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

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1 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 are just one of the ways of solving the problem. Your method is right as long as the final answers match!!) 1) Air at Mach number 1.8, static pressure of 0.8 bar and static temperature of 373 K passes through a normal shock wave. Determine its density after the shock. Answer : (p) ρ y = 1.76 kg/m 3 Given : M x = 1.8; = 0.8 bar; T x = 373 K To find : ρ y From Normal Shock Tables for γ = 1.4, corresponding to M x = 1.8, P y = & T y T x = 1.53 P y =.8904 bar & T y = K ρ y = P y RT y = 1.76 kg/m 3 ) A normal shock occurs in the diverging section of a C-D nozzle. The throat area is one-third of the exit area and the static pressure at the exit is 0.4 times the stagnation pressure at the entry. The flow is throughout isentropic except across the shock. Determine the area of cross-section of the nozzle (in terms of throat area) where the normal shock occurs. Answer : (t).84 times the throat area Given : = 3; P e = 0.4P 0i To find : 1

2 Since the flow is isentropic everywhere except at the shock, = ; = ; A 1 = = ; A y = = = 3 = A A But, P = 0.4 = P 0.4 A 1 = A 1 = = A A A = 1 A = 3 P = = 1. From Isentropic tables, corresponding to P = 1., M = 0.47 From Normal Shock tables, for = 0.466, & 0.4 = P = = = M x =.58

3 From Isentropic tables, for M x =.58, =.84 =.84 3) A C-D nozzle is supplied air kept in a reservoir at pressure 500 kpa. The throat area is 8 cm and exit area is 30 cm. A normal shock stands at a section where the area is 10 cm. Assuming flow to be isentropic everywhere except at the shock, find the static pressure at the exit. Answer : (g) P e = 48 kpa Given : P 0 = 500 kpa; = 8 cm ; A s = 10 cm ; = 30 cm To find : P e = = A 1 = 8 cm Assuming shock is a very thin region, A s = = A y = 10 cm = 10 8 = 1.5 From Isentropic tables, for = 1.5, M x = 1.6 Since flow is isentropic, P 0 = = = 500 kpa From Isentropic Tables, for M x = 1.6, From Normal Shocks tables, for M x = 1.6, = 0.35 = kpa M y = P y =.80 P y = kpa = = kpa = A y A y = cm = A y = cm =

4 From Isentropic tables, for =.505, M e = 0.33 & P e P 0e = P e 48 kpa 4) Find the entropy change across the normal shock for an incoming flow of air with Mach number.5. Answer : (c) s = 137. J/kg K Given : M x =.5 To find : s From Normal Shocks Tables, corresponding to M x =.5, T y T x =.138 & P y = 7.15 s = C P ln ( Ty T x ) Rln ( Py ) s = 137. J/kgK 5) A pitot-static probe is used for measurement in a supersonic flow in a wind tunnel. If the readings of the probe are 600 kpa and 40 kpa, determine the static pressure of the supersonic flow in the wind tunnel. Answer : (k) = 19.3 kpa Given : = 600 kpa; P y = 40 kpa To find : Py = 0.7 From Isentropic tables, corresponding to P y = 0.7, M y = 0.73 From Normal Shocks Tables, corresponding to M y = 0.73, M x = 1.4 & = 3.1 = 19.3 kpa 6) Air from a reservoir at 0 and 500 kpa flows through a duct and forms a normal shock downstream of a throat of area 10 cm. By an odd coincidence it is found that the stagnation pressure downstream of this 4

5 shock exactly equals to the throat pressure. What is the area where shock stands? Answer : (e) A s = 4.7 cm Given : T 0 = 93 K; P 0 = 500 kpa; = 10 cm ; = P t To find : A s = = A 1 = 10 cm Since flow is isentropic everywhere except at the shock, = ; = ; A 1 = = ; A y = Assuming shock is a very thin region, A s = = A y From Isentropic tables, corresponding to M t = 1, P t = 0.58 P t = 64 kpa = = P t = 0.58 From Normal Shock tables, corresponding to = 0.58, M x =.4 From Isentropic tables, for M x =.4, =.448 =.448 = cm 7) A C-D nozzle, discharging a jet of air into the atmosphere ( bar, 300 K), has a ratio of exit area to throat area equal to.5. What should be the reservoir pressure to drive this nozzle if a normal shock stands at an area ratio (area where shock stands to the throat area) of in the divergent section? Answer : (z) P 0 = 161 kpa Given : = bar; T 0 = 300 K; To find : =.5; A s =.0 5

6 Since flow is isentropic everywhere except at the shock, = ; = ; A 1 = = ; A y = A s = = From Isentropic tables, corresponding to =, M x =. From Normal Shocks tables, for M x =., M y = & = = = kpa From Isentropic tables, for M x =., = = kpa 8) A large tank with compressed air is attached to a C-D nozzle at pressure 4 bar and temperature of 35. The nozzle throat area is 3 cm and the exit area is 9 cm. A shock occurs at a location where the cross section area is 6 cm. Calculate the back pressure and temperature of the flow. Answer : (m) P b =.34 bar & T b = 300 K Given : = 4 bar; T 01 = 308 K; = 3 cm ; = 9 cm ; A s = 6 cm To find : P b, T b = = A 1 = 3 cm Since flow is isentropic everywhere except at the shock, = ; = ; A 1 = = ; A y = Assuming shock is a very thin region, A s = = A y = 6 cm A s = = 6 3 = From Isentropic tables, corresponding to =, M x =. 6

7 From Normal Shocks tables, for M x =., M y = & = 0.68 = 0.68 =.51 bar Also, = A y A y = cm = From Isentropic tables, corresponding to = 1.884, M = ; P = ; T T 0 = P.34 bar & T 300 K 9) For the previous problem find the critical values of back pressure P cr1 and P cr3 respectively as defined in the class. Answer : (w) 3.97 bar,1.67 bar Given : = 4 bar; T 01 = 308 K; = 3 cm ; = 9 cm ; A s = 6 cm To find : P cr1, P cr3 = = A 1 = 3 cm = = 9 3 = 3 From Isentropic tables, corresponding to = 3, M,cr1 = & M,cr3 =.64 And, P,cr1,cr1 = & P,cr3,cr3 = P,cr1 3.9 bar & P,cr bar 7

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

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