Introduction to Heat Transfer
Introduction to Heat Transfer
6th Edition
ISBN: 9780470501962
Author: Frank P. Incropera, David P. DeWitt, Theodore L. Bergman, Adrienne S. Lavine
Publisher: Wiley, John & Sons, Incorporated
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Chapter 4, Problem 4.1P

In the method of separation of variables (Section 4.2)for two-dimensional, steady-state conduction, the separationconstant λ 2 in Equations 4.6 and 4.7 must be apositive constant. Show that a negative or zero value of λ 2 will result in solutions that cannot satisfy the prescribedboundary conditions.

Expert Solution & Answer
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To determine

Whether a negative or zero value of λ2 will result in solutions that cannot satisfy the prescribed boundary conditions.

Answer to Problem 4.1P

The boundary condition 4 results in a trivial solution so a negative or zero value of λ2 will result in solutions that cannot satisfy the prescribed boundary conditions.

Explanation of Solution

Formula Used:

The temperature distribution equation is given by,

  θ(x,y)=X(x)Y(x)  .......(I)

The two dimensional steady state conduction equations is given by,

  2θx2+2θy2=0  .......(II)

Calculation:

Substitute Equation (I) in Equation (II).

  2θx2+2θy2=02( XY)x2+2( XY)y2=0Y2Xx2=X2Yy2  .......(III)

Divide Equation (III) by XY .

  YXY2Xx2=XXY2Yy21X2Xx2=1Y2Yy2  .......(IV)

Use separation constant λ2 in Equation (IV).

  1X2Xx2=λ22Xx2+Xλ2=0  .......(V)

  1Y2Yy2=λ22Yy2Yλ2=0  .......(VI)

Case 1:

Assume λ2=k2 .

Substitute λ2=k2 in Equation (V).

  2Xx2Xk2=0

From the above equation, the auxiliary equation is written as,

  m2k2=0m=±k

Here, the roots for the auxiliary equation are unequal and real.

The general solution for the auxiliary equation is given by,

  X(x)=C1ekx+C2ekx  .......(VII)

Substitute λ2=k2 in Equation (VI).

  2Xx2+Yk2=0

From the above equation, the auxiliary equation is written as,

  m2+k2=0m=±ik

The general solution for the auxiliary equation is given by,

  Y(y)C3cosky+C4sinky  .......(VIII)

Substitute (C1ekx+C2ekx) for X(x) and (C3cosky+C4sinky) for Y(y) in Equation (I).

  θ(x,y)=(C1ekx+C2ekx)(C3cosky+C4sinky)  .......(IX)

Apply boundary condition 1: at y=0 , θ(x,0)=0 in Equation (IX).

  θ(x,0)=(C1e kx+C2e kx)[C3cos(k×0)+C4sin(k×0)]0=(C1e kx+C2e kx)(C3)C3=0

Apply boundary condition 2: at x=0 , θ(0,y)=0 in Equation (IX).

  θ(x,y)=(C1e kx+C2e kx)(C3cosky+C4sinky)θ(0,y)=(C1e k×0+C2e k×0)(C3cosky+C4sinky)0=(C1+C2)(0×cosky+C4sinky)C4=0

A trivial solutions results as the constant C4=0 .

Case 2:

Assume λ2=0 .

Substitute 0 for λ2 in Equation (V).

  2Yy2Y(0)=02Yy2=0

Integrate the above obtained equation.

  dYdy=C7

Integrate the above equation.

  Y(y)=C7y+C8  .......(X)

Substitute (C5x+C6) for X(x) and (C7y+C8) for Y(y) in Equation (I).

  θ(x,y)=(C5xC6)(C7y+C8)  .......(XI)

Apply boundary condition 2: at x=0 , θ(0,y)=0 in Equation (XI).

  θ(x,y)=(C5x+C6)(C7y+C8)θ(0,y)=[C5×(0)+C6](C7y+C8)0=(C6)(C7y+C8)C6=0

Apply boundary condition 1: at y=0 , θ(x,0)=0 in Equation (XI).

  θ(x,y)=(C5x+C6)(C7y+C8)θ(x,0)=(C5x+0)[C7(0)+C8]0=(C5x)(C8)C8=0

Apply boundary condition 3: at x=L , θ(L,0)=0 in Equation (XI).

  θ(x,y)=(C5x+C6)(C7y+C8)θ(L,y)=[( C 5×L)+0](C7y+0)0=(C5L)(C7y)C5=0

Apply boundary condition 4: at y=W , θ(x,W)=1 in Equation (XI).

  θ(x,y)=(C5x+C6)(C7y+C8)θ(x,W)=[(0×x)+0][( C 7×W)+0]1=C7×W10

Conclusion:

The boundary condition 4 results in a trivial solution so a negative or zero value of λ2 will result in solutions that cannot satisfy the prescribed boundary conditions.

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