Problem 1
Known:
Dimensions and properties of a cup containing hot coffee as well as outside
thermal conditions
Unknown:
1. Applicability of LTCM
2. Initial rate of heat loss from coffee
3. Time required to cool coffee to 60°C
Schematic
k2=0.13W/m.K
t2 =0.5mm
Ho=83mm
Hot Coffee
ρ=985kg/m3
C=4180J/kg.K
Ti=80°C
hi=50W/m2.K
Lid
k1=0.035W/m.K
t1 =1.5mm
Tsurr=T∞=24°C
h0=10W/m.K
Do=52mm
Large desk
TD=24°C
kD=0.25W/m.K
Assumptions:
1. Constant thermo-physical properties
2. Negligible conduction through the edge of the cup
Properties:
No additional properties are required
Analysis
Uniform temperature will exist if Bi<0.1, the heat is lost from coffee through Lid, Side
walls and Bottom of the cup. These heat transfer processes occur in parallel to each other
and each can be characterized by a different Bi#
In general
Bi =
internal resistance (heat transfer within the coffee cup)
external resistance (from inner surface of the cup to surroundings)
i)
Consider the lid
Because of assumption 2, we ignore the edges, so that one can use internal or external
dimensions of the cup or the average of them. In this solution we will use internal
dimensions of the cup.
BiHeat loss through lead =
1 / hi Ai
=
1 / hi
t2
1
t2 1
+
+
k2 Ai ho Ai
k2 ho
1 / 50
=
0.0005 1
+
0.13 10
= 0.196 > 0.1
So LTCM is not applicable for heat loss through the lid
Consider side wall
1
1 / hi Ai
1 / hi 2πri Li
50 * 0.0245
BiHeat loss t side wall =
=
=
ln(ro / ri )
ln(ro / ri )
ln(26 / 24.5)
1
1
1
+
+
+
0.035
10 * 0.0245
2πk1 Li
ho Ai
ho 2πri Li
2πk1 Li
= 0.141 > 0.1
Also for heat loss through the side wall LTCM is not quite applicable
Consider bottom of the cup
1 / hi Ai
1 / 50 ⋅ π ⋅ 0.0245
BiHeat loss t bottom =
=
0.0015
1
t1
1
+
+
k1 Ai k D S 0.035 ⋅ π ⋅ 0.0245 0.25 ⋅ 4 ⋅ 0.0245
= 0.0063 < 0.1
Where S is shape factor for conduction between a disk (bottom of the cup) to semiinfinite media (case 10 in table 4.1)
LTCM is applicable
Since Bi# in two of three directions is not applicable, the LTCM for the analysis of temp
of coffee as a function of time is not applicable. This means that there will be some temp
gradient within the coffee at any time during the cooling process. This temp gradient will
be largest close to the lid. (the largest Bi#)
In order to proceed with the solution for part 2 and 3, we assume that LTCM is applicable
for part 2 and 3.
ii)
Applying energy balance on the coffee
dT
ρVc
= −qlid − q side − qbottom
dt
 T − T∞ T − T∞ T − TD
dT
= −
+
+
 Rt
dt
R
Rt ,bottom
t
,
,
lid
side


 ρVc


Since T∞=TD the above equation becomes
T − T∞  1
dT
1
1 
=−
+
+
dt
ρVc  Rt ,lid Rt , side Rt ,bottom 
Where
t
1
1
K
Rt ,lid =
+ 2 2 +
= 65.68
2
2
W
hi πri
k 2πri
hoπri
Rt , side =
Rt ,bottom
ln(r0 / ri )
K
1
1
+
+
= 6.60
hi 2πri Li
k1 2πLi
ho 2πri Li
W
t
1
1
K
=
+ 1 2 +
= 74.15
2
kD S
W
hi πri
k1πri
At t = 0, T = Ti = 80°C,
80 − 24
1
1 
dT
 1
=−
+
+


2
dt
985 ⋅ 4180 ⋅ π ⋅ 0.0245 ⋅ 0.081  65.68 6.60 74.15 
= - 0.016 K/s
iii)
Since
 T − T∞
dT
dT
= − A(T − T∞ ) ⇒
= − Adt ⇒ ln(T − T∞ ) TTi = − At ⇒ t = − ln
dt
T − T∞
 Ti − T∞
Where: A =
1  1
1
1
+
+
ρVc  Rt ,lid Rt , side Rt ,bottom

 = 2.866 ⋅ 10 − 4 1

s

 60 − 24 
−4
t = − ln
 2.866 ⋅ 10 = 1542s ≈ 26 min
 80 − 24 

 A

Problem 2
Known:
Initial and final temperatures of a niobium sphere
Diameter and properties of the sphere
Temperature of surroundings and gas flow and convection coefficient associated
with the flow
Unknown:
1. Time required to cool the sphere exclusively by radiation
2. Time required to cool the sphere exclusively by convection
3. Combined effects of radiation and convection
Schematic:
Assumptions:
1. Uniform temperature at any time
2. Negligible effect of holding mechanism on heat transfer
3. Constant properties
4. Radiation exchange is between a small surface and large surrounding
Analysis:
If cooling is exclusively by radiation, the required time is determined from eq. (5.18)
with V = πD 3 / 6, As.r = πD 2 and ε = 0.1
 Tsur + T

 T
T + Ti
− ln sur
+ 2 tan −1 
ln
Tsur − Ti
 Tsur − T
 Tsur

t=
ρVc
3
4εAs ,r σTsur
t=
8600 ⋅ 290 ⋅ 0.01
24 ⋅ (0.1) ⋅ 5.67 ⋅ 10 −8 ⋅ 298 3
= 1190s
 T

 − tan −1  i
 Tsur

 
 
 
 298 + 573

298 + 1173
 573 
−1  1173   
− ln
+ 2 tan −1 
 − tan 
 
ln
298 − 1173
 298  
 298 

 298 − 573
(ε = 0.1)
If ε = 0.6, cooling is six times faster, in which case,
t=199s
If cooling is xexlusively by convection, eq. (5.5) yields
ρcD  Ti − T∞  8600 ⋅ 290 ⋅ 0.01  875 
=
t=
ln
ln

6h  T f − T∞ 
1200
 275 
= 24.1s
With both radiation and convection, the temperature history may be obtained from eq.
(5.15).
dT
4
ρ (πD 3 / 6)c
= −πD 2 h(T − T∞ ) + εσ (T 4 − Tsur
)
dt
Rearrange above equation, yield,
dT
4
= − πD 2 h(T − T∞ ) + εσ (T 4 − Tsur
) ρ (πD 3 / 6)c
dt
[
[
]
]
ρc(πD 3 / 6)
4
4
∫ 2[
] dT
1173 πD h(T − T∞ ) + εσ (T − Tsur )
573
t=
Numerically integrating above equation from Ti=1173 at t=0 to T=573,
Table Numerical integration results
Number of steps
(n)
1
2
3
4
5
6
step length
h
600
300
200
150
120
100
value of
the integral
24.98
20.94
20.05
19.72
19.56
19.48
Error
4.03
0.89
0.33
0.16
0.09
Problem 3
Known:
A ball bearing is suddenly immersed in a molten salt bath
Heat treatement to harden occurs at locations with T > 1000K
Unknown:
Time required to harden outer layer of 1mm
Total energy required to harden outer layer of 1mm
Schematic:
Assumptions:
1. One-dimensional radial conduction
2. Constant properties
3. Fo ≥0.2
Analysis:
Since any location within the ball whose temperature exceeds 1000K will be hardened
the problem is to find the time when the location r=9mm reaches 1000K. then a 1mm
outer layer will be hardened. Begin by finding the Biot number.
hr
5000 ⋅ 0.01
Bi = o =
=1
k
50
Using the one term approximate solution for a sphere, we have
2
T − T∞
θ* =
= C1e −ζ 1 Fo ⋅ f1
Ti − T∞
Where f1 =
sin(ζ 1 r * )
ζ 1r *
This lead to
Fo = −
1
ζ1
2


1
ln θ * C1
sin(ζ 1 r * )
*
ζ 1r


From Table 5.1 with Bi=1, for the sphere find ζ 1 =1.5708 and C1=1.2732. with r* = r/ro =
9/10 = 0.9, substitute numerical values.
Fo = 0.441 > 0.2
So one-term approximate is applicable
From the definition of the Fourier number with α = k / ρc
ro ρc
0.012 ⋅ 7800 ⋅ 500
t = Fo
= 0.441
= 3.4s
50
k
Note that because of the properties of the ball, Fo = 0.441 corresponds to very short time.
2
2)
For a sphere the one-term approximate for the energy loss/gain becomes
Φ = 1 − Biθ 0* = 1 −
θ 0* = C1e −ζ
2
Fo
3[sin ζ 1 − ζ 1 cos ζ 1 ]
ζ1
3
θ 0*
2
= 1.2732e −1.5708 ⋅0.441 = 0.4289
Φ = 0.668
Φ = Q / Q0 ⇒ Q = Φ
πD 3 ρc p (T∞ − Ti )
= 10.9 ⋅ 10 3 J
6
Alternatively Φ can be determined graphically using fig D9
For Bi2*Fo = 0.441 and Bi =1 Î Φ =0.67
Which is very close to the value determined one-term approximation approach.