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

TEST BANK FOR Gas Dynamics 4TH Edition By Ethirajan Rathakrishnan (Solution Manual) Solutions Mannual for the fourth edition of Gas Dynamics Ethirajan Rathakrishnan Contents 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) 7 But cv R = 1 ! − 1 . Thus, ! ! − 1 T2 T1 = # + 1 ! − 1 T2 T1 = 1 + (! − 1)# ! Entropy change for a perfect gas can be written as "s = cp ln ! T2 T1 " − Rln ! p2 p1 " "s R = ln ' 1 # ( 1 + (! − 1)# ! ) ! !−1 * s2 − s1 = Rln ' 1 # ( 1 + (! − 1)# ! ) ! !−1 * Let # = 1+$, where $ # 1. Therefore, "s R = ! ! − 1 ln + 1 + (! − 1)(1 + $) ! , − ln(1 + $) = ! ! − 1 ln + 1 + ! − 1 + $(! − 1) ! , − ln(1 + $) = ! ! − 1 ln + 1 + ! − 1 ! $ , − ln(1 + $) Expanding the RHS, and retaining only upto second order terms, we get, "s R = ! ! − 1 + ! − 1 ! $ − (! − 1)2 2!2 $2 , − ! $ − $2 2 " = $ − ! − 1 2! $2 − $ + $2 2 = − $2 2 + $2 2! + $2 2 = $2 2! Note: Work has been done by the weight which is equal to p2Ap on the gas. The weight has moved by a distance of ds. Therefore,"E = W .ds = p2Ap . ds. 2.6 Since it is an open system, Work done = −cp (T2 − T1) 8 Basic Equations of Compressible Flow = −cp ! T2 T1 − 1 " T1 = −cp -( p2 p1 )!−1 ! − 1 . T1 = −1004.5 × % 21/3.5 − 1 & × 303 = −66.66 kJ/kg 2.7 Work done is given by W = p (V2 − V1) = × 6 (2 − 0.3) = 1.0335MJ since 1 atm = Pa. 2.8 The compression process is given as isentropic. Let subscripts 1 and 2 refer to initial and final states, respectively. By isentropic process relation, we have p1 "! 1 = p2 "! 2 "2 = ! p2 p1 "1/! "1 = ! 690 150 "1/1.3 × 1.5 = 4.85 kg/m3 2.9 As we know, the relation between temperature and pressure for isentropic change of state may be written as T2 T1 = ! p2 p1 "(!−1)/! where subscripts 1 and 2 refer to the initial and final states, respectively. T2 = T1 ! p2 p1 "0.4/1.4 = 298 ! 7 1 "0.286 = 519.9K 9 2.10 By isentropic relation, T2 T1 = ! v1 v2 "(!−1) where subscripts 1 and 2 refer to the initial and final states and v is specific volume. For air ! = 1.4. Therefore, T2 = T1 ! v1 v2 "0.4 = (30+273.15) (30)0.4 = 1181.7K = 908.55"C 2.11 (a) We have cp = ! ! − 1 R, therefore, the gas constant R = ! − 1 ! cp R = 0.4 1.4 × 1000 = 285.7J/(kg K) Also, R = Ru M = 8314 M where M is the molecular weight and Ru is universal gas constant. Thus, M = 8314 285.7 = 29.1 (b) By ideal gas state equation, we have p1V1 = mRT1 p2V2 = mRT2 where subscripts 1 and 2 refer to initial and final states, respectively. But p1 = p2 and therefore, V2 V1 = T2 T1 = 50 + 273.15 200 + 273.15 = 323.15 473.15 = 0.683 2.12 For an ideal gas, the speed of sound a may be expressed as a = / !RT 10 Basic Equations of Compressible Flow where ! is the ratio of specific heats and R is the gas constant. For the given gas, R = Ru/M = 8314/29 = 286.7J/(kg K) Therefore, 400 = / ! × 286.7 × 373.15 ! = 4002 286.7 × 373.15 = 1.5 The specific heat cp and cv can be written as cp = ! ! − 1 R cv = R ! − 1 Therefore, cp = (1.5/0.5) × 286.7 = 860.1J/(kg K) cv = 286.7/0.5 = 573.4J/(kg K) Note: The ratio of specific heats ! = cp/cv. For the present case ! = 1.5 = 860.1/573.4 is correct. This way the answer obtained for cp and cv may be checked. 2.13 At the nozzle exit, V = 390m/s and T = 28+273.15 = 301.15 K. The corresponding speed of sound is a = / !RT = $1.4 × 287 × 301.15 = 347.85 m/s Thus, M = V a = 390 347.85 = 1.12 11 By isentropic relation, we have T0 T = 1+ ! − 1 2 M2 = 1+0.2 × 1.122 = 1.25 T0 = 1.25 × 301.15 = 376.44K = 103.29"C For the flow, the stagnation temperature is T0 = 376.44 K. The static temperature T = 92.5"C = 92.5 + 273.15 = 365.65K T0 T = 376.44 365.65 = 1.03 = 1 + 0.2M2 Thus, M2 = 0.03 0.2 = 0.15 M = 0.387 This is the Mach number at the station where temperature is 92.5"C. 2.14 For hydrogen, the gas constant R = 8314/2.016 = 4124 J/(kgK). By isentropic relation, we have T2 T1 = ! p2 p1 "(!−1)/! = ! 1 7 "0.286 = 0.573 Therefore, T2 = 0.573 × T1 = 0.573 × 300 = 171.9K