Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)
Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)
8th Edition
ISBN: 9781305387102
Author: Kreith, Frank; Manglik, Raj M.
Publisher: Cengage Learning
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Chapter 2, Problem 2.1P

A plane wall, 7.5 cm thick, generates heat internally at the rate of 105 W/m3. One side of the wall is insulated, and the other side is exposed to an environment at 90°C. The convection heat transfer coefficient between the wall and the environment is 500 W/m2 K. If the thermal conductivity of the wall is 12 W/m K, calculate the maximum temperature in the wall.

Expert Solution & Answer
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To determine

The maximum temperature in the wall

Answer to Problem 2.1P

Maximum temperature in the wall is 128.430C at it occurs at x = 0.

Explanation of Solution

Given Information:

Thickness of the wall L = 7.5 cm = 0.075 m

Rate of internal heat generation qg = 105 W/m3

Temperature of environment = 900C

Convective heat transfer coefficient between the wall and the environment h = 500 W/m2K

Thermal conductivity of the wall k = 12W/mK

Assumptions:

  1. One dimensional steady state heat transfer.

  2. Properties of the wall and convective heat transfer coefficient remains constant.

Explanation: For one dimensional heat transfer with internal heat generation we have:

k2Tx2+qg.=ρcTtUnder steady state conditions: Tt= 0kd2Tdx2+qg.=0d2Tdx2=-qg.kdTdx=-qg.kx+C1 ------Equation 1

Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!), Chapter 2, Problem 2.1P

T = -qg.kx22+ C1x + C2 ----------Equation 2At x = 0 ; since it is insulated dTdx=0 From Equation 10=-qg.k(0)+C1C1=0Let T2 be the temperature of the wall at x = LFrom Equation 2,T2 = -qg.kL22+ 0*L + C2T2 = -qg.kL22+ C2

C2=T2+qg.kL22-----------Equation 3All the heat that is generated in the wall gets convected to environment at x = LTotal rate of heat generation in the wall Qg.=(qg.*A*L) WattAt x = L(qg.*A*L)= hA(T2 T)T2=qg.Lh+TSubstitute T2=qg.Lh+T in Equation 3C2=qg.Lh+T+qg.kL22Substitute C2 and C1 values in Equation 2

T = -qg.kx22+qg.Lh+T+qg.kL22 ---------- Equation 4 Equation-4 shows temperature distribution in the wall.From Equation 4 it is observed that , temperature decreases as the value of x increases.Maximum temperature in the wall occurs at x = 0.Substituting x = 0 in Equation - 4 we get

T = qg.Lh+T+qg.kL22Tmax = 105×0.075500+90+105×0.07522×12Tmax = 128.43 0C

Conclusion:

Maximum temperature in the wall occurs at x = 0 and it is 128.43 0C .

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Chapter 2 Solutions

Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)

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