In a manufacturing process, a copper disk is placed in a bath of cooling fluid.
ID: 2997822 • Letter: I
Question
In a manufacturing process, a copper disk is placed in a bath of cooling fluid. The copper disk has a 5 cm radius and a 1 cm thickness. The initial temperature of the copper disk is 450 degrees Celsius. The thermal conductivity, density, and specific heat capacity of copper are 400 W/m-K, 8933 kg/m^3, and 385 J/kg-K, respectively. The heat transfer coefficient between the surface of the disk and the cooling fluid is 100 W/m^2-K. Assume that the reservoir of cooling fluid is significantly large such that the temperature of the cooling fluid remains constant.
a) Can lumped capaciance be assumed?
b) if the disk is to be cooled to 200 degrees Celsius in two minutes, what should the temperature of the cooling fluid be?
c) Would a sphere of copper of similar volume cool faster? Explain
Explanation / Answer
a)
Diameter D = 5*2 = 10 cm = 0.1 m
Biot number Bi = hD/k
= 100*0.1 / 400
= 0.025 which is very low compared to 1.
Sinc Bi << 1, lumped capacitance can be assumed.
b)
Volume V = pi/4*D^2 *thickness
= 3.14/4 *0.1^2 *0.01
= 0.0000785 m^3
Surface area A = pi*D*thickness + 2*pi*R^2
= 3.14*0.1*0.01 + 2*3.14*0.05^2
= 0.01884 m^2
(T-Tamb)/(T0
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.