. 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.
. 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.
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb3b05143-02e3-49df-bca2-429f03e7c8c0%2Fc161d202-9fb3-42a2-a865-a942ae17c6ba%2Flicaw9c_processed.jpeg&w=3840&q=75)
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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