CHAPTER 6
HEAT TRANSFER IN CHANNEL FLOW
6.1 Introduction
• Important factors:
(1) Laminar vs. turbulent flow
transition Reynolds number Re D t is
Re Dt =
where
uD
ν
≈ 2300
(6.1)
D = tube diameter
u = mean velocity
ν = kinematic viscosity
1
(2) Entrance vs. fully developed region
Based on velocity and temperature distribution: two regions:
(i) Entrance region
(ii) Fully developed region
(3) Surface boundary conditions
Two common thermal boundary conditions:
(i) Uniform surface temperature
(ii) Uniform surface heat flux
(4) Objective
Depends on the thermal boundary condition:
(i) Uniform surface temperature. Determine: axial variation of
(1) Mean fluid temperature
(2) Heat transfer coefficient
(3) Surface heat flux
2
(ii) Uniform surface flux. Determine axial variation of:
(1) Mean fluid temperature
(2) Heat transfer coefficient
(3) Surface temperature
6.2 Hydrodynamic and Thermal Regions: General Features
• Uniform inlet velocity V i and temperature Ti
• Developing boundary velocity and thermal boundary layers
• Two regions:
r
(1) Entrance region
(2) Fully developed region
Vi
6.2.1 Velocity Field
(1) Entrance Region (Developing Flow,
0 ≤ x ≤ Lh )
x
uc
u
δ
Lh
fully developed
Fig. 6.1
3
• Hydrodynamic entrance region
• Length Lh : hydrodynamic entrance length
• Streamlines are not parallel ( v r ≠ 0)
• Core velocity uc = uc ( x )
(increasing or decreasing with x?)
• Pressure p = p( x ) , ( dp / dx < 0 )
• δ < D/2
r
Vi
x
uc
δ
Lh
(2) Fully Developed Flow Region
u
fully developed
Fig. 6.1
x > Lh : fully developed flow
Streamlines are parallel ( v = 0)
r
For 2 -D, constant ρ : ∂ u / ∂ x = 0
4
6.2.2 Temperature Field
• Entrance Region (Developing
Temperature, 0 ≤ x ≤ Lt ):
r
• Thermal entrance region
Ts
Tc
T
• Length Lt : Thermal entrance length Vi
• Core temperature Tc is uniform
(T c = T i )
Ti
δt < D / 2
x
Ts
δt
Lt
fully
developed
Fig. 6.2
(2) Fully Developed Temperature Region
• x ≥ Lt fully developed temperature
• T = T ( r , x ) or ∂ T / ∂ x ≠ 0
• Dimensionless temperature φ is invariant with x ( ∂φ / ∂x = 0 )
5
6.3 Hydrodynamic and Thermal Entrance Lengths
Lh and Lt are determined by:
(1) Scale analysis
(2) Analytic or numerical methods
6.3.1 Scale Analysis
(1) Hydrodynamic Entrance Length Lh
• Scaling of external flow:
δ
1
∼
x
Re x
r
(4.16)
Vi
Apply (4.16) to flow in tube: at x = Lh ,
δ ∼D
D
∼
Lh
1
Re Lh
x
uc
u
δ
Lh
fully developed
Fig. 6.1
(a)
6
Express Re L in terms Re D
h
Re Lh
(b) into (a)
=
u Lh u D Lh
L
=
= Re D h
v
v D
D
1/ 2
 Lh / D 


 Re D 
(2) Thermal Entrance Length Lt
~1
(6.2)
Scale for u: u ~ u (all Prandtl numbers)
Scale of δ t : For external flow
δ t ~ LRe L−1/2 Pr −1/2
(4.24)
Apply (4.24) for flow in tube: L = Lt , δ t ∼ D
−
D ~ L t Ret 1/2 Pr
−1/2
7
(a)
Express Re L in terms Re D
t
Re Lt
u Lt u D Lt
Lt
=
= Re D
=
v
v D
D
(b)
(b) into (a)
1/2
 Lt / D 


 Re D Pr 
~1
Lt
~ Pr
Lh
(6.3)
(6.4)
6.3.2 Analytic/Numerical Solutions: Laminar Flow
(1) Hydrodynamic Entrance Length Lh
Lh
= C h Re D
De
(6.5)
8
De = equivalent diameter
4 Af
De =
P
A f = flow area
Table 6.1
Entrance length coefficients
and
Uniform
surface
flux
Uniform surface
temperature
0.056
0.043
0.033
0.090
0.066
0.041
2
0.085
0.057
0.049
4
0.075
0.042
0.054
0.011
0.012
0.008
Geometry
a
b
a
Compare with scaling:
b
1/ 2
C t [1]
Ct
Ch
P = perimeter
C h = coefficient Table 6.1
 Lh / D 


 Re D 
Ch
a
~1
(6.2)
b
Rewrite (6.5)
1/2
 Lh / D 


 Re D 
1/2
= (C h )
(a)
9
Example: Rectangular channel, a / b = 2, , Table 6.1 gives C h = 0.085
Substitute into (a)
1/ 2
 Lh / D 


 Re D 
1/ 2
= (0.085 )
= 0.29
(b)
(2) Thermal Entrance Length Lt
Lt
= Ct PrReD
De
(6.6)
C t is given in Table 6.1
Compare with scaling
1/2
 Lt / D 


 Re D Pr 
~1
(6.3)
Rewrite (6.6):
10
1/2
 Lt / D 


 Re D Pr 
= C t1/2
(b)
Example: Rectangular channel, a / b = 2, , Table 6.1 gives Ct = 0.049 gives
1/2
 Lt / D 


 Re D Pr 
1/ 2
= (0.049 )
(c)
= 0.22
Turbulent flow: Experimental results:
• Lh and Lt are shorter than in laminar flow
Rule of thumb :
Ln
10 <
< 60
D
Ln
40 <
< 100
D
(6.7a)
(6.7b)
11
6.4 Channels with Uniform Surface Heat Flux q ′s′
L
When surface heat flux is uniform
Surface temperature is variable
Tmi
x Tm ( x )
• Section length L
• Inlet temperature: Tmi = Tm ( 0)
• Surface flux q ′s′
q′s′
Fig. 6.3
Determine:
(1) Total heat transfer
(2) Mean temperature variation Tm ( x )
(3) Surface temperature variation Ts ( x )
12
Total heat:
q s = q ′s′ As = q ′s′ P x
(6.8)
As = surface area
P = perimeter
Conservation of energy:
Assumptions:
(1) Steady state
(2) No energy generation
(3) Negligible changes in kinetic and potential energy
(4) No axial conduction
q s = q ′s′ P x = mc p [Tm ( x ) − Tmi ]
or
Tm ( x ) = Tmi
m = mass flow rate
q ′s′ P
+
x
mc p
(6.9)
13
c p = specific heat
(6.9) applies to any region and any flow
(laminar, turbulent or mixed)
Use heat transfer analysis to determine
surface temperature Ts ( x )
Newton’s law of cooling
q ′s′ = h( x )[Ts ( x ) − Tm ( x )]
Solve for Ts ( x )
q ′s′
Ts ( x )= Tm ( x ) +
h( x )
Use (6.9) to eliminate Tm ( x )
14
Ts ( x ) = Tm i
 Px
1 
+ q′s′ 
+

c
h
x
(
)
m
 p

(6.10)
h(x) is needed in (6.10) to determine Ts ( x )
To determine h(x):
(1) Laminar or turbulent flow?
(2) Entrance or fully developed region?
Example 6.2: Maximum Surface Temperature
• Water flows through tube
• Mean velocity = 0.2 m/s
o
• Tmi = 20 C
15
•
L
Tmo = 80 o C
• D = 0.5 cm
Tmi
• Uniform surface heat flux =
0.6 W/cm2
x
q′s′
D
Tm ( x )
0
q′s′
Ts ( x )
• Fully developed flow at outlet
• Nusselt number for laminar fully developed flow
NuD =
hD
= 4.364
k
(A)
Determine the maximum surface temperature
(1) Observations
• Uniform surface flux
•
Ts = Ts ( x ) , maximum at the outlet
16
• Laminar or turbulent flow? Check
Re D
• Is outlet fully developed? Check Lh and
• Uniform Nusselt number (h is constant)
Lt
• Length of tube section is unknown
(2) Problem Definition
(i) Determine L
(ii) Determine
T s (L)
(3) Solution Plan
(i) Apply conservation of energy
(ii) Compute Re D
(iii) Calculate Lh and Lt
(iv) Apply uniform flux analysis
( v ) If applicable use (A) to determine h
17
(4) Plan Execution
(i) Assumptions
• Steady state
• Constant properties
• Axisymmetric flow
• Uniform surface heat flux
• Negligible changes in kinetic and potential energy
• Negligible axial conduction
• Negligible dissipation
(ii) Analysis
Conservation of energy:
π DLq"s = mc p (Tmo − Tmi )
18
(a)
c p = specific heat, J/kg-oC
D = tube diameter = 0.5 cm = 0.005 m
L = tube length, m
m = mass flow rate, kg/s
Tmi = mean temperature at the inlet = 20oC
Tmo = mean temperature at the outlet = 80oC
q ′s′ = surface heat flux = 0.6 W/cm2 = 6000 W/m2
From (a)
Conservation of mass:
mc p (Tmo − Tmi )
L=
π Dq′′
(b)
m = (π / 4) D2 ρ u
(c)
s
where
19
u = mean flow velocity = 0.2 m/s
ρ = density, kg/m3
Surface temperature: Apply (6.10)
Ts ( x ) = Tm i
 Px

1
+ q′s′ 
+

 mc p h( x ) 
(6.10)
h = local heat transfer coefficient, W/m2-oC
P = tube perimeter, m
Ts ( x ) = local surface temperature, oC
x = distance from inlet of heated section, m
Perimeter P:
P=πD
(d)
Maximum surface temperature: set x = L in (6.10)
20
Ts ( L) = Tm i
 PL

1
+ q ′s′ 
+

 mc p h( L) 
Determine h(L): Is flow laminar or turbulent? Compute
Re D =
(e)
Re D
uD
(f)
ν
Properties Tm
Tmi + Tmo
Tm =
2
(g)
( 20 + 80)( o C)
T =
= 50o C
2
For water:
c p = 4182 J/kg-oC
k = 0.6405 W/m-oC
21
Pr = 3.57
-6
ν = 0.5537× 10 m2/s
3
ρ = 988 kg/m
Use (g)
Re D =
0.2(m/s)0.005(m)
−6
2
0.5537 × 10 (m /s)
Compute Lh and
= 1806 , laminar flow
Lt using (6.5) and (6.6)
Lh
= C h Re D
De
Lt
= C t PrRe D
De
C h = 0.056 (Table 6.1)
C t = 0.043 (Table 6.1)
(6.5)
(6.6)
22
Lh = 0.056 × 0.005 (m) × 1806 = 0.506 m
Lt = 0.043 × 0.005 (m) × 1806 × 3.57 = 1.386 m
Is L smaller or larger than
Lh and Lt ?
Compute L using (b). Use (c) to compute m
m = 988(kg/m3) 0.2(m/s)π (0.005)2(m2)/4 = 0.00388kg/s
Substitute into (b)
L=
0.00388(kg/s) 4182(J/kg− o C)(80 − 20)( o C)
2
4
2
2
= 10.33 m
0.005(m) 0.6 (W/cm )10 (cm /m )
L is larger than both Lh and Lt .
Flow is fully developed at the outlet
23
Equation (A) is applicable
NuD =
hD
= 4.364
k
(A)
(iii) Computations. Apply (A)
0.6405(W/m − o C)
h(L) = 4.364
= 559 W/m2-oC
0.005(m)
Use (e)
o
2 
Ts (L) = 20 C + 6000(W/m )

+
o
2 o 
0.00388(kg/s)4182(J/kg− C) 559(W/m − C)
π 0.005(m)10.43(m)
1
Ts ( L) = 90.7o C
(iv) Checking. Dimensional check:
Quantitative checks: (1) Alternate approach: apply Newton’s law at outlet
24
q ′s′ = h ( L )[T s ( L ) − Tmo
]
(i)
solve for T s ( L)
T s ( L) = Tmo
2
4
2
2
q ′s′
0.6(W/cm
)
×
10
(cm
/m
)
+ = 80 (oC) +
= 90.7oC
h
559(W/m 2 − o C)
(2) Compare value of h with Table 1.1
Limiting check: For Tmi
= Tmo , L = 0. Set Tmi = Tmo in (b) gives L = 0.
(3) Comments.
• In laminar flow local h depends on local flow condition: entrance vs.
fully developed
• Check
Re D to determine:
(i) If flow is laminar or turbulent
(ii) Entrance or fully developed
25
6.5 Channels with Uniform Surface Temperature
When surface temperature is uniform,
surface heat flux is variable
• Surface temperature: Ts
Ts
m
• Inlet temperature: Tmi = Tm (0)
• Section length: L
(2) Total heat q s
dx
dqs
Determine
(1) Mean temperature variation Tm ( x )
Tm ( x )
x
Tmi
dTm
Tm +
dx
dx
Tm
dx
Fig. 6.4
26
(3) Surface flux variation q′s′(x)
Analysis
Apply conservation of energy to element dx
Assumptions
(1) Steady state
(2) No energy generation
(3) Negligible changes in kinetic and potential energy
(4) No axial conduction
dq s = m c p dTm
(a)
Newton's law:
dq s = h( x )[Ts − Tm ( x )]Pdx
27
Combine (a) and (b)
dTm
P
=
h( x )dx
Ts − Tm ( x ) m c p
(c)
Integrate from x = 0 ( Tm = Tm (0) = Tmi ) to x ( Tm = Tm ( x ) )
Tm ( x ) − Ts 
P
ln 
=
−

T
−
T
m cp
 mi
s 
x
∫0 h( x )dx
Must determine h(x). Introduce h
1
h=
x
x
∫0 h( x )dx
(6.12)
(6.12) into (6.11)
28
Tm ( x ) = Ts + (Tmi
Ph
− Ts ) exp[−
x]
mcp
(6.13)
(6.13) applies to any region and any flow
(laminar, turbulent or mixed)
To determine h(x):
(1) Is flow laminar or turbulent flow?
(2) Entrance or fully developed region?
Total heat: Apply conservation of energy:
q s = m c p [Tm ( x ) − Tmi ]
29
Surface flux: Apply Newton’s law:
q ′s′ ( x ) = h( x )[Ts − Tm ( x )]
(6.15)
Properties: At mean of inlet and outlet temperatures
Example 6.3: Required Tube Length
• Air flows through tube
• Uniform surface temperature,
r
Ts = 130o C
• Mean velocity = 2 m/s,
u0
• Tmi = 35o C
• D = 1.0 cm
Ts
Ts
D
x
u
L
30
• Nusselt number for laminar fully developed flow
hD
NuD =
= 3.657
k
(A)
Determine: tube length to raise temperature to Tmo = 105o C
(1) Observations
• Laminar or turbulent flow? Check Re D
• Uniform surface temperature
• Uniform Nusselt number (h is constant) for fully
developed laminar flow
• Length of tube is unknown
31
(2) Problem Definition. Determine tube length needed to raise
temperature to specified level
(3) Solution Plan.
• Use uniform surface temperature analysis
• Compute Re D . Laminar or turbulent?
(4) Plan Execution
(i) Assumptions
• Steady state
• Fully developed flow
• Constant properties
32
• Uniform surface temperature
• Negligible changes in kinetic and potential energy
• Negligible axial conduction
• Negligible dissipation
(ii) Analysis
Tm ( x ) = Ts + (Tmi − Ts ) exp[−
Ph
x]
mcp
(6.13)
c p = specific heat, J/kg − o C
h = average h, W/m 2 − o C
m = flow rate, kg/s
P = perimeter, m
33
Tm(x) = mean temperature at x, o C
Tmi = mean inlet temperature = 35 o C
Ts = surface temperature = 130 o C
x = distance from inlet, m
Apply (a) at the outlet (x = L), solve for L
m cp
Ts − Tmi
L=
ln
P h Ts − Tmo
(a)
Tmo = outlet temperature = 105 o C
Properties: at Tm
Tm = Tmi + Tmo
2
(b)
34
P=πD
(c)
D2
m =π
ρu
4
(d)
D = inside tube diameter = 1 cm = 0.01 m
u = mean flow velocity = 2 m/s
ρ = density, kg/m3
For fully developed laminar flow
hD
NuD =
= 3.657
k
(e)
h = heat transfer coefficient, W/m 2 − o C
k = thermal conductivity of air, W/m − o C
35
k
h = h = 3.657 , for laminar fully developed
D
Compute: Reynolds number
uD
Re D =
(f)
(g)
ν
Use (b)
o
(
35
+
105
)(
C)
Tm =
= 70 o C
2
Properties:
c p = 1008.7 J/kg− o C
k = 0.02922 W/m − o C
Pr = 0.707
ν = 19.9 × 10 − 6 m 2 /s
ρ = 11.0287 kg/m 3
36
Use (f)
Re D = 2(m/s)0.01(m) = 1005 , flow is laminar
−6
2
19.9 × 10 (m /s)
(iii) Computations
P = π 0.01(m) = 0.03142 m
(0.01)2 (m 2 )
m =π
1.0287(kg/m 3 )2(m/s) = 0.0001616 kg/s
4
o
0.02922(W
/
m
−
C) = 10.69 W/m 2 − o C
h = 3.657
0.01(m)
Substitute into (a)
L=
0.0001616(kg/s)1008.7(J/kg −o C )
2 o
0.03142(m)10.69(W/m − C)
ln
(130 − 35)(o C)
o
(130 − 105)( C)
= 0.65 m
37
(iv) Checking. Dimensional check
(i) L = 0 for Tmo = Tmi . Set Tmo = Tmi in (a) gives L = 0
(ii) L = ∞ for Tmo = Ts . Set Tmo = Ts in (a) gives L = ∞
Quantitative checks: (i) Approximate check:
(h)
Energy added at the surface = Energy gained by air
Energy added at surface ≈ hπ DL(Ts − Tm )
(i)
Energy gained by air = mc p (Tmo − Tmi )
(j)
(j) and (k) into (i), solve for L
L=
m c p (Tmo − Tmi )
(k)
h πD(Ts − Tm )
38
L=
0.0001616(kg/s)1008.7(J/kg− o C)(105 − 35)( o C)
2
o
o
10.69(W/m − C)π (0.01)(m)(130 − 70)( C)
= 0.57 m
(ii) Value of h is low compared with Table 1.1. Review solution.
Deviation from Table 1.1 are expected
(5) Comments. This problem is simplified by two conditions:
fully developed and laminar flow
39
6.6 Determination of Heat Transfer Coefficient
h( x ) and Nusselt Number Nu D
r
q′s′
Two Methods:
ro
(1) Scale analysis
0
(2) Analytic or numerical solutions
Tm
6.6.1 Scale Analysis
Ts
Fig. 6.5
Fourier’s law and Newton’s law
∂T ( ro , x )
−k
∂r
h=
Tm − Ts
(6.16)
40
Scales:
r ~ δt
(a)
∂T ( ro , x ) Tm − T s
~
δt
∂r
(b)
(a) and (b) into (6.16)
k
h~
(Tm − Ts )
δt
Tm − Ts
or
h~
k
δt
(6.17)
41
Nusselt number:
hD
NuD =
k
(6.17) into the above
NuD ~
D
δt
(6.18)
Entrance region: δ t < D , NuD > 1
Special case: fully developed region
δ t ( x) ~ D
(6.18) gives
NuD ~ 1 (fully developed)
(6.19)
42
δ t in the entrance region: For all Pr
−1 / 2
−1 / 2
x
Pr
Re
δt ~
x
(4.24)
(4.24) into (6.18)
D 1/ 2 1/ 2
Nu D ~ Pr Re x
x
(c)
Express in terms of Re D
ux
uD x
x
Re x =
=
= Re D
ν
ν D
D
(d)
(d) into (c)
1/2
D
NuD ~  
x
Pr 1/2Re1/2
x
(6.20a)
43
or
NuD
1/ 2
 PrRe 


 x/D 
~1
(6.20b)
44
6.6.2 Basic Considerations for the Analytical
Determination of Heat Flux, Heat Transfer
Coefficient and Nusselt Number
(1) Fourier’s law and Newton’s law
∂T ( x , ro )
∂r
Define dimensionless variables
(a)
q′s′ = − k
T − Ts ,
x/D
θ ≡
ξ=
Re D Pr
Ti − Ts
v ∗x
vx
vr
∗
=
, vr =
u
u
,
,
r
R=
ro
Re D =
uD
ν
(6.21)
45
(6.21) into (a)
k
∂ 0(ξ ,1)
q′s′(ξ ) = (Ts − Ti )
ro
∂R
(6.22)
Newton’s law
q"s
h(ξ ) =
Tm − Ts
(6.23)
Combine (6.22) and (6.23)
k (Ts − Ti ) ∂θ (ξ ,1)
k 1 ∂θ (ξ ,1)
h(ξ ) =
=−
ro (Tm − Ts ) ∂R
ro θ m (ξ ) ∂R
(6.24)
Dimensionless mean temperatureθ m :
Tm − Ts
0m ≡
Ti − Ts
(6.25)
46
Nusselt number:
h(ξ ) D h(ξ )2ro
Nu(ξ ) =
=
k
k
(6.26)
(6.24) into (6.26)
− 2 ∂0 (ξ ,1)
Nu(ξ ) =
0 m (ξ ) ∂R
(6.27)
Determine: q ′s′ (ξ ), h(ξ ) and: Nu(ξ )
Find θ (ξ , R ) . Apply energy equation
(2) The Energy Equation
Assumptions
• Steady state
47
• Laminar flow
• Axisymmetric
• Negligible gravity
• Negligible dissipation
• Negligible changes in kinetic and potential energy
• Constant properties
 1 ∂  ∂T  ∂ T 
∂T 
 ∂T ∂ T
+ vz
ρc p  v r
+
 = k
r
+ 2
∂z 
 ∂r ∂θ
 r ∂r  ∂r  ∂z 
2
(2.24)
Replace z by x, express in dimensionless form
4 ∂  ∂0 
1
∂0
∗ ∂0
∗ ∂0
vx
+ 2 Re D Pr v r
=
R +
∂ξ
∂R R ∂R  ∂R  ( Re D Pr )2 ∂ξ 2
2
(6.28)
48
Pe = Re D Pr , Peclet number
(2.29)
• Third term: radial conduction
• Fourth term: axial conduction
• Neglect axial conduction for:
Pe = PrRe D ≥ 100
(6.30)
Simplify (6.28)
v ∗x
∂θ
4 ∂  ∂θ 
∗ ∂θ
+ 2 Re D Pr v r
=
R 
∂ξ
∂R R ∂R  ∂R 
(6.31)
(3) Mean (Bulk) Temperature Tm
Need a reference local temperature. Use Tm ( x )
49
mc pTm =
where
m=
∫0
ro
∫0
ro
ρ c p v xT 2πrdr
(a)
ρ v x 2πrdr
(b)
(b) into (a), assume constant properties
ro
v x T r dr
∫
Tm = 0 r
∫0 v x rdr
(6.32a)
o
In dimensionless form:
Tm − Ts
θm =
=
Ti − Ts
1
∫0
1
v ∗x RdR
∫0
v ∗xθ RdR
(6.32b)
50
6.7 Heat Transfer Coefficient in the Fully
Developed Temperature Region
6.7.1 Definition of Fully Developed Temperature
Profile
Fully developed temperature:
Ts ( x ) − T ( r , x )
φ=
Ts ( x ) − Tm ( x )
Let
x / d > 0.05 Re D Pr
(6.33)
Definition:
For fully developed temperature
φ
is independent of x
51
Therefore
φ = φ (r )
(6.34)
From (6.34)
∂φ
=0
∂x
(6.33) and (6.35):
∂φ
∂ Ts ( x ) − T ( r , x ) 
=0
=


∂x ∂x  Ts ( x ) − Tm ( x ) 
(6.35)
(6.36a)
Expand and use (6.33)
dTs ∂T
 dTs dTm 
− φ (r ) 
=0
−
−

dx 
dx ∂x
 dx
(6.36b)
52
6.7.2 Heat Transfer coefficient and Nusselt number
∂T ( ro , x )
−k
∂r
h=
Tm − Ts
(6.16)
Use(6.33) to form ∂T ( ro , x ) / ∂r , substitute into (6.16)
do/ (ro )
h = −k
dr
(6.37)
Conclusion:
The heat transfer coefficient in the fully developed region
is constant regardless of boundary condition
53
Nusselt number:
hD
dφ ( ro )
NuD =
= −D
h
dr
(6.38)
Entrance region scaling result:
NuD ~ 1 (fully developed)
(6.19)
Scaling of fully developed region:
scale for ∂T ( ro , x ) / ∂r
∂T ( ro , x ) Ts − Tm
~
∂r
D
Substitute into (6.16)
k
h~
D
(6.39)
54
(6.39) into (6.38)
NuD ~ 1 (fully developed)
(6.40)
6.7.3 Fully Developed Region for Tubes at
Uniform Surface flux
r
• Uniform flux
• Determine
• Ts ( x )
• h
Newton’s law:
u
0
q′s′
T
D
x
q′s′
Fig. 6.6
q′s′ = h [Ts ( x ) − Tm ( x )]
(a)
55
Ts ( x ) and Tm ( x ) are unknown
q ′s′ and h are constant
(a) gives:
[Ts ( x ) − Tm ( x )] = constant
Differentiate (b)
(b)
dT s dTm
=
dx
dx
(c)
∂T dT s
=
∂x
dx
(d)
(c) into (6.36b)
Combine (c) and (d)
∂T dTs dTm (for constant q ′′)
s
=
=
∂x
dx
dx
(6.41)
56
To determine h form (6.16) :
Determine: T ( r , r ), Tm ( x ) and Ts ( x )
Conservation of energy for dx
dTm 

′
′
qs Pdx + mc pTm = mc p Tm +
dx 
dx


Simplify
dTm q ′s′ P
=
dx
mc p
(6.42)
= constant
q′s′
(6.42) into (6.41)
∂T dTs dTm q ′s′ P
= constant
=
=
=
∂x
dx
dx mc p
(6.43)
m
dTm
Tm +
dx
dx
Tm ( x )
dx
Fig. 6.7
57
Conclusion: T ( x , r ), Tm ( x ) and T s ( x ) are linear with x
Integrate(6.43)
q′s′P
Tm ( x ) =
x + C1
mc p
(e)
Tm (0) = Tmi
(f)
C 1 = constant
Boundary condition:
Apply (e) to (f)
C1 = Tmi
(e) becomes
Tm ( x ) = Tmi
q′s′P
+
x
mc p
(6.44)
58
Determine T ( r , x ) and Ts ( x )
Apply energy equation (2.23) in the fully developed region
Assumptions
• Negligible axial conduction
• Negligible dissipation
• Fully developed, v r = 0
ρ c pv x
∂T k ∂  ∂T 
=
r

∂x r ∂r  ∂r 
(6.45)
 r2 
v x = 2u 1 − 
2
r

o 
(6.46)
59
However
m = π ro2 ρ u
P = 2π ro
equation (g) becomes
4q′s′  r 2  k ∂  ∂T 
r

1 − 2  =
ro  ro  r ∂r  ∂r 
(6.47)
Boundary conditions:
∂ T ( 0, x )
=0
∂r
Integrate (6.47)
∂T ( ro , x )
k
= q′s′
∂r
4 r 2 r 3 
∂T
q′s′  − 2  = kr
+ f (x)
ro  2 4ro 
∂r
(6.48a)
(6.48b)
(h)
60
f ( x ) = “constant” of integration
Boundary condition (6.48a)
f ( x) = 0
(h) becomes
∂T 4q′s′  r r 3 
=
 − 2
∂r kro  2 4ro 
Integrate
4q′s′
T (r , x ) =
kro
r 2 r4 
 − 2  + g( x )
 4 4ro 
(6.49)
g ( x ) = “constant” of integration
• Boundary condition (6.48b) is satisfied
• Use solution to Tm ( x ) to determine g ( x )
61
Substitute(6.46) and (6.49) into (6.32a)
7 roq′s′
Tm ( x ) =
+ g( x )
24 k
(6.50)
Two equations for Tm ( x ) : (6.44) and (6.50). Equating
g ( x ) = Tmi
7 roq′s′ Pq′s′
−
+
x
24 k
mc p
(6.51)
(6.51) into (6.49)
T ( r , x ) = Tmi
4q′s′
+
kro
r 2
r 3  7 roq′s′ Pq′s′
−
+
x (6.52)
 −
2
mc p
 4 16ro  24 k
Surface temperature Ts ( x ) : set r = ro in (6.52)
62
Ts ( x ) = Tmi
11 roq′s′ Pq′s′
+
x
+
24 k
mc p
(6.53)
T ( r , x ), Tm ( x ) and Ts ( x ) are determined
(6.44), (6.52) and (6.53) into (6.33)
24 1
φ (r ) = 1 −
11 ro2
 2 r 4  24 Pq ′s′
7
x+ x
r − 2  +
11
4ro  11 mc p

(6.54)
Differentiate(6.54) and use (6.38)
48
NuD =
= 4.364 , laminar fully developed
11
(6.55)
63
Comments:
• (6.55) applies to:
• Laminar flow in tubes
• Fully developed velocity and temperature
• Uniform surface heat flux
• Nusselt number is independent of Reynolds and Prandtl
numbers
• Scaling result:
Nu D ~ 1
(6.40)
64
6.7.4 Fully Developed Region for Tubes at
Uniform Surface Temperature
• Fully developed
• Uniform surface temperature Ts
Determine: Nu D and h
Assumptions:
• Neglect axial conduction
• Neglect dissipation
• Fully developed: v r = 0
Energy equation (2.24):
ρ c pv x
∂ T k ∂  ∂T 
=
r

∂x r ∂ r  ∂ r 
(6.45)
65
Boundary conditions:
∂ T ( 0, x )
=0
∂r
T ( ro , x ) = Ts
Axial velocity
 r2 
v x = 2u 1 − 2 
 ro 
(6.56a)
(6.56b)
(6.46)
Eliminate ∂T / ∂x in equation (6.45). Use (6.36a)
∂φ
∂ Ts ( x ) − T ( r , x ) 
=0
=


∂x ∂x  Ts ( x ) − Tm ( x ) 
(6.36a)
66
for Ts ( x ) = Ts, (6.36a) gives
∂T Ts − T ( r , x ) dTm
=
∂x Ts − Tm ( x ) dx
(6.57)
(6.46) and (6.57) into (6.45)
 r 2  Ts − T ( r , x ) dTm k ∂  ∂T 
ρ c p u 1 − 2 
=
r

r ∂r  ∂ r 
 ro  Ts − Tm ( x ) dx
(6.58)
Result: Solution to (6.58) by infinite power series:
NuD = 3.657
(6.59)
67
6.7.5 Nusselt Number for Laminar Fully
Developed Velocity and Temperature
in Channels of Various Cross Sections
• Analytical and numerical solutions
• Results for two classes of boundary conditions:
(1)Uniform surface flux
(2)Uniform surface temperature
• Nusselt number is based on the equivalent diameter
De =
4Af
(6.60)
P
• Results: Table 6.2
68
69
• Compare with scaling
NuD ~ 1 (fully developed)
(6.40)
Example 6.4: Maximum Surface Temperature
in an Air Duct
q′s′
• 4 cm × 4 cm square duct
• Uniform heat flux = 590 W/m 2
Tmo
o
• Heating air from 40 C to 120o C
u
• u = 0.32 m/s
Tmi
L
• No entrance effects (fully developed)
70
Determine: Maximum surface temperature
(1) Observations
• Uniform surface flux
• Variable Surface temperature, Ts ( x ) , maximum at outlet
• Compute the Reynolds number
• Velocity and temperature are fully developed
• The heat transfer coefficient is uniform for fully developed
flow
• Duct length is unknown
• The fluid is air
71
(2) Problem Definition
(i) Find the required length
(ii) Determine surface temperature at outlet
(3) Solution Plan
(i) Apply conservation of energy
(ii) Compute the Reynolds
(iii) Apply constant surface solution
(iv) Use Table 6.2 for h
(4) Plan Execution
(i) Assumptions
• Steady state
72
• Constant properties
• Uniform surface flux
• Negligible changes in kinetic and potential energy
• Negligible axial conduction
• Negligible dissipation
(ii) Analysis
Conservation of energy
P L q′s′ = m c p (Tmo − Tmi )
(a)
c p = specific heat, J/kg − o C
L = channel length, m
m = mass flow rate, kg/s
73
P = perimeter, m
q ′s′ = surface heat flux = 590 W/m 2
Tmi = 40 o C
Tmo = 120 o C
Solve (a) for L
L=
mc p (Tmo − Tmi )
(b)
P q ′′
Find m and P
m = ρ S 2u
(c)
P = 4S
(d)
S = duct side = 0.04 m
u = mean flow velocity = 0.32 m/s
74
ρ = density, kg/m 3
(c) and (d) into (b)
L=
ρ S u c p (Tmo − Tmi )
(e)
4 q′′
Surface temperature: Use solution (6.10)
Ts ( x ) = Tmi
 Px

1

+
+ q′s′

m
c
h
(
x
)
&
p


(f)
h( x ) = local heat transfer coefficient, W/m 2 − o C
Ts ( x ) = local surface temperature,o C
x = distance from inlet, m
75
Maximum surface temperature at x = L
Ts ( L) = Tmi
 4L

1

+
+ q′s′ 

ρ
S
u
c
h
(
L
)
p


(g)
Determine h(L): Compute the Reynolds number
Re De =
u De
(h)
ν
De = equivalent diameter, m
ν = kinematic viscosity, m 2 /s
2
A
S
De = 4 = 4
=S
P
4S
(i)
76
(i) into (h)
Re De =
uS
(j)
ν
Properties: At Tm
Tm = Tmi + Tmo
2
Tm
(k)
(40 + 120)( o C)
= 80 o C
=
2
Properties:
c p = 1009.5 J/kg − o C
k = 0.02991 W/m − o C
Pr = 0.706
77
ν = 20.92× 10− 6 m2/s
ρ = 0.9996 kg/m3
(j) gives
Re De
0.32(m/s)0.04(m)
=
= 611.9 ,
−6
2
20.92 × 10 (m /s)
laminar flow
(6.55) and Table 6.2
Nu De
h=h
h De
=
k
=
3.608
k
h = 3.608
De
(l)
(m)
78
(iii) Computations. Use (e)
3
o
o
0.9996(kg/m ) 0.04(m)0.32(m/s) 1009.5(J/kg- C)(120 − 40)( C)
L=
= 0.4378
2
(4) 590(W/m )
m
Use (m)
0.02991(W/m − o C)
h(L) = h = 3.608
= 2.7 W/m 2 − o C
0.04(m)
Substitute into (g)
Ts ( L) = 40( o C ) +
590( W/m
2


4(0.4378)(m)
1
)
 0.9996(kg/m 3 ) 0.04(m)0.32 (m/s) 1009.5(J/kg − o C) + 2.7(W/m 2 − o C) 


79
Ts ( L) = 338.5 oC
(iv) Checking. Dimensional check:
Quantitative checks: (1) Alternate approach to determining :
Newton’s law at outlet
q′s′ = h [Ts ( L) − Tmo ]
(n)
T s ( L) Solve for
2
′s′
590(W/m
)
q
o
o
T s ( L) = Tmo =
120
(
C)
+
=
338.5
C
=
2 o
2.7(W/m − C)
h
(2) Compare h with Table 1.1
Limiting check: L =0 for Tmo = Tmi . Set Tmo = Tmi
into (e) gives L = 0
80
(5) Comments
(i) Maximum surface temperature is determined by the
heat transfer coefficient at outlet
(ii) Compute the Reynolds number to establish if the flow
is laminar or turbulent and if it is developing or fully
developed
81
6.8 Thermal Entrance Region: Laminar Flow
Through Tubes
6.8.1 Uniform Surface Temperature:
Graetz Solution
• Laminar flow through tube
r
Ts
• Velocity is fully developed
• Temperature is developing
u
• No axial conduction (Pe > 100)
• Uniform surface temperature Ts
T
Ti
0
x
δt
Fig.6.8
Velocity:
vr = 0
(3.1)
82
1 dp 2
vz =
( r − ro2 )
4µ d z
(3.12)
Rewrite (3.1)
vz
= = 2[1 − R 2 ]
u
(3.1) and (6.61) into (6.31)
v*z
1
1 ∂  ∂θ 
2 ∂θ
1− R
=
R 
2
∂ξ R ∂R  ∂R 
(
)
(6.61)
(6.62)
Boundary conditions:
∂θ (ξ ,0)
=0
∂R
θ (ξ ,1) = 0
θ (0, R ) = 1
83
Solution summary: Assume a product solutions
(a)
θ (ξ , R ) = X (ξ )R (R)
(a) into (6.62), separating variables
dX n
+ 2λ2n X n = 0
dξ
(b)
d 2Rn 1 dRn
2
2
2
+
(
1
−
R
)
R
λ
+
n
n =0
2
2
R dR
dR
(c)
• λ n = eigenvalues obtained from the boundary conditions
• Solution X n (ξ ) to (b) is exponential
• Solution Rn ( R ) to (c) is not available in terms of
simple tabulated functions
84
Solutions to (b) and (c) into (a)
∞
θ (ξ , R ) = ∑ C nRn ( R ) exp(−2λ2nξ )
(6.64)
n=0
C n = constant
Surface flux:
k
∂0(ξ ,1)
q′s′(ξ ) = (Ts − Ti )
ro
∂R
(6.22)
(6.64) gives
dRn (1)
∂θ (ξ ,1) ∞
Cn
=
exp(−2λ2nξ )
dRn
∂R
n=0
∑
(d)
Define
C n dR n (1)
Gn = −
2
dR
(e)
85
(e) into (6.22)
∞
2k
q′s′ (ξ ) = − (Ts − Ti ) Gn exp(−2λ2nξ )
ro
n=0
∑
(6.65)
Local Nusselt number: is given by
− 2 ∂0 (ξ ,1)
Nu(ξ ) =
0 m (ξ ) ∂R
(6.27)
(d) gives ∂θ (ξ ,1) / ∂R
Mean temperature θ m (ξ ): (6.61) and (6.64) into (6.32b),
integrate by parts and use(e)
∞
θ m (ξ ) = 8∑
Gn
2
λ
n=0 n
exp(−2λ2nξ )
(6.66)
86
(d), (e) and (6.66) into (6.27)
∞
∑
Nu(ξ ) =
Gn exp(−2λnξ )
n=0
∞
2
2
Gn
∑ λ2
n=0
(6.87)
exp(−2λ2nξ )
n
Average Nusselt number: For length x
h (ξ ) D
Nu(ξ ) =
k
Two methods for determining h (ξ ) :
(f)
(1) Integrate local h(ξ ) to obtain h (ξ )
(2) Use (6.13)
87
Ph
Tm ( x ) = Ts + (Ti − Ts ) exp [−
x]
mcp
(6.13)
Solve for h
mcp
Tm ( x ) − Ts
ln
h =−
Px
Ti − Ts
(g)
(g) into (f), use m = ρ u π D 2 / 4 P = π D and definitions of
ξ , Re D and θ m in (6.21) and (6.25)
1
Nu(ξ ) = − lnθ m (ξ )
4ξ
(6.68)
• Need λ n and G n to compute q ′s′ (ξ ), θ m (ξ ), Nu (ξ ) and Nu (ξ )
• Table 6.3 gives λ n and G n
• (6.67) and (6.68) are plotted in Fig. 6.9 as Nu(ξ ) and Nu88(ξ )
89
Average Nu
Nusselt number
Local Nu
x/D
ReD Pr
Local and average Nusselt number for
tube at uniform surface temepratu re [4]
ξ=
Fig. 6.9
90
Comments
(1) Nu D and NuD decrease with distance from entrance
(2) At any location ξ Nu D > NuD
(3) Asymptotic value (at ξ ≈ 0.05 ) for Nu D and NuD
is 3.657. Same result of fully develop analysis
Nu(∞ ) = 3.657
(6.69)
(4) Properties at
Tmi + Tmo
Tm =
2
(6.70)
(5) Solution by trial and error if Tmo is to be determined
91
Example 6.5: Hot Water Heater
• Fully developed velocity in tube
• Developing temperature
• Uniform inlet temperature Ti = 25 o C
r
• Diameter = 1.5 cm
• Length = 80 cm
• Flow rate = 0.002 kg/s
• Heat water to 75 o C
Determine: Surface temperature
u
Ts
Ti
0
x
δt
(1) Observations
• Uniform surface temperature
92
• Compute Reynolds number: Laminar or turbulent flow?
• Compute Lh and Lt : Can they be neglected?
(2) Problem Definition
(i) Determine Ts
(ii) Determine h
(3) Solution Plan
(i) Apply uniform surface temperature results
(ii) Compute the Reynolds number: Establish if problem is
entrance or fully developed
(iii) Use appropriate results for Nusselt number
(4) Plan Execution
(i) Assumptions
• Steady state
93
• Constant properties
• Uniform surface temperature
• Negligible changes in kinetic and
potential energy
• Negligible axial conduction
• Negligible dissipation
(ii) Analysis
Uniform surface temperature
Tm ( x ) = T s + (Tmi − T s ) exp[−
Ph
x]
mcp
h = average heat transfer coefficient, W/m 2 − o C
m = flow rate = 0.002 kg/s
(6.13)
94
Tmi = mean inlet temperature = 25 o C
Tmo = mean outlet temperature = 75 o C
Apply (6.13) at outlet (x = L) and solve for Ts
Ts =
1
Tmi − Tm ( L) exp( Ph L / mc p )
1 − exp( Ph L / mc p )
[
Properties: at Tm
]
(a)
Tm = Tmi + Tmo
2
Perimeter P
P=πD
D = diameter = 1.5 cm = 0.015 m
95
Determine h : Compute the Reynolds
Re D =
u=
uD
ν
4m
(e)
ρ π D2
Properties: at Tm
c p = 4182 J/kg-oC
o
+
(
20
80
)(
C)
Tm =
= 50 o C
2
k = 0.6405 W/m-oC
Pr = 3.57
ν = 0.5537×10-6 m2/s
96
ρ = 988 kg/m3
Use (e)
u=
4(0.002)(kg/s)
3
2
2
988(kg/m )π (0.015) (m )
= 0.01146 m/s
Use (d) gives
Re D =
0.01146(m/s)0.015(m)
−6
2
0.5537 × 10 (m /s)
= 310.5 , laminar flow
Determine Lh and Lt
Lh
= C h Re D
D
Lt
= C t PrRe D
D
(6.5)
(6.6)
97
C h = 0.056 (Table 6.1)
C t = 0.033 (Table 6.1)
(6.5) and (6.6)
Lh = 0.056 × 0.015 (m) × 310.5 = 0.26 m
Lt = 0.033 × 0.015 (m) × 310.5 × 3.57 = 0.55 m
• Lh and Lt are not negligible, tube length L = 0.8 m
• Use Graetz solution Fig. 6.9 or Table 6.4
Compute ξ
x/D
ξ=
Re D Pr
(f)
98
Nusselt number Nu gives h
k
h = Nu
D
(g)
(iii) Computation. Evaluating ξ at x = L
0.8(m)/).015(m )
= 0.0481
ξ=
310 .5 × 3.57
At ξ = 0.481 Fig. 6.9 gives
Nu ≈ 4.6
Substitute into (g)
0.6405(W/m− o C)
h=
4.6 = 196.4W/m 2 − o C
0.015(m)
99
Equation (a) gives Ts
Compute the exponent of the exponential in (a)
Ph L π (0.015)(m)( 196.4(W/m 2 − o C)0.8(m)
=
= 0.88524
o
mc p
0.002(kg/s)4182(J/kg− C)
Substitute into (a)
Ts =
[
]
1
25( o C) − 75( o C)exp(0.88524) = 110.1o C
1 − exp(0.88524)
(iv) Checking. Dimensional check:
Limiting checks:
(i) For Tmi = T ( L) (no heating) Ts should be equal to Tmi .
Set Tmi = T (L) in (a) gives Ts = Tmi
100
(ii) If L = 0, Ts should be infinite. Set in L = 0 (a) gives Ts = ∞
Quantitative checks:
(i) Approximate check:
Energy added at the surface = Energy gained by water
(h)
Let
Tm = average water temperature in tube
Energy added at surface = hπ DL(Ts − Tm )
(i)
Energy gained by water = mc p (Tmo − Tmi )
(j)
(j) and (k) into (i), solve for Ts
101
Ts = Tm +
m c p (Tmo − Tmi )
(k)
h π DL
(k) gives
Ts = 50(o C) +
0.002(kg/s)4182(J/kg−o C)(75 − 25)(o C)
196.4(W/m2 −o C)π (0.0155(m)(0.8)(m)
= 106.5o C
(ii) Compare computed h with Table 1.1
(5) Comments
• Small error is due to reading Fig. 6.9
• Fully developed temperature model:
Nu D = 3.657, gives h = 156.3 W/m 2 − o C
102
6.8.2 Uniform Surface Heat Flux
• Repeat Graetz entrance problem with uniform
surface heat
• Fully developed inlet velocity
• Laminar flow through tube
• Temperature is developing
• No axial conduction (Pe > 100)
r
Energy equation: Same as
for Graetz problem
u
T
Ti
0
x
q′s′
D
δt
q′s′
Fig. 6.10
103
1
1 ∂  ∂θ 
2 ∂θ
1− R
=
R 
2
∂ξ R ∂R  ∂R 
(
)
(6.62)
Boundary conditions:
∂θ (ξ ,0)
)=0
∂R
∂θ (ξ ,1)
q′s′ro
=
∂R
k (Ti − Ts )
θ ( 0, R ) = 1
(6.71a)
(6.71b)
(6.71c)
Analytic solutions: Based on separation of variables
(1) Local Nusselt number
104

hx  11 1
Nu(ξ ) =
An exp(−2 β nξ )
= −
k
 48 2 n =1

∞
∑
−1
2
(6.72)
2
β
Table 6.6 lists eigenvalues n and constant An
(2) Average Nusselt number
hx  11
Nu(ξ ) =
= −
k  48
2  −1
1
1 − exp(−2 β n ξ )
An

2
2 n =1
2β nξ

∞
∑
(6.73)
Limiting case:
Fully developed: Set ξ = ∞ (6.72) or (6.73)
−1
 11 
Nu(∞ ) =   = 4.364
 48 
Same as fully developed result (6.55).
(6.74)
105
Graphical results: Fig. 6.11
106
Average Nu
Nusselt number
Local Nu
ξ=
x/D
ReD Pr
Fig. 6.11 Local and average Nusselt number for
tube at uniform surface heat flux [4]
107