kJ/kg.
Note that Tsat = TA,sat is the saturation temperature of the water vapor in air at 37°C having a relative
humidity φ = 0.75. That is, Tsat = 305K while Ts = 15°C = 288K. Assuming laminar film condensation
on the horizontal pipe, it follows from Eq. 10.40 that,
1/4
g ρ ( ρ − ρ ) k3 h′
l
l
v
fg
l
hD = 0.729
µl ( Tsat − Ts ) D
1/4
9 . 8 m / s 2 × 997.2kg/m 3 ( 997.2 − 0.04 ) k g / m 3 ( 0.609W/m ⋅ K )3 × 2474 × 10 3 J / k g
h D = 0.729
−6
2
917 × 10 N ⋅ s / m ( 305 − 288 ) K × 0.025m
hD = 7925 W / m 2 ⋅ K.
Hence, the condensate rate is,
2
&′
m = 7925 W / m ⋅ K (π × 0.025m )( 305 − 288 ) K/2474 ×103 J/kg
& ′ = 4.28 ×10−3 kg/s ⋅ m.
m
COMMENTS: The actual dropwise condensation rate exceeds the foregoing estimate.
<
PROBLEM 10.55
KNOWN: Horizontal tube, 50mm diameter, with surface temperature of 34°C is exposed to steam at
0.2 bar.
FIND: Estimate the heat transfer and condensation rates per unit length of the tube.
SCHEMATIC:
ASSUMPTIONS: (1) Laminar film condensation, (2) Negligible non-condensibles in steam.
3
PROPERTIES: Table A-6, Saturated steam (0.2 bar): Tsat = 333K, ρ v = 0.129 kg/m , hfg = 2358
3
kJ/kg; Table A-6, Water, liquid (Tf = (Ts + Tsat )/2 = 320K): ρl = 989.1 kg/m , c p, l = 4180 J/kg⋅K,
-6
2
µl = 577 × 10 N⋅s/m , kl = 0.640 W/m⋅K.
ANALYSIS: From Eqs. 10.32 and 10.33, the heat transfer and condensate rates per unit length of the
tube are
q′ = h D (π D )( Tsat − Ts )
& ′ = q′ / h ′fg
m
where from Eq. 10.26 with Ja = c p,l ( Tsat − Ts ) / h fg,
h ′fg = h fg [1 + 0.68 Ja ] = 2358
kJ
1 + 0.68 × 4180J/kg ⋅ K ( 333 − 307 ) K/2358 ×103 J/kg
kg
h ′fg = 2432 kJ/kg.
For laminar film condensation, Eq. 10.40 is appropriate for estimating h D with C = 0.729,
1/4
g ρ ( ρ − ρ ) k3 h′
l l
v l fg
hD = 0.729
µl ( T − T ) D
sat s
h D = 0.729
9 . 8 m / s 2 × 9 8 9 . 1 k g / m 3 ( 989.1 − 0.129 ) k g / m 3 ( 0 . 6 4 0 W / m ⋅ K )3 × 2432 × 103 J / k g
−6
2
577 × 10 N ⋅ s / m ( 333 − 307 ) K × 0.050m
hD = 6926 W / m 2 ⋅ K.
Hence, the heat transfer and condensation rates are
q′ = 6926 W / m 2 ⋅ K (π × 0.050m )( 333 − 307 ) K = 28.3kW/m
<
& ′ = 28.3 ×103 W / m / 2 4 3 2 ×10 3 J / k g = 1.16 ×10 − 2 kg/s ⋅ m.
m
<
PROBLEM 10.56
KNOWN: Horizontal tube 1m long with surface temperature of 70°C used to condense steam at 1
bar.
FIND: Diameter required for condensation rate of 125 kg/h.
SCHEMATIC:
ASSUMPTIONS: (1) Laminar film condensation, (2) Negligible non-condensibles in steam.
3
PROPERTIES: Table A-6, Water, vapor (1 atm): Tsat = 100°C, ρ v = 0.596 kg/m , hfg = 2257 kJ/kg;
3
Table A-6, Water, liquid (Tf = (Ts + Tsat )/2 = 358K): ρl = 968.6 kg/m , c p, l = 4201 J/kg⋅K, µl =
-6
2
332 × 10 N⋅s/m , kl = 0.673 W/m⋅K.
ANALYSIS: From the rate equation, Eq. 10.33, with A = π D L, the required diameter is
& h ′fg / π L h D ( Tsat − Ts )
D =m
where from Eq. 10.26 with Ja = c p,l ( Tsat − Ts ) / h fg,
(1)
4201J/kg ⋅ K × (100 − 70 ) K
kJ
= 2343kJ/kg.
h ′fg = h fg (1 + 0.68Ja ) = 2257 1 + 0.68
kg
2257 ×103 J/kg
Substituting numerical values, Eq. (1) becomes
125 kg
J
D=
× 2343 ×103 / π ×1m × h
3600 s
kg
(2)
−1
D (100 − 70 ) K = 863.2h D .
(3)
The appropriate correlation for hD is Eq. 10.40 with C = 0.729,
1/4
g ρ ( ρ − ρ ) k3 h′
l
l
v
fg
l
.
hD = 0.729
µ l ( Tsat − Ts ) D
(4)
Substitute Eq. (4) for hD into Eq. (3) and use numerical values,
863.2 D−1 = 0.729 ×
1/4
9.8m/s2 × 968.6kg/m3 (968.6 − 0.596 ) k g / m3 ( 0.673W/m ⋅ K )3 × 2343 × 103 J/kg
332 ×10 −6 N ⋅ s / m2 (100 − 70 ) K× D
863.2 D−1 = 3693.4 D−1 / 4
D = 0.144m = 144mm.
COMMENTS: Note for this situation Ja = 0.06.
<
PROBLEM 10.57
KNOWN: Saturated R-12 vapor at 1 atm condensing on the outside of a horizontal tube.
FIND: Tube surface temperature required for condensation rate of 50 kg/h.
SCHEMATIC:
ASSUMPTIONS: (1) Laminar film condensation, (2) Negligible non-condensibles in vapor.
3
PROPERTIES: Table A-5, R-12 Saturated vapor (1 atm): Tsat = 243K, ρ v = 6.32 kg/m , hfg = 165
3
kJ/kg; Table A-5, R-12 Saturated liquid (Tf ≈ 240K): ρl = 1498 kg/m , c p, l = 892.3 J/kg⋅K, µl =
-2
2
0.0385 × 10 N⋅s/m , kl = 0.069 W/m⋅K.
ANALYSIS: The surface temperature or temperature difference can be written as follows from Eq.
10.33,
& h ′fg / h D π D L
∆T = Tsat − Ts = m
(1)
where A = π D L. To evaluate h ′fg and hD , we require knowledge of Ts or ∆T. Assume a ∆T =
10°C, then Ts = 233K and Tf = (Ts + Tsat )/2 = 240K. From Eq. 10.26 with Ja = c p, l ∆T / h fg , find
h ′fg = h fg (1 + 0.68Ja ) = 165
kJ
J
J
1 + 0.68 × 892.3
× 10K/165× 10 3 = 171kJ/kg.
kg
kg ⋅ K
kg
(2)
The appropriate correlation for hD is Eq. 10.40 with C = 0.729; substitute properties and find hD in
terms of ∆T.
1/4
g ρ ( ρ − ρ ) k3 h′
l l
v l fg
hD = 0.729
µl ( Tsat − Ts ) D
1/4
9.8m/s2 × 1498kg/m3 (1498 − 6.32 ) k g / m3 ( 0.069W/m ⋅ K )3 × 171 ×103 J/kg
h D = 0.729
0.0385 × 10− 2 N ⋅ s / m2 × ∆T × 0.010m
hD = 3082 ∆ T1/4 .
(3)
Substitute Eq. (3) into Eq. (1) for hD , and solve for ∆T,
∆T =
(
)
50
k g / s ×171× 103 J / k g / 3082 ∆T1/4 π ( 0.010m ) ×1m
3600
∆T = 12.9K
or
Ts = 230K.
<
COMMENTS: We used the assumed value of Ts or ∆T only to evaluate properties. Our estimate
for Tf = 240K is to be compared to the calculated value of Tf ≈ 236K. An iteration is probably not
necessary.
PROBLEM 10.58
KNOWN: Saturation temperature and inlet flow rate of R-12. Diameter, length and temperature of
tube.
FIND: Rate of condensation and outlet flow rate.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible concentration of noncondensables in vapor.
3
PROPERTIES: Given, R-12, saturated vapor: ρv = 6 kg/m , hfg = 160 kJ/kg, µv = 150 × 10
-7
2
N⋅s/m . Table A-5, R-12, saturated liquid (Tf = 300 K): ρ " = 1306 kg / m3 , c p," = 978 J / kg ⋅ K,
2
µ" = 0.0254 N ⋅ s / m , k " = 0.072 W / m ⋅ K.
ANALYSIS: The Reynolds number associated with the inlet vapor flow is Re v,i = 4 m v,i / π Dµ v
= 0.04 kg / s / π × 0.025m × 150 × 10
−7
2
N ⋅ s / m = 33, 950 < 35, 000. Hence, the average convection
coefficient may be obtained from Eq. 10.42, where h ′fg = h fg + 0.375 c p," ( Tsat − Ts ) = (1.6 × 10 +
5
5
0.375 × 978 × 20) J/kg = 1.67 × 10 J/kg.
gρ " ( ρ " − ρ v ) k 3" h ′fg
hD = 0.555
µ " ( Tsat − Ts ) D
1/ 4
2
9.8 m / s
≈ 0.555
(
1306 kg / m
3
)(
2
0.072 W / m ⋅ K ) 1.67 × 10
3
2
0.0254 N ⋅ s / m × 20 K × 0.025m
1/ 4
5
J / kg
h D = 297 W / m 2 ⋅ K
The heat rate is then
q = π DL h D ( Tsat − Ts ) = π × 0.025m × 2m × 297 W / m 2 ⋅ K × 20 K = 933 W
and the condensation rate is
cond =
m
q
933 W
=
= 0.0056 kg / s
h ′fg 1.67 × 105
<
The flow rate of vapor leaving the tube is then
v,o = m
v,i − m
cond = ( 0.0100 − 0.0056 ) kg / s = 0.0044 kg / s
m
<
PROBLEM 10.59
KNOWN: Array of condenser tubes exposed to saturated steam at 0.1 bar.
FIND: (a) Condensation rate per unit length of square array, (b) Options for increasing the condensation
rate.
SCHEMATIC:
ASSUMPTIONS: (1) Film condensation on tubes, (2) Negligible non-condensable gases in steam.
PROPERTIES: Table A.6, Saturated water vapor (0.1 bar): Tsat ≈ 320 K, ρv = 0.072 kg/m3, hfg = 2390
kJ/kg; Table A.6, Water, liquid (Tf = (Ts + Tsat)/2 = 310 K): ρ" = 993.1 kg/m3, cp," = 4178 J/kg⋅K, µ"
= 695 × 10-6 N⋅s/m2, k " = 0.628 W/m⋅K.
ANALYSIS: (a) From Eq. 10.33, the condensation rate for a N ×N square array is
′=m
L = h D,N ⋅ N t (π D )(Tsat − Ts ) h′fg
m
where h D,N is the average coefficient for the tubes in a vertical array of N tubes. With Ja = cp," ∆T/hfg
= 4178 J/kg⋅K × (320 - 300)K/2390 × 103 J/kg = 0.035, Eq. 10.26 yields h ′fg = hfg(1 + 0.68 Ja) = 2390
kJ/kg(1 + 0.68 × 0.035) = 2447 kJ/kg.
For a vertical tier of N = 10 horizontal tubes, the average coefficient is given by Eq. 10.41,
1/ 4
gρ ( ρ − ρ ) k 3 h ′
"
"
v
fg
"
h D,N = 0.729
Nµ" (Tsat − Ts ) D
1/ 4
h D,N
9.8 m s 2 × 993.1kg m3 (993.1 − 0.072 ) kg m3 ( 0.628 W m⋅ K )3 × 2447 × 103 J kg
= 0.729
−6
2
10 × 695 × 10 N ⋅ s m (320 − 300 ) K × 0.008 m
h D,N = 6210 W m 2⋅ K .
Hence, the condensation rate for the entire array per unit tube length is
′ = 6210 W m 2⋅ K (100 )π × 0.008 m (320 − 300 ) K 2447 ×103 J kg
m
′ = 0.128 kg s ⋅ m = 459 kg h ⋅ m .
m
<
(b) Options for increasing the condensation rate include reducing the surface temperature and/or the
number of tubes in a vertical tier. By varying the temperature of cold water flowing through the tubes, it
is feasible to maintain surface temperatures in the range 280 ≤ Ts ≤ 300 K. Using the Correlations and
Properties Toolpads of IHT, the following results were obtained for N = 10, 5 and 2, with Nt = 100 in
each case. The results are based on properties evaluated at p = 0.1 bar, for which the Properties Toolpad
yielded Tsat = 318.9 K.
Continued...
PROBLEM 10.59 (Cont.)
Condensation rate, mdot'(kg/s.m)
0.3
0.2
0.1
280
290
300
Surface temperature, Ts(K)
N = 10
N=5
N=2
Clearly, there are significant benefits associated with reducing both Ts and N.
COMMENTS: Note that, since h D,N ∝ N-1/4, the average coefficient decreases with increasing N due
to a corresponding increase in the condensate film thickness. From the result of part (a), the coefficient
for the topmost tube is h D = 6210 W/m2⋅K(10)1/4 = 11,043 W/m2⋅K.
PROBLEM 10.60
KNOWN: Thin-walled concentric tube arrangement for heating deionized water by condensation of
steam.
FIND: Estimates for convection coefficients on both sides of the inner tube. Inner tube wall outlet
temperature. Whether condensation provides fairly uniform inner tube wall temperature approximately
equal to the steam saturation temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible thermal resistance of inner tube wall, (2) Internal flow is fully
developed.
3
PROPERTIES: Deionized water (given): ρ = 982.3 kg/m , cp = 4181 J/kg⋅K, k = 0.643 W/m⋅K, µ =
-6
2
548 × 10 N⋅s/m , Pr = 3.56; Table A-6, Saturated vapor (1 atm): Tsat = 100°C, ρ v = (1/vg) = 0.596
3
kg/m , hfg = 2265 kJ/kg; Table A-6, Saturated water (assume Ts ≈ 75°C, Tf = (75 + 100)°C/2 = 360K):
3
-6
2
ρl = (1/vf) = 967 kg/m , µl = 324 × 10 N⋅s/m , kl = 0.674 W/m⋅K, c p, l = 4203 J/kg⋅K.
ANALYSIS: From an energy balance on the inner tube assuming a constant wall temperature,
(
)
(
hc Tsat − Ts,o = hi Ts,o − Tm,o
)
where h c and hi are, respectively, the heat transfer coefficients for condensation (c) on a horizontal
cylinder and internal (i) flow in a tube.
Condensation. From Eq. 10.40, for the horizontal tube,
1/4
g ρ ( ρ − ρ ) k 3h ′
l l
v l fg
hc = 0.729
µl ( Tsat − Ts ) D
{
where h ′fg = h fg 1 + 0.68c p,l ( Tsat − Ts ) / h fg
}
{
h ′fg = 2265 kJ/kg {1+1.262 ×10 −3 (100 − Ts )}
h ′fg = 2265 kJ/kg 1 + 0.68 × 4203 J / k g ⋅ K (100 − Ts ) /2265 ×103 J/kg
}
hc = 0.729 9.8m/s2 × 967kg/m3 ( 067 − 0.596 ) k g / m3 ( 0.674W/m ⋅ K )3 ×
{
}
1/4
2265 1 + 1.262 ×10 −3 (100 − Ts ) kJ/kg/324 ×10 −6 N ⋅ s / m 2 (100 − Ts ) 0.030 m
Continued …..
PROBLEM 10.60 (Cont.)
1/4
1 +1.262 ×10 −3 (100 − T )
s
hc = 2.843× 104
100 − Ts
.
Internal flow. From Eq. 8.6, evaluating properties at Tm , find
Re D =
&
4m
4 × 5 kg/s
=
= 3.872 ×105
−
6
2
πµ D π × 548 ×10 N ⋅ s / m × 0.030 m
and for turbulent flow use the Colburn equation,
hD
1/3
Nu D = i = 0.023Re0.8
D Pr
k
hi =
)
(
0.8
0.023 × 0.643 W / m ⋅ K
3.872 ×105
( 3.56 )1/3 = 2.22 × 104 W / m 2 ⋅ K.
0.03 m
<
Substituting numerical values into the energy balance relation,
(
1 + 1.262 ×10−3 100 − T
s,o
4
2.843 ×10
100 − Ts,o
1/4
)
(100 − Ts,o ) K
(
)
= 2.22 × 104 W / m2 ⋅ K Ts,o − 60 K
and by trial-and-error, find
Ts,o ≈ 75°C.
With this value of Ts, find that
hc = 1.29 × 104 W / m2 ⋅ K
<
which is approximately half that for the internal flow. Hence, the tube wall cannot be at a uniform
temperature. This could only be achieved if h c ? h i .
PROBLEM 10.61
KNOWN: Heat dissipation from multichip module to saturated liquid of prescribed temperature and
properties. Diameter and inlet and outlet water temperatures for a condenser coil.
FIND: (a) Condensation and water flow rates. (b) Tube surface inlet and outlet temperatures. (c)
Coil length.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions since rate of heat transfer from the module is balanced
by rate of heat transfer to coil, (2) Fully developed flow in tube, (3) Negligible changes in potential and
kinetic energy for tube flow.
PROPERTIES: Saturated fluorocarbon (Tsat = 57°C, given): kl = 0.0537 W/m⋅K, c p, l = 1100
3
3
-3
2
J/kg⋅K, h ′fg ≈ h fg = 84,400 J/kg. ρl = 1619.2 kg/m , ρ v = 13.4 kg/m , σ = 8.1 × 10 kg/s , µl = 440
-6
3
× 10 kg/m⋅s, Prl = 9; Table A-6, Water, sat. liquid ( Tm = 300K ) : ρ = 997 kg/m , cp = 4179 J/kg⋅K,
-6
2
µ = 855 × 10 N⋅s/m , k = 0.613 W/m⋅K, Pr = 5.83.
ANALYSIS: (a) With
q = (q ′′ × A )module = 105 W / m 2 ( 0.1m )2 = 10 3 W
the condensation rate is
& con =
m
q
103 W
=
= 0.0118 kg/s
h′fg 84,400 J / k g
<
and the required water flow rate is
& =
m
(
q
c p Tm,o − Tm,i
)
=
1000 W
= 7.98 ×10 −3 kg/s.
4179 J / k g ⋅ K ( 30K )
<
(b) The Reynolds number for flow through the tube is
Re D =
&
4m
4 × 7.98 ×10 −3 k g / s
=
= 1188.
π Dµ π ( 0.01m ) 855 ×10−6 N ⋅ s / m 2
Hence, the flow is laminar. Assuming a uniform wall temperature,
h i = Nu D k / D = 3.66 ( 0.613 W / m ⋅ K/0.01m ) = 224 W / m 2 ⋅ K.
Continued …..
PROBLEM 10.61 (Cont.)
For film condensation on the outer surface, Eq. 10.40 yields
(
)(
)
1/4
9.8m/s 2 1619.2 k g / m 3 1605.8kg/m3 ( 0.0537 W / m ⋅ K )3 84,400J/kg
h o = 0.729
−
6
440 × 10 k g / m ⋅ s × 0.01m ( Tsat − Ts )
h o = 2150 (57 − Ts )
−1/4
.
From an energy balance on a portion of the tube surface,
h o ( Tsat − Ts ) = h i ( Ts − Tm )
or
2150 ( 57 − Ts )
3/4
= 224 ( Ts − Tm )
At the entrance where ( Tm,i = 285K ) , trial-and-error yields:
<
Ts,i = 50.6°C
and at the exit where ( Tm,o = 315K ) ,
<
Ts,o = 55.4°C
(c) From Eqs. 8.44 and 8.45,
L=
q
hiπ D∆Tlm
where
∆Tl m =
L=
(
( Ts − Tm,i ) − ( Ts − Tm,o ) = 41 −11 = 22.8°C
ln ( Ts − Tm,i ) / ( Ts − Tm,o ) ln ( 41/11)
1000 W
)
224 W / m2 ⋅ K π ( 0.01m ) 22.8°C
= 6.23m.
<
COMMENTS: Some control over system performance may be exercised by adjusting the water
& ( Tm,o − Tm,i ) is reduced for a prescribed q. The value of h i is increased
flow rate. By increasing m,
substantially if the internal flow is turbulent.
PROBLEM 10.62
KNOWN: Saturated ethylene glycol vapor at 1 atm condensing on a sphere of 100 mm diameter
having surface temperature of 150°C.
FIND: Condensation rate.
SCHEMATIC:
ASSUMPTIONS: (1) Laminar film condensation, (2) Negligible non-condensibles in vapor.
3
PROPERTIES: Table A-5, Saturated ethylene glycol, vapor (1 atm): Tsat = 470K, ρ v ≈ 0 kg/m , hfg
= 812 kJ/kg; Table A-5, Ethylene glycol, liquid (Tf = 423K, but use values at 373K, limit of data in
3
-2
2
table): ρl = 1058.5 kg/m , c p, l = 2742 J/kg⋅K, µl = 0.215 × 10 N⋅s/m , kl = 0.263 W/m⋅K.
ANALYSIS: The condensation rate is given by Eq. 10.33 as
(
)
h L π D2 ( Tsat − Ts )
q
& =
m
=
h′fg
h′fg
2
where A = π D for the sphere and h ′fg , with Ja = c p, l ∆T / h fg , is given by Eq. 10.26 as
h ′fg = h fg (1 + 0.68Ja ) = 812
kJ
J
1 + 0.68 × 2742
( 470 − 423) K/812 ×103 J/kg = 900kJ/kg.
kg
kg ⋅ K
The average heat transfer coefficient for the sphere follows from Eq. 10.40with C = 0.815,
1/4
g ρ ( ρ − ρ )k 3 h ′
l l
v l fg
hD = 0.815
µ l ( Tsat − Ts ) D
9 . 8 m / s 2 × 1058.5kg/m 3 (1058.5 − 0 ) k g / m 3 ( 0.263W/m ⋅ K )3× 900 × 10 3 J / k g
h D = 0.815
−2
2
0.215 × 10 N ⋅ s / m ( 470 − 423 ) K × 0.100m
1/4
hD = 1674 W / m 2 ⋅ K.
Hence, the condensation rate is
& = 1674W/m 2 ⋅ K ×π ( 0.100m ) 2 ( 470 − 423 ) K/900 ×103 J/kg
m
& = 2.75 ×10 −3 kg/s.
m
<
COMMENTS: Recognize this estimate is likely to be a poor one since properties were not evaluated
at the proper Tf which was beyond the limit of the table.
PROBLEM 10.63
KNOWN: Copper sphere of 10 mm diameter, initially at 50°C, is placed in a large container filled with
saturated steam at 1 atm.
FIND: Time required for sphere to reach equilibrium and the condensate formed during this period.
SCHEMATIC:
ASSUMPTIONS: (1) Laminar film condensation, (2) Negligible non-condensibles in vapor, (3)
Sphere is spacewise isothermal, (4) Sphere experiences heat gain by condensation only.
3
PROPERTIES: Table A-6, Saturated water vapor (1 atm): Tsat = 100°C, ρ v = 0.596 kg/m , hfg =
3
2257 kJ/kg; Table A-6, Water, liquid (Tf ≈ (75 + 100)°C/2 = 360K): ρl = 967.1 kg/m , c p, l = 4203
-6
2
J/kg⋅K, µl = 324 × 10 N⋅s/m , kl = 0.674 W/m⋅K; Table A-1, Copper, pure ( T = 75° C) : ρ sp =
3
8933 kg/m , cp,sp = 389 J/kg⋅K.
ANALYSIS: Using the lumped capacitance approach, an energy balance on the sphere provides,
E& in − E& out = E& st
& h ′fg = h D A s ( Tsat − Ts ) = ρsp c p,sp Vs
m
dTs
.
dt
(1)
Properties of the sphere, ρ sp and cp ,sp ,. Will be evaluated at Ts = ( 50 +100 ) ° C / 2 = 75° C, while water
(liquid) properties will be evaluated at Tf = ( Ts + Tsat ) / 2 = 87.5 °C ≈ 360K. From Eq. 10.26 with Ja =
c p, l ∆T / h fg where ∆T = Tsat - Ts , find
1 + 0.68 4203 J × 100 − 75 K /2 2 5 7 × 10 3 J / k g = 2328 kJ . (2)
(
)
kg
kg ⋅ K
kg
kJ
h ′fg = h fg (1 + 0.68Ja ) = 2257
To estimate the time required to reach equilibrium, we need to integrate Eq. (1) with appropriate limits.
However, to perform the integration, an appropriate relation for the temperature dependence of hD
needs to be found. Using Eq. 10.40 with C = 0.815,
1/4
g ρ ( ρ − ρ ) k3 h′
l l
v l fg
hD = 0.815
.
µ l ( Tsat − Ts ) D
Substitute numerical values and find,
1/4
9 . 8 m / s 2 × 967.1kg/m 3 ( 967.1 − 0.596 ) k g / m 3 ( 0 . 6 7 4 W / m ⋅ K )3 × 2328 × 10 3 J / k g
h D = 0.815
−6
2
324 × 10 N ⋅ s / m ( Tsat − Ts ) × 0.010m
−1/4
hD = B ( Tsat − Ts )
where
B = 30,707W/m 2 ⋅ ( K )
3/4
. (3)
Continued …..
PROBLEM 10.63 (Cont.)
1
Substitute Eq. (3) into Eq. (1) for hD and recognize Vs / A s = π D 3 / π D 2 = D / 6 ,
6
dT
B ( Tsat − Ts )−1/4 ( Tsat − Ts ) = ρsp c p,sp ( D / 6 ) s .
dt
(4)
Note that d(Ts) = - d(Tsat – Ts); letting ∆T ≡ Tsat – Ts and separating variables, the energy balance
relation has the form
ρsp c p,sp ( D / 6 ) ∆T d ( ∆T )
t
dt = −
(5)
0
∆To ∆T3 / 4
B
where the limits of integration have been identified, with ∆ To = Tsat − Ti and Ti = Ts(0). Performing
the integration, find
∫
t=−
∫
ρsp cp,sp ( D / 6)
B
⋅
1 1/4
∆T
− ∆ T1/4
o .
1− 3 / 4
Substituting numerical values with the limits, ∆T = 0 and ∆To = 100-50 = 50°C,
t=−
8933kg/m3 × 389J/kg ⋅ K ( 0.010m/6)
30,707 W / m 2 ⋅ K 3 / 4
× 4 01/4 − 501 / 4 K1/4
<
t = 2.0s.
To determine the total amount of condensate formed during this period, perform an energy balance on
a time interval basis,
E in − E out = ∆E = E final − E initial
Ein = ρsp cp,sp V ( Tfinal − Tinitial ]
(6)
where Tfinal = Tsat and Tinitial = Ti = Ts(0). Recognize that
E in = M h ′fg
(7)
where M is the total mass of vapor that condenses. Combining Eqs. (6) and (7),
M=
M=
ρsp cp,sp V
h ′fg
[Tsat − Ti ]
8933kg/m3 × 389J/kg ⋅ K (π / 6)( 0.010m )3
2328 × 103 J/kg
[100 − 50 ] K
M = 3.91 ×10 −5 kg.
<
COMMENTS: The total amount of condensate could have been evaluated from the integral,
t
0
t q
t h D A s ( Tsat − Ts ) dt
dt =
0 h ′fg
0
h ′fg
& =∫
M = ∫ mdt
∫
giving the same result, but with more effort.
PROBLEM 10.64
KNOWN: Saturated steam condensing on the inside of a horizontal pipe.
FIND: Heat transfer coefficient and the condensation rate per unit length of the pipe.
SCHEMATIC:
ASSUMPTIONS: (1) Film condensation with low vapor velocities.
3
PROPERTIES: Table A-6, Saturated water vapor (1.5 bar): Tsat ≈ 385K, ρ v = 0.88 kg/m , hfg =
3
2225 kJ/kg; Table A-6, Saturated water (Tf = (Tsat + Ts)/2 ≈ 380K): ρl = 953.3 kg/m , c p, l = 4226
-6
2
J/kg⋅K, µl = 260 × 10 N⋅s/m , kl = 0.683 W/m⋅K.
ANALYSIS: The condensation rate per unit length follows from Eq. 10.33 with A = π D L and has
the form
&′=
m
&
m
= h D (π D )( Tsat − Ts ) / h ′fg
L
where hD is estimated from the correlation of Eq. 10.42 with Eq. 10.43,
1/4
g ρ ( ρ − ρ )k 3 h ′
l l
v l fg
hD = 0.555
µ l ( Tsat − Ts ) D
where
3
J 3
J
h ′fg = h fg + cp,l ( Tsat − Ts ) = 2225 ×103
+ × 4226
( 385 − 373 ) K
8
kg 8
kg ⋅ K
h ′fg = 2244kJ/kg.
Hence,
1/4
kg
kg
3
2
3
9.8m/s × 953.3 3 ( 953.3 − 0.88 ) 3 ( 0.683W/m ⋅ K ) 2244 ×10 J/kg
m
m
hD = 0.555
−
6
2
260 × 10 N ⋅ s / m ( 385 − 373) K × 0.075m
hD = 7127W/m 2 ⋅ K.
It follows that the condensate rate per unit length of the tube is
& ′ = 7127W/m 2 ⋅ K (π × 0.075m )(385 − 373)K / 2225 ×103 J / k g = 9.06 ×10−3 kg / s ⋅ m.
m
<
PROBLEM 10.65
KNOWN: Horizontal pipe passing through an air space with prescribed temperature and relative
humidity.
FIND: Water condensation rate per unit length of pipe.
SCHEMATIC:
ASSUMPTIONS: (1) Drop-wise condensation, (2) Copper tube approximates well promoted
surface.
PROPERTIES: Table A-6, Water vapor (T∞ = 37°C = 310K): pA,sat = 0.06221 bar; Table A-6,
Water vapor (pA = φ⋅pa,sat = 0.04666 bar): Tsat = 305K = 32°C, hfg = 2426 kJ/kg; Table A-6, Water,
liquid (Tf = (Ts + Tsat )/2 = 297K): c p, l = 4180 J / kg ⋅ K.
ANALYSIS: From Eq. 10.33, the condensate rate per unit length is
&′=
m
q ′ h L (π D )(Tsat − Ts )
=
h′fg
h′fg
where from Eq. 10.26, with Ja = c p, l ( Tsat − T s ) / h fg ,
h ′fg = h fg [1 + 0.68Ja ] = 2426
kJ
1 + 0.68 × 4180J/kg ⋅ K ( 305 − 288 ) K/2426k J/kg
kg
h ′fg = 2474kJ/kg.
Note that Tsat is the saturation temperature of the water vapor in air at 37°C having a relative humidity,
φ = 0.75. That is, Tsat = 305K and Ts = 15°C + 288K. For drop-wise condensation, the correlation
of Eq. 10.44 yields
hdc = 51,104 + 2044Tsat
22° C < Tsat < 100°C
2
where the units of h dc and Tsat are W/m ⋅K and °C.
hdc = 51,104 + 2044 ( 32° C) = 116,510 W / m2 ⋅ K.
Hence, the condensation rate is
& ′ = 116,510 W / m 2 ⋅ K ( π × 0.025m )( 305 − 288 ) K/2474× 10 3 J/kg
m
& ′ = 6.288 ×10−2 kg/s ⋅ m
m
<
COMMENTS: From the result of Problem 10.54 assuming laminar film condensation, the
condensation rate was m& ′film = 4.28 ×10 −3 k g / s ⋅ m which is an order of magnitude less than for the
rate assuming drop-wise condensation.
PROBLEM 10.66
KNOWN: Beverage can at 5°C is placed in a room with ambient air temperature of 32°C and
relative humidity of 75%.
FIND: The condensate rate for (a) drop-wise and (b) film condensation.
SCHEMATIC:
ASSUMPTIONS: (1) Condensation on top and bottom surface of can neglected, (2) Negligible noncondensibles in water vapor-air, and (b) For film condensation, film thickness is small compared to
diameter of can.
PROPERTIES: Table A-6, Water vapor (T∞ = 32°C = 305 K): pA,sat = 0.04712 bar; Water vapor
(pA = φ⋅pA,sat = 0.03534 bar): Tsat ≈ 300 K = 27°C, hfg = 2438 kJ/kg; Water, liquid (Tf = (Ts + Tsat)/2
= 289 K): c p," = 4185 J/kg⋅K.
ANALYSIS: From Eq. 10.33, the condensate rate is
=
m
h (π DL )( Tsat − Ts )
q
=
h ′fg
h ′fg
where from Eq. 10.26, with Ja = c p," (Tsat – Ts)/hfg,
h ′fg = h fg [1 + 0.68 Ja ]
h ′fg = 2438 kJ / kg 1 + 0.68 × 4185 J / kg ⋅ K (300 − 278 ) K / 2438 kJ / kg
h ′fg = 2501 kJ / kg
Note that Tsat is the saturation temperature of the water vapor in air at 32°C having a relative
humidity of φ∞ = 0.75.
(a) For drop-wise condensation, the correlation of Eq. 10.44 with Tsat = 300 K = 27°C yields
h = h dc = 51,104 + 2044 Tsat
22°C < Tsat ≤ 100°C
2
where the units of h dc are W/m ⋅K and Tsat are °C,
h dc = 51,104 + 2044 × 27 = 106, 292 W / m 2 ⋅ K
Hence, the condensation rate is
= 1.063 × 105 W / m 2 ⋅ K (π × 0.065 m × 0.125 m )( 27 − 5 ) K / 2501 kJ / kg
m
<
= 0.0229 kg / s
m
Continued …..
PROBLEM 10.66 (Cont.)
(b) For film condensation, we used the IHT tool Correlations, Film Condensation, which is based
upon Eqs. 10.37, 10.38 or 10.39 depending upon the flow regime. The code is shown in the
Comments section, and the results are
Reδ = 24, flow is laminar
<
= 0.00136 kg / s
m
Note that the film condensation rate estimate is nearly 20 times less than for drop-wise condensation.
COMMENTS: The IHT code identified in part (b) follows:
/* Results,
NuLbar
0.5093
Part (b) - input variables and rate parameters
Redelta hLbar
mdot
D
L
Ts
24.05
6063
0.001362 0.065
0.125
278
Tsat
300
*/
/* Thermophysical properties evaluated at Tf; hfg at Tsat
Prl
Tf
cpl
h'fg
hfg
kl
mul
nul
7.81 289
4185
2.501E6 2.438E6 0.5964 0.001109 1.11E-6*/
// Other input variables required in the correlation
L = 0.125
b = pi * D
D = 0.065
/* Correlation description: Film condensation (FCO) on a vertical plate (VP). If Redelta<29,
laminar region, Eq 10.37 . If 31<Redelta<1750, wavy-laminar region, Eq 10.38 . If Redelta>=1850,
turbulent region, Eq 10.26, 10.32, 10.33, 10.35, 10.39 . In laminar-wavy and wavy-turbulent transition
regimes, function interpolates between laminar and wavy, and wavy and turbulent correlations. See
Fig 10.15 . */
NuLbar = NuL_bar_FCO_VP(Redelta,Prl)
// Eq 10.37, 38, 39
NuLbar = hLbar * (nul^2 / g)^(1/3) / kl
g = 9.8
// gravitational constant, m/s^2
Ts = 5 + 273
// surface temperature, K
Tsat = 300
// saturation temperature, K
// The liquid properties are evaluated at the film temperature, Tf,
Tf= (Ts + Tsat) / 2
// The condensation and heat rates are
q = hLbar * As * (Tsat - Ts)
// Eq 10.32
As = L * b
// surface Area, m^2
mdot = q / h'fg
// Eq 10.33
h'fg = hfg + 0.68 * cpl * (Tsat - Ts)
// Eq 10.26; hfg evaluated at Tsat
// The Reynolds number based upon film thickness is
Redelta = 4 * mdot / (mul * b)
// Eq 10.35
// Water property functions :T dependence, From Table A.6
// Units: T(K), p(bars);
x=0
// Quality (0=sat liquid or 1=sat vapor)
hfg = hfg_T("Water",Tsat)
// Heat of vaporization, J/kg; evaluated at Tsat
cpl = cp_Tx("Water",Tf,x)
// Specific heat, J/kg·K
mul = mu_Tx("Water",Tf,x)
// Viscosity, N·s/m^2
nul = nu_Tx("Water",Tf,x)
// Kinematic viscosity, m^2/s
kl = k_Tx("Water",Tf,x)
// Thermal conductivity, W/m·K
Prl = Pr_Tx("Water",Tf,x)
// Prandtl number
PROBLEM 10.67
KNOWN: Surface temperature and area of integrated circuits submerged in a dielectric fluid of prescribed
properties. Height and temperature of condenser plates.
FIND: (a) Heat dissipation by an integrated circuit, (b) Condenser surface area needed to balance heat load.
SCHEMATIC:
ASSUMPTIONS: (1) Nucleate pool boiling in liquid, (2) Laminar film condensation of vapor, (3) Negligible heat
loss to surroundings.
3
-4
PROPERTIES: Dielectric fluid (given, Tsat = 50°C): ρl = 1700kg/m , c p, l = 1005J/kg⋅K, µl = 6.80 × 10
2
5
kg/s⋅m, kl = 0.062W/m⋅K, Prl = 11, σ = 0.013 kg/s , h fg = 1.05 × 10 J/kg, Cs,f = 0.004, n = 1.7.
ANALYSIS: (a) For nucleate pool boiling,
1/2
3
≈ 6.8 × 10 −4 k g / s ⋅ m 1.05 × 10 5 J / k g
C h Pr n
s,f fg l
g ( ρl − ρv )
q ′′s = µ l h fg
σ
1/2
9 . 8 m / s2 ×1 7 0 0 k g / m3
×
2
0.013kg/s
c p, l ∆Te
(
)
3
1005 J / kg ⋅ K × 25K
= 8 4 , 5 3 0 W / m2
5
1.7
0.004 × 1.05 × 10 J / k g × 11
2
q s = A s q ′′s = 8 4 , 5 3 0 W / m × 25 × 10
−6
<
2
m = 2.11W.
1/4
(b) For laminar film condensation on a vertical surface, Nu L
h ′fg = h fg 1 + 0.68
gρ l ( ρl − ρv ) h ′fg L3
= 0.943
µl kl ( Tsat − Ts )
cp, l ∆ T
5
5
= 1.05 × 10 + 0.68 (1005 J / k g ⋅ K × 35K ) = 1.29 ×10 J / k g
h fg
(
)
1/4
2
3 2
5
3
9 . 8 m / s 1700kg/m 1.29 × 10 J / k g ( 0.05 m )
Nu L ≈ 0.943
−4
6.8 × 10
k g / s ⋅ m ( 0.062 W / m ⋅ K )( 35K )
= 703
h L = ( k l / L ) Nu L = ( 0.062 W / m ⋅ K / 0 . 0 5 m ) 703 = 872 W / m ⋅ K
2
( )
q c = h LA c ( Tsat − T c ) = 872 W / m ⋅ K ( 35K ) A c = 30,500A c m
2
2
To balance the heat load, qc = Nq s . Hence
Ac =
500 × 2.11 W
30,500 W / m
2
= 0.0346 m
2
2
<
COMMENTS: (1) With A c = 0.0346m and H = 0.05m, the total condenser width is W = A c /H = 692mm. (2) With
& c / b = Γ = q c / h ′fg W = 1055W/1.29 × 10 5 J / k g × 0.692m = 0.0118kg/s ⋅ m, Reδ = 4 Γ / µ l =
m
-4
4(0.0118kg/s⋅m)/6⋅8 × 10
kg/s⋅m = 69.4. Hence condensate film is in the laminar-wavy regime, and a more
accurate estimate of A c would require iteration.
PROBLEM 10.68
KNOWN: Thin-walled thermosyphon. Absorbs heat by boiling saturated water at atmospheric
pressure on boiling section Lb. Rejects heat by condensing vapor into a thick film which falls length of
condensation section Lc back into boiling section.
FIND: (a) Mean surface temperature, Ts,b, of the boiling surface if nucleate boiling flux is 30%
critical flux, (b) Mean surface temperature, Ts,c of condensation section, and (c) Total condensation
& in thermosyphon. Explain how to determine whether film is laminar, wavy-laminar or
flow rate, m,
turbulent.
SCHEMATIC:
ASSUMPTIONS: (1) Laminar film condensation occurs in condensation section which approximates
a vertical plate, (2) Boiling and condensing section are separated by insulated length Li, (3) Top surface
of condensation section is insulated, (4) For condensation, liquid properties evaluated at Tf = 90°C.
PROPERTIES: Table A-6, Saturated water (100°C): ρ l = 1 / v f = 957.9 k g / m 3 , c p, l = 4217
J/kg⋅K, µ l = 279 × 10−6 N ⋅ s / m 2 , Prl = 1.76, hfg = 2257 kJ/kg, σ = 58.9 × 10 N/m; Saturated vapor
-3
3
(100°C): ρ v = 1/vg = 0.5955 kg/m ; Saturated water (90°C): ρ l = 1 / v f = 964.9kg/m 3 , c p, l = 4207
-6
2
J/kg⋅K, µl = 313 × 10 N⋅s/m , k l = 0.676 W / m⋅ K.
ANALYSIS: (a) The heat flux for the boiling section is 30% the critical heat flux which at
atmospheric pressure is
q′′s,b = 0.30q′′max = 0.30× 1.26 × 106 W / m 2 = 3.78 ×105 W / m 2 .
Using the Rohsenow correlation for nucleate boiling with Tsat = 100°C and typical values for the
surface of Cs,f = 0.0130 and n = 1.0, find
3
1/2
g ( ρl − ρ v ) cp, l Ts,b − Tsat
q′′s,b = µl h fg
σ
Cs,f h fg Prln
(
5
2
3.78 × 10 W / m = 279 × 10
−6
2
)
3
N ⋅ s / m × 2257 × 10 J / k g ×
1/2
9 . 8 m / s 2 ( 957.9 − 0.5955 ) k g / m3
−3
58.9 ×10 N / m
4217 J / kg ⋅ K ( Ts,b − 100 )
3
1.0
0.013 × 2257 × 10 J/kg1.76
3
Continued …..
PROBLEM 10.68 (Cont.)
<
Ts,b = 114.0 °C.
(b) The heat transferred into the boiling section must be rejected by film condensation,
q c = q b = q′′s,b π D 2 / 4 + π DL b
q c = 3.78 × 105 W / m 2 π ( 0.020m ) 2 / 4 + π ( 0.020m ) × 0.020m = 592 W.
The mean surface temperature can be determined from the rate equation
(
q c = h Lc ( π DLc ) Tsat − Ts,c
)
where the convection coefficient is determined from Eq. 10.30,
1/4
gρ ( ρ − ρ ) k 3 h′
l l
v l fg
hLc = 0.943
µ l Tsat − Ts,c Lc
(
)
1/4
h Lc
9 . 8 m / s2 × 964.9kg/m3 ( 964.9 − 0.5955 ) k g / m3 ( 0.676W/m ⋅ K )3 2257 × 103 J / k g
= 0.943
−6
2
313 × 10 N ⋅ s / m (100 − Ts,c ) 0.040 m
{
(
)
where h ′fg = h fg 1 + 0.68c p,l Tsat −T s,c / h fg
3
}
{
(
)
3
h ′fg = 2257 ×10 J / k g 1 + 0.68 × 4207 J / k g ⋅ K 100 − Ts,c /2257 × 10 J / k g
(
hLc = 2.517 × 104 100 − Ts,c
Hence,
} ≈ 2257 ×10
3
J/kg.
)−1/4 .
Using the rate equation, now find Ts,c by trial-and-error,
(
592 W = 2.517 × 104 100 − Ts,c
(
9.358 = 100 − Ts,c
) −1/4 (π × 0.020m× 0.040m) (100 − Ts,c ) K
)0.75
<
Ts,c = 80.3 °C.
(c) The condensation rate in the condenser section is
(
)
m
& = qc / h ′fg = 592W/ 2257 ×103 J / k g = 2.623 ×10 −4 k g / s
and from Eq. 10.35,
Reδ =
&
&
4m
4m
4 × 2.623 ×10−4 k g / s
=
=
= 53.3.
µl b µl (π D ) 313 ×10 −6 N ⋅ s / m 2 ( π × 0.020m )
Since 30 < Reδ < 1800, we conclude the film is laminar-wavy.
<
PROBLEM 10.69
KNOWN: Thermosyphon configuration for cooling a computer chip of prescribed size.
FIND: (a) Chip temperature and total power dissipation when chip operates at 90% of critical heat flux,
(b) Required condenser length.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) Saturated liquid/vapor conditions, (3) Negligible heat transfer
from bottom of chip.
PROPERTIES: Fluorocarbon (prescribed): Tsat = 57°C, c p," = 1100 J/kg⋅K, hfg = 84,400 J/kg, ρ" =
1619.2 kg/m3, ρ v = 13.4 kg/m3, σ = 8.1 × 10-3 kg/s2, µ" = 440 × 10-6 kg/m⋅s, Pr" = 9.01, k " = 0.054
W/m⋅K, ν " = µ" ρ" = 0.272 × 10-6 m2/s.
ANALYSIS: (a) With q′′ = 0.9 q′′max and the critical heat flux given by Eq. 10.7, the chip power
dissipation is
1/ 4
σ g ( ρ" − ρ v )
2
q = 0.9Lc × 0.149h fg ρ v
ρ v2
(
(
)(
)
0.0081kg s 2 9.8 m s 2 1605.8 kg m3
2
3
q = 0.9 ( 0.02 m ) × 0.149 (84, 400 J kg )13.4 kg m
2
13.4 kg m3
)
(
)
1/ 4
q c = 0.9 4 × 10−4 m 2 1.55 × 105 W m 2 = 55.7 W
<
With operation at q′′ = 1.40 × 105 W/m2 in the nucleate boiling region, Eq. 10.5 yields
T = Tsat +
$
T = 57 C +
Cs,f h fg Pr"n
c p,"
0.005 (84, 400 J kg )( 9.01)
1.7
1100 J kg ⋅ K
1/ 3
µ" h fg
q′′
1/ 6
σ
g ( ρ" − ρν )
5
2
1.40 × 10 W m
−4
4.4 × 10 kg m ⋅ s × 84, 400 J
1/ 3
kg
9.8 m
1/ 6
0.0081 kg s
s
2
(
2
1605.8 kg m
3
)
Continued...
PROBLEM 10.69 (Cont.
<
T = 57$ C + 22.4$ C = 79.4$ C
(b) The power dissipated by the chip must be balanced by the rate of heat transfer from the condensing
section. Hence, with A = πDL, Eq. 10.32 yields the requirement that
hLL =
q
πD Tsat − Ts
55.7 W
= 18.5 W m⋅ K
π 0.03 m 32 $ C
$ =
#
%
To determine h L , we combine Eqs. 10.33 and 10.35 to obtain Re δ = 4q µ " bh ′fg , where b = πD = 0.0942
$
#
m and h ′fg = h fg + 0.68c p,1 Tsat − Ts = 84,400 J kg + 0.68 1100 J kg ⋅ K 32 $ C = 108,300 J/kg. Hence, Reδ
= 4(55.7 W)/4.4 × 10 kg/m⋅s(0.0942 m)108,300 J/kg = 49.6 and the condensate film is in the laminarwavy region. Hence, from Eq. 10.38
-4
hL =
k"
ν" g
2
%
1/ 3
Re δ
0.054 W m⋅ K × 0.409
=
1.22
2
1.08 Re δ − 5.2
0.272 × 10 −6 m 2 s 9.8 m s 2
%
1/ 3
= 1130 W m 2⋅ K
in which case,
L=
18.5 W m⋅ K
= 0.0164 m = 16.4 mm
1130 W m 2⋅ K
<
COMMENTS: The chip operating temperature (T = 79.4°C) is not excessive, and the proposed scheme
provides a compact means of cooling high performance chips.
PROBLEM 10.70
KNOWN: Copper plate, 2m × 2m, in a condenser-boiler section maintained at Ts = 100°C separates
condensing saturated steam and nucleate-pool boiling of saturated liquid X.
FIND: (a) Rates of evaporation and condensation (kg/s) for the two fluids and (b) Saturation temperature
Tsat and pressure p for the steam, assuming that film condensation occurs.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Isothermal copper plate.
PROPERTIES: Fluid-X (Given, 1 atm): Tsat = 80°C, hfg = 700 kJ/kg, portion of boiling curve shown
above for operating condition, ∆Te = Ts − Tsat = (100 − 80)°C = 20°C, qs′′ = 5 × 104 W/m2 ; Table A.4,
Water (saturated, 1 atm): Tsat = 100°C, hfg = 2.25 × 106 J/kg; Water (saturated, Tsat): as required in part
(b); Water (saturated, Tf = (Tsat + Ts)/2): as required in part (b).
ANALYSIS: (a) For fluid-X, with ∆Te = Ts − Tsat = (100 − 80)°C = 20 K, the heat flux from the boiling
curve is
q′′s = 50, 000 W m 2
and the heat rate from the copper plate section into liquid-X is
qs = q′′s × As = 50,000 W m 2 × ( 2 × 2 ) m 2 = 200, 000 W
From an energy balance around liquid-X, the evaporation rate for fluid-X is
X = qs h fg,X = 200, 000 W 700, 000 J kg = 0.286 kg s
m
<
The heat rate into the copper plate section from the steam is qs = 200,000 W, and from an energy balance
around the condensate film, the condensation rate for steam (w)
w = qs h′fg,w = 200, 000 W 2.25 × 106 J kg = 0.0889 kg s
m
where we are assuming that Tsat,w is only a few degrees above Ts so that h ′fg′ ≈ h fg .
(b) The condensation heat rate, Eq. 10.32 can be expressed as
qs = h L As ( Tsat − Ts )
and assuming laminar film condensation, Eq. 10.30,
1/ 4
gρ ( ρ − ρ ) k 3 h ′
"
"
ν " fg
h L = 0.943
µ" k " (Tsat − Ts )
Continued...
PROBLEM 10.70 (Cont.)
Recognize that with qs , As and Ts known, this relation can be used to determine Tsat , and from the steam
table, the corresponding psat can be found. The vapor properties (v) are evaluated at Tsat while the liquid
properties ( ) are evaluated at the film temperature Tf = (Tsat + Ts)/2. An iterative solution is required,
beginning by assuming a value for Tsat , evaluate properties and check whether the rate equation returns
the assumed value for Tsat . Using the IHT Correlations Tool, Film Condensation, Vertical Plate for the
laminar region, the results are
Tsat = 381.7 K
psat = 1.367 bar
for which Reδ = 661, so that the flow is wavy-laminar, not laminar. Repeating the analysis but with the
IHT Tool for the laminar, wavy-laminar, turbulent regions, the results with Reδ = 652 are
Tsat = 379.6 K
Psat = 1.27 bar
COMMENTS: A copy of the IHT model for determining Tsat and psat for part (b) is shown below.
// Correlations Tool //Film Condensation, Vertical Plate, laminar, wavy-laminar, turbulent regions
NuLbar = NuL_bar_FCO_VP(Redelta,Prl)
// Eq 10.37, 38, 39
NuLbar = hLbar * (nul^2 / g)^(1/3) / kl
g = 9.8
// Gravitational constant, m/s^2
Ts = 100 + 273
// Surface temperature, K
Tsat = 380
// Saturation temperature, K; explore over range to match q
// The liquid properties are evaluated at the film temperature, Tf,
Tf = Tfluid_avg(Ts,Tsat)
// The condensation and heat rates are
q = hLbar * As * (Tsat - Ts)
// Eq 10.32
As = L * b // Surface Area, m^2
mdot = q / h'fg
// Eq 10.33
h'fg = hfg + 0.68 * cpl * (Tsat - Ts)
// Eq 10.26
// The Reynolds number based upon film thickness is
Redelta = 4 * mdot / (mul * b)
// Eq 10.35
/* Correlation description: Film condensation (FCO) on a vertical plate (VP). If Redelta<29, laminar
region, Eq 10.37 . If 31<Redelta<1750, wavy-laminar region, Eq 10.38 . If Redelta>=1850, turbulent
region, Eq 10.22, 10.32, 10.33, 10.35, 10.39 . In laminar-wavy and wavy-turbulent transition regimes,
function interpolates between laminar and wavy, and wavy and turbulent correlations. See Fig 10.15 . */
// Assigned Variables:
L=2
// Plate height, m
b=2
// Plate width, m
//q = 200000
// Heat rate, W; required heat rate for suitable Tsat
// Properties Tool - Water:
// Water property functions :T dependence, From Table A.6
// Units: T(K), p(bars);
xl = 0
// Quality (0=sat liquid or 1=sat vapor)
pl = psat_T("Water", Tf)
// Saturation pressure, bar
vl = v_Tx("Water",Tf,xl)
// Specific volume, m^3/kg
rhol = rho_Tx("Water",Tf,xl)
// Density, kg/m^3
cpl = cp_Tx("Water",Tf,xl)
// Specific heat, J/kg·K
mul = mu_Tx("Water",Tf,xl)
// Viscosity, N·s/m^2
nul = nu_Tx("Water",Tf,xl)
// Kinematic viscosity, m^2/s
kl = k_Tx("Water",Tf,xl)
// Thermal conductivity, W/m·K
Prl = Pr_Tx("Water",Tf,xl)
// Prandtl number
// Water property functions :T dependence, From Table A.6
// Units: T(K), p(bars);
xv = 1
// Quality (0=sat liquid or 1=sat vapor)
pv = psat_T("Water", Tsat)
// Saturation pressure, bar
vv = v_Tx("Water",Tsat,xv)
// Specific volume, m^3/kg
rhov = rho_Tx("Water",Tsat,xv)
// Density, kg/m^3
hfg = hfg_T("Water",Tsat)
// Heat of vaporization, J/kg
cpv = cp_Tx("Water",Tsat,xv)
// Specific heat, J/kg·K
muv = mu_Tx("Water",Tsat,xv)
// Viscosity, N·s/m^2
nuv = nu_Tx("Water",Tsat,xv)
// Kinematic viscosity, m^2/s
kv = k_Tx("Water",Tsat,xv)
// Thermal conductivity, W/m·K
Prv = Pr_Tx("Water",Tsat,xv)
// Prandtl number
<
PROBLEM 10.71
KNOWN: Thin-walled container filled with a low boiling point liquid (A) at Tsat,A. Outer surface of
container experiences laminar-film condensation with the vapor of a high-boiling point fluid (B).
Laminar film extends from the location of the liquid-A free surface. The heat flux for nucleate pool
boiling in liquid-A along the container wall is given as q′′npb = C(Ts − Tsat)3, where C is a known
empirical constant.
FIND: (a) Expression for the average temperature of the container wall, Ts; assume that the properties of
fluids A and B are known; (b) Heat rate supplied to liquid-A, and (c) Time required to evaporate all the
liquid-A in the container, assuming that initially the container is filled, y = L.
SCHEMATIC:
ASSUMPTIONS: (1) Nucleate pool boiling occurs on the inner surface of the container with liquid-A,
(2) Laminar film condensation occurs on the outer surface of the container with fluid-B over the liquid-A
free surface, y, and (3) Negligible wall thermal resistance.
ANALYSIS: (a) Perform an energy balance on the control surface about the container wall along
locations experiencing boiling (A) and condensation (B) as shown in the schematic above.
E ′′in − E ′′out = 0
(1)
q′′cond − q′′npb = 0
(2)
(
)
(
)
h y (π Dy ) Tsat,B − Ts − (π Dy ) C Ts3 − Tsat,A = 0
(
)
(
h y Tsat,B − Ts = C Ts − Tsat,A
)3
(3)
<
where h y is the average convection coefficient for laminar film condensation over the surface length 0
to y. From Eq. 10.30 and 10.26,
Continued...
PROBLEM 10.71 (Cont.)
1/ 4
gρ ( ρ − ρ ) k 3h ′
"
"
v " fg
h y = 0.943
µ" ( Tsat − Ts ) y
B
(
h′fg = h fg,B + 0.68c p,B Tsat,B − Ts
(3)
)
(4)
where the properties are for fluid-B.
(b) The heat flux supplied to liquid-A is, from Eq. (2), q′′cond = q′′npb . Since h y is a function of y, Ts
and, hence, the heat fluxes will be functions of y, the height of liquid A in the container.
(c) To determine the dry-out time, tf, begin with an energy balance on the inside of the container (fluidA). The heat transfer supplied to liquid-A results in an evaporation rate of liquid-A,
q′′npb (π Dy ) −
dM
h fg = 0
dt
(4)
where M is the mass of liquid-A in the container,
(
)
M = ρ",A π D2 4 y
(5)
Substituting Eq. (5) into (4), separating variables and identifying integration limits, find
(
C Ts − Tsat,A
)3 (π Dy ) = dtd ρ",A (π D2 4 ) y hfg
(
)
ρ",A π D2 4 h fg 0
tf
dy
dt = t f =
3
0
L
Cπ D
Ts − Tsat,A y
∫
∫
(
)
(6)
The definite integral could be numerically evaluated using values for Ts(y) obtained by solving Eq. (3).
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