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S.K.P. Engineering College, Tiruvannamalai VI SEM

Mechanical Engineering Department 1 Gas Dynamics And Jet Propulsion

SKP Engineering College

Tiruvannamalai – 606611

A Course Material

on

Gas Dynamics And Jet Propulsion

By

Dr.J.Kuberan, Dr.K.Venkatesan, Mr .N.Sarankumar,

Mr .G.Karthikeyan

Mechanical Engineering Department



Quality Certificate

This is to Certify that the Electronic Study Material

Subject Code: ME 6604

Subject Name: Gas Dynamics And Jet Propulsion

Year/Sem: III / VI

Being prepared by Dr.J.Kuberan, Dr.K.Venkatesan, Mr .N.Sarankumar,

Mr .G.Karthikeyan and it meets the knowledge requirement of the University curriculum.

Signature of the Author

Name:Dr.J.Kuberan, Dr.K.Venkatesan, Mr .N.Sarankumar, Mr .G.Karthikeyan

Designation: HD, AP,AP,AP

This is to certify that the course material being prepared by Dr.J.Kuberan, Dr.K.Venkatesan, Mr .N.Sarankumar and Mr .G.Karthikeyan is of the adequate quality. He has referred more than five books and one among them is from abroad author.

Signature of HD Signature of the Principal

Name: Dr.J.Kuberan Name: Dr.V.Subramania Bharathi

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ME 6604 GAS DYNAMICS AND JET PROPULSION

Lecture : 3 hrs/Week Internal Assessment: 20 Marks

Tutorial : 1hr/week Final Examination: 80 Marks

Practical : - Credits: 4

TYPE OF COURSE: Required course

ASSESSMENT METHOD: Tutorial classes, 2 internal tests, 1 Model examination and Course end university examination. PREREQUISITE: Fluid Mechanics and Machinery, Applied Thermodynamics, Heat and Mass Transfer and Thermal Engineering COURSE OBJECTIVES:

To study the basic difference between incompressible and compressible flow

To study about the phenomenon of shock waves and its effect on flow.

To gain some basic knowledge about the jet propulsion Rocket propulsion COURSE OUTCOMES: Upon completion of this course the student will be familiar with:

CO1 To understand the basic fundamentals of compressible and incompressible flow concepts

CO2 To analyze the effect of friction (Fanno flow) and heat transfer (Rayleigh flow) in compressible flow and solve Fanno and Rayleigh flow problems

CO3 To develop the concept of shock waves and its effects

CO4 To implement propulsive methods, concept of air-craft propulsion system.

CO5 To execute propulsive methods, concept of rocket propulsion system,

CO-PO MAPPING

CO/PO PO1 PO2 PO3 PO4 PO5 PO6 PO7 PO8 PO9 PO10 PO1

1

PO12

CO1 3 3 3

CO2 3 3 1 3 3

CO3 3 3 3 2 3

CO4 3 3 3 3 3 2 3

CO5 3 3 3 3 3 2 3



SYLLABUS UNIT I BASIC CONCEPTS AND ISENTROPICFLOWS (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. UNITV 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 CONTENT BEYOND SYLLABUS

Axial flow compressor

Alternative propulsion method

Magneto plasma rocket engine

Variable Specific Impulse Magneto plasma Rocket



LEARNINGRESOURCES: TEXT BOOKS:

1. Anderson, J.D., Modern Compressible flow, McGraw Hill, 3rd Edition, 2003. 2. H. Cohen, G.E.C. Rogers and Saravanamutto, Gas Turbine Theory, Longman

Group Ltd., 1980.

3. S.M. Yahya, fundamentals of Compressible Flow, New Age International (P) Limited, New Delhi, 1996..

REFERENCES:

1. P. Hill and C. Peterson, Mechanics and Thermodynamics of Propulsion, Addison –Wesley Publishing company, 1992.

2. N.J. Zucrow, Aircraft and Missile Propulsion, vol.1 & II, John Wiley, 1975. 3. N.J. Zucrow, Principles of Jet Propulsion and Gas Turbines, John Wiley, New

York,1970. 4. G.P. Sutton, Rocket Propulsion Elements, John wiley, 1986, New York. 5. A.H. Shapiro, Dynamics and Thermodynamics of Compressible fluid Flow, , John

wiley, 1953, New York.

6. V. Ganesan, Gas Turbines, Tata McGraw Hill Publishing Co., New Delhi, 1999. 7. PR.S.L. Somasundaram, Gas Dynamics and Jet Propulsions, New Age

International Publishers, 1996.

8. V. Babu, Fundamentals of Gas Dynamics, ANE Books India, 2008. WEB RESOURCES:

1. www.eia.doe 2. www.google advance 3. www.schako.de

ADDITIONAL RESOURCES :

1. NPTEL TUTORIALS (Internal Server) 2. Electronic Books (PDF Formats) 3. Online Objective Questions 4. Videos Materials if any (You tube)



CONTENTS

S.No Particulars Page

1 Unit – I 8

2 Unit – II 63

3 Unit – III 111

4 Unit – IV 146

5 Unit – V 184



UNIT I

BASIC CONCEPTS AND ISENTROPIC FLOWS

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

Machwaves and Mach cone – Effect of Mach number on compressibility – Isentropic

flowthrough variable ducts – Nozzle and Diffusers – Use of Gas tables.

PART A

1. Define mach angle. [ CO1 - L1 – Apr 2011]

The angle between mach line and the direction of motion of the body is known as mach

angle.

2. Define Crocco number. [ CO1 - L1 – Apr 2011]

It is defined as ratio between fluid velocity © and its maximum fluid velocity (Cmax)

CR = C/Cmax

3. Define mach cone [ CO1 - L1 – May 2012]

Tangents drawn from source point on the spheres define a conical surface

referred to as mach cone.

4. Define stagnation enthalpy [ CO1 - L1 – Nov 2009]

Stagnation enthalpy can be defined as enthalpy of gas when it is isentropically

decelerated to zero velocity at zero elevation,

Ho = h + C2/2 5. What are advantages of using M*?[ CO1 - L1 – Nov 2009]

It is more convenient to use M* instead of M due to following reasons

At high velocities M approaches infinity but M* gives a finite value

M* = √((γ+1)/(γ-1))



M is proportional to fluid velocity (c) and sound velocity (a) but M* is proportional

to fluid velocity alone.

6. Define closed and open system [ CO1 - L1 – Apr 2006]

A closed system does not permit and mass number only energy transfer take

place.

In open system both mass and energy

7. Define characteristic of Mach number (M*) [ CO1 - L1 – Apr 2006]

It is defined as ratio between fluid (C) and critical velocity of sound (a*)

M* = C/a*= C/C* (i.e., C* =a*)

8. Define stagnation Pressure [ CO1 - L1 – May 2004]

Stagnation pressure is the pressure of gas when it is isentropically decelerated to

zero velocity at zero elevation.

9. Define Adiabatic Process [ CO1 - L1 – May 2004]

In an adiabatic process there is no heat transfer between system and surrounding.

i.e., Q = 0 10. What is meant by transonic flow? [ CO1 - L1 – Apr 1998]

If the fluid velocity (c) close to speed of sound (a) that type of flow is known as

transonic flow.

Mach number is in between 0.8 to 1.2

11. What is difference between extensive and intensive properties. [ CO1 – L2 – Apr

1998]

Intensive properties

These properties are independent on mass of system.

Ex: Pressure, Temperature, etc., Extensive Properties

These properties are dependent on mass of system.

Ex: Total volume



12. Define mach wave and normal shock [ CO1 - L1 – May 2013]

Mach wave These lines at which pressure difference is concentrated and which generate

cone are called mach number or mach waves. Normal shock

A shock wave is nothing but a steep finite pressure wave. The shock wave is right angle to flow is called normal shock.

13. What is impulse function? [ CO1 - L1 – May 2013]

The sum of pressure force (ρ A) and impulse force (ρ AC2)

14. Define hypersonic flows [ CO1 - L1 – Apr 2011]

In hypersonic flow region fluid velocity (c) is much greater than sound velocity (a)

.In this flow mach number value is always greater than 5

M = C/a > 5 15. Zone of silence is absent in subsonic flow ?why ? [ CO1 - L1 – May 2012]

In sonic flow source of disturbance velocity (c) is equal to the sound velocity (a)

under this conduction sound waves always exit at the present position of point source

and cannot move ahead of it. Therefore zone of lying on left of source of disturbance (S)

is called zone of silence because waves do not reach this zone.

16. What are the basic differences between compressible and incompressible flow?

[ CO1 – L2 – May 2013]

Compressible Incompressible

1. Fluid velocities are appreciable compared with the velocity of sound

2.Density is not constant

3. Compressibility factor is greater

than one.

1. Fluid velocities are small

compared with the velocity of sound

2. Density is constant

3. Compressibility factor is one.



17. Name the four reference velocities that are used in expressing the third velocities

in non- dimensional form. [ CO1 - L1 – May 2013]

1. Local velocity of sound a =√ γRT

2. Stagnation velocity of sound a0= √γRT0

3. Maximum velocity of sound Cmax=a0 2/γ-1

4. Critical velocity of sound / fluid a* = c* = √γRT*

18. Express the stagnation enthalpy in terms of static enthalpy and velocity of flow.

[ CO1 - L1 – Dec 2009]

Stagnation enthalpy, h0=h+C2/2 Where,

h-Static enthalpy-J/kg C-Velocity of fluid-m/s

19. When air is released adiabatically from a tyre, the temperature of air at the nozzle

exit is 370C below that of air inside the tyre. Neglecting irreversibility, calculate

the exit velocity of air. [ CO1 - L1 – Dec 2009]

Exit velocity of air, C= √2Δh = √2cp ΔT = √2x1005x37 C = 272.70 m/s

20. What is subsonic, sonic and supersonic flow with respect to Mach number?

[ CO1 - L1 – May 2011 &2012]

Subsonic flow The subsonic flow region is on the right of the incompressible flow region. The subsonic flow, fluid velocity (c) is less than the sound velocity (a) and the Mach number in this region is always less than unity. M=c/a < 1.

Sonic flow If the fluid velocity (c) is equal to the sound velocity (a) that type of flow is known as sonic flow. In sonic flow Mach number value is unity. M=c/a = 1, c = a Supersonic flow The supersonic region is on the right of the transonic flow region. In supersonic flow, fluid velocity (c) is more than the sound velocity (a) and the Mach number in this region is always greater than unity.

M=c/a > 1.



21. How the area and velocity vary in subsonic flow of nozzle and diffuser.

[ CO1 – L2 – May/June 2011&2014]

Nozzle A nozzle is a device that increases the velocity of a fluid at the expense of pressure. Diffuser A diffuser is a device that increases the pressure of a fluid by slowing it down.

For nozzle: Area - Decrease

Velocity - Increases For diffuser: Area - Increases

Velocity – Decrease

22. Explain the meaning of stagnation state with example. [ CO1 – L2 – May/June 2014]

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

Eg: Fluid in a reservoir (or) in a settling chamber.

23. Distinguish between static and stagnation pressures. [ CO1 – L2 – May 2016]

In stagnation pressure state, the velocity of the flowing fluid is zero whereas in the static pressure state, the fluid velocity is not equal to zero.

24. What is meant by isentropic flow with variable area? [ CO1 - L1 – May 2016]

A steady one dimensional isentropic flow in a variable area passages is called “variable area flow”. The heat transfer is negligible and there is no other irreversibility due to fluid friction, etc.



25. Write the equation for efficiency of the diffuser. [ CO1 - L1 – May 2013]

Diffuser efficiency =

=

26. What is impulse function and give its uses? [ CO1 - L1 – May 2013]

Impulse function is defined as the sum of pressure force and inertia force. Impulse function F = Pressure force ρA + inertia force ρAc2

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 flowing fluid between two sections can be obtained by using change in impulse function.

27. What is chocked flow? State the necessary conditions for this flow to occur in

anozzle. [ CO1 - L1 – Nov 2007]

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 ratio must be equal to the critical pressure ratio where the mach number M = 1.

28. Derive the maximum fluid velocity. [ CO1 – L2 – Nov 2007]

The fluid velocity (Cmax) corresponding the condition of h = o, C = Cmax, is the maximum velocity that would be achieved by the fluid when it is accelerated to absolute zero temperature.

29. Define Mach number? [ CO1 - L1 – Apr/May-2015]

The Mach number is an index of the ratio between inertia force and elastic force.

M ² = Inertia force ⁄ Elastic force

It is also defined as the ratio of the fluid velocity (c) to the velocity of sound (a).

M = c/a

2 1

'

2 1

P P

P P



30.Define the critical velocity of sound and prove that c* = {2 Cp (To – T*)1/2

[ CO1 - L1 – Apr/May-2015]

The critical velocity of a fluid is its velocity at a Mach number of unity. M-critical = c*/a*; c* = a* = 1

CpTo= CpT* + ½ c*2

Therefore, c* = {2 Cp (To – T*)1/2

31. What do you understand by adiabatic energy equations? Give the equations.

[ CO1 - L1 – May 2012]

The adiabatic energy equation is the equation derived from the energy equation for a flow process with q = 0. The heat transfer during the process such as expansion in gases and vapours in turbines are negligibly small.

h1 + ½ c1 = h2 + ½ c2

32. Distinguish between Mach wave and Normal shock. [ CO1 – L2 – Nov/Dec 2013] Mach wave The lines at which the pressure difference is concentrated and which generate the cone are called mach lines or Mach waves Normal shock Shockwave is nothing but a steep finite pressure wave. When the shock wave is right angle to the flow, it is called normal shock. 33. Find the sonic velocity through air at 00C [ CO1 - L1 – Nov/Dec 2013] At 0°C is ρ0 = 1.293 kg/m3, Z0 = 428 N·s/m3, and c0 = 331 m/s , The speed of sound in air c is determined by the air itself and is not dependent upon the amplitude, To get the speed of sound the temperature is important, not the barometric pressure. in air in m/s vs. temperature ϑ (theta) in degrees Celsius 34. Differentiate nozzle and diffuser? [ CO1 – L2 – May 2014& Nov 2007]

Nozzle

It is a device which is used to increase the velocity with pressure drop in the

fluids.

Diffuser It is a device which is used to increase the pressure with velocity drop in

the fluids.



35. When does the maximum mass flow occur for an isentropic flow with variable

area duct? What type of nozzle used for sonic flow and supersonic flow?

[ CO1 - L1 – May/June-14 & Nov/Dec-2014 ]

Mass flow rate will be maximum at throat section where the mach number is

one.

Convergent and divergent nozzles are used for supersonic flows

36. Name the different regions of compressible fluid flow. [ CO1 - L1 – Nov/Dec-2014] 1.Incompressible flow region 2.Subsonic flow region 3.Transonic flow region 4.Supersonic flow region 5.Hypersonic flow region

37.Find the maximum possible velocity in a medium of air when static temperature is 200°C and velocity is 120 m/s. [ CO1 – L2 – Dec 2014]

Solution: T=200°C+273=473 K

C= 120m/s

2

2

2

max

3

max

max

1.4 287 473 1 1120

1.4 1 2 2

1482.32 10

2

982.16

c

c

mc

s

38.What is meant by Gas dynamics? [ CO1 - L1 – Apr/May-2015]

Gas dynamics deals with the study of motion of gases and its effects.



PART-B

1. Air ( γ = 1.4 , R 287 J/Kg k) 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.

Stagnation temperature

Maximum velocity

Mass flow rate

Area of cross section at exit [ CO1 - H1 – Apr/May-2008]

Given data:

γ= 1.4 R = 287 J/Kg K T1 = 300 k P1 = 3.45 bar C 1= 150 m/s P2 = 2.058 bar = 2.058 * 10 5 N/m2 C2 = 260 m/s A1 = 500 cm2 = 500 * 10 -4 m 2

To find:

1. Stagnation temperature

2. Maximum velocity

3. Mass flow rate

4. Area of cross section at exit

Solution:

Sound velocity at inlet, a1 = √( γRT1) = √(1.4*287*300) = 347.19 m/s Inlet mach number, M1 = C1/a1 = 150/347.19

M1 = 0.432 Refer isentropic flow table for γ=1.3 and M1 = 0.432



T1/T01 = 0.964 To 1 = T1/0.964 = 311.2 K

Stagnation temperature, T01 = 311.2 k

For isentropic flow, stagnation temperature remains constant To = To 1 = To 2 =311.2 K

From stagnation enthalpy equation we know

h0 = a2/(γ-1) + (1/2 C2) = ½ Cmax

2 = (ao

2/( γ-1))

½ Cmax2=( ao

2/( γ-1)) Cmax

2= 2ao2/ γ-1

= 2* (γ R T0) / (γ-1) = (2*1.4*287*311.2)/(1.4-1)

Cmax2 = 625.2 * 103

Cmax =760.69 Mass flow rate,

.

m = ρ A C= ρ1 A1 C1= ρ2 A2 C2

= ρ1 A1 C1

= (3.45*10^5)/ (287*300) * 500 * 10^-4 * 150 = 30.05 kg/s

.

m = ρ2 A2 C2

= P2/RT2 * A2*C2

30.05 =( 2.058*10^5)/(287*277)* A2 * 260

A2 = 0.0446 m2

Result: 1. To = 311.2 K

2. Cmax = 760.69 m/s

3. .

m = 30.05 kg/s

4. A2 = 0.0446 m2

5.



2. An air jet (γ = 1.4, R = 287 J/Kg k) at 400 K has sonic velocity

[ CO1 - H1 – Apr/May-2015]

Determine 1) Velocity of sound at 400 K

2) Velocity of sound at stagnation condition

3) Maximum velocity of jet

4) Stagnation enthalpy

5) Crocco number

Given:

γ = 1.4

R= 287 J/Kg k

T = 400 K

At sonic condition,

Mach number, M =1

M = C/a = 1

C=a

To find:

1) Velocity of sound at 400 K (a)

2) Velocity of sound at stagnation condition (a0)

3) Maximum velocity of jet (Cmax)

4) Stagnation enthalpy (ho)

5) Crocco number (C r)

Solution:

We know that,



velocity of sound, a = √( γRT1)

= √(1.4*287*400)

= 400.8 m/s

Stagnation temperature – mach number relation,

(To/T) = 1+( (γ-1)/2 M2)

(To/T) = 1+( (γ-1)/2 (1)2)

(To/400) = 1+((1.4-1)/2)

To = 480 K\

Velocity of sound at stagnation condition

ao = √( γRT0)

= √(1.4*287*480)

ao= 439.16 m/s

From stagnation enthalpy equation we know that

h0 = a2/(γ-1) + (1/2 C2)

= ½ Cmax2 = (ao

2/( γ-1))

½ Cmax2=( ao

2/( γ-1))

Cmax2= 2ao

2/ γ-1

= (2* (43916)2/ (1.4-1))

= 963.10^3

Maximum fluid velocity,Cmax = 981.9 m/s

From stagnation enthalpy equation,

h0=½ Cmax2

= (1/2) (981.9)2

= 482.06 * 103 J/Kg



Crocco number, Cr = C/Cmax = (400.8/981.9) (at sonic condition )

M=1

M= C/a= 1

C = a

Result

1. a = 400.8 m/s

2. a0 = 439.17 m/s

3. Cmax= 981.9

4. ho = 482.06 * 10 ^3

5. Cr = 0.408

A.

3. The jet of gas at 593 K (γ 1.3, R = 469 J /Kg K) has a mach number of 1.2.

Determine for local and stagnation condition of velocity of sound and enthalpy.

What is maximum attainable velocity of this jet

[ CO1 – H1 – Apr/May-2011]

Given:

T = 593 K

γ = 1.3

R = 469J/Kg K

To find:

a) Local and stagnation velocity of sound and enthalpy a, a0, h, h0

b) Maximum velocity of this jet, Cmax.

Solution:

1. Local velocity of sound (a)

Velocity of sound, a = √(γRT) = √(1.4*469*593) = 601.29 m/s 2. Stagnation temperature – mach number relation

(To/T) = 1+( (γ-1)/2 M2) (To/T) = 1+( (γ-1)/2 (1.2)2)



(To/593) = 1+((1.3-1)/2) To = 721.08 K

To be calculated by using gas tables,

Refer isentropic flow table for, γ= 1.3, M = 1.2

(T/To)= 0.822

To= 721.4 K


a0 = √( γRT0)

= √ (1.3* 469* 721.08)

a0 = 663.05 m/s

3. Static enthalpy,

h= CpT

= ((γR)/(γ-1)) * T

= ((1.3*469)/(1.3-1))* 593

h = 1205*10^3 J/Kg

4. Static enthalpy, ho= CpTo

= ((γR)/(γ-1)) * To

= ((1.3*469)/(1.3-1))* 721.08

ho= 1465*10^3 J/ Kg

From stagnation enthalpy equation, we know that

h0= a2/(γ-1) + (1/2 C2)

= ½ Cmax2

= (ao2/( γ-1))

Cmax2 = 2ao

2/ γ-1

= (2* (663.05)2/ (1.3-1))

Maximum velocity,Cmax = 1711.9 m/s



4. Discuss various regions of flow using adiabatic energy equation for a perfect gas

in C-a plot

[ CO1 –H2 – Apr/May-2011]

Adiabatic energy equation for a perfect gas is derived in terms of velocity of fluid

(c) and velocity of sound (a) this is then discussed graphically by differentiating various regions of flow on the c –a plot.

WKT,

Stagnation enthalpy, ho= h + (1/2)C2 Static enthalpy, h =CpT

= ((γR)/( γ – 1)) * T ( i.e., Cp=(γR)/( γ – 1)) ho= (γRT)/ ( γ – 1)

Velocity of sound, a = √(γRT) a2 = (γRT) Substitute (γRT) value in equation, ` h =a 2/ ( γ – 1) Substitute “ h “ value in stagnation enthalpy

ho=a 2/ ( γ – 1) + (1/2 C2) -----------------------------------(1) At case 1

T = 0, h = 0 , c = Cmax Substitute these values in stagnation equation

ho= (1/2 Cmax2) ---------------------------------------(2)

At case 2 C = 00, a = a0

Substitute these values in stagnation equation ho = h = a 2/ ( γ – 1) ho= ao

2/ ( γ – 1) --------------------------------------(3)

5. The jet, of a gas at 598K (γ=1.3,R=469 J/kg-K) has a Mach number of 1.2.

Determine for local and stagnation conditions the velocity of sound and

enthalpy. What is the maximum attainable velocity of this jet?

[ CO1 –H1 – Apr/May-2011]



Given: T = 598 K γ = 1.3 R = 469 J/Kg K M = 1.2 To Find:

1. Local and stagnation velocity of sound and enthalpy

i.e., a, a0, h, h0

2. Maximum velocity of this jet, Cmax

Solution:

1. Local velocity of sound

a = √(γRT)

= √(1.3 * 469 * 598)

a = 603.82 m/s

2. Stagnation temperature – Mach number relation,

(To/T) = 1+( (γ-1)/2 M2)

(To/598)= 1+( (1.3-1)/2 (1.2)2)

(To/598)= 1.216

Stagnation temperature, T0 = 727.168 K

Stagnation temperatures can be calculated by using gas tables

Refer Isentropic flow table for γ = 1.3 and M = 1.2

(T/T0) = 0.822

T0 = T/0.822

= 598/ 0.822




a0 = √( γRT0) = √(1.3 * 469 * 727.168)

a0 = 665.85

m/s

3. Static enthalpy, h = cpT

= ((γR)/(γ – 1)) * T

= (( 1.3 * 469 )/ (1.3 – 1) )* 598

h = 1215 * 103

J/Kg

4. Static enthalpy, h0 = cpT0

= ((γR)/(γ – 1)) * T0

= (( 1.3 * 469 )/ (1.3 – 1) )* 727.168

h0 = 1477 * 103

J/Kg

From Stagnation enthalpy equation, we know that

h0 = ( a2 /γ – 1) + ½ C 2 = ½ C max

2 = a02 / γ – 1

½ C max2 = a0

2 / γ – 1

C max2 = 2a0

2 / γ – 1

T0 = 727.49 K



C max=√(2a02 / γ – 1)

= √(2 * (665.85) 2 / 1.3 – 1)

Maximum Velocity, C max= 1717.02

m/s

Result

1. a = 601. 29 m/s

a0 = 665.85 m/s h = 1215 * 10 3 J/Kg h0= 1477 * 10 3 J/Kg

2. C max= 1717.02 m/s

6. Air is discharged from a reservoir at P0=6.91 bar and t0=3250C through a nozzle to

an exit pressure of 0.98 bar. If the flow rate is 3600 kg/hr, determine throat area,

pressure and velocity at the throat, exit area, exit Mach number and maximum

velocity .Consider the flow is isentropic.

[ CO1 - H1 – Apr/May-2012]

Given:

P0=6.91 bar

=6.91 x105 N/m2

T0= 3250 +273

=598K

P2=0.98 bar

= 0.98 x105 N/m2

.

m = 3600 kg/h

= 3600 kg /3600

=1 kg/s



To find:

1. Area, pressure and velocity at throat (A*,p*,c*). 2. Area and Mach number at exit (A2, M2). 3. Maximum possible velocity,cmax

Solution:

We know that at throat (*) section,M=1.

From Isentropic flow table for M=1 and γ=1.4.

=0.834

= 0.528

(From gas table‟s fifth edition page no.31)

T*=T0 x 0.834

= 598 x 0.834

T*= 498.73 K

P* = 0.528 x p0

P* = 0.528 x 6.91 x105

P* = 3.65 x 105 N/m2

We know that,

C*=a* =√γRT*

= √1.4 x 287 x 498.73

C*= 447.65 m/s

Mass flow rate, m= ρ* A* c*

A*=m/ρ*c*

=m/ p*/RT* xc*



=5

1

3.66 10447.65

287 498.73

XX

X

A*= 8.76 X10-4m2

From given data,

5

2 2

5

0 02

0.98 100.142

6.91 10

p p X

p p X

(p0=p01=p02)

Refer isentropic flow table for 2

02

0.142 1.4p

andp

.

M = 1.93 (From gas tables page no.33)

2

02

2

*

2 *

4

4 2

2

0.573

1.593

1.593

8.76 10 1.593

13.9 10

T

T

A

A

A A X

X X

A X m

Maximum velocity, max

*

1

1

c

c

*

max

1

1c c

1.4 1447.65

1.4 1

Cmax = 1096.51 m/s.

Result:

1. A*= 8.76X10-4m2

P*= 3.65X105N/m2



C*= 447.65 m/s

2. A2= 13.9X10-4m2

M2= 1.93

3. Cmax=1096.51 m/s.

7. Write short notes on :(1) Mach number (2) Critical velocities (3) Maximum

velocities. [ CO1 – L2 – Apr/May-2011]

1) Mach number:

The Mach number is an index of the ratio between inertia force and elastic force.

M ² = Inertia force ⁄ Elastic force

It is also defined as the ratio of the fluid velocity (c) to the velocity of sound (a).

M = c/a

2) Critical velocities:

The critical velocity of a fluid is its velocity at a Mach number of unity.

Mcritical = c*/a*; c* = a* = 1

CpTo= CpT* + ½ c*2 Therefore, c* = {2 Cp (To – T*)1/2

3) Maximum velocities : Maximumvelocity is the highest possibly speed an object can travel before the forces acting on it reach an equilibrium and it is no longer able to accelerate

8. A supersonic nozzle expands air from P0 = 25 bar and T0 = 1050 K to an exit

pressure of 4.35 bar. The exit area of the nozzle is 100 cm2. Determine 1. Throat

area, 2. Pressure and Temperature at the throat, 3. Temperature at exit, 4. Exit

velocity as friction of the maximum attainable velocity, 5. Mass flow rate.

[ CO1 - H1 – Apr/May-2011]

Given:

P0=25bar =25x105N/m2



T0=1050 K

P2=4.35 bar =4.35x105N/m2

A2=100Cm2=100x10-4m2

To find:

1. A* =?

2. P*=?

3. T*=?

4. T2=?

5. m =?

6. 2

max

c

c=?

Solution:

5

2

5

02

4.35 100.174

25 10

p X

p X 0 01 02p p p

2

0 2

0.174p

p

Refer Isentropic flow table for 2

02

0.174p

p and γ=1.4

M=1.418

2

02

0.612T

T (From gas table page no-33)



2

*

2 02

1.418

0.612

A

A

T XT

=0.612 X 1050

T2=642.6K (T0=T01=T02)

* 2

4* 3 2

* 3 2

1.418

100 107.05 10

1.418

7.05 10

AA

XA X m

A X m

At throat section (*),M=1

Refer Isentropic flow table for M=1 and γ=1

*

0

*

0

*

0

0.834

0.528

0.834

T

T

P

P

T T X

1050 0.834X

*

*

0

875.7

0.528

T K

P XP

50.528 25 10X X

* 5 213.2 10 /P X N m

Mass flow rate ,.

m =ρ*A*C*



** *

*

53 *

*

13.2 107.05 10

287 875.7

0.03702

0.037

0.037 1.4 287 875.7

PXA XC

RT

XX X XC

X

C

RT

X X

* * *c a RT

.

m 21.96 / seckg

We know that ,

2 2 2

max 02 p

c M a

c C T

2 2 2, max 02 pc M a C c T

2

0

1.78

2

1.78 1.4 287 642.6 904.47

1452.752 1005 1050

p

X RT

C T

X X X

X X

2

max

0.622c

c

Result:

* 5 2

* 5 2

*

2

Re

7.05 10

13.5 10 /

875.7

642.6

sult

A X m

P X N m

T K

T K

.

m = 21.96 kg/sec



2

max

0.622c

c 96kg/sec

9. A conical diffuser has entry and exit diameters of 15 cm and 30 cm respectively.

The pressure, temperature and velocity of air at entry are 0.69 bar, 340 K and 180

m/s respectively. Determine 1. The exit pressure, 2. The exit velocity and 3. The

force exerted on the diffuser walls. Assume isentropic flow, γ = 1.4, Cp = 1.00

kJ/kg-K. [ CO1 – H2 – Apr/May-2011]

Solution:

A1=

d1

2

= 0.785x152

= 176.8 cm2

A2=

d2

2

= 0.785x302

= 706 cm2

T01= T1+2

1

2 p

c

c

= 340+

= 340+16.2

= 356.2 K

=

= 0.952

From isentropic flow tables at this temperature ratio

M1 = 0.5,

= 0.843,

*

1

A

A



= 1.34, *

1

F

F

= 1.203

=

= 0.843

= =

= 0.818 bar

*

1A =

*

1A

= 1.34

*

1A = *

2A = 176.8/134

= 132 cm2

*

1 010.528P p

* *

1 2

* * * *

1 2 1 1

* 5

1

* *

1 2

2

*

2

0.528 0.818 0.432

1

0.432 10 0.0132 1 1.4

1370

7065.37

132

p p X bar

F F p A

F X X

F F N

A

A

From isentropic tables at this area ratio

M2=0.107,

2 2

*

02 2

0.992, 4.30p F

p F

(a) 2 02

2

0.992 0.992 0.818

0.810

p Xp X

p bar



(b)

2 2 1 1

12 1

2

2

2

176.8180

706

45 /

A c A c

Ac c

A

c X

c m s

(c) The force exerted on the diffuser walls is equal to the thrust of the

flow in the backward direction; this is equal to the change in the

impulse function .

2 1

* *

2 1

*

1

4.3 1.203

3.097

3.097 1370

4243

F F

F F

XF

X

N

10. A supersonic diffuser diffuses air in an isentropic flow from a mach number of 3

to a mach number of 1.5. The static conditions of air at inlet are 70 KPa and -70C.

If the mass flow rate of air is 125 kg/s, determine the stagnation conditions, areas

at throat and exit, static conditions (pressure, temperature, velocity) of air at exit.

[ CO1 - H1 – Apr/May-2012]

Given:

M1=3 M2=1.5 P1= 70kpa =70X103 pa = 70 x103N/m2 T1=-7°+273 =266 K

.

m = 125 kg/s

To find:

1. Stagnation conditions,(p0, T0)

2. Area at throat and exit,(A*,A2)

3. Static condition at exit,(p2,T2,c2,ρ2)



Solution:

Refer isentropic flow table for M1=3 and γ = 1.4

= 0.357 (From gas tables page no .37)

= 0.0272

=4.235

=

=

= 745.09 K

= = = 745.09 K =

[For isentropic flow, = = ]

3

101

70 10

0.0272 0.0272

PP

P01= 25.73 x103N/m2=p02=p0

(For Isentropic flow ,p0=p01=p02)

Mach number at entry, M1=1

1

c

a

1 1 1c M Xa

13

3 1.4 287 266

X RT

X X X

C1= 980.77 m/s

Mass flow rate, .

m = ρ1A1c1

1 1

1

pXA Xc

RT



3

1

1

70 10980.77

287 266

899.29

XXA X

X

A

.

m = 899.29 x 4.235x A* 1

*4.235

A

A

125= 3808.49 A*

Throat area, A* = 0.0328 m2

Refer Isentropic flow table for γ= 1.4 and M2 = 1.5

2

02

2

02

2

*

2 02

0.689

0.272

1.176

0.689

T

T

P

P

A

A

T XT

0.689 745.09X

T2= 513.37 K

P2= 0.272 x P02

= 0.272 x 25.73x103

P2= 6.99 x 105 N/m2

A2 = A* x 1.176

= 0.0328 x 1.176

A2= 0.0385 m2

Mach number at exit, M2 = 2

2

c

a

C2=M2xa2



21.5

1.5 1.4 287 513.37

X RT

X X X

= 681.26 m/s

Density at exit ,

22

2

5

2

3

2

6.99 10

287 513.37

4.74 /

p

RT

X

X

kg m

Result:

1. P0=25.73x105 N/m2

T0= 745.09 K

2. A* = 0.0328 m2

A2= 0.0385 m2

3. P2= 6.99 x 105 N/m2

4. T2= 513.37 K

5. C2= 681.26 m/s

6. 3

2 4.74 /kg m

11. Prove that 2 40 1 1 .............

4 40

M MP P (May-13)

Effect Of Mach Number On Compressibility (OR) Derive the expression for pressure coefficient equation for compressible equation. [ CO1 – H2 – Apr/May-2013&Nov/Dec 2014]

From Bernoulli equation, we know that stagnation pressure for incompressible flow is

2

0

0

2

1

2

11

2

p p c

p p

c

This equation shows the value of pressure co-efficient, cp(sometimes referred to as compressibility factor) is unity.



But for compressible flow the value of the pressure co-efficient deviates from

unity.

For isentropic compressible flow , the relationship between stagnation pressure

and stagnation temperature is given by

10 0

120 1

1 (1.42)2

p T

p T

pM

p

Expanding this equation as Taylor series i.e.,

2 3

1 1 21 1 .....

2! 3!

n n n n n nx nx x x

Here

21

2

1

x M

n

2 31

2 2 2 2

1 1 21 1 1 1 11 1 1 1

1 12 1 2 2! 2 3! 2

M M X M X M

2 3

2 4 6

1 1 2 1

1 1 1 1 11 11

2 2 4 6 8M X M X XM

2 3

2 4 6

1 1 2

1 11 1 1 1 11

2 2 4 6 8M X M X XM



2 4 6

12 2 4 6

21

2 8 48

211 1

2 2 8 48

M M M

M M M M

Substitute this value in equation (1.42)

2 4 60

2 4 60

2 4 60

21

2 8 48

21

2 8 48

2

2 8 48

pM M M

p

pM M M

p

p pM M M

p

Divided by 2

2M

2

40

2

21 (1.43)

4 24

2

p p MM

pX M

We know that,

Mach number,

2

2

2

22

cM

a

cM

a

cM

RT

a RT



22

22

22

2 2

2 2

2 2

cM X

RT

cM

RT

pcpX M

RT

RT

PV RT p P RTV

2

2

RTc

RT

2

2

2 2

cpX M

------------------------------------------------1.44)

Substitute 2

2PX M

value in Equation (1.43)

2

40

2

21.43 1

4 24

2

P p MM

c

Substitute γ=1.4

2 4

0

2

1 (1.45)1 4 40

2

P P M M

c

This is pressure co-efficient equation for compressible flow.

12. Air (Cp=1.05kJ/kg K, γ=1.38 ) at P1=3x105N/m2 and T1=500K flows with a

velocity of 200m/s in a 30cm diameter duct available. Calculate 1) mass flow

rate, 2) Stagnation temperature, 3) Mach number, 4) Stagnation pressure

values, Assuming the flow is compressible and incompressible respectively.

[ CO1 - H1 – Apr/May&Dec-2013]



Given:

3

5 2

1

1

1

1

1.05 / 1.05 10 /

1.38

3 10 /

500

200 /

30 0.30

pc kJ kgK X J kgK

p X N m

T K

c m s

d cm m

To find:

1. Mass flow rate, .

m

2. Stagnation temperature,T0

3. Mach number,M1

4. Stagnation pressure,p0(compressible)

5. Stagnation pressure,p0 (Incompressible)

Solution:

We know that,

Gas constant, R=cp-cv

p p

v

v

c cc

c

3

3 1.05 101.05 10

1.38

p

p

cc

XX

289.13 /R J kgK

Mass flow rate, .

m 1 1 1 2 2 2Ac Ac A c



.

m

1 1 1

211 1

1

52

4

3 10(0.30) 200

289.13 500 4

Ac

pX d Xc

RT

XX X

X

.

m 29.33 /kg s

Inlet Mach number,M1

1

1

1

1

200

1.38 289.13 500

c

a

c

RT

X X

M1=0.44

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

1

01

1

01

101

0.963

0.876

0.963

T

T

p

p

TT

500

0.963

01 519.21T K

We know that, for isentropic flow stagnation temperature remains constant

0 01 02 519.21T T T K



From tables,

1

01

101

0.876

0.876

p

p

pp

5 2

01 3.42 10 /p X N m

For isentropic flow, stagnation pressure remains constant.

5 2

0 01 02 3.42 10 /p p p X N m

From Bernoulli equation, for incompressible flow, stagnation pressure

equation is

2

0

2

01 1 1 1

1

2

1

2

p p c

p p c

211 1

1

525

1

2

1 3 103 10 200

2 289.13 500

pp Xc

RT

XX X

X

5 2

01

5 2

0 01 02

3.41 10 /

3.41 10 /

p X N m

p p p X N m

Result: 1. .

m =29.33kg/s

2. T0=519.21K

3. M=0.44

4. p0(Compressible flow)= 3.42X105N/m2

P0(Incompressible flow)= 3.41x105N/m2



13. The pressure, temperature and velocity of air at the entry of a diffuser are 0.7

bar, 345 K and 190m/s respectively. The entry diameter of a diffuser is 15 cm

and exit diameter is 35 cm. determine the following, 1) Exit pressure, 2) exit

velocity and 3) Force exerted on the diffuser walls. Assuming isentropic flow

and take γ=1.4,Cp=1005J/kgK [ CO1 - H1 – Nov/Dec-2013]

Given:

5 2

1

1

1

1

2

0.7 0.7 10 /

345

190 /

15 0.15

35 0.35

1.4

1005 /p

p bar X N m

T K

c m s

d cm m

d cm m

c J kgK

To find :

1. Exit pressure,P2

2. Exit velocity,c2

3. Force exerted on the diffuser walls,(F2-F1).

Solution:

We know that,

Velocity of sound at inlet, a1 1RT

1.4 287 345X X

1 372.32 /a m s

Mach number at inlet, 11

1

cM

a

190

372.32

1 0.510M



From Isentropic flow table for 11.4 0.510andM

1

01

1

01

1

*

1

1

*

1

0.951

0.837

1.321

1.190

T

T

p

p

A

A

F

F

(From gas tables page no-29)

1

01

01 02

345362.78

0.951 0.951

362.78

TT K

T K T

01 02,ForIsentropicflow T T

101

01 02

0.70.836

0.837 0.837

0.836

pp bar

p bar p

(For Isentropic flow ,p01=p02)

1

*

1

2

1* 1

1

1.321

41.321 1.321

A

A

dA

A

2

0.154

1.321

* *

1 20.01337A A



Area at exit, A2

2

2

2

4

0.354

d

2

2 0.096A m

2

*

2

0.0967.180

0.01337

A

A

In this problem,d2>d1=>A2>A1. So, this is divergent type diffuser. For divergent

type diffuser mach number value is less than unity.

From Isentropic flow table,for 2

*7.1800 7.262 1.4

Aand

A

2

2

02

2

02

2

*

2

0.08

0.995

0.998

5.753

M

p

p

T

T

F

F

(Note : For 2

*7.180

A

A value, we can refer gas tables page no-28 and page no-

36 But we have to take mach number less than 1 corresponding values because it is

divergent type diffuser)



2

02

2

2

0.995

0.995 0.836

0.832

p

p

p X

p bar

Exit pressure,5 2

2 0.832 10 /p X N m

2

02

02 02

0.998

0.998

T

T

T XT

0.998 362.78X

2 362.05T K

Mach number,

22

2

2 2 2

cM

a

c M Xa

.. .2a RT

2 2

0.08 1.4 287 362.05

M X RT

X X X

Exit velocity, 2 30.51 /c m s

At throat (*) section,M=1,γ=1.4

*

0

*

0

0.528

0.528 0.528 0.836

p

p

p Xp X

(From gas tables page no-31



* 0.441p bar 0 01 02p p p

* * * 5 2

1 2 0.441 10 /p p p X N m

Force excerpted on the diffuser walls is equal to the thrust of the flow

2 1F F

* *

2 1

*

1

*

1

* *

1 1

5

5.753 1.190

5.753 1.190

4.563

4.563 1

4.563 0.441 10 0.01337 1 1.4

F XF

F

F

p XA

X X

* *

1 2

* * * 1

F F

F p A

6457N

Result:

5 2

2

2

0.832 10 /

30.51 / sec

6457

P X N m

C m

N

14. Derive the expression for massflow rate in terms of Mach number

[ CO1 – H2 – Nov/Dec-2014]

Since A/A* is a unique function of Mach number in an isentropic flow, its value from

equation,

……….(1)



The non-dimensional form of the mass flow parameter is obtained by simplifying

and rearranging

.……..(2)

The expression can be reduced to the non-dimensional form by rearranging term



15. A nozzle in a wind tunnel gives a test-section Mach number of 2.0. Air enters the nozzle from a large reservoir at 0.69 bar and 310 K. The cross-sectional area of the throat is 1000 cm2. Determine the following quantities for the tunnel for one dimensional isentropic flow:

a. Pressures, temperatures and velocities at the throat and test sections,

b. Area of cross-section of the test section,

c. Mass flow rate and Power required to drive the compressor.

[ CO1 - H1 – Nov/Dec-14]



Solution:



15. Air flows through a nozzle which has inlet areas of 0.001. m2. If the air has a

velocity of 80 m/s, a temperature, of 301 K and a pressure of 700 kPa at the inlet section and a pressure of 250 kpa at the exit, find the mass flow rate through the nozzle and assuming one-dimensional isentropic flow, the velocity at the exit section of the nozzle. [ CO1 – H2 – May-2014]

Given: Po =700 kpa

To = 301K

P2 =250 kpa

C = 80m / s

To find:

The velocity at the exit section of the nozzle (C max&A2)

Solution:

P2/Po2 = (7.5 x 10 5)/ (12 x 105)

= 0.625

Refer isentropic flow table for P2/Po2 = 0.625 and γ = 1.4

M2 = 0.85

T2/ To2 = 0.874

T2 = 0.874 x To2

= 0.874 x 793

= 693.08 K

Mach number at exit, M2 = C2 / a2

0.85 = C2 / √(γ RT2)

C 2 = 448.55 m/s

Type of nozzle is convergent because mach number value is less than unity

So,

T2 = T* = 693.08 K

C2 = C* = 448.55 m/s



P2 = P* = 7.5 x 10 105 N/ m2

Maximum possible velocity, Cmax / C* = √((γ +1 )/ (γ – 1 ))

C max = C* √((γ +1 )/ (γ – 1 ))

= 448.55 x √(2.4/0.4)

C max= 1098.72 m/s

Mass flow rate, m = ρ2 x A2 x C2

(assume m=1.4 for one-dimensional isentropic flow)

= (P2/R×T2)x A2 x C2

1.4 = (7.5 x 10 5) / (287 x 693.08) x A2 x 448.55

A2 = 8.28 x 10 -4 m2 = A* (A2= A*)

Result:

1. Thevelocity,C max = 1098.72 m/s

2. The exit section of the Nozzle,A2 = 8.28 x 10 -4 m2

17. Typical cruising speeds and altitudes for three commercial aircraft are Dash 8: Cruising speed- 500 km/hr at an altitude of 4500m. Boeing 747: Cruising speed: 978 km/hr at an altitude of 9500 .Find the Mach number of the aircraft when flying at these cruise conditions. [ CO1 - H1 – May-2014] Given: Dash8- v= 500 km/hr Boeing 747 v= 978 km/hr Solution:

(i) Tofind the Mach number of the aircraft

M= c/a



1.4 287 258.90

322.53

a RT

a

a

M=138.88/322.53

M=0.43

(ii) To find the Mach number of the aircraftwhen flying at these cruise

conditions.

M= c/a

1.4 287 226.40

301.60

a RT

a

a

M=271.66/301.60

M=0.90

Result:

(i) To find the Mach number of the aircraft

M=0.43

(ii) To find the Mach number of the aircraft when flying at these cruise

conditions.

M=0.90

18.An aircraft flies at a velocity of 700 kmph in an atmosphere where the

pressure is 75 kPa and temperature is 5˙C. Calculate the mach number

and stagnation properties. [ CO1 - H1 – Apr/May-2015]



19.Air expands isentropically through the convergent nozzle fromconstant inlet

conditions P0 = 4br ,T0 = 550 K. Exit area of nozzle is 1000 cm3. Determine the

exit velocity and the mass flow rate for the following two cases at exit.

[ CO1 – H2 – Apr/May – 2015]

a. M = 1

b. M = 0.85



20. What is difference between the transonic flow and hypersonic flows?

[ CO1 – L2 – Nov/Dec – 2014]

Transonic flow:

If the fluid velocity (c) close to speed of sound (a) that type of flow is known as

transonic flow.

Mach number is in between 0.8 to 1.2

Hypersonic flow: In hypersonic flow region fluid velocity (c) is much greater than sound velocity (a)

.In this flow mach number value is always greater than 5

M = C/a > 5



UNIT II

FLOW THROUGH DUCTS

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

(Fanno flow) – variation of flow properties – Use of tables and charts – Generalized gas

dynamics.

PART -A

1. Differentiate adiabatic and isentropic processes. [ CO2 – L2 – Apr/May – 2003]

Adiabatic process

In an adiabatic process there is no heat transfer from the fluid to the surrounding

or from the surrounding to the fluid.

Q= 0

Isentropic process

In an isentropic process entropy remains constant and it is reversible. During this

process there is no heat transfer from the fluid to surroundings or from the surroundings

to the fluid. Therefore an isentropic process can be stated as reversible adiabatic

process.

S= 0

Q = 0

2. Write down the ratio of velocities between any two sections in terms of their

Machnumbers in a fanno flow [ CO2 – L1 – Nov – 2004]

3. Differentiate nozzle and diffuser?(Nov-07) [ CO2 – L2 – Nov – 2007&May 2014]

Nozzle

It is a device which is used to increase the velocity with pressure drop in the

fluids.



Diffuser

It is a device which is used to increase the pressure with velocity drop in the

fluids.

4. What is impulse function? [ CO2 – L1 – Apr – 2003]

The sum of pressure (pA) and impulse force (ρAC2)gives impulse function (F)

F = pA + ρAc2

Diabatic flow

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

diabatic flow (Rayleigh Flow)

Adiabatic flow

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

adiabatic flow (Fanno Flow).

5. Define one dimensional steady flow. [ CO2 – L1 – May – 2006]

If the flow parameters do not vary in directions normal to the flow direction and

do not vary with time, then the flow is said to be one dimensional steady flow.

6. Explain the difference between fanno flow and isothermal flow. [ CO2 – L2– Nov

– 2004&2014]

Sl.No Fanno flow Isothermal flow

1. Flow in a constant area duct with

friction and without heat transfer

Flow in a constant area duct with

friction and with heat transfer

2. Static temperature is not constant Static temperature remains constant

7. Give any two assumptions regarding Fanno flow. [ CO2 – L1 – Nov –

2005&2014]

1. One dimensional steady flow

2. Flow takes place in constant sectional area

3. There is no heat transfer

4. The gas is perfect with constant specific heats.



8. State the two governing equations used in plotting Rayleigh line

[ CO2 – L2 – Nov – 2005]

1. Continuity equation

2. Momentum Equation.

9. What are stagnation properties? [ CO2 – L1 – May – 2006]

Stagnation state is obtained by decelerating a gas isentropically to zero velocity

at zero elevation. The properties of the fluid at stagnation state are called the stagnation

properties.

10. Explain Chocking in Fanno flow. [ CO2 – L2 – May – 2007]

In a fanno flow, in sub sonic flow region, the effect of friction is to increase the

velocity and Mach number and to decrease the enthalpy and pressure of the gas.

In supersonic flow region, the effect of friction will decrease the velocity/mach

number and to increase the enthalpy and pressure of the gas.

In both cases entropy increases up to limiting state where the mach number is

one M=1, So the mass flow rate is maximum at M = 1 and it is constant afterwards. At

this point flow is said to be a choked flow.

11. What is Fanno line? [ CO2 – L1 – Nov – 2006]

Flow in a constant area duct with friction and without heat transfer is described

by a curve which is known as Fanno line orFanno curve.

12. What is chocked flow through a nozzle? [ CO2 – L1 – May – 2007]

The mass flow rate of nozzle is increased by decreasing the back pressure. The

maximum mass flow conditions are reached when the throat pressure ratio achieves

critical value. After that there is no further increase in mass flow with decrease in back

pressure. This condition is called chocking condition.



13.What type of nozzle used for sonic flow and supersonic flow? [ CO2 – L2 – Nov

– 2004]

Constant area duct nozzle is used for sonic flow and divergent nozzle is

used for supersonic flow

14. When does the maximum mass flow occur for an isentropic flow with variable

area duct? [ CO2 – L2 – Nov 2009]

Mass flow rate will be maximum at throat section where the mach number

is one.

16. Draw the shape of the nozzle for the expansion of air from 1Mpa to 700 Kpa [ CO2 - L1 – Nov 2009]

16. Give two practical examples where the fanno flow occurs[ CO2 - L1 – Nov

2008]

Flow in air breathing engines

Flow in refrigeration and air conditioning

Flow of fluids in long pipe

17. Define fannings co efficient of skin friction[ CO2 - L1 – Apr 2008]

It is the ratio between wall shear stress and dynamic head

f = Wall shear stress

Dynamic head

18. What are the three equations governing fanno process[ CO2 - L1 – Apr 2012]

1. Energy equation

2. Continuity equation

3. Equation of state



19. Write down the ration of density between any two section in terms of their

Mach number in a fanno flow. [ CO2 - L1 – Nov 2012]

ρ2/ρ1 = M1/M2 ={ [1+ (γ-1)/2 M12] /[1+ (γ-1)/2 M2

2]}1/2

20. Write down the ration of density between any two section in terms of

therirmach number in a fanno flow. [ CO2 - L1 – Apr 2009]

P2/P1 = M1/M2={ [1+ (γ-1)/2 M12] /[1+ (γ-1)/2 M2

2]}1/2

21. Give the expression forT/T0andT *Tfor isentropic flow through variable area in terms of Mach number. [ CO2 – L2 – Apr 2016]

22. Sketch the isentropic and adiabatic expansion process in P-V and T-S diagram}.[ CO2 – L2 – Apr 2016]



23. Represent the adiabatic flow through a diffuser on T-S diagram. Label the

different states, the initial and final points. [ CO2 – L2 – Apr 2013]

24. Air from a reservoir is discharged through a nozzle. Show the variation of pressure along the axis of the nozzle. [ CO2 – L2 – Apr 2013]



25. Write the Fliegner‟s formula. [ CO2 – L2 – Apr 2010]

26. Write the equation for efficiency of the diffuser. [ CO2 – L1 – Apr 2010]

27. Define critical condition in Fanno flow [ CO2 – L1 – May 2014]

In sub sonic flow region, the effect of friction is to increase the velocity and Mach

number and to decrease the enthalpy and pressure of the gas.

In supersonic flow region, the effect of friction will decrease the velocity/mach

number and to increase the enthalpy and pressure of the gas.

In both cases entropy increases up to limiting state where the mach number is

one M=1, So the mass flow rate is maximum at M = 1 and it is constant afterwards. At

this point flow is said to becritical condition in Fanno flow.

28. Give assumptions made on Rayleigh flow [ CO2 – L1 – May 2014]

Unsteady flow

Flow takes place in variable sectional area

Heat transfer takes place



29. Draw the variation ofP/P0along the length of a convergent divergent device when it functions as (a) diffuser, (b) nozzle and (c) venturi. [ CO2 – L2 – Nov 2014]

29. 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 and mass flow rate.

[ CO2 – H1 – Apr/May-2015 & Nov/Dec-2014] Given:

Po =10 bar To =400K D = 50 mm



f = 0.002 γ = 3 To find:

1. The length of the pipe and

2. Mass flow rate



31.What is meant by stagnation pressure? [ CO2 – L1 – Apr/May-2015]

Stagnation is the pressure of fluid which is attained when it is decelerated to zero

velocity at zero elevation in a reversible adiabatic (isentropic) process.

32.What is „Fanno flow‟? [ CO2 – L1 – Apr/May-2015]

Flow in a constant area duct with friction in the absence of work andheat transfer

is known as fanno flow.

PART – B

1. The cross section at the entry of a diffuser is 0.24 sq. meter. Mach number is

1.5 and temperature if air is 340 k. If the exit Mach number is 0.78, determine the

velocity and temperature of air and the area of cross section at exit for isentropic

flow. [ CO2 –H1 – Apr/May-2011]

Given:

A1 = 0.24 m2

M1 = 1.5

T1 = 340 K

M2 = 0.78

For air R = 287 J/Kg K

To find,

1. Velocity and temperature of air at exit (C 2, T 2 )

2. Area of cross section at exit (A 2)

Solution

Refer isentropic flow table for γ = 1.4, M 1 = 1.5

T 1/ T0 1 = 0.689,

A 1/ A* = 1.176

T o1 = T 1 / 0.689



= 340/ 0.689

= 493.47 K

To1=To2=493.72 K

A* = A1/1.176

A* = 0.204 m 2

Refer isentropic flow table for γ = 1.4, M 2 = 0.78

T 2 /To2 = 0.892,

A 2 / A * = 1.047

T 2 = 0.892 x T o 2

= 0.892 x 493.47

T 2 = 440.18 K

A 2 = 1.047 x A *

= 1.047 x 0.204

= 0.2135 m2

Fluid velocity at exit C2= M2 x a2

= M 2 x √ (γRT2 )

= 0.78 x √ (1.4 x 287 x 440.18)

C2 = 328.03 m/s

Result:

1. C2 = 328.03 m/s

2. T2 = 440.18 K

3. A2 = 0.2135 m2



2. An ambient air (Po = 1 bar, To = 285 K) is sucked by a blower through a

convergent nozzle. The throat diameter is 12 cm. If the air velocity at the throat

reaches the sonic value, determine

1. Pressure and temperature at throat

2. Maximum mass flow rate [ CO2 – H1 – Apr/May-2011]

Given:

Po = 1 bar

To = 285 K

Throat diameter, d * = 12 cm = 0.12 m

To find:

1. Pressure and temperature at the throat, (p*, T*)

2. Maximum mass flow rate, M max

Solution:

We know that the throat (*) = M = 1

Refer isentropic flow table for = 1.4, M = 1

T*/ To = 0.834,

P*/ Po = 0.528

T* = 0.834 x 285

= 237.69 K

P*/ 1x 10 5 = 0.528

P* = 0.528 x 10 5 N/ mm2

Throat area, A* = π/4 (d*) 2

= π/4 (0.12) 2

= 0.0113 m 2

Maximum mass flow rate,



1

2 1max 0

0

M T R 2

A P 1

Substitute γ= 1.4, R = 287 J/ Kg K, To = 285, Po = 1.013 x 10 5 N / mm2

1.4 1

2 1.4 1max

5

M 285 287 2

0.113 1 10 1.4 1.4 1

Mmax = 2.70 kg/s

Result

1. P* = 0.528 x 10 5 N/mm2

2. T* = 237.69 K

3. Mmax= 2.70 Kg / sec

3. A conical air diffuser has an inlet diameter of 40 cm and an exit diameter of 80

cm. Air enters the diffuser with a static pressure of 200 kPa, static temperature of

37oC and velocity of 265 m/sec calculate

i) Mass flow rate

ii) Properties at exit [ CO2 – H2 – Nov-2007]

Given:

d1 = 40 cm = 0.40 m

d2 = 80 cm = 0.8 m

P1 = 200 kPa = 200 KN/m2 = 200 x 10 3 N/ m2

T1 = 37oC +273 = 310 K

C1 = 265 m/s

To find:

1. Mass flow rate, m

2. P2, T2, C2, ρ2.



Solution:

Area at inlet, A1 = π/4 (d1)2

= π/4 (0.4)2

= 0.126 m2

Mass flow rate, m = ρ 1 x A1 x C1

= P1/RT1 x A1 x C1

= (200 x 1000)/ (287 x 310) x 0.126 x 265

= 75.05 kg/sec

Mach number at entry, M1 = C1/a1

= C1 / √(γRT1)

= 265 / √(1.4 x 287 x 310 )

= 0.75

Refer isentropic flow table for M1 = 0.75, γ = 1.4

T 1 / To1 = 0.899

P1/ Po1 = 0.688

A1/A1* = 1.062

To1 = T1/0.899

= 310/0.899

To1 = 344.83 K = To2

Po1 = P1/0.688

= (200 x 1000) / 0.688



= 2.906 x 105 N/m2 = Po2

A1/ A1* = 1.062

A1* = A1/1.062

= 0.126/1.062

= 0.1186 m2

Area at exit, A2 = π/4 (d2)2

= π/4 (0.80)2

= 0.5206 m2

A2/A2* = 0.5206 / 0.1186

A2/A* = 4.238

In this problem d2 > d1 => A2> A1 so this divergent type of diffuser. For divergent type

diffuser Mach number value is less than unity

Refer isentropic flow table for A2/A* = 4.238 and γ = 1.4

M2 = 0.14

P2/Po2 = 0.986

T2/ To2 = 0.996

We have to take the Mach number less than unity corresponding values because it is divergent type diffuser.

P2/ Po2 = 0.986

P2 = 0.986 x Po2

= 0.986 x 2.906 x 10 5

P2 = 2.86 x 10 5 N/m2



T2/To2 = 0.996

T2 = 0.996 x To2

= 0.996 x 344.83

= 343.45 K

Mach number at exit, M2 = C2/a2

C2 = M2 x a2

= M2 x √(γRT2)

= 0.14 x √(1.4 x 287 x 343.35)

= 52 m/s

Density at exit, ρ2 = P2 / RT2

= (2.86 x 10 5) / (287 x 343.35)

= 2.90 Kg / m3

Result: 1. m = 75.05 kg / s

2. P2 = 2.86 x 10 5 N/m2

3. T2 = 343.45 K

4. C2 = 52 m/s

5. ρ2 = 2.90 kg / m 3-

4. An air is isentropically expanded from Po = 12 bar and To = 520 oC in a nozzle

to an exit pressure of 7.5 bar. If the rate of flow of the air is 1.4 kg/s. Calculate the

following

1. Pressure, temperature and velocity at the nozzle throat and exit

2. Maximum possible velocity

3. Type of nozzle

4. Throat area

Take γ = 1.4, R = 287 J / kg k [ CO2 –H1 – Apr/May-2012]

Given

Po = 12 bar = 12 x 10 5 N/m2



To = 520 oC +273 = 793 K

P2 = 7.5 bar = 7.5 x 105N/m2

m = 1.4 kg / s

Solution,

P2/Po2 = (7.5 x 10 5)/ (12 x 105)

= 0.625

Refer isentropic flow table for P2/Po2 = 0.625 and γ = 1.4

M2 = 0.85

T2/ To2 = 0.874

T2 = 0.874 x To2

= 0.874 x 793

= 693.08 K

Mach number at exit, M2 = C2 / a2

0.85 = C2 / √(γ RT2)

C 2 = 448.55 m/s

Type of nozzle is convergent because mach number value is less than

unity

So,

T2 = T* = 693.08 K

C2 = C* = 448.55 m/s

P2 = P* = 7.5 x 10 105 N/ m2

Maximum possible velocity, Cmax / C* = √((γ +1 )/ (γ – 1 ))



C max = C* √((γ +1 )/ (γ – 1 ))

= 448.55 x √(2.4/0.4)

= 1098.72 m/s

Mass flow rate, m = ρ2 x A2 x C2

= P2/RT2 x A2 x C2

1.4 = (7.5 x 10 5) / (287 x 693.08) x A2 x 448.55

A2 = 8.28 x 10 -4 m2 = A*

5. Derive an expression for the acoustic velocity of a compressible fluid

flow in terms of its temperature. [ CO2 –H2 – Apr/May-2012]

DERIVATION OF ACOUSTIC VELOCITY (or) SOUND VELOCITY (a)

Sound waves are infinitely small pressure disturbances. The speed with which

sound propagates in a medium is called speed of sound and is denoted by „a‟.

If an infinitesimal disturbance is created by the piston, as shown in fig., the wave

propagates through the gas at the velocity of sound relative to the gas into which the

disturbance is moving.



In the case the stagnant gas at pressure „p‟ on the right appears to flow towards

the left with a velocity a. When the flow has passed through the wave to the left its

pressure raised to p + dp and the velocity is lowered to a – dc.

Apply momentum equation for this process.

A p p dp m a dc a

A p p dp Aa a dc a m Ac m Aa

Adp Aa dc

dp adc

From continuity equation for the two sides of the wave



m Aa ( d )(a dc)A

Aa A[a dc ad d dc]

a [ a dc ad d dc]

The term d dc is negligible

a a dc ad

dc ad

Substituting Equation (B) in Equation (A)

(A) dp ad a

dp (ad ) a

2a d

2dpa

d

dpa

d

For Isentropic flow

dp

RTd

dp

a RTd

Sound velocity (or) acoustic velocity a RT



6. starting from the continuity equation derive the expression for the area

variation in terms of Mach number and velocity variation and hence obtain the

shape for both sub sonic and supersonic nozzles and diffusers. [ CO2 –H1 – Nov-

2009]

AREA RATIO AS AFUNCTION OF MACH NUMBER

Like temperature, pressure and density ratios, area ratio is also a useful quantity.

We know that,

Mass flow rate, m Ac *A*c*

A * c*

A* c

Characteristic

Mach number

Substituting c*

c and

*

values

1 121

2 2

1 1

1 22

A 2 1 1 2 1M M

A* 1 1 M 1 1

A 1 2 1M

A* M 1 1

2 1

2 121 2 1

MM 1 1

2

2

2

1M

2M*

11 M

2



1

2 12A 1 2 1

MA* M 1 1

We know that

120p 1

1 Mp 2

1

2 12

0 12

1

2 12

0 12

A p 1 2 1 1M

A* p M 1 11

1 M2

1 2 1M

M 1 1A p

A* p1

1 M2

1 1

2 1 2 12

12

1 2 11 M

M 1 2

11 M

2



`

1

2 121

2 1

12

1 1

2 1 2 1 12 2

1 1

2 1 2 12

11 M

21 2

M 11

1 M2

1 2 1 11 M 1 M

M 1 2 2

1 2 11 M

M 1 2

1

1 1 1

2 1 1 2 121 2 1

1 MM 1 2



1 1 1 2

2 1 1 22

1 1 1

2 1 1 22

1 1

2 1 22

1

2 1

2

1 2 11 M

M 1 2

1 2 11 M

M 1 2

1 2 11 M

M 1 2

1 2

M 1

11 M

2

1

2

1

2 1

1

0 22

1 2

M 1A p

A* p1

1 M2

7. Describe how the properties change in a fanno flow with suitable diagram.

[ CO2 –L2 – Nov 2006]



Fanno flow equations

We know that

Mass flow density, G = ρc

ln(G) = ln[ρc]

ln(G) =ln ρ + ln c

Differentiating,

d dc

0c

[Mass flow density G = constant]

d dc

c

2

2

d d[c ]

2c

2

2

2

2

d c 2cdc

d[c ]dc

2c

dc d[c ]

c 2c

Gas equation

pv = RT

1

p RTv

p = ρRT

ln[p] = ln[ρRT]

ln[p] = ln[ρ] + ln[R] + ln[T]

Differentiating



dp d dT

p T

[ R = constant]

dp d dT

p T

Mach number, c

Ma

c

RT

a RT

22

22

2 2

cM

RT

c[M ]

RT

[M ] c RT

ln ln

ln ln ln

2 2

2 2

[M ] c R T

[M ] c T

ln ln ln ln ln

ln ln ln

Differentiating

2 2

2 2

dM dc dT

M c T

,R cons tan t

Stagnation enthalpy, 2

01h h c

2

2

01h c h cons tan t

2

Differentiating



p

2cdcdh 0

2

c dT cdc 0

2

p

cc dT d 0

2

p ph c T,dh c dT

2R 1dT dc 021

2R T 1dT dc 021 T

p

Rc

1

2

22

RT dT 1 dc 021 T

a dT 1 dc 021 T

Multiply throughout by 1 a RT

2 2

1dTa dc 0

T 2

2

2

2

1c dTdc 0

M T 2

cM

a

22

2

22

2

dT 1 Mdc 0

T 2 c

dT 1 dcM 0

T 2 c

Fanning‟s co-efficient of skin friction (f)

= ---------------------- Wall shear stress

Dynamic

head



w

2

T

1 c2

w

2

Tf

1 c2

2

w1T f c

2

The area of the duct is given by

dAw = perimeter x length

dAw = p x dx

The hydraulic mean diameter of the duct is given by

4AD

p

4Ap

D

Substituting perimeter p value

w

4AdA dx

D

The momentum equation between state (1) and state (2) is given by

pA mc (p dp)A m(c dc)

Considering shear stress

w w

w w

w w

pA mc (p dp)A m(c dc) T dA

pA mc pA Adp mc mdc T dA

Adp mdc T dA 0



w wmdc Adp T dA

Substituting Tw and dAw values in above equation

2

2

2

2

4A1mdc Adp f c dx2 D

f c 4mdc A dp dx

2 D

m c dxdc dp 4f

A 2 D

m pM dxdc dp 4f

A 2 D

2 2p

c cRT

2 2

2

2

pM a

RT

pM RT

RT

pM



We know that

m Ac

dc dcA A

cdc

2 2

2

c d[c ]

2c

2d[c ] 2cdc

2 2

2

pM d[c ]

2c

2 2c pM

2 2

2

m pM d[c ]dc

A 2c

Substitute m

dcA

value

2 2 2

2

pM d[c ] pM dxdp 4f

2c 2 D

Divide by p



2 2 2

2

2 2 2

2

pM dc dp M dx4f 0

c p 2 D

dp M dc M dx4f 0

p 2 c 2 D

2 2 2

2

dp M dc M dx4f 0

p 2 c 2 D

Stagnation pressure – Mach number relation

120

120

2

0

p 11 M

p 2

p 11 M

p 2

1p p 1 M

1 2

ln ln

ln ln ln

Differentiating

2

0

20

1d 1 M

dp dp 2

1p p 11 M

2



2

0

20

2

0

20

2 2

0

220

2 2

0 0

220 0

10 dM

dp dp 2

1p p 11 M

2

dp dp dM

1p p2 1 M

2

dp dp M dM

1p p M2 1 M

2

dp dp M dM

1p p M2 1 M

2

8. An ambient air (Po = 1.013 bar, To = 288 K) is sucked by a blower through a

convergent nozzle. The throat diameter is 10 cm. If the air velocity at the throat

reaches the sonic value, determine

1. Pressure and temperature at throat

2. Mass flow rate through nozzle and

3. Maximum mass flow rate [ CO2 –H2 – Nov 2006]

Given:

P0 = 1.013 bar

T0 = 288 K

Throat diameter, d* = 10 cm = 0.10 m

To find:

1. Pressure and temperature at the throat, (p*, T*)

2. Maximum mass flow rate, M max

Solution:

We know that the throat (*) = M = 1

Refer isentropic flow table for = 1.4, M = 1

T*/ To = 0.834,



P*/ Po = 0.528

T* = 0.834 x 288

= 240.192 K

P*/ 1.013x 10 5 = 0.528

P* = 0.5348 x 10 5 N/ mm2

Throat area, A* = π/4 (d*) 2

= π/4 (0.10) 2

= 0.00785 m 2

Maximum mass flow rate,

1

2 1max 0

0

M T R 2

A P 1

Substitute γ= 1.4, R = 287 J/ Kg K, To = 288, Po = 1.013 x 10 5 N / mm2

1.4 1

2 1.4 1max

5

M 288 287 2

0.00785 1.013 10 1.4 1.4 1

Mmax = 1.893 kg/s

Result:

1. P* = 0.5348 x 10 5 N/mm2

2. T* = 240.192 K

3. Mmax= 1.893 Kg / sec



9. The area of cross-section at the entry of a diffuser is 0.25 m2. Mach number 1.6

and temperature of air 350 K. if the exit Mach number is 0.80 calculate the velocity

and temperature of air and the area of cross section at exit for isentropic flow.

Draw the sketch of the diffuser indicating the directions of the flow and the thrust.

[ CO2 –H2 – Nov 2007]

Given:

A1 = 0.25 m2

M1 = 1.6

T1 = 350 K

M2 = 0.8

For air R = 287 J/Kg K

To find:

1. Velocity and temperature of air at exit (C2, T2)

2. Area of cross section at exit (A2)

Solution:

Refer isentropic flow table for γ = 1.4, M1 = 1.6

T 1/ T0 1 = 0.661, A 1/ A* = 1.250

T o1 = T 1 / 0.661

= 350/ 0.661

= 529.5 K

To1=To2=529.5 K

A* = A1/1.250

A* = 0.2 m2



Refer isentropic flow table for γ = 1.4, M1 = 0.8

T 2 /To2 = 0.886, A 2 / A * = 1.038

T 2 = 0.886 x T o 2

= 0.886 x 529.5

T 2 = 469.137 K

A 2 = 1.038X A *

= 1.038 x 0.2

A 2 = 0.2076 m2

Fluid velocity at exit C2= M2 x a2

= M 2 x √(γRT2 )

= 0.8 x √(1.4 x 287 x 469.137)

C2 = 347.331 m/s

Result:

1. C2= 347.331m/s

2. T2 = 0.886 x 529.5

3. A2 = 0.2076 m2

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

entry pressure and temperature are 3.45 bar and 38oC respectively and the

coefficient of friction 0.005. if the Mach number at entry is 0.15, determine



(a) The diameter of the duct,

(b) Length of the duct,

(c) Pressure and temperature at the exit,

(d) Stagnation pressure loss. [ CO2 –H2 – Nov 2007]

Solution:

T1 = 273 +38 = 311 K

(a)To find the diameter of the duct,

531

1

1

1 1

1

1 1 1

p 3.45 103.865 kg / m

RT 287 311

a RT

a 1.4 287 311 353.486 m / s

c M a 0.15 353.496 53.02m / s

1 1Ac m

2

1 1

2 2

2

m 8.25A 0.0403m

c 3.865 53.02

D 403cm4

D 22.66cm

From isentropic tables for γ = 1.4 M1 = 0.15

P1/p01 = 0.984

101

p 3.45p 3.506bar

0.984 0.984



M p/p* T/T* p0/p0* c/c* 4fLmax/D

M1 = 0.15

M2 = 0.50

7.319

2.138

1.1945

1.143

3.928

1.340

0.164

0.534

28.354

1.069

(b)To findLength of the duct,

1 2

max max

M M

L L4fL4f 4f

D D D

4fL28.354 1.069 27.285

D

L 27.285 0.2266 / 4 0.005

L 309meters

(c)To findPressure and temperature at the exit,

*

22 1*

1

2

2

p / p 2.138p p 3.45

7.319p / p

p 1.008bar

1.143T 311 297.59K

1.1945



(d)To findStagnation pressure loss.

*

02 0202 01*

01 01

02 01

p / pp p

p / p

1.343.506 1.196bar

3.928

p p 3.506 1.196 2.31bar

11. The Mach number at inlet and exit for a Rayleigh flow are 3 and 1.5

respectively. At inlet static pressure is 50 KPa and stagnation temperature is

295 K. Consider the fluid is air. Find

(a) The static pressure, static temperature and velocity at exit,

(b) Stagnation pressure at inlet and exit,

(c) Heat transferred,

(d) Maximum possible heat transfer,

(e) Change in entropy between the two sections,

(f) Is it a cooling or heating process? [ CO2 –H2 –Apr/May 2007]

Given:

M1 = 3, M2 = 1.5

P1 = 50 kPa = 50 x 103 N/m2

T01 = 295 K

For air γ = 1.4 and R = 287 J/kg-K

Solution:

Refer Isentropic flow table for M1 = 3 and γ = 1.4

T/T01 = 0.357, P1/P01 = 0.0272

T1 = T01 x 0.357

= 295 x 0.357



T1 = 105.31 K

P01 = P1 / 0.0272 = 50 x103 / 0.0272

P01 = 18.38 x 105 M/m2

Refer Raleigh flow table for M1 = 3 and γ = 1.4

1

*

1

01

*

01

01

*

01

1

*

1

3* 5 21

1

* 5 2 *

1 2

P0.176

P

P3.24

P

T0.654

T

c1.588

c

P 50 10P 2.84 10 N / m

0.176 0.176

P 2.84 10 N / m P



`

* 0101

55 2

* 5 2

01

* 11

* *

1 2

* 0101

* *

01 02

* 11 1 11

*

1

PP

3.424

18.38 105.36 10 N / m

3.424

P 5.36 10 N / m

T 105.41T

0.281 0.281

T 374.76 K T

T 295T

0.654 654

T 451.07 K T

M RTc M ac

1.588 1.588 1.588

3 1.4 287 105.31

1.588

c 388

*

2.60 m / s c

Refer Raleigh flow table for M2 = 1.5 and γ = 1.4



2

*

2

02

*

02

02

*

02

2

*

2

* 5

2 2

5 2

2

* 5

02 02

5 2

02

*

2 2

2

*

02 02

P0.578

P

P1.122

P

T0.909

T

c1.301

c

P P 0.578 2.84 10 0.578

P 1.64 10 N / m

P P 1.122 5.36 10 1.22

P 6.01 10 N / m

T T 0.753 374.76 0.753

T 282.19K

T T 0.909 451.07 0.909

2

*

2 2

2

T 410.02K

c c 1.301 388.60 1.301

c 505.56 m / s



12. Prove that the variation of flow parameter and the maximum possible heat

transfer

22

max 2

1

2 1p

MQ C T

M

[ CO2 –H2 –Apr/May 2007]

Flow properties at M = M* = 1 are used as reference v I for non-dimensional

various properties at any section of the duct Property ratios in Equations.

This is done by making the following substitutions in the abovementioned

equations:

Following expressions are obtained after the above procedure



22

max 2

1

2 1p

MQ C T

M

It may be pointed out here that the final sonic state ( M - 1,T=T* etc.) may not

be reached in a given process. However, since this is the limiting state for both subsonic

and supersonic heatingprocesses it is used besides non-dimensional, to give a

comparativeidea of the magnitudes of various flow parameters.

13. The condition of a gas in combustor at entry is

1 1 10.343 , 310 , 60 / .p bar T K c m s To determine the Mach number, pressure,

temperature and velocity at the exit if the increase in stagnation enthalpy of the

gas between entry and exit is 1172.5 KJ/kg. Takecp=1.005 KJ/kgK,

=1.4 [ CO2 –H2 –Apr/May 2016]



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

pressure and temperature are 3.45 bar and 38°C respectively and the coefficient

of friction 0.005. If the Mach number at entry is 0.15, determine

(a) The diameter of the duct,

(h) Length of the duct,

(c) Pressure and temperature at the exit,

(d) Stagnation pressure loss, and

(e) Verify the exit Mach number through exit velocity and temperature.

[ CO2 –H2 –Apr/May 2015]



15.The stagnation temperature of air is raised from 85˙C to 376 ˙C in a heat

exchanger. If the inlet Mach number and percentage drop in pressuer.

[ CO2 –H1 –Nov/Dec 2015]



UNIT-3

NORMAL AND OBLIQUE SHOCKS

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

shocks – Prandtl – Meyer relations – Use of table and charts – Applications.

PART-A

1. What is meant by shock wave? [ CO3 –L1 –Apr/May 2007]

A shock wave is nothing but a steep finite pressure wave. The shock wave may

be described a compression wave front in a supersonic flow field across while there is

abrupt change in flow properties.

2. What is normal shock? [ CO3 –L1 –Apr/May 2007]

When the shock wave is at right angle to the flow, it is called normal shock.

3. What is oblique shock wave?[ CO3 –L1 –Apr/May 2015, Nov/Dec-14&May/Jun

2004]

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

shock.

4. What is Prandtl – Meyer relation? [ CO3 –L1 –Apr/May 2015]

Prandtl – Meyer relation which is the basis of other equation for shock waves. It

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

the critical velocity of sound.

2

* *

x y

*

x y

M M 1

c c a

5. Define strength of shock wave. [ CO3 –L1 –Apr/May 2010]

It is defined as the ratio of difference in downstream and upstream shock

pressures (py - px) to upstream shock pressure (px). It is denoted by ζ.

y x

x

p p

p



6. How the Mach number before and after the occurrence of a normal shock are

related? [ CO3 –L1 –Apr/May 2010]

Mach number after the normal shock

2

2

y2

x

2M

1M

2M 1

1

7. What are applications of moving shock wave? [ CO3 –L1 –Apr/May 2009]

The applicationsof moving shock waves are

1. Jet engines

2. Shock tubes

3. Supersonic wind tunnel

4. Practical admission turbines.

8. Write the equation for efficiency of a diffuser. [ CO3 –L1 –Apr/May 2009]

1

0y01

1 0x

D2

1

pT1

T p

1M

9. Shock waves cannot develop in subsonic flow. Why? [ CO3 –L1 –Apr/May 2008]

In subsonic flow, the velocity of fluid is less than the velocity of sound. Due to this

reason, deceleration is not possible in subsonic flow. So, shock waves cannot develop

in subsonic flow.

9. Give the expression for y

x

T

T across the normal shock.

[ CO3 –L1 –Apr/May 2008]

2 2

x x

y

22xx

2 1M 1 1 M

T 1 2

MT1

2 1



11. Define compression and rarefaction shocks? Is the latter possible.

[ CO3 –L1 –Nov/Dec 2008]

A shock wave which is at a higher pressure than the fluid in to which it is moving

is called compression shock wave.

A shock wave which is at a lower pressure than the fluid into which it is moving is

called an expansion shock wave or rarefaction shock wave. It is not possible.

12. State the necessary conditions for a normal shock to occur in compressible

flow. [ CO3 –L1 –Nov/Dec 2008]

1. The compression wave is to be right angle to the compressible flow.

2. Flow should be supersonic.

3.

13. Write down the Rankine-Hugniot equation. [ CO3 –L2 –Nov/Dec 2006]

x

yx

xy

y

p11

1 p

p1

1 p

17. Is the flow through a normal shock an equilibrium one? [ CO3 –L1 –Nov/Dec 2014]

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

during normal shock.

15. Write down the static pressure ratio expression for a normal shock. [ CO3 –L1

–Nov/Dec 2012]

2xx

y

p 2 1M

p 1 1



16. Give the difference between Normal and Oblique shocks. [ CO3 –L2 –Nov/Dec

2014]

17. What are properties changes across a normal shock?) [ CO3 –L1 –Nov/Dec

2014]

1. Stagnation pressure decreases.

2. Stagnation temperature remains constant.

3. Static temperature and static pressure increases.

18. What are the beneficial and adverse effects of shock waves? [ CO3 –L1 –

May/Jun 2012]

Some useful applications of the shock waves are in the shock tubes and

supersonic compressors. A strong moving shock wave is utilized to accelerate the flow

to a high Mach number in shock tube where flow behavior at high Mach numbers can

be studied.

Shocks are undesirable because they interfere with the normal flow behavior.

Other undesirable forms of the shock waves are the sonic boom created by supersonic

aircrafts and the blast waves generated by an explosion. These wave have a damaging

effect on human life and buildings.

Normal shock Oblique shock

1. Shock wave is right angle to the flow

2. One dimensional flow

Shock wave is inclined at an angle to

the flow

Two dimensional flow



19. What are the assumptions used for oblique shock flow? [ CO3 –L1 –Nov/Dec

2013]

The following assumptions are used for oblique shock flow.

1. Flow is steady, adiabatic and frictionless.

2. The gas is perfect with constant specific heats.

3. Absence of work transfer across the boundaries.

4. Absence of body forces.

20. Why the efficiency of the machine, expressing shock wave is considerably

low? [ CO3 –L3 – May/June 2014]

The efficiency of the machine, expressing shock wave is considerably low

because of 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 waves are abrupt. The normal

shock wave is perpendicular to the one dimensional flow.

21. What is the use of Pitot tube in supersonic flow? [ CO3 –L3 – May/June 2014]

The Pitot tube can also be used for the measurement of gas velocity in

supersonic flow.

A pilot tube along with a wall tapping can be used to determine the Mach number of a

supersonic stream.



PART-B

1. Prandtl – Mayer Relation [ CO3 –H2 – Apr/ May 2011]

The fundamental relation between the gas velocities before and after the normal

shocks and the critical velocity of sound is derived. This is known as Prandtl-Mayer

relation which is basis of other equations for shock waves.

Applying the adiabatic energy equation to the flow before and after the shock

wave we get

2

22y2 2 *x

x y

aa 1 1 1 1c c a

1 2 1 2 2 1

. . . (1)

First part of this equation gives

2

2

2 * 2

x x

2 *

xx

x x

1 1a a c

2 2

a 1 a 1c

c 2 c 2

. . . (2)

Similarly the other part of equation (1) gives

22 *y

y

y y

a 1 a 1c

c 2 c 2

. . . (3)

Momentum equation gives

x y y x

Ap p c c

m . . . (4)



Continuity equation for shock waves, x x x y y ym c A c A . . . (5)

Substituting, A

min equation (4),

yxy x

x x y y

yxy x

x x y y

y2 2xx y

x y

ppc c

c c

pp1 1c c

c c

ppBut a and a

Therefore 22yx

y x

x y

aac c

c c . . . (6)

Introduction of Equations (2) and (3) in (6), gives

2 2

2

2

2

* *

x y y x

x y

*

y x y x

x y

y x*

y x y x

x y

*

x y x y

1 a 1 1 a 1c c c c

2 c 2 2 c 2

1 1 1 1a c c c c

2 c c 2

c c1 1a c c c c

2 c c 2

1 1a c c c c

2 2

Multiplying by 2



2

2

2

2

2

2

*

x y x y

*

x y x y

*

x y x y x y

*

x y x y

*

x y

x y*

1 a 1 c c 2 c c

1 a 2 c c 1 c c

1 a 2 c c c c c c

1 a c c c c

1 a c c 1

c c 1a

1

2*

x ya c c . . . (7)

This is equation is known as Prandtl – Mayer relation.

* *

x y

yx

* *

* *

x y

a a c c

cc1

a a

1 M M

* *

x yM M 1 . . .(8)

This is another useful form of Prandtl – Mayer relation.

2. The ratio of the exit to entry area in a subsonic diffuser is 4.0. The Mach

number of a jet of approaching the diffuser at P0 = 1.013 bar, T = 290 K is 2.2.

There is a standing normal shock wave just outside the diffuser entry. The flow in

the diffuser is isentropic. Determine at the exit of the diffuser, (i) Mach number,



(ii) Temperature and pressure, (iii) Stagnation pressure loss between the initial

and final states of the flow. [ CO3 –H2 – May / Jun 2013]

Fig. Normal shock at the diffuser entry

The given data is

p0x = 1.013 bar, Tx = 290 K, Mx = 2.2, AE = 4Ay

From Isentropic tables for γ = 1.4, at M = Mx = 2.2

x x

0 0x 0 0x

p pp T0.0935, T 0.508

p p T T

px = 0.0935 x 1.013 = 0.0947 bar

0x

290T 570.87 K

0.508

From normal shock tables for γ = 1.4, at Mx = 2.2

My = 0.547, y y 0y

x x 0x

p T p5.48, 1.857, 0.628

p T p

Ty = 1.857 x 290 = 538.53 K

py = 5.48 x 0.0947 = 0.519 bar



From Isentropic tables at M = My = 0.547

*

y

* *

y E

y y

* *

y E

E E E

* *

yE E

A1.26

A

Now, A A

A ATherefore,

A A

A A A1.26 4 5.04

AA A

Parameters at various points in the flow are show in fig.

From Isentropic tables at A/A* = AE/A* = 5.04

M=ME = 0.116

E

0 0E

0x 0y 0E

E

TT0.997

T T

T T T 570.87K

T 0.997 570.87 569.15K

The continuity equation between section Y and e gives,

y y y E E E

y Ey y y E E E

y E

y y y E E E

y E

y yE E

y E E y

c A c A

p pM RT A M RT A

RT RT

p M A p M A

T T

M Ap T

p M A T



E

y

p 0.547 1 569.151.212

p 0.116 4 538.53

E

0y 0x

0y

0x

p 1.212 0.519 0.629bar

p 0.628p 0.628 1.013

p 0.636bar 636 mbar

p 1013 mbar

Stagnation pressure loss = 1013 – 636 = 377 mbar

Alternatively, from Isentropic tables for γ = 1.4 at My = 0.547

py /p0y = 0.816, therefore,

p0E = p0y = yp 0.519

0.636 bar0.816 0.816

From Isentropic tables, for γ = 1.4 at ME = 0.116

pE/p0E = 0.990, therefore,

pE = 0.990 x p0E = 0.990 x p0y

pE = 0.990 x 0.636 = 0.629 bar.

3. Derive the equation for Mach number in the downstream of the normal shock

wave. [ CO3 –H2 – May / Jun 2012]

Generally the upstream Mach number (Mx) in a given problem is known and it is

desired to determine the Mach number (My) downstream of the shock wave.

It will be seen in this that the only independent parameter required for a given

gas (of cp/cv) to determine the downstream Mach number is the upstream Mach

number.

For adiabatic flow of a perfect gas gives,

2*

0

2a RT

1

… (1)



From Prandtl Mayer relation, 2

x yc .c a … (2)

From equation (1) & (2), x y 0

2c .c RT

1

… (3)

But x x x x x

y y y y y

c M a M RT

c M a M RT

Substitute these values in Equation (3)

x x y y 0

2M RT .M RT RT

1

Rearranging and squaring on both sides

2

2 2 0 0x y

x y

T T2M .M

1 T T

Substituting the values of T0/Tx and T0/Ty for adiabatic flow

2

2 2 2 2

x y x y

2 1 1M .M 1 M 1 M

1 2 2

This is a convenient form to solve of My2. After some algebraic steps it gives,

2

x2

y2 2 2

x x

4 2 1 MM

11 M 2 1 1 M

2

This on rearrangement and further simplification yields a simple relation as follows.



2

x2

y2

x

2M

1M

2M 11

4. The velocity of a normal shock wave moving into stagnant air (p = 1.0 bar, t =

17oC) is 500 m/s. If the area of cross section of the duct is constant, determine

pressure, temperature, velocity of air, stagnation temperature and Mach number

imparted upstream of the wave front. [ CO3 –H1 – May / Jun 2012]

Fig.2 Flow of gas through a stationary shock wave (moving coordinate system)

Refer to Fig.2, the shock is first considered in the moving coordinate system,

here the shock is stationary and the stagnant air on the right of the wave appears to be

following through the wave towards the left.

Tx = 273 + 17 = 290 K, px = 1.0 bar

cx = 500 m/s

x x

x

xx

x

a RT

a 1.4 287 290 341.35 m / s

c 500M 1.465

a 341.35

From normal shock tables at Mx = 1.465



y y

y

x x

y x

'

y y

y x

'

y y

y y

y y y

'

y x y

'

y x y

p TM 0.715, 2.335, 1.297

p T

p 2.335 p 2.335 1 2.335 bar

p p 2.335 bar

T 1.297 T 1.297 290 376.13K

T T 376.13K

a RT 1.4 287 376.13 388.75 m / s

c M .a 0.715 388.75 277.95 m / s

c c c

c c c 50

'

y

0 277.95

c 222.05 m / s

Now the shock wave is considered as moving,

'

y'

y

y

c 222.05M 0.571

a 388.75

From Isentropic tables, at M = My‟ = 0.571

y

0 0y

y

0y

T 'T0.939

T T '

T ' 376.13T ' 400.56 K

0.939 0.939

5. Air approaches a symmetrical wedge (angle of deflection δ = 15o) at a mach

number of 2. Consider strong waves conditions. Determine the wave angle,

pressure ratio, density ratio, temperature ratio and downstream Mach number.

[ CO3 –H1 – May / Jun 2012]



2 2

1

2 2 2

1 1

M sin 1tan 2cot

2 M M 1 2sin

Here δ = 15o and M1 = 2.0

2o

2

2

2

4sin 1tan15 2cot

2 1.4 4 4 1 2sin

4sin 10.1339 cot

11.6 8sin

The right hand side for σ = 80o, 79o, 80.3o, 78.8o gives 0.132, 0.142, 0.128 and

0.134 respectively. This iteration which is quite tedious gives σ =78.8o.

σstorng =78.8o

2 221

1

22

1

2

1

p 2 1M sin

p 1 1

p 2 1.4 1.4 12sin 79.8

p 1.4 1 1.4 1

p4.353

p

2

1

2

1

tan tan79.8

tan tan 79.8 15

2.615

2

2 1

21

1

2

1

pT p 4.353

T 2.615

T1.665

T



1

2 2 22

1

2

2 2

2

2

2

2

p1

pM sin

1

114.3531.4 M sin 79.8 15 0.4774

2.615 1

1.4 0.8187M 0.4774

M 0.4774

We proceed in a similar manner for the weak shock wave.

σweak =15o

22

1

2

1

p 2 1.4 1.4 12sin 43.3

p 1.4 1 1.4 1

p2.191

p

2

1

2

1

tan tan45.3

tan tan 45.3 15

1.729

2

1

1

2 22 2

1

2

T 2.1911.267

T 1.729

p1p 1

Msin

1



2

2 2

2

11

12.191M1.729 1 1.4sin 30.3

M 1.448

6. Derive the equation for static pressure ratio across the oblique shock waves.

[ CO3 –H2 – May / Jun 2012]

2 22 11

1 2

p1 M sin 1

p

…(1)

Density ratio is given by,

2 2

12

2 2

1 1

1 M sin

2 1 M sin

… (2)

Substitute equation (2) in (1)

2 2

12 221 2 2

1 1

2 1 M sinp1 M sin 1

p 1 M sin

After simplifying and rearranging,

2 221

1

2 221

1

p 21 M sin 1

p 1

p 2 11 M sin

p 1 1



7. Derive the Prandtl‟ equation for flow through an oblique shock [ CO3 –H2 –

May / Jun 2013]

2* 2

t n1 n2

1a C C C

1

2 2 2 22 1n1 t1 n2 t2

2 1

p p 1 1c c c c

1 2 2

This is on rearranging gives,

2 2 2 21 2n1 t1 n1 t1

1 2

p p1 1c c c c

1 2 1 2

Using velocity triangles,

2 2 21n1 t1 1 1 0

1

p 1 1c c h c h

1 2 2

…. (1)

2*

0

1 1h a

2 1

… (2)

From equations (2) and (1), we get,

2

2

2 2 *1n1 t1

1

* 2 2

1 1 n1 t1

p 1 1 1c c a

1 2 2 1

1 1p a c c

2 1

2* 2 2

1 1 n1 t1

1 1p a c c

2 2

… (3)

Similarly, 22 2 *2

n2 t2

2

p 1 1 1c c a

1 2 2 1

… (4)



2* 2 2

2 2 n2 t2

1 1p a c c

2 2

… (5)

Momentum equation in normal direction is,

2 2

1 n1 2 n2 2 1c c p p … (6)

Substitute Equations (3) and (5) in (6), gives

22 2 * 2 2 2 2

1 n1 2 n2 2 1 1 n1 t1 2 n2 t2

1 1c c a c c c c

2 2

Using ct1 = ct2 =ct and rearranging we get,

2

2

2

2

* 2 2 2

2 1 1 n1 2 n2 2 1 t

* 2 2 2

2 1 1 n1 2 n2 2 1 t

2 2* 2 1 n1 2 n2

t

2 1

* 2 n1 n2 n1 n2t 1 2

2 1 n2 n1

1 1 1a 1 c c c

2 2 2

1 1 1a c c c

2 2 2

c c1a c

1

c c c c1a c

1 c c

However, n1 2

n2 1

c

c

Therefore, 2* 2

t n1 n2

1a c c c

1

This is known as Prandtl‟s equation for oblique shocks. This reduces for normal

shocks by making the following substitutions:

ct = 0 cn1 = c1 = cx

cn2 = c2 = cy



8. A normal shock occurs in the diverging section of a convergent - divergent air

nozzle. The throat area is 1/3 times exit area and the static pressure at exit is 0.4

times the stagnation pressure at the entry. The flow is throughout isentropic

except through the shock. Determine:

(i) Mach numbers M. and My

(ii) The static pressure and

(iii) The area of cross section of the nozzle at the section of nozzle where the

normal shock occurs. [ CO3 –H1 – Apr/May 2014]

Solution:

The Following relations may be noted

* * *

1

* *

2

0

*

t t x

y

A A A A

A A

m Tconst

A Po

The nozzle is shown in Figure .Following relations may be noted:



9. A gas (y = 1.3) at p, = 345 mbar, T, = 350 K and M, = 1.5 is to be isentropically

expanded to 138 mbar.

Determine

(i) The deflection angle.

(ii) Final Mach number

(iii) The temperature of the gas.(May-14) [ CO3 –H1 – Apr/May 2014&2015]



10. Derive the expression for Rankine-Hugoniot equations (Density ratio across

the shock). [ CO3 –H2 – Nov/Dec 2014]

Equation of state for a perfect gas gives



11. An oblique shock wave occurs at the leading edge of a symmetrical. wedge.

Air has a Mach number of 2.0 and deflection angle (8) of 15°. Determine the

following for strong and weak waves.

1. Wave angle.

2. Pressure ratio,

3. Density ratio,

4. Temperature ratio and

5. Downstream Mach number. [ CO3 –H2 – Nov/Dec 2014]



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

thriugh a normal shock wave, determine the following for downstream of the

shock.

1. Mach number

2. Pressure

3. Temperature

4. Speed of sound

5. Jet velocity [ CO3 –H2 – Apr/May 2015]

6. Density



UNIT 4

JET PROPULSION

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

Operation principle, cycle analysis and use of stagnation state performance of ram jet,

turbojet, turbofan and turbo prop engines.

PART-A

1. What is meant by a jet propulsion system? [ CO4 –L1 –Apr/May 2016]

It is the system for the propulsion of a jet aircraft or missiles by the reaction of jet

coming outwith high velocity. The jet propulsion in used when the oxygen is obtained

from thesurrounding atmosphere.

2. How will you classify propulsive engines? [ CO4 –L1 –Apr/May 2016]

The jet propulsion engines are classified into:

i. Air breathing engines and

ii. Rocket engines which do not use atmospheric air.

3. What is the difference between shaft propulsion and jet propulsion? [ CO4 –L2

–Apr/May 2008]

SHAFT PROPULSION JET PROPULSION

a) The power to the propeller is transmitted through a reduction gear

b) At higher altitude, the performance is c) Poor. Hence it is suitable for lower

altitudes. d) With increasing speeds and size of the e) aircrafts, the shaft propulsion engine f) Becomes too complicated. g) Propulsive efficiency is less.

a) There is no reduction gear.

b) Suitable for higher altitudes. c) Construction is simpler. d) More.

4. List the different types of jet engines. [ CO4 –L1 –Apr/May 2013]

i. Turbo-jet

ii. Turbo-prop engine,

iii. Ram jet engine,

iv. Pulse jet engines.



5. Define the principle of Ram jet engine. [ CO4 –L1 –Apr/May 2013]

The principle of jet engine is obtained from the application of Newton‟s law of

motion. We know that when a fluid is accelerated, a force is required to produce this

acceleration is the fluid and at the same time, there is an equal and opposite reaction

force of the fluid on the engine is known as the thrust, and therefore the principle of jet

propulsion is based on the reaction principle.

6. When can a ram jet engine be used? [ CO4 –L1 –Apr/May 2011]

In ramjet, no rotating machinery is used and compression is achieved by the

intake and diffuser. As such they require high speed to compress air enough that good

efficiency can be achieved. Ramjets are inefficient at subsonic speeds and they can be

used at supersonic speeds.

7. Give the components of a turbo jet. [ CO4 –L1 –Apr/May 2013]

i. Diffuser

ii. Mechanical compressor,

iii. Combustion chamber,

iv. Turbine and

v. Exhaust nozzle.

8. Give the difference between pulse jet and ram jet engine. [ CO4 –L2 –Apr/May 2012]

PULSE JET RAM JET

a) Mechanical valve arrangements are used during combustion.

b) The stagnation temperature at the diffuser exit is comparatively less.

a) Works without the aid of any mechanical device and needs no moving parts.

b) Since the Mach number in Ram jet engine is supersonic, the stagnation temperature is very high.

http://en.wikipedia.org/wiki/subsonic



9. Give the difference between turbojet and ram jet engine. [ CO4 –L2 –Apr/May

2012]

TURBO JET RAM JET

a) Compressor and turbine are used. b) Lower thrust and propulsive

efficiency at lower speeds. c) Construction cost is more.

a) Compressor and turbine are not used but diffuser and nozzle are used.

b) It provides high thrust per unit weight.

c) In the absence of rotating machines, the construction is simple and cheap.

10. What is specific impulse? [ CO4 –L1 –Nov/Dec 2012]

Specific impulse is the thrust developed per unit weight flow rate through

thepropulsive device. It is a useful performance parameter in aerospace propulsion

systems.

I spe = F/ W

11. Give the difference between Jet propulsion and Rocket propulsion. [ CO4 –L2

–Nov/Dec 2012]

JET PROPULSION ROCKET PROPULSION

a) Oxygen is obtained from the surrounding atmosphere for combustion purposes.

b) The jet consists of air plus combustion products.

c) Mechanical devices are also used.

a) The propulsion unit consists of its own oxygen supply for combustion purposes.

b) Jet consists of the exhaust gases only.

c) Mechanical devices are not used.



12. What is the difference between turbo prop engine and turbo jet engine. [ CO4

–L2 –Nov/Dec 2011]

TURBO – PROP TURBO - JET

a) The specific fuel consumption based on thrust is low.

b) Propulsive efficiency within the range of operation is higher.

c) On account of higher thrust at low speeds the take-off role is short and requiring shorter runway.

d) Use of centrifugal compressor stages increases the frontal area.

e) Higher weight per unit thrust.

a) TSFC is comparatively higher at lower speeds and altitudes.

b) Propulsive efficiency is low. c) Take – off role is longer and

requiring longer run way. d) Lower Frontal area. e) Lower weight per unit thrust.

13. Write down the formula for propulsive efficiency and define the same. [ CO4 –

L1 –Nov/Dec 2011]

The force which propels the aircraft forward at a given speed is called thrust

(or)propulsive force.

Propulsive efficiency is defined as the ratio between propulsive power (or) thrust

Power to the power output of the engine.

ηP= Thrust power (or) Propulsive power/ Power output of the engine = F x u/ P

ηP = 2/(1+ σ)

Where σ= u/ Cj

σ = Effective speed ratio (or) flight to jet velocity

u=flight velocity

Cj = jet velocity

14. What is ram effect? [ CO4 –L1 –Nov/Dec 2006]

When an aircraft flies with high velocity, the incoming air is compressed to high

Pressure without external work at the expense of velocity energy is known as “ram

effect”.

15. Explain specific thrust as applied to jet engines. [ CO4 –L1 –Nov/Dec 2006]

Specific thrust is defined as the thrust produced per unit mass flow rate through

the propulsive device.

Specific thrust Fsp F/ m

Where, F = thrust and m = mass flow rate



16. Differentiate between pressure thrust and momentum thrust. [ CO4 –L2 –

Nov/Dec 2005]

Pressure thrust mainly depends on the difference in pressure between the

nozzleexit pressure and the ambient pressure and is given by

Pressure thrust = (Pe – Pa) A

Momentum thrust depends on the difference in velocity between the aircraft

velocityand jet velocity is given by

Momentum thrust = m (cj – u) where,

Pe = nozzle exit pressure

Pa = ambient pressure

A = Area of cross section at the nozzle exit

Cj = jet velocity and

u = forward speed of aircraft

17. What is “thrust augmentation”? [ CO4 –L1 –May/Jun 2012]

To achieve better take-off performance, higher rates of climb and

increasedperformance at altitude during combat maneuvers, there has been a demand

for increasing thethrust output of aircraft for short intervals of time. This is achieved by

during additional fuelin the tail pipe between the turbine exhaust and entrance section of

the exhaust nozzle. Thismethod of thrust increases the jet velocity is called “Thrust

Augmentation”.

18. Why after burners are used in turbojet engine? [ CO4 –L1 –May/Jun 2011]

Exhaust gases from the turbine have large quantity of oxygen, which can support

thecombustion of additional fuel. Thus if a suitable burner is installed between the

turbine andexhaust nozzle, a considerable amount of fuel can be burned in this section

to producetemperatures entering the nozzle as high as 1900°C. The increased

temperature greatly augments the exhaust gas velocity, and hence provides the thrust

increase.

19. What is after burning in turbo jet engines? [ CO4 –L1 –May/Jun 2007]

An afterburner (or reheat) is a combustor located immediately upstream of the engine final nozzle, where fuel is burnt to raise the nozzle entry gas temperature, thereby increasing the net thrust of the engine. Because of the relatively high fuel consumption associated with afterburning, the system has to be used sparingly. A variable area nozzle is normally fitted to accommodate the increased gas volume flow when the afterburner is alight.

http://en.wikibooks.org/w/index.php?title=Jet_Propulsion/Afterburner&action=edit&redlink=1



20. Why a ram jet engine does not require a compressor and a turbine?

[ CO4 –L3 –Nov/Dec 2013]

In general, the speed of a ram jet engine is supersonic (the range of Mach

number) isvery high. At this flight speed the contributions of the compressor to the total

static pressurize is insignificant. Hence, arm jet engine does not require compressor

and turbine.

21. Briefly explain thrust augmentation and any two methods of achieving it.

[ CO4 –L1 –May/Jun 2009]

The method of increasing Jet thrust by different methods is called thrust

augmentation.

The thrust can be increased by injecting additional fluids and it is then called wet

thrust. Water is injected at the air compressor inlet or the diffuser to cool the

compressing air which permits an increase in pressure for higher burning.

Today's military combat engines use an afterburner for increased thrust. An

afterburner or "reheat jet pipe" is a device added to the rear of the jet engine. It provides

a means of spraying fuel directly into the hot exhaust, where it ignites and boosts

available thrust significantly;

22. Compare the propulsive efficiency of jet engine with that of rocket engine.

[ CO4 –L1 –Nov/Dec 2006]

Propulsive efficiency ( ): how much of the energy of the jet ends up in the

vehicle body rather than being carried away as kinetic energy of the jet is the Propulsive

efficiency .

The exact formula for air-breathing engines moving at speed with an exhaust

velocity is given in the literature as

And for a rocket

http://en.wikipedia.org/wiki/Afterburner

http://en.wikipedia.org/wiki/Kinetic_energy



23. Define specific impulse. [ CO4 –L1 –Nov/Dec 2006]

Specific impulse is defined as the thrust (N) divided by the fuel weight flow rate (N/s).

The resulting measure is usually quoted in seconds and defines the weight fraction that

is necessary to give a particular delta V for a rocket or range for an aircraft with a given

lift to drag ratio.

24. What is TWR? [ CO4 –L1 –Nov/Dec 2013]

Thrust-to-weight ratio is a ratio of thrust to weight of a rocket, jet

engine, propeller engine, or a vehicle propelled by such an engine. It is a dimensionless

quantity and is an indicator of the performance of the engine or vehicle.

The instantaneous thrust-to-weight ratio of a vehicle varies continually during

operation due to progressive consumption of fuel or propellant, and in some cases due

to a gravity gradient.

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

velocity is 900 m/s? [ CO4 –L1 –Nov/Dec 2013&2014]

Given:

Jet velocity, cj = 500 m/s

Flight velocity, u = 900 m/s

Solution:

Propulsive efficiency,

j

p2 2

j

c u

1c u

2

2 2

p

500 900

1500 900

2

450000

530000

0.8490 84.9%

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

http://en.wikipedia.org/wiki/Weight

http://en.wikipedia.org/wiki/Rocket

http://en.wikipedia.org/wiki/Jet_engine


http://en.wikipedia.org/wiki/Propeller_(aircraft)

http://en.wikipedia.org/wiki/Dimensionless_quantity

http://en.wikipedia.org/wiki/Dimensionless_quantity



26. A turbojet engine having a flight velocity of 800 km/hr at an ambient pressure

of 60 kPa, the properties of gas entering the nozzle are 300 kPa and 200oC. The

mass flow rate of air is 20 kg/s. Assuming for air Cp/Cv = 1.4 and R = 0.287 kJ/kgK.

Find the thrust power of the engine. [ CO4 –L1 –May/Jun 2012]

Given:

Fight velocity, u = 800 km/hr = 222.22 m/s

Ambient pressure, = 60 kPa

Pressure, p = 300 kPa = 300 x 103 Pa

Temperature, T = 200oC = 473 K

Mass flow rate, ̇ = 20 kg/s

Cp/Cv = γ = 1.4, R = 0.287 kJ/kgK

Solution:

patm =pe= 60 kPa

p1 = 300 kPa

1

1 1

1

298 6

2

2 1005 473 298 6

592 06

e e

e

j p e

T p

T p

T . K

C C T T

.

. m / s

Thrust force, F = ma(cj-u)

= 7397.3 N

Thrust power, = F x u

= 1643.69 kW



27. Define thrust power and propulsive efficiency of aircraft engine.

[ CO4 –L1 –May/Jun 2014 & 2015]

Thrust power:

The force that propels the rocket at a given velocity is known as the thrust.Rocket

thrust is employed to overcome the drag and gravitational forces besides providing the

acceleration.

Propulsive Efficiency:

The turbo propulsive system is preferable in the lower range o f speed.

Performance of turbojet and turbofan engines continues to improve with speed.

The propulsive efficiency is given by

Propulsive efficiency ηp= ( )

28. Why a ram jet engine does not require a compressor and turbine?

[ CO4 –L1 –May/Jun 2014&Nov/Dec 2014]

The flight speed of a turbojet engine is very high , say in the range of Mach

number 2 to 4 the pressure rise in the diffuser (ram pressure) is very high ; at this flight

speed the contribution of the compressor to the total static pressure rise is in significant.

Therefore, it can be removed from the engine along with its prime mover the turbine. In

this manner a thrust producing device (engine) is obtained which does not have

compressor and turbine.

29.Define Propulsive efficiency. [ CO4 –L1 –Apr/May 2015]

Propulsive efficiency is defined as the ratio between propulsive power (or) thrust

Power to the power output of the engine.



30.What is the type of compressor used in turbo jet? [ CO4 –L1 –Apr/May 2015]

There are two main types of compressors used in jet engines.

They are,

1. centrifugal compressor

2. axial compressor

PART-B

1. With a neat sketch explain the principle of operation of turbo-jet engine also

list its important applications. [ CO4 –H3 –Apr/May 2015,2011&2012]

TURBO-JET ENGINE:

The turbojet is the oldest kind of general-purpose air breathing jet engine. Turbojets consist of an air inlet, an air compressor, a combustion chamber, a gas turbine (that drives the air compressor) and a nozzle. The air is compressed into the chamber, heated and expanded by the fuel combustion and then allowed to expand out through the turbine into the nozzle where it is accelerated to high speed to provide propulsion.

Air intake

Preceding the compressor is the air intake (or inlet). It is designed to be as

efficient as possible at recovering the ram pressure of the air stream tube approaching

the intake. The air leaving the intake then enters the compressor.

Compressor

The compressor is driven by the turbine. The compressor rotates at a very high

speed, adding energy to the airflow and at the same time squeezing (compressing) it

into a smaller space. Compressing the air increases its pressure and temperature.

Several types of compressor are used in turbojets and gas turbines in general: axial,

centrifugal, axial-centrifugal, double-centrifugal, etc.

https://www.grc.nasa.gov/www/K-12/airplane/caxial.html

http://en.wikipedia.org/wiki/Airbreathing_jet_engine

http://en.wikipedia.org/wiki/Air_compressor

http://en.wikipedia.org/wiki/Turbine

http://en.wikipedia.org/wiki/Nozzle

http://en.wikipedia.org/wiki/Energy

http://en.wikipedia.org/wiki/Pressure

http://en.wikipedia.org/wiki/Temperature

http://en.wikipedia.org/wiki/Gas_turbines



After leaving the compressor section, the compressed air enters the combustion

chamber.

Figure-1

Combustion chamber

In a turbojet the air and fuel mixture passes unconfined through the combustion

chamber. As the mixture burns its temperature increases dramatically, but the pressure

actually decreases a few percent.

The fuel-air mixture must be brought almost to a stop so that a stable flame can

be maintained. This occurs just after the start of the combustion chamber. Because of

the shape of the combustion chamber the flow is accelerated rearwards. Some pressure

drop is required, as it is the reason why the expanding gases travel out the rear of the

engine rather than out the front. Less than 25% of the air is involved in combustion, in

some engines as little as 12%, the rest acting as a reservoir to absorb the heating

effects of the burning fuel.

Jet engines run a very lean mixture, so lean that it would not normally support

combustion. A central core of the flow (primary airflow) is mixed with enough fuel to

burn readily. The unburned air (secondary airflow) mixes into the burned gases to bring

the temperature down to something a turbine can tolerate.

Turbine

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

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



Hot gases leaving the combustor are allowed to expand through the turbine.

Turbines are usually made up of high temperature metals such as Inconelto resist the

high temperature, and frequently have built-in cooling channels.

The turbine's rotational energy is used primarily to drive the compressor. Some

shaft power is extracted to drive accessories, like fuel, oil, and hydraulic pumps.

Because of its significantly higher entry temperature, the turbine pressure ratio is much

lower than that of the compressor. In a turbojet almost two-thirds of all the power

generated by burning fuel is used by the compressor to compress the air for the engine.

Nozzle

After the turbine, the gases are allowed to expand through the exhaust nozzle to

atmospheric pressure, producing a high velocity jet in the exhaust plume. In a

convergent nozzle, the ducting narrows progressively to a throat. The nozzle pressure

ratio on a turbojet is usually high enough for the expanding gases to reach Mach 1.0

and choke the throat. Normally, the flow will go supersonic in the exhaust plume outside

the engine.

If, however, a convergent-divergent de Laval nozzle is fitted, the divergent (increasing

flow area) section allows the gases to reach supersonic velocity within the nozzle itself.

This is slightly more efficient on thrust than using a convergent nozzle. There is,

however, the added weight and complexity since the convergent-divergent nozzle must

be fully variable in its shape to cope with changes in gas flow caused by engine

throttling.

Applications: Turbojets are quite inefficient if flown below about Mach and very noisy. Most

modern aircraft use turbofans instead for economic reasons. Turbojets are still very

common in medium range cruise missiles, due to their high exhaust speed, low frontal

area and relative simplicity.

THE TURBOFAN ENGINE: The turbofan engine is a propulsive mechanism to combine the high thrust of a

turbojet with the high efficiency of a propeller.

Basically, a turbojet engine forms the core of the turbofan; the core contains the

diffuser, compressor, burner, turbine, and nozzle.

http://en.wikipedia.org/wiki/Inconel

http://en.wikipedia.org/wiki/Angular_momentum

http://en.wikipedia.org/wiki/De_Laval_nozzle

http://en.wikipedia.org/wiki/Jet_engine#Noise

http://en.wikipedia.org/wiki/Turbofan

http://en.wikipedia.org/wiki/Cruise_missile



However, in the turbofan engine, the turbine drives not only the compressor,

but also a large fan external to the core.

The fan itself is contained in a shroud that is wrapped around the core.

The flow through a turbofan engine is split into two paths.

One passes through the fan and flows externally over the core; this air is

processed only by the fan, which is acting in the manner of a sophisticated,

shrouded propeller.

The propulsive thrust obtained from this flow through the fan is generated with

an efficiency approaching that of a propeller.

The second air path is through the core itself.

The propulsive thrust is obtained from the flow through the core is generated

with an efficiency associated with a turbojet.

The overall propulsive efficiency of a turbofan is therefore a compromise

between that of a propeller and that of a turbojet.

This compromise has been found to be quite successfulthe vast majority of

jetpropelled airplanes today are powered by turbofan engines.



An important parameter of a turbofan engine is the bypass ratio, defined as the

mass flow passing through the fan, externally to the core divided by the mass

flow through the core itself.

Everything else being equal, the higher the bypass ratio, the higher the

propulsive efficiency.

For the large turbofan engines that power airplanes such as the Boeing 747,

for example, the Rolls-Royce RB211 and the Pratt & Whitney JTBD, the bypass

ratios are on the order of 5.

Typical values of the thrust specific fuel consumption for these turbofan engines

are 0.6 lb/(lb h) almost half that of a conventional turbojet engine.

2. Differentiate turbojet and turboprop propulsion engines with suitable diagrams.

(M/J - 2012) [ CO4 –H1 –Apr/May 2012]

Turbojet

A turbojet is an air breathing gas turbine engine executing an internal combustion

cycle during the operation. It also belongs to the reaction engine type of the aircraft

propulsion engines.

Turboprop

Turboprop engines are another variant built on the turbojet engine, and use the

turbine to produce shaft work to drive a propeller. Also, turboprop engines can be

seen as a turbo-shaft engine with propeller connected to the shaft through a

reduction gear mechanism.

Turbojet Engine

Cold air entering through the intake is compressed to high pressure in the

successive stages of an axial flow compressor. The pressurization of the air also

increases the temperature, and when mixed with the fuel produces a combustible gas

mixture.



Combustion of this gas increases the pressure and temperature to a very high

level (1200 oC and 1000 kPa) and the gas pushes through the blades of the turbine.

In the turbine section, gas exerts force on the turbine blades and rotates the turbine

shaft; in a common jet engine, this shaft work drives the compressor of the engine.

Figure - 2

Then the gas is directed through a nozzle, and this produce a large amount of

thrust, which can be used to power an aircraft. At the exhaust, the speed of the gas can

be well above the speed of sound. The operation of the Jet engine is ideally modeled by

the Brayton cycle.

The turbojets are inefficient at low speed flight, and optimal performance lies

beyond Mach 2. Another disadvantage of the turbojets is that the turbojets are

extremely noisy. However, they are still used in the mid-range cruise missiles because

of the simplicity of production and low speed

Turboprop Engine

Turboprop engine is an advanced version of the turbojet engine, where the shaft work

is used to drive a propeller through a reduction gear mechanism attached to the turbine

shaft. In this form of jet engines, majority thrust is generated by the propeller reaction

and the exhaust generates a negligible amount of usable energy; hence mostly not

used for thrust.

http://files.differencebetween.com/wp-content/uploads/2012/08/Turbojet-Engine1.png



Figure - 3

The propellers in turboprop engines are usually a constant speed (variable pitch)

type, turboprop engines usually contain at least one stage of centrifugal compression.

Propellers lose efficiency as aircraft speed increases, but very efficient at flight

speeds below 725 km/h. Hence turboprops are normally not used on high-speed

aircraft and are used to power small subsonic aircraft.

The difference between Turbojet and Turboprop Engine

Turbojets were the first air breathing gas turbine engine for aircraft, while

turboprop is an advanced variant of turbojet, using the gas turbine to drive a

propeller to generate thrust.

Turbojets show good performance at supersonic speed, while turboprops

show good performance at subsonic speeds.

Turbojets are used in specific military applications at present, but turboprops

are widely used in both military and commercial aircraft.

3. With the help of a neat sketch describe the working of a ramjet engine.

[ CO4 –H3 –Apr/May 2007&2013]

A ramjet sometimes referred to as a stovepipe jet, is a form of air breathing jet engine using the engine's forward motion to compress incoming air, without a rotary compressor. Ramjets cannot produce thrust at zero airspeed, thus cannot move an aircraft from a standstill.



http://files.differencebetween.com/wp-content/uploads/2012/08/Turboprop-Engine1.png



Ramjets therefore require some other propulsion system to accelerate the vehicle to a speed where the ramjet begins to produce thrust. Ramjets work most efficiently at supersonic speeds around Mach 3. This type of engine can operate up to speeds of Mach 6.

Figure-4

A ramjet is designed around its inlet. An object moving at high speed through air

generates a high pressure region upstream. A ramjet uses this high pressure in front

of the engine to force air through the tube, where it is heated by combusting some of

it with fuel. It is then passed through a nozzle to accelerate it to supersonic speeds.

This acceleration gives the ramjet forward thrust.

A ramjet is sometimes referred to as a 'flying stovepipe', a very simple device

comprising an air intake, a combustor, and a nozzle. Normally, the only moving parts

are those within the turbo pump, which pumps the fuel to the combustor in a liquid-

fuel ramjet. Solid-fuel ramjets are even simpler.

http://en.wikipedia.org/wiki/Supersonic

http://en.wikipedia.org/wiki/Mach_number

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


http://en.wikipedia.org/wiki/Turbopump



Inlet

Ram jets try to exploit the very high dynamic pressure within the air approaching

the intake lip. An efficient intake will recover much of the free stream stagnation

pressure, which is used to support the combustion and expansion process in the

nozzle.

Most ramjets operate at supersonic flight speeds and use one or more conical (or

oblique) shock waves, terminated by a strong normal shock, to slow down the airflow

to a subsonic velocity at the exit of the intake. Further diffusion is then required to get

the air velocity down to a suitable level for the combustor.

Combustor

As with other jet engines, the combustor's job is to create hot air, by burning a

fuel with the air at essentially constant pressure. The airflow through the jet engine is

usually quite high, so sheltered combustion zones are produced by using 'flame

holders' to stop the flames from blowing out.

Since there is no downstream turbine, a ramjet combustor can safely operate at

stoichiometricfuel: air ratios, which implies a combustor exit stagnation temperature

of the order of 2400 K for kerosene.

Nozzles

The propelling nozzle is a critical part of a ramjet design, since it accelerates

exhaust flow to produce thrust.

For a ramjet operating at a subsonic flight Mach number, exhaust flow is

accelerated through a converging nozzle. For a supersonic flight Mach number,

acceleration is typically achieved via a convergent-divergent nozzle.

Performance and control

Although ramjets have been run from as low as 45 m/s (162 km/h) upwards,

below about Mach 0.5, they give little thrust and are highly inefficient due to their low

pressure ratios.

http://en.wikipedia.org/wiki/Dynamic_pressure

http://en.wikipedia.org/wiki/Stagnation_pressure

http://en.wikipedia.org/wiki/Stagnation_pressure

http://en.wikipedia.org/wiki/Supersonic

http://en.wikipedia.org/wiki/Cone_%28geometry%29

http://en.wikipedia.org/wiki/Shock_wave

http://en.wikipedia.org/wiki/Flame_holder

http://en.wikipedia.org/wiki/Flame_holder

http://en.wikipedia.org/wiki/Stoichiometry

http://en.wikipedia.org/wiki/Stagnation_temperature

http://en.wikipedia.org/wiki/Propelling_nozzle


http://en.wikipedia.org/wiki/De_Laval_nozzle



Above this speed, given sufficient initial flight velocity, a ramjet will be self-

sustaining. Indeed, unless the vehicle drag is extremely high, the engine/airframe

combination will tend to accelerate to higher and higher flight speeds, substantially

increasing the air intake temperature. As this could have a detrimental effect on the

integrity of the engine and/or airframe, the fuel control system must reduce engine

fuel flow to stabilize the flight Mach number and, thereby, air intake temperature to

reasonable levels.

Due to the stoichiometric combustion temperature, efficiency is usually good at

high speeds (Mach 2-3), whereas at low speeds the relatively poor pressure ratio

means the ramjets are outperformed by turbojets, or even rockets.

4. With the help of a neat sketch describe the working of a pulse jet engine. [ CO4

–H3 –Apr/May 2013]

A pulse jet engine (or pulsejet) is a type of jet engine in which combustion occurs

in pulses. Pulsejet engines can be made with few or no moving parts, and are capable

of running statically. Pulse jet engines are a lightweight form of jet propulsion, but

usually have a poor compression ratio, and hence give a low specific impulse.

Function

Fig. 5.

Pulse jet schematic. First part of the cycle: air flows through the intake (1), and is

mixed with fuel (2). Second part: the valve (3) is closed and the ignited fuel-air mix (4)

propels the craft.

http://en.wikipedia.org/wiki/Drag_%28physics%29

http://en.wikipedia.org/wiki/Mach_number

http://en.wikipedia.org/wiki/Turbojets

http://en.wikipedia.org/wiki/Rocket


http://en.wikipedia.org/wiki/Pulse

http://en.wikipedia.org/wiki/Moving_parts

http://en.wikipedia.org/wiki/Compression_ratio

http://en.wikipedia.org/wiki/Specific_impulse

http://en.wikipedia.org/wiki/File:Pulse_Jet_schematic.svg



The combustion cycle comprises five or six phases: Induction, Compression, (in

some engines) Fuel Injection, Ignition, Combustion, and Exhaust.

Starting with ignition within the combustion chamber, a high pressure is raised by the

combustion of the fuel-air mixture. The pressurized gas from combustion cannot exit

forward through the one-way intake valve and so exits only to the rear through the

exhaust tube.

The inertial reaction of this gas flow causes the engine to provide thrust, this force

being used to propel an airframe or a rotor blade. The inertia of the traveling exhaust

gas causes a low pressure in the combustion chamber. This pressure is less than the

inlet pressure (upstream of the one-way valve), and so the induction phase of the cycle

begins.

In the simplest of pulsejet engines this intake is through a venturi which causes fuel

to be drawn from a fuel supply. In more complex engines the fuel may be injected

directly into the combustion chamber. When the induction phase is under way, fuel in

atomized form is injected into the combustion chamber to fill the vacuum formed by the

departing of the previous fireball; the atomized fuel tries to fill up the entire tube

including the tailpipe. This causes atomized fuel at the rear of the combustion chamber

to "flash" as it comes in contact with the hot gases of the preceding column of gas. This

resulting flash "slams" the reed-valves shut or in the case of valve less designs, stops

the flow of fuel until a vacuum is formed and the cycle repeats.

There are two basic types of pulsejets. The first is known as a valved or traditional

pulsejet and it has a set of one-way valves through which the incoming air passes.

When the air-fuel is ignited, these valves slam shut which means that the hot gases can

only leave through the engine's tailpipe, thus creating forward thrust.

The cycle frequency is primarily dependent on the length of the engine. For a small

model-type engine the frequency may be around 250 pulses per second, whereas for a

larger engine such as the one used on the German V-1 flying bomb, the frequency was

closer to 45 pulses per second. The low-frequency sound produced resulted in the

missiles being nicknamed "buzz bombs."

http://en.wikipedia.org/wiki/Venturi_pump

http://en.wikipedia.org/wiki/V-1_%28flying_bomb%29



The second type of pulsejet is known as the valve less pulsejet. Technically the term

for this engine is the acoustic-type pulsejet, or aerodynamically valved pulsejet. The

advantage of the acoustic-type pulsejet is simplicity. Since there are no moving parts to

wear out, they are easier to maintain and simpler to construct.

5. The diameter of the propeller of an aircraft is 2.5 m. It flies at a speed of 500

kmph at an altitude of 8000 m. For a flight to jet speed ratio of 0.75 determine: (i)

The flow rate of air through the propeller, (ii) Thrust produced, (iii) Specific thrust,

(iv) specific impulse and (v) The thrust power. [ CO4 –H1 –Apr/May 2013]

Solution:

Area of cross-section of the propeller disc

2 2 2A d 2.5 4.908m4 4

Air density at Z = 8000 m is

ρ = 0.525 kg/m3

Flight speed u = 500 kmph = 138.89 m/s

σ = u/cj = 0.75

cj = 138.89/0.75 = 185.18 m/s

(a) Velocity of air flow at the propeller disc is

j

1c u c2

c 0.5 138.89 185.18 162.035m / s

Theoretical value of the flow rate is given by

a

a

m Ac 0.525 4.908 162.035

m 417.516 kg / s

(b) a jF m c u

http://en.wikipedia.org/wiki/Moving_parts



F = 417.516 (185.18 – 138.89) x 10-3

F = 19.3268 kN

(c) s

a

F 19326.8F 46.29 N / kg / s

m 417.516

(d) s

s a

F F 46.29I 4.718 s

w m g 9.81

(e) Thrust power is, P = F x u = 9.3268 x 138.89 = 2684.3 kW.

6. 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 of 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 the jet, (ii) Resistance or Drag of the plane, (iii) Overall efficiency of the plane and (iv) Thermal efficiency. [ CO4 –H2 –Nov/Dec 2013]

Given:

No of jets, n = 2,

Diameter of jet, dj = 0.25 m,

Net Power at turbine, PE = 3000 kW,

Fuel consumption per kWh, mf = 0.42 kg/kWh

= 0.42 x 3000/3600

= 0.35 kg/s,

cv = 49 MJ/kg,

Velocity of plane, u = 300 m/s,

Density, ρi = 0.168 kg/m3,

Air fuel ratio, a

f

m53

m



Solution:

Mass flow rate of air fuel mixture, a fm m m

af

f

mm 1m

0.35 53 1

m 18.9 kg / s

We know that,

Mass flow rate, j j

m nA c

j j

2

j

2

j

j

m 2 A c

m 2 d c4

18.9 2 0.25 c 0.1684

c 1145.91 m / s

Absolute velocity of jet, cabs = cj – u

= 1145.91 – 300

cabs = 845.91 m/s

Thrust or Resistance or Drag, j a

F mc m u

a fm m m 18.9 0.35 18.55 kg / s

F = 18.9 x 1145.91 – 18.55 x 300

F = 16.09 x 103 N



Overall efficiency, j

0

f

m c u u

m Calorific Value

0 6

0

18.9 1145.91 300 300

0.35 49 10

0.279 or 27.9%

Thermal efficiency,

2 2

j

t

f

1m c u2

m Calorific value

2 2

6

118.9 1145.91 300

20.35 49 10

t 0.673 (or) 63.7%

7. A turbojet engine operates at an altitude of 11 km and a mach number of 0.82. The data for an engine is given below: Stagnation temperature at the turbine inlet = 1220 K, Stagnation temperature rise through the compressor = 170 K, Calorific value of the fuel = 42 MJ/kg, Compressor efficiency = 0.75, Combustion chamber efficiency = 0.97, Turbine efficiency = 0.83, Exhaust nozzle efficiency = 0.96 and Specific impulse = 20 seconds. Determine: (i) Air fuel ratio, (ii) Compressor pressure ratio, (iii) Turbine pressure ratio, (iv) Exhaust nozzle pressure ratio and (v) Mach number of exhaust gas.

[ CO4 –H2 –Nov/Dec 2013]

Given:

Altitude, z = 11 km = 11000 m

Inlet Mach number, M1 = 0.82

Stagnation temperature at turbine inlet, T03 = 1220 K

Stagnation temperature rise through the compressor, T02 – T01 = 170 K



Calorific value of the fuel, C.V = 42 MJ/kg = 42 x 103 J/kg

Compressor efficiency, ηC =0.75

Combustion chamber efficiency, ηB = 0.97

Turbine efficiency, ηT = 0.83

Nozzle efficiency, ηN = 0.96

Specific impulse, Isp = 20 seconds

Solution:

From gas table at z = 11000 m

Ti = 216.65 K

ai = 295.20 m/s

Stagnation Temperature – Mach Number Relation

20T 1

1 MT 2

At inlet, 20ii

i

T 11 M

T 2

20i

i

0i

i

0i i

0i

0i 01

T 1.4 11 0.82

T 2

T1.134

T

T 1.134 T

1.134 216.65

T 245.78 K

T T 245.78 K

From the given data,

T02 – T01 = 170 K

T02 = 170 + T01



= 170 + 245.78

T02 = 415.78 K

We know that combustion efficiency

ap 03 p 02 p 03

fB

mc T c T c T

m

C.V

a

f

6

3 6a

f

6

6 3a

f

a

f

a

f

m1005 1220 1005 415.78 1005 1220

m0.97

42 10

m808 10 1.22 10

m0.97

42 10

m39.52 10 808 10

m

m48.9

m

m0.020

m

Air fuel ratio, a

f

m0.020

m



Compressor efficiency,

1

01 0c

c

02 01

1

0c

1

0c

1

0c

10c

1.4

1.4 1

0c

T R 1

T T

245.78 R 1

0.75170

R 1 0.5187

R 1.5187

R 1.5187

1.5187

R 4.30

Compressor pressure ratio 0cR 4.30

We know that,

Compressor pressure required = Power supplied by the turbine

p 02 01 p 03 04

02 01 03 04

03 04

04 03

04

m c T T m c T T

T T T T

170K T T

T T 170K

T 1220 170K

04T 1050K

Turbine efficiency,

03 04T

03 1

0c

T T

1T 1

R



1

0T

1

0T

1

0T

1

0T

1

0T

1.4 1

1.40T

0T

1700.83

11220 1

R

11 0.1679

R

10.8320

R

1R

0.8320

R 1.2

R 1.2

R 1.903

Turbine pressure ratio, 0303

04

pR 1.903

p

Entry Mach number, Mi = u/ai

0.82= u/295.2

Velocity of aircraft, u = 242.064 m/s

We know that,



Specific impulse,

sp

u 1I 1

g

242.064 120 1

9.81

10.810 1

0.552

Effective speed ratio, j

u

c

j

j

241.930.552

c

c 438.27 m / s

Velocity of jet, cj = 438.27 m/s

Stagnation temperature at exit of nozzle,

2

e0e e

p

2

j

04 e 04 0e e j

p

2

e

cT T

2c

cT T T T ;c c

2c

438.271050 T

2 1005

Exit temperature, Te = 954.43 K

Nozzle efficiency,

04 eN

03 1

0N

T T

1T 1

R



1

0N

1

0N

1

0N

1

0N

0N

1050 954.430.96

11050 1

R

11 0.0948

R

R 1.10

R 1.10

R 1.417

Nozzle pressure ratio, 040N

e

pR 1.417

p

Mach number at exhaust exit, e j

e

e

c or cM

a

e

e

e

e

438.27M

RT

438.27M

1.4 287 954.37

M 0.707

8. Turbojetengine operating at a Mach number of 0.8 and the altitude is 10 km has

the following data. Calorific value of the fuel is 42,800 kJ/kg. Thrust force is 50

kN, mass flow rate of air is 45 kg/s, mass flow rate of fuel is 2.65 kg/s.

Determine the specific thrust, thrust specific fuel consumption, jet velocity,

thermal efficiency, propulsive efficiency and overall efficiency. Assuming the

exit pressure is equal to ambient pressure. [ CO4 –H2 –May/Jun 2012]

Given:



Mach number, M = 0.8

Altitude, z = 10 km

Calorific value, CF = 42,800 kJ/kg

Thrust power, F = 50 kN = 50,000 N

Mass flow rate of air, am 45 kg / s

Mass flow rate of fuel, fm 2.65 kg / s

Solution:

a f

m m m

= 47.65 kg/s

From Gas Tables at z = 10 km,

a = 299.6 m/s

T = 233.15 K

Flight speed, u = a. M

= 299.6 x 0.8 = 239.68 m/s

Thrust, j

F m c u

50 x 103 = 47.65 [Cj – 239.68]

Cj = 1275.67 m/s

0 1878j

u.

c

Specific thrust, Fs =

350 10

47 65

1049 32

F

.m

. N / kgof jet / s

Thrust specific fuel consumption = fm

F

3

5

2 65

50 10

5 3 10

.

. kg / N s

Propulsive power = 2 21

2j

mC mu



= 2 21

47 65 1275 67 47 65 239 682

. . . .

= 37.48 MW

Heat supplied = mf x Cv

= 2.65 x 42800 x 103

= 113. 42 MJ

Propulsive efficiency, 2

1p

= 31.97 %

Thermal efficiency, t

Pr opulsive power

Heat supplied

= 33.04%

Overall efficiency, 0 p th

= 10.57%

9. Write the equations to calculate propulsive efficiency and thermal efficiency if

an aircraft. [ CO4 –H1 –May/Jun 2014 & 2012]

Solution:


p

Pr opulsive power or thrust power

Power output of the engine

Substituting for thrust in terms of the effective jet velocity,

j

p2 2 jj

pj

p

m c u u 2u

1 c um c u2

2 2

c 111

u

2

1

Thermal efficiency,



th

2 2

i

th

f f

2

j 2

th

f


power input to the engine through fuel

1m c u2mQ

c11

2 fQ

10. 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 bar. Find the following: (i) Thrust,

(ii) Thrust power, (iii) Propulsive efficiency. [ CO4 –H1 –Nov/Dec 2013]

Given:

Air craft speed, u = 440 kmph = 122.22 m/s

Diameter of the propeller, d = 4.1 m

Speed ratio, σ = 0.8

Ambient temperature, T = 255 K

Pressure, p = 0.55 bar = 0.55 x 105 N/m2

To find:

(i) Thrust (F)

(ii) Thrust power (P)

(iii) Propulsive efficiency (ηp)

Solution:

Area of the propeller disc, 2

4A d



2 24 1 13 20

4A . . m

Speed ratio, j

u

c

122 22

0 8j

..

c

Velocity of jet cj = 152.77 m/s

Velocity of air flow at the propeller

1

2j

c u c

1

122 22 152 772

. .

c = 137.49 m/s

Mass flow rate of air fuel mixture, m A c

50 55 10

13 20 137 49287 255

1363 9

pA c

RT

.. .

m . kg / s

W.k.t a f

m m m

Since mass flow rate of fuel f

m

is not give, let us take

a

m m



1363 9am . kg / s

Thrust, 1363 9 152 77 122 22

41 667

a jF m c u

. . .

F , N

Thrust power, P = Thrust (F) x Flight speed (u)

= 41,667 x122.22

P = 5.09 x106 W

Propulsive Efficiency,

2 2 122 22

152 77 122 22

0 88 88

p

j

p

u .

c u . .

. or %

11.An aircraft flies at 960 km/hr. One of its turbojet engines takes in 401g/s of air

and expands the gases to the ambient pressure. The air-fuel ratio is 50 and the

lower calorific value of the fuel is 43 MJ/kg. For maximum thrust power

determines:

1. Jet velocity

2. Thrust

3. Specific thrust

4. Thrust power

5. Propulsive, thermal and overall efficiencies and TSFC.

[ CO4 –H2 –Apr/May 2014] Solution: u = 960 x 1000/3600 = 266.7 m/s

(a) For maximum thrust power

σ = u/cj= 0.5



cj= 266.7 / 0.5

= 533.4 m/s



12.An aircraft propeller flies at aspeed of 440 kmph.The diameter of the propeller

is 4.1m and the speed ratio is 0.8. The ambient conditions of air at the flight

altitude are T = 255 K and P = 0.55bar.Find the following: Thrust,

Thrust Power and Propulsive efficiency. [ CO4 –H2 – Apr/May-2015]



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

1. What is rocket propulsion? Why a rocket is called a non-air breathing engine?

[ CO5 –L1 – May/June - 2007]

If the propulsion unit contains its own oxygen supply for combustion purpose, the

system is known as “Rocket propulsion”. Because rockets carry their own oxidizers with

them, they are called a non-air breathing engine.

2. What are the types of rocket engines? [ CO5 –L1 – May/June - 2007]

Rocket engines are classified in the following manner,

(a) On the basis of source of energy employed

i. Chemical rockets,

ii. Solar rockets

iii. Nuclear rockets and

iv Electrical rockets

(b) On the basis of propellants used

i. Liquid propellant

ii. Solid propellant

iii. Hybrid propellant rockets.

3. Mention any four applications of rockets. [ CO5 –L1 – May/June - 2012]

The applications of rockets are

1. Space propulsion systems

2. Missiles

3. Rocket assisted take off for aircrafts

4. Positioning and course correction of satellite in their orbits.



4. Define thrust for a rocket engine and how it is produced. [ CO5 –L1 – May/June

- 2012]

The trust that propels the rocket at a given velocity is known as thrust. This is

produced due to the change in momentum flux of the outgoing gasses as well as the

difference between the nozzle exit pressure and the ambient pressure.

5.What is a mono propellant? Give examples. [ CO5 –L1 – May/June – 2013&2015]

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

is known as a monopropellant. It is stable at normal ambient conditions and liberates

thermo-chemical energy on heating.

6. What are the types of rocket engines based on source of energy employed?

[ CO5 –L1 – May/June – 2013&2015]

The types of rocket engines are

(i) Chemical rocket engine

(ii) Solar rocket engine

(iii) Nuclear rocket engine

(iv) Electrical rocket engine

7. A rocket flies at 10080 km/hr with an effective exhaust jet velocity of 1400 m/s,

and the propellant flow rate of 5 kg/s. Find the propulsion efficiency and

propulsion power of the rocket. [ CO5 –L1 – May/June – 2012]

Solution:



3

j

p

j

p 2

p j

10080 10u

3600

2800 m / s

C 1400 m / s

m 5 kg / s

uσ 2C

2ση 80%

σ 1

Pr opulsive power m .C .u

19.6MW

8. Mention any four specific applications of rocket. [ CO5 –L1 – May/June – 2012]

The applications of rocket are

(i) Military

(ii) Space

(iii) Aircraft

(iv) Communication

9. What are the types of liquid propellants used in rocket engines? [ CO5 –L1 –

May/June – 2007]

1. Mono propellants

2. Bi-propellants

10. Give two examples for liquid propellants. [ CO5 –L1 – May/June – 2007]

Liquid fuels: Liquid hydrogen, UDMH, hydrazine

Solid fuels: Polymers, plastics and resin material



11. Compare solid and liquid propellant rockets. [ CO5 –L2 – Nov/Dec – 2007]

Solid propellant rockets Engine

Liquid propellant rockets engine

a) Solid fuel and oxidizers are used in

rockets engines

b) Generally stored in combustion

chamber (both oxidizer and fuel).

c) Burning in the combustion chamber

is uncontrolled rate.

a) Liquid fuels and oxidizers are

usedrockets engines

b) Separate oxidizers and fuel tanks

are used for storing purpose.

c) Controlled rate.

12. Distinguish between monopropellant and bipropellant. [ CO5 –L2 – Nov/Dec –

2007]

Monopropellant Bipropellant

a) Fuels and oxidizers are combined in

a single chemical.

b) Require only one transfer pump.

a) Liquid fuels and oxidizers are stored

in separate units.

b) Separate oxidizer and fuel transfer

pumps are required.

13. What is bi-propellant? Give example. [ CO5 –L1 – Nov/Dec – 2008]

A liquid propellant which contains the fuel and oxidizer in separate units is known as bi-propellant.

14. Name some oxidizers and fuels used in rockets. [ CO5 –L1– Nov/Dec – 2008]

OXIDIZERS:

a) Liquid oxygen b) Hydrogen peroxide c) Nitrogen tetroxide d) Nitric acid



FUELS:

a) Gasoline b) Liquid hydrogen c) UDMH

15. What are the advantages and disadvantages of liquid propellant rockets?

[CO5 –L1 – May/June – 2011]

Advantages:

Liquid propellant can be reused or recharged. Hence it is economical.

Increase or decrease of speed is possible when it is in operation.

Storing and transporting is easy as the fuel and oxidizer are kept separately.

Specific impulse is very high.

Disadvantage:

They require more complex engine systems to transfer the liquid propellants

to the combustion chamber.

Some liquid propellants need extremely low temperature-to remain liquid.

Hence, it is very difficult to handle and store.

16. What are inhibitors? [ CO5 –L1 – May/June – 2011]

Inhibitors are material which are used to regulate (or prevent) the burning of

propellant at some sections.

17. Give the important requirement of rocket engine fuels.

What are the disadvantages of rocket engine. [ CO5 –L1– Nov/Dec – 2014]

It must be able to produce a high chamber temperature.

It should have a high calorific value per unit of propellant.

It should not chemically react with motor system including tanks, piping, valves

and injection nozzles.



18. What is meant by restricted burning in rockets? [ CO5 –L1– Nov/Dec – 2014]

In restricted burning, the inhibition material (or) restrictions prevent the

propellant grain from burning in all directions. The propellant grain burns only at some

surfaces while other surfaces are prevented from burning.

18. What is hydrazine? [ CO5 –L1– Apr/May – 2014]

Hydrazine is a family of compounds like Hydrzine, Monomethyl hydrazine (MMH)

and Unsymmetrical dimethyl hydrzine (UDMH) that are used in the rocket industry as

monopropellants.

It is used as a monopropellant because hydrazine will decompose (ignite) when

placed in contact with platinum group metal catatlysts. Hydrazine is popular with

satellite builders because the monopropellant properties save space, complexity, and

weight. It saves space because there is no need to carry up a separate oxidizer. It save

weight and simplifies the system because there is only one set of plumbing and no

ignition system.

20. What is specific impulse of a rocket engine? [ CO5 –L1– Apr/May – 2014]

A prime criterion for rating rocket performance is specific impulse, which provides

an index of the efficiency with which a rocket uses its supply of propellant or working

fluid for thrust production. Specific impulse is the change in momentum per unit mass

for rocket fuels.

The specific impulse is:

ISP = ueq/ge

Where;

ISP = specific impulse

Ueq = total impulse/mass of expelled propellant

ge = acceleration at Earthˊs surface (9.8 m/s2)

21. List any two advantages and disadvantages of rocket compared to air

breathing engines.(AU Nov 2009) [ CO5 –L1– Nov/Dec – 2009]



Advantages:

They do not depend on atmospheric air for operation

They are compact in design

Disadvantages:

They are not normally reusable

Costlier in compared to the air breathing engines

22. What is the need for multi stage rockets? [ CO5 –L1– Nov/Dec – 2006]

Upon completion of the operation of a particular stage of a multi-stage rocket, the

same can be detached from the rocket so that the total load on the propulsion system is

reduced and hence the thrust requirements are also reduced.

23. What is meant by hypergolic propellant? [ CO5 –L1– Apr/May – 2010]

Hypergolic propellant do not require ignition.

24. What is bypass engine and define bypass ratio? [ CO5 –L1– Apr/May – 2010 &

Nov/Dec-2007]

Bypass engine: Turbofan engines are usually described as bypass engine. In this type

of engine a portion of the total flow of air bypass part of the compressor.

Bypass ratio: The ratio of the mass flow rates of cold air (mc) and the hot air (mh) is

known as Bypass ratio.

25. What is thrust augmentation and any two methods of achieving it. [ CO5 –L1–

Nov/Dec – 2009]

To achieve better take-off performance, additional fuel is burnt in the tail pipe

between the turbine exhaust section and entrance section of the exhaust nozzle. This

method of thrust augmentation increases the jet velocity and is known as after burning.

It is used for fast and easier take off.

The two methods of achieving thrust augmentation.

1. Momentum thrust

2. Pressure thrust



26. What is after burning in turbojet engines? [ CO5 –L1– Apr/May – 2009]

Large quantity of oxygen is available in the exhaust gas which can support the

combustion of additional fuel. When the thrust of the engine is desired to be increased

without changing the physical dimensions of the compressor, turbine, etc., additional

quantity of fuel can be burnt in a section of the jet pipe to increase the velocity of the jet.

This process is called reheating or after burning.

27. Why rocket engines are called non-breathing engine? [ CO5 –L1– Apr/May –

2009]

In rocket engine, the thrust required for the propulsion of the rocket is produced

by the high velocity of gases leaving from the nozzle which is similar to jet propulsion.

In air breathing engine, combustion takes place by using atmospheric air. But in

rocket engines, combustion takes place by using its own oxygen supply. So , it is called

as non-air breathing engines.

28. Name three commonly used aircraft engines. [ CO5 –L1– Nov/Dec – 2009]

1. Turbojet engine.

2. Turboprop engine.

3. Turbofan engine.

29. Name some propellants for space application. [ CO5 –L1– Nov/Dec – 2009]

1. Hydrogen peroxide

2. Hydrazine

3. Nitroglycerine and

4. Nitromethane, etc.

30. What is terminal velocity? [ CO5 –L1– Nov/Dec – 2010]

The terminal velocity of an object falling towards the earth is the velocity at which

the gravitational force pulling it downwards is equal and opposite to the air resistance-

pushing it upwards.

31. When a ram jet engine can be used? [ CO5 –L1– Nov/Dec – 2009]

Ramjet produces very high thrust with high efficiency at supersonic speeds. So, it

is best suitable for high speed aircrafts.



32. Why rocket is called as non-breathing engine? Can rocket work at vacuum?

[ CO5 –L1– Apr/May – 2014]

Rocket is called as non-breathing engine because of the rockets which do not

take air from outside for propulsion.

33. What is the use of inhibitors in solid propellants? [ CO5 –L1– Apr/May – 2014]

Inhibitors or restrictors are also used to regulate or prevent burning the propellant

grain or some section so fits surface.

34. Define characteristic velocity. (Dec-14) [ CO5 –L1– Nov/Dec – 2014]

Characteristic velocity:

Rocket performance is also frequently expressed in term s of a characteristic

velocity which is defined as Characteristic velocity is equal to the effective jet velocity for

a thrust coefficient of unity.

V* = Effective jet velocity / thrust coefficient

V* = cj/CF

35. What are the advantages and disadvantages of liquid propellant rocket

engine? [ CO5 –L1– Nov/Dec – 2014]

Advantages

Cryogenic propellants are in the gaseous state at normal temperatures

and require extremely low temperature to maintain them in the liquid state;

They release much higher heat energy in the thrust chamber and offer

better performance compared to the storable type .

Some commonly used cryogenic propellants are liquid oxygen, hydrogen,

fluorine and ammonia.



Disadvantages

Both “storable” and cryogenic propellants are used in liquid rocket

engines.

Storable propellants (fuels and oxidizers) can be stored in the liquid state

at normal ambient temperatures whereas cryogenic Propellants (at very

low temperature) require special arrangements for storage and handling.

PART-B

1. Write short notes on:

(a) Solid-propellants

(b) Liquid propellants[ CO5 –H2– Apr/May – 2016]

SOLID-PROPELLANTS

Two general types of solid propellants are in use. The first, the so called double-

base propellant, consists of nitrocellulose and notroglycerine, plus additives in small

quantity. There is no separate fuel and oxidizer. The molecules are unstable, and upon

ignition break apart and rearrange themselves, liberating large quantities of heat. These

propellants lend themselves well to smaller rocket motors. They are often processed

and formed by extrusion methods, although casting has also been employed

The other type of solid propellant is the composite. Here, separate fuel and oxidizer

chemicals are used, intimately mixed in the solid grain. The oxidizer is usually

ammonium nitrate, potassium chlorate, or ammonium chlorate, and often comprises as

much as four-fifths or more of the whole propellant mix. The fuels used are

hydrocarbons, such as asphaltic-type compounds, or plastics. Because the oxidizer has

no significant structural strength, the fuel must not only perform well but must also

supply the necessary form and rigidity to the grain. Much of the research in solid

propellants is devoted to improving the physical as well as the chemical properties of

the fuel.



Ordinarily, in processing solid propellants the fuel and oxidizer components are

separately prepared for mixing, the oxidizer being a powder and the fuel a fluid of

varying consistency. They are then blended together under carefully controlled

conditions and poured into the prepared rocket case as a viscous semisolid. They are

then caused to set in curing chambers under controlled temperature and pressure.

Solid propellants offer the advantages of minimum maintenance and instant readiness.

However, the more energetic solids may require carefully controlled storage conditions,

and may offer handling problems in the very large sizes, since the rocket must always

be carried about fully loaded. Protection from mechanical shocks or abrupt temperature

changes that may crack the grain is essential.

LIQUID BIPROPELLANTS

Most liquid chemical rockets use two separate propellants: a fuel and an

oxidizer. Typical fuels include kerosene, alcohol, hydrazine and its derivatives, and

liquid hydrogen. Many others have been tested and used.

Oxidizers include nitric acid, nitrogen tetroxide, liquid oxygen, and liquid fluorine,

which exist as liquids only at very low temperature; this adds greatly to the difficulty of

their use in rockets. Most fuels, with the exception of hydrogen, are liquids at ordinary

temperatures.

Certain propellants combinationsare hypergolic; that is, they ignite

spontaneously upon contact of the fuel and oxidizer. Others require an igniter to start

them burning, although they will continue to burn when injected into the flame of the

combustion chamber.

LIQUID MONOPROPELLANTS

Certain unstable, liquid chemicals which, under proper conditions, will decompose and

release energy, have been tried as rocket propellants. Their performance, however, is

inferior to that of bipropellants or modern solid propellants, and they are of most interest

in rather specialized applications, like small control rockets. Outstanding examples of

this type of propellant are hydrogen peroxide and ethylene.



In general, the liquid propellants in common use yield specific impulses superior to

those of available solids. On the other hand, they require more complex engine system

to transfer the liquid propellants to the combustion chamber.

2. Explain the construction and operation of a solid propellant rocket engine. Also

name any four solid propellants and state its advantages and disadvantages.

[ CO5 –H2– Nov/Dec – 2004&2016]

SOLID PROPELLANT ROCKET

In the solid-chemical rocket, the fuel and oxidizer are intimately mixed together

and cast into a solid mass, called a grain, in the combustion chamber (fig). the

propellant grain is firmly cemented to the inside of the metal or plastic case, and is

usually cast with a hole down the center. This hole, called the perforation, may be

shaped in various ways, as star, gear, or other more unusual outlines, the perforation

shape and dimension affects the burning rate or number of pounds of gas generated

per second and, thereby, the thrust of the engine.

After being ignited by a pyrotechnic device, which is usually triggered by an

electrical impulse, the propellant grain burns on the entire inside surface of the

perforation. The hot combustion gasses pass down the grain and are ejected through

the nozzle to produce thrust.

The propellant grain usually consist of 1 of 2 types of chemical. One type is the

double-base, which consists largely of nitroglycerine and nitrocellulose. It resembles

smokeless gunpowder. The second type, which is now predominant, is the composite

propellant, consisting of an oxidizing agent, such as ammonium nitrate or ammonium

perchlorate intimately mixed with an organic or metallic fuel. Many of the fuels used are

plastics, such as polyurethane.



A solid propellant must not only produce a desirable specific impulse, but it must

also exhibit satisfactory mechanical properties to withstand ground handling and the

flight environment. Should the propellant grain develop a crack, for example, ignition

would cause combustion to take place in the crack, with explosion as a possible result.

Fig.1 Solid Rocket Propulsion Engine



It can be seen from figure that the case walls are protected from the hot gas by the

propellant itself. Therefore, it is possible to use heat-treated alloys or plastics for case

construction. The production of light-weight, high-strength cases is a major development

problem in the solid – rocket field.

Since nozzles of solid rockets are exposed to the hot gas flowing through them, they

must be of heavy construction to retain adequate strength at high temperature. Special

inserts are often used in the region of the nozzle throat to protect the metal from the

erosive effects of the flowing gas.

For vehicle guidance it is necessary to terminate thrust sharply upon command. This

may be accomplished with solid rockets by blowing off the nozzle or opening vents in

the chamber walls. Either of these techniques causes the pressure in the chamber to

drop and, if properly done, will extinguish the flame.

The specific impulse of various solid-propellant rockets now falls in the range 175 to

250 seconds. The higher figure of 250 applies to ammonium perchlorate-biased

propellants.

Some solid propellants:

Oxidizer-binder combination (solid):

Oxidizer Binder

Potassium perchlorate Thiokol or asphalt

Ammonium perchlorate Thiokol

Rubber

Polyurethane

Nitropolymer

Ammonium nitrate Rubber

Polyurethane

Nitropolymer



Double base

Boron metal components and oxidant

Lithium metal components and oxidant

Aluminum metal components and oxidant

Magnesium metal components and oxidant perfluoro-type propellants

Advantages:

They require simple engine systems

Propellants do not need extremely low temperatures

Disadvantages:

Solid propellants can be recharged.

Increase or decrease of speed is not possible when it is in operation.

Storing and transportation requires extra caution as fuel and oxidizer are already

in mixed condition. Should the propellant grain develop a crack, for example,

ignition would cause combustion to take place in the crack, with explosion as a

possible result

Specific impulse is low compared to liquid propellant engines.

3. A rocket engine has the following data. Combustion chamber pressure is 38

bar, Combustion chamber temperature is 3500 K, Propellant flow rate is 41.67

Kg/s and the properties of exhaust gases are γ = 1.3, R = 287J/Kg K. the

expansion takes place to the ambient pressure of 0.0582 bar. Calculate the nozzle

throat area, thrust, thrust Co-efficient, exit velocity of the exhaust and maximum

possible exhaust velocity. [ CO5 –H1– Apr/May – 2012]

Given Data:

Combustion chamber pressure Po = 38 bar



= 38𝘹105 N/m2

CC Temperature T0 = 3500 K

mp= 41.67 kg/s

= 1.3

R = 287 J/Kg K

Pe=Pa = 0.0582 bar = 0.0582 𝘹105 N/m2

To Find:

A*= ?

F = ?

CF= ?

Cj= ?

Cjmax= ?

Solution:

We know that,

mp = 𝜌AC = 𝜌*A*C*

𝜌* = P*/RT* and C* = a* = √ RT*

Assuming the flow to have reached critical condition at the Throat corresponding

to a nozzle pressure ratio (Pe/Po),

P*/PO = (2/ +1)/ +1

= (2/1.3+1)1.3/1.3+1

P*/38𝘹105 = (2/2.3)1.3/0.3

= 0.546



P*= 38𝘹105𝘹0.546

= 20.74𝘹105 N/m2

Now , for isentropic flow,

T*/T0=(P*/P0) -1/

T*=T0(P*/P0)

-1/

= 3500 (20.74𝘹105/38𝘹105)(1.3-1/1.3)

= 3043.1 K

𝜌* = P*/RT*

= 20.74𝘹105/287𝘹3043.1

= 2.375 kg/m3

As ,mp = 𝜌*A*C*

= 𝜌*A*√ RT*

A* =mp/𝜌*√ RT*

= 41.67/ 2.375𝘹√1.3𝘹287𝘹3043.1

= 0.01647 m3

We know that,

Thrust F=mp𝘹Cj

Where Cj= 2

1

RT0{1-(Pe/P0)

-1/ }

= 2 1.3

287 35001.3 1

{1-(0.0582/28)1.3-1/1.3}



= 3264.4 m/s

∴F =41.67 𝘹3264.4

F = 136.03 𝘹 103N

Thrust coefficient CF = F/P0 A*

= 136.03 𝘹 103./38 𝘹105𝘹0.01647

CF= 2.17

4. A rocket flies at 10,000 kmph with an effective exhaustive jet velocity of 1350

m/s and propellant flow rate of 4.8 kg/s. If the heat of reaction of the propellant is

6600 kJ/kg of the propellant mixture determine, (1) the propulsive efficiency and

power, (2) engine output and thermal efficiency and (3) overall efficiency. [ CO5 –

H1– Apr/May – 2013]

Given:

Rocket speed, u = 10000 kmph = 2777.7 m/s

Jet velocity, Cj = 1350 m/s

Heat produced, Q = 6600 kJ/kg

Calorific value (C.V) = 6600 x 103 J/kg

Propellant flow rate, mp = 4.8 kg/s

Solution:

W.k.t. Speed ratio, j

u 2777.7σc 1350



σ = 2.06


p 2 2

2σ 2 2.06η 0.785

1 σ 1 2.06

ηp = 78.5 %

Thrust, F = mp x Cj

= 4.8 x 1350

F = 6480 N

Propulsive power, P = F x u

= 6480 x 2777.7

P = 17.9 x 106 W

6

6

Pr opulsive powerPr opulsive efficiency


17.9 100.785


Engine output 22.9 10 W

t

6

t

p

6

3

t

o t p

o

Power output of the engineThermal Efficiency,η

Power input of the engine

22.9 10η

m C.V

22.9 10

4.8 6600 10

η 0.722 72.2%

Overallefficiency,η η η 0.722 0.785 0.567

η 56.7%



5. What are the advantages of liquid propellant rocket engines?

[ CO5 –H2– Apr/May – 2013 &Nov/Dec 2014]

Advantages of liquid propellant rocket engines:

(i) Liquid propellant engines can be reused after recovery. So it is

economical.

(ii) Combustion process is controllable i.e. it is easy to stop the

combustion by closing the fuel (or) oxidizer valve.

(iii) Speed regulation i.e., increase and decrease of speed is possible.

(iv) High specific impulse.

(v) More economical for long range operation.

(vi) Malfunctions and accidents can be rectified.

6. A rocket has the following data: Propellant flow rate = 5 kg/s, Nozzle exit

diameter = 10 cm, Nozzle exit pressure = 10.2 bar, Ambient pressure = 1.013 bar,

Thrust chamber pressure = 20 bar, Thrust = 7 kN. Determine the effective jet

velocity, actual jet velocity, specific impulse and the specific propellant

consumption. Recalculate the values of thrust and specific impulse for an altitude

where the ambient pressure is 10 m bar. [ CO5 –H1– Apr/May – 2012]

Given:

mp = 5 kg/s

Nozzle exit diameter de = 10 cm = 0.1 m

Nozzle exit pressure Pe= 1.02 bar

= 1.02 x 105 N/m2

Ambient pressure Pa= 1.013 bar

= 1.03x105N/m2

Thrust chamber pressure Po = 20 bar

= 20x105 N/m2

Thrust F = 7 KN

To find:



1. Effective jet velocity Cj

2. Actual jet velocity Ce

3. Specific impulse Isp

4. Specific propellent consumption (Spc)

Solution:

We know that,

F = mp×Cj

7x103 = 5xCj

Cj= 1400 m/s

Effective jet velocity Cj= 1400 m/s

Now ,Thrust F = mp×Ce+(Pe- Pa)Ae (∵Ae=

de

2)

7x103 = 5xCe+(1.02x105-1.03x105)

(0.12)

∴ Actual jet velocity Ce = 1399.9 m/s

We know,

Specific impulse Isp= F/Wp = F/mpxg

=

Isp= 142.71 S

Specific propellent consumption (Spc)

Spc= F/Wp

= 1/Isp = 1/142.71

∴Spc= 7x 10-3 S-1

Result:



1. Cj= 1400 m/s

2. Ce = 1399.9 m/s

3. Isp= 142.71 S

4. Spc= 7x 10-3 s-1

7. List out the important properties of solid propellants. [ CO5 –L1– Apr/May –

2013& Nov/Dec-14]

a) It should release large amount of heat during combustion.

b) Physical and chemical properties should not change during processing.

c) It should have high density.

d) It should not be poisonous and hazardous.

e) It should be cheap and easily available.

f) It should be non-corrosive and non-reactive with components of the engine.

g) Storage and handling should be easy.

8. The effective jet velocity from a rocket is 2700m/s. the forward flight velocity is

1350m/s and the propellant consumption is 78.6 Kg/s. Calculate: thrust, thrust

power and propulsive efficiency. [ CO5 –H2– Apr/May – 2013]

Given:

Cj = 2700 m/s

u=1350 m/s

mp = 78.6 kg/s

To find :

1) Thrust F

2) Thrust power

3) Propulsive efficiency𝜼p

Solution:

We know that



F = mp×Cj

= 78.6 x 2700

F = 212.22x103 N

Thrust power P = Fxu

= 212.22x103x1350

P = 286.49x106 W

Speed ratio σ = u/Cj

=

σ = 0.5

We know that,

Propulsive efficiency p =

=

𝜼p= 80 % Result :

1) F = 212.22x103 N

2) P = 286.49x106 W

3) 𝜼p= 80 %

9. The specific impulse of a rocket is 1255 and the flow rate of propellant is 44

Kg/s. the nozzle throat area is 18 cm2 and the pressure in the combustor is 25 bar.

Determine the thrust coefficient, propellant flow coefficient, specific propellant

consumption and characteristic velocity. [ CO5 –H2– Apr/May – 2013]

Data given

Specific impulse Isp= 125 S



Propellant flow rate mp= 44 Kg/s

Nozzle throat area A* = 18 cm2

= 18 x10-4m2

Combustor pressure P0=25 bar

= 25x105 N/m2

To find

1. Thrust coefficient CF

2. propellant flow coefficient

3. specific propellant consumption (Spc)

4. characteristic velocity C*

Solution

We know that,

Specific impulseIsp= F/Wp

= F/mpxg

F= Ispx mpx g

Thrust F= 125 x 44 x 9.81

Thrust F= 53.95 x 103N

Thrust Coefficient CF =

=3

5 4

53.95 10

25 10 18

CF= 11.99

Propellant Weight flow coefficient



Cw=

=

= 5 425 10 18 10

Cw= 95.92×10-3

Specific propellant consumption (Spc) =

pm g

F

=

Spc=8 10-3 S-1

Thrust F = mp×Cj

53.95×103 = 44 × Cj

Effective jet velocity Cj= 1226.13 m/s

Characteristic velocity C* =

=

C* = 102.26 m/s

Result

1. CF= 11.99

2. Cw= 95.92×10-3

3. Spc=8 10-3 S-1

4. C* = 102.26 m/s



10. Draw a neat sketch explaining the general working of the hybrid propellant

Rocket. [ CO5 –H3– Nov/Dec – 2013]

Hybrid rocket engines employ a combination of liquid and solid propellants in a

large number of experiments and comparatively fewer actual flights of hybrid rockets the

fuel is solid and the oxidizer a liquid. This type of rocket combines the advantages of

both the liquid and solid propellant rockets.

Figure shows the arrangement employed in a hybrid rocket. The liquid oxidizer is

injected into the central region of the solid fuel shell where the chemical reaction

between the two propellants takes place. For some combinations ignition is not required

to start the reaction; merely a contact between the fuel and the oxidizer starts the

requires reaction. Such propellants are known as hypergolic. Rocket engines with

hypergolic propellants are easy to fire, and the reaction can be started and stopped as

desired in a given operation.

Fig.2 Hybrid propellant rocket engine



Some fuel-oxidizer combination for hybrid propellant rockets are:

Beryllium hydride (Be-H2) - Fluorine (F2)

Lithium hydride (Li H) - Chlorine Trifluoride (CIF3)

Lithium hydride (Li H) - Nitrogen tetroxide (N2O4)

Hydrocarbon (CH2)n - Nitrogen tetroxide (N2O4)

Hybrid rockets would r preferred over other types in view of their simplicity and

other advantages some of which are given below:

1. Thrust control is comparatively easier because only the flow of the liquid

oxidizer need be regulated.

2. Since the fuel and the oxidizer are kept separate the chemical deterioration

that occurs in solid propellant rockets is absent here.

3. Hybrid rockets are lighter compared to the corresponding liquid propellant

type on account of less elaborate propellant pumping equipment and higher

fuel density.

4. There is greater choice in the selection of fuel grain configuration compared

to the solid propellant rockets.

5. In case of an accident or crash the explosion (if any) is less destructive

compared to the liquid propellant rocket engines.

11. A rocket operating at an altitude of 19 km with the following data: Propellant

flow rate = 1 kg/s, Thrust chamber pressure = 28 x 103 N/m2. Thrust chamber

temperature = 2500 K and Nozzle area ratio = 10.12. Calculate: (i) Effective jet

velocity and (ii) Specific impulse. Take γ=1.3 and R=355 J/kgK. [ CO5 –H1–

Nov/Dec – 2013]

Given:

Altitude, z = 19 km = 19,000 m

Propellant flow rate, mp = 1 kg/s

Thrust chamber pressure, p0 = 28 x 105 N/m2

Thrust chamber temperature, T0 = 2500 K



Nozzle area ratio, 10 12

1 3 355

eA

.A *

. , R J / kgK

To find:

(i) Thrust, F

(ii) Effective jet velocity, cj

(iii) Specific impulse, Isp

Solution:

Refer gas tables at z = 19,000 m

pa = 0.0641 bar

pa = 0.0641 x 105 N/m2

Refer isentropic flow table for 10 12eA

.A *

= 10.2 and γ = 1.3

0

0

0 0093

0 339

3 60

e

e

e

p.

p

T.

T

M .

pe= 0.0093 x 28 x 105

pe=0.260 x 28 x 105

Te = 0.339 x T0

Te = 0.993 x 2500

Te =847.5 K

We know that,



3 60

3 601 3 355 847 5

2251 42

ee

e

e

e

e

e

cExit Mach number M

a

c.

RT

c.

. .

Actual jet velocity ,c . m / s

5

5 2

5 5 5

0 260 101 2251 42

355 847 5

5 13 10

1 2251 42 0 260 10 0 0641 10 5 13 10

2351 9

2351 9 1

2351 9

p e e e

ep e e

e

e

e

p e e a e

p j

j

j

Pr opellant flow rate, m A c

pm A c

RT

.A .

.

A . m

Thrust,F m c p p A

. . . .

F . N

We know that

Thrust,F m c

. c

c .



2351 9

1 9 81

239 746

sp

p

sp

FSpecific Impulse,I

W

F

m g

.

.

I . s

12. Calculate the orbital and escape velocities of a rocket at mean sea level and

an altitude

of 300 km from the following data

Radius of the earth at mean sea level = 6341.6 km

Acceleration due to gravity at mean sea level = 9.81 m/s2. [ CO5 –H2– May/Jun –

2013]

Solution:

At Z = 0,

orb o o

orb

esc orb

esc

o

oorb o

o

orb

esc

u g R 9.809 6341.6 1000

u 7887 m / s.

u 2 u 2 7887

u 11,154 m / s.

At Z 300km,

R R Z 6341.6 300 6641.6km

g 9.809u R 6341.6 1000

R Z 6641.6 1000

u 7706.8 m / s.

u 2 7706.8 10,899.96 m / s.



14.List the main components of Liquid propellant rockets engine and explain.

[ CO5 –H2– Apr/May-2015]

Liquid propellant rockets engine

Construction:

The construction of Liquid propellant rockets engine is shown in figure.

Liquid fuel (refine petrol ,liquid hudrogen,hydrazine,etc.)

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