Exam (elaborations) TEST BANK FOR Gas Dynamics 4TH Edition By Ethirajan Rathakrishnan (Solution Manual)

1 SomePreliminaryThoughts 1 2 BasicEquations ofCompressibleFlow 3 3 Wave Propagation 23 4 One-DimensionalFlow 25 5 NormalShockWaves 79 6 ObliqueShock andExpansionWaves 119 7 PotentialEquationforCompressibleFlow 157 8 SimilarityRules 161 9 TwoDimensionalCompressibleFlows 165 10 Prandtl-Meyer Flow 169 11 Flow with Friction and Heat Transfer 173 12 MOC 205 13 Measurements in Compressible Flow 207 iii Chapter 2 Basic Equations of Compressible Flow 2.1 In the reservoir, the air is at stagnation state. So, the entropy relation would be s2 − s1 = cp ln ! T02 T01 " − Rln ! p02 p01 " But, T01 = T02 for adiabatic process. Therefore, "s = Rln ! p01 p02 " = Rln ! p01 12 p01 " = Rln 2 = 198.933 J/(kg K) Note: It should be noted that, for entropy only subscripts 2 and 1 are used; since entropy is not defined like static or stagnation entropy. 2.2 Let the initial state be denoted by subscript 1 and expanded state by subscript 2. (a) Since the cylinder is insulated, preventing any heat transfer what-so-ever, the process is adiabatic. The governing equation for this process is given by p1V! 1 = p2V! 2 = constant (1) Also, from ideal gas state equation p1V1 T1 = p2V2 T2 = R (2) 3 4 Basic Equations of Compressible Flow From Eqs. (1) and (2), we get p1 p2 = ! V2 V1 "! = ! T1 T2 "!/(!−1) Therefore, T2 = T1 # 10(!−1) = 557.35K "T = −842.65K (b) Work = $ pdv = $ dh − $ du − $ vdp (3) Also, pv! = constant from equation (1) Differentiating equation (1), we have, p!v!−1dv + v!dp = 0 Dividing throughout by v!−1 and integrating, we get $ p!dv + $ vdp = 0 $ vdp = −!w (4) Substituting equation (4) in equation (3) and simplifying, we get (1 − !)w = R"T w = R"T 1 − ! = 287 × (−842.65) (−0.4) = 6.04 × 105 J/kg Note: Since the process undergone is expansion from a high pressure, the work removed is positive, i.e, work is done by the gas. (c) Also, from equation (1) p1 p2 = ! V2 V1 "! = 101.4 = 25.1189 Therefore, The pressure ratio = 25.1189 5 2.3 p1v! 1 = p2v! 2, where v is specific volume, i.e. volume per unit mass = V/m. Therefore, p1 ! V1 m1 "! = p2 ! V2 m2 "! Also, V1 = V2 = V = volume of the tank. p2 = p1 ! m2 m1 "! = 5× 105 × ! 1 2 "1.4 = 1.8946 × 105 Pa From equation of state for a calorically perfect gas, p1 p2 = "1 "2 T1 T2 T2 = ! p2 p1 "! m1 m2 " T1 = ! 1.8946 5 " × 2 × 500 = 378.92K 2.4 p1 p2 = ! T1 T2 "!/(!−1) (a) Therefore, T2 = ! p2 p1 "(!−1)/! T1 = 61/3.5 × 290 = 483.868K The change in the temperature is "T = T2 − T1 = 483.868 − 290 = 193.868K (b) By first law of thermodynamics, we have du + d(pe) + d(ke) = dq + dw 6 Basic Equations of Compressible Flow Here, velocity changes are neglected. Therefore, d(ke) = 0 Also, assuming d(pe) = 0 The first law of thermodynamics reduces to du = dq + dw But the process is isentropic, thus dq = 0. Therefore, du = dw = cv"T = 717.5 × 193.868 = 1.39 × 105 J/kg (c) The work done is negative, i.e. work is done on the gas. It has been computed in (b) above. 2.5 Work done by the weight on the piston goes towards increasing the internal energy of the gas. From the first law of thermodynamics E2 − E1 = Q +W where, E, Q, and W are respectively the internal energy, heat transfered, and work done. Since no heat is transfered, Q = 0. Therefore, E2 − E1 = W = $ F .ds where, F is force and ds is distance. At the new equilibrium position, the force acting on the piston face is F = p2Ap, Ap is the area of the piston face. The distance traveled by the piston is ds = (V1 − V2)/Ap, V1 and V2 are the initial and final volumes. Thus we have, E2 − E1 = p2 .Ap(V1 − V2)/Ap = −p2(V2 − V1) For unit mass, e2 − e1 = −p2(V2 − V1) For calorically perfect gas, e = cvT. Therefore cv(T2 − T1) = −p2 ! RT2 p2 − RT1 p1 " cv R ! T2 T1 − 1 " = − T2 T1 + p2 p1 T2 T1 % 1 + cv R & = cv R + # (where # = p2/p1