Suppose that two plates (thickness t=0.027 m and Area A=1.2 m^2 for each plate)
ID: 1861542 • Letter: S
Question
Suppose that two plates (thickness t=0.027 m and Area A=1.2 m^2 for each plate) are mated together. The thermal conductivity of the first plate is that of common brick (k=0.72 W/m*K), while the second plate is made of hard rubber (k=0.15 W/m*K). The temperature difference between the left-hand surface of the first plate and the right hand surface of the second plate is 8.5 C.
(a) Calculate the rate of heat transfer through the plates.
(b) What is the temperature at the point where the two plates are in contact?
(c) What is the temperature at a point that is 0.012 m from the left-hand face of the first plate?
Explanation / Answer
Thermal resistance R = t/(kA)
R1 = 0.027 / (0.72*1.2) = 0.03125 K/W
R2 = 0.027 / (0.15*1.2) = 0.15 K/W
Total thermal resistance = R1 + R2 = 0.18125
a)
Heat transfer Q = (T1 - T2) / R
Q = 8.5 / 0.18125 = 46.9 W
b)
Q = (T1 - T) / R1 = (T - T2) / R2
T1 - T = Q*R1
T - T2 = Q*R2
Subtracting, T1 + T2 - 2T = Q(R1 - R2)
T1 + T2 - 2T = 46.9*(0.03125 - 0.15)
T = (T1 + T2)/2 + 2.784
c)
Q = k1*A1*(T1 - T) / t1
Rate of temerature change = (T1 - T)/t1
(T1 - T) / 0.027 = (T1 - Tx) / 0.012
Tx = T1 - 12/27*(T1 - T)
Tx = T1 - 12/27*(T1 - ((T1 + T2)/2 + 2.784))
Tx = 0.777*T1 + 0.222*T2 + 1.237
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.