solutions 11

8/3/2019 Solutions 11

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ENGN 2810, Fall 2011

Homework 11 Solutions

Problem 2 For convencience, we shift the frame of reference from a shock moving at speed, us, to a stationary shock with

oncoming velocity, u1 = us. We then apply conservation of mass, momemtum, and energy across the shock and the equationof state to get the desired relation for the shock speed.

State:p = RT

Mass:1u1 = 2u2

Momentum:p1 + u

2

1= p2 + 2u

2

2

Using mass to replace 2u2 with 1u1:

p1

p2

1u1= u2 u1

Energy:

h1 +1

2U21

= h2 +1

2U22

Assuming a calorically perfect gas, we take h = cpT.

cpT1 +1

2u21 = cpT2 +

1

2u22

Given the relationship between cp, , and R: cp =R1

we have:

R

1

T1 +1

2

u21 =

R

1

T2 +1

2

u22

Using the equation of state to replace temerature in favor of density and pressure:

1

p11

+1

2u21 =

1

p22

+1

2u22

Using the combination of momentum and mass from above, the previous equation can be rewritten as:

p11p22

= 1

2

2u1 +

p1 p21u1

p1 p21u1

Using mass:

p1 u2

u1p2 =

1

22 + p1 p2

1u2

1 (p1 p2)

Using the momemtum-mass combination to eliminate u2:

p1

1 +

p1 p21u21

p2 =

1

2

2 +

p1 p21u21

(p1 p2)

Now introducing the definition of the speed of sound: c2 = p

:

p1

1 +

p1 p2p1c21

u21

p2 =

1

2

2 +

p1 p2p1c21

u21

(p1 p2)


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Forming the Mach number, M= uc

and rewriting:

(p1 p2)

1

p2p1M21

=

1

2

2 +

1 p2p1

M21

(p1 p2)

M21 =p2p1 1

2

p2p1

+ 1

2

Simplifying:

u2s = c2

1

1

2++ 1

2p2p1

us = c1

1

2++ 1

2

p2p1

Calculate the speed if the temperature of the ambient air is 30C and the pressure ratio, p2/p1, is 3.0.

c1 =RT1

us =

(1.4)(280)(303)

1.4 1

2(1.4)+

1 + 1.4

2(1.4)(3) = 567.8

m

s

Problem 4 1.) Find the ratio of the exit area to the throat area which is necessary for the supersonic exhaust to be correctly

expanded.

For correctly expanded flow, the ratio of the exit area to the throat area, AEA

, will be given in terms of the exit Machnumber, ME.

AEA

2=

1

M2E

2

+ 1

1 +

1

2M2E

+11

Assuming isentropic flow, we calculate the exit Mach number based on the properties in the reservoir.

p0pE

=

1 +

12

M2E 1

M2E =2

1

p0pE

1

1

ME =

2

1.4 1

(37)

1.411.4 1

= 3.005

The area ratio is then: AEA

2=

1

(3.005)2

2

1.4 + 1

1 +

1.4 1

2(3.005)2

1.4+11.4

= 18.097

AE

A = 4.2542.) Find the Mach number of the exit flow under correctly expanded conditions.

From part (1):ME = 3.005

3.) Find the lowest pressure ratio at which the same nozzle would be choked.

Since we are looking at the same nozzle, we fix the ratio of the exit area to the throat area. For isentropic flow, a given arearatio leads to two correctly expanded flows with the flow sonic at the throat: one that is supersonic at the exit and one thatis subsonic at the exit. If the pressure ratio, p0

pE, is lower than that for the subsonic solution, the flow is no longer choked. For


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these lower pressure ratios, the flow is subsonic throughout the nozzle and the mass flow rate out of the nozzle varies withpressure ratio. Thus, the pressure ratio that corresponds to the correctly-expanded subsonic solution demarks the boundarybetween choked and unchoked flow. From the isentropic tables with AE

A= 4.254 the exit Mach number is 0.1379. At this

Mach number the pressure ratio would be:

p0pE

=

1 +

1.4 1

2(0.1379)2

1.41.41

= 1.013

4.) Find the pressure ratio at which there would be a normal shock wave at the exit.

As the pressure ratio is increased from the correctly-expanded subsonic to supersonic solution, a shock that forms in theincreasing-area portion of the nozzle moves from the throat (where it is exceedingly weak) to the exit. The shock standsdirectly at the exit for the correctly-expanded supersonic solution (ME = 3.005). The pressure upstream of the shock isgiven from isentropic flow.

p0p1

= 37

The pressure then drops across the shock according to the normal shock relations:

p2p1

= 1 +2

+ 1

M2

1 1

= 1 +2(1.4)

1.4 + 1((3.005)2 1) = 10.367

Noting that the pressure downstream of the shock is equal to the exit pressure (p2 = pE), the pressure ratio at which therewould be a normal shock wave at the exit is:

p0pE

=p0p1

p1pE

=37

10.37= 3.569


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solutions 11

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