CP502 Advanced Fluid Mechanics

CP502 Advanced Fluid Mechanics

Compressible Flow

Part 02_Set 02:Steady, quasi one-dimensional, isentropic

compressible flow of an ideal gas in a variable area duct

(continued)

R. Shanthini 23 Feb 2012

Show that the steady, one-dimensional, isentropic, compressible flow of an ideal gas with constant specific heats can be described by the following equations:

Problem 6 from Problem Set #2 in Compressible Fluid Flow:

(2.6)20

2

11 M

T

T

(2.5)

120

2

11

M

p

p

1

1

20

2

11

M

(2.7)

(2.8)

0

100

T

T

p

p

where p0, T0 and ρ0 are the stagnation (where fluid is assumed to be at rest)

properties, p, T and ρ are the properties at Mach number M and γ is the specific heat ratio assumed to be a constant.


T0

p0

u0=0Stagnationproperties

T

p

u

M

Ideal gas satisfies

Isentropic flow of an ideal gas satisfies p

RTp

Using the above two equations, we can easily prove

(2.5)

0

100

T

T

p

p


Start with the energy balance for steady, adiabatic, inviscid, quasi one-dimensional, compressible flow:

dh + udu = 0

T0

p0

Stagnationproperties

Using dh = cp dT, which is applicable for ideal gas, in the above, we get

cp dT + udu = 0

Integrating the above between the two cross-sections, we get

cp (T – T0) + (u2 – 0)/2 = 0

(2.2)

T

p

u

M

u0=0


Using the definition of M, we get

cp (T – T0) + M2 γRT/2 = 0

Since cp = γR/(γ -1) for an ideal gas, the above can be written as

γR (T – T0) /(γ -1) + M2 γRT/2 = 0

which can be rearranged to give the following:

The above equation relates the stagnation temperature (at near zero velocity) to a temperature at Mach number M for steady, isentropic, one-dimensional, compressible flow of an ideal gas in a variable area duct.

(2.6)20

2

11 M

T

T


Using (2.6) in (2.5), we can easily get the following equations relating the stagnation properties (at near zero velocity) to properties at Mach number M for steady, isentropic, one-dimensional, compressible flow of an ideal gas in a variable area duct:

120

2

11

M

p

p

1

1

20

2

11

M

(2.7)

(2.8)



Show that the mass flow rate in a steady, one-dimensional, isentropic, compressible flow of an ideal gas with constant specific heats is given by the following equations:

RTAMpm

)1(2

1

200

2

11

1

MRTAMp

(2.9)

(2.10)


A large air reservoir contains air at a temperature of 400 K and a pressure of 600 kPa. The air reservoir is connected to a second chamber through a converging duct whose exit area is 100 mm2. The pressure inside the second chamber can be regulated independently. Assuming steady, isentropic flow in the duct, calculate the exit Mach number, exit temperature, and mass flow rate through the duct when the pressure in the second chamber is (i) 600 kPa, (ii) 500 kPa, (iii) 400 kPa, (iv) 300 kPa and (v) 200 kPa.


T0 = 400 K

p0 = 600 kPa

Air

Ae = 100 mm2

pb kPa is given

Assumptions: Steady, isentropic flow

Determine the following at the exit of the converging duct: Mach Number Me = ? Temperature Te = ? Mass flow rate = ?

pb is known as the back pressure


(i) For pb = 600 kPa, there will be no flow since p0 = 600 kPa as well

(ii) For pb = 500 kPa, assume pe is the same as pb.

Using (2.5), we get4.1

4.01

00 600

500) K400(

p

pTT e

e = 379.7 K

eeee RT

pMART

AMpm

= (100 x10-6 m2) (0.517) (500,000 Pa)(8314/29)(379.7) J/kg

0.5

( ) 1.4

= 0.0927 kg/s

= 0.517Using (2.7), we get4.0

4.1

2

2

4.01

500

600

eM

2.0

1)5/6( 4.1/4.0 eM

Using (2.9), we get


Results summarized:Back pressure,

pb (in kPa)Exit pressure,

pe (in kPa)Exit Mach

number, Me

Exit temperature, Te (in K)

Mass flow rate (in kg/s)

600 600 0 400 0

500 500 0.517 379.7 0.0927

400 400 0.784 356.2 0.1161

300 300 1.046 328.1 0.1211

200 200 1.358 292.2 0.1110

T0 = 400 K

p0 = 600 kPa

Air

Ae = 100 mm2

pb kPa(given)

Is there a problem with the above results?





number, Me



600 600 0 400 0

500 500 0.517 379.7 0.0927

400 400 0.784 356.2 0.1161

300 300 1.046 > 1 328.1 0.1211

200 200 1.358 > 1 292.2 0.1110

Yes, there is. Supersonic Mach numbers have been reached from stagnation condition in a converging duct?

It can not happen!

T0 = 400 K

p0 = 600 kPa

Air

Ae = 100 mm2

pb kPa(given)


Maximum Mach number at the exit of a converging duct must be Me = 1.

Corresponding pe (denoted by p*) could be calculated using (2.7) as follows:

4.0

4.1

2

2

4.01

500

600

eM = 317 kPap* = 600

Using (2.5), we get = 333.3 K4.1

4.01

00 600

317) K400(

**

p

pTT

Using (2.9), we get the mass flow rate as 0.1213 kg/s.





number, Me



600 600 0 400 0

500 500 0.517 379.7 0.0927

400 400 0.784 356.2 0.1161

300 317 1 333.3 0.1213

200 317 1 333.3 0.1213

Flow chocks at the throat of the converging duct at an exit pressure of 317 kPa at a maximum flow rate of 0.1213 kg/s?

T0 = 400 K

p0 = 600 kPa

Air

Ae = 100 mm2

pb kPa(given)


T0 = 400 K

p0 = 600 kPa

Air

Ae = 100 mm2

pe kPa(given)

Pressure at the throat cannot be less than the limiting pressure (317 kPa in this case) even if we keep a lower pressure in the second chamber (known as the back pressure).

Mass flow rate cannot be increased above the maximum mass flow rate ( 0.1213 kg/s in this case) even if we increase the driving force by decreasing the back pressure in the second chamber.


Air at 900 kPa and 400 K enters a converging nozzle with a negligible velocity. The throat area of the nozzle is 10 cm2. Assuming isentropic flow, calculate and plot the exit pressure, the exit Mach number, the exit velocity,

and the mass flow rate versus the back pressure pb for 900 ≤ pb ≤ 100 kPa.



100

200

300

400

500

600

700

800

900

100 200 300 400 500 600 700 800 900

Back pressure (in kPa)

Exi

t pre

ssur

e (in

kP

a)

Limiting pressure = 475.45 kPa

Results:


0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1

100 200 300 400 500 600 700 800 900


Exi

t Mac

h nu

mbe

r

Sonic condition M = 1

Results (continued):


0

50

100

150

200

250

300

350

400

100 200 300 400 500 600 700 800 900


Exi

t vel

ocity

(in

m/s

)

Maximum velocity 365.77 m/s



0

2

4

6

8

10

12

14

16

18

20

100 200 300 400 500 600 700 800 900


Mas

s flo

w r

ate

(in k

g/s)

Maximum mass flow rate 18.1981 kg/s



Consider a converging-diverging duct with a circular cross-section for a mass flow rate of 3 kg/s of air and inlet stagnation conditions of 1400 kPa and 200oC. Assume that the flow is isentropic and the exit pressure is 100 kPa. Plot the pressure and temperature of the air flow along the duct as a function of M.Plot also the diameter of the duct as a function of M.


p0 = 1400 kPaT0 = (273+200) KM0 ≈ 0

m = 3 kg/s

pe = 100 kPa


(2.6)20

2

11 M

T

T

120

2

11

M

p

p(2.7)

p0 and T0 are known.

Use the equations below to calculate p and T for different values of M.

Use (2.9) to calculate the exit are of the duct as follows:


0

200

400

600

800

1000

1200

1400

1600

0 0.5 1 1.5 2 2.5 3

Mach number along the duct

Pressure along the duct (in kPa)

Temperature along the duct (in K)

Diameter of the duct (in mm)


0

50

100

150

200

250

0 0.5 1 1.5 2 2.5 3


Diameter of the duct (in mm)

Enlarged version


-250

-200

-150

-100

-50

0

50

100

150

200

250

0 0.5 1 1.5 2 2.5 3


Shape of the converging-diverging duct


Air at approximately zero velocity enters a converging-diverging duct at a stagnation pressure and a stagnation temperature of 1000 kPa and 480 K, respectively. Throat area of the duct is 0.002 m2. The flow inside the duct is isentropic, and the exit pressure is 31.7 kPa. For air, γ = 1.4 and R = 287 J/kg. Determine (i) the exit Mach number, (ii) the exit temperature, (iii) the exit area of the duct, and (iii) the mass flow rate through the duct.


p0 = 1000 kPaT0 = 480 KM0 ≈ 0

Athroat = 0.002 m2

pe = 31.7 kPaMe = ?Ae = ?


(i) Rearrange (2.7) to determine Me as follows:

11

21

0

ee p

pM

2.0

17.311000 4.1

4.0

= 2.9

(ii) Rearrange (2.5) to determine Te as follows:

4.1

4.01

00 1000

7.31) K480(

p

pTT e

e = 179 K


(iii) Using (2.10), the exit area of the duct can be calculated as follows:

)1(2

1

200

)1(2

1

200

2

11

1

2

11

1

e

ee

t

tt

MRTpMA

MRTpMAm

Since the exit Mach number is 2.9, the flow is supersonic in the diverging section. It is not possible unless sonic conditions are achieved at the throat. Therefore Mt = 1 in the above expression.

Therefore, we get

= 0.0077 m2

3

2

3

2.01

1

2.1

1

eeet M

MAA

32

2.1

2.01

e

e

te

M

M

AA

Since Me = 2.9 and At = 0.002 m2, the above expression gives


(iv) Use (2.9) to get the mass flow rate through the duct as follows:

= 3.695 kg/s

J/kg) 791x287(

4.1)kPa 1000x7.31)(9.2)(m 0077.0( 2

eeee RT

pMAm

CP502 Advanced Fluid Mechanics

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