EBK FUNDAMENTALS OF AERODYNAMICS
EBK FUNDAMENTALS OF AERODYNAMICS
6th Edition
ISBN: 9781259681486
Author: Anderson
Publisher: MCGRAW HILL BOOK COMPANY
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Chapter 8, Problem 8.7P

The flow just upstream of a normal shock wave is given by p 1 = 1 atm , T 1 = 288 K , and M 1 = 2.6 . Calculate the following properties just downstream of the shock: p 2 , T 2 , ρ 0 , M 2 , p 0 , T 0 , 2 , , and the change in entropy across the shock.

Expert Solution & Answer
Check Mark
To determine

The downstream pressure.

The downstream temperature.

The downstream density.

The Mach number of downstream shock.

The pressure p02 .

The temperature T02 .

The entropy change.

Answer to Problem 8.7P

The downstream pressure is 7.72atm .

The downstream temperature is 658.66K .

The downstream density is 4.138kg/m3 .

The Mach number of downstream shock is 0.503 .

The pressure p02 is 9.1759atm .

The temperature T02 is 692K .

The entropy change is 22.898J/kgK .

Explanation of Solution

Given:

The upstream Mach number is M1=2.6 .

The upstream static pressure is p1=1atm .

The upstream temperature is T1=288K .

Formula used:

The expression for the downstream pressure is given as,

  p2p1=1+2.8γ+1(M121)

Here, γ is the adiabatic index of air and its value is 1.4 .

The expression for the downstream temperature is given as,

  T2=p2Rρ2

The expression for the downstream Mach number is given as,

  M22=1+( γ12)M12γM12( γ1)2

The expression for the downstream density is given as,

  ρ2ρ1=(γ+1)M122+(γ1)M12

The expression for the pressure p02 is given as

  po2p2=(1+ ( γ1 )2M22)γγ1

The expression for the temperature T02 is given as,

  To2T2=( p o2 p 2 )γ1γ

The expression for the entropy change is given as,

  Δs=Rlnpo2po1

Here, R is the ideal gas constant and its value is 287J/kgK .

The expression for the pressure po1 is given as,

  po1p1=(1+ ( γ1 )2M12)γγ1

Calculation:

The downstream pressure can be calculated as,

  p2p1=1+2.8γ+1(M121)p2=1atm(1+ 2.8 1.4+1( 2.6 2 1))p2=7.72atm

The downstream density can be calculated as,

  ρ2ρ1=( γ+1)M122+( γ1)M12ρ2=1.2kg/m3( ( 1.4+1 ) ( 2.6 ) 2 2+( 1.41 ) ( 2.6 ) 2 )ρ2=4.138kg/m3

Here, at pressure 1atm and temperature 288K the standard density is ρ1=1.2kg/m3 .

The temperature of the downstream can be calculated as,

  T2=p2Rρ2T2=( 7.72atm×101325)Pa287J/kgK×4.138kg/ m 3T2=658.66K

The Mach number at the downstream can be calculated as,

  M22=1+( γ1 2 )M12γM12 ( γ1 )2M22=1+( 1.41 2 ) ( 2.6 )21.4× ( 2.6 )2 ( 1.41 )2M2= 2.352 9.264M2=0.503

The pressure po2 can be calculated as,

  p o2p2=(1+ ( γ1 ) 2 M 2 2)γ γ1po2=7.72atm(1+ ( 1.41 ) 2 ( 0.503 ) 2) 1.4 1.41po2=9.1759atm

The pressure po1 can be calculated as,

  p o1p1=(1+ ( γ1 ) 2 M 1 2)γ γ1po1=1atm(1+ ( 1.41 ) 2 ( 2.6 ) 2) 1.4 1.41po1=19.95atm

The temperature T02 can be calculated as,

  T o2T2=( p o2 p 2 ) γ1γT o2658.66K=( 9.1759atm 7.72atm) 1.41 1.4To2=692K

The change in entropy can be calculated as,

  Δs=Rlnp o2po1Δs=287J/kgln9.176atm19.95atmΔs=222.898J/kg

Conclusion:

Therefore, the downstream pressure is 7.72atm .

Therefore, the downstream temperature is 658.66K .

Therefore, the downstream density is 4.138kg/m3 .

Therefore, the Mach number of downstream shock is 0.503 .

Therefore, the pressure p02 is 9.1759atm .

Therefore, the temperature T02 is 692K .

Therefore, the entropy change is 22.898J/kgK .

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