Fundamentals of Gas Dynamics (NOC16 - ME05) Assignment - 10 : Solutions

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1 Fundamentals of Gas Dynamics (NOC16 - ME05) Assignment - 10 : Solutions Manjul Sharma & Aswathy Nair K. Department of Aerospace Engineering IIT Madras April 18, 016 (Note : The solutions discussed below are just one of the ways of solving the problem. Your method is right as long as the final answers match!!) 1) Air at M 1 = and P 1 = 100 kpa undergoes isentropic expansion to a downstream pressure of 50 kpa. What is the desired turn angle in degrees? Answer : (q) θ = 11.4 Given : M 1 = ; P 1 = 100 kpa; P = 50 kpa To find : θ ( P 01 = P 1 1+ γ 1 ) γ γ 1 M 1 P kpa We have, P = = From Isentropic tables, for P = , M =.44 for M =.44, ν = for M 1 =.0, ν 1 = 6.38 θ = ν ν ) A large tank at 500 K and 165 kpa feeds air to a converging nozzle. The back pressure outside the nozzle exit is sea-level standard ( kpa). What is the approximate exit area if the desired mass flow rate is 7 kg/hour? Answer : (n) A e = m 1

2 Given : T 0 = 500 K; P 0 = 165 kpa; P b = kpa; ṁ = 7 kg/h To find : A e P e = = > 0.58 nozzle is not choked P ṁ A e P throat = P atm = kpa RT0 P 0 = γ γ 1 ( ) Pe P 0 γ [ 1 P e P 0 A e m ]γ 1 γ 3) Airflow at M 1 = 3., P 1 = 1 bar (see figure below) passes through a 5 oblique shock deflection. What isentropic expansion turn is required to bring the flow back to M 1? Answer : (v) θ = 31.9 Given : M 1 = 3.; P 1 = 1 bar; θ 1 = 5 To find : θ for which M 3 = M 1 From θ β M chart, for M 1 = 3. & θ 1 = 5, From Normal Shocks tables, for M n1 =.16, β = 4.56 M n1 = M 1 sinβ =.16 M = M n = Since M 3 = M 1 = 3., from Prandtl-Meyer function tables, M n sin(β θ 1 ) = 1.83 for M 3 = 3., ν 3 = for M = 1.83, ν = 1.59 θ = ν 3 ν = 31.9

3 4) What would be the isentropic turn required in problem 3 to bring the flow back to P 1 condition? Answer : (t) θ = 6.3 Given : M 1 = 3.; P 1 = 1 bar; θ 1 = 5 To find : θ for which P 3 = P 1 From Normal Shocks tables, for M n1 =.16, M n = & P P 1 = 5.76 P 5.3 bar We have, P 3 = P 1 = 1 bar, ( = P 1+ γ 1 M ) ( γ γ 1 ) = bar Since state to state 3 is an expansion fan, it is isentropic. Hence, P 03 = = bar P 3 = 1 P From Isentropic tables, for P 3 P 03 = , M 3 =.906 for M 3 =.906, ν 3 = 47.9 for M = 1.83, ν = 1.59 θ = ν 3 ν = 6.3 5) Air flows through a variable area duct, where areas of different sections of duct are: A 1 = 4cm, A = 18cm and A 3 = 3cm. A normal shock stands at section. The flow properties at section 1 are M 1 =.5, P 1 = 40 kpa and T 1 = 30. Compute the mass flow rate through the duct. Answer : (g) ṁ = 0.96 kg/s Given : A 1 = 4 cm ; A = 18 cm ; A 3 = 3 cm ; M 1 =.5; P 1 = 40 kpa; T 1 = 30 A normal shock at section To find : ṁ a 1 = γrt 1 = 349 m/s v 1 = a 1 M 1 = 87 m/s ρ 1 = P 1 RT 1 = 0.46 kg/m 3 3

4 ṁ = ρ 1 A 1 v kg/s 6) What would be the Mach number and stagnation pressure at section 3 in problem 5? Answer : (y) M 3 = 0.7 & P 03 = 435 kpa Given : A 1 = 4 cm ; A = 18 cm ; A 3 = 3 cm ; M 1 =.5; P 1 = 40 kpa; T 1 = 30 A normal shock at section To find : M 3 & P 03 From Isentropic tables, for M 1 =.5, A 1 A 1 =.64 A 1 =.4 = 9.1 cm.64 Since flow is isentropic from section 1 to (before the shock), A x = A 1 = 9.1 cm A x A x = From Isentropic tables, for A x A x = 1.98, M x =.18 We have, ( P 01 = 1+ γ 1 ) ( γ M 1 P 1 P kpa Since flow is isentropic from section 1 to (before the shock), x = P 01 = 683 kpa From Normal shocks tables, for M x =.18, γ 1) M y = 0.55 & y x = y = 435 kpa Since flow is isentropic from section (after the shock) to section 3, y = P 03 = 435 kpa From Isentropic tables, for M y = 0.55, A y A y = 1.55 A y 14.3 cm Since flow is isentropic from section (after the shock) to section 3, A y = A 3 = 14.3 cm From Isentropic tables, for A 3 A 3 =.4, A 3 A 3 = =.4 4

5 M ) Air is flowing downstream a ramp as shown in figure. If the initial pressure and Mach number are 400 kpa and 1.8, determine the pressure of air on the ramp. Answer : (a) P = kpa Given : P 1 = 400 kpa; M 1 = 1.8 To find : P for M 1 = 1.8, ν 1 = 0.75 θ = ν ν 1 ν = θ +ν 1 = M = 3.05 From Isentropic tables, for M 1 = 1.8, P 1 P 01 = P 01 = kpa Since flow is isentropic, = P 01 = kpa From Isentropic tables, for M = 3.05, P = P = kpa 8) Air at pressure 34.5 kpa, temperature 350 K and Mach number 1.5 is to be isentropically expanded to 13.8 kpa using a Prandtl-Meyer expansion fan. Determine the deflection angle. Answer : (j) θ = Given : P 1 = 34.5 kpa; T 1 = 350 K; M 1 = 1.5; P = 13.8 kpa To find : θ 5

6 From Isentropic tables, for M 1 = 1.5, Since flow is isentropic, = P 01 = kpa P 1 P 01 = 0.7 P 01 = kpa P = = From Isentropic tables, for P = , M =.10 From Prandtl-Meyer functions tables, for M 1 = 1.5, ν 1 = for M =.10, ν = θ = ν ν 1 = ) Air is flowing through the configuration as shown in figure below. Determine the final Mach number M 3 if a weak oblique shock was observed at the first turn. Answer : (d) M 3 = 1.97 Given : M 1 =.0; θ 1 = 14 ; weak oblique shock at first turn To find : M 3 From θ β M chart, for M 1 =.0 & θ 1 = 14, From Normal Shocks tables, for M n1 = 1.389, β = 44 M n1 = M 1 sinβ = M n =

7 M n M = sin(β θ 1 ) = From the figure, for the flow to be in the same direction as the flow at state 1, θ = 14 for M = 1.488, ν = θ = ν 3 ν ν 3 = θ +ν = M 3 = ) A compressed air reservoir which stores air discharges it through a small opening to the atmosphere. If flow is assumed to be isentropic, what is the maximum pressure at which air can be stored in the reservoir such that it would not result in the choked flow at the opening? Answer : (m) P 01 = atm Given : P = 1 atm To find : maximum P 01 such that opening is not choked Since flow is isentropic, = P 01 The critical pressure ratio at which flow becomes choked (i.e. M e = 1) is, P = 0.58 = atm P 01 = atm 7

Fundamentals of Gas Dynamics (NOC16 - ME05) Assignment - 8 : Solutions

Fundamentals of Gas Dynamics (NOC16 - ME05) Assignment - 8 : Solutions Manjul Sharma & Aswathy Nair K. Department of Aerospace Engineering IIT Madras April 5, 016 (Note : The solutions discussed below