Heat Transfer Through a Plane Wall

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SUMMARY

The discussion focuses on calculating the rate of heat transfer through a plane wall with specific parameters: thickness L = 0.4m, thermal conductivity k = 1.8 W/(m*K), surface area A = 30 m², and temperatures T1 = 90°C and Ts = 25°C. The calculated heat transfer rate is 7,387 W, which differs from the book's answer of 7,079 W. Participants confirmed the calculations using thermal resistance methods, indicating that the book may contain an error. The discussion also clarifies that temperature can be expressed in Celsius for temperature differences.

PREREQUISITES
  • Understanding of heat transfer principles, specifically conduction and convection.
  • Familiarity with thermal resistance calculations.
  • Knowledge of differential equations related to temperature distribution.
  • Basic concepts of temperature measurement in Celsius and Kelvin.
NEXT STEPS
  • Study the derivation of the heat transfer equation for plane walls.
  • Learn about thermal resistance in heat transfer applications.
  • Explore the differences between using Celsius and Kelvin in thermal calculations.
  • Review the textbook "Heat and Mass Transfer Fundamentals & Applications 4th Ed" by Cengel and Ghajar for additional context.
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Students and professionals in mechanical engineering, thermal engineering, and anyone involved in heat transfer analysis and problem-solving.

bob1182006
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Homework Statement


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.

Homework Equations


\dot{Q}_{wall} = -kA\frac{dT(0)}{dx}
\frac{d^2T}{dx^2} = 0
T(0) = 90
-k\frac{dT(L)}{dx} = h[T(L) - Ts]

The Attempt at a Solution



Solving the differential equation and applying B.C.:
T(x) = xC_1 + C_2
T(0) = C_2 = 90
-kC_1 = hLC_1+hC_2-25h]
C_1 = -\frac{h(C_2-25)}{k+hL}

Plugging numbers in:
C_2 = 90
C_1 = -136.8
T(x) = 90-136.8x
\dot{Q}_{wall} = -1.8*30*(-136.8) = 7,387 W

Did I make a mistake or is the book's answer wrong?
 
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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 = 7.41e-3 K/W and the equivalent thermal resistance of the convection effect as 1/hA = 1.39e-3 K/W, giving total thermal resistance of 8.80e-3 K/W so dQ/dt = (90 - 25)/8.80e-3 = 7387W.

Just curious - what is your textbook?
 
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.
 
arent we suppose to take temp. as kelvin here?
 
spree said:
arent we suppose to take temp. as kelvin here?

It's OK to interchange K and C as long as you're dealing with temperature differences.
 
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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