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.

Question

Want to see more full solutions like this?

Answer 1

Textbook Question

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

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,

$∂2θ∂x2+∂2θ∂y2=0$ .......(II)

Calculation:

Substitute Equation (I) in Equation (II).

$∂2θ∂x2+∂2θ∂y2=0∂2( X⋅Y)∂x2+∂2( X⋅Y)∂y2=0−Y∂2X∂x2=X∂2Y∂y2$ .......(III)

Divide Equation (III) by $XY$ .

$−YXY∂2X∂x2=XXY∂2Y∂y2−1X∂2X∂x2=1Y∂2Y∂y2$ .......(IV)

Use separation constant $λ2$ in Equation (IV).

$−1X∂2X∂x2=λ2∂2X∂x2+Xλ2=0$ .......(V)

$1Y∂2Y∂y2=λ2∂2Y∂y2−Yλ2=0$ .......(VI)

Case 1:

Assume $λ2=−k2$ .

Substitute $λ2=−k2$ in Equation (V).

$∂2X∂x2−Xk2=0$

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

$m2−k2=0m=±k$

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

The general solution for the auxiliary equation is given by,

$X(x)=C1e−kx+C2ekx$ .......(VII)

Substitute $λ2=−k2$ in Equation (VI).

$∂2X∂x2+Yk2=0$

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 $(C1e−kx+C2ekx)$ for $X(x)$ and $(C3cosky+C4sinky)$ for $Y(y)$ in Equation (I).

$θ(x,y)=(C1e−kx+C2ekx)⋅(C3cosky+C4sinky)$ .......(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).

$∂2Y∂y2−Y(0)=0∂2Y∂y2=0$

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).

$θ(x,y)=(C5x+C6)⋅(C7y+C8)θ(x,W)=[(0×x)+0]⋅[( C 7×W)+0]1=C7×W1≠0$

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Complete the following problems. Show your work/calculations, save as.pdf and upload to the assignment in Blackboard. missing information to present a completed program. (Hint: You may have to look up geometry for the center drill and standard 0.5000 in twist drill to know the required depth to drill). 1. What are the x and y dimensions for the center position of holes 1,2, and 3 in the part shown in Figure 26.2 (below)? 6.0000 Zero reference point 7118 1.0005 1.0000 1.252 Bore 6.0000 .7118 Cbore 0.2180 deep (3 holes) 2.6563 1.9445 Figure 26.2 026022 (8lot and Drill Part) (Setup Instructions--- (UNITS: Inches (WORKPIECE NAT'L SAE 1020 STEEL (Workpiece: 3.25 x 2.00 x0.75 in. Plate (PRZ Location 054: ' XY 0.0 - Upper Left of Fixture TOP OF PART 2-0 (Tool List ( T02 0.500 IN 4 FLUTE FLAT END MILL #4 CENTER DRILL Dashed line indicates- corner of original stock ( T04 T02 3.000 diam. slot 0.3000 deep. 0.3000 wide Intended toolpath-tangent- arc entry and exit sized to programmer's judgment…

A program to make the part depicted in Figure 26.A has been created, presented in figure 26.B, but some information still needs to be filled in. Compute the tool locations, depths, and other missing information to present a completed program. (Hint: You may have to look up geometry for the center drill and standard 0.5000 in twist drill to know the required depth to drill).

We consider a laminar flow induced by an impulsively started infinite flat plate. The y-axis is normal to the plate. The x- and z-axes form a plane parallel to the plate. The plate is defined by y = 0. For time t <0, the plate and the flow are at rest. For t≥0, the velocity of the plate is parallel to the 2-coordinate; its value is constant and equal to uw. At infinity, the flow is at rest. The flow induced by the motion of the plate is independent of z. (a) From the continuity equation, show that v=0 everywhere in the flow and the resulting momentum equation is მu Ət Note that this equation has the form of a diffusion equation (the same form as the heat equation). (b) We introduce the new variables T, Y and U such that T=kt, Y=k/2y, U = u where k is an arbitrary constant. In the new system of variables, the solution is U(Y,T). The solution U(Y,T) is expressed by a function of Y and T and the solution u(y, t) is expressed by a function of y and t. Show that the functions are identical.…

Answer 2