
#1
Sep1611, 05:17 PM

P: 492

1. The problem statement, all variables and given/known data
Consider a large plane wall of thickness L = 0.4m, thermal conductivity k = 1.8W/(m*K), and surface area A = 30m2. The left side of the wall is maintained at a constant temperature of T1 = 90 C while the right side looses heat by convection to the surrounding air at Ts = 25 C with a heat transfer coefficient of h = 24 W(m2*K). Assuming constant thermal conductivity and no heat generation in the wall evaluate the rate of heat transfer through the wall. Answer: 7079 W My answer doesn't match the book's answer. 2. Relevant equations [tex]\dot{Q}_{wall} = kA\frac{dT(0)}{dx}[/tex] [tex]\frac{d^2T}{dx^2} = 0[/tex] [tex]T(0) = 90[/tex] [tex]k\frac{dT(L)}{dx} = h[T(L)  Ts][/tex] 3. The attempt at a solution Solving the differential equation and applying B.C.: [tex]T(x) = xC_1 + C_2[/tex] [tex]T(0) = C_2 = 90[/tex] [tex]kC_1 = hLC_1+hC_225h][/tex] [tex]C_1 = \frac{h(C_225)}{k+hL}[/tex] Plugging numbers in: [tex]C_2 = 90[/tex] [tex]C_1 = 136.8[/tex] [tex]T(x) = 90136.8x[/tex] [tex]\dot{Q}_{wall} = 1.8*30*(136.8) = 7,387 W[/tex] Did I make a mistake or is the book's answer wrong? 



#2
Sep1711, 06:40 AM

HW Helper
Thanks
PF Gold
P: 4,429

I used a somewhat different approach (underlying equations are of course the same) and got the same answer you did.
I computed the thermal resitance of the wall as L/kA = 7.41e3 K/W and the equivalent thermal resistance of the convection effect as 1/hA = 1.39e3 K/W, giving total thermal resistance of 8.80e3 K/W so dQ/dt = (90  25)/8.80e3 = 7387W. Just curious  what is your textbook? 



#3
Sep1711, 01:58 PM

P: 492

Awesome thanks, I didn't want to waste any more time redoing this problem looking for a mistake, and my class is just starting the thermal resistance chapter so if I see more problems that have wrong answers I'll do them that way to check.
We're using Heat and Mass Transfer Fundamentals & Applications 4th Ed by Cengel and Ghajar. 



#4
Oct1811, 02:58 PM

P: 5

Heat Transfer Through a Plane Wall
arent we suppose to take temp. as kelvin here?




#5
Oct1811, 03:33 PM

HW Helper
Thanks
PF Gold
P: 4,429





#6
Oct1811, 03:45 PM

P: 5

oh ure right. new to the subject, just trying to learn sorry :) btw I personally know professor cengel and i will inform him about this. I will let you know about it. if you suspect any other mistake in the book feel free to ask please.



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