up

Discussion, MAE 91, 8/11/2004

by Nasser M. Abbasi

Contents

1 Problem 8.95
 1.1 Statment
 1.2 Assumptions
 1.3 Laws
 1.4 Steps
  1.4.1 Numerical
 1.5 Steps
  1.5.1 Numerical

1 Problem 8.95

1.1 Statment

see book.

1.2 Assumptions

air is ideal gas

constant specific heat coefficients

1.3 Laws

1Q2 - W  = m (u2 - u1)

du = Cv0 dT

            k-1
       (P2) k--
T2 = T1 P1    for adiabatic process

PV =  mRT

1.4 Steps

Control mass energy equation gives

1Q2 - W  = m  (u2 - u1)

Since adiabatic, then  Q  = 0
1 2  , hence

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But du = Cv0 dT  , hence above becomes

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Need to find T2

Since adiabatic, then

       ( P ) k-k1
T2 = T1  -2-
         P1

Where     C
k = Cpv00

So equation (1) becomes

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Mass m  can be found from ideal gas law. P V = mRT  , hence m  = PV-
     RT  , so equation (2) becomes

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1.4.1 Numerical

For air, from table A.5, use Cv0 = 0.717  KJ/Kg-K,K  = 1.4  , R  = 0.287  , so equation (3) becomes

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NOw To find the length.

1.5 Steps

V2 = L A

Hence

    V2-
L = A

But P1V1 = mR
 T1  and P2V2= mR
 T2  , so P1V1-= P2V2
T1     T2   , hence V  = P1T2V
 2   P2T1 1

So

            (       )
    V2-   1- P1T2-
L =  A =  A  P2T1 V1

but        (P2) k-k1
T2 = T1 P1   hence

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1.5.1 Numerical

Given A = 5cm2   , V1 = 10cm3   so

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