. A steel ball [c-0.46 kJ/kg C, k-35 W/m. C] 5.0 cm in diameter and initially at a uniform temperature of 450-C is suddenly placed in a controlled environment in which the temperature is maintained at 100°C. The convection heat-transfer coefficient is 10 W/m². C. Calculate the time required for the ball to attain a temperature of 150°C.

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Chapter3: Transient Heat Conduction
Section: Chapter Questions
Problem 3.30P
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Example:
.
A steel ball [c-0.46 kJ/kg C, k-35 W/m. C] 5.0 cm in diameter and
initially at a uniform temperature of 450°C is suddenly placed in a
controlled environment in which the temperature is maintained at 100°C.
The convection heat-transfer coefficient is 10 W/m². C. Calculate the
time required for the ball to attain a temperature of 150°C.
Solution:
We anticipate that the lumped-capacity method will apply because of the low value of
h and high value of k.We can check by using Equation (3.9):
hLc
B₁ =
k
3
Lc =
T 0.025
=-=
= 0.00833
A₂
4mr² 3
3
hle
10-0.00833
B₁ =
=
= 0.0023 < 0.1
k
35
The lumped system capacitance method is valid and can used. Therefore, may can use
equation 3.4:
t=
pvc In
0₁
hAs 0
92
HEAT TRANSFER
t = pvc In
pvc Ti-Too
hAs
T-Too
T = 150 °C, p = 7800 kg/m³.
Too-100 °C, h=10 w/m².C.
T₁-450 °C, c-460 J/kg. "C.
pvc In
Ti-Too
t =
hAs T-Too
t=
Pr³c
In Ti-Too
h4r² T-T
t= 1
In Ti-To
pre
3h T-T
V
-=
=
7800+0.025-460
In
3.10
450-100
150-100
= 5818.3 s= 1.62 h
6
Transcribed Image Text:Example: . A steel ball [c-0.46 kJ/kg C, k-35 W/m. C] 5.0 cm in diameter and initially at a uniform temperature of 450°C is suddenly placed in a controlled environment in which the temperature is maintained at 100°C. The convection heat-transfer coefficient is 10 W/m². C. Calculate the time required for the ball to attain a temperature of 150°C. Solution: We anticipate that the lumped-capacity method will apply because of the low value of h and high value of k.We can check by using Equation (3.9): hLc B₁ = k 3 Lc = T 0.025 =-= = 0.00833 A₂ 4mr² 3 3 hle 10-0.00833 B₁ = = = 0.0023 < 0.1 k 35 The lumped system capacitance method is valid and can used. Therefore, may can use equation 3.4: t= pvc In 0₁ hAs 0 92 HEAT TRANSFER t = pvc In pvc Ti-Too hAs T-Too T = 150 °C, p = 7800 kg/m³. Too-100 °C, h=10 w/m².C. T₁-450 °C, c-460 J/kg. "C. pvc In Ti-Too t = hAs T-Too t= Pr³c In Ti-Too h4r² T-T t= 1 In Ti-To pre 3h T-T V -= = 7800+0.025-460 In 3.10 450-100 150-100 = 5818.3 s= 1.62 h 6
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