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ME 1303 GAS DYNAMICS

AND JET PROPULSION

Presented by

G.Kumaresan

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Mass Flow rate

Mass flow rate

In terms of Pressure ratio In terms of Area ratio

In terms of Mach number

Non dimensional parameter (Numerical

Value)

Mass flow rate in terms of Mach Number

From m.f.r in terms of Area ratio

A

A

RT

P

2

1

1

2

A

m

0

o

1

(1)

----------------------------------------------------------------------------------------

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Cont..We already derived that area ratio(A/A*) is a unique

function of Mach number in an isentropic flow 121

2M

1

1

1

2

M

1

A

A

(2)

Substitute eqn. 2 in 1

1212

121

o

o

M1

1

1

2

M

1

2R

AP

Tm

This on rearranging gives the value of the m.f.

parameter in terms of the local Mach number.

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Cont..

This relation can also be obtainedfrom m.f.r in terms of pressure ratio

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Cont..Substitute Po/P relation in the above

expression 2

1

o

1

2o M2

11

P

PM

2

11

P

P

After substituting Po/P, the equationmodified as

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Cont..

1212o

o

M2

11

MRAPTm

(3)

For maximum mass flow conditions,

A=A* and Mach no.=1

Substituting these values in eqn. 3 121

o

omax

1

2R

PA

Tm

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Numerical value of the non-dimensional max.mass flow parameter

(4)

The above expression is already obtainedfor mass flow parameter in terms of Pr.

ratio and Area ratio

Eqn. 4 show that for a given gas (i.e. R and ) massflow per unit area at a given mach number is directly

proportional to the initial stagnation pressure and

inversely proportional to the square root of the

121

o

omax

1

2R

PA

Tm

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Cont..

The value of the maximum mass flow

parameter is fixed by the properties of the gas.Therefore for the same gas various

combinations ofPo and To can be used to give

maximum mass flow rate from the following

expression.

For air, R= 287 J/kg K and =1.4. Nowsubstituting this value in the above equation

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Cont..

0404.0PA

Tm

o

omax

Where max 02 2

/ , - K

A - m , - N/mo

m kg s T

P

Equation (5) is known as Fliegner'sFormula on the name of one of the first

engineers who observed experimentally

the choking phenomenon

(5)

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Problem

Air is allowed to flow from a reservoir with

temperature of 21oC and with pressure of

5 MPa through a tube. It was measured

that air mass flow rate is 1 kg/sec. At some

point on the tube static pressure was

measured to be 3 MPa. Assume thatprocess is isentropic and neglect the

velocity at the reservoir, calculate the

Mach number, velocity and the crosssection area at that point where the static

pressure was measured. Assume that the

ratio of specific heat is 1.4

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Problem

Solution:

The stagnation conditions at the reservoirwill be maintained through out tube

because the process is isentropic.

Hence the stagnation temperature and

pressure both of them constant (the

condition at the reservoir).

For the point where the static pressure is

known, the Mach number can be

calculated utilizing the pressure ratio.

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Cont..

With known Mach number, the temperature,

and velocity can be calculated.

Finally, the cross section can be calculated

with all these information.

From the Given data

6.05

3

P

p

o

At 0.6 refer isentropic gas table, note down

all the values

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Cont..

M M* T/T0 P/P0 A/A* F/F*

0.88 0.897 0.869 0.604 1.013 1.006 0.612

0P

P

A

A

Velocity at measured at that point

282m/sv

5.2552874.188.0RTMv;a

vM

From the above table

K5.255294869.0T;869.0T

T

o

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Cont..

25

36

m106.828241

1

v

mAAvm

m/kg415.255287

103

RT

pRTp

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Problem2(without using Gas Table)

A reservoir whose temperature can be varied in a

wide range of temperature receives air at aconstant pressure of 1.5 bar. The air is expanded

isentropically in a nozzle to an exit pressure of

1.015 bar. Determine the values of the

temperature to be maintained in the reservoir toproduce the following velocities at the nozzle exit:

(a)100 m/s (b) 250 m/s

What are the values of the mach numbers in the

two cases

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Problem2(without using Gas Table)

Solution:

From Energy Eqn.

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Cont..

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Cont..

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Cont..

Mach number is same in the two casesbecause the pressure ratio is same