EBK THERMODYNAMICS: AN ENGINEERING APPR
EBK THERMODYNAMICS: AN ENGINEERING APPR
8th Edition
ISBN: 8220102809444
Author: CENGEL
Publisher: YUZU
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Chapter 17.7, Problem 151RP
To determine

The expressions for the ratio of the stagnation pressure after a shock wave to the static pressure before the shock wave as a function of k and the Mach number upstream of the shock wave Ma1.

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Answer to Problem 151RP

The expressions for the ratio of the stagnation pressure after a shock wave to the static pressure before the shock wave as a function of k and the Mach number upstream of the shock wave Ma1 is obtained and shown in Equation (VI).

Explanation of Solution

Write the Equation 17-38 as in text book (the relation between the pressures after shock and before shock for an ideal gases).

P2P1=1+kMa121+kMa22 (I)

Here, the specific heat ratio is k, the Mach number is Ma, the subscript 1 and 2 indicates the states of before and after shocks.

Write the relation between the stagnation pressure (P0) and static pressure (P) for ideal gas at isentropic flow.

P0P=[1+(k12)Ma2]kk1 (II)

Here, the subscript 0 indicates the stagnation state.

Write the Equation 17-39 as in text book (the expression for Mach number after shock).

Ma22=Ma12+2/(k1)2Ma12k/(k1)1 (III)

Conclusion:

Rearrange the Equation (I) to obtain P2.

P2=P1(1+kMa121+kMa22)

Express the Equation (II) for state 2 i.e. after shock.

P02P2=[1+(k12)Ma22]kk1 (IV)

Substitute P1(1+kMa121+kMa22) for P2 in Equation (IV)

P02P1(1+kMa121+kMa22)=[1+(k12)Ma22]kk1P02P1=(1+kMa121+kMa22)[1+(k12)Ma22]kk1 (V)

Refer Equation (III).

Substitute Ma12+2/(k1)2Ma12k/(k1)1 for Ma22 in Equation (V).

P02P1=(1+kMa121+k(Ma12+2/(k1)2Ma12k/(k1)1))[1+(k12)(Ma12+2/(k1)2Ma12k/(k1)1)]kk1=(1+kMa12[2Ma12k/(k1)1]+[kMa12+k2/(k1)]2Ma12k/(k1)1)[1+(k1)[Ma12+2/(k1)]2(2Ma12k/(k1)1)]kk1=((1+kMa12)(2Ma12k/(k1)1)2Ma12k/(k1)1+kMa12+k2/(k1))[1+(k1)Ma12+22(2Ma12k/(k1)1)]kk1=((1+kMa12)(1k1)(2Ma12k1(k1))2Ma12k/(k1)1+kMa12+k2/(k1))[1+2[(k1)Ma12/2+1]2(2Ma12k/(k1)1)]kk1

=((1+kMa12)(2Ma12kk+1)(k1)[2Ma12k/(k1)1+kMa12+k2/(k1)])[1+(k1)Ma12/2+12Ma12k/(k1)1]kk1=((1+kMa12)(2Ma12kk+1)2Ma12k1(k1)+(k1)kMa12+2k)[1+(k1)Ma12/2+12Ma12k/(k1)1]kk1=((1+kMa12)(2Ma12kk+1)2Ma12kk+1+k2Ma12kMa12+2k)[1+(k1)Ma12/2+12Ma12k/(k1)1]kk1=((1+kMa12)(2Ma12kk+1)2Ma12k+k2Ma12kMa12k+1+2k)[1+(k1)Ma12/2+12Ma12k/(k1)1]kk1

=((1+kMa12)(2Ma12kk+1)kMa12(2+k1)+k+1)[1+(k1)Ma12/2+12Ma12k/(k1)1]kk1=((1+kMa12)(2Ma12kk+1)kMa12(k+1)+k+1)[1+(k1)Ma12/2+12Ma12k/(k1)1]kk1 (VI)

Thus, the expressions for the ratio of the stagnation pressure after a shock wave to the static pressure before the shock wave as a function of k and the Mach number upstream of the shock wave Ma1 is obtained and shown in Equation (VI).

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Chapter 17 Solutions

EBK THERMODYNAMICS: AN ENGINEERING APPR

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