Fluid Mechanics
Fluid Mechanics
8th Edition
ISBN: 9780073398273
Author: Frank M. White
Publisher: McGraw-Hill Education
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Chapter 4, Problem 4.40P
To determine

The temperature distribution T(y) for a constant wall temperature tw.

Expert Solution & Answer
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Answer to Problem 4.40P

The temperature distribution T(y) for a constant wall temperature tw is Tw+μU23k(1y4h4).

Explanation of Solution

Given information:

The velocity component of laminar flow between parallel plate in x direction, y direction and z direction is u=U(1y2h2), v=0, w=0 respectively Here, the velocity in x direction is u, the velocity in y direction is v, the velocity in z direction is w, the distance between the parallel plate is h, the distance from the centreline is y.

Write the expression for energy equation for the flow.

ρcvuTt=k2Ty2+μ(uy)2        …… (I)

Here, the specific volume of the fluid is cv, thermal conductivity is k and dynamic viscosity of the fluid is μ, density of the fluid is ρ, the rate of change of velocity component u with respect to distance from the centre line is uy.

Write the expression for change of temperature with respect to time

Tt=Tt+uTx+vTy+wTz        …… (II)

Here, the rate of change of temperature with respect to x is uTx, the rate of change of temperature with respect to y is vTy, the rate of change of temperature with respect to z is wTz, the rate of change of temperature with respect to t is Tt.

Since there is no change in temperature in y and z Substitute 0 for vTy, 0 for wTz in Equation (II).

Tt=Tt+uTx+0+0Tt=Tt+uTx

Substitute Tt+uTx for Tt in Equation (I).

ρcvu(Tt+uTx)=k2Ty2+μ(uy)2        …… (III)

Substitute 0 for Tt Equation (III).

ρcvu(uTx)=k2Ty2+μ(uy)2        …… (IV)

Since temperature only varies in y axis, Substitute 0 for Tx in Equation (IV)

ρcvu(u(0))=k2Ty2+μ( u y)20=k2Ty2+μ( u y)2k2Ty2=μ( u y)2        …… (V)

Calculation:

Substitute U(1y2h2) for u in Equation (V).

k2Ty2=μ( [ U( 1 y 2 h 2 )] y)2k2Ty2=μ(U y( 1 y 2 h 2 ))2k2Ty2=μ(U( 2y h 2 ))2k2Ty2=4μU2y2h4

2Ty2=4μU2y2kh4

Further integrating the equation.

Ty=( 4μ U 2 y 2 k h 4 )dy= 4μ U 2 h 4 k( y 2 )dy=4μU2y33h4k+c        …… (VI)

Substitute the boundary equation 0 for y, 0 for Ty in Equation (VI).

0=4μU2(0)3h4k+c0=c

Substitute 0 for c in Equation (VI)

Ty=4μU2y33h4k+(0)Ty=4μU2y33h4k

Further integrating the equation.

T y= 4μ U 2 y 3 3 h 4 kdyT(y)=4μU23h4ky3dyT(y)=4μU23h4k[y44]+cT(y)=μU2y43h4k+c        …… (VII)

Substitute the boundary equation Tw for T, h for y in Equation (VII).

Tw=μU2h43h4k+cTw=μU23k+cTw+μU23k=c

Substitute Tw+μU23k for c in Equation (VII).

T(y)=μU2y43h4k+(Tw+μ U 23k)=Tw+μU23k(1 y 4 h 4)

Conclusion:

The temperature distribution T(y) for a constant wall temperature tw is Tw+μU23k(1y4h4).

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