me6604 gas dynamics and jet propulsion l t p c 3 0 0 3 ... · me6604 gas dynamics and jet...

REGULATION : 2013 ACADEMIC YEAR : 2018-19

JIT-JEPPIAAR/MECH/Dr.S.BOOPATHI/IIIrd Yr/SEM 05/ME6604/GAS DYNAMICS AND JET PROPULSION/UNIT 1-5/QB+Keys/Ver1.0

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ME6604 GAS DYNAMICS AND JET PROPULSION L T P C

3 0 0 3

OBJECTIVES:

• To understand the basic difference between incompressible and compressible flow.

• To understand the phenomenon of shock waves and its effect on flow. To gain some basic knowledge

about jet propulsion and Rocket Propulsion.

(Use of Standard Gas Tables permitted)

UNIT I BASIC CONCEPTS AND ISENTROPIC FLOWS 6

Energy and momentum equations of compressible fluid flows – Stagnation states, Mach waves and Mach

cone – Effect of Mach number on compressibility – Isentropic flow through variable ducts – Nozzle and

Diffusers

UNIT II FLOW THROUGH DUCTS 9

Flows through constant area ducts with heat transfer (Rayleigh flow) and Friction (Fanno flow) – variation

of flow properties.

UNIT III NORMAL AND OBLIQUE SHOCKS 10

Governing equations – Variation of flow parameters across the normal and oblique shocks – Prandtl –

Meyer relations – Applications.

UNIT IV JET PROPULSION 10

Theory of jet propulsion – Thrust equation – Thrust power and propulsive efficiency – Operating

principle, cycle analysis and use of stagnation state performance of ram jet, turbojet, turbofan and turbo

prop engines.

UNIT V SPACE PROPULSION 10

Types of rocket engines – Propellants-feeding systems – Ignition and combustion – Theory of rocket

propulsion – Performance study – Staging – Terminal and characteristic velocity – Applications – space

flights.

TOTAL: 45 PERIODS

OUTCOMES:

Upon completion of this course, the students can able to successfully apply gas dynamics principles in

the Jet and Space Propulsion



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TEXT BOOKS:

1. Anderson, J.D., "Modern Compressible flow", 3rd Edition, McGraw Hill, 2003.

2. Yahya, S.M. "Fundamentals of Compressible Flow", New Age International (P) Limited, New Delhi,

1996.

REFERENCES:

1. Hill. P. and C. Peterson, "Mechanics and Thermodynamics of Propulsion", Addison – Wesley

Publishing company, 1992.

2. Zucrow. N.J., "Aircraft and Missile Propulsion", Vol.1 & II, John Wiley, 1975.

3. Zucrow. N.J., "Principles of Jet Propulsion and Gas Turbines", John Wiley, New York, 1970.

4. Sutton. G.P., "Rocket Propulsion Elements", John wiley, New York,1986,.

5. Shapiro. A.H.," Dynamics and Thermodynamics of Compressible fluid Flow", John wiley, New York,

1953.

6. Ganesan. V., "Gas Turbines", Tata McGraw Hill Publishing Co., New Delhi, 1999.

7. Somasundaram. PR.S.L., "Gas Dynamics and Jet Propulsions", New Age International Publishers,

1996.

8. Babu. V., "Fundamentals of Gas Dynamics", ANE Books India, 2008.

9. Cohen. H., G.E.C. Rogers and Saravanamutto, "Gas Turbine Theory", Longman Group Ltd., 1980.



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Subject Code : ME6604 Year/Semester : III/ 06

Subject Name : Gas Dynamics and Jet Propulsion Subject Handler : Dr. S. Boopathi

UNIT I BASIC CONCEPTS AND ISENTROPIC FLOWS

Energy and momentum equations of compressible fluid flows – Stagnation states, Mach waves and Mach

cone – Effect of Mach number on compressibility – Isentropic flow through variable ducts – Nozzle and

Diffusers

PART * A

Q.No. Questions

1. Write any two applications of nozzle and diffuser. (Nov/Dec 2015) BTL1

Aircraft (jet engines) and rocket applications.

2

"Higher the velocity of supersonic flow, smaller the angle of Mach cone" comment on the

validity of this statement. (Nov/Dec 2015) BTL4

Sin of Mach angle is equal to the reciprocal of Mach number, therefore the higher the velocity of

supersonic flow, smaller the angle of Mach cone.

3 What is meant by gas dynamics? (April/May 2015) BTL1

G s dynamics deals with the steady of motion of gasses and its effects.

4

Define Mach number. (April/May 2015) BTL1

The Mach number is an index of the ratio between the inertia force and elastic force or It is also

defined as the ratio of the fluid velocity (C or U) to the velocity of sound (a).

5

Find the maximum possible velocity in a medium of air when static temperature is 200ºC and

velocity is 120 m/s. (Nov/Dec 2014) BTL3

C max = 982.16 m/s.

6

When does the maximum mass flow occur for an isentropic flow with variable area and what

type of nozzle used for sonic flow and supersonic flow? (Nov/Dec 2014) BTL4

Mass flow rate will be maximum at throat section where Mach number is one. Constant area duct

nozzle is used for supersonic flow.

7

Distinguish between nozzle and diffuser. (May /June 2014) BTL4

Nozzle: varying cross sectional area through which pressure energy decreases and kinetic energy

increases. Diffuser: varying cross sectional area through which pressure energy increases and

kinetic energy decreases.

8 When does maximum flow occur for an isentropic flow with variable area duct? (May /June

2014) BTL3



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When the pack pressure ratio P/Po = (2/ν+1) power of (ν/ν-1) , mass flow rate will be maximum,

M=1 at throat section.

9

What are the basic difference between compressible and incompressible flows? (May /June

2013) BTL1

Compressible flow Density of the fluid changes from point to point, Incompressible density remains

constant.

10

Name the four reference velocities that are used in expressing the third velocities in non-

dimensional form. (May /June 2013) BTL1

Local velocity of sound (a), Stagnation velocity (ao), Maximum velocity of fluid Cmax, Critical

velocity of sound a* or c*.

11

Define Isentropic process. BTL1

An isentropic process is one in which there is no change in entropy. Such a process is a reversible

adiabatic process, for this process in the absence of shaft work, there is no change in eh stagnation

pressure and temperature.

P V = Constant

12

Define stagnation state. BTL1

The state of a fluid attained by isentropically decelerating it to zero velocity at zero elevation is

referred to as the stagnation state. Stagnation enthalpy h0 = h1 + ½ C2

13

Define stagnation temperature. BTL1

Stagnation temperature is the temperature of the gas when it is adiabatically decelerated to zero

velocity at zero elevation.

14

Define stagnation velocity of sound. BTL1

Stagnation velocity of sound is defined as the velocity of sound at stagnation temperature.

a0 = (r – 1 ) h0 = r R T0

15 Define Crocco Number. BTL1

It is defined as the ratio of fluid velocity to the maximum fluid velocity.

16

What is the effect of Mach number of compressibility? BTL1

In a compressible flow the value of pressure co-efficient deviates from unity and this deviation

increases with Mach number of the flow.

17 Relate the classification of flow based on Mach number value. BTL2

The classification of flow based on Mach number values are,



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M < < 1 Incompressible, M < 1 Sub Sonic , M 1 Transonic, M = 1 Sonic

M > 1 Supersonic, M > > 1 Hypersonic

18

Define Critical velocity of sound. BTL1

Critical velocity of sound is the velocity of fluid at unity Mach number.

C* = a* = r Rt*

19

Define Mach cone. BTL1

Mach cone is defined as the conical boundary obtained by drawing the tangent to the circles of sonic

wave fronts such that C > a is the velocity of fluid which is greater than velocity of sound.

20

Define Mach angle. BTL1

The angle between the Mach line and the direction of motion of the body is called Mach angle.

Sin = a/c = 1/M, where - Mach angle

Part B

1

An aircraft flies at a velocity of 700 kmph in an atmosphere where the pressure is 75 kPa and

temperature is 5 ºC. Calculate the Mach number and stagnation properties. (April/May 2015)

(13 M) BTL4

Answer: Page 1.76- Dr.S.Senthil

Air velocity C= 700 x103/3600 =194.44 m/s, P=75 x103 N/m2, T=278 K (4M)

a=√𝛾𝑅𝑇=334.22 m/s, (2M)

M=C/a = 0.58 (2M)

Refer isentropic flow table for 𝛾 = 1.4 𝑎𝑛𝑑 𝑀 = 0.58

T0=296.69 K, Stagnation pressure P0=0.942 x105 N/m2 (2M)

Stagnation density = 1.106 kg/m3 (3M)

2

The pressure, temperature and velocity of air at the entry of a diffuser are 0.7 bar, 345 K and

190 m/s respectively. The entry diameter of a diffuser is 15 cm and exit diameter is 35 cm.

Determine the following: (i) Exit Pressure (ii) Exit velocity and (iii) Force exerted the diffuser

walls. Assuming isentropic flow and take ᵞ = 1.4, Cp=1005 J/kgK. (Apr 2018) (13 M) BTL5


Velocity of sound at inlet, a1=372.32 m/s

Mach number at inlet, M1= 0.51 (2M)



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From isentropic flow table for ᵞ=1.4 and M1=0.51

T01=362.78 K=T02, P01=0.836 bar=P02, A1*=0.01337=A2*, A2= 0.096 m2, (4M)

A2/A2*=7.180 convergent type diffuser

From Isentropic flow table A2/A2*=7.180 and ᵞ=1.4

P2=0.832 bar, T2= 362.05 K, C2= 30.51 m/s, P*=P1*=P2*=0.441 bar (4M)

Force exerted on the diffuser walls is equal to the thrust of the flow τ= F2-F1= 6457 N (3M)

3

An air jet at 300 K has sonic velocity. Determine the following, (i) Velocity of sound at 300 K,

(ii) Velocity of sound at stagnation condition, (iii) Maximum jet velocity, (iv) Stagnation

enthalpy and (v) Crocco number Take ᵞ = 1.4 and R = 287 J/kgK. (May/June 2016) (13 M)

BTL3


M=1, a=√𝛾𝑅𝑇=347.18 m/s, (3M)

Stagnation temperature T0= 1 +𝛾−1

2𝑀2 , T0=360 K, (3M)

a0=√𝛾𝑅𝑇0=380.32 m/s, (3M)

From stagnation enthalpy equation, 𝐶𝑚𝑎𝑥 = 850.42 𝑚/𝑠 ,

ℎ0 = 361.6 𝑥 103 𝐽

𝑘𝑔 (2M)

Crocco number 𝐶𝑟 =𝐶

𝐶𝑚𝑎𝑥= 0.408 (2M)

4

Air enters the nozzle from a large reservoir at 0.7 bar and 300 K. The cross sectional area of

the throat is 1200 cm2 and test section Mach number is 1.98. Calculate the following for one

dimensional isentropic flow (i) Pressure, temperature and velocity at the throat, (ii) Pressure,

temperature and velocity at test section, (iii) Mass flow rate, (iv) Power required to drive the

compressor and (v) Area of cross section at test section. (13 M) BTL5


At throat (*) section, M=1, Refer isentropic flow table 𝛾 = 1.4 𝑎𝑛𝑑 𝑀 = 1 (2M)

T*=250.2 K, P*=0.369 bar, c*=a* = √𝛾𝑅𝑇∗=317.06 m/s, (2M)

From Isentropic flow table 𝛾 = 1.4 𝑎𝑛𝑑 𝑀𝑡 = 1.98 (3M)

Tt=168.3 K, Pt=0.0924 bar, At=0.199 m2, Ct=𝑀𝑡 ∗ √𝛾𝑅𝑇𝑡= 514.88 m/s, (3M)

Mass flow rate �̇� = 𝜌𝐴𝐶𝜌 ∗A*C* =19.55 kg/s,Power required P=�̇�𝑃(𝑇0 − 𝑇𝑡)=2587 x103 W. (3M)



5-7

5

An air jet (ᵞ=1.4, R=287 J/kgK) at 400 K has sonic velocity. Determine (i) Velocity of sound at

400 K (ii) Velocity of sound at the stagnation conditions (iii) Maximum velocity of the jet (iv)

Stagnation enthalpy and (v) Crocco number. (May/June 2016) (13 M) BTL4


M=1, a=√𝛾𝑅𝑇=400.8 m/s, (3M)

Stagnation temperature T0= 1 +𝛾−1

2𝑀2 , T0=480 K, (3M)

a0=√𝛾𝑅𝑇0=439.16 m/s, From stagnation enthalpy equation,

𝐶𝑚𝑎𝑥 = 981.9 𝑚/𝑠 , (3M)

ℎ0 = 482.06 𝑥 103 𝐽

𝑘𝑔 (2M)

Crocco number 𝐶𝑟 =𝐶

𝐶𝑚𝑎𝑥= 0.408 (2M)

6

Air is discharged from a reservoir at P0=6.91 bar and T0=325ºC through a nozzle to an exit

pressure of 0.98 bar. If the flow rate is 3600 kg/hr. Determine for isentropic flow (i) Area,

pressure and velocity at throat, (ii) Area and Mach number at exit, and (iii) Maximum possible

velocity. (May-June 2012) (Nov-Dec 2009) (Nov-Dec 2003) (13 M) BTL5


At throat (*) section, M=1, Refer isentropic flow table 𝛾 = 1.4 𝑎𝑛𝑑 𝑀 = 1 (1M)

T*=498.73 K, P*=3.65 bar, c*=a* = √𝛾𝑅𝑇∗=447.65 m/s, (2M)

Mass flow rate �̇� = 𝜌𝐴𝐶 = 𝜌 ∗A*C* A*=8.76 x 10-4 m2 (4M)

P2/P02 = 0.142, Refer Isentropic flow table 𝛾 = 1.4

M2=1.93, A2=13.9 x 10-4 m2 (4M)

Maximum velocity Cmax = √𝛾+1

𝛾−1= 1096.51 𝑚/𝑠 (2M)

7

A certain quantity of air at a pressure of 3.344 bar and temperature 627ºC is flowing through

a C-D nozzle. The exit pressure is 1.05 bar. Determine the temperature, velocity and density of

air at exit. Also determine the pressure, temperature, density and velocity of air at exit if the

divergent portion is to act as diffuser. Assume isentropic flow in both cases. (13 M) BTL4


Case (i) P2/P02=0.3139, For Isentropic flow table 𝛾 = 1.4,

M2 = 1.40, T2/T02=0.718, A2/A*=1.115, T2=646.2 K,



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M2 =C2/a2 C2=713.37 m/s, Density at exit,𝜌 =0.566 kg/m3 (8M)

Case (ii) A2/A*=1.115≈ 1.118 𝑎𝑛𝑑 𝛾 = 1.4

T2=825.3 K, P2=2.47 x 105 N/m2, C2=385.82 m/s, 𝜌2=1.143 kg/m3 (5M)

8

Oxygen at 200 kPa flows at a velocity of 50 m/s. Find the Mach number at a point where its

density is 2.9 kg/m3. Molecular weight of oxygen is 32 and 𝜸 = 𝟏. 𝟒. (13 M) BTL4

Answer: Refer class notes

Gas constant R = universal gas constant / Molecular weight = 8314/32 =259.81 J/kg K (3M)

Density = P/RT T= 265.44 K (3M)

a=√𝛾𝑅𝑇=310.72 m/s, (2M)

Mach number M= C/a = 0.16 (5M)

9

An air jet (𝜸=1.4, R=287 J/kgK) enters a straight axis symmetric duct at 300 K, 3.45 bar and

150 m/s and leaves it at 277 K, 2.058 bar and 260 m/s. The area of cross section at entry is 500

cm2, assuming adiabatic flow determine (i) Stagnation temperature (ii) maximum velocity (iii)

Mass flow rate (iv) Area of cross section at exit. (13 M) BTL4


Sound of velocity at inlet 𝑎1=√𝛾𝑅𝑇1=347.19 m/s,

Inlet Mach number M1=0.432 (1M)

Refer Isentropic flow table for 𝛾=1.3 and M1=0.43

T01=311.12 K=T01=T02= T0 , (2M)

From stagnation enthalpy equation, ℎ0 =𝑎2

𝛾−1+

1

2𝐶2 =

1

2𝐶𝑚𝑎𝑥

2 =𝑎0

2

𝛾−1 𝐶𝑚𝑎𝑥 = 790.69

𝑚

𝑠 (4M)

From Mass flow rate �̇� = 𝜌𝐴𝐶 = 𝜌1𝐴1𝐶1=30.05 kg/s, (3M)

�̇� = 𝜌2𝐴2𝐶2=30.05 kg/s 𝐴2 = 0.0446 𝑚2 (3M)

10

A supersonic nozzle air expands from P0=24 bar and T0=1000 K to an exit pressure of 4.3 bar.

If the exit area of the nozzle is 110 cm2, Calculate the following (i) Throat area (ii) Pressure and

temperature at the throat (iii) Temperature at exit (iv) Mass flow rate (v) Exit velocity as

fraction of the maximum attainable velocity. (13 M) BTL5

Answer: Page 2.166- Dr.S.Senthil 2018

𝑃2/P02 =0.179, (2M)

Refer Isentropic flow table, 𝛾=1.4 and 𝑃2/P02 =0.179,



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M2=1.78, T2=612 K, A*=7.75 x10-3 m2 (4M)

At throat section (*) M=1, Refer Isentropic flow table 𝛾=1.4 and M=1

T*=834 K, P* = 12.6 x 105 N/m2 , (2M)

Mass flow rate �̇� = 𝜌 ∗A*C* �̇� =23.5 kg/s (2M)

𝐶2

𝐶𝑚𝑎𝑥=

𝑀2𝑎2

√2𝐶𝑝𝑇0= 0.622 (3M)

11

A conical diffuser has an inlet area of 0.11 m2 and an exit area of 0.44 m2. Air enters the diffuser

with a static pressure of 0.18 MPa, static temperature of 37C and velocity of 267 m/s. Calculate

(i) The mass flow rate of air through the diffuser (ii) The Mach number, static temperature and

static pressure of the air leaving diffuser and (iii) The net thrust acting upon the diffuser due to

diffusion. (13 M) BTL5


From Mass flow rate �̇� = 𝜌𝐴𝐶 = 𝜌1𝐴1𝐶1=59.42 kg/s,

Mach number M1= C1/a1 = 0.75

Refer Isentropic flow table for 𝛾=1.4 and M1=0.75

T01=344.82 K=T02, P01=2.61 bar=P02, A1*=0.1035=A2*, A2/A*= 4.25 m2, (2M)

In this problem A2>A1, So, this divergent type diffuser. For divergent diffuser, Mach number value

is less than unity. (2M)

Refer Isentropic flow table for A2/A*= 4.25 ≈ 4.182 and 𝛾=1.4

M2=0.14, P2=2.57 x 105 N/m2, T2=343.44 K, (4M)


P* = P1* = P2* = 1.51 x 105 N/m2 (2M)

Force exerted on the diffuser walls is equal to the thrust of the flow τ= F2-F1= 86.72 x 103 N. (3M)

Part * C

1

(i) Discuss the change of Mach number in CD nozzle under various back pressure. (8)

(ii) An airplane is travelling while you are observing from the ground. How will you know

whether it is subsonic or supersonic? Explain? (7).



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The pressure distribution in the nozzle

A plot of the pressure distribution along the

nozzle (figure 1) provides a good way of

summarizing its behavior. To understand how

the pressure behaves you have to remember only

a few basic rules

• When the flow accelerates (sub or

supersonically) the pressure drops

• The pressure rises instantaneously across

a shock

• The pressure throughout the jet is always

the same as the ambient (i.e. the back pressure)

unless the jet is supersonic and there are shocks

or expansion waves in the jet to produce pressure

differences. (8M)

Figure- Mach number in CD nozzle under

various back pressure

• The pressure falls across an expansion wave.

• M<1 Subsonic, M>1 Supersonic (7M)

2

Air (Cp=1.05 kJ/kgK, 𝜸 = 𝟏. 𝟑𝟖) at P1= 3x 105 N/m2 and T1=500 K flows with a velocity of 200

m/s in a 200 m/s in a 30 cm diameter duct available. Calculate (i) mass flow rate (ii) stagnation

temperature (iii) Mach number (iv) stagnation pressure values assuming the flow is

compressible and incompressible respectively. (15M) BTL5


R= Cp-Cv = Cp-Cp/𝛾 = 289.13 J/kgK (3M)

�̇� = 𝜌𝐴𝐶 = 𝜌1𝐴1𝐶1=𝜌2𝐴2𝐶2=29.33 kg/s, (3M)

Inlet Mach number M1=C1/a1 = 0.44 (2M)

Refer Isentropic flow table for 𝛾=1.38≈ 1.4 and M1=0.44

T01=519.21 K=T01=T02= T0 , P01=3.42 x 105 N/m2= P02= P0 (4M)

Incompressible flow, from Bernoulli equation, stagnation pressure equation P0=P+1

2𝜌𝐶2

P01=𝑃1+1

2𝜌𝐶1

2= P02= P0 =3.41x105 N/m2 (3M)

3

Air flows in a duct with a velocity of 215 m/s. The temperature of air measured at a point along

the duct is 30ºC and the air pressure is 5 bar. Determine (i) stagnation pressure (ii) Mach

number at that point. (7 M) BTL5 Answer: Page 1.105- Dr.S.Senthil 2018,



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(ii) Carbon dioxide expands isentropically through a nozzle from a pressure of 3.2 bar to 1 bar.

If the initial temperature is 475 K, determine the final temperature, the enthalpy drop and the

change in internal energy. (8 M) BTL5 Answer: Page 1.106- Dr.S.Senthil

(i) a=√𝛾𝑅𝑇=348.92 m/s,

Mach number M= C/a = 0.16

Refer Isentropic flow table for 𝛾=1.4 and M=0.616≈ 0.62

P/P0 =0.772, P0=6.47 x 105 N/m2 (7M)

(ii) Refer gas tables for gas dynamic properties of carbon dioxide at various temperature.

At T1=475 K≈ 473 𝐾

Cp=993 /kgK, Cv=804 J/kgK, 𝛾 = 1.237, R=189 J/kgK

𝑇2

𝑇1= (

𝑃2

𝑃1)

𝛾−1𝛾

==> 𝑇2 = 380.11 𝐾

∆ℎ = 𝐶𝑝∆𝑇 = 𝐶𝑝(𝑇2 − 𝑇1) = 94.2 𝑥 103𝐽

𝑘𝑔

Change in internal energy

∆𝑢 = 𝐶𝑣∆𝑇 = 𝐶𝑣(𝑇1 − 𝑇2) = 79.29𝑥 103 𝐽

𝑘𝑔 (8M)

4

Air is discharged from a receiver at P0=6.91 bar and T0=325ºC through a nozzle to a pressure

of 0.98 bar. If the flow rate is 3600 kg/h. Determine for isentropic flow (i) Area, Pressure and

velocity at throat (ii) Area and Mach number at exit and (iii) Maximum possible velocity. (15

M) BTL5



T*=498.73 K, P* = 3.65 x 105 N/m2 , C*=√𝛾𝑅𝑇∗=447.65 m/s (5M)

Mass flow rate �̇� = 𝜌 ∗A*C*, A*=8.76 x10-4 m2, (2M)

From given data P2/P0=0.142,

Refer Isentropic flow table 𝛾=1.4 and P2/P0=0.142

M2=1.93, A2=13.9 x10-4 m2, (5M)

𝐶𝑚𝑎𝑥

𝐶∗ = √𝛾+1

𝛾−1 𝐶𝑚𝑎𝑥 = 1096.51 𝑚/𝑠 (3M)



5-12



UNIT II FLOW THROUGH DUCTS

Flows through constant area ducts with heat transfer (Rayleigh flow) and Friction (Fanno flow) – variation

of flow properties.

PART * A

Q.No. Questions

1.

What do you mean by friction chocking? (Nov/Dec 2015) BTL1

Flow with friction in the Fanno flow, the restriction flow occurs due to chocking, is called friction

chocking.

2

Complete the following table with increases, decreases, remains constant for a flow through a

constant area duct with heat transfer:

Static Temperature-

Subsonic flow: Heating------- Cooling---------

Supersonic flow: Heating-------------- Cooling-------------. (Nov/Dec 2015) BTL4

Static Temperature- Subsonic flow : Heating:-remains constant-Cooling: Decreases

Supersonic flow: Heating-Increases- Cooling-Decreases.

3

What is meant by stagnation pressure? (April/May 2015) BTL1

Stagnation pressure is the pressure of the gas when it is isentropically decelerate to zero velocity at

zero elevation. Po/P = (To/T) power of (ν/ν-1).

4 What is Fanno flow? (April/May 2015) BTL1

Flow in a constant area duct with friction and without heat transfer is known as Fanno flow.

5

Air at Po=10 bar and To=400K is supplied to a 50 mm diameter pipe, the friction factor for

the pipe surface is 0.002. If the Mach number changes from 3 at entry to 1 at the exit.

Determine the length of the pipe. (Nov/Dec 2014) BTL5

4fL/D = 0.522 (M1=3, ν=1.4) , 4fL/D = 0 (M2=1, ν=1.4) , L= 3.26m

6

What are the assumptions made for Fanno flow? And give the two practical examples where

the Fanno flow occurs. (Nov/Dec 2014) BTL1

One dimensional, steady flow, flow takes place in constant sectional area. Flow in air breathing

engine, flow of fluids in long pipes

7 Give assumption made on Rayleigh flow. (May /June 2014) BTL3



5-13

Fluid is perfect gas, flow is one dimensional steady and duct area is constant, effect of friction is

negligible, No external work and change due to elevation.

8

Define critical condition in Fanno flow. (May /June 2014) BTL1

Due to friction in subsonic or supersonic flow in a constant area duct, flow will reach the critical

condition (*) where the Mach number M=1 and the mass flow reach the maximum value. If the

length of pipe is increased further, will not give any effect on mass flow rate.

9 What is impulse function and give its uses? (May /June 2013) BTL1

The sum of force (pA) and Impulse force (ρAc2) F=pA+ρAc2.

10

Give the expression for T0/T and T*/T for isentropic flow through variable area in terms of

Mach number. (May /June 2013) BTL1

To/T = 1+ (ν-1)/2 * M2, T*/T=2/ν+1 + (ν-1)/(ν+1)M2.

11

What is meant by isentropic flow with variable area? BTL2

A steady one dimensional isentropic flow in a variable area passages is called “variable area flow”.

The heat transfer is negligible and there are no other irreversibility due to fluid friction, etc.

12 What is the effect of pressure in diffusers and nozzle? BTL3

Diffusers => increase in pressure Nozzles => decrease in pressure

13

Describe the function of a nozzle and a diffuser. BTL2

The main function a nozzle is to accelerate the flow i.e. increase in velocity and decrease in pressure

of the flowing fluid. Whereas, the function of a diffuser is to decelerate the flow i.e, increase in

pressure and decrease in velocity of the flowing fluid.

14 Is the flow through a nozzle an irreversible process? BTL4

Yes. In actual practice, all the processes are irreversible due to friction.

15

When will the convergent-divergent nozzle have the maximum discharge rate? BTL3

From continuity equation

m Ac = Constant

mAc

A = discharge rate

The discharge rate will be maximum only at the throat section. Where Mach number M=1.

16

Write the Fliegner’s formula. BTL1

1

2 10Max

0

0max

0

Tm 2For air 1.4 and R 287 J/Kg K SI units

A P R

Tm 0.0404 Fliegner 's formula

A P



5-14

17

What is impulse function and give its uses? BTL2

Impulse function is defined as the sum of pressure force and inertia force. Impulse function

F = Pressure force A + inertia force Ac2.

Since the unit of both the quantities are same as unit of force, it is very convenient for solving

jet propulsion problems. The thrust exerted by the following fluid between two sections can be

obtained by using change in impulse function.

18

What is chocked flow? State the necessary conditions for this flow to occur in a nozzle. BTL1

When the back pressure is reduced in a nozzle, the mass flow rate will increase. The maximum mass

flow conditions are reached when the back pressure is equal to the critical pressure. When the back

pressure is reduced further, the mass flow rate will not change and is constant. The condition of flow

is called “chocked flow”. The necessary conditions for this flow to occur in a nozzle is : the nozzle

exit pressure ration must be equal to the critical pressure ration where the Mach number M=1.

19 Give the important uses of nozzle and diffuser. BTL3

1. Used in all types of jet propulsion engines, turbo-jet, turboprop, ram jet engines, etc.

2. Used in rocket propulsion systems.

20

Give the important difference between nozzle and venturi. BTL4

NOZZLE VENTURI

1. The flow is accelerated continuously i.e.,

Mach number and velocity increases

continuously.

1. The flow is accelerated up to M=1 and then

Mach number is decreased.

2. Used to increase velocity and Mach

number.

2. Used for flow measurement. (discharge)

3. Generally convergent portion is short. 3. Convergent and divergent portions are equal.

21

What is the pressure ratio between the inlet and throat for isentropic flow of air? BTL1

y

y 1

2

P y 1

(y 1)P2 1 M

2

Part B

1

Given adiabatic flow (Rayleigh flow) of dry air having of some section a Mach number is equal

to 3 and a stagnation temperature of 300 K, while the static pressure is 0.5 bar. For some other

section where Mach number is 1.5. Find (i) Stagnation temperature, (ii) Stagnation Pressure,

(iii) Static pressure and (iv) Amount of heat transferred that caused the reduction in Mach

number. (13 M) (Nov 95) BTL5


(i) From Isentropic flow table, P01=18.38 x105 N/m2 (3M)

(ii) Refer Rayleigh flow table 𝛾 = 1.4 𝑎𝑛𝑑 𝑀1 = 3

P1*=2.84 x105 N/m2 =P2

*, P01* =5.36 x105 N/m2 =P02

*, T01*= 458.72 K=T02

* (5M)



5-15

(iii) Refer Rayleigh flow table 𝛾 = 1.4 𝑎𝑛𝑑 𝑀1 = 1.5

P2=1.64 x105 N/m2 , P02 =6.013 x105 N/m2 , T02= 416.97 K

Heat transfer Q = m Cp( 𝑇02 − 𝑇01)=1.17 x105 J/kg (5M)

2

The data for a gas (ᵞ = 1.3, Cp=2.144 kJ/kgK) at entry of combustion chamber are C1=150 m/s,

P1=4 bar and T1=395K, if the exit Mach number is 0.78, calculate the following (i) The initial

Mach number, (ii) Final pressure, temperature and velocity of the gas and (iii) Stagnation

pressure loss. (13 M) (April-May 2016) BTL5


Mach number M1=0.305

(i) Refer isentropic flow table 𝛾 = 1.3 𝑎𝑛𝑑 𝑀1 = 0.3

T01=400.20 K, P01=4.23 x105 N/m2 , (3M)


P1*=1.94 x105 N/m2 =P2

*, P01* =3.55 x105 N/m2 =P02

*, T1*= 1034.032 K=T2

*

, T01*= 1191.07 K=T02

* , C1*=810.81 m/s =C2

* (5M)


P2=2.49 x105 N/m2 , P02=3.63 x105 N/m2 , T2=1037.13 K, T02=1133.89 K, C2=633.24 m/s

Stagnation pressure loss ∆𝑃0 = 𝑃02 − 𝑃01 = 0.6 x 105 N/m2

Stagnation enthalpy rise ∆ℎ0 = 𝐶𝑝(𝑇02 − 𝑇01) =𝐶𝑎𝑙𝑜𝑟𝑖𝑓𝑖𝑐 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑓𝑢𝑒𝑙

𝐴𝑖𝑟 𝑓𝑢𝑒𝑙 𝑟𝑎𝑡𝑖𝑜+1

Air fuel ratio = 25.70: 1 (5M)

3

A circular duct passes 8.25 kg/s of air at an exit Mach number of 0.5. The entry pressure and

temperature are 3.5 bar and 38ºC respectively and coefficient of friction is 0.005. If the Mach

number at entry is 0.15, determine (i) Diameter of the duct, (ii) Length of the duct, (iii) Pressure

and temperature at the exit, and (iv) Stagnation pressure loss. (13 M) (May-June 2016) (May-

June 2012) BTL5


(i) Refer isentropic flow table 𝛾 = 1.4 𝑎𝑛𝑑 𝑀1 = 0.15

T01=312.256 K, P01=3.55 x105 N/m2, (3M)

(ii) Refer Fanno flow table 𝛾 = 1.4 𝑎𝑛𝑑 𝑀1 = 0.15

P1*=0.478 x105 N/m2 =P2

*, P01* =0.9037 x105 N/m2 =P02

*, T1*= 260.35 K=T2

*,

T01*= 1191.07 K=T02

* , C1*=323.32 m/s =C2

* (5M)



5-16

(iii) Refer Fanno flow table 𝛾 = 1.4 𝑎𝑛𝑑 𝑀1 = 0.5

P2=1.021 x105 N/m2 , P02=1.210 x105 N/m2 , T2=297.58 K,

Diameter of the duct

Mass flow rate �̇� = 𝜌𝐴𝐶 = 𝜌1𝐴1𝐶1 D1= D2=D=0.224 m,

Length of the duct, 4�̅�𝐿

𝐷= (

4�̅�𝐿𝑀𝑎𝑥

𝐷)

𝑀1

− (4�̅�𝐿𝑀𝑎𝑥

𝐷)

𝑀2

= 27.285 L=305.59 m

Stagnation pressure loss ∆𝑃0 = 𝑃02 − 𝑃01 = 2.34 x 105 N/m2 (5M)

4

Air at an inlet temperature of 60ºC flows with subsonic velocity through an insulated pipe

having inside diameter of 50 mm and a length of 5 m. The pressure at the exit of the pipe is 101

kPa and the flow is chocked at the end of the pipe. If the friction factor 4f = 0.005, determine

the inlet Mach number, the mass flow rate and the exit temperature. (13 M) (Dec 2003) BTL5


Flow is chocked at the end of the pipe. It means at the end of the pipe Mach number value is one.

M2=1,

(i) Refer Fanno flow table 𝛾 = 1.4 𝑎𝑛𝑑 𝑀2 = 1

P2 = 1.01 x105 N /m2 =P2*, (5M)

4�̅�𝐿

𝐷= (


𝐷)

𝑀1

− (4�̅�𝐿𝑀𝑎𝑥

𝐷)

𝑀2

, (4�̅�𝐿𝑀𝑎𝑥

𝐷)

𝑀1

= 0.5 refer Fanno flow table, M1=0.6,

P1=1.78 x105 N /m2 , T1*= 297.58 K=T2

* (5M)

Mass flow rate �̇� = 𝜌𝐴𝐶 = 𝜌1𝐴1𝐶1 = 0.802 𝑘𝑔/𝑠 (3M)

5

Air is heated in a constant area duct from a Mach number of 0.2 to 0.8. The inlet stagnation

conditions are 2 bar and 93ºC. Determine the stagnation conditions of air at exit, the amount of

heat transferred per unit flow and change in entropy. (13 M) (Nov 1996) BTL5


Refer isentropic flow table 𝛾 = 1.4 𝑎𝑛𝑑 𝑀1 = 0.2

T1= 363.07 K, P1=1.94 x105 N /m2 (3M)

(ii) Refer Rayleigh flow table 𝛾 = 1.4 𝑎𝑛𝑑 𝑀1 = 0.2

P1*=0.853 x105 N/m2 =P2

*, P01* =1.619 x105 N/m2 =P02

*, T1*= 1753.96 K=T2

*, T01*= 2103.44K=T02

*

. (4M)



5-17


P2=1.079 x105 N/m2 , P02=1.649 x105 N/m2 , T2=1797.80 K, T02=2027.71 K,

Stagnation pressure loss ∆𝑃0 = 𝑃02 − 𝑃01 = 0.6 x 105 N/m2

Heat transfer Q = m Cp( 𝑇02 − 𝑇01)=1.67 x106 J/kg

Change in entropy =∆𝑠 = 𝐶𝑝𝑙𝑛 ((

𝑇2𝑇1

)

(𝑃2𝑃1

)

𝛾−1𝛾

)=1776.17 J/kgK (6M)

6

Air enters a constant area duct at M1=3, P1=1 atm. and T1=300 K. Inside the heat added per

unit mass is Q=3x105 J/kg. Calculate the flow properties M2, P2, ρ2, T2, T02 and P02 at the exit.

(13 M) (Dec 2003) BTL5


Refer isentropic flow table 𝛾 = 1.4 𝑎𝑛𝑑 𝑀1 = 3

T01= 840.34 K, P01=36.76 x105 N /m2 (3M)


P1*=5.68 x105 N/m2 =P2

*, P01* =10.73 x105 N/m2 =P02

*, T1*= 1067.61 K=T2

*, T01*= 1284.92=T02

*

ρ1*=1.844 kg/m3 = ρ 2

*

Heat transfer Q = m Cp( 𝑇02 − 𝑇01)=1.67 x106 J/kg==> 𝑇02 =1138.85 K (6M)

𝑇02

𝑇01= 0.886 (iii) Refer Rayleigh flow table 𝛾 = 1.4 𝑎𝑛𝑑

𝑇02

𝑇01= 0.886

P2=2.97 x105 N/m2 , P02=12.61 x105 N/m2 , T2=1797.80 K, ρ1=1.376kg/m3 (4M)

7

Air at M1= 0.2 is expanded from stagnation properties of 200 kPa and 30 through a convergent

nozzle for which the exit Mach number is 0.8. Assume the nozzle is isentropic and flow through

the pipe of diameter 25 mm is adiabatic. Find the length of the pipe and percentage of decrease

in stagnation pressure. Also find the change in entropy for the friction factor 4f=0.005. (13 M)

(Apr 98) BTL5


(i) Refer Fanno flow table 𝛾 = 1.4 𝑎𝑛𝑑 𝑀1 = 0.2

P01* =67.47 x103 N/m2 =P02

*, P02 = 70.03 x103 N/m2 (3M)

Length of the pipe L



5-18

4�̅�𝐿

𝐷= (


𝐷)

𝑀1

− (4�̅�𝐿𝑀𝑎𝑥

𝐷)

𝑀2

= 27.285 L=72.3 m (4M)

Percentage of decrease in stagnation pressure loss % ∆𝑃0 = (𝑃02 − 𝑃01)/𝑃01 ∗ 100 =64.9 % (3M)

Change in entropy S2-S1/R = ln(𝑃01

𝑃02) = 301.17 𝐽/𝑘𝑔𝐾 (3M)

8

The friction factor for a 50 mm diameter steel pipe is 0.005. At the inlet to the pipe velocity is

70 m/s, temperature is 80ºC and the pressure is 10 bar. Find the temperature, pressure and

Mach number at exit if the pipe is 25 m long. Also determine the maximum possible length. (13

M) (Dec 2005) BTL5


Mach number M1=0.185

Refer Fanno flow table for 𝛾 = 1.4 𝑎𝑛𝑑 𝑀1 = 0.18

P1*=1.648 x105 N/m2 =P2

*, T1*= 296.14 K=T2

*, (5M)

L max = 46.357 m

4�̅�𝐿

𝐷= (


𝐷)

𝑀1

− (4�̅�𝐿𝑀𝑎𝑥

𝐷)

𝑀2

==> (4�̅�𝐿𝑀𝑎𝑥

𝐷)

𝑀2

= 8.543 (4M)

Refer Fanno flow table, for 𝛾 = 1.4 𝑎𝑛𝑑 (4�̅�𝐿𝑀𝑎𝑥

𝐷)

𝑀2

= 8.543

P2=7.186 x105 N/m2 , T2=350.92 K (4M)

9

The pressure, temperature and Mach number of air in a combustion chamber are 4 bar, 100ºC

and 0.2 respectively. The stagnation temperature of air in a combustion chamber is increased 3

times its initial value. Calculate (i) The Mach number, pressure and temperature at the exit (ii)

Stagnation pressure loss (iii) Heat supplied per kg of air. (13 M) (Apr 2014) BTL5



T01= 376 K, P01=4.11 x105 N /m2

Refer Rayleigh flow table 𝛾 = 1.4 𝑎𝑛𝑑 𝑀1 = 0.2

P1*=1.759 x105 N/m2 =P2

*, P01* =3.327 x105 N/m2 =P02

*, T1*= 1801.93 K=T2

*, T01*= 2160.91=T02

*

ρ1*=1.844 kg/m3 = ρ 2

* (5M)

Refer Rayleigh flow table 𝛾 = 1.4 𝑎𝑛𝑑𝑇02

𝑇02∗ = 0.529



5-19

M2=0.40, P2=3.449 x105 N/m2 , P02=3.849 x105 N/m2 , T2=1108.18 K (3M)

Stagnation pressure loss ∆𝑃0 = 𝑃01 − 𝑃02 = 0.261 x 105 N/m2 (2M)

Heat transfer Q = m Cp( 𝑇02 − 𝑇01)=755.7 x103 J/kg (3M)

10

The condition of a gas in a combustion chamber at entry are T1=375K, P1=0.50 bar, C1=70

m/s. The air-fuel ratio is 29 and the calorific value of the fuel is 42 MJ/kg. Calculate (i) The

initial and final Mach number (ii) Final pressure, temperature and velocity of the gas (iii)

Percentage of stagnation pressure loss (iv) Maximum stagnation temperature. Take 𝜸 =𝟏. 𝟒 𝒂𝒏𝒅 𝑹 = 𝟎. 𝟐𝟖𝟕 𝑱/𝒌𝒈𝑲 (13 M) () BTL5


Mach number at entry M1=0.180,


T01= 377.26 K, P01=0.5112 x105 N /m2 (2M)


P1*=0.218 x105 N/m2 =P2

*, P01* =0.4119x105 N/m2 =P02

*, T1*= 2192.98 K=T2

*, T01*= 2638.18=T02

*

C1*= 945.95 m/s = C 2

* (4M)

Stagnation enthalpy rise due to the combustion of one kg of fuel is 1.4 x106 J/kg, 𝑇02

𝑇02∗ = 0.67


𝑇02∗ = 0.67

M2=0.48, P2=0.3956 x105 N/m2 , P02=0.462 x105 N/m2 , T2=1662.27 K (3M)

% of Stagnation pressure loss ∆𝑃0 =𝑃01−𝑃02

𝑃01𝑥 100 = 9.62% (2M)

Maximum stagnation temperature = 𝑇0𝑚𝑎𝑥 = 𝑇01∗ = 𝑇02

∗ = 𝑇0∗ = 2638.18 𝐾 (2M)

Part * C

1

(i) Derive the expression between the stagnation to static temperature ratio and the Mach

number of the flow. (7)

(ii) Draw the Rayleigh curve and explain the effect of heat transfer on supersonic flow. (8).

1D form of energy conversion equation is



5-20

Here subscript 1 stands for initial state of the fluid and subscript 2 stands for final decelerated

state of fluid. Since, V2=0, lets represent T2=T0 is above equation. Then,

Dividing the above equation by

(7M)

(ii) Explanation and diagram

Figure- T-s diagram (8M)

2 A combustion chamber in a gas turbine plant receives air at 350K, 0.55 bar and 75 m/s. The

air-fuel ratio is 29 and the calorific value of the fuel is 41.87 MJ/kg. Taking 𝜸 = 𝟏. 𝟒 𝒂𝒏𝒅 𝑹 =



5-21

𝟎. 𝟐𝟖𝟕 𝑱/𝒌𝒈𝑲 for the gas, determine (i) The initial and final Mach number (ii) Final pressure,

temperature and velocity of the gas (iii) Percentage of stagnation pressure loss (iv) Maximum

stagnation temperature. (May-June 2016) (April-May 2011) (15 M) BTL5


Mach number at entry M1=0.199,


T01= 352.82 K, P01=0.565 x105 N /m2 (5M)


P1*=0.2419 x105 N/m2 =P2

*, P01* =0.457x105 N/m2 =P02

*, T1*= 1690.82 K=T2

*,

T01*= 2027.70 K =T02

* , C1*= 824.17 m/s = C 2

* (5M)

Stagnation enthalpy rise due to the combustion of one kg of fuel is 1.4 x106 J/kg, 𝑇02

𝑇02∗ = 0.8588


𝑇02∗ = 0.859

M2=0.64, P2=0.368 x105 N/m2 , P02=0.48 x105 N/m2 , T2=1611.35 K, C2= 515.106 m/s

% of Stagnation pressure loss ∆𝑃0 = 𝑃01 − 𝑃02 = 0.085 𝑥105𝑁/𝑚2

Maximum stagnation temperature = 𝑇0𝑚𝑎𝑥 = 𝑇01∗ = 𝑇02

∗ = 𝑇0∗ = 2027.70 𝐾 (5M)

3

The stagnation temperature of air is raised from 85ºC to 376ºC in a heat exchanger. If the inlet

Mach number is 0.4, determine the final Mach number and percentage drop in pressure. Page

(15 M) (Apr 1998) BTL5


Refer Rayleigh flow table for 𝛾 = 1.4 𝑎𝑛𝑑 𝑀1 = 0.4 (3M)

T01*= 676.75 K =T02

* 𝑇02

𝑇02∗ = 0.958 (3M)

Refer Rayleigh flow table for 𝛾 = 1.4 𝑎𝑛𝑑𝑇02

𝑇02∗ = 0.955 (3M)

M2=0.78, P2/P2* =1.296, Note inlet Mach number , M1<1, so M2<1 (3M)

Percentage drop in pressure =

𝑃1𝑃1

∗−𝑃2

𝑃2∗

𝑃1𝑃1

∗

x 100 = 33.9 % (3M)



5-22



UNIT III NORMAL AND OBLIQUE SHOCKS

Governing equations – Variation of flow parameters across the normal and oblique shocks – Prandtl –

Meyer relations – Applications. (https://nptel.ac.in/courses/112103021/8)

PART * A

Q. No. Questions

1.

How shock condensation of a shock wave is defined? (Nov/Dec 2015) BTL2

Condensation Shock wave is a special form of shock wave. It occurs in an accelerating supersonic

gas flow as a result of the condensation of vapors contained in the gas. A condensation shock is

usually observed in a supersonic nozzle when the accelerated motion of the gas is accompanied by a

monotonic decrease in the gas’s temperature and a corresponding increase in its relative humidity.

2

What is compression corner? (Nov/Dec 2015) BTL1

Compression corners and shocks impinging on boundary layers are common in scramjet inlets and

isolators. Due to the simplicity of the configuration, compression ramps.

3 What is Oblique shock? (April/May 2015) BTL2

When the shock wave is inclined at an angle to the flow, it is called Oblique shock.

4

What is Prandtl Meyer Relation? (April/May 2015) BTL2

It gives the relationship between the gas velocities before and after the normal shock and the critical

velocity of sound. Mx* X My* =1 or Cx X Cy = a*2.

5

What are the properties changes across a normal shock? Is the flow through a normal shock

an equilibrium one? (Nov/Dec 2014) BTL2

Stagnation pressure decreases, stagnation temperature remains constant.

No, since the fluid properties like pressure, temperature, and density are changed during the normal

shock.

6

Give the difference between normal and oblique shocks. (Nov/Dec 2014) BTL4

Normal Shock: Shock wave is right angle to the flow, one dimensional flow.

Oblique shock: Shock wave is inclined at an angle to the flow, two dimensional flow.

7

Why the efficiency of a machine, experiencing shock wave is considerably low? (May /June

2014) BTL4

Shock may cause boundary layer separation and deviation of flow from its designed direction.

There will be a loss in stagnation pressure and increase in entropy across the shock wave.



5-23

8

What is the use of pitot tube in supersonic flow? (May /June 2014) BTL1

Introduction of pitot tube produces a curved shock a little distance upstream of its mouth. Therefore

it measures the stagnation pressure downstream the shock wave.

9 State assumption made to drive the iso-thermal flow equation. (May /June 2013) BTL3

One dimensional, Constant area duct, Frictional flow at constant temperature, The gas is perfect.

10 Difference between Fanno flow and Rayleigh flow. (May /June 2013) BTL4

Fanno flow-with friction without heat transfer, Rayleigh flow-With heat transfer.

11 Define shock wave. BTL2

A shock wave is a special kind of wave referred to as a steep finite pressure wave. The changes

in the flow properties across such a wave are abrupt.

12

Define Normal shock. BTL1

Normal shock wave is defined as the abrupt jump in pressure when they are at right angles to the

direction of flow.

The normal shock wave is perpendicular to the one dimensional flow.

13

List out some effects of shock in a flow. BTL4

The effects are:

1. Shocks may cause boundary layer separation.

2. Shock derivate the flow from its designated direction.

14

List out the changes in properties before and after the normal shock. BTL4

The change in properties before and after the normal shock is,

i. There is an increase in entropy.

ii. There is stagnation pressure loss.

iii. There is an increase in temperature after the shock.

15

Define Prandtl – Meyer relation. BTL2

The fundamental relation used to compare gas velocities before and after normal shock and to

derive critical velocity of sound is known as Prandtl Meyer relation.

The Prandtl meyer relation of normal shock wave is,

Cx Cy = a*2 i.e Mx My = 1

16

Define Rankine – Hugoniot equation. BTL2

The equation which relates pressure and density ratio across a shock wave in a perfect gas is

known as Rankine – Hugoniot equation.

For normal shock wave,

1 11

1 1 (1) (2)

1 1

1 1

y y

y yx x

y yx x

x x

P

P

17

Define strength of a shock wave. BTL1

The strength of a shock wave is defined as the ratio of pressure increase due to shock to the

initial or upstream pressure.

Py - Px

Strength = _________

Px



5-24

18 What are the effects of normal shocks in a flow through convergent divergent passages? BTL2

In case of convergent divergent passages the normal shock leads to drop in fluid velocity

instantaneously from supersonic to subsonic as the pressure rise across the shock wave front.

19 Define compression shock. BTL2

A shock which occurs in a stream that is supersonic in that upstream from the shock plane and

accompanied by an increase in entropy is known as compression shock.

20

Define Rarefaction shock. BTL2

A shock which is opposite to the compression shock and would accelerate the flow from subsonic to

supersonic velocity is known as rarefaction shock.

21

Give expression for change in entropy across the normal shock in terms of stagnation pressure.

BTL2

Poy

S = - R ln _______

Pox

S – Change in entropy, Poy – Stagnation pressure at downstream,

Pox - Stagnation pressure at upstream of shock, R – Gas constant

22

List out the applications of propagating steep waves (or) shock waves. BTL3

Propagating steep waves of finite amplitude finds applications in,

i. Shock tubes

ii. Supersonic gas ducts

iii. Partial admission turbines

iv. Pulse jet engines.

v. Complex engines

23

List out the applications of supersonic diffuser. BTL3

Supersonic diffusers have applications in

i. Supersonic wind tunnels.

ii. Compressors

iii. Supersonic inlets of aircraft engines and missiles.

24

Give two stages in which diffusion occurs. BTL2

i. Through a normal shock at the entry section

ii. Through isentropic deceleration of the subsonic flow after the shock in the diverging passage.

25

Define Diffuser efficiency. BTL2

Diffuser efficiency is defined as the ratio of change of enthalpy in reversible diffusion to the

change of enthalpy in actual diffusion.

Change of enthalpy in reversible diffusion

D = ____________________________________________________

Change of enthalpy in actual diffusion

h2s – h1

D = ____________

h2 – h1



5-25

26

Define oblique shock wave. BTL2

When the direction of flow is inclined at an oblique angle to the shock wave it si known as

oblique shock wave. It is also known as two dimensional plane shock wave.

27

What is the wave angle of an oblique shock wave? BTL

The wave angle of an oblique shock wave is less than 90o degrees.

28

When oblique shock wave occurs? BTL4

Oblique shock wave occurs when a supersonic flow is deflected inwards (anti clockwise

direction).

29

Give some practical examples of oblique shock waves. BTL4

The practical examples are:

a. At the exit of turbine blade passages with supersonic flow.

b. At the entry of supersonic diffuser of aircraft engines.

c. Drag due to shock on an aerodynamic body.

30

Define wave angle. BTL2

The angle which the oblique shock waves makes with the initial direction of the flow is called

wave angle ().

31

List out the assumptions made in the analysis of oblique shock wave. BTL4

i. Perfect gas,

ii. Steady, adiabatic and friction less flow through stationary oblique shock wave.

iii. Absence of external work and iv. Absence of body forces

32

Give the expression for deflection angle of an oblique shock wave. BTL5

M12 Sin 2 - 2 Cot

tan = ______________________________

2 + M12 ( r – Cos 2 )

33

Give the expression of Mach angle () of an oblique shock wave. BTL5

The expression for Mach angle () of an oblique shock wave is,

= Mach wave = Sin-1 1/M

34

Define shock polar. BTL2

The graphical representation of oblique shock properties is known as shock polar.



5-26

35

Define Prandtl Meyer flow. BTL2

The isentropically accelerating flow over the convex wall is known as Prandtl Meyer flow.

Part B

1

Air flows adiabatically in a pipe. A normal shock wave is formed. The pressure and temperature

of air before the shock are 150 kN/m2 and 25ºC respectively. The pressure just after the normal

shock is 350 kN/m2. Calculate (i) Mach number before the shock, (ii) Mach number, static

temperature and velocity of air after the shock wave, (iii) Increase in density of air, (iv) Loss of

stagnation pressure of air and (v) change in entropy. (13 M) (May 2004) BTL5


Py/Px=2.333,

Refer Normal shock tables for Py/Px=2.333, 𝛾 = 1.4 (3M)

𝑇𝑦 = 385.61 𝐾, 𝑃0𝑦 = 489.75 𝑥 103 𝑁/𝑚2, 𝑃0𝑥 = 519.9 𝑥 103 𝑁/𝑚2

Stagnation pressure loss ∆𝑃0 = 𝑃0𝑥 − 𝑃0𝑦 = 30.15 x 103 N/m2

Mach number after the shock My=Cy/ay Cy=281.83 m/s,

Density before shock 𝜌𝑥 =𝑃𝑥

𝑅𝑇𝑥= 1.753 𝑘𝑔/𝑚3 (4M)

Density after the shock 𝜌𝑦 =𝑃𝑦

𝑅𝑇𝑦= 3.162 𝑘𝑔/𝑚3 (3M)

Increase in density = 1.409 kg/m3

Change in entropy, ∆𝑆 = 𝑅 𝑙𝑛 (𝑃0𝑥

𝑃0𝑦) = 17.148 𝐽/𝑘𝑔𝐾 (3M)

2

When a convergent divergent nozzle is operated at off-design condition a normal shock occurs

at a section where the cross sectional area is 18.75 cm2 in the diverging portion. At inlet to the

nozzle the stagnation state is given as 0.21 MPa and 36ºC. The throat area is 1.25 cm2 and exit

area is 25 cm2. Estimate the exit Mach number, exit pressure and loss in stagnation pressure

for flow through nozzle. (13 M) (Dec 2005) BTL5


𝐴𝑥

𝐴𝑦= 1.5,

Refer Isentropic flow table for 𝐴𝑥

𝐴𝑦= 1.5, 𝛾 = 1.4 Mx=1.86 (2M)

Note upstream Mach number, Mx value is always greater than 1

Refer Normal shock Table for Mx=1.86 and 𝛾 = 1.4, My=0.604, 𝑃0𝑦 = 1.65 x 105 N/m2 (3M)



5-27

Refer Isentropic flow table for My=0.604 and 𝛾 = 1.4 , 𝐴𝑦∗ = 1.578 𝑥 10−3𝑚2 (3M)

𝐴2

𝐴𝑦∗ = 1.584, (1M)

Refer Isentropic flow table , 𝐴2

𝐴𝑦∗ = 1.584 and 𝛾 = 1.4

M2=0.4, P2=1.476 x 105 𝑁/𝑚2

We Know that, Stagnation pressure loss, ∆𝑃 = 𝑃0𝑥 − 𝑃0𝑦 = 0.45 𝑥 105𝑁/𝑚2 (4M)

3

A gas at a pressure of 340 mbar, temperature of 355 K and entry Mach number of 1.4 is

expanded isentropically to 140 mbar. Calculate the following (i) Deflection angle (ii) Final Mach

number and (iii) Final temperature of the gas. (13 M) (April-May 2015) (May-June 2014)(April-

May2011) BTL5


i) Refer isentropic flow table 𝛾 = 1.3 𝑎𝑛𝑑 𝑀1 = 1.4

T01=459.24 K= T02, P01=1.039 x105 N/m2 , P2/P02 =0.134 (3M)

Refer Isentropic flow table for P2/P02 =0.135 and 𝛾 = 1.3

M2=1.98, T2=288.86 K

Refer Prandtl-Meyar function table for M2=1.4 and 𝛾 = 1.3 𝜔(𝑀1) = 9.542 (3M)

Refer Prandtl-Meyar function table for M2=1.98 and 𝛾 = 1.3 𝜔(𝑀2) = 28.06 (3M)

Deflection angle 𝛿 = 𝜔(𝑀1) − 𝜔(𝑀2) = −18.51º (4M)

4

A pitot tube kept in a supersonic wind tunnel forms a bow shock, ahead of it. The static pressure

upstream of the shock is 16 kPa and the pressure at the mouth is 70 kPa. Estimate the Mach

number of the tunnel. If the stagnation temperature is 300ºC, calculate the static temperature

and total pressure upstream and downstream of tube. (13 M) (Dec 2003) (Dec 2005) BTL5


𝑃0𝑦

𝑃𝑥= 4.375 (2M)

Refer Normal shocks table for 𝑃0𝑦

𝑃𝑥= 4.375 and 𝛾 = 1.4 (1M)

Mx=1.74, Pox=0.834 x105 N/𝑚2 (2M)

𝑇0

𝑇= 1 +

𝛾−1

2𝑀2,

𝑇0𝑥

𝑇𝑥= 1 +

𝛾−1

2𝑀𝑥

2 Tx=356.89 K, (4M)

From Table, 𝑇𝑦

𝑇𝑥= 1.487, ===> 𝑇𝑦 = 350.69 K (4M)



5-28

5

An oblique shock wave occurs at the leading edge of a symmetrical wedge. Air has a Mach

number of 2.1 and deflection angle of 15º. Determine the following for strong and weal waves.

(i) Wave angle (ii) Pressure ratio (iii) Density ratio, (iv) Temperature ratio and (iv) Downstream

Mach number. (13 M) (Nov 2015) BTL5


Deflection angle, tan𝛿 = 2 𝑐𝑜𝑡𝜎𝑀1

2𝑆𝑖𝑛2𝜎−1

2+𝛾𝑀12+𝑀1

2(1−2𝑆𝑖𝑛2𝜎)====> 𝜎 = 80.8°, 𝑎𝑛𝑑 𝜎 = 43°

The angle being closer to 90° is the angle of strong wave. Similarly the angle much lesser than 90°

is the angle of weak wave.

For strong shock wave

Upstream Mach number of Normal shock Mx= M1 Sin 𝜎 = 2.07 (2M)

Refer normal shock table for Mx=2.07 and 𝛾 = 1.4, (2M)

My = 0.565, M2 = 0.619, Density at exit / Density at entry = 2.769 (3M)

For Weak shock wave,

Upstream Mach number of Normal shock Mx= M1 Sin 𝜎 = 1.432 (3M)

Refer normal shock table for Mx =1.432 and 𝛾 = 1.4,

My = 0.727, M2 = 1.548, Density at exit / Density at entry = 1.741 (3M)

6

A jet approaches a symmetrical wedge of a Mach of 2.4 and wave angle of 60º. Determine the

following (i) Deflection angle (ii) Pressure ratio (iii) Temperature ratio and (iv) Final Mach

number. (13 M)(April-May 2002) BTL6




2+𝛾𝑀12+𝑀1

2(1−2𝑆𝑖𝑛2𝜎)====> 𝛿 = 28.08° (3M)

We know that, Entry Mach number, M1= Mx/Sin 𝜎 Mx = 2.07 (2M)

Refer normal shocks table for Mx = 2.07 and 𝛾 = 1.4,

My = 0.565, Py/Px=4.832 = P2/P1, Ty/Tx= 1.745 =T2/T1 (4M)

Exit Mach number, M2 = My/ Sin(𝜎 − 𝛿) = 1.068 (4M)

7

A convergent divergent duct is operating under off design conditions as it conducts air from a

high pressure tank where, P0=210 kPa and T0 = 37ºC. A normal shock is present in the diverging

section of nozzle. Find the exit pressure, loss in stagnation pressure, and increase in entropy for

the following areas. (i) Ax* =13 cm2, (ii) Aexit = 26 cm2 and (iii) Ax = Ay = 19.5 cm2. (13 M) (Oct

1996) BTL5



5-29


𝐴𝑥

𝐴𝑥∗ = 1.5 ,

Refer Isentropic flow table for 𝐴𝑥

𝐴𝑥∗ = 1.5 , 𝛾 = 1.4 Mx=1.86 (2M)

Note upstream Mach number, Mx value is always greater than 1

Refer Normal shock Table for Mx=1.86 and 𝛾 = 1.4, My=0.604, 𝑃0𝑦 = 1.65 x 105 N/m2

Refer Isentropic flow table for My=0.60 and 𝛾 = 1.4 , 𝐴𝑦∗ = 1.64 𝑥 10−3𝑚2 (3M)

𝐴2

𝐴𝑦∗ = 1.584 ≈ 1.59,

Refer Isentropic flow table , 𝐴2

𝐴𝑦∗ = 1.59 and 𝛾 = 1.4

M2=0.4, P2=1.476 x 105 𝑁/𝑚2 (3M)

We Know that, Stagnation pressure loss, ∆𝑃 = 𝑃0𝑥 − 𝑃0𝑦 = 0.45 𝑥 105𝑁/𝑚2 (2M)

Change in entropy, ∆𝑆 = 𝑅 𝑙𝑛 (𝑃0𝑥

𝑃0𝑦) = 69.2 𝐽/𝑘𝑔𝐾 (3M)

8

An oblique shock wave at an angle of 33º occurs at the leading edge of a symmetrical wedge.

Air has a Mach number of 2.1 upstream temperature of 300 K and upstream pressure of 11

bar. Determine the following (i) Downstream pressure (ii) Downstream temperature (iii) Wedge

angle (iv) Downstream Mach number. (13 M) (May 2014) BTL5


Mach number of oblique shock, 𝑀1 =𝑀𝑥

sin 𝜎𝑀𝑥 = 1.143 (3M)

Refer Normal shock table for 𝑀𝑥 = 1.143 𝑎𝑛𝑑 𝛾 = 1.4

My=0.882, Ty=327 K, Py=14.83 x 105 𝑁/𝑚2 (3M)



2+𝛾𝑀12+𝑀1

2(1−2𝑆𝑖𝑛2𝜎)====> 𝛿 = 5.44° (3M)

Downstream Mach number of Oblique shock, M2 = My/ Sin(𝜎 − 𝛿) = 1.91 (4M)

9

A jet of air at 270 K and 0.7 bar has an initial Mach number of 1.9. If it passes through a normal

shock wave, determine the following for downstream of the shock. (i) Mach number (ii)

Pressure (iii) Temperature (iv) Speed of sound (v) Jet velocity (vi) Density. (13 M) (April 2004)

BTL6




5-30

Refer normal shock Mx=1.9 and 𝛾 = 1.4

My=0.596, Py=2.831 x 10 5 N/m2, (4M)

Ty=434.16 K, ay= 417.66 m/s, (5M)

Cy= 248.93 m/s, 𝜌𝑦 = 2.27 𝑘𝑔/𝑚3 (4M)

10

A convergent divergent nozzle is designed expand air from a reservoir in which the pressure is

800 kPa and temperature is 40C to give a Mach number at exit of 2.5. The throat area is 25 cm2

find (i) Mass flow rate (ii) Exit area (iii) When a normal shock appears at a section where the

area is 40 cm2, determine the pressure and temperature at exit. Throat. (13 M) BTL6.



T2=138.97 K, P2=0.468 x105 N/m2, A2=6.59 x 10-3 m2

Mass flow rate �̇� = 𝜌𝐴𝐶 = 𝜌1𝐴1𝐶1 = 𝝆𝟐𝑨𝟐𝑪𝟐 = 4.568 𝑘𝑔/𝑠 (3M)

Normal shock appears at Ax = 40 x 10-4 m2,

Ax/Ax*=1.6 Refer isentropic flow table for Ax/Ax*=1.6 , and 𝛾 = 1.4

Upstream Mach number Mx always greater than 1, Mx=1.94 (2M)

Refer normal shock table for Mx=1.94 and 𝛾 = 1.4

My=0.588, P0y = 5.992 x 105 N/m2, (2M)

Refer Isentropic flow table for My=0.59 and 𝛾 = 1.4 Ay/Ay*=1.2 Ay*=3.33 x 10-3 m2

A2/Ay*=1.977 , (3M)

Refer Isentropic flow table for A2/Ay*=1.976 and 𝛾 = 1.4

M2=0.31, P2=5.60x105 N/m2, T2=307.05 K (3M)

Part *C

1

(i) Derive the equation for Mach number in the downstream of the normal shock wave.

(ii) Derive the equation for static pressure ratio across the shock wave. (15 M) BTL3


A normal shock wave is one of the situations where the flow properties change drastically in one

direction. The shock wave stands perpendicular to the flow as shown in Fig. 4.4.4. The quantitative

analysis of the changes across a normal shock wave involves the determination of flow properties.

All conditions of are known ahead of the shock and the unknown flow properties are to be determined



5-31

after the shock. There is no heat added or taken away as the flow traverses across the normal shock.

Hence, the flow across the shock wave is adiabatic.

Fig. Schematic diagram of a standing normal shock wave.

The basic one dimensional compressible flow equations can be written as below;

For a calorically perfect gas, thermodynamic relations can be used,

The continuity and momentum equations can be simplified to obtain,

Since, and , the energy equation is written as,

Both can now be expressed as,

Substitute and solve for



5-32

Recall the relation for and substitute

Substitute and solve for

(9M)

The pressure ratio can be obtained by the combination of momentum and

continuity equations i.e.

(2M)

Substituting the ratio and simplifying for the pressure ratio across the

normal shock, we get, (4M)

2

Derive Rankine-Hugoniot equation. (15 M) BTL3


Hugoniot equation:

Shock wave compression is an adiabatic and irreversible process; hence shock waves can be treated

as irreversible adiabatic compressors. Let’s calculate the work required for this compression

process.



5-33

(15M)

3

A jet of air entering the subsonic diffuser at P0=1 bar and T=280 K. The entry Mach number

is 2 and the ratio of the exit to entry area of the diffuser is 4. If there is a normal shock wave

just outside the diffuser entry, determine the following for exit (i) Mach number (ii)

temperature (iii) Pressure (iv) Stagnation pressure loss. (15 M) BTL6


Refer isentropic flow table Mx=M1=2 and 𝛾 = 1.4

Tox=504.50 K, Px= 0.128 x 105 𝑁/𝑚2, (2M)

Refer normal shock table Mx = 2 and 𝛾 = 1.4

My=0.577, Py=0.576 x 105 𝑁/𝑚2, Ty=472.36 K (3M)

Refer isentropic flow table for My=0.58 and 𝛾 = 1.4

P0y=0.723 x 105 𝑁/𝑚2, A2/Ay*=4.852, (5M)

Refer isentropic flow table A2/Ay*=4.852, and 𝛾 = 1.4, take M<1 since it is subsonic

M2=0.12, T2=503.03 K, P2=0.715 x105 𝑁/𝑚2,

Stagnation Pressure loss ∆𝑃 = 𝑃0𝑥 − 𝑃0𝑦 = 0.277𝑥 105𝑁/𝑚2 (5M)

Figure- Rankine Hugoniot Curve



5-34



5-35



UNIT IV JET PROPULSION

Theory of jet propulsion – Thrust equation – Thrust power and propulsive efficiency – Operating principle,

cycle analysis and use of stagnation state performance of ram jet, turbojet, turbofan and turbo prop engines.

PART * A

Q.No. Questions

1.

What is the significance of low and high TSFC in jet propulsion? (Nov/Dec 2015) BTL2

Thrust specific fuel consumption (TSFC) or sometimes simply specific fuel consumption, SFC, is

an engineering term that is used to describe the fuel efficiency of an engine design with respect to

thrust output. SFC varies with throttle setting, altitude and climate. For jet engines, flight speed also

has a significant effect upon SFC; SFC is roughly proportional to air speed.

2

Draw T-s diagram of ideal and actual Brayton cycle and bring out the difference between

them. (Nov/Dec 2015) BTL3

T-S Diagram, Constant pressure heat addition and constant pressure heat rejection, remaining are

isentropic processes (S=Constant).

3 Define propulsive efficiency. (April/May 2015) BTL2

Propulsive efficiency = (Propulsive power / Power output)

4 What is the type of compressor used in turbo jet? (April/May 2015) BTL3

Rotary compressor

5

Why ramjet engine does not require a compressor and a turbine? (Nov/Dec 2014) BTL4

In ramjet engine due to supersonic and subsonic diffuser, the static pressure of air is increased to

ignition pressure, so there is no need of compressor and turbine.

6

Find the optimum propulsive efficiency when the jet velocity is 500 m/s and flight velocity is

900 m/s. (Nov/Dec 2014) BTL4

Propulsive efficiency = 2u/Cj+u = (2*900)/500+900 = 12.8%

7

Define thrust power and propulsive efficiency of aircraft engine. (May /June 2014) BTL2

Thrust power is the power required to propel the aircraft engine forward at constant velocity. The

product of thrust force and speed of aircraft is called thrust power.

8 Why a ram jet engine does not require a compressor and turbine? (May /June 2014) BTL4



5-36

The speed of the ram jet engine is supersonic, therefore the pressure rise is due to the ram effect, so

compressor and a turbine not necessary.

9

List out the different types of jet engines. At assumption made to drive the iso-thermal flow

equation. (May /June 2013) BTL4

Ram jet, pulse jet, turbo jet, turbo propeller, turbo shaft.

10 Give the components of a turbo jet. (May /June 2013) BTL2

Diffuser, Rotary compressor, Combustion chamber, Turbine and Exhaust nozzle

11

Define Jet propulsion. BTL2

When oxygen is obtained from the surrounding atmosphere for combustion process, the

system is called as Jet propulsion.

12

List out the components of aircraft gas turbine. BTL4

The components of aircraft gas turbines are:

i. Compressor

ii. Combustion chamber

iii. Turbine

iv. Tail pipe (or) Nozzle

v. After burner

13

List the different types of jet engines. BTL4

i. Turbo jet engine

ii. Turbo prop engine

iii. Ram jet engine

iv. Pulse jet engine

v. Turbo fan engine.

14

Distinguish between Rocket and jet propulsion. BTL3

S.N

o

Rocket Propulsion Jet Propulsion

1. The exhaust jet consists of exhaust

gases only.

Jet consists of air and combustion products.

2.

Propulsion unit consist of its own

oxygen supply for combustion

purpose.

Oxygen obtained from the surrounding

atmosphere for combustion purpose.

3. No altitude limitation Altitude limitation

4. Thrust improves with altitude Thrust decreases with altitude.

5. Mechanical devices not used. Mechanical devices are also used.

15

Give the expression for propeller thrust. BTL5

F = ma (Cj – u)

Where

F – Thrust

U – Flight speed

Cj – jet velocity.



5-37

16

Give the expression for jet thrust. BTL4

The expression for jet thrust is given by

Net thrust = momentum thrust + Pressure thrust

F = Fmom + Fpr

F = ma (CQ – u) + (Pe – Pa) Ac

17 Define Ram effect. BTL2

The pressure rise takes place due to the conversion of kinetic energy of incoming air into

pressure energy by the diffuser. This type of compression is called as Ram effect.

18

Define pulse jet engine. BTL2

Pulse jet engine is defined as the thrust producing device without turbine and compressor.

19

Why the engine is called as pulse jet engine? BTL3

The short burst of expansion of gases create a vacuum in the combustion chamber, due to

which the thrust is unsteady and produced in short pulses. Hence it is termed as pulse jet engine.

20

List out the components of pulse jet engine. BTL4

i. Inlet diffuser

ii. Inlet valve under automatic control.

iii. Combustion chamber

iv. Exit nozzle

Part B

1

An aircraft flies at a speed of 520 kmph at an altitude of 8000 m. The diameter of the propeller

of an aircraft is 2.4 m and flight to jet speed ratio is 0.74. Find the following: (i) The rate of air

flow through propeller, (ii) Thrust produced, (iii) Specific thrust, (iv) specific impulse and (v)

thrust power. (13 M) (Nov-Dec 2013) BTL5


Area of the propeller disc = 4.52 m2

Velocity of the jet Cj= 195.19 m/s (1M)

Velocity of air flow at the propeller, C= 169.81 m/s (1M)

Mass flow rate of air-fuel mixture �̇� = 𝜌𝐴𝐶 = 402.96 𝑘𝑔/𝑠̇ (2M)

The rate of air flow through propeller =�̇�𝑎= 402.96 𝑘𝑔/𝑠̇ (2M)

Thrust produced=F=�̇�𝑎(𝐶𝑗 − 𝑈) = 20.45 𝑥 103 𝑁 (2M)



5-38

Specific thrust 𝐹𝑠𝑝 =𝐹

𝑚 ̇= 50.75 𝑁/(

𝑘𝑔

𝑠) (1M)

Specific impulse = 𝐼𝑠𝑝 =𝐹

�̇�𝑔= 5.17 𝑆 (2M)

Thrust power P = Thrust (F) x Flight speed (U) = 2.95 x 106 W (2M)

2

A turbojet propels an aircraft at a speed of 900 km/h while taking 3000 kg of air per minute.

The isentropic enthalpy drop in the nozzle is 200 kJ/kg and the nozzle efficiency is 90 %. The

air-fuel ratio is 85 and the combustion efficiency is 95 %. The calorific value of the fuel is 42,000

kJ/kg. Calculate, (i) Propulsive power, (ii) Thermal efficiency and (iii) Propulsive efficiency. (13

M) (Dec 2003) (Dec 2004) BTL5


Mass flow rate of air –fuel mixture �̇� = �̇�𝑎 + �̇�𝑓 = 50.58𝑘𝑔

𝑠 (1M)

Mass flow rate of fuel �̇�𝑓 = 0.58𝑘𝑔

𝑠 (1M)

Velocity of jet 𝑐𝐽 = √2 𝑥 ɳ𝑁𝑥 ∆ℎ0 = 600 m/s (1M)

Thrust F =�̇� 𝐶𝑗 − �̇�𝑈 = 17.84 𝑥 103 𝑁 (2M)

Thrust power or propulsive power P= F x u = 4.46 x 106 W (2M)

Propulsive efficiency ɳ𝑃 =2𝑈

𝐶𝑗+𝑢 = 0.588 = 58.8 % (3M)

Thermal efficiency, ɳ𝑡 =1

2 𝑚 (𝐶𝑗

2−̇ 𝑈2)

ɳ𝑏 𝑥 �̇�𝑓 𝑥𝐶.𝑉 =0.325= 32.5% (3M)

3

A turbojet plane has two jets of 250 mm diameter and the net power at the turbine is 3000 kW.

The fuel consumption per kWhr is 0.42 kg with a fuel calorific value 49 MJ/kg, when flying at

a speed of 300 m/s in atmospheric having a density of 0.168 kg/m3. The air-fuel ratio is 53.

Calculate, (i) Absolute velocity of jet, (ii) Resistance or Drag of the plane, (iii) Overall efficiency

of the plane and (iv) Thermal efficiency. (13 M) (Oct 1995) BTL5



𝑠 (2M)

Velocity of Jet 𝑐𝐽 = 1145.91 m/s (1M)

Absolute velocity of the jet 𝐶𝑎𝑏𝑠 = 𝐶𝑗 − 𝑢 = 845.91 𝑚/𝑠 (2M)

Thrust or Resistance or Drag = F = �̇� 𝐶𝑗 − �̇�𝑈 = 16.09 𝑥 103 𝑁 (2M)

Overall efficiency ɳ𝑂 =�̇� (𝐶𝑗−̇ 𝑢)

�̇�𝑓 𝑥 𝐶.𝑉= 0.279 = 27.9 % (3M)



5-39


2 𝑚 (𝐶𝑗

2−̇ 𝑈2)

ɳ𝑏 𝑥 �̇�𝑓 𝑥𝐶.𝑉 =0.673= 67.3 % (3M)

4

A turbojet engine takes in 50 kg/s of air and propels an aircraft with uniform flight speed of

880 km/hr. Isentropic enthalpy change for nozzle is 188 kJ/kg and velocity coefficient is 0.96.

The fule air ratio is 1.2 %. Combustion efficiency is 95 %. Calorific value of fuel is 44,000 kJ/kg.

Find out, (i) Thermal efficiency of the engine, (ii) Fuel flow in kg/hr, (iii) Propulsive efficiency

and (iv) Overall efficiency. (13 M) (April 1997) BTL5



𝑠

Mass flow rate of fuel �̇�𝑓 = 0.6 𝑘𝑔

𝑠 = 2160 kg/h (3M)

Velocity of jet 𝑐𝐽 = √2 𝑥 ∆ℎ0 = 613.18 m/s, (2M)

Velocity coefficient= 0.96, Velocity of jet 𝑐𝐽 = 0.96x613.18 = 588.66 m/s (3M)


𝐶𝑗+𝑢 = 0.5868 = 58.68 % (2M)


2 𝑚 (𝐶𝑗

2−̇ 𝑈2)

ɳ𝑏 𝑥 �̇�𝑓 𝑥𝐶.𝑉 =0.289= 28.9 % (2M)

Overall efficiency ɳ𝑂= ɳ𝑃 𝑥 ɳ𝑡 = 0.169 = 16.9 % (1M)

5

The flight speed of a turbojet is 800 km/h at 10,000 m altitude. The density of air at that altitude

is 0.17 kg/m3. The drag force for the plane is 6.8 kN. The propulsive efficiency of the jet is 60

%. Calculate the SFC, Air-fuel ratio, Jet velocity. Assume the calorific value of fuel is 45000

kJ/kg and overall efficiency of the turbojet plane is 18%. (13 M) (April 2003) (Dec 2005) BTL5



𝐶𝑗+𝑢 ==== 𝐶𝑗 = 518.51 𝑚/𝑠 (3M)

Overall efficiency ɳ𝑂= 𝑇ℎ𝑟𝑢𝑠𝑡 𝑝𝑜𝑤𝑒𝑟

𝑃𝑜𝑤𝑒𝑟 𝑖𝑛𝑝𝑢𝑡 =

𝐹 𝑥 𝑈

�̇�𝑓𝑥 𝐶𝑉= �̇�𝑓 = 0.186

𝑘𝑔

𝑠 (3M)

Thrust F = �̇� 𝐶𝑗 − 𝑚𝑎̇ 𝑈𝑚𝑎̇ = 22.62𝑘𝑔

𝑠 (3M)

Air fuel ratio 𝑚𝑎̇

𝑚𝑓̇= 121.61 (2M)

Thrust specific fuel consumption TSFC of SFC = 𝑚𝑓/𝐹̇ =2.735 x 10−5 kg/s-N (2M)

6 An aircraft propeller flies at a speed of 440 kmph. The diameter of the propeller is 4.1 m and

the speed ratio is 0.8. The ambient conditions of air at the flight altitude are T=255 K and P=0.55



5-40

bar. Find the following: (i) Thrust (ii) Thrust power (iii) Propulsive efficiency. (13 M) (Nov-Dec

2013) BTL5


Area of the propeller disc = 13.20 m2

Velocity of the jet Cj= 152.77 m/s

Velocity of air flow at the propeller, C= 137.49 m/s (2M)

Mass flow rate of air-fuel mixture �̇� = 𝜌𝐴𝐶 = 1363.9 𝑘𝑔/𝑠̇ (2M)

The rate of air flow through propeller =�̇�𝑎= 1363.9 𝑘𝑔/𝑠̇ (2M)

Thrust produced=F=�̇�𝑎(𝐶𝑗 − 𝑈) = 41.667𝑥 103 𝑁 (2M)

Thrust power P = Thrust (F) x Flight speed (U) = 5.09 x 106 W (3M)


𝐶𝑗+𝑢= 0.88 = 88 % (2M)

7

An aircraft takes 45 kg/s of air from the atmosphere and flies at a speed of 950 kmph. The air

fuel ratio is 50 and the calorific value of the fuel is 42 MJ/kg. For maximum thrust power, find,

(i) Jet velocity (ii) Thrust (iii) Specific thrust (iv) Thrust power (v) Propulsive efficiency (vi)

Thermal efficiency (vii) Overall efficiency (viii) Thrust specific fuel consumption (TSFC). (13

M) (April-May 2014) BTL5


(i) Jet velocity, Cj= 527.76 m/s

(ii) Thrust, F=12.35 x 103 N (2M)

(iii) Specific thrust, 𝐹𝑠𝑝 = 269.06𝑁𝑘𝑔

𝑠

(2M)

(iv) Thrust power, P = 3.2 x 106 W (3M)

(v) Propulsive efficiency = 0.66=66 %

(vi) Thermal efficiency=0.126=12.6 % (3M)

(vii) Overall efficiency =0.083= 8.3 %

(viii) Thrust specific fuel consumption (TSFC)=7.29 x 10-5 kg/s-N (3M)

8

A turbo jet has a speed of 750 km/h while flying at an altitude of 10000 m. the propulsive

efficiency of the jet is 50% and overall efficiency of the turbine plant is 16%. The density of air

at 10000 m altitude is 0.173 kg/m3. Calculate (i) absolute velocity of the jet (ii) diameter of the

jet (iii) power output of the unit in kW. (13 M) (April-May 2012) BTL5




5-41


𝐶𝑗+𝑢 ==== 𝐶𝑗 = 624.99 𝑚/𝑠 (2M)

Velocity of jet 𝑐𝑎𝑏𝑠 = 416.66 m/s, (3M)

Thrust produced=F=�̇�(𝐶𝑗 − 𝑈) ==> �̇� = 15 𝑘𝑔/𝑠 (3M)

Diameter of the jet dj=0.42 m (2M)

Power output = 2.6 x106 W (3M)

Part * C

1

Derive the expression for the thrust, propulsive efficiency, thermal efficiency, overall efficiency

and the optimum value of flight to jet speed ratio for a turbojet engine. (15 M) BTL5


Thrust: The exit pressure is only equal to free stream pressure at some design condition. We must,

therefore, use the longer version of the generalized thrust equation to describe the thrust of the system.

If the free stream pressure is given by p0, the thrust F equation becomes:

F = m dot * Ve + (pe - p0) * Ae (5M)

Propulsive efficiency: In aircraft and rocket design, overall propulsive efficiency is the efficiency

with which the energy contained in a vehicle's propellant is converted into kinetic energy of the

vehicle. (5M)

Overall propulsive efficiency is the efficiency with which the energy contained in a vehicle's

propellant is converted into kinetic energy of the vehicle, to accelerate it, or to replace losses due to

aerodynamic drag or gravity. le, to accelerate it. (5M)

2

Explain with neat sketches the principle of operation of (i) Turbofan engine and (ii) Turbojet

engine. (15 M) BTL5


Turbofans were invented to circumvent an awkward feature of turbojets, which was that they were

inefficient for subsonic flight. To raise the efficiency of a turbojet, the obvious approach would be to

increase the burner temperature, to give better Carnot efficiency and fit larger compressors and

nozzles. However, while that does increase thrust somewhat, the exhaust jet leaves the engine with

even higher velocity, which at subsonic flight speeds, takes most of the extra energy with it, wasting

fuel. Instead, a turbofan can be thought of as a turbojet being used to drive a ducted fan, with both of

those contributing to the thrust. Whereas all the air taken in by a turbojet passes through the turbine

(through the combustion chamber), in a turbofan some of that air bypasses the turbine.

https://www.grc.nasa.gov/www/k-12/airplane/thrsteq.html

https://en.wikipedia.org/wiki/Aircraft_design_process

https://en.wikipedia.org/wiki/Propellant

https://en.wikipedia.org/wiki/Propellant

https://en.wikipedia.org/wiki/Aerodynamic_drag

https://en.wikipedia.org/wiki/Carnot_efficiency

https://en.wikipedia.org/wiki/Ducted_fan

https://en.wikipedia.org/wiki/Thrust

https://en.wikipedia.org/wiki/Turbojet

https://en.wikipedia.org/wiki/Combustion_chamber



5-42

Figure-Turbofan Engine (3M)

Because the turbine has to additionally drive the fan, the turbine is larger and has larger pressure and

temperature drops, and so the nozzles are smaller. This means that the exhaust velocity of the core is

reduced. The fan also has lower exhaust velocity, giving much more thrust per unit energy (lower

specific thrust). The overall effective exhaust velocity of the two exhaust jets can be made closer to a

normal subsonic aircraft's flight speed. In effect, a turbofan emits a large amount of air more slowly,

whereas a turbojet emits a smaller amount of air quickly, which is a far less efficient way to generate

the same thrust. (4M)

Turbojet engine:

Figure (3M)

All jet engines, which are also called gas

turbines, work on the same principle. The

engine sucks air in at the front with a fan. A

compressor raises the pressure of the air. The

compressor is made with many blades

attached to a shaft. The blades spin at high

speed and compress or squeeze the air. The

compressed air is then sprayed with fuel and

an electric spark lights the mixture. The burning gases expand and blast out through the nozzle, at the

back of the engine. As the jets of gas shoot backward, the engine and the aircraft are thrust forward.

As the hot air is going to the nozzle, it passes through another group of blades called the turbine. The

turbine is attached to the same shaft as the compressor. Spinning the turbine causes the compressor

to spin. (5M)

3

A ramjet engine operate at M=1.2 at an altitude of 6500 m. The diameter of inlet diffuser at

entry is 50 cm and the stagnation temperature at the nozzle entry is 1500 K. The calorific value

of the fuel is 40 MJ/kg. The properties of the combustion gases are same those of air (𝜸 =𝟏. 𝟒, 𝑹 = 𝟐𝟖𝟕 𝑱/𝒌𝒈𝑲). The velocity of the air at the diffuser exit is negligible. Calculate, (i) The

efficiency of the ideal cycle. (ii) Flight speed (iii) Air flow rate (iv) Diffuser pressure ratio (v)

Fuel air ratio (vi) Nozzle jet Mach number. The efficiencies of the diffuser =0.9, combustor

=0.98 and the nozzle =0.96. (13 M) BTL5

Figure- Turbo Jet Engine

https://en.wikipedia.org/wiki/Specific_thrust

https://www.grc.nasa.gov/www/k-12/airplane/turbine.html

https://www.grc.nasa.gov/www/k-12/airplane/turbine.html



5-43

Answer: Refer Class notes

(i) The efficiency of the ideal cycle =22.3 % (3M)

(ii) Flight speed, 377.4 m/s (3M)

(iii) Air flow rate, 46.228 kg/s (2M)

(iv) Diffuser pressure ratio, 2.239 (2M)

(v) Fuel air ratio=0.0303 (2M)

(vi) Nozzle jet Mach number=1.110 (3M)

4

Calculate the thrust and specific thrust of a jet propulsion unit whose data are as follows: Total

head isentropic efficiency of the compressor= 80%, total head isentropic efficiency=85%, total

pressure ratio including combustor pressure loss = 4:1, Combustion efficiency =98 %,

Mechanical transmission efficiency=99%, Nozzle efficiency =90%, Maximum cycle

temperature=1000K, Air rate of flow =220 N/s, For air Cp=1005 J/kgK, 𝜸 = 𝟏. 𝟒, For gases

Cp=1153 J/kgK, 𝜸 = 𝟏. 𝟑, Ambient temperature and pressure are 15ºC and 1 bar. Neglect the

weight of fuel. (15 M) BTL5


From Compressor efficiency T02=462.43 K (1M)

From Work input of the compressor T04 = 846.42 K (2M)

From Turbine efficiency P04=1.689 bar (3M)

From Exit velocity of the gas Cj= 447.40 m/s (3M)

Thrust of a jet propulsion =10.03 x 103 N (3M)

Specific thrust of a jet propulsion=447.40 N/(kg/s) (3M)

`



5-44



UNIT V SPACE PROPULSION

Types of rocket engines – Propellants-feeding systems – Ignition and combustion – Theory of rocket

propulsion – Performance study – Staging – Terminal and characteristic velocity – Applications – space

flights.

PART * A

Q.No. Questions

1.

What is bi-propellant? Give example. (Nov/Dec 2015) BTL2

If the fuel and oxidizer are different from each other in its chemical composition, then the propellant

is called bipropellant. Liquid Oxygen-Gasoline, Hydrogen peroxide-Hydrazine.

2

What is the role of inhibitors in rocket propulsion system? (Nov/Dec 2015) BTL3

They are used to regulate the burning of propellant.

Inhibitors or restrictors are used to regulate or prevent burning of the propellant grain or some

sections of its surface.

3

What is mono propellant? (April/May 2015) BTL2

A liquid propellant which contains both the fuel and oxidizer in a single chemical is known as

monopropellant.

4 Classify the rocket engines. (April/May 2015) BTL3

Chemical rocket engine, Solar rocket engine, nuclear rocket engine and Electrical rocket engine.

5

What are the properties of solid propellants? (Nov/Dec 2014) BTL2

It should have high density. It should not be poisonous and hazardous. It should be cheap and easily

available, storage and handling should be easy.

6 Define characteristic velocity. (Nov/Dec 2014) BTL4

C* = Cj/CF = Effective jet velocity / Thrust coefficient

7

Why rocket is called as non-air breathing engine? Can rocket work at vacuum? (May /June

2014) BTL5

Rocket engine carries its entire propellant (Oxidizer and Fuel) with it. Its performance does not

depend on atmospheric air like aircraft engines. Therefore they are called non-air breathing engines.

It can even work in vacuum.



5-45

8 What is the use of inhibitors in solid propellants? (May /June 2014) BTL2

They are used to regulate the burning of propellant.

9

What is mono-propellant? Give example. (May /June 2013) BTL2

A liquid propellant which contains both the fuel and oxidizer in a single chemical is known as

monopropellant. Example: 1.Nitroglycerine 2.Nitromethane

10

What are the types of rocket engines based on source of energy employed? (May /June 2013)

BTL3

Chemical rocket engine, Solar rocket engine, nuclear rocket engine and Electrical rocket engine.

11

Give the classification of rockets on the basis of number of stages (ii) basis of size and range.

BTL4

i. On the basis of number of stages:

a. Single stage rockets.

b. Multi stage rockets.

ii. On the basis of size and range

a. Short range small rockets

b. Long range large rockets

12

What are the components of rocket engine? BTL2

a. Container

b. Combustion

c. Exhaust nozzle

d. Control & navigational equipment

e. Pay load.

13

Give some liquid fuels used in Rocket propulsion system. BTL4

The liquid fuels are:

1. Liquid hydrogen

2. Unsymmetrical dimethyl – hydrazine (UDMH).

3. Hydrazine

4. Alcohol.

14

List out some oxidizers used in liquid propellant rockets. BTL4

a. Liquid oxygen

b. Red fuming Nitric acid (RFNA).

c. Liquid fluorine

d. WFNA

15

List out the types of fuels and oxidizers used in solid propellant rockets. BTL4

Fuels – Plastic or resin material

Oxidizer – Nitrates, perchlorates.

16

Name some fuel – oxidizer combination for hybrid propellant rockets. BTL4

Beryllium hydride (Be – H2) – Flourine (F2)

Lithium hydride (LiH) – Clorine tri fluoride (Cl F3)



5-46

Lithium Hydride (LiH) – nitrogen tetroxide (N2O4)

Hydro carbon (CH2) – Nitrogen tetroxide (N2O4)

17

Define Fin effectiveness. [Nov. ’96 M.U. Apr. 2001 M.U.] BTL2

Fin effectiveness is the ratio of heat transfer with fin to that without fin

Fin effectiveness =

with fin

without fin

Q

Q

18

Define hypergolic propellants. BTL2

For some combinations of fuels and oxidizer, ignition is not required to start the reaction;

merely the contact between the fuel and oxidizer starts the required reaction. Such propellants are

known as hypergolic propellants.

19

List out the main components of liquid propellant rocket engine. BTL4

The main components of liquid propellant rocket engine are:

i. Fuel and oxidizer tanks

ii. Gas pressure or turbo pump feed system

iii. Injectors

iv. Combustion chamber

v. Exhaust nozzle

20

Define cryogenic propellants. BTL2

The propellants which are in the gaseous state at normal temperature and require extremely

low temperature to maintain them in liquid state are known as cryogenic propellants.

21

List out some cryogenic propellants. BTL4

i. Liquid oxygen

ii.Hydrogen

iii. Flourine

iv. Ammonia

22

Define propellant grain. BTL1

A well – mixed fuel and oxidizer already present in thrust chamber is known as propellant grain.

23

List out the types of burning of solid propellants grain. BTL2

The types of burning of sold propellants are:

i. Regressive burning

ii. Neutral burning

iii. Progressive burning



5-47

24

Distinguish between solid and liquid propellants. BTL5

S.No Solid Propellants Liquid Propellants

1. Solid propellants are generally stored in

combustion chamber.

Liquid propellants are stored in separate fuel tanks and

supplied to cc.

2. Solid propellants are burnt at

uncontrolled rate

Liquid propellant are burnt at controlled rate.

25

Define (i) Weight flow coefficient (ii) Thrust co-efficient. BTL5

(i) Weight flow co-efficient:

It is the ratio of gas or propellant flow rate and force. (P0A*)

Wp

Cw = ________

P0A*

ii. Thrust co-efficient: ratio of thrust and force (p0A*)

F

Cw = ________

P0A*

26

Define Impulse to weight ratio. BTL2

Impulse to weight ratio is the ratio of total impulse of the rocket to the total weight of rocket

vehicle system.

It

IWR = _____

Wt

27

Define characteristics velocity. BTL2

Characteristics velocity is defined as the ratio of effective jet velocity to thrust coefficient.

Effective jet velocity Cj

V* = __________________________ = _____

Thrust coefficient Cf

28

Define booster in rockets. BTL2

The First stage lifts off the entire rocket vehicle system is known as booster.

29 Define sounding rockets. BTL2

Rockets meant for taking instruments to high altitudes for meteorological measurement are



5-48

called sounding rocket.

30

List out any 3 properties of good propellant. BTL4

i. It must have high calorific value.

ii. It must have high density

iii. It should be non – corrosive.

31

Name few merits of liquid propellants over solid propellants. BTL4

i. Liquid propellant engines can be reused after recovery.

ii. Higher values of specific impulse are obtained

iii. More flexible and greater control over the thrust.

iv. Easy to stop the operation in case of an impending catastrophe.

v. It is much easier to run an auxiliary power plant.

Part * B

1

A rocket nozzle has a throat area of 18 cm2 and combustor pressure of 25 bar. If the specific

impulse is 127.42 sec. and the rate of propellant is 44.145 N/s, determine the thrust coefficient,

propellant weight flow coefficient, specific propellant consumption and characteristic velocity.

(13 M) (May 2004) BTL5



�̇�𝑝𝑇ℎ𝑟𝑢𝑠𝑡 𝐹 = 5624.95 𝑁 (3M)

Thrust coefficient, 𝐶𝐹 =𝐹

𝑃0𝐴∗ = 1.25 (2M)

Propellant weight flow coefficient, 𝐶𝑤 =𝑊𝑃

𝑃0𝐴∗ = 9.81 𝑥 10−3 (2M)

Specific Propellant consumption = SPC = 𝑊𝑃

𝐹=

1

𝐼𝑆𝑃= 7.84 𝑥 10−3 𝑆−1 (2M)

Thrust F = �̇�𝑝𝐶𝑗 = 𝐶𝑗 = 1249.98𝑚

𝑠 (2M)

Characteristics velocity 𝐶∗ = 1000 𝑚/𝑠 (2M)

2

The specific impulse of a rocket is 125 s and the flow rate of propellant is 44 kg/s. The nozzle

throat area is 18 cm2 and the pressure in the combustion chamber is 25 bar. Determine the

throat coefficient, propellant flow coefficient, specific propellant consumption and

characteristic velocity. (13 M) (May 2005) BTL5



�̇�𝑝𝑇ℎ𝑟𝑢𝑠𝑡 𝐹 = 53950 𝑁 (2M)



5-49

Thrust coefficient, 𝐶𝐹 =𝐹

𝑃0𝐴∗ = 11.99 (2M)

Propellant weight flow coefficient, 𝐶𝑤 =𝑊𝑃

𝑃0𝐴∗ = 95.92 𝑥 10−3 (2M)


𝐹=

1

𝐼𝑆𝑃= 8 𝑥 10−3 𝑆−1 (3M)

Thrust F = �̇�𝑝𝐶𝑗 = 𝐶𝑗 = 1226.13𝑚

𝑠 (2M)

Characteristics velocity 𝐶∗ = 102.6 𝑚/𝑠 (2M)

3

In rocket engine, propellant flow rate is 5.2 kg/s, nozzle exit diameter is 9 cm, nozzle exit

pressure is 1.02 bar, atmospheric pressure is 1.013 bar, thrust chamber pressure is 22 bar and

thrust is 7.2 kN. Calculate the following (i) Effective jet velocity (ii) Actual jet velocity (iii)

Specific impulse (iv) Specific propellant consumption. (13 M)(Nov-Dec 2003) BTL5

Answer: Page 5.26- Dr.S.Senthil 2018.

Thrust F=�̇�𝑃 𝑥 𝐶𝑗 = ==> 𝐸𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑗𝑒𝑡 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝐶𝑗 = 1384.62 𝑚/𝑠 (3M)

Thrust F=�̇�𝑝𝐶𝑒 + (𝑃𝑒 − 𝑃𝑎)𝐴𝑒 Actual jet velocity 𝐶𝑒 = 1383.76 m/s (3M)


�̇�𝑝𝑔= 141.14 𝑆 (3M)


𝐹=

1

𝐼𝑆𝑃= 7.08 𝑥 10−3 𝑆−1 (4M)

4

A rocket flies at a speed of 10,000 kmph with an effective exhaust jet velocity of 1350 m/s and

the heat produced by the propellant is 6600 kJ/kg. If the propellant flow rate is 4.8 kg/s,

determine, (i) Propulsive efficiency (ii) Propulsive power (iii) Engine output (iv) Thermal

efficiency (v) Overall efficiency. (13 M) (May-June 2013) BTL5


Rocket speed = 2777.7 m/s, C.V = 6600 x 103 J/kg (1M)

Speed ratio 𝜎 =𝑢

𝐶𝑗= 2.06 (1M)

Propulsive efficiency = ɳ𝑃 =2𝜎

1+𝜎2= 0.785 = 78.5% (1M)

Thrust F = �̇�𝑝𝐶𝑗= 6480 N (1M)

Propulsive power P= F x u = 17.9 x 106 W (1M)

Propulsive efficiency ɳ𝑃 =𝑃𝑟𝑜𝑝𝑢𝑙𝑠𝑖𝑣𝑒 𝑝𝑜𝑤𝑒𝑟

𝑃𝑜𝑤𝑒𝑟 𝑜𝑢𝑡𝑝𝑢𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑛𝑔𝑖𝑛𝑒 (2M)

𝑃𝑜𝑤𝑒𝑟 𝑜𝑢𝑡𝑝𝑢𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑛𝑔𝑖𝑛𝑒 = 22.9 𝑥 106 𝑊 (2M)



5-50

Thermal efficiency ɳ𝑡 =𝑃𝑜𝑤𝑒𝑟 𝑜𝑢𝑡𝑝𝑢𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑛𝑔𝑖𝑛𝑒

𝑃𝑜𝑤𝑒𝑟 𝑖𝑛𝑝𝑢𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑛𝑔𝑖𝑛𝑒 = 0.722= 72.2 % (2M)

Overall efficiency ɳ𝑂 = ɳ𝑡 𝑥ɳ𝑝 = 0.567 = 56.7 % (2M)

5

A rocket engine has the following data: Effective jet velocity =1200 m/s, Flight to jet speed ratio

= 0.82, Oxidizer flow rate = 3.4 kg/s, Fuel flow rate = 1.2 kg/s. Heat of reaction per kg of the

exhaust gases =2520 kJ/kg, Calculate the following, (i) Thrust (ii) Specific impulse (iii)

Propulsive efficiency (iv) Thermal efficiency (v) Overall efficiency. (13 M) (April-May 2015)

BTL5


Propellant flow rate �̇�𝑃 = �̇�𝑂 + �̇�𝑓= 4.6 kg/s (1M)


𝐶𝑗= 2.06 Flight speed, u = 984 m/s, (2M)



�̇�𝑝𝑔= 122.32 𝑆 (2M)


1+𝜎2 = 0.98 = 98.0 % (2M)

Thermal efficiency ɳ𝑡 =𝐶𝑗

2+𝑢2

2 𝑥 𝐶𝑉= 0.477 = 47.7 % (2M)


6

A rocket has the following data: Propellant flow rate = 5 kg/s, Nozzle exit diameter = 10 cm,

Nozzle exit pressure = 1.02 bar, Ambient pressure = 1.013 bar, Thrust chamber pressure = 20

bar, Thrust = 7 kN. Determine (i) Effective jet velocity (ii) Actual jet velocity (iii) Specific

impulse (iv) Specific propellant consumption. (13 M) (May-June 2016) (May-June 2012) (Nov-

Dec 2009) BTL5


Thrust F=�̇�𝑃 𝑥 𝐶𝑗 = ==> 𝐸𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑗𝑒𝑡 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝐶𝑗 = 1400 𝑚/𝑠 (3M)

Thrust F=�̇�𝑝𝐶𝑒 + (𝑃𝑒 − 𝑃𝑎)𝐴𝑒 Actual jet velocity 𝐶𝑒 = 1399.9 m/s (4M)


�̇�𝑝𝑔= 142.71 𝑆 (3M)


𝐹=

1

𝐼𝑆𝑃= 7 𝑥 10−3 𝑆−1 (3M)

7

The effective jet velocity from a rocket is 2700 m/s. The forward flight velocity is 1350 m/s and

the propellant consumption is 78.6 kg/s, Calculate the thrust, thrust power, propulsive

efficiency. (13 M) (Nov 199) BTL5



5-51


Thrust F = �̇�𝑝𝐶𝑗= 212.22 N (3M)

Propulsive power P= F x u = 286.49 x 106 W (3M)


𝐶𝑗= 2.06 𝜎 = 0.5 (3M)


1+𝜎2 = 0.80 = 80 % (4M)

8

A rocket engine has the following data: Effective jet velocity =1250 m/s, Flight to jet speed ratio

= 0.8, Oxidizer flow rate = 3.5 kg/s, Fuel flow rate = 1 kg/s. Heat of reaction per kg of the exhaust

gases =2500 kJ/kg, Calculate the following, (i) Thrust (ii) Specific impulse (iii) Propulsive

efficiency (iv) Thermal efficiency (v) Overall efficiency. (13 M) (April-May 2015) BTL5


Propellant flow rate �̇�𝑃 = �̇�𝑂 + �̇�𝑓= 4.5 kg/s (1M)


𝐶𝑗= 0.8 Flight speed, u = 1000 m/s, (2M)



�̇�𝑝𝑔= 127.42 𝑆 (2M)


1+𝜎2 = 0.975 = 97.5 % (2M)

Thermal efficiency ɳ𝑡 =𝐶𝑗

2+𝑢2

2 𝑥 𝐶𝑉= 0.5125 = 51.25 % (2M)


9

Determine the escape and orbital velocities of a rocket for the given conditions. (a) at mean sea

level (b) at an altitude 650 km. Take radius of earth at mean sea level is 6341.6 km. Acceleration

due to gravity at mean sea level =9.809 m/s2. (13 M) (May-June 2013) BTL5


At, Z=0, 𝑈𝑜𝑟𝑏 = √𝑔0𝑅0 =7887 m/s (4M)

𝑈𝑒𝑠𝑐 = √2 𝑈𝑜𝑟𝑏 = 11153.902 𝑚/𝑠 (3M)

At, Z=650, 𝑈𝑜𝑟𝑏 = 𝑅0√𝑔0

𝑅0+𝑍 =7511.435 m/s (3M)

𝑈𝑒𝑠𝑐 = √2 𝑈𝑜𝑟𝑏 = 10622.773 𝑚/𝑠 (3M)

Part *C



5-52

1

Write short note on: (i) Theory of rocket propulsion (ii) Types of rocket engine. (15 M) (May-

June 2016) BTL4


Thrust is the force that propels a rocket or spacecraft and is measured in

pounds, kilograms or Newton’s. Physically speaking, it is the result of

pressure which is exerted on the wall of the combustion chamber.

Figure shows a combustion chamber with an opening, the nozzle,

through which gas can escape. The pressure distribution within the

chamber is asymmetric; that is, inside the chamber the pressure varies

little, but near the nozzle it decreases somewhat. The force due to gas

pressure on the bottom of the chamber is not compensated for from the

outside. The resultant force F due to the internal and external pressure

difference, the thrust, is opposite to the direction of the gas jet. It

pushes the chamber upwards.

To create high speed exhaust gases, the necessary high

temperatures and pressures of combustion are obtained by using a

very energetic fuel and by having the molecular weight of the

exhaust gases as low as possible. It is also necessary to reduce the pressure

of the gas as much as possible inside the nozzle by creating a large section ratio. The

section ratio, or expansion ratio, is defined as the area of the exit Ae divided by the area of the throat

At.

The thrust F is the resultant of the forces due to the pressures exerted on the inner and outer walls by

the combustion gases and the surrounding atmosphere, taking the boundary between the inner and

outer surfaces as the cross section of the exit of the nozzle. As we shall see in the next section,

applying the principle of the conservation of momentum gives

where q is the rate of the ejected mass flow, Pa the pressure of the ambient atmosphere, Pe the pressure

of the exhaust gases and Ve their ejection speed. Thrust is specified either at sea level or in a vacuum.

. (6M)

https://www.sciencelearn.org.nz/resources/393-types-of-chemical-rocket-engines

Solid propellant rocket engines: In a solid fuel engine, the fuel and oxidizer are already mixed

together and set as a solid inside the combustion chamber. This solid is called the propellant grain.

The rate at which the chemical reaction takes place depends on the type of fuel chosen and the surface

area of the exposed grain. The inside length is normally hollow section to increase the amount of

grain exposed and available to react. A star-shaped hollow section is often used to maintain a steady

burn with even thrust. (3M)

Liquid propellant rocket engines

Figure-Rocket engine

https://www.sciencelearn.org.nz/resources/393-types-of-chemical-rocket-engines



5-53

Liquid propellant rocket engines use a liquid fuel (such as liquid hydrogen or kerosene) and liquid

oxidizer (such as liquid oxygen). These are stored in separate tanks and then pumped into the

combustion chamber as required. As they are sprayed into the combustion chamber through injection

nozzles, they rapidly mix together and react before being ejected. (3M)

Hybrid propellant rocket engines

A hybrid propellant system has the fuel as a solid inside the combustion chamber. The liquid oxidizer

is stored in a separate tank. The simplest hybrid system is to have the oxidizer under pressure in its

tank. When a valve is opened, this oxidizer is released into the combustion chamber. It then reacts

with the solid fuel before being ejected. (3M)

2

Draw neat sketches and explain the multi stage rocket vehicle system. (15 M) (April-May 2003)

BTL4


A multistage rocket, or step rocket is a launch vehicle that uses two

or more rocket stages, each of which contains its own engines and

propellant. A tandem or serial stage is mounted on top of another

stage; a parallel stage is attached alongside another stage. The

result is effectively two or more rockets stacked on top of or

attached next to each other. Two-stage rockets are quite common,

but rockets with as many as five separate stages have been

successfully launched. By jettisoning stages when they run out of

propellant, the mass of the remaining rocket is decreased. Each

successive stage can also be optimized for its specific operating

conditions, such as decreased atmospheric pressure at higher

altitudes. This staging allows the thrust of the remaining stages to

more easily accelerate the rocket to its final speed and height.

In serial or tandem staging schemes, the first stage is at the bottom

and is usually the largest, the second stage and subsequent upper

stages are above it, usually decreasing in size. In parallel staging

schemes solid or liquid rocket boosters are used to assist with

launch. These are sometimes referred to as "stage 0". In the typical

case, the first-stage and booster engines fire to propel the entire

rocket upwards. When the boosters run out of fuel, they are detached from the rest of the rocket

(usually with some kind of small explosive charge) and fall away. The first stage then burns to

completion and falls off. This leaves a smaller rocket, with the second stage on the bottom, which

then fires. Known in rocketry circles as staging, this process is repeated until the desired final velocity

is achieved. In some cases with serial staging, the upper stage ignites before the separation—the inter

stage ring is designed with this in mind, and the thrust is used to help positively separate the two

vehicles. (15M)

3

A rocket has the following data: Combustion chamber pressure = 36 bar, combustion chamber

temperature = 3600 K, Oxidizer flow rate = 41 kg/s, Mixture ratio= 5, Ambient pressure = 585

N/m2, Determine (i) Nozzle throat area, (ii) Thrust, (iii) Thrust coefficient (iv) Characteristic

velocity (v) Exit velocity of exhaust gases. Take, 𝜸 = 𝟏. 𝟑 𝒂𝒏𝒅 𝑹 = 𝟐𝟖𝟕 𝑱/𝒌𝒈𝑲. (13 M)(Nov-Dec

2014) BTL6

Figure- Liquid propellant rocket

engine

https://en.wikipedia.org/wiki/Launch_vehicle

https://en.wikipedia.org/wiki/Rocket

https://en.wikipedia.org/wiki/Rocket_engine

https://en.wikipedia.org/wiki/Rocket_propellant

https://en.wikipedia.org/wiki/Newton%27s_First_Law_of_Motion

https://en.wikipedia.org/wiki/Solid_rocket_booster

https://en.wikipedia.org/wiki/Liquid_rocket_booster

https://en.wikipedia.org/wiki/Explosive



5-54


(i) Nozzle throat area , A*= 0.0129 m2 (3M)

(ii) Thrust, F=137.68 x 103 N (3M)

(iii) Thrust coefficient, 𝐶𝐹 = 2.96 (3M)

(iv) Characteristic velocity, C*= 945.45 m/s (3M)

(v) Exit velocity of exhaust gases, Ce= 2798.56 m/s (3M)

4

With the help of neat sketches explain any two arrangements used for fuel feeding in liquid

propellant rocket systems. (15 M) (April-May 2011) BTL4


Figure (5M)

The common liquid rocket is bipropellant; it

uses two separate propellants, a liquid fuel and

liquid oxidizer. These are contained in separate

tanks and are mixed only upon injection into

the combustion chamber. They may be fed to

the combustion chamber by pumps or by

pressure in the tanks Propellant flow rates must

be extremely large for high-thrust engines,

often hundreds of gallons per second. Pump-

fed systems may require engines delivering

several thousand horsepower to drive the

pumps.4 This power is usually developed by a

hot gas turbine, supplied from a gas generator

which is actually a small combustion chamber.

The main rocket propellants can be used for the gas generator. (10M)

5

Describe with a schematic diagram the principle of working of a liquid propellant rocket. (15

M) (April-May 2011) BTL4


Figure - liquid propellant rocket (5M)

A liquid-propellant rocket or liquid rocket is a rocket engine that uses liquid propellants. Liquids are

desirable because their reasonably high density allows the volume of the propellant tanks to be

https://history.nasa.gov/conghand/propulsn.htm#REF5-4

https://en.wikipedia.org/wiki/Rocket_engine

https://en.wikipedia.org/wiki/Liquid_rocket_propellant



5-55

relatively low, and it is possible to use lightweight centrifugal turbo pumps to pump the propellant

from the tanks into the combustion chamber, which means that the propellants can be kept under low

pressure. This permits the use of low-mass propellant tanks, resulting in a high mass ratio for the

rocket. (5M)

An inert gas stored in a tank at a high pressure is sometimes used instead of pumps in simpler small

engines to force the propellants into the combustion chamber. These engines may have a lower mass

ratio, but are usually more reliable, and are therefore used widely in satellites for orbit maintenance.

Liquid rockets can be monopropellant rockets using a single type of propellant, bipropellant rockets

using two types of propellant, or more exotic tri propellant rockets using three types of propellant.

Some designs are throttle able for variable thrust operation and some may be restarted after a previous

in-space shutdown. Liquid propellants are also used in hybrid rockets, in which a liquid oxidizer is

generally combined with a solid fuel. (5M)

https://en.wikipedia.org/wiki/Turbopump

https://en.wikipedia.org/wiki/Mass_ratio

https://en.wikipedia.org/wiki/Monopropellant_rocket

https://en.wikipedia.org/wiki/Tripropellant_rocket

https://en.wikipedia.org/wiki/Rocket_engine_throttling

https://en.wikipedia.org/wiki/Hybrid_rocket

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