Chemical Engineering - Heat Transfer - Discussion
Discussion Forum : Heat Transfer - Section 7 (Q.No. 40)
40.
A metal ball of radius 0.1 m at a uniform temperature of 90°C is left in air at 30°C. The density and the specific heat of the metal are 3000 kg/m3 and 0.4 kJ/kg.K respectively. The heat transfer co-efficient is 50 W/m2.K Neglecting the temperature gradients inside the ball, the time taken (in hours) for the ball to cool to 60°C is
Discussion:
6 comments Page 1 of 1.
Anil kardam said:
1 year ago
(T-Ta)/(To-Ta)= e^-(hAt/pCpV).
ln(T-Ta/To-Ta)= -3ht/pCpr.
Put the value in this formula and find the value of time t.
That the answer = 554.23s = 0.15 hrs.
ln(T-Ta/To-Ta)= -3ht/pCpr.
Put the value in this formula and find the value of time t.
That the answer = 554.23s = 0.15 hrs.
Mujahid said:
6 years ago
Good explanation. Thanks for giving it.
Aniche F. said:
6 years ago
Thanks for explaining.
Vijay vala said:
9 years ago
Mass of metal ball:
Mass = Density*volume.
= Density*(4*3.14*r^3/3).
(There for volume of ball = 4*3.14r^3/3).
= 3000(kg/m^3)*(4*3.14*0.1^3/3)(m^3).
= 3000*(4*3.14*0.001/3).
= 3000*(0.0041866).
Mass = 12.56kg.
Mass = Density*volume.
= Density*(4*3.14*r^3/3).
(There for volume of ball = 4*3.14r^3/3).
= 3000(kg/m^3)*(4*3.14*0.1^3/3)(m^3).
= 3000*(4*3.14*0.001/3).
= 3000*(0.0041866).
Mass = 12.56kg.
Yousaf said:
10 years ago
12.56 kg how?
A.Ali said:
10 years ago
First find the mass of metal ball:
Mass = Density*Volume = 12.56 kg.
Now use formula for uniform cooling process or lumped capacitance model.
t = (m*Cp/hA)*ln(T-Ta/t-Ta).
Plug in the values and solve:
t = 554.23 s = 0.15 hrs.
Mass = Density*Volume = 12.56 kg.
Now use formula for uniform cooling process or lumped capacitance model.
t = (m*Cp/hA)*ln(T-Ta/t-Ta).
Plug in the values and solve:
t = 554.23 s = 0.15 hrs.
(2)
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