Vijayamangalam – 638 056Department of Mechanical Engineering Effective Study Material ME 6604 – Gas Dynamics and Jet Propulsion Prepared by C.Ravindiran. ME., Assistant Professor, Department of Mechanical Engineering, Sasurie College of Engineering. Vijayamangalam – 638 056 Department of Mechanical Engineering Subject Name: Gas Dynamics and Jet Propulsion UNIT I BASIC CONCEPTS AND ISENTROPIC FLOWS Energy and momentum equations of compressible fluid flows – Stagnation states, Mach waves and Mach cone – Effect of Mach number on compressibility – Isentropic flow through variable ducts – Nozzle and Diffusers PART-A 1. Define compressible flow and Mach number. (April/May 2015) (April/May 2010) Key Hint: Compressible flow: The density of the fluid changes from point to point (ρ ≠ Constant) (1 mark) Mach number: The ratio of the fluid velocity to the velocity of sound M= c a (1 mark) 2. Define stagnation state. (Nov/Dec 2012) (April/May 2010) Key Hint: It is obtained by decelerating a gas isentropically to zero velocity at zero elevation. (2 marks) 3. Distinguish between nozzle and diffuser. (May/June 2014) Key Hint: Nozzle: It is used to increase the velocity and decrease the pressure of fluids. (1 mark) Diffuser: It is used to increase the pressure and decrease the velocity of fluids. (1 mark) 4. When does maximum flow occur for an isentropic flow with variable area duct? (May/June 2014) (Nov/Dec 2014) Key Hint: Mass flow rate will be maximum at throat section where the Mach number is one. (2 marks) 5. Define zone of action and zone of silence. (Nov/Dec 2013) Key Hint: Zone of action: The region inside the Mach cone. Zone of silence: The region outside the Mach cone. (1 mark) (1 mark) 6. Name the different regions of compressible fluid flow. (Nov/Dec 2013) (May/June 2012) Key Hint: Incompressible flow region Subsonic flow region Transonic flow region Supersonic flow region Hypersonic flow region (2 marks) 7. What are the basic differences between compressible and incompressible flows? (May/June 2013) Key Hint: Compressible flow: The density of the fluid changes from point to point (ρ ≠ Constant) (1 mark) In Compressible flow: The density of the fluid is constant (ρ ≠ Constant) (1 mark) 8. What is the cross section of the nozzle required to increase the velocity of compressible fluid flow from (a) Subsonic to supersonic, (b) Subsonic to sonic. (May/June 2012) Key Hint: The cross section of the nozzle is decided based on the equation, dA dc 2 A = c (M – 1) (a) Subsonic to supersonic: Convergent - Divergent Subsonic to sonic: Convergent (1 mark) (1 mark) 9. What is subsonic, sonic and supersonic flow with respect to Mach number? (April/May 2011) (April/May 2010) Key Hint: Subsonic: Fluid velocity is less than the sound velocity. Sonic: Fluid velocity is equal to the sound velocity. Supersonic: Fluid velocity is more than the sound velocity (2 marks) 10. How the area and velocity vary in supersonic flow of nozzle and diffuser? (April/May 2011) Key Hint: For nozzle: Area – Decreases Velocity – Increases (1 mark) For Diffuser: Area – Increases Velocity – Decreases (1 mark) 11. Zone of silence is absent in subsonic flow. Why? (May/June 2012) Key Hint: It is observed that the wave fronts move ahead of the source of disturbance and therefore the zone of silence is absent. (2 marks) 12. Name the four reference velocities that are used in expressing the third velocities in nondimensional form. (May/June 2013) Key Hint: 1. Local velocity of sound. 2. Stagnation velocity of sound. 3. Maximum velocity of fluid. 4. Critical velocity of fluid. (2 marks) 13. What is mean by gas dynamics? (April/May 2015) Key Hint: Gas dynamics deals with the study of motion of gases and its effects. (2 marks) 14. Express the stagnation enthalpy in terms of static enthalpy and velocity of flow. (Nov/Dec 2009) Key Hint: 1 h0=h+ c 2 2 (2 marks) 15. What is the advantage of using M* (second kind of Mach number) instead of M (Local Mach Number) in some cases? (May/June 2009) (Nov/Dec 2008) Key Hint: At high velocities M approaches infinity but M* gives a finite value. M is proportional to the fluid velocity and sound velocity, but M * is proportional to the fluid velocity alone. (2 marks) Part – B 1. A conical diffuser has entry and exit diameters of 15cm and 30cm respectively. The pressure, temperature and velocity of air at entry are 0.69bar, 340 K and 180 m/s respectively. Determine (i)the exit pressure, (ii)the exit velocity and (iii) the force exerted on the diffuser walls assume isentropic flow, γ=1.4, C p=1.00kJ/kg-K. (Nov/Dec 2013) (April/May 2011) (May/Jun 2009) (Nov/Dec 2008) (MK) (16) Key Hint: T1 =0.952 From isentropic tables at this temperature ratio, T 01 M1 = 0.5, P1 A F =0.843, ¿ =1.34, ¿ =1.203 P01 A F A2 ¿ =5.37 A2 from isentropic tables at this area ratio, M 2=0.107, p2 F =0.992, 2¿ =4.3 p02 F2 (i) (ii) (iii) p2 = 0.992xp02 p2 =0.81145bar A2c2= A1c1 c2=45m/s τ=F2-F1 τ=4243 N (2 marks) (2 marks) (4 marks) (4 marks) (4 marks) 2. The pressure, temperature and Mach number at the entry of a flow passage are 2.45bar, 26.50C and 1.4 respectively. If the exit Mach number is 2.5, determine the following for adiabatic flow of a perfect gas (γ = 1.3, R = 0.469 kJ/kg K). (i) Stagnation temperature, (ii) Temperature and velocity of gas at exit, (iii) The flow rate per square meter of the inlet cross section. (Nov/Dec 2012) (April/May 2010) (MK) (16) Key Hint: From isentropic table, for γ = 1.3, M1 = 1.4 T1 =0.773,T 01=T 02=T 0=387.5 K T 01 (4 marks) From isentropic table, for γ = 1.3, M2 = 2.5 T2 =0.516,T 2=199.95 K T 02 marks) (4 M1=0. M=1 . exit area.45 bar and 150 m/s and leaves it at 277 K. pressure and velocity at the throat.2K a 20 1 2 c max= 2 γ−1 (4 marks) cmax = 790. the area of cross section at entry is 500cm2. (16) Key Hint: At throat M = 1 From isentropic table. If the flow rate is 3600kg/hr. (ii) Maximum velocity.69 m/s (4 marks) m = ρAc m = 30.05 kg/s (4 marks) m = ρ2A2c2 A2 = 0.432 T1 =0.98 bar.M2= c2 a2 c2 = 872.4.3.91bar and T0=3250C through a nozzle to an exit pressure of 0. Air (γ=1. γ = 1.0446 m2 (4 marks) 4. 2.964 T 01 T01=T02=311. Determine throat area. (May/Jun 2012) (Nov/Dec 2009) (MK).89 m/s (4 marks) m = ρAc m =1040. exit Mach number and maximum velocity. 3. (i) Stagnation temperature.432 From isentropic table.4. (iii) Mass flow rate. γ = 1. Assuming adiabatic flow determine.96 kg/ s m2 A (4 marks) 3. Consider the flow is isentropic. Air is discharged from a reservoir at p0 = 6. (Nov/Dec 2013) (April/May 2010) (MK) (16) Key Hint: a1= √ γR T 1 . R=287J/kgK) enters a straight axisymmetric duct at 300K. (iv) Area of cross section at exit. M 1= c1 a1 = 0.058bar and 260m/s. 51m/ s (4 marks) 5. (Nov/Dec 2014) (May/Jun 2013) (May/June 2009) (MK) (16) Key Hint: From Bernoulli equation.65 bar c =a =√ γRT ¿ ¿ (2 marks) ¿ c* = 447.834.593 A A2 = 13.93 A2 ¿ =1.76x10-4 m2 (2 marks) p2 =0.12. (2 marks) m = ρ*A*c* A* = 8. γ=1. Derive the expression for pressure co-efficient equation for compressible flow.4 p0 M2 = 1. =0.9x10-4 m2 γ +1 c max =c ¿ γ−1 (6 marks) √ c max =1096.¿ ¿ T p =0. γ [1+ γ (2−γ ) 6 γ −1 2 γ −1 γ γ M ] =1+ M 2+ M 4 + M 2 2 8 48 (4 marks) .528 T0 p0 p* = 3. 1 2 p0= p+ ρ c 2 γ p0 γ−1 2 γ−1 =[1+ M ] p 2 (4 marks) By applying Taylor series.65 m/s. maximum velocity of jet. a0 =√ γRT 0 a0 = 439.89 m/s At stagnation conditions. velocity of sound at stagnation condition. Determine: velocity of sound at 400K. An air jet at 400K has sonic velocity. stagnation enthalpy. T0 γ −1 2 =1+ M T 2 (4 marks) T 0 =480 K .16 m/s a 20 1 2 c = 2 max γ−1 (4 marks) cmax = 982 m/s (4 marks) 1 2 h0= c max 2 h0=482162 J /kg (4 marks) . (Nov/Dec 2012) (Nov/Dec 2008) (GK) (16) Key Hint: T = 400K. M=1 a=√ γRT a = 400.p0− p M 2 (2−γ) 4 =1+ + M γ 2 4 24 p× M 2 (4 marks) M= p× c a 2 γ 2 M ( 2−γ ) 4 M =1+ + M 2 4 24 p0 − p M2 M4 =1+ + 1 2 4 40 ρc 2 (2 marks) (2 marks) 6. 7 K T 02 c 2=M 2 × a2 .834.58 (4 marks) From isentropic table.69K (4 marks) p =0. γ = 1. (April/May 2015) (GK) (16) Key Hint: M= c a . Calculate the Mach number and stagnation properties. M=0.58 T =0.942 bar ρ 0= (4 marks) p0 RT 0 ρ0=1. exit area of nozzle is1000cm2.7. T0 = 550 K.937 T0 T0 = 296.85 (April/May 2015) (BK) (16) Key Hint: (a) M2 = 1.T 2=458. (b) M = 0. An aircraft flies at a velocity of 700kmph in an atmosphere where the pressure is 75 kPa and temperature is 50C.4. a=√ γRT M = 0. γ = 1. Air expands isentropically through the convergent nozzle from constant inlet conditions p 0 = 4bar. Determine the exit velocity and mass flow rate for the following two cases at exit.796 p0 p0 = 0.106 kg/m3 (4 marks) 8.4 T2 =0. (a) M = 1. γ = 1.874. (2 marks) . Give Assumptions made on Rayleigh Flow.58 kg/s (4 marks) UNIT II FLOW THROUGH DUCTS Flows through constant area ducts with heat transfer (Rayleigh flow) and Friction (Fanno flow) – variation of flow properties.4 T2 =0.3 m/s (4 marks) m = ρ2A2c2 m = 68. What is meant by stagnation pressure? (April/May 2015) Key Hint: It is the pressure of the gas when it is isentropically decelerated to zero velocity at zero elevation. flow will reach the critical condition where M = 1. Absence of work transfer across the boundaries.c 2 = 429.87 kg/s (4 marks) (b) M2 = 0. (2 marks) 3. The gas is perfect. 1. Define critical condition in Fanno flow. (May/June 2014) (Nov/Dec 2013) (May/June 2012) Key Hint: 1-D steady flow.85. (2 marks) 2.7 K T 02 c 2=M 2 × a2 c 2 = 373.T 2=480. (May/June 2014) Key Hint: Due to friction in subsonic or supersonic flow in a constant area duct.56 m/s (4 marks) m = ρ2A2c2 m = 67. Flow in constant area. (1 mark) 8. The gas is perfect with constant specific heats. April/May 2011) Key Hint: Fanno flow: Flow in air breathing engines. (Nov/ Dec 2012. Assumptions: 1-D steady flow Flow takes place in constant sectional area. Flow in heat exchangers.4. Give two practical examples for Fanno flow and Rayleigh flow analysis. There is no heat transfer. Where are the convergent nozzles and convergent – divergent nozzles used? (April/May 2010) Key Hint: Convergent nozzles: . (1 mark) Rayleigh flow: Flow in combustion chamber. What is impulse function and give its uses? (May/June 2013) Key Hint: The sum of pressure force and impulse force gives impulse function. (May/June 2013) Key Hint: T0 γ −1 2 =1+ M T 2 (1 mark) T¿ 2 γ −1 2 = + M T γ +1 γ +1 (1 mark) 6. 5. (1 mark) (1 mark) 7. Define Fanno flow and what are the assumptions made for Fanno flow? (April/May 2015) (Nov/ Dec 2014) (Nov/ Dec 2013) (Nov/ Dec 2012) (April/May 2011) Key Hint: Fanno flow: Flow in a constant area duct with friction and without heat transfer. Give the expression for T0 T (2 marks) ¿ and T T for isentropic flow through variable area in terms of Mach number. Flow in refrigeration and air conditioning. (May/June 2012) Key Hint: Mass density. Specific volume Specific weight Temperature Specific gravity 12. After that there is no further increase in mass flow with decrease in back pressure. (Nov/Dec 2013) Key Hint: (2 marks) 11. Show the Rayleigh line in h-s diagram and give the different Mach number regions for heating and cooling. Subsonic and sonic flows. What is chocked flow through a nozzle? (Nov/Dec 2013) Key Hint: The maximum mass flow conditions are reached when the throat pressure ratio achieves critical value. Give the assumption made in Isothermal flow. (2 marks) 10. (2 marks) . It is used as flow measuring and flow regulating devices. (1 mark) (1 mark) 9. Convergent – divergent nozzles: Supersonic flow. Key Hint: One dimensional flow. List some flow properties. What are the three equation governing Fanno Process? Key Hint: Energy equation Continuity equation Equation of state. (2 marks) 13.15. (2 marks) Part – B 1. The gas is perfect. Key Hint: It is the ratio between the wall shear stress and dynamic head. Define fanning’s coefficient of skin friction. (ii) (4 marks) Length 4 ´f Lmax 4 ´f L 4 ´f Lmax =[ ] −[ ] D D D M M 1 2 .5bar and 380C respectively and co-efficient of friction is 0. the entry pressure and temperature are 3. (2 marks) 15. (ii) Length of the duct. (i) Diameter of the duct.25 kg/s of air at an exit Mach number of 0. (Nov/Dec 2012) (May/Jun 2012) (April/May 2012) (May/Jun 2009) (MK) (16) Key Hint: (i) Diameter m = ρ1A1c1 D1 = 0. If the Mach number at the entry is 0. Frictional flow at constant temperature. Key Hint: 2 p2 1+γ M 1 = p1 1+γ M 22 (2 marks) 14. (iii) Pressure and temperature at the exit. determine.224 m.5. Write down the expression for the pressure ratio of two sections in terms of Mach number in Rayleigh flow. Constant area duct. (iv) Stagnation pressure loss. A circular duct passes 8.005. 369 bar (4 marks) .64 p0 p ¿ =1. (iv) stagnation pressure loss in the combustion chamber. ¿ =1.34 ¯¿ (4 marks) 2. (i) The initial and final Mach numbers.143 T2 T2 = 297. Assuming γ=1.59 m (iii) (4 marks) From Fanno flow table.5 p2 =2.021 bar T2 ¿ =1. and 75 m/s.58 K (iv) (4 marks) ∆ p= p 02−p 01 Δp=2. 0. M2 = 0.525.138 p¿2 p2 = 1. temperature and velocity of the gas.4 .55 bar. γ = 1.859 T ¿0 M2 = 0. the air-fuel ratio is 29 and the calorific value of the fuel is 42 MJ/kg. (ii) Final pressure. Air is supplied to a combustion chamber in a gas turbine plant at 350K. (Nov/Dec 2013) (May/Jun 2013) (April/May 2011) (Nov/Dec 2007) (MK) (16) Key Hint: c c M= = a √ γRT M1 = 0.4 and R=287J/kgK for the gas. (iii) the maximum stagnation temperature attainable.2 ∆ h 0=c p ( T 02−T 01 ) T 02 =0.061 p p0 p2 = 0.L=305. determine. 003. If the Mach number at exit is 0. (ii) Length of the pipe (April/May 2015) (May/Jun 2013) (April/May 2011) (MK) (16) Key Hint: m = ρ1A1c1 T1 p =0. 1 =0.2 (i) static pressure and temperature. Air at P0=11bar.953 T¿ T2 = 1611.357.001.4.7K (4 marks) 3. (iv) distance of this section from the inlet and (v) mass flow rate of air.8.5 . (Nov/Dec 2013) (Nov/Dec 2008) (April/May 2008) (MK) (16) Key Hint: γ = 1.T =0. stagnation temperature 310 K and static pressure 0.11 m/s (4 marks) T0 max = T02*= T01*= T01* T0 max = 2027. Air enters the pipe at a Mach number of 2. (ii) stagnation pressure and temperature (iii) velocity of air. Determine for a section at which the Mach number reaches 1.449 D L = 5. A long pipe of 25.507 bar.0272 T 01 p01 m = 0.625 c c2 = 515.5.4 mm diameter has a mean co-efficient of friction of 0.05m (8 marks) 4.813 kg/s 4 ´f Lmax 4 ´f L 4 ´f Lmax =[ ] −[ ] D D D M M 1 (8 marks) 2 4 ´f L =0. M1 = 2. T0=420K enters a 45mm diameter pipe at the Mach number of 3 and the friction co-efficient for the pipe surface is 0. Determine (i) Mass flow rate.35K c ¿ =0. 5 T ¿02 (4 marks) . 1 =0.444.25.987.676K. and M1 = 0.16 m/s (2 marks) 4 ´f Lmax 4 ´f L 4 ´f Lmax =[ ] −[ ] D D D M M 1 2 L = 0. p1 = 1.4. The stagnation temperature of air in a combustion chamber is increased to 3. ¿ =1.4.0585 T 01 p01 T1 = 240.396 bar (4 marks) T01 = 310K.4 and M2 = 0.252 p1 p0 T1 T0 Refer Rayleigh flow table for γ = 1. Pressure and temperature at the exit. 105 0C and a Mach number of 0.68 T 02 =0.957 T 01 p01 Refer Rayleigh flow table for γ = 1. and M1 = 0. (2 marks) m = ρ1A1c1 m = 0. ¿ =0. If the air at entry is at 5bar.(April/May 2010) (April/May 2008) (MK) (16) Key Hint: Refer isentropic flow table for γ = 1.305.386 bar (4 marks) M= c a c2 = 373.217.25. p1 p0 T1 T0 ¿ =2.207.25. (iii) the heat supplied per kg of air.T1 p =0. 1 =0. ¿ =0. determine: (i) the Mach number.842 m. (ii)Stagnation pressure loss. p01 = 3. T1 p =0.5 times its initial value.382 kg/s (4 marks) 5. T1 p A =0.252 p1 p0 T1 T0 p2 = 3. and M1 = 1. (Nov/Dec 2013) (GK) (16) Key Hint: Refer isentropic flow table for γ = 1. (ii) Diameter of the nozzle throat and (iii) Pressure and temperature at the pipe exit.8 and 1. T2 = 9440C (4 marks) ∆ p0= p 01−p 02 ∆ p0=0. ¿ =1.607.0025.217.3bar. 1 =0.305.5 cm (4 marks) 4 ´f Lmax 4 ´f L 4 ´f Lmax =[ ] −[ ] D D D M M 1 Lmax =968 cm 2 (4 marks) .8.3 kJ /kg (4 marks) 6. ¿ =0.735 ¯¿ (4 marks) Q=m c p (T 02−T 01) Q=962. A convergent – divergent nozzle is provided with a pipe of constant cross section at its exit.439 T 01 p01 A1 ¿ A 1=0.0 respectively.207. the Mach numbers at the entry and exit of the pipe are 1.174.p1 p0 T1 T0 ¿ =2. Determine (i) The length of the pipe. 1¿ =1. The mean co-efficient f friction for the pipe is 0.8735 m ¿ A 1= 2 (4 marks) π ¿2 D 4 D ¿ =33. ¿ =0. the stagnation pressure and temperature of air at the nozzle entry are 12 bar and 600 K.4. The exit diameter of the nozzle and that of the pipe is 40cm. (iii) heat transferred. (iv) Maximum possible heat transfer. Find (i) the static pressure. ¿ =1. (vi) is it a cooling or a heating process.728 p T1 p¿ =4.4.955 T ¿02 M2 = 0. (ii) stagnation pressure at inlet and exit. determine the final Mach number and percentage drop in pressure. ¿ =0.5 respectively.615.8 p1 T1 ¿ =0. (April/May 2015) (GK) (16) Key Hint: Rayleigh flow table. M1 = 1. and M1 = 3. Consider the fluid is air.405 ¯¿ T ¿1 =500.275 K (4 marks) 7.4and (4 marks) T 02 =0. (v) change in entropy between the two sections.γ = 1.529 p1 p0 T1 T0 For γ = 1. ¿ =0.From Fanno table. (May/Jun 2012) (BK) (16) Key Hint: Refer isentropic flow table for γ = 1. The stagnation temperature of air is raised from 85 0C to 3760C in a heat exchanger.474.157. temperature and velocity at exit.4 and M1 = 0. ¿ =0.9 % (4 marks) (8 marks) 8.4 p1 p0 T1 T0 ¿ =1. The Mach number at inlet and exit for a Rayleigh flow are 3 and 1. γ = 1.4. If the inlet Mach number is 0.4.961.78 Percentage drop in pressure = p1 p2 − p¿1 p¿2 ×100 p1 p ¿1 =33. . At inlet static pressure is 50kPa and stagnation temperature is 295K. ¿ =0.4 and M1 = 3.122. p1 p0 T1 T0 ¿ =0.281. γ = 1.59 kJ/kg.19 K .176.38bar p02=6. ¿ =0.01bar (3 marks) Q=m c p (T 02−T 01) Q = 115.4 and M2 = 1.424.753.5. 1 =0. ¿ =3. (3 marks) 2 2 [1−M 1 ] Qmax =c p T 1 2 2(1+ γ ) M 1 Qmax =156.T1 p =0. ¿ =0. ¿ =0. ¿ =1. c 2=505. p2 p0 T2 T0 ¿ =0.654 p1 p0 T1 T0 Rayleigh flow table.909 p2 p0 T2 T0 p2=1.578.0272 T 01 p01 Rayleigh flow table.64 ¯.357.0272 p01 p01=18.56 m/s (3 marks) p1 =0. T 2=282. γ = 1.79 kJ /kg p2 p1 ¿ ¿ γ −1 ¿ γ ¿ T2 T1 ¿ s 2−s1=c p ln¿ (2 marks) . There will be a loss in stagnation pressure and increase in entropy across the shock wave. What is Oblique shock? (April/May 2015) (Nov/Dec 2012) (April/May 2011) Key Hint: When the shock wave is inclined at an angle to flow.51 J /kgK (3 marks) The heat transfer rate is positive. experiencing shock wave is considerably low? (May/June 2014) Key Hint: Shock may cause boundary layer separation and deviation of flow from its designed direction.s 2−s1=649. (2 marks) 4. (2 marks) 2. Write the Prantl – Meyer relation? (April/May 2015) (Nov/Dec 2012) Key Hint: ¿2 a =c x c y ¿ Mx ¿ × M y =1 (2 marks) 3. Why the efficiency of a machine. it is cooling process (2 marks) UNIT – III NORMAL AND OBLIQUE SHOCKS Governing equations – Variation of flow parameters across the normal and oblique shocks – Prandtl – Meyer relations – Applications. What is the use of pitot tube in supersonic flow? (May/June 2014) Key Hint: . PART A 1. Why the shock waves cannot be developed in subsonic flow? (April/May 2011) Key Hint: Velocity of fluid is less than the velocity of sound. (2 marks) 8. Mention the useful applications of shock wave. Adverse effects: It cause undesirable interference with normal flow behavior. Introduction of the pitot tube produces a curved shock a little distance upstream of its mouth. 1 2 Normal shock Shock wave is right angle to the flow Oblique shock Shock wave is inclined at an angle to the flow One dimensional flow Two dimensional flow (2 marks) . No. What are the situations where shocks are undesirable? (April/May 2010) Key Hint: They interfere with the normal flow behavior.(Nov/Dec 2014) Key Hint: S. Due to this reason deceleration is not possible in subsonic flow. (2 marks) 6. Give the difference between normal and oblique shocks. The sonic boom created by supersonic aircraft and the blast waves generated by an explosion. What are the beneficial and adverse effects of shock waves? (May/June 2012) Key Hint: Beneficial effects: A strong wave is utilized to accelerate the flow to a high Mach number in a shock tube. Therefore it measures the stagnation pressure downstream the shock wave. Used in supersonic compressor to obtain considerably high pressure ratio in one stge. (2 marks) 7. Therefore. (2 marks) 9. It create sonic flows in supersonic aircraft and damage the flow passage. Thus the efficiencies of turbo machines experiencing shock waves are considerably low. the efficiency of turbo machineries decreases. (2 marks) 5. (April/May 2010) Key Hint: Jet engines Shock tubes Supersonic wind tunnel Practical admission turbines. ( Nov/Dec 2013) Key Hint: Compression shocks: A shock wave which is at a higher pressure than the fluid into which it is moving. Key Hint: The ratio of difference in downstream and upstream shock pressures to upstream shock pressure. (1 mark) No. (2 marks) 15. (1 mark) 12. Define compression and rarefaction shocks. The gas is perfect with constant specific heats. adiabatic and frictionless. What are the assumptions used for oblique shock flow? (Nov/Dec 2013) Key Hint: Flow is steady. Stagnation temperature remains constant. State the necessary conditions for a normal shock to occur in compressible flow. What are the properties changes across a normal shock? Is the flow through a normal shock an equilibrium one? (Nov/Dec 2014) Key Hint: Stagnation pressure decreases. Absence of body forces. temperature and density are changed during normal shock. Calculate the strength of shock wave when normal shock appears at M = 2. Since the fluid properties like pressure. . Key Hint: i. Static temperature and static pressure increases. The compression wave is to be at right angle to the compressible flow. It is not possible. (1 mark) 11. Flow should be supersonic.10. Absence of work transfer across the boundaries. (2 marks) 13. ii. (2 marks) 14. Define the strength of shock wave. (1 mark) Rarefaction shocks: A shock wave which is at a lower pressure than the fluid into which it is moving. γ = 1.5 px ξ=3.4. (Nov/Dec 2014) (Nov/Dec 2008) (MK) (16) Key Hint: ρ= p RT ρ y py T x = × ρx p x T y M 2x = (2 marks) γ +1 p y γ−1 + 2 γ px 2γ ( ) [ p y γ−1 p y × +1 T y p x γ +1 p x = Tx p y γ −1 + p x γ +1 ] (4 marks) .5 (1 mark) Part – B 1. py =4.Key Hint: ξ= p y− p x px (1 mark) Refer normal shocks table for Mx = 2. Derive the expression for Rankine – Hugoniot Equations. (iii) Final temperature of the gas. Calculate the following: (i) Deflection Angle. T1 p =0. and M1 = 1.98 (6 marks) p2 =0. A gas at a pressure of 340 mbar.3. and M2 = 1.86K (4 marks) .3 (April/May 2015) (May/Jun 2014) (April/May 2011) (MK) (16) Key Hint: Refer isentropic flow table for γ = 1.3. (ii) Final Mach number. temperature of 355K and entry Mach number of 1. Take γ = 1.773.327 T 01 p01 δ = w(M1) – w(M2) δ = -18.4.135 p02 (6 marks) T2 =0.4 is expanded isentropically to 140mbar.p y γ −1 + py Tx px γ + 1 × = px T y γ−1 p y 1+ × γ +1 p x γ −1 p y × +1 ρ y γ +1 p x = ρx γ−1 p y × γ +1 p x (6 marks) ρ y γ +1 −1 p y ρ x γ−1 = px γ +1 ρ y − γ −1 ρ x [ ] (4 marks) 2.629 T 02 T2 = 288. 1 =0.510 Refer isentropic flow table for γ = 1. M x =M 1 sin σ p y p2 = =4.3.8 0 0 σ weak =4 3 (4 marks) For strong shock wave. (i) Wave angle. (iii) Density ratio. Determine the following for strong and weak waves.745 T x T1 marks) M 2= My sin ( σ−δ ) (2 . An oblique shock wave occurs at the leading edge of a symmetrical wedge. (Nov/Dec 2014) (Nov/Dec 2014) (May/Jun 2012) (MK) (16) Key Hint: 2 1−2 sin ❑ σ 2+ γ M 21+ M 21 ¿ M 21 sin ❑2σ tan δ =2 cot σ ¿ σ strong=80. (v) Downstream Mach number.1 and deflection angle of 150.832 p x p1 (2 marks) ρ= p RT ρ2 =2.769 ρ1 T y T2 = =1. (ii) Pressure ratio. Air has a Mach number of 2. (iv) Temperature ratio. (May/Jun 2013) (April/May 2011) (April/May 2010) (MK) (16) Key Hint: From isentropic tables γ = 1.013bar.741 ρ1 T y T2 = =1. (ii) Temperature and pressure.4. M x =M 1 sin σ p y p2 = =2. The Mach number of a jet of air approaching the diffuser at p0 = 1. Determine at the exit of the diffuser.274 T x T1 (2 marks) M 2= My sin ( σ−δ ) M 2=1.M 2=0. The flow in the diffuser is isentropic.0.548 (2 marks) 4.219 p x p1 (2 marks) ρ= p RT ρ2 =1. There is a standing normal shock wave just outside the diffuser entry.2. (iii) Stagnation pressure loss between the initial and final states of the flow. (i) Mach number. The ratio of the exit to the entry area in a subsonic diffuser is 4. Mx = 2. .2.619 (2 marks) For strong weak wave. T = 290 K is 2. The entry conditions of the gas (γ=1.0935 T 0x p0x From normal shock tables.628 px Tx p0x A =5.Tx p =0.4.857. y =1. (4 marks) A supersonic nozzle is provided with a constant diameter circular duct at its exit. 0 y =0.287 kJ/kgK) are p 0 = 10 bar. Mx = 2. Calculate the static pressure.508. Nozzle exit cross-section is three times that of its throat.4 A .04 A¿ ME = 0. γ = 1. Mach number and the velocity of the gas in the duct: (i) when the nozzle operates at its design condition. x =0.629 bar (4 marks) ∆ p0= p 0 x − poy ∆ p0=377 ¯¿ 5. R=0. The duct diameter is same as the nozzle exit diameter.116 (4 marks) TE =0. (April/May 2010) (April/May 2008) (MK) (16) Key Hint: From isentropic table.212 Py pE = 0.997 T 0E TE = 569. (ii) when a normal shock occurs at its exit. γ = 1.48.15K (4 marks) ρAc= ρ y A y c y PE =1.547. My = 0.2.4. A2 ¿ =3 . T0 = 600K. py T p =5. 2K p2 = 0.75 bar c y =M y ×a y (2 marks) c y =239.417.05 m/s (4 marks) From isentropic table.64. T2 P =0. A2 Ax = =3 .417. h0= a 2 c 2 1 γ + 1 ¿2 + = a γ−1 2 2 γ −1 ( ) Before shock wave.0471 T0x P0x (2 marks) M y =0.471 bar c 2=M 2 × a2 (2 (2 marks) c 2=837.2K py = 3. 2 =0.32 m/s (4 marks) 6. (2 marks) .4 A ¿ A x¿ M x =2.5 Ty = 570. Starting from energy equation derive Prandtl-Mayer equation.0471 T 02 P 02 marks) T2 = 250. x =0.64. γ = 1. Tx P =0.M 2=2. (May/Jun 2013) (Nov/Dec 2012) (GK) (16) Key Hint: Stagnation Enthalpy relation. a2y γ +1 a¿2 γ −1 = − c cy 2 cy 2 y ( ) (2 marks) m =ρx c x = ρ y c y A γ px γ p y − =γ (c y −c x ) ρx c x ρ y c y (4 marks) a2x a2y − =γ ( c y −c x ) cx c y a¿ 2=c x c y ¿ (4 marks) ¿ M x × M y =1 (2 marks) 7. (i) Derive the equation for static pressure ratio across the shock waves. (May/Jun 2012) (Nov/Dec 2009) (GK) (8) Key Hint: F = p+ρc2 p y 1+ γ M 2x = p x 1+γ M 2y (4 marks) py = px (1+ γ M 2x ) 2γ ( γ−1 ) M −1 γ+1 γ M 2x + 1 ] [ γ −1 2 x .a2x γ +1 a¿2 γ −1 = − c cx 2 cx 2 x ( ) (2 marks) After shock wave. determine pressure. =1. py Ty M y =¿ =2. T = 17 0C)is 500 m/s.56 K mark) (1 .297 0. velocity of air. p x Tx (2 marks) p y =2.05 m/s (2 marks) c 'y M = ay ' y M 'y =0.0bar.571 From isentropic tables (1 mark) M 'y =0. stagnation temperature and Mach number imparted upstream of the wave front. (May/Jun 2012) (GK) (8) Key Hint: M x= cx =1.13 K (2 marks) c y =M y ×a y c 'y =c x −c y c 'y =222.939 T0y T '0 y =400.571 Ty =0.335 ¯ ¿ T y =376.715.335. temperature.465 ax From normal shock tables.py 2γ γ −1 = M 2− p x γ +1 x γ +1 ( ) (4 marks) (ii) The velocity of a normal shock wave moving into stagnant air (p=1. if the area of cross section of the duct is constant. 272 T 0x p0x (2 marks) T x =252. Air enters this nozzle with a stagnation pressure of 6. A converging-diverging nozzle has an exit area to throat area ratio of 2. Define propulsive efficiency.5.4. (2 marks) .4.345 ¯¿ T y =332.458. My = 0.86 K (6 marks) UNIT – IV JET PROPULSION Theory of jet propulsion – Thrust equation – Thrust power and propulsive efficiency – Operating principle. The divergent section of the nozzle acts as a supersonic nozzle. turbojet. Tx p =0.5. PART-A 1. cycle analysis and use of stagnation state performance of ram jet.701.8.5. γ = 1. turbofan and turbo prop engines. Determine the exit plane of the nozzle. Assume that a normal shock stands at a point M = 1.25cm2.320 px Tx (2 marks) (2 marks) p y =4. (May/Jun 2009) (BK) (16) Key Hint: From isentropic tables γ = 1.17 K px =1.5bar and a stagnation temperature of 93 0C. py T =2. x =0.689. (April/May 2015) Key Hint: The ratio of propulsive power to the power output of the engine. The throat area is 6. the static pressure and temperature and Mach number. Mx = 1. y =1. Mx = 1.768 ¯¿ (4 marks) From normal shock tables. List out the different types of jet engines. (2 marks) 3. Give the components of a turbo jet. What are the benefits of thrust augmentation in a turbojet engine? (May/June 2012) (April/May 2011) Key Hint: Short take-off distance. Define thrust power and propulsive efficiency of aircraft engine. the static pressure of air is increased to ignition pressure. (May/June 2013) Key Hint: Diffuser Rotary compressor Combustion chamber Turbine Exhaust nozzle (1 mark) (1 mark) (1 mark) (1 mark) 7. (2 marks) 5. (May/June 2014) Key Hint: Thrust power: The product of thrust and flight speed. (May/June 2013) Key Hint: Ramjet engine Pulse jet engine Turbo jet engine Turboprop engine Turbo fan engine. (1 mark) Propulsive efficiency: The ratio of propulsive power to the power output of the engine. Why a ram jet engine does not require a compressor and turbine? (May/June 2014) (Nov/Dec 2014) (Nov/Dec 2013) Key Hint: Due to supersonic and subsonic diffuser. 6. (1 mark) 4.2. (1 mark) . What is the type of compressor used in turbo-jet? (April/May 2015) Key Hint: Rotary compressor is used due to its high thrust and high efficiency. It is used for fast and easier take off. (1 mark) 8. What is ram effect? Key Hint: In ramjet engine the subsonic and supersonic diffusers are used to convert the kinetic energy of the entering air into pressure energy. What is thrust augmentation? (April/May 2011) Key Hint: To achive better take-off performance. (2 marks) 11. What is a bypass engine and define bypass ratio. This energy transformation is called ram effect.28 (1 mark) 12. (April/May 2011) Key Hint: Bypass engine: Turbo fan engine. (2 marks) 13. Thrust augmentation increases the jet velocity and is known as after burning. Define speed ratio in propulsive system ( Nov/Dec 2012) Key Hint: The ratio of flight speed to the jet velocity (2 marks) 10. (2 marks) 14. High climb rate to very high altitude. Give the expression for the thrust developed by a turbojet engine. additional fuel is burnt in the tail pipe between the turbine exhaust section and entrance section of the exhaust nozzle. Write down the principle of jet engine (Nov/Dec 2012) Key Hint: Every action there is an equal and opposite reaction. (Nov/Dec 2014) Key Hint: 2u η p= (1 c j+u mark) ηp = 1. A portion of the total flow of air bypass part of the compressor. Key Hint: F=m ´ c j− m ´ au (2 marks) . Find the optimum propulsive efficiency when the jet velocity is 500 m/s and flight velocity is 900 m/s. (1 mark) Bypass ratio: The ratio of the mass flow rates of cold air and the hot air is known as Bypass ratio. (1 mark) 9. propulsive efficiency. (2 marks) PART – B 1.15. (Nov/Dec 2014) (May/Jun 2014) (MK) (16) Key Hint: Jet thrust Mass flow rate of nozzle at exit of the nozzle = m´ a + m´ f Net thrust = Momentum thrust +Pressure Thrust m´ + m´ c − m´ a ×u Net thrust(F) = ( a f ) e (3 marks) Propeller thrust m´ + m´ c − m´ a ×u Net thrust considering mass of fuel (F) = ( a f ) e c e =c j F= m(¿ c j−u) (or) ´¿ . overall efficiency and the optimum value of flight to jet speed ratio for a turbo jet engine. thermal efficiency. Derive the expression for the jet thrust. What is scram jet? Key Hint: A supersonic combustion ramjet engine is known as scramjet. propeller thrust. V ηt =¿ 2 2 (3 marks) Overall efficiency: Power input Propulsive power ( ¿ ) Thrust power η0 = theengine ¿ ¿ ´ m(¿ c j−u)× u m´ f ×C . V η 0=¿ η0=η p × ηt (2 marks) .´ F = m a (c j −u) (3 marks) Propulsive efficiency: η p= η p= Propulsive power ( ¿ ) Thrust power Power output of the engine 2u c j+u η p= 2σ 1+σ (3 marks) Thermal efficiency: Power input Power output of the engine ηt = the engine through fuel ¿ ¿ ´ 1 m(¿ c j−u ) 2 m´ f ×C . The thrust developed by the secondary air is at lower velocity. (2 marks) . turbine and exhaust nozzle. Air from the atmosphere enters into turbofan engine. Combustion takes place in the combustion chamber and the thrust is produced in the opposite direction. Explain with neat sketches the principle of operation of (i) Turbo fan engine and (ii) Turbo jet Engine. The air after passing through the fan is divided into two streams. employing a low pressure ducted fan. namely primary air and secondary air.Effective speed ratio σ= σ= Flight speed Jet velocity u cj (2 marks) 2. The primary air flow through the turbofan engine consisting of compressor. The total thrust produced in this engine is the sum of thrust produced by the primary air and the secondary air. combustion chamber. (3 marks) Air and the thrust developed by the primary air is at much higher velocity. This total thrust propels the aircraft. (May/Jun 2015) (Nov/Dec 2013) (Nov/Dec 2012) (April/May 2011) (MK) (16) Key Hint: Turbo fan engine: Combination of turboprop and turbo jet engine. Heat is supplied at constant pressure.The function of the nozzle is to convert the pressure energy of the combustion gases into kinetic energy.The highly heated products of combustion gases are then enters the turbine and partially expanded. the fuel is injected by suitable injectors and the air-fuel mixture is burnt. Explain the working principle of the ramjet engines with neat sketch and state its advantages and disadvantages. In the combustion chamber. (May/Jun 2013) (April/May 2010) (May/Jun 2009) (MK) (16) Key Hint: (4 marks) .(3 marks) Turbo jet Engine: (i) Diffuser . (3 marks) (3 marks) 3.The function of the diffuser is to convert the kinetic energy of the entering air into pressure energy.High pressure air flows into the combustion chamber. (iv) Turbine . (2 marks) (iii) Combustion chamber . (v) Exhaust nozzle. (ii) Rotary compressor – The air passes through the rotary compressor in which the air is further compressed. Determine. The function of nozzle is to convert pressure energy of gas into kinetic energy. (3 marks) Advantages: Ramjet engine is very simple and does not have any moving part. (v) Thrust power. There is no upper limit to the flight speed. (ii) Thrust. This energy transformation is called ram effect and the pressure rise is called the ram pressure. a reaction or thrust is produced in the opposite direction. Velocity of the gases at the exit of the jet pipe = 505 m/s.28 bar.74 m. Diameter at inlet section = 0. (3 marks) Disadvantages Since the take-off thrust is zero. This thrust propels the air craft. Pressure at the exit of the jet pipe = 0. (3 marks) 4. (iv) Specific impulse. A turbo jet flies at a speed of 870kmph at an altitude of 10000m. In the combustion chamber. the data for the engine is . the fuel is injected by suitable injectors and the air fuel mixture is burnt. (vi) TSFC (April/May 2011) (Nov/Dec 2012) (May/Jun 2007) (MK) (16) Key Hint: π A i= d 2i 4 m=ρ ´ i Aiu m= ´ m ´ a +m ´f . The combustion chan:ber required flame holder to stabilize the combustion due to high speed of air . Air – fuel ratio = 42. (iii) Specific thrust. Due to high velocity of gases coming out from the unit. Cost is low Less maintenance The specific fuel consumption is better than other gas turbine power plants at high speed.The function of supersonic and subsonic diffusers are to convert the kinetic energy of the entering air into pressure energy. (i) Air flow rate through the engine. The high pressure air flows into the combustion chamber. It has low thermal efficiency. (3 marks) The highly heated products of combustion gases are then allowed to expand in the exhaust nozzle section. it is not possible to start a ramjet engine without an external launching device. It is very difficult to design a diffuser which will give good pressure recovery over a wide range of speeds. 9 kg/s 1 c= ( u+ c j ) = 137.8.41 s (2 marks) P=Fxu P = 2. The diameter of the propeller is 4.78 x 106 W TSFC= (2 marks) m ´f F TSFC=0. An air craft propeller flies at a speed of 440 kmph. (ii) Thrust power. (3 marks) F=m ´ c j −m ´ au F sp = F m ´ F sp=268. (iii) Propulsive efficiency.95 N / I sp = ( kgs ) (3 marks) F mg ´ I sp=27.1m and the speed ratio is 0.55 bar.m ´ a=41. the ambient conditions of air at the flight altitude are T = 255 K and p = 0.9 kg/ s (3 marks) F = 11538 N. (May/Jun 2015) (May/Jun 2013) (Nov/Dec 2008) (MK) (16) Key Hint: ´ F = m a (c j −u) m ´ a=m= ´ ρAc = 1363.49 kg/s 2 . Find the following: (i) Thrust.312 kg h−N (3 marks) 5. 09 x 106 W η p= (5 marks) 2u c j+u η p=0. thermal and overall efficiencies. one of its turbo jet engines takes in 40 kg/s of air expands the gases to the ambient pressure.F = 41667 N (6 marks) P=Fxu P = 5.4 m/s (3 marks) F=11. An air craft flies at 960 km/hr. determine: (i) Jet velocity. (v) Propulsive. The air-fuel ratio is 50 and the lower calorific value of the fuel is 43 MJ/kg. (ii) Thrust.88 ( ¿ ) 88 (5 marks) 6.094 kN (3 marks) F=m ´ c j −m ´ au F sp = F m ´ F sp=277.35 N / ( kgs ) (3 marks) P=Fxu P = 2. for maximum thrust power. (iv) Thrust power. (May/Jun 2014) (GK) (16) Key Hint: σ= u cj c j =533.958 x 106 W η p= 2u c j+u (3 marks) . (iii) Specific thrust. (i) Differentiate turbo jet and turboprop engines with suitable diagrams. (3 marks) (3 marks) Turboprop Engine: .42 (4 marks) 7.V ηt =¿ 2 2 ηt =12.6 ´ 1 m(¿ c j−u ) 2 m´ f ×C . (v) Exhaust nozzle. (ii) Rotary compressor – The air passes through the rotary compressor in which the air is further compressed.The highly heated products of combustion gases are then enters the turbine and partially expanded.The function of the nozzle is to convert the pressure energy of the combustion gases into kinetic energy.η p=66. (iv) Turbine . Heat is supplied at constant pressure.The function of the diffuser is to convert the kinetic energy of the entering air into pressure energy. (May/June 2012) (GK) (12) Key Hint: Turbo jet Engine: (i) Diffuser . (iii) Combustion chamber .High pressure air flows into the combustion chamber. the fuel is injected by suitable injectors and the air-fuel mixture is burnt. In the combustion chamber.65 η0=η p × ηt η0=8. a reaction (or) thrust is produced in the opposite direction. This total thrust propels the air craft. Propeller is used to increase the flow rate of air which results in better fuel economy. The total thrust produced in this engine is the sum of the thrust produced by the propeller and the thrust produced by the nozzle. But the propeller cannot run at higher angular velocity. Due to high velocity of gases coming out from the unit. the turbine drives the compressor and propeller. So a reduction gear box is provided before the power is transmitted to the propeller.The angular velocity of the shaft is very high. (3 marks) (3 marks) (ii) Write the equations to calculate propulsive efficiency and thermal efficiency of an aircraft. (May/June 2012) (GK) (4) Key Hint: Propulsive efficiency: η p= Propulsive power ( ¿ ) Thrust power Power output of the engine η p= 2u c j+u marks) (2 . 3 ×10−5 kg s−N (3 marks) . Thrust force is 50 kN.V ηt =¿ (2 marks) 8.32 N / TSFC= ( kgs ) (3 marks) m ´f F TSFC=5. mass flow rate of fuel is 2. Determine the specific thrust. (May/June 2012) (BK) (16) Key Hint: m= ´ m´ a + m´ f ´ F = m(c j −u) c j =1275. mass flow rate of air is 45 kg/s. Calorific value of the fuel is 42800 kJ/kg. A turbo jet engine operating at a Mach number of 0.8 and the altitude is 10km has the following data. thrust specific fuel consumption.67 F sp = m s (4 marks) F m ´ F sp=1049. propulsive efficiency and overall efficiency. Assuming the exit pressure is equal to ambient pressure.Thermal efficiency: Power input Power output of the engine ηt = the engine through fuel ¿ ¿ ´2 2 1 m(¿ c j−u ) 2 m´ f ×C .65 kg/s. thermal efficiency. jet velocity. 57 (2 marks) .V ηt =¿ η0=η p × ηt η0=10.97 (2 marks) ηt =33.04 (2 marks) ´2 2 1 m(¿ c j−u ) 2 m´ f ×C .η p= 2u c j+u η p=31. Why rocket is called as non – breathing engine? Can rocket work at vacuum? (May/June 2014) Key Hint: Combustion takes place by using its own Oxygen supply instead of atmospheric air. What are the types of rocket engines based on source of energy employed? (April/May 2015) (May/June 2013) Key Hint: Chemical rocket engine Solar rocket engine (1 mark) Nuclear rocket engine Electrical rocket engine (1 mark) . (2 marks) 4. It can work at vacuum. What is mono propellant? Give Examples. (2 marks) 3. PART A 1. Nitro methane. (April/May 2015) (May/June 2013) (April/May 2011) Key Hint: A liquid propellant which contains both the fuel and the oxidizer in a single chemical. What is the use of inhibitors in solid propellants? (May/June 2014) Key Hint: To regulate the burning of propellant.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. (1 mark) Example: Nitroglycerine. (1 mark) 2. (2 marks) . It should be non-corrosive and non-reactive with components of the engine. (1 mark) 10. Explain chemical rocket propulsion system. Physical and chemical properties should not change during processing.5.Cj. (Nov/Dec 2012) Key Hint: High pressure and high temperature gases are produced in combustion chamber due to chemical reaction. Mention any four applications of rocket.u = 19. Define characteristic velocity. (May/June 2012) Key Hint: η p= 2σ σ 2+1 σ= u C j =2 ηp = 80% (1 mark) propulsive power = mp. A rocket flies at 10080 km/hr with an effective exhaust jet velocity of 1400 m/s. Find the propulsion power of the rocket. (2 marks) 9. 6.6 MW. (1 mark) (1 mark) (1 mark) 7. What is meant by hypergolic propellant? (April/May 2010) Key Hint: Hypergolic propellant do not require ignition (2 marks) 8. (Nov/Dec 2014) Key Hint: The ratio between effective jet velocity and thrust coefficient. What are the properties of solid propellants? (Nov/Dec 2014) Key Hint: It should release large amount of heat during combustion.(1 mark) It should have high density. and the propellant flow rate of 5kg/s. (May/June 2012) Key Hint: Military Space Aircraft Communication. Key Hint: Nitroglycerine. Hydrogen peroxide. Name some propellants for space application. Lighter than liquid propellant rocket High fuel density. (2 marks) 14. Define specific propellant consumption. There is no chemical deterioration between fuel and Oxidizer. (1 mark) I sp = F Wp (1 mark) 13. (1 mark) (1 mark) 12. then the propellant is called bipropellant. (1 mark) SPC = WP F (1 mark) 15. (2 marks) PART – B 1. List the main components of liquid propellant rocket engine and explain.11. What is bipropellant? Key Hint: If the fuel and oxidizer are different from each other in its chemical nature. Key Hint: The propellant consumption rate per unit thrust is known as specific propellant consumption. (April/May 2015) (MK) (16) Key Hint: . Nitromethane. What are the advantages of a hybrid rocket? (Nov/Dec 2013) Key Hint: Speed regulation is possible High load capacity. What is specific impulse of a rocket? Key Hint: The thrust developed per unit weight flow rate of the propellant is known as specific impulse. Hydrazine. When the combustion takes place in the combustion chamber. So it is economical. Nozzle is used to increase the velocity and decrease the pressure of the gases. increase and decrease of speed is possible. (4 marks) Advantages Liquid propellant engines can be reused after recovery. There are additional handling and safety problems if the propellants are poisonous and corrosive. hydrazine. Liquid fuel (refined petrol. a force (or) thrust is produced in the opposite direction. liquid hydrogen. (3 marks) Disadvantages Construction is more complicated compared to solid propellant rocket. High specific impulse. The preheated fuel-oxidizer mixture is injected into the combustion chamber through suitable injector and combustion takes place. The highly heated products of combustion gases are then allowed to expand in the nozzle section.. .e.. (2 marks) (4 marks) Liquid fuel and liquid oxygen are pumped separately into a combustion chamber through control valves.e. Combustion process is controllable i. very high pressure and very high temperature gases are produced. it is easy to stop the combustion by closing the fuel valve (or) oxidizer valve. Speed regulation i. Manufacturing cost is high. Liquid fuel and liquid oxygen are stored separately in two different tanks. etc) and liquid oxygen are used in this engine. Preheater is used to heat the fuel and oxidizer. Due to high velocity of gases coming out from the unit. This thrust propels the rocket. (v) Overall efficiency (April/May 2015) (MK) (16) Key Hint: F=m´ p c j m´ p=m´ 0 + m´ f m´ p=4.V ηt =0.82. Flight to jet speed ratio = 0.32 s (3 marks) η p= 2σ 2 1+σ η p=98 2 ηt = (3 marks) 2 c j +u 2×C . Heat of reaction per kg of the exhaust gases = 2520 kJ/kg. (iv) Thermal efficiency. (ii) Specific impulse. High vibration.8 (3 marks) . Oxidizer flow rate = 3. Calculate the following: (i) Thrust.4 kg/s. A rocket engine has the following data: Effective jet velocity = 1200 m/s. (3 marks) 2.6 kg s F=5520 N I sp = (4 marks) F Wp I sp =122. fuel flow rate = 1. (iii) Propulsive efficiency. The size and weight of the engine is more compared to solid propellant rocket.2 kg/s.477 (3 marks) η0=η p × ηt η0=46. (i) Injection (ii) Atomization (iii) Mixing (iv) Vaporization (v) Ignition (vi) Chemical reaction between fuel and oxidizer (3 marks) The propellants are injected into the combustion chamber through fine orifices for proper atomization. steam and oxygen are generated. the gas turbine. Because of the third liquid. (May/Jun 2014) (April/May 2010) (MK) (16) Key Hint: In this system. Liquid fuel and liquid oxidizer are forced into the combustion chamber at high pressure by the fuel and oxidizer pumps. the pumps and additional lines are necessary.3. So the pump pressurization system is considerably more complex than gas pressurization system. Liquid hydrogen peroxide (H2O2) from the tank is decomposed by a catalyst such as calcium or sodium permanganate. Various methods are followed to atomize and mix the fuel and oxidizer. . Gas turbine is used to operate the fuel and oxidizer pumps. This steam is used to drive the turbine. Due to this. Explain the working of a turbo – pump feed system used in a liquid propellant rocket. Design of pump is a greatest problem that will handle the liquids safely and without leaks. (6 marks) (4 marks) Combustion of a liquid propellant (fuel and oxidizer mixture) in the combustion chamber requires the following basic processes . liquid fuel and the liquid oxidizer are stored in a separate tank at low pressure. It should be non-corrosive and non-reactive with components of the engine. (3 marks) 4. (8 marks) PROPERTIES OF SOLID PROPELLANTS It should release large amount of heat during combustion. Storage' and handling should be easy. the combustion rate and the gas flow rate through the exhaust nozzle. It should be non-corrosive and non-reactive with components of the engine. The combustion starts with the arrival of one of the propellants. (8 marks) 5. In order to obtain a low value of the oxidizer fuel ratio. Physical and chemical properties should not change during processing. It should have low values of vapor pressure and viscosity. Describe the important properties of liquid and solid propellants desired for rocket propulsion (May/Jun 2013) (April/May 2011) (April/May 2010) (MK) (16) Key Hint: PROPERTIES OF LIQUID PROPELLANTS Propellant should have high calorific value. It should not be poisonous and hazardous. (Nov/Dec 2013) (Nov/Dec 2012) (MK) (16) Key Hint: . It should not be poisonous and hazardous. It should have high density. It should have higher specific heat and thermal conductivity . the fuel jet is allowed to enter the combustion chamber first. It density should be high. Energy released during combustion per unit mass of the propellant combination should be high. It should be cheap and easily available. Products of combustion should have low molecular weight to produce high jet velocity. The combustion pressure and temperature depends on the flow rates of the propellants. It should be cheap and easily available. Draw a neat sketch explaining the general working of the hybrid propellant rocket. It should be easily ignitable. Due to high velocity of gases coming out from the unit. Solid fuel is packed in the combustion chamber and the liquid oxidizer is stored in the separate tank. solid fuel along with liquid oxidizer is used as a propellant. (2 marks) Working The liquid oxidizer which is stored in the separate tank is injected into the combustion chamber. In this type. very high pressure and very high temperature gases are produced. a force (or) thrust is produced in the opposite direction. pressure energy of the gas is converted into kinetic energy. This thrust propels the rocket. So the gases coming out from the unit with very high velocity. (4 marks) Advantages Speed regulation is possible by regulating the supply of oxidizer. In the nozzle. . High load capacity. combustion takes place automatically. When the combustion takes place in the combustion chamber. The highly heated products of combustion gases are then allowed to expand in the nozzle section.(4 marks) The hybrid rocket engines combine the advantages of both solid and liquid propellant rockets. When liquid oxidizer mixes with solid fuel in the combustion chamber. Space: (a) Military purposes – reconnaissance.lunar. Hybrid rockets are lighter when compared to the liquid propellant type rockets. (iii) Space. signal rockets. (iv) Scientific. longer. (4 marks) Disadvantage Nozzle erosion cannot be avoided. ascend and achievement of orbit (2) altitude control.German Mel 163 fighter used in II World war X. (4 marks) iii. .solar systems (e) Rocket engines are use to (1) take-off from earth. other military payloads (or mail carriers to reach remote villages in mountain areas) (b) Guided missiles:. (2 marks) 6. assisted take-off.g. reentry. interplanetary. (4 marks) Military: (a) Rocket projectiles (unguided missile) .ground to ground. ground to air (aircraft). (i) Aircrafts. (4 marks) Scientific.1 Research engine (first to break the sonic barrier) X . attainment of lunar and planetary orbits and landing. (b) Commercial purposes .:. In case of an accident or crash the explosion is less destructive compared to the liquid propellant rocket engines. (b) For throwing life line to ships. there is no chemical deterioration between fuel and oxidizer. (May/Jun 2014) (GK) (16) Key Hint: i.15 supersonic research aircraft ii. Describe briefly the important applications of rocket propulsion in the following fields. payloads explosive charges. (ii) Military. Aircrafts: (a) Primary power plants e. trajectory connections. iv.similar to rocket projectiles but bigger in size and the trajectory is controlled . Higher fuel density.not accurate. (a) Rockets which carry instruments to measure meteorological and scientific data at high altitudes may be guided or unguided. ship to air and ship to ship missiles . maneuvers etc. Since the fuel and oxidizer are kept separately. air to air. air to ground. (b) Auxiliary power plants for super performance or improving the performance (speed.payloads are atomic weapons. separation of vehicle stages.communication and weather studies scientific purposes . rate of climb). trains. smoke charge. antitank rockets. space and planets (c) Manned and unmanned space vehicles (d) Near earth. under water rockets etc. V η0=η p × ηt . (ii) engine output and thermal efficiency and (iii) overall efficiency.(c) Miscellaneous . (May/Jun 2013) (GK) (16) Key Hint: u σ = =2 cj η p= 2σ 1+σ 2 η p=80 (3 marks) Power=19.38 (3 marks) η0=0.ejection or crew escape capsules and stores personnel “propulsion belts”.5 MW (3 marks) ηt =75.6 MW (3 marks) Engine output =24. if the heat of reaction of the propellant is 6500 kJ/kg of the propellant mixture determine (i) the propulsive efficiency and power.603 (4 marks) F=m´ p c j F=7000 N Power=F ×u Engine output = ηt = Power ηp c 2j +u 2 2×C .0 kg/s. A rocket flies at 10080 kmph with an effective exhaust jet velocity of 1400 m/s and propellant floe rate of 5. (4 marks) 7. 96 c¿= (4 marks) (4 marks) cj cF c ¿ =945. Ambient pressure = 585 N/m2.0129 m2 (6 marks) F=m ´ p × ce F=137. Combustion chamber temperature = 3600 K.25× 10−5 po Ae =285.68× 103 N cF = F ¿ p0 A c F =2. Oxidizer flow rate = 41 kg/s. (ii) Thrust. (Nov/Dec 2014) (May/Jun 2012) (BK) (16) Key Hint: From isentropic table. (iv) Characteristic velocity.45 m/s (2 marks) . Determine: (i) Nozzle throat area. (iii) Thrust co-efficient. Mixture ratio = 5.694 m2 A ¿ =0.8.3 A¿ A e =3. A rocket has the following data: Combustion chamber pressure = 36 bar. pe =16. (v) Exit velocity of exhaust gases.