Numerical Methods for Engineers
Numerical Methods for Engineers
7th Edition
ISBN: 9780073397924
Author: Steven C. Chapra Dr., Raymond P. Canale
Publisher: McGraw-Hill Education
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Chapter 27, Problem 1P

A steady-state heat balance for a rod can be represented as

d 2 T d x 2 0.15 T = 0 T

Obtain an analytical solution for a 10-m rod with ( 0 ) = 240  and  T ( 10 ) = 150 .

Expert Solution & Answer
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To determine

To calculate: An analytical solution for a 10-m rod with T(0)=240 and T(10)=150, where a steady-state heat balance for a rod is represented as d2Tdx20.15T=0.

Answer to Problem 1P

Solution:

The solution is T(x)=3.016944×e0.15 x+236.9831×e0.15 x.

Explanation of Solution

Given Information:

A steady-state heat balance for a rod is represented as d2Tdx20.15T=0 where the rod is 10-m long with T(0)=240 and T(10)=150.

Calculation:

Consider the steady-state heat balance of rod given as d2Tdx20.15T=0.

Suppose T=eλx be the solution of the above differential equation. Then,

T=λeλxT=λ2eλx

Substitute the values of T and T in the given differential equation.

d2eλxdx20.15eλx=0λ2eλx0.15eλx=0eλx(λ20.15)=0

Since, eλx0. Then,

λ20.15=0λ2=0.15λ=±0.15

Therefore, the general solution of the differential equation is,

T(x)=A×e0.15 x+B×e0.15 x

The constant A and B can be evaluated by substituting each of the boundary conditions to generate two equations with two unknown constants.

Firstly, T(0)=240. Thus,

T(0)=A×e0.15×0+B×e0.15×0

Solve further to get,

240=A+B.. .. .. (1)

The second value is T(10)=150. Thus,

T(10)=A×e0.15×10+B×e0.15×10

Solve further to get,

150=48.08563A+0.0207962B.. .. .. (2)

Now to evaluate equations (1)and(2)

240=A+B150=48.08563A+0.0207962B

Multiply equation(1) by 48.0856 on both sides then subtract equation (2) from equation(1).

Then,

A=3.016944 and B=236.9831

Now put the value of A and B in general solution of the equation.

T(x)=3.016944×e0.15 x+236.9831×e0.15 x

Substitute the values of x=0,,10

At x=0

T(x)=3.016944×e0.150+236.9831×e0.150=240

At x=1

T(1)=3.016944×e0.151+236.9831×e0.151=165.329

At x=2

T(2)=3.016944×e0.152+236.9831×e0.152=115.769

At x=3

T(3)=3.016944×e0.15×3+236.9831×e0.15×3=83.7924

At x=4

T(4)=3.016944×e0.154+236.9831×e0.154=64.5426

At x=5

T(5)=3.016944×e0.155+236.9831×e0.155=55.0957

At x=6

T(6)=3.016944×e0.156+236.9831×e0.156=54.0171

At x=7

T(7)=3.016944×e0.157+236.9831×e0.157=61.1428

At x=8

T(8)=3.016944×e0.158+236.9831×e0.158=77.5552

At x=9

T(9)=3.016944×e0.159+236.9831×e0.159=105.747

At x=10

T(10)=3.016944×e0.1510+236.9831×e0.1510=150

The values of T generated corresponding to x are shown in the below table:

xT02401165.3292115.729383.7924464.5426555.0957654.0171761.1428877.55529105.74710150

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Chapter 27 Solutions

Numerical Methods for Engineers

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