Fundamentals of Heat and Mass Transfer
Fundamentals of Heat and Mass Transfer
7th Edition
ISBN: 9780470501979
Author: Frank P. Incropera, David P. DeWitt, Theodore L. Bergman, Adrienne S. Lavine
Publisher: Wiley, John & Sons, Incorporated
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Chapter 10, Problem 10.25P
To determine

The expression for the maximum temperature in the stainless tube and in the boron nitride (bn), and to express the results in terms of geometric parameters (r1,r2,r3,r4) , thermal conductivities (Kss,kbn) and the boiling parameters (C,Tsat,Ts) .

The value of the maximum temperature of the stainless steel tube and boron nitride tube is q˙2kss(r12r22)+(q˙r122k ss)ln(rr2)+Ts+(R23+R34)C(TsT sat)3 .

Given:

The given diagram is shown in Figure 1

  Fundamentals of Heat and Mass Transfer, Chapter 10, Problem 10.25P , additional homework tip  1

Figure 1

Calculation:

The required diagram for the resistance of the concentric cylinder circuit is shown in Figure 2

  Fundamentals of Heat and Mass Transfer, Chapter 10, Problem 10.25P , additional homework tip  2

Figure 2

In the above diagram the thermal resistive circuit is represented by R23 and resistance of the boron nitride is represented by R34 .

The expression for the heat generation for the boiling and the evaporation rate is given by,

  q˙boil=qxAx

The expression for the heat generation rate is evaluated as,

  q˙boil=qxAxq˙(r22r12)=(2πr4)C( T s T sat)3q˙=( 2 r 4 r 2 2 r 1 2 )C( T s T sat)3 ...... (1)

The expression for the maximum temperature of the stainless steel and boron nitride is evaluated as,

  T(r)=q˙2kssr2+C1lnr+C2  ........... (2)

The first boundary condition are given by,

  r=ηdTdr=0

Apply above boundary condition in equation (2).

  q˙2k ss2r1+C1η+0=0C1=±qr12k ss

The second boundary conditions are given by,

  r=r2T(r2)=T2

Apply above boundary equation in equation (2).

  T2=q˙2k ss2r22+q˙r12k sslnr2+C2=0C2=T2+q˙2k ssr22q˙r12k sslnr2

Rewrite equation (2) in terms of the above equations.

  T(r)=q˙2k ssr2+( q ˙ r 1 2 k ss )lnr+T2+( q ˙ 2 k ss )r22q˙r12k sslnr2=q˙2k ss(r2r22)+( q ˙ r 1 2 2 k ss )ln(r r 2 )+T2

From above and from equation (1) the final temperature equation is calculated as,

  T2TsR23+R34=(2πr4)C(TsT sat)3

The maximum temperature at inner stainless steel tube is calculated as,

  Tmax=q˙2kss(r12r22)+(q˙r122k ss)ln(rr2)+Ts+(R23+R34)C(TsT sat)3

The thermal resistance for he boron and the stainless steel is given by,

  R23=ln( r 3 r 2 )2πk bnR34=ln( r 4 r 3 )2πk ss

Conclusion:

Therefore, the value of the maximum temperature of the stainless steel tube and boron nitride tube is q˙2kss(r12r22)+(q˙r122k ss)ln(rr2)+Ts+(R23+R34)C(TsT sat)3 .

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Chapter 10 Solutions

Fundamentals of Heat and Mass Transfer

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