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Solutions Mannual for the fourth edition of Gas Dynamics Ethirajan Rathakrishnan Preface This manual gives the detailed solution for all the problems given at the end of different chapters of the 3rd edition of Gas Dynamics. My sincere thanks to my doctoral and masters students who helped me in checking and keying in the solutions of this manual. My sincere thanks to the Continuing Education Centre of Indian Institute of Technology Kanpur for the financial support to prepare this manual. E Rathakrishnan i ii Contents 1 Some Preliminary Thoughts 1 2 Basic Equations of Compressible Flow 3 3 Wave Propagation 23 4 One-Dimensional Flow 25 5 Normal Shock Waves 79 6 Oblique Shock and Expansion Waves 119 7 Potential Equation for Compressible Flow 157 8 Similarity Rules 161 9 Two Dimensional Compressible Flows 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 1 Some Preliminary Thoughts 1 2 Some Preliminary Thoughts Chapter 2 Basic Equations of Compressible Flow 2.1 In the reservoir, the air is at stagnation state. So, the entropy relation would be ! " ! " T02 p02 s2 − s1 = cp ln − R ln T01 p01 But, T01 = T02 for adiabatic process. Therefore, ! " p01 ∆s = R ln p02 ! " p01 = R ln 1 = R ln 2 2 p01 = 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 p1 Vγ1 = p2 Vγ2 = constant (1) Also, from ideal gas state equation p1 V1 p2 V2 = =R T1 T2 3 (2) 4 Basic Equations of Compressible Flow From Eqs. (1) and (2), we get p1 = p2 ! ! V2 V1 "γ T1 # 10(γ−1) = 557.35 K = T1 T2 "γ/(γ−1) Therefore, T2 = ∆T = (b) Work = Also, $ − 842.65 K pdv = $ pv γ = constant dh − $ du − $ vdp (3) from equation (1) Differentiating equation (1), we have, pγv γ−1 dv + v γ dp = 0 Dividing throughout by v γ−1 and integrating, we get $ $ pγdv + vdp = 0 $ vdp = −γw (4) Substituting equation (4) in equation (3) and simplifying, we get (1 − γ) w = R∆T w = R∆T 287 × (−842.65) = 1−γ (−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 Therefore, ! V2 V1 "γ = 101.4 = 25.1189 The pressure ratio = 25.1189 5 2.3 p1 v1γ = p2 v2γ , where v is specific volume, i.e. volume per unit mass = V/m. Therefore, ! "γ ! "γ V1 V2 p1 = p2 m1 m2 Also, V1 = V2 = V = volume of the tank. ! "γ m2 p 2 = p1 m1 ! "1.4 1 = 5 × 105 × 2 1.8946 × 105 Pa = From equation of state for a calorically perfect gas, p1 p2 = T2 = ρ1 T1 ρ2 T2 ! "! " p2 m1 T1 p1 m2 ! " 1.8946 × 2 × 500 5 = 378.92 K = 2.4 p1 = p2 (a) Therefore, T2 ! T1 T2 "γ/(γ−1) = ! = 61/3.5 × 290 = 483.868 K p2 p1 "(γ−1)/γ T1 The change in the temperature is ∆T = = T2 − T1 = 483.868 − 290 193.868 K (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 − E 1 = 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 = p2 Ap , 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 = cv T . Therefore ! " RT2 RT1 cv (T2 − T1 ) = −p2 − p2 p1 " ! c v T2 T2 p2 −1 = − + R T1 T1 p1 cv & T2 % cv 1+ = + λ (where λ = p2 /p1 ) T1 R R 7 But 1 cv = . Thus, R γ−1 γ T2 γ − 1 T1 = T2 T1 = λ+ 1 γ−1 1 + (γ − 1)λ γ Entropy change for a perfect gas can be written as ! " ! " T2 p2 ∆s = cp ln − R ln T1 p1 ' ( ) γ * 1 1 + (γ − 1)λ γ−1 ∆s = ln R λ γ ' ( ) γ * 1 1 + (γ − 1)λ γ−1 s2 − s1 = R ln λ γ Let λ = 1 + $, where $ # 1. Therefore, + , 1 + (γ − 1)(1 + $) ∆s γ = ln − ln(1 + $) R γ−1 γ + , 1 + γ − 1 + $(γ − 1) γ ln = − ln(1 + $) γ−1 γ + , γ−1 γ ln 1 + $ − ln(1 + $) = γ−1 γ Expanding the RHS, and retaining only upto second order terms, we get, , ! + " ∆s γ γ−1 (γ − 1)2 2 $2 = $− $ − $ − R γ−1 γ 2γ 2 2 = $− = $2 2γ $2 $2 $2 γ−1 2 $2 $ −$+ =− + + 2γ 2 2 2γ 2 Note: Work has been done by the weight which is equal to p2 Ap on the gas. The weight has moved by a distance of ds. Therefore, ∆E = W . ds = p2 Ap . ds. 2.6 Since it is an open system, Work done = −cp (T2 − T1 ) 8 Basic Equations of Compressible Flow −cp = ! " T2 − 1 T1 T1 -( p2 p1 ) γ−1 γ . = −cp − 1 T1 = % & −1004.5 × 21/3.5 − 1 × 303 − 66.66 kJ/kg = 2.7 Work done is given by W = = p (V2 − V1 ) = 101325 × 6 (2 − 0.3) 1.0335 MJ since 1 atm = 101325 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 ργ1 ρ2 = p2 ργ2 = ! = p2 p1 "1/γ ρ1 = ! 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 "(γ−1)/γ where subscripts 1 and 2 refer to the initial and final states, respectively. T2 = = T1 ! p2 p1 "0.4/1.4 519.9 K ! "0.286 7 = 298 1 9 2.10 By isentropic relation, T2 = T1 ! v1 v2 "(γ−1) where subscripts 1 and 2 refer to the initial and final states and v is specific volume. For air γ = 1.4. Therefore, T2 ! "0.4 v1 v2 = T1 = (30 + 273.15) (30) = 0.4 = 1181.7 K 908.55◦ C γ−1 γ R, therefore, the gas constant R = cp γ−1 γ 0.4 × 1000 = 285.7 J/(kg K) R= 1.4 2.11 (a) We have cp = Also, Ru 8314 = M M where M is the molecular weight and Ru is universal gas constant. Thus, R= M= 8314 = 29.1 285.7 (b) By ideal gas state equation, we have p1 V1 = mRT1 p2 V2 = mRT2 where subscripts 1 and 2 refer to initial and final states, respectively. But p1 = p2 and therefore, V2 V1 = 323.15 T2 50 + 273.15 = = T1 200 + 273.15 473.15 = 0.683 2.12 For an ideal gas, the speed of sound a may be expressed as / a = γRT 10 Basic Equations of Compressible Flow where γ is the ratio of specific heats and R is the gas constant. For the given gas, R = Ru /M = 8314/29 = 286.7 J/(kg K) Therefore, 400 = γ = / γ × 286.7 × 373.15 4002 = 1.5 286.7 × 373.15 The specific heat cp and cv can be written as cp = γ R γ−1 cv = R γ−1 = (1.5/0.5) × 286.7 = 860.1 J/(kg K) Therefore, cp cv = 286.7/0.5 = 573.4 J/(kg K) Note: The ratio of specific heats γ = cp /cv . For the present case γ = 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 = 390 m/s and T = 28 + 273.15 = 301.15 K. The corresponding speed of sound is / √ a = γRT = 1.4 × 287 × 301.15 = 347.85 m/s Thus, M = V 390 = a 347.85 = 1.12 11 By isentropic relation, we have T0 T T0 γ−1 2 M 2 = 1+ = 1 + 0.2 × 1.122 = 1.25 = 1.25 × 301.15 = 376.44 K 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 = 365.65 K 376.44 T0 = = 1.03 = 1 + 0.2 M 2 T 365.65 Thus, M2 = 0.03 = 0.15 0.2 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/(kg K). By isentropic relation, we have T2 T1 = = ! "(γ−1)/γ p2 p1 ! "0.286 1 = 0.573 7 Therefore, T2 = 0.573 × T1 = 0.573 × 300 = 171.9 K By energy equation, we have h2 + V22 = h1 , since V1 = 0 2 V22 = 2 (h1 − h2 ) 12 But Basic Equations of Compressible Flow h = cp T , therefore, V2 = 0 cp 2cp (T1 − T2 ) = 1.4 γ R= × 4124 γ−1 0.4 = 14434 J/(kg K) Thus, V2 = / = 2 × 14434 (300 − 171.9) 1923 m/s The speed of sound is given by / √ a2 = γRT2 = 1.4 × 4124 × 171.9 = 996.23 m/s Thus, M2 = V2 1923 = 1.93 = a2 996.23 The mass flow rate is given by ṁ = ρ2 A2 V2 ρ2 = p2 , by state equation RT2 ρ2 = 101325 , since 1 atm =101325 Pa 4124 × 171.9 = 0.143 kg/m3 Thus, ṁ = = 0.143 × 10 × 10−4 × 1923 0.275 kg/s 2.15 The given process is a polytropic process with index n = 1.32. Since air is given as an ideal gas with constant specific heats, we have from isentropic relations the change of entropy as ! " ! " T2 p2 s2 − s1 = cp ln − R ln T1 p1 13 For a polytropic process, T2 = T1 ! p2 p1 " n−1 n Combining these two equations, we obtain + ! " , ! " n−1 p2 s2 − s1 = cp − R ln n p1 since cp = γ R for an ideal gas, the above equation may be written as γ−1 ! " p2 (n − γ) R s2 − s1 = ln n (γ − 1) p1 With the given data, R= γ−1 0.4 cp = × 1004 = 287 J/(kg K) γ 1.4 Therefore, s2 − s1 = (1.32 − 1.4) 287 1100 ln 1.32 × 0.4 101 = −103.8 J/(kg K) Note: Since the entropy of the gas decreases for this internally reversible process, heat must be removed from the gas. This is why the cylinder used for such a compression process is usually water jacketed. Also, we know that the entropy of the system and surrounding cannot decrease. But in this problem, the entropy decreases. It should be noted that, what decreases is entropy of the system alone and not the combined entropy of the system and surrounding. 2.16 For oxygen, molecular weight M = 32. The gas constant R = 8314/32 = 259.8 J/(kg K). Therefore, cp = cv = γ R = 909.3 J/(kg K) γ−1 cp = 649.5 J/(kg K) γ The increase in internal energy is %u = = cv (T2 − T1 ) = 649.5 (125 − 25) 64950 J/(kg K) 14 Basic Equations of Compressible Flow The increase in enthalpy is %h = cp (T2 − T1 ) = 909.3 (125 − 25) = 90930 J/(kg K) 2.17 Let the subscripts 1 and 2 refer to the inlet and exit states. At state 1, p1 = 100 kPa, ρ1 = 1.175 kg/m3 . Therefore, T1 = At state 2, p1 100 × 103 = 296.5 K = Rρ1 287 × 1.175 p2 = 500 kPa, ρ2 = 5.875 kg/m3 . Therefore, T2 = p2 500 × 103 = 296.5 K = Rρ2 287 × 5.875 Assuming air to be a perfect gas, the enthalpy difference is given by h2 − h1 = = cp (T2 − T1 ) = cp (296.5 − 296.5) 0 2.18 The entropy change is given by ds = δq T where δq is the reversible heat addition per unit mass. For an ideal gas δq = dh − vdp where dh is the enthalpy change and dh = cp dT δq = cp dT − vdp Using the above relation, we get ds = cp By state equation v dT − dp T T pv = RT , R v = T p 15 Therefore, ds = cp dT R − dp T p Taking log of state equation and differentiating, we get dT dp dv = + T p v Substituting for dT is ds expression, we obtain T ! " dp dv dp ds = cp + −R p v p ds = cp But cp − cv = R, dv dp + (cp − R) v p cp − R = cv . Thus, ds = cp dv dp + cv v p For an isentropic change of state, we have 0 = dv v cp dv cv v cp But cp dv dp + cv v p = −cv = − dp p dp p cp = γ. Therefore, cv γ dp dv =− v p Integrating both sides, we get γ ln v = − ln p + constant ln p + ln v γ = constant ln (pv γ ) = constant or pv γ = constant 16 Basic Equations of Compressible Flow 2.19 For an isothermal process, the change in entropy %s is given by ! " p1 %s = s2 − s1 = R ln p2 where the subscripts 1 and 2 stand for the initial and final states, respectively. By state equation p1 V1 = mRT1 p2 V2 = mRT2 But T1 = T2 , therefore, p1 V1 = p2 V2 = mRT2 . Hence, p2 = p1 V1 0.7 × 106 × 0.014 = V2 0.084 = 0.117 MPa Thus, the change of entropy is %s = R ln ! p1 p2 " = 287 ln ! 0.7 0.117 " 513.4 J/(kg K) = 2.20 Let subscripts 1 and 2 refer to initial and final states, respectively. For process 1 p2 = 700 kPa. By state equation, we have p1 V = mRT1 p2 V = mRT2 Therefore, T2 = p2 700 × 308.15 = 616.3 K × T1 = p1 350 The change in entropy is given by %s = cp ln ! T2 T1 " − R ln ! p2 p1 " For air cp = 1004.5 J/(kg K) and R = 287 J/(kg K). Thus, ! " ! " 616.3 700 %s = 1004.5 ln − 287 ln 308.15 350 = 497.33 J/(kg K) 17 For 0.3 kg of air, %s = 0.3 × 497.33 = 149.2 J/K For process 2 pV1 = mRT1 pV2 = mRT2 Therefore, T2 = V2 × T1 V1 Initial volume is given by V1 = 0.3 × 287 × 308.15 mRT1 = = 0.0758 m3 p 350 × 103 T2 = 0.2289 × 308.15 = 930.55 K 0.0758 The entropy change is given by %s ! T2 T1 " = cp ln = 1110.163 J/(kg K) = 1004.5 ln ! 930.55 308.15 " For 0.3 kg of air, %S = 0.3 × 1110.163 = 333 J/K 2.21 Given flow is adiabatic and frictionless and therefore, isentropic. By energy equation, we have h1 + V12 V2 = h2 + 2 2 2 For perfect gas h = cp T , thus, c p T1 + V12 V2 = cp T2 + 2 2 2 For air cp = 1004.5 J/(kg K). Therefore, 1004.5 × 273.15 + 9002 3002 = 1004.5 T2 + 2 2 18 Basic Equations of Compressible Flow Solving this, we get T2 = 631.54 K. Therefore, the temperature increase becomes = T2 − T1 = 631.54 − 273.15 %T = 358.39 K By isentropic relation, we have p2 = p1 ! T2 T1 γ " γ−1 Thus, p2 = p1 = %p T2 T1 "3.5 = 140 ! 631.54 273.15 "3.5 = 2631 kPa 2.631 MPa p2 − p1 = = 2.22 For nitrogen, ! 2.491 MPa R = Ru /M = 8314/28 = 297 J/(kg K). For reversible process, the change in entropy may be expressed as ! " ! " T2 p2 %s = cp ln − R ln T1 p1 But T2 = T1 , therefore, %s = = − R ln ! p2 p1 " = − 297 ln ! 300 100 " − 0.3263 kJ/(kg K) Note: We know that the entropy of the system and surrounding cannot decrease. But in this problem, the entropy decreases. It should be noted that, what decreases is entropy of the system alone and not the combined entropy of the system and surrounding. 2.23 By isentropic relation, p2 = p1 ! T2 T1 γ " γ−1 19 where subscripts 1 and 2 refer to initial and final states. That is, T2 T1 T2 = = = ! p2 p1 " γ−1 γ ! " 0.4 p2 1.4 T1 p1 ! "0.286 550 300 = 475.4 K 110 The change in enthalpy is (h2 − h1 ), and h = cp T . Therefore, h2 − h1 = cp (T2 − T1 ) Also, cp = γ R γ−1 and R = 287 J/(kg K), for air. cp = 1.4 × 287 = 1004.5 J/(kg K) 0.4 Thus, h2 − h1 = = 1004.5 (475.5 − 300) 176.19 kJ/kg 2.24 Let the initial and final states of air are designated by subscripts 1 and 2, respectively. By perfect gas state equation, we have Dividing one by other, we get p1 V = mRT1 p2 V = mRT2 p1 T1 = p2 T2 The change in entropy for a perfect gas is given by ! " ! " T2 p2 %s = cp ln − R ln T1 p1 ! " T2 = (cp − R) ln T1 20 Basic Equations of Compressible Flow Given, T1 = 50 + 273.15 = 323.15 K T2 = 125 + 273.15 = 398.15 K cp = γ R = 1004.5 J/(kg K) γ−1 %s = (1004.5 − 287) ln = 149.75 J/(kg K) Thus, ! 398.15 323.15 " 2.25 By energy equation, we have h0 = h + V2 2 where subscript 0 refers to stagnation condition. Assuming air to be a perfect gas, we can express h = cp T . Therefore, c p T0 = cp T + T0 − T = V2 2 cp V2 2 From standard atmosphere table, at 10000 m, T = 223.15 K. Therefore, the speed of sound becomes / √ a = γ R T = 1.4 × 287 × 223.15 = 299.436 m/s The flight speed is V = M a = 2 × 299.436 = 598.872 m/s Thus, the temperature becomes ∆T = T0 − T = = 178.52 598.8722 2 × 1004.5 21 2.26 Let subscripts 1 and 2 refer to the initial and final states. Given, T1 = 15◦ C = 288.15 K, v1 = 0.06 m3 , v2 = 0.12 m3 , By perfect gas state equation, pv = RT Therefore, T1 T2 = v1 v2 since p1 = p2 . Thus, T2 = T1 288.15 × 0.12 = 576.3 K v2 = v1 0.06 = 303.15◦ C p 1 = p2 22 Basic Equations of Compressible Flow Chapter 3 Wave Propagation 23 24 Wave Propagation Chapter 4 One-Dimensional Flow 4.1 Total temperature at “1” is T01 = 300 K. This much temperature is required as static temperature at the test-section. Therefore, TT = 300 K T0T TT = 1+ γ−1 2 M 2 T0T 300 = 1+ 0.4 × 2.52 2 T0T = 675 K Hence, the temperature rise required, ∆T = = T0T − TT = 675 − 300 375 K 4.2 Let the test-section conditions be denoted by subscript 2, and the sonic conditions by superscript *. γ−1 2 M 2 T02 T2 = 1+ 373 T2 = 1 + 0.2 × 22 T2 = 207.2 K 25 26 One-Dimensional Flow Further, since we know that T∗ T0 = 0.8333 T∗ = 373 × 0.8333 = 310.82 K a∗ = √ p0 = 3.12 × 101325 = 3.16 × 105 ρ0 = p0 RT0 = 3.16 × 105 287 × 373 = 2.95 kg/m3 ρ∗ ρ0 = 0.634 ρ∗ = 1.8703 kg/m3 ṁ = ρ∗ A∗ V ∗ = = 1.4 × 287 × 310.82 = 353.4 m/s = V ∗ 1.8703 × 80 × 10−4 × 353.4 5.288 kg/s 4.3 Let subscripts 0, 1, and e refer to stagnation state and states at the nozzle entrance and exit, respectively. We know that, Te = 0.8333 T0 This gives, T0 = 293 = 351.6 K 0.8333 Also, T0 T1 = 1+ γ−1 2 M 2 27 T1 = 351.6 1 + 0.2 × 9 = 125.57 K pe p0 = 0.5283 p0 = 0.8 = 1.5143 atm 0.5283 = ! p0 p1 = p1 1 γ−1 2 M 1+ 2 1 + 0.2 × 32 = 1.5143 36.73 = 0.04123 atm γ " γ−1 23.5 √ 1.4 × 287 × 293 = 343.11 m/s Ve = ae = Ae = 40 × 10−4 m2 ρe = pe RTe = 0.8 × 101325 = 0.964 kg/m3 287 × 293 ṁ = ρe Ae Ve = 0.964 × 40 × 10−4 × 343.11 = 1.323 kg/s Note: It should be noted that the calculation made with equations for pressure, density, and temperature in the problem can also be done using gas tables. In fact that procedure will result in considerable time saving. 4.4 Given, A1 = 0.6 × 0.4 = 0.24 m2 28 One-Dimensional Flow A∗ = 0.3 × 0.4 = 0.12 m2 A2 = h2 × 0.4 = 0.4 h2 m2 M2 = 2.5 T2 = − 10◦ C p2 = 0.15 atm For M2 = 2.5, isentropic table gives A2 = 2.6367 A∗ and T2 = 0.4444 T02 This gives, Thus, A2 = 2.6367 × 0.12 = 0.3164 m2 h2 T02 = 0.3164 0.4 = 0.791 m = 263 0.4444 = 592 K A1 0.24 =2 = A∗ 0.12 A1 From subsonic part of isentropic table, for A ∗ = 2, we get M2 = 0.3 and T1 = 0.9823 T01 For isentropic flow, T01 = T02 . Therefore, T1 = = V1 = = 0.9823 × 592 581.5 K / M1 a1 = 0.3 γ R T 145 m/s 29 4.5 For M1 = 0.5, from isentropic table, we have p1 p0 = 0.84 Therefore, p2 p0 = p2 p1 1.0 × 0.84 × = p1 p0 6.5 = 0.129 For p/p0 = 0.129, from isentropic table, we have M2 = 2.0 A∗ A2 = 0.6 Mass flow rate is given by ṁ = = = ρ1 V1 A1 = p1 M1 A1 3 γ RT1 6.5 × 101325 × 0.5 × 0.016 × 3 1.4 287.4 × 440 17.54 kg/s 4.6 If p0 is the stagnation pressure and A∗ is the critical throat area, p∗ p0 = A A∗ = 1 1 + 0.2M 2 1 M ! 2−3.5 5 + M2 6 "3 With these relation we obtain, p1 p2 = A1 A2 = ! "−3.5 1 + 0.2M12 1 + 0.2M22 ! "3 M2 5 + M12 M1 5 + M22 where subscripts 1 and 2 refer to entrance and throat of venturi. Substituting proper values into the above two relations, we get 30 One-Dimensional Flow 1.5 1.2 = 4 3 = ! "−3.5 1 + 0.2M12 1 + 0.2M22 ! "3 M2 5 + M12 M1 5 + M22 ! "3.0 M2 1 + 0.2M12 M1 1 + 0.2M22 = These two simultaneous equations can be solved to get M1 and M2 ! 4 3 Therefore, ! 1.5 1.2 1.5 1.2 "3.0/3.5 "3.0/3.5 M2 M1 = 5 4 = From above two equations, we get, = ! = M1 M2 4 3 ! 1 + 0.2M12 1 + 0.2M22 "−3.0 ! "6/7 5 = 1.61 4 "−3.5 1 + 0.2M12 1 + 0.2M22 M1 = 0.46 M2 = 0.74 4.7 p0 is the total pressure and p∞ is the static pressure. 1 2 ρV q∞ = 2 p0 − p∞ = 490 mm of Hg = 0.653 × 105 Pa p∞ = (0.35 + 1.0132) × 105 = 1.3632 × 105 Pa p0 = (0.653 + 1.3632) × 105 = 2.0162 × 105 Pa T0 = 25◦ C = 298 K 31 Therefore, T∞ ! "0.286 p∞ T0 p0 ! "0.286 1.3632 298 × = 266 K 2.0162 = = But, 1 2 2 T∞ 1 + 0.2M∞ T0 = 2 M∞ = T0 − T∞ 298 − 266 = 0.602 = 0.2T∞ 0.2 × 266 M∞ = 0.776 a∞ = / √ γRT∞ = 1.4 × 287 × 266 = 326 m/s V∞ = 253 m/s 4.8 pi = 1.0 atm pf = 6.0 atm T1 = 290 K For isentropic compression, we have γ ! "γ ! " γ−1 ρ2 T2 p2 = = p1 ρ1 T1 (a) Tf = Ti ! pf pi γ " γ−1 0.4 = (6) 1.4 = 1.67 Tf = 290 × 1.67 = 484 K ∆T = 484 − 290 = 194 K (b) Change in the internal energy is given by, de = 1.39 × 105 (N m)/kg = cv dT = 287 × 194 0.4 32 One-Dimensional Flow (c) Since the process is isentropic, from first law of thermodynamics, we have de = δq + δw, and δq = 0. Hence, work imparted to the air becomes, δw = δe = 1.39 × 105 N m/kg 4.9 a = M = / γRT = 347.21 m/s 180 = 0.518 347.19 The maximum pressure that can be achieved is the isentropic stagnation pressure. Therefore, p0 p = ! p0 = 1.013 × 105 × 1.201 Pa γ−1 2 1+ M 2 γ " γ−1 = 1.201 1.217 × 105 pa = 4.10 (a) The critical pressure ratio for the sonic condition at the nozzle exit is p∗ = 0.528 p0 If the nozzle is choked, then p2 = = p∗ = 0.528 × 6.895 × 105 3.64 × 105 Pa since, p2 > patm , the flow is choked. Hence, the pressure in the exit plane is p2 = p∗ (b) The minimum stagnation pressure for choked flow occurs when p2 = p∗ = patm Thus, p0min = patm p∗ = = 1.92 × 105 Pa 0.528 0.528 33 (c) For p0 = 1.724 × 105 Pa, p∗ = 0.528 × p0 = 0.91 × 105 Pa. Since p∗ < patm flow will not be choked. Further, subsonic flow is always correctly expanded, hence, p2 = patm = 1.014 × 105 Pa 4.11 Let the reaction force acting on the diffuser be R. Then, R + p1 A1 − p2 A2 = ṁV2 − ṁV1 R = ṁ(V2 − V1 ) − (p1 A1 − p2 A2 ) where subscripts 1 and 2 refer to diffuser inlet and exit, respectively. Given, p1 = 0.35 × 105 Pa, V1 = 200 m/s, T1 = 230 K, ṁ = 25 kg/s From the given data, we get ρ1 = p1 0.35 × 105 = 0.53 kg/m3 = RT1 287 × 230 A1 = ṁ 25 = 0.236 m2 = ρ1 V1 0.53 × 200 200 = 0.66 1.4 × 287 × 230 From isentropic table for M1 = 0.66, we have M1 = √ p1 T1 = 0.7465, = 0.9199 p01 T01 Thus, p01 = 0.35 × 105 = 0.468 × 105 Pa 0.7465 230 = 250 K 0.9199 For M2 = 0.2, from isentropic table, we have T2 p2 = 0.9725, = 0.9921 p02 T02 T01 p2 = = 0.9725 p01 (since for isentropic flow p01 = p02 ) = 0.45 × 105 Pa T2 = 0.9921 T01 = 248 K V2 = √ M2 a2 = 0.2 1.4 × 287 × 248 = 63 m/s 34 One-Dimensional Flow Therefore, R = = 25 × (63 − 200) − (0.35 × 105 × 0.236 − 0.45 × 105 × 0.5) −10.815 kN 4.12 Let subscripts 1 and 2 refer to entrance and exit of the tank. By energy equation we have, 1 cp T1 + u21 2 = 1 cp T2 + u22 2 T1 − T2 = u22 − u21 4 × 104 − 1 × 104 = 2cp 2 × 1004 ≈ 15◦ T2 = T1 − 15◦ 4.13 Work u1 u2 Figure S4.13 Schematic of the work delivering machine. Work delivered per unit mass is given by u21 − u22 + cp (T1 − T2 ) = work delivered 2 Given, T1 = 373 K u1 = 200 m/s T2 = 288 K cp = 1004 (N m)/(kg K) Using this we get, 20000 − u22 + 1004 × 85 = 2 100000 35 20000 − u22 + 85340 2 = u2 = 100000 103.3 m/s When the machine is idling, u22 2 = u2idling = 20000 + 85340 = 105350 459 m/s 4.14 Let jets “1” and “2” be denoted by the subscripts 1 and 2, and let T0 denote the temperature in the reservoir. For q = 0, adiabatic energy equation gives, ! 2 " ! 2 " u2 u1 + c p T2 + m + cp T1 = 2mcp T0 m 2 2 This gives, T0 = u2 + u22 T1 + T2 + 1 2 4cp = 300 + = (1 + 9) × 104 4 × 1004 324.9 K 4.15 Let T0 denote the temperature in the tyre. Since the process is adiabatic, we have u2 2 = c p T0 cp (T0 − T ) = u2 2 u2 = 2cp (T0 − T ) cp T + u = = √ 2 × 1004 × 37 272.57 m/s 36 One-Dimensional Flow 4.16 At 15000 m, we have Speed of sound T = −56.5◦ C = 216.5 K p = 1.206 × 104 Pa a = / γRT = √ = 294.94 m/s 1.4 × 287 × 216.5 Speed of the airplane is, u = 800 km/hr = 800 × = 222.22 m/s 1000 3600 This gives the Mach number as M = 222.22 u = a 294.86 = 0.753 (a) Maximum possible temperature of the airplane skin will be the stagnation temperature, at the nose of the airplane. Thus, it is the total temperature of the air, T0 T = 1.1134 T0 = 1.1134 × 216.5 = 241.05 K (b) Maximum possible pressure that can be felt by the airplane cannot exceed the stagnation pressure. This will be felt at the place where air comes to complete rest, i.e., at the nose of the airplane and other similar places. Thus, p0 = 1/0.6866 p p0 = 1.206 × 104 0.6866 = 1.756 × 104 Pa 37 (c) Critical velocity of air relative to the airplane is 3 2 a0 a∗ = γ+1 √ √ γRT0 γRT0 ∗ √ = √ a = 1.2 1.2 = 284.1 m/s (d) Vmax = = Vmax = = 3 3 2 ao γ−1 γ+1 ∗ a γ−1 √ 6a∗ 695.9 m/s 4.17 M = 0.6 Area = A Figure S4.17 Schematic of convergent channel. The mass flow rate is given by ṁ = = = ρAV / p A M γRT RT 3 3 γ p T0 1 √ MA p0 R p0 T T0 From the problem and theory we know that, 38 One-Dimensional Flow T0 = 550 K p p0 = 0.7840 p0 = 2 × 105 Pa T T0 = 0.9328 = 1.0354 = 4 3 T0 T 3 γ R ṁ = 29.188 kg/s M = 0.6 = = / √ γ cp γ−1 γ γ cp (γ − 1) 1.4 = 0.0694 1017 × 0.4 This gives A = = /γ p R p 0 p0 ṁ 0 T0 √1 M T T0 0.10125 m2 4.18 1 m2 2m2 Ae = 4 m 2 Figure S4.18 Schematic of convergent channel. At the mouth of the duct p = patm = 1.013 × 105 Pa T0 = 288 K (at sea level) 39 (a) Maximum mass flow rate is given by ṁmax = p 1 √ 0 A∗ 24.741 T0 = 1 1.013 × 105 √ ×1 24.741 288 = 241.26 kg/s (b) Ae A∗ = 4, A∗ = 0.25 Ae For this area ratio, from isentropic table, we get pe = 0.0298 p0 Me = 2.94, pe = 0.0298 × 1.013 × 105 Pa = 3018.7 Pa 4.19 A∗ 1 m2 p, T, ρ M =2 Figure S4.19 Schematic of convergent channel. Here we have to find ṁ, p∗ , T ∗ , ρ∗ , A, p, T , and ρ. ρ0 ṁ = p0 7 × 105 kg/m3 = RT0 287 × 313 = 7.8 kg/m3 = 0.6847 × p0 A∗ √ RT0 = 1599.133 kg/s 40 One-Dimensional Flow Fluid properties at the throat T∗ = 0.8333 T0 =⇒ T ∗ = 261 K p∗ = 0.5283 p0 =⇒ p∗ = 3.7 × 105 Pa ρ∗ = 0.6339 ρ0 =⇒ ρ∗ = 4.944 kg/m3 For the test-section Mach number 2, from isentropic table, we have T = 0.5556 T0 =⇒ T = 173.9 K p = 0.1278 p0 =⇒ p = 8.946 × 104 Pa ρ = 0.2300 ρ0 =⇒ ρ = 1.794 kg/m3 A 1 = A∗ 0.5926 =⇒ A = 1.6875 m2 4.20 60 m/s 245 m/s 1 2 Figure S4.20 Schematic of divergent channel. V1 p1 ṁ = = = 245 m/s 1 × 105 Pa 13.6 kg/s (a) ρ1 = = a1 = p1 1 × 105 = RT1 287 × 300 1.1614 kg/m3 / γRT1 = 347.2 m/s V2 T1 = = 60 m/s 300 K 41 M1 = 245 = 0.7056 ≈ 0.706 347.2 p1 p0 = 0.7171 =⇒ T1 T0 = 0.909 =⇒ A1 = ṁ 13.6 m2 = 0.0478 m2 = ρ1 V1 1.614 × 245 A1 = 1 πD12 4 = 0.2467 m =⇒ p0 = 1.39 × 105 Pa T0 = 330 K D1 (b) At the outlet, the energy equation is c p T2 + V22 = c p T0 2 Therefore, T2 = T0 − a2 = / M2 = V22 602 = 328.2 K = 330 − 2cp 2 × 1004 γRT2 = 363.1 m/s V2 = 0.165 a2 For M1 = 0.706, A1 /A∗ = 1.09 and for M2 = 0.165, A2 /A∗ = 3.57. Thus gives A2 = 0.1566 m2 and D2 = 0.4465 m (c) The rise in static temperature is T2 − T1 = 28.2 K 4.21 p02 = 2.0 × 105 Pa p1 = 0.15 × 105 Pa By Rayleigh supersonic pitot formula, &1/(γ−1) % 2γ γ−1 2 M − 1 γ+1 γ+1 p1 = 1 γ+1 2 2γ/(γ+1) p02 M 2 1 42 One-Dimensional Flow p02 p1 = = ! γ+1 2 Mi 2 γ+1 2 M1 2 "γ/(γ−1) ' % 1 2γ 2 γ+1 M1 γ+1 2 2 M1 2γ γ−1 2 γ+1 M1 − γ+1 − γ−1 γ+1 1 * γ−1 Select M1 and calculate p02 for given p1 . Check that &1/(γ−1) p1 p01 is less than p1 p02 . Trial and error method gives M1 = 3.16 . Also, check p02 /p01 calculated with normal shock relation and isentropic relation agree exactly. For this problem p02 /p01 = 0.286. 4.22 ṁmax = 241 kg/s A = 3 A∗ For this area ratio from isentropic table, M = 2.64. This is the Mach number just upstream of the shock. Let the conditions just upstream and downstream of the shock be represented by subscripts 1 and 2. From Normal shock table, for M1 = 2.64, we have p02 p2 M2 = 0.5 = 0.4452 = 7.9645 p01 p1 Therefore, p02 = 0.4452 × p01 = 0.4452 × 1.0133 × 105 = 45110 Pa For M2 = 0.5, from isentropic table, we have A2 = 1.3398 A∗2 Therefore, A∗2 = 3 A2 = = 2.239 m2 1.3398 1.3398 Ae A∗2 = 4 = 1.786 2.239 From isentropic table, for Ae = 1.786, we have A∗2 Me = 0.35 pe p0e = 0.9187 43 But p0e = p02 . Therefore, pe 0.9187 × 45110 = = 41442.5 Pa 4.23 p1 = 0.7 × 105 Pa T1 = 300 K A1 = 0.15 m V1 = 240 m/s 2 V2 = 120 m/s (a) ṁ p1 A1 V1 RT1 = ρ1 A1 V1 = = 0.7 × 105 × 0.15 × 240 287 × 300 = 29.26 kg/s (b) a1 = M1 = / γRT1 = 347.5 m/s V1 = 0.69 A1 Thus, p1 p01 = 0.72735 T1 T01 = 0.9131 p01 = 0.9624 × 105 Pa T01 = 328.6 K Thus, stagnation pressure at the exit p02 = p01 = 0.9624 × 105 Pa. (c) From (b) above, the stagnation temperature at the exit is T02 = T01 = 328.6 K 44 One-Dimensional Flow (d) 1 cp T2 + V22 2 = c p T0 T2 = T0 − A2 = / M2 = V2 = 0.334 A2 p2 p02 = 0.9257 V22 1202 ) = 321.4 K = 328.6 − 2cp 2 × 1004 γRT2 = 359 m/s Therefore, the static pressure at the exit is p2 = 8.91 × 105 Pa (e) Entropy change across the diffuser is zero since the flow is isentropic. (f ) For M1 = 0.69, the exit area, A1 A∗ = 1.1018, and for M2 = 0.334, A2 = A2 A∗ = 1.850. This gives A2 A∗ A1 = 0.252 m2 A∗ A1 4.24 The pressure, density, and temperature in the settling chamber can be taken as stagnation quantities. Therefore, p0 = 1.014 × 105 Pa, ρ0 = 1.44 kg/m3 , T0 = 35 + 273 = 308 K Since the effects of viscosity is neglected, the flow in the working section can be treated as isentropic. For M = 0.8, from isentropic table, we have p = 0.656, p0 ρ = 0.740, ρ0 T = 0.886 T0 Hence, p = = ρ = = 0.656 p0 = 0.656 × 1.014 × 105 0.665 × 105 Pa 0.740 ρ0 = 0.740 × 1.144 0.847 kg/m2 45 T = = 0.886 T0 = 0.886 × 308 273 K 4.25 Shock 1 2 M1 = 3 Ath Ae Figure S4.25 Schematic of convergent-divergent channel. Given, Ae = 11.91, M1 = 3.0 At M1 = 3.0 gives, M2 = 0.4752 downstream of the normal shock, and pp02 = 01 0.3281. This gives, p02 = 2.2981 × 105 Pa. Since the flow is adiabatic, T02 = T01 = 500 K. And since the flow is isentropic between the downstream of the normal shock and the nozzle exit, T0e = T02 , and p0e = p02 . A1 A1 A2 = 4.2346, and for M2 = 0.4752, A For M1 = 3.0, A ∗ = A ∗ = 1.390(from t 1 2 isentropic table). Also, since the shock is very thin, the area before and after the shock can be taken as the same, i.e. A1 = A2 . Therefore, ! " A Ae = ∗ A e A∗2 = Ae At A2 At A1 A∗2 = 3.9094 For this area ratio, from isentropic table, we get Me = 0.15 Te T0e = 0.9955 pe p0e = 0.9844 46 One-Dimensional Flow Thus, pe = 2.2623 × 105 Pa Te = 497.8 K 4.26 M = 2.5 p0 = Ats = 1 m2 T0 = 7 × 105 Pa 27 + 273 = 300 K (a) For M = 2.5 we have, A A∗ = A∗ = 2.637 0.38 m2 (b) T∗ T0 = 0.833 T∗ = 0.833 × 300 = 249.9 K = − 23.1 ◦ C (c) M = a = V = V a / γRT = 20.04 × √ T = 231.5 m/s M a = 578.75 m/s (d) ṁ A = p0 1 √ × A 24.743 T0 A∗ = 1 7 × 105 √ × 2.637 24.743 × 300 47 = 620 ṁ = 620 kg/s 4.27 By definition, we have the pressure coefficient as CP = CP∗ = = = = = p − p∞ q∞ " ! ∗ p −1 p∞ p∞ q∞ " ! ∗ p 2p∞ −1 p∞ 2 γp∞ M∞ " ! ∗ p p0 2 −1 2 γM∞ p0 p∞ '! * 1 2 γ γ "− γ−1 2 γ−1 2 + (γ − 1)M∞ γ+1 2 −1 γ 2 γM∞ 2 2 γ−1 1 2 γ 2 γ−1 2 2 + (γ − 1)M∞ − 1 γ 1 γ+1 2 γ−1 γ 2 γM∞ × 2 γ−1 2 = 91 2 : γ 2 2 + (γ − 1)M∞ /(γ + 1) γ−1 − 1 2 /2 γM∞ 4.28 p0 = 500 kPa, T0 = 30◦ C = 30 + 273.15 = 303.15 K 101325 The nozzle has to choke since pe /p0 = = 0.203, which is well below 500 × 103 the critical pressure ratio of 0.528. Therefore, ṁ = ṁmax = 0.6847 p0 ∗ √ A RT0 where R = 287 m2 /(s2 K) for air. Thus, ṁ = 0.6847 × 500 × 103 √ × 0.5 × 10−4 287 × 303.15 = 0.058 kg/s 48 One-Dimensional Flow 4.29 The pressure ratio 0.1429, we get pe pe 1 = = 0.1429. From isentropic table, for = p0 7 p0 Me = 1.93, Te = 0.57307 T0 T0 = 180 + 273.15 = 453.15 K Therefore, Te = 259.7 K. √ The speed of sound ae = γ RTe = 323 m/s. The exit velocity is Ve = Me ae = 1.93 × 323 = 623.4 m/s 4.30 The pressure ratio p/p0 across the nozzle is 1/5. This is well below the critical pressure ratio and therefore the flow is chocked at the exit. The flow is adiabatic and frictionless, therefore, the maximum mass flow rate is given by ṁ = 0.6847 × 5 × 105 √ × 6.5 × 10−4 287 × 288.15 = 0.774 kg/s Thus, from isentropic table, for M = 1. we get T T0 = T = 0.83333 240.12 K 4.31 The mass flow rate may be expressed as ṁ = 0.6847 p0 √ Ath RT0 where p0 and T0 are the stagnation pressure and temperature, respectively. The pressure ratio we get pe pe 91.4 = 0.90495. From isentropic table, for = = 0.905, p0 101 p0 Me = 0.38 and Ae = 1.6587 Ath 49 Therefore, Ath = 0.033 = 0.0199 m2 . Thus, 1.6587 ṁ = 0.6847 × 101 × 103 × 0.0199 √ kg/s 287 × 293.15 = 4.74 kg/s At the section with A = 0.022 m2 , the area ratio is A 0.022 = 1.1055 = Ath 0.0199 The corresponding Mach number is M = 0.69 and p = 0.72735 p0 Thus, the pressure at A = 0.022 m2 is p = 73.5 kPa 4.32 Nozzle is adapted, therefore the exit pressure pe = p16000 at 16000 m altitude. From standard atmospheric table, we get p16000 = 10.299 kPa = pe Therefore, pe p0 = 10299 , since 1 atm = 101325 Pa 15 × 101325 = 0.0067762 By isentropic relation, p0 = pe 1 + 0.2 ! γ−1 Me2 1+ 2 Me2 = ! Me = 3.98 γ " γ−1 1 0.0067762 "0.286 For Me = 3.98, from isentropic table, we get Ae Te = 10.53 , = 0.23992 ∗ A T0 Te = = 0.23992 × (2600 + 273.15) 689.33 K 50 One-Dimensional Flow The exit velocity becomes Ve = M e a e = Me / γRTe √ = 3.98 1.4 × 287 × 689.33 = Thrust 2094.6 m/s = ṁe Ve = (ρe Ae Ve ) Ve = ρe Ae Ve2 9000 By state equation, we have pe 10299 = 0.052 kg/m3 = RTe 287 × 689.33 ρe = Therefore, Ae = Thus, 9000 0.052 × (2094.6) A∗ = Ath = 2 = 0.0394 m2 0.0394 = 0.00374 m2 10.53 4.33 Given, V = 200 m/s and T = 15 + 273.15 = 288.15 K. The speed of sound is given by / √ a = γRT = 1.4 × 287 × 288.15 = 340.3 m/s The Mach number becomes M= From isentropic table, for p p0 = V = 0.59 a M = 0.59, we get 0.79013, T ρ = 0.93491, = 0.84514 T0 ρ0 Thus, p0 = 101 0.79013 51 T0 ρ0 = 127.8 kPa = 288.15 0.93491 = 308.2 K = p 1 ρ = × 0.84514 RT 0.84514 = 101 × 103 287 × 288.15 × 0.84514 = 1.445 kg/m3 ρ0 is also given by ρ0 = p0 127.8 × 103 = R T0 287 × 308.2 = 1.445 kg/m3 Note: This problem may also be solved using the isentropic relations directly, instead of table. 4.34 The sound speed at station 1 is / √ γRT1 = 1.4 × 287 × 303.15 = 349 m/s a1 = M1 = V1 90.5 = 0.26 = a1 349 For this flow with M1 = 0.26 to choke, the area ratio tropic table) is A1 required (from isenA∗ A1 = 2.317 A∗ The present area ratio 2.317 = 1.60. 6.9 A2 = 0.69. Therefore, = A1 10 A2 A2 A1 = = 0.69 × ∗ A A1 A∗ For this area ratio, from isentropic table, we get M2 = 0.4 , p2 T2 = 0.89561, = 0.96899 p02 T02 For M1 = 0.26, from isentropic table, we get p1 T1 = 0.95408, = 0.98666 p01 T01 52 One-Dimensional Flow p01 = T01 = For isentropic flow, T01 = T02 100 = 104.8 kPa 0.95408 303.15 = 307.25 K 0.98666 and p01 = p02 . Therefore, p2 = 0.89561 × 104.8 = 93.86 kPa T2 = 0.96888 × 307.25 = 297.72 K 4.35 For CO2 the molecular weight is 44 and γ = 1.3. The gas constant for CO2 is 8314 = 189 J/(kg K) R= 44 The stagnation pressure p0 = 6 atm and the back pressure pa = 1 atm. Therefore, the pressure ratio pa 1 = = 0.167 p0 6 which is well below the critical pressure ratio of 0.54573, required for the flow to choke. Hence, the flow is choked at the orifice. That is, M = 1 at the orifice. From the isentropic table, for M = 1, we get T = 0.86957 T0 Thus, T = = 0.86957 × (30 + 273.15) = 263.6 K −9.55◦ C This is the temperature with which CO2 comes out of the orifice. The mass flow rate ṁ = ρ AV . By state equation, ρ = 0.54573 × 6 × 101325 p = RT 189 × 263.6 = 6.66 kg/m3 A = V = = 1 22 π 1 × 10−3 π = × 10−6 m2 4 4 / √ M a = M γ R T = 1 × 1.3 × 189 × 263.6 254.5 m/s 53 Thus, ṁ = = 6.66 × π × 10−6 × 254.5 4 0.00133 kg/s Aliter: The mass flow rate (for the gas with γ = 1.3) can also be expressed as ṁ = 0.6672 × p0 × Ath √ R T0 = 0.6672 × 6 × 101325 × √ 189 × 303 = 0.00133 kg/s π × 10−6 4 4.36 Let subscripts 1 and 2 refer to inlet and exit of the nozzle. V1 = ṁ 0.7 = ρ1 A1 8 × 12 × 10−4 = 72.92 m/s M1 = V1 V1 =√ a1 γ RT1 = √ 72.92 = 0.18 1.4 × 287 × 400 p1 = ρ1 RT1 = 8 × 287 × 400 = 918.4 kPa From isentropic table, for M1 = 0.18, we get p1 p01 = 0.97765 T1 T01 = 0.99356 p01 = 918.4 = 939.4 kPa 0.97765 T01 = 400 = 402.59 K 0.99356 54 One-Dimensional Flow p2 ρ2 RT2 = 4 × 287 × 300 = = p2 p2 = , p02 p01 344.4 kPa since the flow is isentropic. Therefore, p2 344.4 = 0.3666 = p01 939.4 p2 = 0.3666, we get p01 From isentropic table, for M2 = 1.29 Therefore, V2 / = M2 a 2 = M2 = √ 1.29 1.4 × 287 × 300 = γ R T2 447.87 m/s Mass flow rate is ṁ = ρ2 A2 V2 A2 = ṁ 0.7 = ρ2 V2 4 × 447.87 = 3.9 cm2 4.37 From energy equation, we have T0 = T1 + V12 2 cp where the subscripts 0 and 1 refer to stagnation and inlet conditions. T0 = 400 + = 405 K 1002 2 × 1004.5 55 By isentropic relation, we have γ ! " γ−1 T0 p0 = p1 T1 3 ! 405 400 p0 = 200 × 10 pe p0 = 150 = 0.718 209 "3.5 = 209 kPa By isentropic relation, we have p0 pe = 1 + 0.2Me2 = 1 1 + 0.2 Me2 1 23.5 (1.393) 3.5 Solving we get, The speed of sound Me = 0.7 T0 Te = 1+ Te = 405 = 368.85 K 1.098 ae = γ−1 2 Me = 1.098 2 √ γRTe = 385 m/s. Thus, the exit velocity is Ve = = Me ae = 0.7 × 385 269.5 m/s By continuity, ρ1 A1 V1 = ρe Ae Ve . or p1 pe A1 V1 = Ae Ve RT1 RTe Ae = p1 T e V 1 A1 pe T 1 V e = 1 22 π 75 × 10−3 100 200 368.85 × × × 150 400 269.5 4 = 0.00202 m2 56 One-Dimensional Flow Thus, de 3 = = 4 × 0.00202 = 0.0507 m π 50.7 mm The mass flow rate is ṁ = ρ1 A1 V1 = 22 1 p1 π 75 × 10−3 × 100 RT1 4 = π × 752 × 10−6 200 × 103 × × 100 287 × 400 4 = 0.77 kg/s With the exit conditions, ṁ = ρe Ae Ve = 150 × 103 × 0.00202 × 269.5 287 × 368.85 = 0.77 kg/s 4.38 The flow process is described by the relation pVγ = constant. Per unit mass of air, pv γ = constant. Therefore, ! By state equation, we have v1 p1 V1γ = p2 V2γ "γ = p1 p2 v2 v1 pv = RT . Therefore, = R T1 287 × (400 + 273.15) = p1 3 × 106 = 0.0644 m3 /kg Thus, v2 = v1 ! p1 p2 " γ1 = 0.0644 ! " 1 3 1.4 0.5 57 T2 = 0.2316 m3 /kg = 0.5 × 106 × 0.2316 p2 v 2 = R 287 = 403.4 K (a) By energy equation h2 + V22 2 = h1 + V12 2 where V1 and V2 are the velocity at the entrance and exit of the nozzle. Also, it is reasonable to assume that V1 is very small and hence can be taken as zero. Thus, V2 h2 + 2 = h1 2 Further, for air being a perfect gas, h = cp T . Therefore, V22 = 2 (h1 − h2 ) = 2cp (T1 − T2 ) = 2 × 1004.5 (673.15 − 403.4) V2 = 736 m/s a2 = / = 402.6 m/s γRT2 = √ 1.4 × 287 × 403.4 Therefore, M2 = V2 736 = a2 402.6 = 1.83 (b) Since the flow at the nozzle exit is supersonic, at the throat the flow is choked. That is, at the throat Mth = 1. From isentropic table, for M = 1, we get p∗ p0 = 0.52828 58 One-Dimensional Flow T∗ T0 = 0.83333 p∗ = 0.52828 × 3 × 106 = 1.58484 MPa T∗ = 0.83333 × 673.15 = 560.96 K a∗ = / ṁ = ρ∗ Ath a∗ ρ∗ = p∗ 1.584 × 106 = ∗ RT 287 × 560.96 = 9.8 kg/m3 γRT ∗ = 474.76 m/s Thus, Ath (c) The mass flow rate is = 15.89 9.8 × 474.76 = 3415 mm2 ṁ = ρ2 A2 V2 . ρ2 = 1 = 4.318 kg/m3 v2 Therefore, ṁ = = 4.318 × 5000 × 10−6 × 736 15.89 kg/s 4.39 First of all, let us check whether the flow is isentropic. The change in entropy (assuming air to be calorically perfect gas) is ! " ! " T2 p2 s2 − s1 = cp ln − R ln T1 p1 ! " ! " 290 101 = 1004.5 ln − 287 ln 300 200 = 162 J/(kg K) 59 Hence, the flow is nonisentropic. Consider the inner surface of the nozzle as the control volume. Let F be the force acting on the nozzle, due to the flow. By momentum balance, we have p1 A1 − p2 A2 − F = ṁ (V2 − V1 ) V1 = ṁ ρ1 A1 ρ1 = p1 200 × 103 = RT1 287 × 300 = 2.32 kg/m3 = 5 = 10.78 m/s 2.32 × 0.2 V1 By energy equation, we have h1 − h2 = V2 V22 − 1 2 2 2 cp (T1 − T2 ) = V22 − V12 = V22 − 10.782 2 × 1004.5 (300 − 290) + 10.782 = V22 V2 = 142.15 m/s The exit area is given by A2 = ṁ ρ2 V2 where ρ2 A2 = p2 101 × 103 = RT2 287 × 290 = 1.214 kg/m3 = 5 = 0.029 m2 1.214 × 142.15 Using these values in the momentum equation, we get 200 × 103 × 0.2 − 101 × 103 × 0.029 − F = 5 (142.15 − 10.78) Thus, F = 40000 − 2929 − 656.85 60 One-Dimensional Flow = 36414.15 N 4.40 Let us solve the problem by using relations and also by using gas tables. The speed of sound at the flight altitude is given by / / a = γRT = 1.4 × 287 × (−15 + 273.15) = 322.1 m/s The flight speed V = M a = 0.8 × 322.1 = 257.68 m/s. The stagnation enthalpy h0 is given by h0 = h + For an ideal gas h0 = cp T + V2 2 V2 and therefore, 2 c p T0 = c p T + or T0 = T + V2 2 V2 2cp Thus, 2 T0 = 258.15 + = 291.2 K (257.68) 2 × 1004.5 By isentropic relation, we have p0 p = p0 = = Using gas tables " γ T0 γ − 1 T ! "3.5 291.2 44 258.15 ! 67.08 kPa 61 For M = 0.8, from isentropic table, we get p T = 0.65602 = 0.88652 p0 T0 Thus, p0 = 44 = 67.07 kPa 0.65602 T0 = 258.15 = 291.2 K 0.88652 Note: From the above solution it is seen that, use of the gas tables has tremendous advantage over using equations directly. 4.41 Static temperature T = 216 K. Total temperature T0 = 40 + 273.15 = 313.15 K. From isentropic relation, we have T0 T 1.45 γ−1 2 M 2 = 1 + 0.2 M 2 = 1+ M2 = 0.45 = 2.25 0.2 M = 1.5 The pressure ratio p0 /p given by isentropic relation is p0 p = = p0 = ! 9 γ−1 2 1+ M 2 1 + 0.2 × 1.52 γ " γ−1 :3.5 = 3.671 (3.671)(0.55) = 2.019 atm Note: This problem can be solved by using gas tables instead of solving the isentropic relations, as follows. For T = 0.689655, from isentropic tables, we get T0 M = 1.5 and p/p0 = 0.2724 Therefore, p0 = 0.55 = 2.019 atm 0.2724 62 One-Dimensional Flow 4.42 By energy equation, we have T01 V12 2cp = T02 = T1 + = (70 + 273.15) + = 348.13 K 1002 2 × 1004.5 Similarly, T2 T02 − = 4002 = 268.49 K 2 × 1004.5 / √ γRT2 = 1.4 × 287 × 268.49 348.13 − = a2 V22 2cp = = 328.45 m/s = V2 400 = a2 328.45 = 1.218 Therefore, M2 Vmax / = = 2 cp T01 = √ 2 × 1004.5 × 348.13 836.3 m/s By isentropic relation, Thus, p01 = M1 = p01 = p02 = 1 23.5 p1 1 + 0.2 M12 V1 = 0.269 γRT1 23.5 1 2.5 1 + 0.2 × 0.2692 = 2.63 atm √ % &3.5 2 p2 1 + 0.2 × (1.218) = 1.24 atm p02 1.24 = 0.471 = p01 2.63 63 4.43 Given, p0 = 350 kPa, T0 = 420 K Ae = 0.22 m2 , Ve = 525 m/s By energy equation we have T0 = Te + Te = T0 − = Ve2 2cp Ve2 5252 = 420 − 2 cp 2 × 1004.5 282.8 K √ Therefore, the speed of sound at nozzle exit ae = γRTe . That is, √ ae = 1.4 × 287 × 282.8 = 337.01 m/s Thus, Me pe ρe ṁ = Ve 525 = ae 337.01 = 1.56 = p0 (1 + 3.5 0.2Me2 ) = 350 4.007 = 87.35 kPa = pe 87.35 × 103 = RTe 287 × 282.8 = 1.076 kg/m3 = = ρe Ae Ve = 1.076 × 0.22 × 525 124.3 kg/s From isentropic table, for Me = 1.56, we get Ae = 1.219 A∗ Thus, A∗ = 0.22 = 0.18 m2 1.219 64 One-Dimensional Flow 4.44 The speed of sound at 1 is / √ a1 = γRT1 = 1.4 × 287 × 315 = 355.76 m/s The Mach number at 1 is M1 = V1 150 = 0.42. = a1 355.76 Now, for M1 = 0.42, the corresponding area ratio A1 /A∗ may be calculated using the area–Mach number relation for isentropic flow, or the value of A1 /A∗ may be read directly from isentropic table. From isentropic table for M1 = 0.42, we get A1 /A∗ = 1.52891 Thus, A1 = 1.52891 × 25 = 38.22 cm2 The mass flow rate ṁ = ρ1 A1 V1 ρ1 = p1 152 × 103 = RT1 287 × 315 = 1.681 kg/m3 Therefore, ṁ = 1.681 × 38.22 × 10−4 × 150 = 0.9637 kg/s 4.45 T0 = 40◦ C = 40 + 273.15 = 313.15 K, √ The exit velocity Ve = 200 m/s = Me γRTe . γ = 1.4 By isentropic relation we have 23.5 1 p0 = 1 + 0.2 Me2 pe By energy relation, we have Ve2 2cp T0 = Te + Te = 313.15 − = / = 343.25 m/s The speed of sound is ae 2002 = 293.24 K 2 × 1004.5 γRTe = √ 1.4 × 287 × 293.24 65 The Mach number at the exit is Me p0 pe = Ve 200 = ae 343.25 = 0.584 1 = 1 + 0.2 × 0.5832 23.5 = 1.259 For the convergent nozzle, pexit has to be equal to the atmospheric pressure, since the subsonic flow exiting a convergent nozzle will always be correctly expanded. pe Thus, = ρgh = 13.6 × 103 × 9.81 × 22.7 × 10−3 = 3028.54 Pa (gauge) = 101325 + 3028.54 = 104353.54 Pa (absolute) pa = 104353.54 Pa The storage pressure is given by p0 = 1.259 × 104353.54 = 131381.11 Pa = 224.7 mm of Hg (gauge) Note: The standard atmospheric pressure is patm = 101325 Pa = 760 mm of Hg. 4.46 Let the first and second sections are represented by subscripts 1 and 2, respectively. At section 1, M1 V1 365 =√ γRT1 1.4 × 287 × 305.15 = √ = 1.04 From isentropic table, for M1 = 1.041, we get p1 p0 = 0.50389 T1 T0 = 0.82215 66 One-Dimensional Flow Therefore, p0 = 158.76 kPa, T0 = 371.16 K. At station 2, p2 p0 From isentropic table, for 120 = 0.7559 158.76 = p2 /p0 = 0.7559, we get T2 = 0.92428 T0 M2 = T2 = 343.06 K = 69.91◦ C a2 = / Therefore, 0.64 γRT2 = 371.27 m/s V2 = M2 a2 = 237.6 m/s 4.47 For M = 3.0, from isentropic table, we get p = 0.027224, p0 T = 0.35714, T0 A = 4.23456 A∗ Thus, A∗ = 0.05 = 0.0118 m2 4.23456 p0 = 0.2 × 101325 = 744.4 kPa 0.027224 T0 = 300 = 840 K 0.35714 The mass flow rate is given by ṁ = 0.6847 p0 A∗ √ RT0 = 0.6847 × 7.444 × 105 × 0.0118 √ 287 × 840 = 12.25 kg/s 4.48 The velocity at nozzle entrance is very low. Hence, the pressure and temperature at the entrance can be taken as the stagnation pressure and stagnation temperature. That is, p0 = 1 MPa, T0 = 300 K 67 From isentropic table, for M = 2, we have p = 0.1278, p0 T = 0.55556, T0 ρ = 0.23005 ρ0 Therefore, p = 0.1278 × 1 × 106 = 127.8 kPa T = 0.55556 × 300 = 166.67 K ρ = 0.23005 × ρ0 = (0.23005) (p0 /RT0 ) " 106 = 2.672 kg/m3 287 × 300 √ √ a = γRT = 1.4 × 287 × 166.67 = 258.78 m/s. There0.23005 V = M a = 517.56 m/s ṁ = ρAV = 2.672 × 0.15 × 517.56 The speed of sound fore, = 4.49 Given, ! = 207.43 kg/s p0 = 3.5 MPa, T0 = 500◦ C = 773.15 K pe = 0.7 MPa, ṁ = 1.3 kg/s At the nozzle exit, the pressure ratio is pe 0.7 = 0.2 = p0 3.5 From isentropic table, for this pressure ratio, we get Me = 1.71 , Te Ae = 0.63099, = 1.3471 T0 Ath Therefore, Te = 0.63099 × 773.15 = 487.85 K The corresponding speed of sound ae is / √ ae = γRTe = 1.4 × 287 × 487.85 = 442.74 m/s 68 One-Dimensional Flow Ve = Me ae = 757 m/s The exit velocity ρe = pe , by state equation RTe Therefore, 0.7 × 107 = 4.9 kg/m3 287 × 487.85 ρe = The mass flow rate ṁ = ρe Ae Ve . Thus, Ae Ath 4.50 Given, = ṁ 1.3 = ρe Ve 4.9 × 757 = 3.5 cm2 = 3.5 Ae = 1.3471 1.3471 = 2.6 cm2 T0 = 25◦ C = 298.15 K, Ae = 15 cm2 . Applying momentum analysis to the control volume considered, we get F = 100 N = ρe Ae Ve2 Assuming air to be an ideal gas, Ae ρe Ve2 = Ae pe 2 Ae γpe 2 Ve = V = Ae γpe Me2 RTe γRTe e Also, γ Ae pe Me2 = 100 N The nozzle exit flow is subsonic and hence has to be correctly expanded with pe = patm = 101325 Pa. Thus, Me2 = 100 = 0.47 1.4 × 15 × 10−4 × 101325 Me = 0.69 From isentropic table, for Me = 0.69, we get Te = 0.91306, T0 pe = 0.72735 p0 69 Thus, Te p0 Ve = 298.15 × 0.91306 = 272.23 K = −0.92◦ C = 101325 = 139.3 kPa 0.72735 = 1.37 atm = Me a e = Me = 0.69 × = / γRTe √ 1.4 × 287 × 272.23 228.2 m/s 4.51 Given, ṁ = 1 kg/s, pe = 101.325 kPa. The mass flow rate may be expressed as ṁ = 0.6847 p0 ∗ √ A RT0 Thus, ∗ A = = = At the exit, √ RT0 0.6847 p0 √ 287 × 325 0.6847 × 700 × 103 6.37 cm2 pe 101325 = = 0.14475 p0 700 × 103 From isentropic table, the corresponding Mach number Me and temperature ratio are the following. Me = 1.92 Te T0 = 0.57561 Te = 0.57561 × 325 = 187 K 70 One-Dimensional Flow The speed of sound ae = Thus, √ Ve γ RTe = 274.1 m/s. Me ae = 1.92 × 274.1 = = 526.27 m/s 4.52 Assuming air to be a perfect gas, we have γ = 1.4 and R = 287 J/(kg K). The speed of sound at nozzle inlet is / √ γRT1 = 1.4 × 287 × 300 a1 = = 347.2 m/s The inlet Mach number is M1 = V1 100 = 0.29 = a1 347.2 From isentropic table, for M1 = 0.29, we get p1 p01 = 0.94329 T1 T01 = 0.98346 p01 = 101325 = 107416.6 Pa 0.94329 T01 = 300 = 305 K 0.98346 The flow is supersonic at the exit, therefore at the throat M ∗ = 1. For M = 1, from isentropic table we have p∗ p0 = 0.52828 T∗ T0 = 0.8333 Thus, p∗ = = 0.52828 × 107416.6, 56746 Pa since p∗0 = p01 71 = T∗ 0.56 atm 0.8333 × 305 = = 254 K Since the flow is isentropic, T02 = T01 = 305, K p02 = p01 = 107416.6 Pa = 1.06 atm The mass flow rate is ṁ = ρ1 A1 V1 = ρt At Vt = p1 101325 × 5 × 10−4 × 100 A1 V1 = RT1 287 × 300 = 0.0588 kg/s At = 0.0588 ρt Vt ρt = pt 56746 = RTt 287 × 254 = 0.78 kg/m3 = at = = 319.5 m/s Thus, Vt Therefore, At = / γRTt = √ 1.4 × 287 × 254 0.0588 = 2.36 cm2 0.78 × 319.5 4.53 Let subscripts 01 and 02 refer to properties at reservoirs 1 and 2, respectively. Pressure p02 is the back pressure for the nozzle. Therefore, the nozzle pressure ratio (NPR) becomes p02 3 = = 0.5 p01 6 72 One-Dimensional Flow Also, ! p∗ = p01 2 γ+1 "(γ)γ−1 = 0.528 where p∗ is the pressure corresponding to sonic speed. The NPR is less than the critical pressure ratio of 0.528. Therefore, the nozzle must experience sonic condition at the exit. Thus, the mass flow through the nozzle ṁ is ṁ = ρ∗ V ∗ A∗ where ρ∗ and V ∗ are the density and velocity at M = 1. From isentropic table, for M = 1, we get T∗ T01 = 0.8333 Thus, T∗ = 0.8333(30 + 273.15) = 252.6 K V∗ = a∗ = = 318.58 m/s / γRT ∗ = √ 1.4 × 287 × 252.6 From state equation, we have ρ∗ = p∗ 6 × 101325 × 0.528 = ∗ RT 287 × 252.6 = 4.428 kg/m3 = ṁ 0.5 = ρ∗ V ∗ 4.428 × 318.58 = 0.0003544 m2 = 3.544 cm2 Thus, A∗ Aliter: This problem can also be solved without going into the details at the nozzle exit, as follows. We know that in terms of reservoir pressure and temperature, the mass flow rate ṁ is given for a gas with γ = 1.4 as 0.6847 p01 A∗ ṁ = √ RT01 73 Thus, A∗ = / ṁ RT01 0.6847 p01 = 0.0003543 m2 = 3.543 cm2 4.54 The overall pressure ratio pe /p0 = 100 × 103 = 0.1. 106 This is well below the critical pressure ratio of 0.528. Therefore, the flow is a choked flow. The area ratio Ae /Ath = 13.5/8 = 1.6875. From isentropic table, for Ae /Ath = 1.6875, the corresponding exit Mach number Me = 2.0 4.55 Given, p0 = pa = 101325 Pa and T0 = 288.15 K. For the flow to choke, pb /p0 ≤ 0.528. Therefore, the pressure in the tank pb has to be pb ≤ 0.528 × 101325 ≤ 53.5 kPa This implies that pbmax = 53.5 kPa. The mass flow rate is given by ṁ = 0.6847 p0 Ath √ RT0 = 0.6847 × 101325 π √ × (0.04)2 4 287 × 288.15 = 0.303 kg/s 4.56 Given, Ae /Ath = 4 and Me > 1. Therefore, the supersonic solution of the area Mach number relation is the answer for the problem. That is, the supersonic value of M given by the following equation is the exit Mach number. Ae 1 = Ath M ! 5 + M2 6 "3 This equation may be solved directly to get Me or the value of Me corresponding to Ae /Ath = 4 can be read directly from isentropic table. From isentropic table, we get 74 One-Dimensional Flow Me = 2.94 and pe = 0.029795 p0 This is the pressure ratio required. 4.57 (a) For Me = 1.63, from isentropic table, we have Ae pe = 1.275 = 0.22501 A∗ p0 For correct expansion, pe = pb . Therefore, pb = 0.22510 × p0 = 0.22501 × 10 × 101325 Pa since 1 atm = 101325 Pa. Thus, pb = 228 kPa (b) The flow will remain supersonic at the exit for all back pressures below 228 kPa (c) For choking, M = 1 at the throat. After choking the flow can expand as a subsonic flow at the divergent portion of the nozzle. From isentropic table, for Ae /A∗ = 1.275, we have Me = 0.54 p/p0 = 0.82 (approx) Therefore, the nozzle will remain choked for all back pressures below pb = 0.82 × 10 × 101325 = 830.87 kPa 4.58 From normal shock theory, we have 1 2 2 M12 − 1 ρ2 = 1+ 2 ρ1 M1 (γ − 1) + 2 M2 = '1 2 1 2 *1/2 γM12 + 1 − M12 − 1 2 (γM12 + 1) − (γ + 1) where subscripts 1 and 2 refer to states upstream and downstream of the shock, respectively. Given, M1 is very large. Also, γ = 1.4 for air. The density ratio can be written as & % 1 2 1 − ρ2 M12 =1+ ρ1 (γ − 1) + M22 1 when M1 → ∞, we get 75 ρ2 ρ1 Similarly, = 1+ 1 2 =1+ γ−1 1.4 − 1 = 1+ 2 =1+5 0.4 = 6 % & % & 1/2 γ + M12 − 1 − M12 1 & % 1 M2 = γ+1 1 2 γ + M2 − M2 1 In the limit M1 → ∞, it reduces to M2 = = 3 γ−1 = 2γ 1 3 0.4 2.8 0.378 4.59 From isentropic table, for M = 2.0, we get p = 0.1278 p0 Therefore, p 0.1278 × 3 × 101325 = 38848 Pa = since, 1 atm = 101325 Pa. Given, 101325 Pa = 760 mm Hg, therefore, p = 38848 × 760 101325 = 291.384 mm of Hg This is absolute pressure. The gauge pressure, shown by the manometer, will be pg = = pabs − patm = 291.384 − 760 − 468.61 mm The negative sign indicates that the measured pressure is below the atmospheric pressure or subatmospheric. 76 One-Dimensional Flow 4.60 The difference between the measured pressures is 500 mbar. That is, ∆p = 0.5 bar We know that, 1 bar = 105 Pa Therefore, ∆p = 0.5 × 105 Pa Also, ∆p = ρair g h where h is the vertical height climbed. Therefore, h = 0.5 × 105 9.81 × 1.1 = 4633.49 m 4.61 Let subscript ‘0& refer to stagnation state. Given that the total pressure of air is p0 . For maximum velocity, the limiting pressure is zero. Assuming the flow as incompressible, by Bernoulli equation, we have 1 p + ρV 2 = p0 2 For Vmax , p = 0, thus, 1 2 ρV = p0 2 Also, for incompressible flow, ρ = ρ0 , thus, 3 p0 Vmax, incomp. = 2 ρ0 For compressible flow, the Bernoulli equation, we have Therefore, γ p0 γ p V2 + = γ−1ρ 2 γ − 1 ρ0 Vmax, comp. = That is, Vmax, comp. = = = 3 3 3 2 γ p0 γ − 1 ρ0 γ Vmax, incomp. γ−1 1.4 Vmax, incomp. 1.4 − 1 1.87 Vmax, incomp. 77 That is, the error in treating this compressible flow as incompressible is 87 percent. 4.62 Let the inlet and the exit of the nozzle be denoted by subscripts 1 and 2, respectively. (a) By energy equation, we have h1 + V12 2 = h2 + V22 2 3025 × 103 + 602 2 = 2790 × 103 + V22 = 473600 V2 = = ρAV = A1 V1 v1 = 0.1 × 60 0.19 = 31.58 kg/s 688.2 m/s (b) The mass flow rate is ṁ (c) Mass flow rate is also given by ṁ = ρ2 A2 V2 = A2 V2 v2 Thus, A2 = ṁ v2 V2 = 31.58 × 0.5 688.2 = 0.0229 m2 V22 2 78 One-Dimensional Flow Chapter 5 Normal Shock Waves 5.1 Vg 500 − Vg 500 m/s (a) Stationary observer (b) Observer moving with the shock Figure S5.1 Mach number of the air stream, M1 is given by 500 M1 = √ = 1.51 1.4 × 287 × 273 From Normal shock table, for M1 = 1.51, we have p2 p1 = 500 m/s 2.493, T2 = 1.327 T1 Therefore, p2 = 1.745 atm T2 = 362.27 K V2 V1 = 1 500 − Vg = 500 1.879 500 − Vg = 266.1 m/s Vg = Also, 233.9 m/s 79 80 Normal Shock Waves Since the velocity of the observer does not affect the static properties, pb = p2 = 1.745 atm Tb = T2 = 362.3 K The Mach number of the flow behind the shock wave is Mb Vg 233.9 = 381.5 γRTb = √ = 0.613 From isentropic table, for M = 0.613, we have p pt = 0.7760, T = 0.9301 Tt Therefore, after passage of shock the stagnation pressure is p tb T tb = 1.745 0.7760 = 2.25 atm = 362.3 0.9301 = 389.5 K Note: • For a stationary observer the stagnation temperature after passage of the wave is greater than that before passage of the wave. • For an observer sitting on the wave, however, there is no change of stagnation temperature across the wave. 81 5.2 up u2 = C s − up Cs 2 u1 = C s 1 Figure S5.2 Shock wave motion in a tube. / a1 = γRT1 = 347 m/s u1 u2 = Cs (γ + 1)M12 = Cs − u p (γ − 1)M12 + 2 M1 = Cs a1 Therefore, Cs Cs − u p = Cs2 +2 a21 % &2 s (γ + 1) C a1 % &2 s (γ − 1) C +2 a1 = (γ + 1) (γ − 1)Cs2 + 2a21 = (γ + 1)Cs2 − (γ + 1)up Cs 2Cs2 − (γ + 1)up Cs − 2a21 = 0 up Cs − a21 = 0 γ + 1 up M1 − 1 2 a1 = 0 (γ − 1) Cs2 − ! γ+1 2 M12 − " M1 = = Cs (Cs − up ) a21 1 γ + 1 up ± 2 2 a1 1.19 4! γ + 1 up 2 a1 "2 + 4 Positive sign is taken here, since M1 cannot be less than 1. Hence, Cs = M1 a1 = 413 m/s . From Normal shock table, for M1 = 1.19, pp21 = 1.485. Thus, the pressure on the face of the piston is p2 = 1.485 × 1.0133 × 105 = 1.505 × 105 Pa 82 Normal Shock Waves 5.3 up 2 1 Figure S5.3a The flow field. (a) p2 p1 = a1 = ! " 2γ γ − 1 |up | γ−1 1− 2 a1 / √ γRT1 = 20.04 300 = 347 m/s Therefore, p2 p1 = ! p2 = 0.606p1 = 120 1 − 0.2 347 "7 = 0.606 0.606 atm (b) up u2 = C s − up Cs 2 1 Figure S5.3b Schematic of the flow field. u1 u2 = Cs Cs − u p = (γ + 1)M12 (γ − 1)M12 + 2 u1 = C s 83 M12 − (γ − 1)M12 + 2 = (γ + 1)M12 − (γ + 1) γ + 1 up M1 − 1 2 a1 = 0 γ + 1 up 2 a1 = 1.2 × up M1 a1 120 = 0.415 347 Therefore, M12 − 0.415 M1 − 1 = 0. Solving for positive value of M1 , we get M1 = 1.228 p2 p1 = 1+ = 1.595 2 2γ 1 2 M1 − 1 γ+1 Therefore, p2 = 1.595 atm 5.4 Cs up Cs u1 = C s + up Gas at rest u2 = 0 u2 = C s Figure S5.4 Schematic of the flow field. The velocity of the wave relative to the pipe = Cs . Velocity of air entering the normal shock wave relative to the shock wave is u1 = Cs + up M1 = Cs + u p a1 u1 u2 = (γ + 1)M12 2 + (γ + 1)M12 84 Normal Shock Waves Cs = Cs + u Cs = M1 a 1 − u p From this we get (γ + 1)M12 2 + (γ − 1)M12 M12 γ+1 − 2 ! up a1 " = M1 a 1 M1 a 1 − u p = M1 u M1 − a1p M1 − 1 = 0 a1 = / γRT1 = 347 m/s Solving for M1 , we get M1 = 1.2924, taking only the positive sign, since M1 is supersonic. Hence, Cs = M1 a1 − up = 1.2924 × 347 − 150 = 298.5 m/s From shock tables, for M1 = 1.294, we have p2 p1 T2 = 1.185 T1 = 1.775, = 1.775 × 1.5 × 105 Therefore, p2 = T2 = = 2.66 × 105 Pa 1.185 × 300 355.5 K Also, since the gas is at rest, 5.5 p02 = p2 T02 = T2 85 Vp 1 u2 = C s u1 = Cs + Vp Cs 2 Figure S5.5 Schematic of the flow field. (γ + 1)M12 2 + (γ − 1)M12 u1 u2 = u1 = Cs M1 Cs a1 = = Cs − Vp u2 Therefore, (γ + 1) % 2 + (γ − 1) Cs a1 % &2 Cs a1 = &2 = + , Cs Cs Vp (γ + 1) − a1 a1 a1 Solving for Cs a1 , Cs a1 Cs a1 − Vp a1 2 + (γ − 1) ! Cs a1 "2 we get, Cs = a1 In the limit as = Cs Cs − u p Vp a1 ! γ+1 4 → ∞, Cs a1 " ' ! "2 ! "2 *1/2 Vp Vp 1 γ+1 + 1+ a1 4 2 a1 → ∞. In the limit as Vp a1 5.6 a4 Cs 3 up = 300 m/s 4 → 0, Cs a1 → 1. 86 Normal Shock Waves Figure S5.6a Schematic of the flow field. (a) up = p3 p4 = 300 m/s ! " 2γ γ − 1 |up | γ−1 1− 2 a4 ! " 2.8 300 0.4 1 − 0.2 × 360 ! "7 5 6 ! "7 5 × 1 = 0.2791 atm 6 + , γ − 1 |up | 5 1− = 2 a4 6 = = Therefore, a3 = 5 6 p3 = a3 a4 = × 360 = 300 m/s. Slope of the terminating characteristic is dx dt = C3 = a4 − = 360 − γ+1 |up | 2 2.4 × 300 = 0 2 For the pressure on the face of the piston, ! . 12 "2γ p2 γ+1 −1 p2 γ−1 p3 p3 + γ+1 up = a3 γ up = a3 = 300 m/s Therefore, ! p2 p3 "2 ! "2 p2 −1 p3 = γ − 3.68 p2 + 0.72 p3 = 0 2 Therefore, p2 p3 = 3.473 p3 = 0.2791 atm ' p2 p3 + γ−1 γ+1 2γ γ+1 * 87 Therefore, pressure at the piston face p2 = 3.473 × 0.2791 = 0.969 atm (b) The velocity of the shock with respect to a stationary observer is C s = a3 + γ − 1 γ + 1 p2 + 2γ 2γ p3 , 12 = 529.88 m/s Therefore, time for the shock to hit the terminating characteristic after the piston has stopped is 30 = 0.0566 s t1 = 529.88 (c) t 0 Piston path 300 m/s dx/dt = 0 t a4 ock 300 m/s Pis ton dx /dt pa th =| up | Terminating characteristic dx /d t= Sh Cs x 30 m Figure S5.6c 5.7 Given, MS = 5.0, T1 = 27c ircC = 300 K, p1 = 0.01 atm, T4 = 300 K. From normal shock table, for MS = 5, we have p2 p1 p4 p1 = 29.0 = 4 ", −2γ + ! γ4 −1 2γ1 MS2 − (γ1 − 1) 1 γ4 − 1 1− MS − γ1 + 1 γ4 + 1 MS = 2.26 × 106 Cs 0 p2 p3 88 Normal Shock Waves p4 = 2.26 × p1 = 2.26 × 104 atm Note: Instead of above equation for p4 /p1 , equation (5.57) also can be used for solving this problem. 5.8 (a) p2 /p1 = 29. Therefore, static pressure behind the shock is, p2 = 29p1 = 0.29 atm. From normal shock table for M1 = 5, we get T2 = 5.8 T1 Therefore, static temperature behind the shock T2 is, T2 = 5.8 × 300 = 1740 K (b) Particle speed u2 is given by, u2 a2 M2 = ! " 2 1 a 1 MS − γ1 + 1 MS = 1388.71 m/s = 20.04 × = 836 m/s = 1388 836 = 1.66 / T2 From isentropic tables, for M2 = 1.66, we have T2 p02 = 0.6447 p2 p02 = 0.215 Thus, T02 = 2699 K and p02 = 1.349 atm (c) Testing time available is ∆t = t2 − t1 = 8 8 − u2 Cs 89 u2 Cs = 1388.71 m/s Ms a1 = 5 × 347 = 1735.9 m/s = t dx /d t = u 2 t2 t1 dt = dx/ Cs x 8m Figure S5.8c Therefore, ∆t = 8 8 − 1388 1735 = 1.15 × 10−3 s (d) µ sin−1 = = 1 1 = sin−1 M2 1.66 37 deg 5.9 up 2 up 2 2 Cs 1 uR up + uR 5 uR 5 90 Normal Shock Waves Figure S5.9 Schematic of the flow field. (a) uR Cs = uR up up Cs up + uR uR = ρ5 , ρ2 = ρ5 ρ1 × ρ1 ρ2 = ζ η from continuity Therefore, up uR ζ −η η = Also, (refer to Example 5.5) up Cs = 1 (1 − ) η From this, we get uR Cs = ! " 1 η × 1− ζ −η η = η−1 ζ −η (b) p2 p1 = = " ρ1 1+ 1− ρ2 ! " 1 1 + γ1 MS2 1 − η γ1 MS2 ! Similarly, p5 p2 = = ! " 2 (up + uR ) ρ2 × 1 − a22 ρ5 ! "2 ! " a2 u p + u R η 1 + γ2 12 1− a2 a1 ζ 1 + γ2 91 "2 ! " up uR Cs η + 1− a1 Cs a1 ζ ! "2 ! " η−1 ζ −η γ1 p1 ρ2 η−1 MS + MS = 1 + γ2 ρ1 γ2 p2 ζ − η η ζ a21 a22 = 1 + γ1 p1 (η − 1)2 ζ M2 η p2 η2 ζ −η S Therefore, p5 p1 = = = = p5 p2 p2 p1 (η − 1)2 ζ p2 + γ1 MS2 p1 η ζ −η ! " (η − 1)2 ζ 1 1 + γ1 MS2 1 − + γ1 MS2 η η ζ −η ! " η−1 (η − 1)ζ 1 + γ1 MS2 1+ η (ζ − η) = 1 + γ1 MS2 η − 1 η(ζ − 1) η ζ −η = 1 + γ1 MS2 (η − 1)(ζ − 1) (ζ − η) (c) Therefore, ! = 1 + γ2 h2 h1 = h5 h2 = ! " γ1 − 1 2 1 MS 1 − 2 1+ 2 η "2 ! " ! γ2 − 1 up + uR ρ2 1+ 1 − 22 2 a2 ρ5 = γ2 − 1 a21 2 (η − 1)2 ζ2 1+ M S 2 a22 η2 (ζ − η)2 h2 = a22 γ2 − 1 h1 = a21 γ1 − 1 a21 a22 = γ 1 − 1 h2 γ 2 − 1 h1 ! η2 1− 2 ζ " 92 Normal Shock Waves Therefore, γ1 − 1 h1 2 (η − 1)2 ζ + η M 2 h2 S η 2 ζ −η h5 h2 = 1+ h5 h1 = h5 h2 h2 h1 = = = h2 γ1 − 1 2 (η − 1)2 ζ + η MS + h1 2 η2 ζ −η ! " γ1 − 1 2 η − 1 ζ +η 1+ MS 2 (η + 1) + (η − 1) 2 η ζ −η 1 + (γ1 − 1)MS2 (η − 1)(ζ − 1) η(ζ − η) 5.10 Given that, the stagnation pressure and temperature are p0 = 1 atm = 101325 Pa, T0 = 15◦ C = 288.15 K From isentropic table, for M = 3.0, we have p = 0.02722 p0 Therefore, the static pressure at the test-section is p = 0.02722 p0 = 0.02722 × 101325 = 2758 Pa The pitot pressure measured by a pitot tube placed in the test-section is the pressure behind a normal shock. From normal shock table, for M = 3.0, we have p02 = 0.3283 p01 Thus, the pressure that a pitot tube at the test-section will measure is p02 = 0.3283 p01 = 0.3283 × 101325 = 33.265 kPa 93 5.11 From normal shock table, for M1 = 2.5, we have p2 = 7.125, p1 T2 = 2.1375, T1 ρ2 = 3.3333, ρ1 p02 = 8.5261 p1 Therefore, M2 = p2 = (7.125)(1) = 7.125 atm ρ2 = (3.3333)(1.225) = 4.083 kg/m3 p02 = (8.5261)(1) = 8.5261 atm T2 = (2.1375)(T1 ) 0.51299 From state equation, we have T1 = 101325 p1 = ρ1 R 1.225 × 287 = 288.2 K Therefore, T2 = (2.1375)(288.2) = 616.03 K a2 = / V2 = M2 a2 = (0.51299)(497.51) = γ R T2 = 497.51 m/s 255.22 From isentropic table, for M2 = 0.51299, we have T2 = 0.950 T02 Therefore, T02 = 616.03 T2 = 0.950 0.950 = 648.45 K 94 Normal Shock Waves 5.12 For nitrogen, molecular weight is 28.02 and γ = 1.4. Thus, the gas constant is γ R = 1.4 R = 8314/28.02 = 297 J/(kg K) and cp = γ−1 0.4 × 297 = 1039.5 J/(kg K). The speed of sound upstream of the shock is / √ a = γ R T = 1.4 × 297 × 303 = 354.95 m/s Therefore, the upstream Mach number M1 is M1 V1 923 = a1 354.95 = = 2.6 Also, since the flow process across a normal shock is adiabatic, T01 = T02 . Now, from normal shock table, for M1 = 2.6, we have p2 = 7.72 p1 M2 = 0.504 T2 = 2.2383 T1 Therefore, p2 = 7.72 × 300 = 2.316 MPa a2 V2 = / = 531.03 m/s = M2 a2 = 0.504 × 531.03 = γRT = √ 1.4 × 297 × 2.2383 × 303 267.64 m/s When the flow slows down isentropically from V1 to V2 , by energy relation, we have T2 = 678.32 K V12 − V22 2cp = T1 + = 303 + 9232 − 267.642 2 × 1039.5 95 By isentropic relation, we have p2 p1 = ! T2 T1 γ " γ−1 = ! = 16.786 678.32 303 "3.5 Thus, p2 = 5.036 MPa 5.13 For blunt nosed model at Mach 3, there will a detached bow-shock standing in front of the nose. This shock can be approximated to a normal shock at the nose of the model around the stagnation point. Therefore, pressure at the stagnation point is the total pressure behind the normal shock. From normal shock table, for M1 = 3, we have p02 = 0.32834 p01 Thus, p02 = 0.32834 p01 = 0.32834 × 10 = 3.2834 atm = 332.69 kPa The flow process across the shock is adiabatic. Hence, T02 = T01 . Therefore, T02 = 315 K After the normal shock, the flow decelerates isentropically to stagnation condition at the nose. Hence, the stagnation density ρ02 can be expressed as ρ02 = p02 332690 = R T02 287 × 315 = 3.68 kg/m3 5.14 Let us make the shock stationary and look at the field. The flow field with stationary shock will look like that shown in Fig. s5.14. 96 Normal Shock Waves T1 = 300 K p1 = 101 kPa p2 = 5000 kPa V2 V1 Shock Figure S5.14 p2 p1 = 49.5 = 1+ = 1 + 1.167 M12 − 1.167 M12 = 49.5 + 0.167 = 42.55 1.167 M1 = 6.52 2 2γ 1 2 M1 − 1 γ+1 Thus, The velocity becomes V1 = M1 a1 = 6.52 = 2374.16 m/s √ 1.4 × 287 × 330 By normal shock relation, we have M2 = ' = 0.4 γ−1 2 2 M1 M12 − γ−1 2 1+ γ *1/2 1 2 2 2(γ − 1) γ M12 + 1 1 2 48.41 × 41.5 = 9.205 M1 − 1 = 1 + 2 2 (γ + 1) M1 244.86 T2 T1 = 1+ T2 = 9.205 × 330 = 3037.65 K The speed of sound downstream of the shock is / √ γ R T2 = 1.4 × 287 × 3038.34 a2 = = 1104.9 m/s 97 Thus, the velocity V2 is V2 Vg = M2 a2 = 0.4 × 1104.8 = 441.92 m/s = Cs − Vg = 2374.16 − 441.92 = 1932.24 m/s Aliter The above problem can also be solved using gas tables, as follows. From normal shock tables, for pp21 = 49.5, we have M1 ≈ 6.5 M2 ≈ 0.4 T2 T1 ≈ 9.2 T2 ≈ 3036 K a2 = 1104.47 m/s V2 = M2 a2 = 0.4 × 1104.47 = 441.8 m/s = √ 1.4 × 287 × 330 = 364.13 m/s = M1 a1 = 6.5 × 364.13 = 2366.84 m/s a1 V1 98 Normal Shock Waves = 2366.84 − 441.8 Vg = 1925.04 m/s 5.15 Upstream of the shock, the speed of sound a1 is / √ γ RT1 = 1.4 × 287 × 300 a1 = = The Mach number M1 = 347.2 m/s V1 412 = 1.19. = a1 347.2 From isentropic table, for M1 = 1.19, we get p1 T1 = 0.41778, = 0.77929 p01 T01 Therefore, p01 T01 = 92 p1 = 0.41778 0.41778 = 220.21 kPa = 300 T1 = 0.77929 0.77929 = 384.96 K From normal shock table, for M1 = 1.19, we get p2 T2 p02 = 1.4854, = 1.1217, = 0.99372, M2 = 0.84846. p1 T1 p01 Therefore, p2 T2 = 1.4854 × 92 = 136.66 kPa = 1.1217 × 300 = 336.51 K 99 T02 = T01 = 384.96 K , since flow process across a shock is adiabatic p02 = 0.99372 × 220.21 = 218.83 kPa The speed of sound is a2 = fore, V2 √ = = γRT = √ 1.4 × 287 × 336.51 = 367.7 m/s. There- M2 a2 = 0.84846 × 367.7 311.98 m/s The entropy rise across a normal shock is given by ! " p01 ∆s = R ln p02 ! " 220.21 = 287 ln 218.83 = 1.8042 J/(kg K) 5.16 Let the subscripts 1 and 2 refer to conditions upstream and downstream of the normal shock, respectively. From normal shock table, for M1 = 2.5, we get p2 p02 = 7.1250, = 8.5261. p1 p1 The static pressure in the flow just downstream of the shock is, p2 = 7.125 × 1.0 = 7.125 atm If a normal shock has to be positioned at the nozzle exit, the back pressure to which the nozzle discharges has to be equal to the total pressure downstream of the shock. The total pressure downstream of the shock is p02 = = 8.5261 × 1.0 = 8.5261 atm 863.91 kPa i.e. the back pressure has to be 863.91 kPa to position a normal shock at the nozzle exit. 100 Normal Shock Waves 5.17 From isentropic table, for Me = 2.5, we have pe = 0.058528, p0e Ae = 2.637 A∗ where subscripts 0, e, and ∗ refer to stagnation, exit and throat, respectively. The throat area is A∗ = 4 Ae = 2.637 2.637 1.517 cm2 1 = 17.09 atm p0e = 0.058528 For the normal shock, the upstream Mach number is 1.5. From isentropic table, for M1 = 1.5, = A1 = 1.176, A∗ p1 = 0.2724, p01 T1 = 0.68966 T01 Thus, A1 = 1.176 × 1.517 = 1.784 cm2 p1 = 0.2724 × 17.09 = 4.655 atm T1 = 0.68966 × 500 = 344.83 K From normal shock table, for M1 = 1.5, p2 = 2.40583, p1 T2 = 1.3202 T1 p02 = 3.4133, p1 M2 = 0.70109 The back pressure required is p02 , thus, p02 = 3.4133 × 4.655 = 15.89 atm Downstream of the shock, the flow is isentropic upto the nozzle exit. However, for this flow the effective throat area is not the same as A∗ , since p02 < p01 . Let the effective throat area downstream of shock to be A∗2 . From isentropic table, for M2 = 0.70, A2 = 1.09437 A∗2 where A2 is the area at the shock location, which is same as A1 . Thus, A∗2 = 1.784 = 1.63 cm2 1.09437 Ae A∗2 = 4 = 2.454 1.63 101 From isentropic table for Ae A2 ∗ = 2.454, we have T2 = 0.98, Te Me = 0.245 Thus, Te = 0.98 × 500 = 490 K The speed of sound is ae = √ 1.4 × 287 × 490 = 443.71 m/s Ve = Me × ae = 0.245 × 443.71 The flow velocity is = 108.71 m/s 5.18 Since the Mach number upstream of the shock is 2.32, the area ratio corresponding to this Mach number will give the area at the shock location. (a) For M1 = 2.32, from isentropic table, A1 = 2.23 A∗ Thus, area at shock location is A1 = 2.23 × 5 = 11.15 cm2 (b) The Mach number downstream of the shock M2 given by normal shock table for M1 = 2.32 is M2 = 0.53 For M2 = 0.53, from isentropic table, we have A2 = 1.29 A∗ Since A1 = A2 = area at the shock location, we have A∗2 = 11.15 A2 = = 8.64 cm2 1.29 1.29 Therefore, 12.5 Ae = 1.447 = A∗2 8.64 From isentropic table, for Ae = 1.447, the exit Mach number Me = 0.45 A∗2 102 Normal Shock Waves (c) For the given nozzle, the area ratio Ae is Ath Ae Ae 12.5 = 2.5 = ∗ = Ath A 5 Ae = 2.5, we have A∗ p2 = 0.064261 Me = 2.44 and p02 From isentropic table, for For complete isentropic flow, p02 = p0 = 700 kPa. Thus, 44.98 kPa. p2 = 0.064261×700 = The back pressure range for the flow to be completely isentropic is pb ≤ 44.98 kPa . 5.19 Given, T1 = 22 K and T01 = 400 K, where subscripts 1 and 2 refer to conditions ahead of and behind the shock, respectively. For T1 22 = 0.055, from isentropic table, = T01 400 M1 = 9.2 From normal shock table, for M1 = 9.2, we have M2 = T2 T1 = T2 = 0.3893 17.4 382.8 K 5.20 Upstream of the normal shock, the Mach number M1 is M1 = V1 500 =√ a1 1.4 × 287 × 300 = 1.44 From normal shock table for M1 = 1.44, we get M2 = 0.72345, p2 = 2.2525, p1 T2 = 1.2807, T1 Thus, p2 = 2.2525 × 100 = 225.25 kPa p02 = 0.9476 p01 103 T2 = 1.2807 × 300 = 384.21 K V2 = M2 a2 = 0.72345 × = √ 1.4 × 287 × 384.21 284.25 m/s The entropy increase is ∆s = R ln ! p01 p02 " = 287 ln (1.0552) 15.432 J/(kg K) = 5.21 The flow velocity at nozzle entrance is low. Therefore, the pressure and temperature of the flow at the entry can be treated as the stagnation quantities. Thus, T01 = 300 K p01 = 1 MPa, From isentropic table, for M1 = 2, we have p1 = 0.1278 p01 and T1 = 0.55556 T01 Therefore, p1 = (0.1278)(1) = 0.1278 MPa T1 = (0.55556)(300) = 166.67 K From normal shock table, for M1 = 2, we get p2 T2 p02 = 4.5, = 1.6875, = 0.72087 p1 T1 p01 and M2 = 0.57735 Thus, p2 = 4.5 × 0.1278 = 0.5751 MPa T2 = 1.6875 × 166.67 = 281.3 K p02 = 0.72087 × 1 = 0.72087 MPa a2 = √ 1.4 × 287 × 281.3 = 336.19 m/s 104 Normal Shock Waves V2 M2 a2 = 0.57735 × 336.19 = = 194.1 m/s 5.22 Given, p0 = 101 kPa and T0 = 30 + 273.15 K. A normal shock at the nozzle exit implies that the entire nozzle flow is isentropic and also the flow is choked at the throat. The area ratio is Ae 0.0724 = 2.896 = ∗ A 0.025 From isentropic table, for Ae A∗ = 2.896, we have p1 = 0.050115, p0 M1 = 2.6, T1 = 0.42517 T0 This is the Mach number upstream of the shock. Thus, M1 = p1 = 0.050115 × 101 = 5.06 kPa T1 = 0.42517 × 303.15 = 128.9 K p01 = 2.6 101 kPa Let the conditions downstream of the shock be referred to by subscript 2. From normal shock table, for M1 = 2.6, we get M2 p2 p1 = 0.504 T2 = 2.2383, T1 = 7.72, p2 = 39.06 kPa T2 = 288.5 K p02 = 46.47 kPa p02 = 0.46012 p01 105 5.23 Given, p0 = 200 kPa and T0 = 350 K. Let subscripts 1 2 and 3 refer to locations upstream and downstream of the shock wave and the nozzle exit, respectively. Ath = A∗1 = 0.2 m2 At the shock location, A1 = 0.6 m2 , thus, A1 0.6 = 3.0 = ∗ A1 0.2 From isentropic table, for A1 = 3.0, we get A∗1 M1 = 2.64, p1 = 0.04711 p01 Up to the shock the flow in the nozzle is isentropic and therefore, p01 = p0 . Thus, p01 = p1 = 200 kPa 200 × 103 × 0.04711 = 9.422 kPa where, p1 and p01 are the static and total pressures, respectively, ahead of the shock. Let subscript 2 refer to condition behind the shock. Now, from normal shock table, for M1 = 2.64, we have p2 = 7.9645, p1 M2 = 0.50, p02 = 0.44522 p01 Thus, p2 = = p02 = = 7.9645 × 9.422 75.04 kPa 0.44522 × 200 89.04 kPa Also, from normal shock theory we have (Section 4.6) p01 A2 ∗ = = 2.25 p02 A1 ∗ 106 Normal Shock Waves Thus, A∗2 = 2.25 × 0.20 = 0.45 m2 A∗2 may also be obtained from M2 . From isentropic table, for M2 = 0.5, A2 = 1.34 A∗2 Thus, 0.6 = 0.448 m2 1.34 This A∗2 is the equivalent throat area for the flow downstream of the shock. Therefore, at the nozzle exit, A2 ∗ = A3 0.8 = 1.786 = ∗ A3 0.448 Now, from isentropic table, for M3 = 0.35, A3 = 1.786, A∗3 p3 = 0.91877, p03 T3 = 0.97609 T03 Here, p03 = p02 and T03 = T02 = T0 . Thus, p3 = p03 = T3 = 0.91877 × 89.04 = 81.81 kPa 89.04 kPa 0.97609 × 350 = 341.63 K For area ratio A3 /A∗3 the subsonic solution from the isentropic table was used since after a normal shock the flow becomes subsonic and this flow is further decelerated in the divergent portion of the duct. 5.24 Given, p01 = 5 atm and p02 = 3.6 atm. Therefore, p02 3.6 = 0.72 = p01 5 Assuming γ = 1.4, from normal shock table, for M1 = 2.0, p02 = 0.72, we have p01 p2 = 4.5 p1 Now, from isentropic table, for M1 = 2.0, we get p1 = 0.1278 p01 107 Hence, the pressure just behind the normal shock at the nozzle exit is p2 = = 4.5 p1 = 4.5 × 0.1278 × 5 2.8755 atm 5.25 The pitot tube will read the actual total pressure in a subsonic stream. But in a supersonic flow, the pressure measured by a pitot probe is the total pressure downstream of a detached shock which stands at the nose of the pitot tube. Therefore, it is essential to find out whether the flow is subsonic or supersonic. It can be easily seen from the isentropic relations that for M = 1, the pressure 0.95 p = = 1.8 atm. ratio pp0 = 0.528. Hence, p0 = 0.528 0.528 Thus, when p0 < 1.8 atm, the flow is subsonic, and when p0 > 1.8 atm, the flow is supersonic. (i) p0 = 1.1 atm. The flow is subsonic and hence the pitot tube is measuring the actual total pressure of the flow. p 0.95 = 0.8636 = p0 1.1 From isentropic table, for p = 0.8636, we get p0 M = 0.465 (ii) p0 = 2.5 atm. The flow is supersonic and the pitot tube measures p02 behind a normal (detached) shock. p02 2.5 = 2.63 = p1 0.95 p02 From normal shock table, for = 2.63, we get p1 M = 1.275 (iii) p0 = 10 atm. The flow is supersonic. p02 10 = 10.526 = p1 0.95 p02 From normal shock table, for = 10.526, we have p1 M = 2.79 108 Normal Shock Waves 5.26 The shock will be only at the divergent portion of the nozzle, since only after the throat the flow becomes supersonic. The Mach number M1 just upstream of the shock will be given by the area ratio Ashock . Ath Ashock 2000 =2 = Ath 1000 From isentropic table, for area ratio 2, we have M1 = 2.2 , p1 = 0.093522 p01 Up to the shock, the stagnation pressure does not change and therefore, p01 = 200 kPa p1 = 200 × 0.093522 = 18.7 kPa Now, from normal shock table, for M1 = 2.2, we get p2 = 5.48, p1 M2 = 0.55, p02 = 0.62814 p01 where subscript 2 refer to condition just downstream of the shock. Therefore, p2 = 5.48 × 18.7 = 102.48 kPa p02 = 0.62814 × 200 = 125.63 kPa For M2 = 0.55, from isentropic table, we get A2 = 1.255 Ath2 where Ath2 is the throat area required for the flow downstream of the shock to choke. A2 Ath2 = 1.255 2 But A2 = A1 = Ashock = 2000mm , therefore, Ath2 = Thus, 2000 = 1593.63 mm2 1.255 Ae 3000 = 1.8825 = Ath2 1593.63 For this area ratio, from isentropic table (subsonic part), we get Me = 0.325 109 Therefore, 1 p02 pe = pe = 125.63 p02 = 1.076 1.076 = 116.76 1 + 0.2 × 0.3252 23.5 = 1.076 The pressure loss occurs only across the shock and the loss of pressure ∆p0 is ∆p0 = = p01 − p02 = (200 − 125.63) 74.37 kPa 5.27 Let the subscripts i, 1, 2 and e refer to the inlet, just upstream and just downstream, and the nozzle exit, respectively. It can be shown for this flow that, (1) p01 A∗1 = p02 A∗2 For Mi = 2.0, from isentropic table, we get Ai = 1.6875 A∗i Therefore, A1 A∗1 = A1 Ai Ai A∗1 = A1 Ai , Ai A∗i = 2 × 1.6875 = 3.375 since A∗1 = A∗i For this area ratio, from isentropic table, we get M1 = 2.76 For M1 = 2.76, from normal shock table, we have p02 = 0.40283 p01 Using this in Eq. (1), we get A∗1 = 0.40283 A∗2 110 Normal Shock Waves Ae A∗2 = Ae Ai A∗1 Ai A∗1 A∗2 = 4 × 1.6875 × 0.40283 = 2.7191 For this area ratio, from isentropic table (subsonic solution), we get Me = 0.22 The exit pressure pe may be expressed as pe = pe p02 p01 pi p02 p01 pi = pe p02 p0i pi , p02 p01 pi = 0.96684 × 0.40283 × = since p01 = p0i 1 × 80 0.1278 243.8 kPa Thus, the back pressure required is 243.8 kPa. 5.28 When a pitot tube is placed in a supersonic stream, there will be a detached shock standing at its nose. At the nose where the pressure tap is located, the shock may be treated as a normal shock and hence what the pitot tube measures is the pitot pressure p02 downstream of the shock. The wall pressure measured by a pressure tap may be treated as the actual static pressure of the stream. Thus we may take the static pressure upstream of the shock as p1 = 112 kPa. Thus, p02 2895 = 25.848 = p1 112 p02 = 25.848, we get p1 T2 M1 = 4.44 , = 4.7706 T1 Now from isentropic table, for M1 = 4.44, we get From normal shock table, for T1 T01 = 0.2023 T1 = 0.2023 × 500 = 101.15 K a1 = / γ R T1 = 201.6 m/s 111 Thus, V1 = M1 a1 = 895.1 m/s 5.29 At 10,000 m altitude, from standard atmospheric table, we have p = 26.452 kPa, T = 223.15 K During steady–state operation, mass flow through the test–section is given by ṁ / p A M γRT RT = ρ AV = = √ π(0.25)2 26.452 × 103 × × 2.4 1.4 × 287 × 223.15 287 × 223.15 4 = 14.57 kg/s The stagnation temperature during steady–state operation is T0 = T0 T T For, M = 2.4, T0 1 = = 2.152 T 0.46468 Thus, T0 = 2.152 × 223.15 = 480.22 K For the present geometry of fixed angle diffuser, the optimum condition for steady state operation is a normal shock at the diffuser throat. The diffuser throat area is A∗ A A∗ = A From normal shock table, for M1 = 2.4, M2 = 0.52. From isentropic table, for M2 = 0.52, we have A2 = 1.3 A∗2 Here A∗2 is the area of the second throat and A is the area of test-section. A2 Therefore, looking in to the corresponding supersonic Mach number for ∗ = A2 1.3, we get M = 1.66 112 Normal Shock Waves This is the Mach number just upstream of the second throat with a shock. From normal shock table, for M1 = 1.66, we have p02 = 0.872 p01 This pressure loss must be compensated by the compressor. (a) The power required for the compressor is given by Power = h0 − hi = Cp (T0 − Ti ) where the subscripts 0 and i refer to outlet and inlet conditions. For an isentropic compression, ! " γ−1 p0 γ T0 = Ti pi * '! " γ−1 p0 γ T0 − Ti = Ti −1 pi '! * "0.286 1 = 480.22 − 1 = 19.18 K 0.872 Thus, the power input required for mbox mass of air becomes Power = 1004.5 × 19.18 = 19.266 kJ/kg The total power required for the compressor is ṁ 19266 Power = 746 = 376.3 hp This is the running power required for the compressor. (b) During start-up, M1 = 2.4 in the test-section. The corresponding total pressure ratio across the normal shock is p02 = 0.5401 p01 The isentropic work required for the compressor during start-up to drive the shock out of the test-section per mbox mass of air is '! * "0.286 1 Power = Cp T0i −1 0.5401 '! * "0.286 1 = 1004.5 × 480.22 −1 0.5401 = 92929.87 J/kg 113 The power required to start the tunnel is Power = ṁ × 92929.87 14.57 × 92929.87 = 746 746 = 1815 hp 5.30 Let subscripts 1 and 2 refer to conditions just upstream and downstream of the normal shock. The Mach number just upstream of the shock is 1.85. From isentropic table, for M1 = 1.85, we get p1 A1 = 0.1612, ∗ = 1.495 p01 A1 p1 = 0.1612 × 400 = 64.48 kPa A1 = 1.495 × 10 = 14.95 cm2 Now from normal shock table, for M1 = 1.85, we have p02 = 0.79023 p01 M2 = 0.60, Again from isentropic table, for M1 = 0.6, we get A2 = 1.1882 A∗2 But A2 = A1 = 14.95 cm2 . Therefore, A∗2 = 14.95 = 12.58 cm2 1.1882 We want the Mach number and pressure at a state 3 where A3 = 2 A1 = 2 × 14.95 = 29.9 cm2 . Downstream of the shock, A∗2 = A∗3 . Thus, A3 A∗3 = 29.9 = 2.377 ≈ 2.4 12.58 From isentropic table, for this area ratio (from subsonic solution), M3 = 0.25 and p3 = 0.95745 p03 But p03 = p02 = 316.1 kPa. Therefore, p3 = 0.95745 × 316.1 = 302.65 Note: This problem has been solved using isentropic table and hence the results obtained are only approximate. For exact results, we have to use the actual relations. 114 Normal Shock Waves 5.31 (a) Given, ṁ = 0.9 kg/s, p0 = 4×101325 = 405300 Pa, T0 = 30+273 = 303 ∆p = 2. K and p1 The pressure ratio across the shock is p2 − p 1 = p1 p2 p1 2 = 2+1 = 3 p2 For = 3, from normal shock table, p1 M1 = 1.65, M2 = 0.654, T2 = 1.4228 T1 From isentropic table, for M1 = 1.65, T1 A = 0.6475, ∗ = 1.292 T0 A Therefore, T1 T2 = 0.6475T0 = 0.6475 × 303 = 196.19 K = 1.4228T1 = 1.4228 × 196.19 = 279.14 K The mass flow rate is 0.9 kg/s, therefore, ṁ = Ath = = = 0.9 = 0.6847 p0 Ath √ RT0 √ 0.9 RT0 0.6847 p0 √ 0.9 287 × 303 0.6847 × 405300 9.56 cm2 115 Therefore, the exit area becomes 1.292 Ath = 1.292 × 9.56 = Ae 12.35 cm2 = (b) The exit velocity is V2 = M2 a 2 = M2 = 0.654 × = √ / γRT2 1.4 × 287 × 279.14 219 m/s 5.32 a) Given, (p2 −p1 )/p1 = 3.5, where p1 and p2 are the pressures ahead of and behind the shock. Therefore, the pressure ratio across the shock is p2 /p1 = 4.5. For p2 /p1 = 4.5, from normal shock table, M2 = 2, ρ2 = 2.67 ρ1 Therefore, the shock speed becomes Cs = M1 a 1 = 2 × = 2× = 694.4 m/s √ / γRT 1.4 × 287 × 300 The piston speed is given by Vp = = = = = V1 − V2 = V1 ! ! V2 1− V1 " ρ1 Cs 1 − ρ2 ! " 1 694.4 × 1 − 2.67 694.4 × (1 − 0.375) 434 m/s " 116 Normal Shock Waves (b) Given (p2 − p1 )/p1 = 7. Therefore, p2 /p1 = 8. For p2 /p1 = 8, from normal shock table, ρ2 M2 = 2.65, = 3.5047 ρ1 Therefore, the shock speed becomes √ Cs = 2.65 × 1.4 × 287 × 300 = 920.05 m/s = ! 920.05 × 1 − = 920.05 × 0.7147 Thus, Vp = 1 3.5047 " 657.56 m/s 5.33 (a) Let subscripts 1 and 2 refer to states ahead of and behind the shock, respectively and subscript e refer to the nozzle exit. Given, p01 − p02 × 100 = 12.4 p01 1− p02 p01 = 0.124 p02 p01 = 0.876 From normal shock table, for p02 /p01 = 0.876, M1 = 1.65 , M2 = 0.654, T2 = 1.4228 T1 (b) For M1 = 1.65 from isentropic table, we have T1 T01 = 0.6475 T1 = 0.6475 × T01 = 0.6475 × 330 = 213.7 K 117 Therefore, T2 = 1.4228 × 213.7 = 304 K By energy equation, we have he + Ve2 2 = h0e c p Te + Ve2 2 = cp T0e But the flow process across the shock is adiabatic, therefore, T01 = T02 = T0e . Hence the flow speed at the nozzle exit becomes Ve = = = 0 / 2 × cp (T0e − Te ) 2 × 1004.5 × (330 − 300) 245.5 m/s The flow speed behind the shock is V2 = M2 a 2 = M2 × = 0.654 × = √ / γ R T2 1.4 × 287 × 304 228.57 m/s (c) The mass flow rate is ṁ = 0.6847 p01 Ath √ RT01 = 0.6847 × (5 × 101325) × (5 × 10−4 ) √ 287 × 330 = 0.5636 kg/s 118 Normal Shock Waves Chapter 6 Oblique Shock and Expansion Waves 6.1 Given, M1 = 2 and β = 40◦ , therefore, M1n = 2.0 sin 40 = 1.29. From normal shock tables, for M1n = 1.29, we have p2 p1 = 1.775 T2 p1 = 1.185 and M2n = 0.7911 Therefore, p2 T2 = 0.5 × 1.775 × 105 = 0.8875 × 105 Pa = = 1.185 × 273 323.5 K For the adiabatic process, Tt1 = Tt2 . From isentropic tables, for Mach 2, we have T1 /Tt = 0.556. Thus, T1t = T2t = 273 = 491 K 0.556 From isentropic table, for T2 /T2t = 0.659, we have M2 = 1.61 119 120 Oblique Shock and Expansion Waves Also, u2 w2 = 0.791 0.791 = 0.4913 = M2 1.61 β−θ = 29.43 =⇒ θ = 10.57 deg sin (β − θ) = Thus, the wedge angle = 2θ = 21.14◦ 6.2 For M1 = 2.0, from isentropic table, we have ν1 = 26.38◦ . Prandtl-Meyer function after expansion is ν2 = ν1 + θ = 26.38 + 5 = 31.38 deg Therefore, for ν2 = 31.38◦ the corresponding Mach number from isentropic table is M2 ≈ 2.18 . From isentropic tables for M2 = 2.18, we have p2 p02 T2 T02 ρ2 ρ02 = 0.0965; = 0.5127; = 0.1882; p2 = p1 × = T2 ρ2 6.3 (a) p1 p01 T1 T01 ρ1 ρ01 0.0965 0.0965 = 98 × 0.1278 0.1278 74 kPa = 300 × 0.5127 0.5556 = 276.8 K = 74 × 103 287 × 276.8 = 0.9315 kg/m3 = 0.1278 = 0.5556 = 0.2300 121 pe = 1 atm Me = 3 p0 = 70 × 105 kPa Figure S6.3a M1 = 3.0 =⇒ p1 = 0.02722 p01 With p1 = pe = patm , p01 = 1.01325 × 105 = 37.2 × 105 Pa. 0.02722 Therefore, the supply pressure for which oblique shock wave will first appear in the exhaust jet is ≤ 37.2 × 105 Pa (b) 0.45 D p0 = 70 × 105 kPa 0.1 D 0.45 D D 97 D 1.0 0.45 D D Figure S6.3b sin β = 0.45 = 0.41 1.097 M1n = M1 sin β = 3 × 0.41 = 1.23 p2 p1 = 1.5984; p2 = pe 122 Oblique Shock and Expansion Waves Therefore, p1 M1 = atm pe = 1.5984 1.5984 = 0.634 × 105 Pa = 3.0 =⇒ p1 = 0.02722 p01 Therefore, p01 = 0.634 × 105 0.02722 = 23.3 × 105 Pa Therefore, minimum supply pressure for obtaining the desired test region is 23.3 × 105 Pa. (c) pe = 1 atm M1 = 3 p0 = 70 × 105 kPa 1 2 Figure S6.3c M1 p2 p1 p1 p01 = 3.0 p1 = = 10.33 p2 = = 0.02722 p01 = patm = 0.098105 Pa 10.333 pe = patm 0.098 × 105 = 3.6 × 105 Pa 0.02722 Therefore, the minimum supply pressure for which a normal shock will appear at the nozzle exit is 3.6 × 105 Pa Note: The static pressure after the shock has to be equal to the back pressure, namely the atmospheric pressure. This is because, subsonic jets are always correctly expanded. Thus, the total pressure of this subsonic flow is higher than 123 its static pressure. Hence, the flow will move some distance downstream of the nozzle exit before coming to rest. 6.4 The relation between θ and β is 1 2 2 cot β M12 sin2 β − 1 tan θ = M12 (γ + cos 2β) + 2 For θ = 15◦ , M1 = 2.0, solution by iteration yields β = 79.8◦ . This is the strong shock solution. (a) β = 79.8 deg (b) p2 p1 = 2γ γ−1 M 2 sin2 β − γ+1 1 γ+1 = 4.354 (c) T2 T1 = p2 ρ2 / p1 ρ1 = 1.662 (d) β−θ = 64.8◦ ρ2 ρ1 = tan β tan (β − θ) = 2.615 (e) M12 sin2 (β − θ) = γ−1 2 2 2 M1 sin β γ−1 M12 sin 2 β − 2 1+ γ = 1 + 0.2 × 4 × (sin 79.8◦ × sin 79.8◦ ) 1.4 × 4 × (sin 79.8◦ )2 − 0.2 = 1.775 = 0.34 5.224 124 Oblique Shock and Expansion Waves M22 = 0.34 0.34 = 0.415 = (sin 64.8)2 0.819 M2 = 0.644 Weak shock (a) Solving θ − β − M relation we can obtain βweak = 45.3◦ (b) p2 p1 = 2γ γ−1 M 2 sin2 β − γ+1 γ+1 = 2.191 (c) T2 T1 = 1.267 (d) β−θ = 45.3 − 15 = 30.3 deg ρ2 ρ1 = tan 45.3◦ tan 30.3◦ = 1.729 (e) M2 = 1.446 Aliter: Using Normal shock tables: M1n = M1 sin β , 1.97 Strong shock solution lution Weak shock so- 125 γ = 1.4 M1n = 1.97 M1n = 2sin 45.3 deg = 1.42 γ = 1.4 p2 p1 = 4.361 M2n = 0.7314 T2 T1 = 1.663 p2 p1 = 2.186 M2n = 0.583 ρ2 ρ1 = 1.724 ρ2 ρ1 = 2.622 T2 T1 = 1.268 M2n = M2 sin (β − θ) M2 = M2 = M2 n sin (β − θ) M2n sin (β − θ) = 0.7314 = 1.45 sin 30.3 = 0.583 = 0.644 sin(79.8 − 15) Note: The solution obtained with oblique shock relations may also be obtained using oblique shock tables, which will result in considerable time saving. 6.5 2 1 Figure S6.5 From isentropic tables, for M1 = 2.0, we have p1 = 0.1278. Since, pt1 0.75 = 0.0599. 0.1278 × 1.60 pt1 = pt2 for this isentropic expansion, For this pressure ratio, from isentropic table, we get M2 = 2.48 . p2 = pt2 126 Oblique Shock and Expansion Waves From isentropic table, the Prandtl-Meyer function for Mach 2.0 and 2.48, respectively are ν1 = 26.38◦ ν2 = 38.655◦ Therefore, the flow turning angle becomes ν12 = ν2 − ν1 = 12.275◦ 6.6 (a) Angles for which the oblique shock remains attached to the wedge (from oblique shock table) are At At M1 θmax = = 2.0 22◦ M1 θmax = 3.0 = 34◦ θd 15◦ 25◦ M1min 1.65 2.11 40◦ 4.45 (b) 6.7 M1 = 3.5 θ M2 45◦ θ Figure S6.7 M1 = 3.5 β = 45 deg 127 28.158◦ =⇒ θ = M1n = M1 sin β = 3.5 sin 45 = 2.47 M2n = 0.51592 M2 = M2n sin (β − θ) = 1.78 6.8 M2 = 4.0 ν2 = 65.785◦ ν2 = ν1 + |∆θ| ν1 = ν2 − |∆θ| Therefore, (a) |∆θ| = 60 deg −15 deg = 45 deg Therefore, ν1 = =⇒ M1 = 65.785 deg −45 deg = 20.785 deg 1.8022 (b) |∆θ| = 60 deg −30 deg = 30 deg Therefore, ν1 = =⇒ M1 = 65.785 deg −30 deg = 35.785 deg 2.360 128 Oblique Shock and Expansion Waves (c) |∆θ| = 0 deg Therefore, nu1 = ν2 = 65.785 deg For this value of Prandtl Meyer function, from isentropic table, we get M1 = 4.0 . (d) |∆θ| = 60 deg +15 deg = 75 deg ν1 = 65.785 deg −75 deg = − 9.215 deg A negative ν is not possible. The flow downstream can exist only upto |∆θ| = 65.785 deg for which ν1 = 0 =⇒ M1 = 1.0. 6.9 p1 , M 1 p2 , M2 θ Figure S6.9 For M1 = 2.0, from isentropic table, we have p1 /p0 = 0.1278. Therefore, p2 p1 1 p2 = × = × 0.1278 = 0.0639 p0 p1 p0 2 For p2 /p0 = 0.0639, from isentropic table, we have M2 = 2.444 . M1 = 2.0 =⇒ ν1 = 26.38 deg M2 = 2.444 =⇒ ν2 = 37.81 deg ν2 = ν1 + |∆θ| = ν1 + θ2 θ2 = ν2 − ν1 = 37.81 − 26.38 = 11.43 deg 129 6.10 M 1 , p1 p0 θ Me Figure S6.10 (a) =⇒ and M1 = 3.0 p1 p0 = 0.02722 pe p0 = 1.01325 × 105 70 × 105 = 0.0145 < p1 p0 Hence, there will be expansion at the nozzle exit to reduce the pressure from p1 to pe . From isentropic table, for pp0e = 0.145 =⇒ Me = 3.431. M1 = 3.0 =⇒ ν1 = 49.75◦ Me = 3.431 =⇒ νe = 57.42◦ For expansion, νe = ν1 + |∆θ| Therefore, |∆θ| = 57.42 − 49.75 = 7.67◦ (b) For θe = 0, pe = p1 = 1.0133 × 105 Pa pe p1 = = 0.02722 p0 p0 Therefore, stagnation pressure required for θe = 0 is νe = ν1 , p0 = = 1.0133 × 105 0.02722 37.226 × 105 Pa 130 Oblique Shock and Expansion Waves 6.11 From isentropic table, for M1 = 1.5, we have p1 p01 = 0.2724 T1 T01 = 0.6897 Therefore, p2 p01 = p2 p1 × p1 p01 = 0.13 × 0.2724 = 0.118 0.30 =⇒ M2 = 2.05 T2 T02 = 0.5433 (a) From Prandtl-Meyer function table, we have M1 = 1.5 =⇒ νM1 = 12◦ M2 = 2.05 =⇒ νM2 = 27.75◦ The deflection angle required is θ = νM1 − νM2 = −15.75◦ (b) M2 = 2.05 (c) T2 6.12 = T2 T01 0.5433 × 350 × × T1 = T02 T1 0.6897 = 275.7K 131 #1 1 #2 2 3 A1 #3 A2 4 A3 5◦ 10◦ 15◦ A4 Figure S6.12 Flow deflection in regions (2), (3) and (4) as measured from flow deflection in region (1) are − 5 deg, − 15 deg and − 30 deg. For expansion, ν2 = ν1 + |∆θ| Region (1) (2) (3) (4) |∆θ| (deg) 0 5 10 15 ν (deg) 26.38 31.38 41.38 56.38 M 2.000 2.187 2.60 3.370 µ (deg) 29.928 27.200 22.600 17.260 Fan angle (deg) ξ1 = µ1 − (µ2 − ∆θ) 7.73 14.60 20.34 A A∗ 1.688 1.980 2.896 6.012 Therefore, A1 : A2 : A3 : A4 = = 1.688 : 1.980 : 2.896 : 6.012 1 : 1.173 : 1.716 : 3.562 6.13 From oblique shock table, for M1 = 2.4 and θ = 7 deg, we get β = 30.25 deg. M1n = = M1 sin β 2.4 sin 30.25 = 1.21 From normal shock table for M1n = 1.21, we have p2 p1 T2 T1 a2 a1 p02 p01 = 1.539 = 1.134 = 1.065 = 0.9918 132 Oblique Shock and Expansion Waves M2n = M2 = From normal shock table, for p3 p2 T3 T2 a3 a2 p03 p02 M3 p2 T2 p3 T3 0.830 M2n = 2.12 sin (β − θ) M2 = 2.12, we have = 5.077 = 1.7875 = 1.337 = 0.6649 = = 0.5583 1.57 × p1 = 0.77 × 105 Pa 1.134 × 280 = 317.52 K 4.881 p2 = 3.91 × 105 Pa 1.7875 × 317.52 = 567.57 K = = = (a) ṁ = ρ3 A3 V3 Therefore, Ai = ρ3 = V3 = = = A3 = ṁ ρ3 V3 p3 RT3 2.4 kg/m3 M3 a3 = 0.5583 × 477.6 266.64 m/s Hence, Ai = = 20 2.385 × 266.64 0.03125 m2 (b) At the exit, Me = Ve /ae . By energy equation, we have Ve2 a2e a2 + = 0e 2 γ−1 γ−1 133 ae = a0e = 0 a20e − 0.2Ve2 / a03 = a01 = 20.04 T01 From isentropic table, for M1 = 2.4, 0.061 T1 T01 = T01 = Thus, a0e = Hence ae = Me = 0.4647 280 = 602.5K 0.4647 √ 20.04 602.5 = 492 m/s / 4922 − 0.2 × 302 = 492 m/s 30 = 0.061 492 For Me = 0.061, pe p03 Te T03 For M3 p3 p03 T3 p03 = 0.9975 = 0.9993 = 0.5583 = 0.809 = 0.941 Therefore, pe = = = Te = ρe = = Hence, Ae = pe p03 p3 p03 p3 0.9975 × 3.91 × 105 0.809 4.82 × 105 Pa 0.9993 × 567.57 = 602.7 K 0.940 pe 4.75 × 105 = RTe 287 × 595 2.79 kg/m3 20 = 0.240 m2 2.78 × 30 for Me = 134 Oblique Shock and Expansion Waves 6.14 (a) For normal shock diffuser, at M1 = 3.0, p02 p01 = 0.3283 (b) For the wedge shaped diffuser, M1 = 3.0, θ = 8◦ Therefore, β = 25.5◦ M2 θ = = 2.6 8◦ = = 29.0◦ , 2.26 For M2 = 2.6 and θ = 8◦ , we have β M3 Therefore, M1n M2n = = M1 sin β 3.0 sin 25.5 = = 1.3 2.6 sin 29 = 1.26 From Normal shock table, for M1n = 1.3, p02 p01 M2n p03 p01 M3 p04 p03 = 0.9794, = 1.26 = 0.986, = 2.26 = 0.60105 Therefore, the overall stagnation pressure ratio is p04 = 0.9794 × 0.986 × 0.60105 = 0.5804 p01 Note: The first diffuser with a single normal shock, the pressure loss is 67.2% but for the wedge shaped diffuser it is only 41.96%. 135 6.15 The wave pattern over the plate will be as shown in the Fig. S6.15a. (a) M1 1 2 θ=α β= 42 ◦ 3 Figure S6.15a M1 = 2.0 β = 42 deg =⇒ θ = α = 12.3◦ (b) 1 2 4 M1 β δ Slipstream 3 5 Figure S6.15b M1n = M1 sin β = 1.338 For M1 = 1.338, from normal shock table, for M1 = 2.0, p3 p1 M3n Therefore, p3 = 1.928 = = 0.7664 1.928 atm From isentropic table, M3 = ν1 |∆θ| = = For expansion, ν2 = = The corresponding Mach number from isentropic table is M3n = 1.55 sin (β − θ) 26.38◦ 12.3◦ ν1 + |∆θ| 26.38 + 12.3 = 38.68◦ 136 Oblique Shock and Expansion Waves M2 For M1 p1 p0 = = 2.48 2.0 = 0.1278 For M2 p2 p0 Therefore, p2 = 2.48 = 0.0604 = p2 p0 p1 p0 p1 = 0.473 atm (c) Trial M2n 1 2 Trial δ β = M2 sin β 12 deg 34 1.387 12.5 deg 34.5 1.405 M4 p4 p1 p4 0.744 0.740 2.09 2.12 0.989 1.003 p5(1) 0.041 ν5 = ν3 + |∆θ| M5 p5 p0 p3 p5 p5 p5(2) ∆p p4(2) p4(1) 12◦ 12.5◦ δ Fig. S6.15c ∆p − 0.041 δ − 12 For ∆p δ For which p4 = p5 = 1 atm 6.16 = = = −0.002 − 0.041 0.5 0 12.5◦ 1 2 25.38 1.96 25.88 1.98 0.136 0.132 0.534 0.521 1.03 1.005 137 1 2 M1 15◦ Figure S6.16 For θ = 15◦ , M1 = 3.0 =⇒ β = 32.2 deg. M1n = M1 sin β = 1.60 M2n = 0.6684 p2 p1 = 2.82 ρ2 ρ1 = 2.03 T2 T1 = 1.388 p02 p01 = 0.8952 From isentropic table, for M1 = 3.0, we have p1 p01 = 0.0272 T1 T01 = 0.3571 Therefore, p1 = 1.904 × 105 Pa T1 = 107.13 K ρ1 = a1 = p1 = 6.19 kg/m3 RT1 / γRT1 = 207.5 m/s Using ratios of flow properties, we obtain, 138 Oblique Shock and Expansion Waves (a) p2 = 5.37 × 105 Pa ρ2 = 12.57 kg/m3 T2 = 148.7 K p02 = 62.66 × 105 Pa (b) (c) M2 a2 V2 6.17 λu = = = λl = = CL = = = M2n = 2.26 sin (β − θ) / = γRT2 = 244.4 m/s = = M2 a2 = 552.3 m/s ! dz dx " u 1 − α 0 ≤ x ≤ 0.3c 3 1 − − α 0.3c ≤ x ≤ c !7 " dz dx l −α $ c 2 − (λu + λl ) dx βc 0 " +$ 0.3c ! 1 2 − − α dx βc 0 3 " $ c ! 1 + − − α dx 7 0.3c , $ c + −αdx 0 " +! 2 1 − − α 0.3c βc 3 139 − 2 [−2αc] βc 4α β = π 4 = 0.049365 α = 2 deg, CL = − √ × 8 90 α = 0 deg, CL = 0 and for CD = = = = = CD = λu λl = = λc = = = λc = = Cmac = 2 βc $ " 1 + α 0.7c − αc ] 7 − = For ! c 0 1 2 λ2u + λ2l dx "2 $ 0.3c ! 1 2 [ − α dx βc 0 3 "2 $ c ! $ c 1 + −α2 dx ] − − α dx + 7 0.3c 0 ! "2 1 2 π [ − 0.3c βc 3 90 ! "2 % π &2 1 π + + 0.7c + c] 7 90 90 2 [0.02672 + 0.02212 + 0.0012185] β 2 × 0.05006 β 0.0354 λt + λc − α −λt + λc − α (λu + λl ) +α 11 2 2 3 −α−α +α 2 1 6 1 1 2 −7 − α − α +α 2 1 − 14 4 βc2 $ c λc xdx 0 0 ≤ x ≤ 0.3c 0.7c ≤ x ≤ c 140 Oblique Shock and Expansion Waves = = = = xcp c = = xcp = $ c $ 0.3c 1 1 4 xdx − xdx ] [ βc2 0 6 14 0.3c + <, 1 4 1 ; 2 2 2 (0.3c) − − (0.3c) c βc2 12 28 , + 1 4 0.09 √ − (1 − 0.09) 28 8 12 −0.0354 Cmac 1 + CL 2 (−0.0354) 1 + − 0.049365 2 − 1.217 c 6.18 1 2 M1 = 3 3 1" 2 " 3" Figure S6.18 Upper Surface For M1 = 3.0, from isentropic table, we have ν1 = 49.757 deg Therefore, ν2 = 49.757 + 12 = 61.757 deg For ν2 = 61.757◦ , from isentropic table, we get M2 = 3.71 Flow Mach number normal to the oblique shock is M2n = M2 sin β where β is shock angle. From oblique shock chart for M2 = 3.71 at θ = 12 deg, we have β2 = 26 deg 141 Therefore, = M2n 3.71 sin 26 = 1.626 From normal shock table, for M2n = 1.63, we get M3n = M3 = 0.6596 Therefore, 0.6596 M3n = sin (β − θ) sin 14 = 2.726 Lower Surface For M1 = 3.0, and θ = 12◦ , from oblique shock chart, we get M 1# n = M1 sin β β 1# = 29.25 deg Therefore, = M 1# n 3 sin 29.25 = 1.47 From normal shock table, for M1# n = 1.47, we have M 2# n = 0.712 Therefore, M 2# 0.712 = 2.4 sin (29.25 − 12) = From isentropic table, for M2# = 2.4, we have ν 2# = 36.75 deg Therefore, ν 3# = 36.75 + 12 = 48.75 deg The corresponding Mach number is M 3# 6.19 = 2.95 142 Oblique Shock and Expansion Waves 1 2 12◦ M1 = 3 3 10◦ 4 2" 3" 4" Figure S6.19 θ12 ν2 = = M2 p2 p01 θ23 = 3.105 = 0.02326 = 20 deg ν3 M3 p3 p01 θ34 β34 = = 71.75 deg 4.493 = 0.003484 = = 22 deg 33.5 deg M3n4 p4 p3 p4 p01 M4n = 2.48 = 7.009 = 0.0245 = M4 = θ12# $12# = = 0.5149 0.5149 = 2.58 sin11.5 22 deg 40 deg Comparing 6.20 p4 p01 and 2 deg 51.75 deg p4 # p01 , M1n p 2# p1 p 2# p01 p02# p01 M2# n1 = 1.93 = 4.179 = 0.115 = 0.7535 = M 2# = ν 2# = = 0.5899 0.5899 = 1.91 sin18 23.85 20 ν 3# M 3# p3 p02# p 3# p01 θ 3# 4# ν 4# = = 43.87 deg 2.71 = 0.0423 = 0.0319 = = 2 deg 45.87 deg M4 p 4# p02# p 4# p01 = 2.8058 θ 2# 3# = 0.0366 = 0.0276 it can be seen that, the slipstream is very weak. 143 p2 M1 p3 10◦ p1 Figure S6.20 M1 p1 = = 3.0 1.0133 × 105 Pa θ β = = M2 p2 p1 ν2 = 2.505 = 2.054 = 39.24◦ θ3 ν3 10◦ 27.38◦ M3 M2 p2 p02 M3 p3 p03 = 20◦ = ν 2 + θ3 = = 39.24◦ + 20◦ 59.24◦ = = 3.545 2.505 = 0.05808 = 3.545 = 0.0123 Therefore, p3 p1 = = L = CL = = = = = = p2 p3 p02 p1 p03 p2 0.01230 = 0.435 2.055 × 0.05808 ; ; c< c< −p2 1 × − p3 1 × + p1 [1 × c] 2 2 L q1 c L γ 2 2 M1 p1 c 2 L γM12 cp1 , + ! " p2 2 p3 1 + 1 + − γM12 p1 p1 2 , + 2 1 1 − (2.055 + 0.435) 1.4 × 9 2 − 0.0389 144 Oblique Shock and Expansion Waves Drag, D = = CD = = = = c c p2 tan 10◦ − p3 tan 10◦ 2+ 2 , p1 c p2 p3 − tan 10◦ 2 p1 p1 D 1 2 2 γM1 p1 c + , p2 1 p3 − tan 10◦ γM12 p1 p1 1 (1.62 × 0.1763) 1.4 × 9 0.02267 6.21 Let subscripts 1 and 2 refer to states upstream and downstream of the expansion fan. We know that the flow across the Prandtl-Meyer expansion fan is isentropic. Therefore, p01 = p02 . From isentropic table, for M1 = 2.0, we have p1 = 0.1278, µ1 = 30◦ , and ν1 = 26.37◦ p01 Therefore, p2 = 0.5 = p1 p2 p02 p1 p01 Thus, p2 = 0.5 × 0.1278 = 0.0639 p02 p2 Again from isentropic table, for = 0.0639, we have p02 M2 = 2.44, µ2 = 24.19◦ , and ν2 = 37.71◦ Thus, the flow turning angle is θ = ν2 − ν1 = (37.71 − 26.37) = 11.34◦ For the first Mach line the angle relative to the freestream is µ1 = 30◦ . For the last Mach line the angle relative to the freestream is µ2 − θ = 24.19 − 11.33 = 12.86◦ 6.22 From oblique shock table, for M1 = 2.2 and θ = 5◦ C, we have β = 31.09719 and p2 = 1.3397 p1 145 M1n = M1 sin β = 1.136. From normal shock table, for M1n = 1.136, we have p02 = 0.99726 p01 6.23 Assuming air to be a perfect gas, γ = 1.4. From oblique shock table, for M1 = 3.0 and θ = 8◦ , from gas tables, we have the shock angle as β = 25.61◦ p2 = 1.7953, where p2 is the pressure behind the shock, which is also the p1 pressure at the cone surface. Thus, Also, p2 = 0.05 × 1.7953 = 0.0898 atm = 0.0898 × 101.325 = 9.1 kPa 6.24 Basically the given stream passes through an expansion fan and a oblique shock at the convex and concave corners, respectively, as shown in Fig. S6.24. 1 2 3 15◦ 15◦ Figure S6.24 From isentropic table, for M1 = 3.0, we have ν1 = 49.757◦ , Given, θ1 = 15◦ , thus, p1 = 0.02722 p01 ν2 = ν1 + 15◦ = 64.757◦ For ν2 = 64.757◦ , from isentropic tables, we have M2 = 3.92, p2 = 0.0073316 p02 From oblique shock chart, for M2 = 3.92 and θ2 = 15◦ , we have β = 28◦ . Therefore, M2n = M2 sin β = 3.92 sin 28 = 1.84 146 Oblique Shock and Expansion Waves From normal shock table, for M2n = 1.84, we have p3 M3n = 0.6078, = 3.7832 p2 Thus, M3 = 0.6078 M3n = sin (β − θ) sin (28 − 15) = 2.7 p3 = p3 p2 p1 p2 p1 Now, p2 p1 = p2 p02 p01 p02 p01 p1 = p2 p01 p02 p1 since p01 = p02 across an expansion fan. Therefore, p2 p1 = 0.0073316 0.02722 = 0.269346 Thus, p3 = = (3.7832)(0.269346)(1) 1.019 atm 6.25 By Ackerets theory, CL = / Also, CL = 1 2 4α 2 −1 M∞ L γ M2 A p where A = 1 × 1 = 1 m2 is the wing area. Thus, / 2 −1 C L M∞ α = 4 √ 0.228 1.62 − 1 = 4 = 0.0712 radians = 4◦ 147 Again by Ackerets theory, CD = / Thus, 4 α2 = α CL 2 −1 M∞ CD = (0.228)(0.0712) = 0.0162 The aerodynamic efficiency of the plate is CL 0.228 = 14 = CD 0.0162 6.26 For M1 = 2.4 and β = 33◦ , from oblique shock table, we get θ = 10◦ M1n = M1 sin β = 1.31 From normal shock table, for M1n = 1.31, we have p2 = 1.8354 p1 T2 T1 = 1.1972 M2n = 0.78093 M2 = 0.78093 M2n = sin (β − θ) sin 23◦ = 2.00 6.27 The pressure ratio across an oblique shock is given by 2 p2 2γ 1 2 2 M1 sin β − 1 =1+ p1 γ+1 That is, p2 −1 p1 = p2 − p1 = = p2 − p1 ρ1 = 2 2γ 1 2 2 M1 sin β − 1 γ+1 ! " 2γ 1 p1 M12 sin2 β − 2 γ+1 M1 ! " 2γ u21 1 u2 2 p1 2 sin β − 2 , since M12 = 21 γ+1 a1 M1 a1 ! " 2 γ p1 u21 1 sin2 β − 2 γ + 1 ρ1 a21 M1 148 But Oblique Shock and Expansion Waves γp ρ = a2 , from isentropic relations. Therefore, p2 − p1 4 = 1 2 γ + 1 ρ u 2 1 1 " ! 1 sin2 β − 2 M1 6.28 Given, p2 /p1 = 5. From normal shock table, for p2 /p1 = 5, we get the Mach number normal to the oblique shock at the compression corner as M1n = 2.1. But, M1n = M1 sin β where β is the shock angle. Thus, β = = " M1n M1 ! " 2.1 −1 sin = 44.427◦ 3.0 sin−1 ! From oblique shock chart, for M1 = 3.0 and β = 44.427◦ , we get θ = 25.5◦ 6.29 From oblique shock table, for M1 = 2 and θ = 10◦ , we get β = 39.3◦ Therefore, the Mach number normal to the shock becomes M1n = M1 sin β1 = 1.27 From normal shock table, for M1n = 1.27, we get M2n = 0.80167 Therefore, M2 = M2n = 1.64 sin (β1 − θ) For M2 = 1.64 and θ = 10◦ , from oblique shock chart, we get β3 = 49.5◦ Thus, the angle of the reflected shock relative to the flat wall is φ = β3 − θ = 49.5 − 10 = 39.5◦ M2n = M2 sin 49.5 = 1.25 149 From normal shock table, for M2n = 1.25, we have M3n = 0.81264 Thus, M3 = 0.81264 M3n = sin (β3 − θ) sin 39.5◦ = 1.28 6.30 For M1 = 2 and θ = 7◦ , from oblique shock table, we get M2 = 1.75 For M2 = 1.75, from oblique shock table, we have θmax = 18◦ This is the maximum of θ up to which the second shock will remain attached. 6.31 The given flow is through an oblique shock. Therefore, there are two solutions possible for the flow. One is called the weak solution, for which the flow downstream of the shock will continue to be supersonic with a reduces Mach number. The second is strong solution for which the Mach number downstream of the shock will be subsonic. I. Weak solution Let subscripts 1 and 2 refer to conditions upstream downstream of the shock. For M1 = 2 and flow turning angle θ = 15◦ , from oblique shock table, we have β = 45.34 deg and M2 = 1.45 The Mach number normal to the shock M1n = M1 sin β. M1n = 2 sin 45.34 = 1.42 Now, the shock may be treated as a normal shock with upstream Mach number 1.42. From normal shock relations, the change in entropy across the shock is given by ! " p01 ∆s = R ln p02 From normal shock table, for M1n = 1.42, we have p01 = 1.049 p02 150 Oblique Shock and Expansion Waves Thus, %s = 287 ln (1.049) = 13.73 J/(kg K) II. Strong solution For M1 = 2 and flow turning angle θ = 15◦ , from oblique shock table, we have β = 79.83◦ M2 = 0.64 M1n = 2 sin (79.83◦ ) = 1.97 From normal shock table, for M1n = 1.97, we get p01 = 1.36 p02 Thus, %s = 287 ln (1.36) = 88.25 J/(kg K) The shock will be attached up to θmax = 22.97◦ 6.32 Since the flow is turned at a compression corner, the problem is effectively getting the flow field downstream of a oblique shock. From oblique shock table(or chart), for M1 = 3.0 and θ = 10◦ , we have β = 27.38◦ The Mach normal to the shock is M1n = M1 sin β = 1.38 From normal shock table, for M1n = 1.38, we get M2n = 0.74829, p2 = 2.0551, p1 T2 = 1.2418, T1 Thus, M2 = M2n sin (β − θ) p02 = 0.96304 p01 151 p2 = 0.74829 sin (17.38◦ ) = 2.5 = (2.0551)(1) = T2 2.0551 atm = (1.2418)(200) = 248.36 K For M1 = 3, from isentropic table, p1 = 0.027224, p01 p02 = p02 p01 p1 p01 p1 = (0.96304) (1/0.027224) (1) = T02 T1 = 0.35714 T01 35.37 atm T01 T1 T1 = T01 = = 200 0.35714 = 560 K 6.33 Note that for a given Mach number and turning angle, the oblique shock can have a weak and a strong solutions. Further, the Mach number downstream of the shock is supersonic for weak solution and subsonic for the strong solution. Let us solve this problem with oblique shock table as well as with oblique shock charts. From oblique shock table, for γ = 1.4 and M1 = 3.0 and θ = 10◦ , we have For weak solution 152 Oblique Shock and Expansion Waves β = 27.38256◦ M2 = 2.505 , p2 = 2.0545 p1 For strong solution β = 86.40836◦ M2 = 0.48924 , p2 = 10.292 p1 From oblique shock chart I, for M1 = 3.0 and θ = 10◦ , we have β = 27.4◦ as weak solution β = 86.3◦ as strong solution Using oblique shock chart II, the pressure ratio across and Mach number downstream of the shock are obtained as For weak solution M2 = 2.5, p2 = 2.05 p1 M2 = 0.49, p2 = 10.3 p1 For strong solution From the above solutions the elegance of oblique shock tables is obvious. Further, the solutions obtained with oblique shock charts are only approximate whereas, the oblique shock table gives accurate results. 6.34 (a) From oblique shock table, for M1 = 2.2 and θ = 6◦ , we have β = 31.98◦ Therefore, the Mach number normal to the shock is M1n = M1 sin β = 1.16 From normal shock table, for M1n = 1.16, we have p02 = 0.99605, p01 Therefore, M2 = M2n = 0.86816 M2n = 1.98 sin (β − θ) From normal shock table, for M2n = 1.98, we have p03 = 0.73021 p02 153 Thus, p03 p02 = p03 p02 p02 p01 = (0.73021)(0.99605) = 0.727 Thus, Pressure loss = (1 − 0.727) × 100% = 27.3 per cent (b) From oblique shock table, for M2 = 1.98 and θ = 6◦ , β = 35.8◦ Therefore, the Mach number normal to the shock is M2n = M2 sin β = 1.1582 ≈ 1.16 From normal shock table, for M2n = 1.16, we have p03 = 0.99605, p02 Therefore, M3 = M3n = 0.86816 M3n = 1.7468 sin (β − θ) From normal shock table, for M3 = 1.75, we have p04 = 0.83457 p03 Thus, p04 p01 = p04 p03 p02 p03 p02 p01 = (0.83457)(0.99605)(0.99605) = 0.8279 Thus, Pressure loss = = (1 − 0.8279) × 100% 17.2 per cent 6.35 For M1 = 2.4 and β12 = 30◦ , from oblique shock chart, we have θ12 = 6.2◦ 154 Oblique Shock and Expansion Waves The Mach number normal to the incident shock is M1n = M1 sin β12 = 2.4 × sin 30◦ = 1.2 From normal shock table, for M1n = 1.2, we have M2n = 0.84217 Therefore, M2 = M2n sin (β − θ) = 0.84217 sin (30 − 6.2) = 2.09 For M2 = 2.09 and θ23 = 6.2◦ , from oblique shock chart, we have β23 = 34◦ The Mach number normal to the reflected shock is M2rn = M2 sin β23 = 2.09 sin 34◦ = 1.17 From normal shock table, for M2rn = 1.17, we get M3n = 0.86145 M3 = M3n sin (β − θ) = 0.86145 sin (34 − 6.2) = 1.84 Hence the Mach numbers upstream and downstream of the reflected shock are 2.09 and 1.84, respectively. 6.36 Let the state ahead of and behind the first shock be represented by subscripts 1 and 2, respectively, and those behind the second shock by 2 and 3, respectively. Given M1 = 2.3 and θ1 = 8◦ . For this Mach number and turning angle, from oblique shock table, we get β1 = 32.42◦ , M2 = 1.99 155 Therefore, M1n = M1 × sin β = 2.3 × sin (32.42◦ ) = 1.23 From normal shock table for Mn1 = 1.23, p02 = 0.9896 p01 The pressure loss caused by the second shock is " ! 1 1 p02 (1 − 0.9896) = 1− 2 p01 2 = Thus, 0.0052 p03 = 1 − 0.0052 = 0.9948 p02 For p03 /p02 = 0.9948, from normal shock table, M2n = 1.18 Thus, β2 ! = sin−1 = 36.37◦ M2n M2 " = sin−1 ! 1.18 1.99 " For M2 = 1.99 and β2 = 36.37◦ , from oblique shock chart I, θ2 ≈ 7◦ 156 Oblique Shock and Expansion Waves Chapter 7 Potential Equation for Compressible Flow 7.1 (a) We know from Eq. (7.6) that, T ds dn uζ + = uζ + cp T ds = i.e., T0 ds T0 ds dn ds dn = dT0 dn (1) 1 dh − dp ρ 1 dp0 ρ0 = dh0 − = cp dT0 − = cp Multiplying Eq. (1) by T0 dh0 dn = 1 dp0 ρ0 dT0 1 dp0 − dn ρ0 dn T0 , we get T T0 T0 dT0 uζ + cp T T dn (3) From Eqs. (2) and (3), we get cp 1 dp0 dT0 − dn ρ0 dn (2) = T0 T0 dT0 uζ + cp T T dn 157 158 Potential Equation for Compressible Flow i.e., − 1 dp0 ρ0 dn T0 uζ + T = ! = 1+ ! " T0 dT0 − 1 cp T dn " γ−1 2 γ − 1 2 dT0 M uζ + M cp 2 2 dn 7.2 For supersonic flows,by Eq. (7.37), we have 1 or 2 ∂ 2 φ 1 ∂φ ∂ 2 φ 2 − 2 =0 M∞ −1 − ∂x2 r ∂r ∂r ∂ 2 φ 1 ∂φ − β2 = 0 + ∂r2 r ∂r / 2 − 1. The solution for the above equation, valid over a slender where, β = M∞ body of revolution with a closed nose and arbitrary (smooth) meridional section (Ref. Liepmann and Roshko) is, ! " $ x 2 U U S && (ξ) ln (x − ξ)dξ φ(x, r) = − S & (x) ln − 2π βr 2π 0 where, S(x) = πR2 is the cross sectional area of the body at x and ξ = x − βr. Therefore, ! " , + $ x 2 ∂φ S & (x) 1 d & = U∞ − ln S (ξ) ln (x − ξ)dξ u = U∞ − ∂x 2π βr 2π dx 0 ∂φ R dR and, v = U∞ = U∞ i On the body, r = R, the pressure coefficient is ∂r r dx given by S && (x) Cp = ln π ! 2 λR " 1 d + π dx For the given body, R = $x3/2 $ x 0 && S (x) ln (x − ξ)dξ − 0≤x≤1 Therefore, S = πR2 = π$2 x3 S & (x) = dS(x) = 3π$2 x2 dx S && (x) = 6π$2 x dR dx = 3 1/2 $x 2 ! dR dx "2 159 Substituting the above relation into the CP expression, we obtain Cp = Cp $2 = Now consider the term 6 d dx $ ! " $ x 2 9 2 d ξ ln (x − ξ) dξ − $2 x + 6$ dx 0 4 β$x3/2 + ! " , $ x 2 9 d 6x ln ξ ln (x − ξ) dξ − x − ln x3/2 + 6 β$ dx 0 4 6$2 x ln =x d dx x ξ ln (x − ξ)dξ 0 = Now use the rule, ξln (x − ξ) dξ. Express this as 0 i6 $ , + $ x $ x d (x − ξ) ln (x − ξ) dξ + x ln (x − ξ) dξ − dx 0 0 a 0 f (a − t)dt = $ a f (t)dt 0 Therefore, we have 6 d dx $ x ξ ln (x − ξ) dξ 0 −6 = d −6 dx = −6 = lim ξ → 0 both since d 6 dx $ ξ2 2 +$ x 0 +! ξ ln ξ dξ − x $ " 2 x ξ2 ξ ln ξ − 2 4 0 x ln ξ dξ 0 , − x (ξ ln ξ − ξ ln (x − ξ)dξ , + d x2 x2 ln x − − x2 ln x + x2 dx 2 4 = & % x x −6 x ln x + − − x − 2x ln x + 2x 2 2 = 6x ln x − 6x Hence, Cp $2 = = Cp $2 = x ξ)0 ln ξ and ξ ln ξ tend to zero. Therefore, x 0 d dx + ! " , 2 9 6x ln − ln x3/2 + 6x ln x − 6x − x β$ 4 ! " 2 33 3 6x ln x − 6x ln x + 6x ln x − β$ 2 4 6x ln % 2 β% & − 3x ln x − 33 4 x , 160 Potential Equation for Compressible Flow Drag Coefficient The drag coefficient can be expressed as, CD S(L) = S& = ! 3 $ L Cp (x)S & (x)dx 0 3π$2 x2 Therefore, CD S(L) 3π$4 $ = 0 + " 2 33 3 3 − 3x ln x − x dx 6x ln β$ 4 ! 4 " ,L 6 4 x x4 2 33 x4 x ln −3 ln x − − 4 β$ 4 16 4 4 0 + , L L 11 3 3 2 L L ln − ln L + − 2 β$ 2 8 8 = = S(x) L π$2 x3 = Therefore, S(L) = π$2 L3 Thus, But, β = √ CD π$2 L3 3π $4 = + , L L 11 3 3 2 L L ln − ln L + − 2 β$ 2 8 8 With L = 1, we get, CD 3$2 = CD = + , 10 3 2 − ln 2 β$ 8 + , 2 15 2 3 ln − 3$ 2 β$ 8 M 2 − 1 = 1 and $ = 0.1. Therefore, CD = = 0.03 [1.5 ln 20 − 1.875] 0.0786 Chapter 8 Similarity Rules 8.1 By Prandtl-Glauert rule, we have ! dCL dα dCL dα " (dCL /dα)inc / 2 1 − M∞ ! " dCL = 0.108 deg dα M∞ =0.2 = = inc Therefore, dCL dα M∞ 0.2 0.3 0.4 0.5 0.6 0.7 0.75 1 dC 2 L dα / = 1 dC 2 0.108 per degree 2 1 − M∞ L measured 0.108 0.113 0.115 0.124 0.130 0.127 0.100 dα P −G rule 0.1080 0.1132 0.1178 0.1247 0.1350 0.1512 0.1633 8.2 For hypersonic flow, M∞ tan (θ + α) ≥ 0.5. When (θ + α) is small, M∞ (θ + α) ≥ 0.5. where, θ is the half angle at the vertex of the model and α is the angle of attack. For hypersonic similarity, M∞1 (θ1 + α1 ) M∞1 = = M∞2 (θ2 + α2 ) 10 161 162 Similarity Rules α1 θ1 = = 3 deg 3 deg M∞1 (θ1 + α1 ) = 10 × (3 + 3) × π >1 180 Hence the flow is hypersonic. When the test Mach number is small, the angle involved will be large. (a) M∞2 θ2 = = 3.0 12 deg M∞1 tan (θ1 + α1 ) M∞2 tan (θ2 + α2 ) = = 10 tan 6 deg = 1.05 3 tan (12 + α2 ) = 1.05 = = 0.35 19.3 tan (12 + α2 ) 12 + α2 Therefore, α2 = 7.3 deg (b) M∞3 = 3.0 θ3 M∞3 tan (θ3 + α3 ) tan (3 + α3 ) = = = 3 deg 3.0 tan (3 + α3 ) = 1.05 0.35 3 + α3 = 19.3 Therefore, α3 = 16.3 deg 8.3 Hypersonic similarity factor K = M θ. In order to apply the results of model to the missile, K should be same for the model and the missile. That is, K 1 = K2 that is, M 1 θ 1 = M2 θ 2 Given, M1 = 12, θ1 = 4 deg and M2 = 2.5. Therefore, θ2 = M1 θ 1 M2 163 = 12 × 4 2.5 = 19.2 deg That is, the semi-vertex angle of model = 19.2deg 8.4 For supersonic flow past thin wedge with semi-wedge angle δ, the linearized theory yields, ! " 2 dz CPu = β dx u = 2 2 tan δ ≈ δ β β Therefore, (γ + 1)1/3 CP δ 2/3 = = or 2δ 1/3 (γ + 1)1/3 / 2 −1 M∞ 4 2/3 [(γ + 1) δ] 2 2 −1 M∞ 9 : 2 1/3 CP (γ + 1) M∞ δ 2/3 where, χ = = 2 χ 2 M∞ −1 2 ]3/2 [δ(γ+1)M∞ ∗ 8.5 (a) Given M∞ = 0.3. Therefore, by Equation (8.57), we have the minimum Cp over the profile as '! * "3.5 2 + 0.2.32 2 Cp = −1 1.4 × 0.32 2.4 '! * "3.5 2.018 = 15.873 × −1 2.4 = −7.22 ∗ (b) Given M∞ = 0.4. Therefore, Cp = 2 1.4 × 0.42 '! 2 + 0.2.42 2.4 "3.5 * −1 164 Similarity Rules = 8.929 × = −3.94 '! 2.032 2.4 "3.5 * −1 Chapter 9 Two Dimensional Compressible Flows 9.1 The linearized perturbation velocity potential equation for two-dimensional supersonic flow Eq.(9.1) is β 2 φxx − φyy = 0 (1) / 2 − 1 and φ is the perturbation velocity potential. Equation where β = M∞ (1) may also be written as, ∂2φ 1 ∂2φ − =0 2 − 1 ∂y 2 ∂x2 M∞ (2) 2 Keeping in mind that (M∞ − 1) > 0 for supersonic flow, it can be visualized that equation (2) is also of the form of the classical wave equation. Hence, a solution to equation (2) can be expressed as, / / 2 − 1y) + g(x + 2 − 1y) M∞ (3) φ(x, y) = f (x − M∞ For the problem under consideration, only left running waves are present and therefore, / 2 − 1y) = 0 g(x + M∞ Thus, φ(x, y) = f (x − and / 2 − 1y) M∞ < ∂φ ; # = f (x − βy) (−β) ∂y By boundary condition, at the wall ∂φ ∂y w = U∞ dy dx x x x dyw = k(1 − ) − k = k − 2k dx l l l 165 (4) (5) 166 Two Dimensional Compressible Flows Therefore, % ∂φ x& = U∞ k − 2k ∂y l Substituting this into Eq.(5), we get, % # x& = −βf (x − βy) U∞ k − 2k l # U∞ % x& f (x) = − k − 2k β l (6) (7) Integrating Eq.(7) with respect to its argument, [Note that the argument is (x − βy), but with y = 0], we have ! " U∞ x2 f (x) = − kx − k + constant (8) β l At x = 0, f = 0, which gives constant = 0. Therefore, ! " U∞ x2 f (x) = − kx − k β l (9) Since f (x) is defined throughout the flow, not just at the wall, and because it has the form of Eq.(8a), where x represents the argument of f , Eq.(4) can be written as & % x−βy k φ(x, y) = f (x − βy) = − U∞ β (x − βy) 1 − l 9.2 λu = = λL = CL = = = ! dz dx " = u 1 −α 7 0 ≤ x ≤ 0.7c 1 − −α 0.7c ≤ x ≤ c 3 ! " dz = −α dx l $ c 2 − (λu + λL ) dx βc 0 " " , +$ 0.7c ! $ c ! $ c 1 2 1 − − α dx + −αdx − − α dx + βc 0 7 3 0.7c 0 " ! " , +! 1 2 1 − α 0.7c − + α 0.3c − αc − βc 7 3 167 = β = α = 4α β / M 2 − 1 = 2.29 2 deg = 2π = 0.0349 180 Thus, CL CD = 4 × 0.0349 2.29 = 0.06096 = 2 βc $ c 0 '! 1 2 λ2u + λ2L dx 1 −α 7 "2 ! 1 +α 3 "2 = 2 βc = 2 [0.0081583 + 0.0406787 + 0.001218] 2.29 = 2 × 0.050055 2.29 = 0.7c + 2 0.3c + α c * 0.04372 9.3 0 ≤ x ≤ 0.3c λu = = λl 0.3c ≤ x ≤ c " 0.1 −α 0.3 0.333 − α " ! 0.03 −α 0.3 0.1 − α ! = = α = 2 deg = 2 β = √ λu = = λl = = π ≈ 0.035 radian 180 8 = 2.83 " 0.1 −α 0.7 −0.143 − α " ! 0.03 −α − 0.7 −0.043 − α ! − 168 Two Dimensional Compressible Flows φ(x, y) = f (x − βy) ∂φ | ∂y y=0 = −βf & (x) = U∞ λ Therefore, U∞ λ β f & (x − βy) f & (x) = φx (x, y) = φx (x, 0) = f & (x) = − Cp = −2 − U∞ λ β φx λ =2 U∞ β For 0 ≤ x ≤ 0.3c Cpu = 2 = C pl = 0.298 × 2 λu = β 2.83 0.211 2 = 0.065 × 2 λl = β 2.83 0.046 For 0.3 ≤ x ≤ c C pu = 2 − 0.1258 = C pl = = 0.178 × 2 λu =− β 2.83 2 0.078 × 2 λl =− β 2.83 −0.0551 Chapter 10 Prandtl-Meyer Flow 10.1 Let subscripts 1 and 2 refer to conditions upstream and downstream of the Prandtl-Meyer fan. p0 = 33.5 × 105 Pa M1 = 2.0 From isentropic table, for M1 = 2.0, ν1 = 26.38 deg p1 p0 = 0.1278 p2 p0 = 1.0133 = 0.03025 33.5 where subscript 2 refer to conditions downstream of the expansion fan. From p2 isentropic table, for = 0.03025, p0 M2 = 2.93 ν2 = 48.388 deg But ν2 = ν1 + ∆θ Therefore, ∆θ = = 48.388 − 26.38 22 deg 169 170 Prandtl-Meyer Flow That is, after initial expansion, the flow direction is 22 deg with respect to nozzle axis. 10.2 For M1 = 2.3 and θ = 12 deg, from oblique shock chart, β = 36.5 deg. M1n = M1 sin β = 2.3 × sin 36.5 deg = 1.368 For M1n = 1.368, normal shock table gives: M2n = 0.7527, and Thus, M2 T2 = M2n sin (β − θ) = 1.815 = 1.235 × 500 = 617.5K T2 T1 = 1.235. For M2 = 1.815, from isentropic table, ν2 = 21 deg. Also, |θ| = 24 deg. Therefore, ν3 = ν2 + |θ| = 21 + 24 = 45 deg For ν3 = 45 deg, isentropic table gives, M3 = 2.764 , and fore, T3 = 0.396. ThereT0 T3 T3 T02 = T2 T02 T2 But T02 = T03 , thus, we have T3 T2 = 0.396 = 0.657 0.603 T3 = 0.657 × 617.5 = 405.7 K For M3 = 2.764 and θ = 12 deg, oblique shock chart gives, β = 32 deg, and M3n = 2.764 × sin 32 = 1.465. Now, for M3n = 1.465, the normal shock table 171 gives M4n = 0.7157, and T4 = 1.2938. Thus, T3 M4 = T4 = 0.7157 = 2.09 sin 20 deg 524.9 K Note: In this problem, oblique shock table may be used in places where oblique shock chart is used. 10.3 Let the subscripts 1 and 2 refer to conditions upstream and downstream of expansion. M1 = 2.0 p1 p01 = 0.1278 For isentropic compression p02 = p01 . Therefore, ! " 760 − 240 p2 = 0.1278 × = 0.0642 p02 760 + 275 =⇒ M2 = 2.44 =⇒ ν1 = 26.38 deg =⇒ ν2 = 37.71 deg or, the flow has been turned through ν2 − ν1 = 11.33 deg . Note: Compare this problem with 6.8. 172 Prandtl-Meyer Flow Chapter 11 Flow with Friction and Heat Transfer 11.1 M∗ = 1 M 1 , p1 p∗ p01 p02 Figure S11.1 This problem has to be solved by trial and error. The pressure ratio written as Given, p1 can be p∗ p1 p01 p02 p1 = p∗ p01 p02 p∗ p01 p02 = 10. We want ∗ = 1.893 (critical pressure ratio). Thus, p02 p p1 p∗ = 18.93 p1 p01 or p1 /p∗ p1 /p01 = 18.93 p1 p1 and for different M1 and check for the ratio to be 18.93. For ∗ p p01 that, use the relations Calculate p1 p∗ = 1 M1 ! 1.2 1 + 0.2M12 173 " 12 174 Flow with Friction and Heat Transfer p1 p01 = Trial ! 1+ M1 1 2 3 4 5 0.1 0.06 0.055 0.059 0.058 γ "− γ−1 γ−1 2 M1 2 % & p1 p1 p∗ / p01 10.944/0.99 18.25/0.977 19.90/0.998 18.56/0.998 18.88/0.998 = = = = = 11.05 18.29 19.95 18.60 18.92 It is seen that M1 = 0.058 is the required Mach number. Now use the relation ' * γ+1 2 M 4f¯L∗ 1 − M12 γ+1 1 2 = ln + 2 D γM12 2γ 1 + γ−1 2 M1 to calculate L∗ , 4f¯L∗ D L∗ ! 0.0040368 1.0006728 = 211.6 + 0.857 ln = 211.6 − 4.72 = 206.88 = 206.88 × 0.1 4 × 0.005 = 1034.4 m p1 = T1 p2 ṁ = = = " 11.2 L = f¯ = 3.5 × 105 mboxP a 300 K 1.4 × 105 mboxP a 0.09 kg/s 600 m 0.004 ṁ = ρ1 V1 A π p1 / γRT1 M1 D2 RT1 4 = 0.090 √ 3.5 × 105 π × 20.04 300 × × M1 D2 287 × 300 4 = 0.090 M1 D 2 = 8.12 × 10−5 175 For an isothermal flow with friction, it can be shown that ' ! "2 * ! "2 p2 p2 4f L 1 = 1− + ln 2 D γM1 p1 p1 ' * ! ! "2 "2 1.4 1.4 4 × 0.004 × 600 1 = 1 − + ln D γM12 3.5 3.5 9.6 D = 0.6 1 − 1.832 M12 From Eqs. (1) and (2), we obtain 9.6 D = 9.6 D = 9.10 × 107 D4 − 1.832 0.6 D4 − 1.832 (8.12)2 × 10−10 Solving equation (3), We obtain D = 0.0402 m . 11.3 Energy equation gives, h+ V2 = h0 = constant 2 Also, for a perfect gas, V2 a2 + γ−1 2 ! " γ−1 2 2 M a 1+ 2 ! " γ−1 2 M T1 1 + 2 = a20 γ−1 = a20 = T0 From continuity equation for a perfect gas, G = = = ṁ = ρ1 V1 = ρ1 a1 M1 A p1 / γRT1 M1 RT1 3 γ 1 √ p1 M1 R T1 From Eq. (2), we have T1 = γp21 M12 RG2 176 Flow with Friction and Heat Transfer Using this in Eq. (1), we get ! " γ p21 M12 γ−1 2 M1 1+ R G2 2 ! " γ−1 2 M1 M12 1 + 2 M1 ! γ−1 2 M1 1+ 2 = = " 12 = T0 R G2 T0 2 γ p1 4 R G T0 γ p1 Now, G = ṁ ṁ∗ = A A = Γ A∗ √ p0 γRT0 A where Γ = γ ! 2 γ+1 γ+1 " 2(γ−1) Using Eq. (4) in Eq. (3), we get, M1 ! γ−1 2 M1 1+ 2 " 12 = 4 = Γ p0 A∗ γ p1 A1 R Γ A∗ 1 T0 p0 γ γRT0 A1 p1 i.e. ! "1 γ γ−1 2 2 M1 1 + M1 Γ 2 p1 p0 = A A∗ = = p0 A∗ p1 A1 2.38 × 105 = 0.0354 67.3 × 105 ! "2 ! "2 D 0.0127 = = 2.082 = 4.33 D∗ 0.0061 Therefore, p1 A p0 A∗ = 0.0354 × 4.33 = 0.1533 177 =⇒ M1 ! =⇒ 4f L∗ D " = 2.51 = 0.4341 = 4.85 × 105 = 0.072 67.3 × 105 1 p2 p0 Therefore, =⇒ p2 A × ∗ p0 A = 0.072 × 4.33 = 0.312 =⇒ M2 = 1.53 = 0.147 ! 4f L∗ D " 2 Therefore, ! " = 0.4341 − 0.147 ∆L D = 29.60 − 1.75 = 27.85 4f L∗ D Hence, the average friction coefficient is f¯ = = 0.2871 4(27.85) 0.002577 11.4 (a) Ae /A∗ = 2 1 p0 T0 Figure S11.4a pb Shock M =1 A∗ L 178 Flow with Friction and Heat Transfer Normal shocks can exist only for M1 > 1. Therefore, the interpretation that a normal shock stands in the throat implies M = 1 at the throat and downstream is subsonic. Therefore, At = A∗ . A∗ Ae 0.50 M1 0.306 ρ1 ρ0 T1 T0 a1 a0 p1 p0 0.9547 0.9816 0.9907 0.9371 p0 7 × 105 = 8.13 kg/m3 = RT0 287 × 300 / √ γRT0 = 1.4 × 287 × 300 = 347 m/s ρ0 = a0 = ρ1 = (0.9547)ρ0 = (0.9547) × (8.13) = 7.7617 kg/m3 a1 = (0.9907)a0 = (0.9907) × (347) = 343.8 m/s p1 = (0.9371)p0 = (0.9371) × (7 × 105 ) = 6.5597 × 105 Pa % ¯ ∗& 4f L p1 M1 D p∗ M1 0.306 5.0776 3.547 p1 = 1.849 × 105 Pa 3.547 Therefore, p∗ < pB =⇒ duct length L < L∗ and p2 = pB at L. p∗ p2 p∗ = pB 2.8 × 105 = = 1.514 p∗ 1.849 × 105 % ¯ ∗& = p2 p∗ 1.514 Therefore, 4f¯L D 4f L D M2 0.220 = ! ¯ " ! ¯ " 4f L 4f L − D M1 D M2 = 5.0776 − 0.220 = G = = = 4.8576 ρ1 V1 = ρ1 a1 M1 (7.7617)(343.8)(0.306) 816.5 kg/m2 s 179 (b) 1 Shock M =1 A∗ L Figure S11.4b Upstream of normal shock wave A∗ Ae p p0 ρ ρ0 T T0 a a0 M 0.5 0.09396 0.1847 0.5088 0.7133 2.197 p = ρ = = T a = = (0.09396)(7 × 105 ) = 0.6577 × 105 Pa (0.1847)(8.13) 1.5 kg/m3 (0.5088)(300) = 152.6 K (0.7132)(347) = 247.5 m/s Downstream of normal shock wave This zone is referred by subscript 1. M 2.197 p1 p 5.465 ρ1 ρ 2.947 T1 T 1.854 a1 a 1.362 M1 0.5475 Therefore, p1 ρ1 = = (5.465)(0.6577 × 105 ) = 3.5943 × 105 Pa (2.947)(1.5) = 4.4205 kg/m3 T1 a1 = = (1.854)(152.6) = 282.9 K (1.362)(247.5) = 337.1 m/s % ¯ ∗& 4f L p1 M1 D p∗ M1 0.5475 0.7451 1.9434 180 Flow with Friction and Heat Transfer Thus, 3.5943 × 105 p1 = = 1.85 × 105 Pa 1.9434 1.9434 Since p∗ < pB =⇒ duct length L < L∗ and p2 = pB at L. p∗ = p2 pB 2.8 × 105 = = = 1.514 p∗ p∗ 1.85 × 105 % ¯ ∗& p2 p∗ 4f L D 1.514 Therefore, 4f¯L D = = = G M2 0.220 ! ¯ " ! ¯ " 4f L 4f L − D M1 D M2 0.7451 − 0.220 0.5251 = ρ1 V1 = ρ1 a1 M1 = (4.4205)(337.1)(0.5475) = 815.9 kg/m2 s (c) 1 Shock M =1 A∗ Figure S11.4c Properties at station 1 are the same as properties in Part-b. Upstream of normal shock wave Therefore, p1 ρ1 = = 0.6577 × 105 Pa 1.5 kg/m3 T1 a1 M1 = = = 152.6 K 247.5 m/s 2.197 181 M1 2.197 Thus, p∗ = % 4f¯L∗ D & p1 p∗ M1 0.36011 0.35566 p1 = 1.849 × 105 Pa. 0.35566 Therefore, p∗ < pB =⇒ duct length L < L∗ and p2 = pa at L downstream of normal shock wave. From the diagram in Fig. s11.4d, it is clear that the Mach number Mx at point x jumps. h pb = 0 28 a kP * 1 s Figure S11.4d The normal shock pressure jump and the Fanno line equation are given by = 2γ γ−1 Mx2 − γ+1 γ+1 px p∗ = 1 Mx 2.8 × 105 1.849 × 105 = Let = p3 px f (Mx ) - 1+ γ+1 2 γ−1 2 2 Mx .1/2 2 1 1 1.1667Mx2 − 0.1667 Mx 2 1 1 1.1667Mx2 − 0.1667 Mx ! ! 1.2 1 + 0.2Mx2 1.2 1 + 0.2Mx2 Use Secant method to find Mx as follows. ! " xj − xj−1 xj+1 = xj − f (xj ) f (xj ) − f (xj−1 ) Trial 1 x1 = 1.6 "1/2 "1/2 − 1.5143 182 Flow with Friction and Heat Transfer = x0 1.5 x2 = 1.6 − 0.0559 x2 = 1.53 f (x2 ) = 0.000752 ! 1.6 − 1.5 0.0559 − (−0.0236) x3 = 1.53 − 0.000752 x3 = 1.53, ! 1.53 − 1.6 0.000752 − 0.0553 Therefore, Mx = 1.53. M2 ! ¯ ∗" 4f L D M2 1.53 0.14699 Therefore, 4f¯L D = ! ¯ " ! ¯ " 4f L 4f L − D M1 D M2 = 0.36011 − 0.14699 = G 0.21312 = ρ1 V1 = ρ1 a1 M1 = (2.197)(247.5)(1.5) = 815.6 kg/(m2 s) 11.5 The speed of sound a1 is given by, a1 = = M1 = = / " √ γRT1 = 20.04 333.3 366 m/s V1 73.15 = a1 366 0.2 " 183 From tables Rayleigh flow tables, for M1 = 0.2, we have T01 T0∗ T1 T∗ p01 p∗0 p1 p∗ V1 V∗ = 0.17355 = 0.20611 = 1.2346 = 2.2727 = 0.09091 From isentropic tables, for M1 = 0.2, we have p1 = 0.97250 p01 Therefore, p01 = = T1 T01 = 0.5516 × 105 0.9725 5.672 × 104 Pa 0.99206 Therefore, T01 = 333.3 = 336 K 0.99206 (a) T02 ∆h = = stagnation temperature after combustion cP (T02 − T01 ) Therefore, = T02 = = ∆h cP 1395.5 × 103 336 + 1004.5 1725.2 K T01 + (b) T02 T01 = 1725.2 = 5.1345 336.0 184 Flow with Friction and Heat Transfer T02 T0∗ =⇒ M2 T2 T∗ p02 p∗0 p2 p∗ V2 V∗ = T01 T02 T0∗ T01 = (0.17355)(5.1345) = 0.89109 = 0.68 = 0.98144 = 1.0489 = 1.4569 = 0.67366 (c) T2 = T2 T ∗ T1 T ∗ T1 = 0.98144 × 333.3 0.20661 = 1583.2 K (d) p02 ∆p0 = p02 p∗ p01 p∗ p01 = 1.0489 × 5.672 × 104 = 4.819 × 104 Pa 1.2346 = p02 − p01 = (4.819 − 5.672) × 104 −8530 Pa = (e) s2 − s1 = = = cp ln ! T02 T01 " − R ln ! p02 p01 " 1004.5 ln (5.1345) − 287 ln 1690.1 J/kg.K ! 4.819 5.672 " 185 (f ) V2 = V2 V ∗ V1 V ∗ V1 = 0.67366 × 73.15 0.09091 = 542.1 m/s (g) The initial conditions can be maintained till the flow is choked at the duct T01 exit after combustion. That is, M2 = 1 and TT02∗ = 1 and since ∗ = 0.17355, 0 T0 and T01 = 336K, we have T02 = T02 T0∗ T01 T0∗ T01 = 336.0 0.17355 = 1936 K Maximum heat of reaction q is given by q = = cP ∆T0 = 1004.5(1936 − 336) 1607.2 kJ/kg 11.6 The flow process is adiabatic and therefore, it can be treated as a Fanno flow. The velocity at station 1 is / V 1 = M1 a 1 = M1 γ R T 1 = √ 0.2 1.4 × 287 × 300 = 69.44 m/s From Fanno flow table, for M1 = 0.2, we have p1 = 5.4554, p∗ T1 = 1.1905, T∗ V1 = 0.21822, V∗ p01 = 2.9635 p∗0 Again, from Fanno flow table, for M2 = 0.5, we have p2 = 2.1381, p∗ T2 = 1.1429, T∗ V2 = 0.53452, V∗ p02 = 1.3398 p∗0 186 Flow with Friction and Heat Transfer Thus, p2 T2 V2 p02 = p2 p∗ 2.1381 × 5 × 101325 p1 = p∗ p1 5.4554 = 198.558 kPa = T2 T ∗ 1.1429 × 300 T1 = ∗ T T1 1.1905 = 288 K = V2 V ∗ 0.53452 × 69.44 V1 = V ∗ V1 0.21832 = 170.09 m/s = p02 p∗0 1.3398 × 520.951 p01 = p∗0 p01 2.9635 = 235.52 kPa 11.7 For, M1 = 0.2, from isentropic table, we have, T1 = 0.99206 T01 Therefore, 72 + 273.15 T1 = = 347.9 K 0.99206 0.99206 Since the tube is perfectly insulated, T01 = T02 , thus, T01 = T02 = 347.9 K The initial density is ρ1 = p1 2 × 101325 = R T1 287 × 345.15 = 2.046 kg/m3 since, 1 atm = 101325 Pa. Thus, the mass flow rate through the tube is ṁ = ρ1 A1 V1 = ρ1 A1 M1 a1 = 2.046 × (0.1 × 0.1) × 0.2 × / γ R T1 187 = √ 0.004092 1.4 × 287 × 345.15 = 1.524 kg/s Now, assuming a control volume between the sections 1 and 2, we can write the force balance equation as (p1 − p2 ) A + F = ρ1 A1 V1 (V2 − V1 ) where F is the frictional drag and A is the cross-sectional area of the tube. For M2 = 0.76, from isentropic table, we have T2 = 0.89644 T02 Therefore, T2 = 0.89644 T02 = 0.89644 × 347.9 = 311.87 K a2 = √ 1.4 × 287 × 311.87 = 354 m/s V2 = M2 a2 = 0.76 × 354 = 269.04 m/s For obtaining p2 , let us use the Fanno flow table. For M1 = 0.2 and M2 = 0.76, respectively, from Fanno flow table, we have p1 = 5.4554, p∗ p2 = 1.3647 p∗ Therefore, p2 = p2 p∗ p1 p∗ p1 = 1.3647 × 2 = 0.5 atm 5.4554 Thus, F = ρ1 A1 V1 (V2 − V1 ) − (p1 − p2 ) A 188 Flow with Friction and Heat Transfer = 1.524(269.04 − 74.52) − (2 − 0.5) × 101325 × (0.1 × 0.1) = 296.45 − 1519.88 = − 1223.43 N Hence, Drag = 1223.43 N 11.8 Given, L/D = 50, V1 = 195 √ m/s, T1 = 310 K and Me = 1. The speed of sound at the entrance is a1 = γRT1 . For carbon dioxide, γ = 1.3 and molecular weight is 44, thus, R 8314 Ru = M 44 188.95 J/(kg K) = = Therefore, a1 = = M1 = = √ 1.3 × 188.95 × 310 275.95 m/s V1 195 = a1 275.95 0.71 From Fanno flow table for γ = 1.3, for M1 = 0.71, we have 4 f Lmax D = 0.20993 Thus, f = = 0.20993 50 × 4 0.00105 11.9 From Fanno flow table, for M2 = 0.8, we have 4 f L∗2 D p2 p∗ T2 T∗ 4 f L∗1 D = 0.07229 = 1.2893 = 1.0638 = = 4 f L∗2 4 fL + D D 4 × 0.005 × 51 0.07229 + = 40.87229 0.025 189 From Fanno flow table, for 4 f L∗ 1 D = 40.87229, we have M1 = 0.13 p1 p∗ = 8.457 T1 T∗ = 1.1959 p1 = p1 p∗ 8.457 ×1 p2 = ∗ p p2 1.2893 = 6.56 atm = T1 T ∗ 1.1959 × 270 T2 = T ∗ T2 1.0638 = 303.52 K T1 11.10 By energy equation, we have V12 2 Treating air as a perfect gas, we can express h = cp T and hence h0 = h1 + c p T0 = c P T1 + V12 2 For air cp = 1004.5 J/(kg K). Therefore, T1 a1 = T0 − = ! = = V12 2cp " 1352 K = 349.9 K 2 × 1004.5 / √ γRT1 = 1.4 × 287 × 349.9 359 − 374.95 m/s Therefore, M1 = V1 135 = 0.36 = a1 374.95 (a) For M1 = 0.36, from Fanno flow table, we have 4f Lmax = 3.1801 D 190 Flow with Friction and Heat Transfer Thus, Lmax = 3.1801 × 5 × 10−2 4 × 0.02 = 1.99 m This is the minimum length of the tube for the flow to choke. (b) For L2 = 0.6 m, ! 4f L D " = 2 ! 4f L D " 1 − ! 4f L D " 12 where 1 and 2 stands for the inlet and exit of the tube ! " 4f L 4 × 0.02 × 0.6 = 3.1801 − D 2 5 × 10−2 3.1801 − 0.96 = 2.22 = The corresponding Mach number, from Fanno flow table, is M2 ≈ 0.41 and For M1 = 0.36, from Fanno flow table, p02 p02 = 1.5587 p∗0 p02 = 1.7358. Therefore, p∗0 = p02 p∗0 p01 p∗0 p01 = 1.5587 × 135 1.7358 = 121.23 kPa For M1 = 0.36, from isentropic table, p1 p01 = 0.91433 p1 = 0.91433 × 135 = 123.43 kPa ρ1 = p1 123.43 × 103 = RT1 287 × 349.9 = 1.229 kg/m3 191 Thus, the mass flow rate through the tube ṁ is ṁ ρ1 A1 V1 = 1.229 × 135 × = = π × 52 × 10−4 4 0.326 kg/s 11.11 For hydrogen, M = 2.016 and γ = 1.4 R= 8314 = 4124 J/(kg K) 2.016 At the tube inlet, the Mach number is M1 = V1 V1 200 =√ =√ a1 γRT1 1.4 × 4124 × 303 = 0.15 From Fanno flow table, for M1 = 0.15, we get 4f¯Lmax D p1 p∗ = 27.932 = 7.2866 Thus, the tube length required for the flow to choke is Lmax pexit = 27.932D 4f¯ = 27.932 × 25 × 10−3 4 × 0.03 = 5.82 m p1 7.286 = p∗ = = 250 7.2866 = 34.31 kPa 11.12 At the pipe entrance, Mach number M1 = V1 = 200 m/s V1 a1 , where 192 Flow with Friction and Heat Transfer and a1 = / = 349 m/s γRT1 = √ 1.4 × 287 × 303.15 Thus, 200 = 0.57 349 From Fanno flow table, for M1 = 0.57, we have M1 = 4f Lmax = 0.62288 D Therefore, Lmax = 0.62288 × 20 × 10−3 = 0.156 m 4 × 0.02 Thus, the length of the pipe at which the flow would be sonic is 15.6 cm 11.13 For methane, γ = 1.3, R M1 = 8314 Ru = M 16.04 = 518 J/(kg K) = V1 V1 =√ a1 γRT1 = √ = 0.0538 25 25 = 464.2 1.3 × 518 × 320 From Fanno flow equations, we have 4f¯L∗1 D = = = 1 − M12 (γ + 1) M12 γ+1 2 ln 1 + 2 2 γM1 2γ 2 1 + γ−1 2 M1 ! " 0.00666 0.997 + 0.8846 ln = 265.16 − 5.046 0.00376 2 260.114 The pipe length at which the flow chokes is L∗ = 260.114 × 25 × 10−3 4 × 0.004 = 406.4 m 193 Again by Fanno flow relations, we have ' * 12 p1 1 γ+1 2 2 1 = p∗ M1 2 1 + γ−1 M1 2 = 1 0.0538 = 19.93 ! 2.3 2 " 12 Thus, p∗ T1 T∗ = 1 × 106 p1 = 19.93 19.93 = 50.18 kPa = γ+1 2.3 2= 1 2 2 2 1 + γ−1 M 1 2 = 1.15 Thus, T∗ V∗ = 320 T1 = 1.15 1.15 = 278.16 K = a∗ = = / γRT ∗ = √ 1.3 × 518 × 278.26 432.87 m/s 11.14 For argon, γ = 1.67. The given flow is a Fanno flow. From Fanno flow table, for M1 = 0.6, we have " ! 4f Lmax = 0.4908 D 1 p1 p∗ = 1.76336 T1 T∗ = 1.1194 For the given duct, 4f L 4 × 0.02 × 1.1194 = = 0.2984 D 0.3 194 Flow with Friction and Heat Transfer Also, 4f L D = = ! 4f Lmax D " 1 − ! 4f Lmax D " 2 0.2984 where subscript 2 refers to duct exit. From the above equation, we have ! " 4f Lmax = 0.4908 − 0.2984 D 2 For ! 4f Lmax D " = 0.1924 = 0.1924, from Fanno flow table, we get 2 M2 = 0.73 p2 p∗ = 1.426 T2 T∗ = 1.084 Thus, p2 T2 = p2 p∗ 1.426 × 90 p1 = ∗ p p1 1.76336 = 72.81 kPa = T2 T ∗ 1.084 × 300 T1 = T ∗ T1 1.1194 = 290.62 K 11.15 %L = D 4f -! 4f l D " M1 − ! = D (14.533 − 0) 4f = 50 × 10−3 × 14.533 4 × 0.006 = 30.277 m 4f l D " M2 . 195 11.16 A = π × 0.12 = 78.539 × 10−4 m2 4 ρ = p = 1.94 kg/m3 RT Therefore, the inlet velocity becomes V = ṁ = 177.2 m/s ρA M = 0.49 From Fanno flow table, for M = 0.49, we have 4f l D = 1.1539 p p∗ = 2.1838 T T∗ = 1.145 Thus, L p∗ T∗ = 1.1539 × 40.1 4 × 0.006 = 4.8 m = 1.8 × 105 2.1838 = 82.4 kPa = 323.15 = 282.23 K 1.145 = 9.08◦ C At half way before the chocking location, L = 2.4 m. Thus, 4f L 4 × 0.006 × 2.4 = = 0.576 D 0.1 From Fanno flow table, for 4f L D p p∗ = 0.576, we get = 1.8282 196 Flow with Friction and Heat Transfer T T∗ = p = 150.6 kPa T = 44.19◦ C 1.1244 11.17 (a) From Fanno flow table, for M = 0.2, we have 4f Lmax = 14.533 D where Lmax the distance from the Mach 0.2 location at which the Mach number becomes unity. Therefore, Lmax = 14.533 × 0.05 4 × 0.00375 = 48.44 m (b) This problem has to be solved by finding the chocking location for initial Mach numbers 0.2 and 0.6. LM =0.6 = (Lmax )M =0.2 − (Lmax )M =0.6 From Fanno flow table, for M = 0.6, we have 4f Lmax D = 0.49082 Lmax = 0.49082 × 0.05 = 1.64 m 4 × 0.00375 Thus, LM =0.6 = 48.44 − 1.64 = 46.8 m 11.18 (a) Given, T0 = 380 K. By energy equation, we have h0 = h1 + V12 2 where h0 is the stagnation enthalpy and h1 and V1 are the static enthalpy and velocity, respectively, at the duct entrance. Treating air to be a perfect gas, we 197 have, h0 = cp T0 and h1 = cp T1 . Therefore, the energy equation becomes, T0 = T1 + V12 2 cp T1 = T0 − V12 302 = 380 − 2 cp 2 × 1004.5 = 379.6 K since cp = 1004.5 J/(kg K) for air. The speed of sound is / a1 = γ R T1 = 390.54 im/s M1 = V1 30 ≈ 0.08 = a1 390.54 From Fanno flow table, for M1 ≈ 0.08, we have ! " 4f L = 106.72 D 1 Thus, D = 4 × 0.02 × 55 m 106.72 = 4.12 cm (b) The inlet velocity V1 = 90 m/s. Therefore, T1 = 380 − a1 = / M1 = 902 = 376 K 2 × 1004.5 γ R T1 = 388.7 m/s 90 = 0.23 388.7 From Fanno flow table, for M1 = 0.23, we have 4f L D = 10.416 D = 4 × 0.02 × 55 1046 = 0.422 m 198 Flow with Friction and Heat Transfer (c) The inlet velocity V1 = 425 m/s. Therefore, T1 = 380 − a1 = / 4252 = 290.1 K 2 × 1004.5 γ R T1 = 341.4 m/s 425 = 1.24 341.4 From Fanno flow table, for M1 = 1.24, we have M1 = 4f L D = 0.04547 D = 4 × 0.02 × 55 0.04547 = 96.77 m Note: For the supersonic flow at the duct entrance, the diameter comes out to be more than the length, for the present data. 11.19 The hydraulic diameter of the duct is Dh = 4 × 0.03 × 0.03 4 × cross-sectional area = perimeter 4 × 0.03 = 0.03 m At the duct entrance the flow velocity is V1 = 1000 m/s. The local speed of sound is / √ a1 = γRT1 = 1.4 × 287 × 350 = 375 m/s Thus, the Mach number at duct entrance is M1 = V1 1000 = 2.67 == a1 375 For M1 = 2.67, from Fanno flow table, we have ! " 4f Lmax = 0.46619 D 1 Therefore, the duct length required to decelerate the flow to Mach 1.0 is Lmax = 0.46619 × 0.03 4 × 0.0025 = 1.40 m 199 11.20 For mass flow rate ṁ to be maximum, the exit Mach number M2 should be unity, i.e. the flow is choked. For the given duct, 4f L D = 4 × 0.023 × 0.25 0.025 = 0.92 4f L D , For this value of if the flow at the exit has to choke, from Fanno flow table, we have M1 = 0.52. From isentropic table, for M1 = 0.52, we have p1 = 0.83165 p01 T1 T01 = 0.94869 p1 = 0.83165 × 50 × 101325 = 4.21 MPa T1 = 0.94869 × 320 = 303.58 K Therefore, ṁmax = ρ1 A1 V1 = ρ1 A1 M1 a1 = 48.32 × = / 22 π1 25 × 10−3 × 0.52 × γRT1 4 √ 0.012334 1.4 × 285 × 303.58 = 4.3 kg/s Further, p∗ p1 = 1 = 0.487 2.0519 Thus, p∗ = 0.487 × 4.21 = 2.05 MPa Therefore, the mass flow rate will remain maximum for the back pressure range 0 < pb < 2.05 MPa 11.21 (a) Let subscripts 1 and 2 refer to duct entry and exit conditions. The duct length ∆L required to accelerate the from Mach 0.2 to 0.5 can be determined from ! ∗" ! ∗" fL fL f ∆L = − D D M =0.2 D M =0.5 200 Flow with Friction and Heat Transfer where f is friction factor, D is duct diameter and M is Mach number. From Fanno flow table, we have ! ∗" fL = 14.533 D M =0.2 ! ∗" fL = 1.0691 D M =0.5 Thus, 0.025∆L 30 × 10−3 ∆L = = 14.533 − 1.0691 16.157 m (b) From Fanno flow table, we have ! ∗" fL D M =1 ≈ 0 Thus, f ∆L D = 14.533 ∆L = 14.533 × 30 × 10−3 0.025 = 17.44 m 11.22 For the given pipe, we have 4f Le De = 4 × 0.02 × 18 5 × 10−2 = 28.8 For 4fDLe e = 28.8, from Fanno flow table, Me = 0.15. The speed of sound at pipe exit is √ ae = 1.4 × 287 × 468 = 433.74 m/s The exit velocity and density are Ve = 0.15 × 433.74 = 65.062 m/s ρe = pe 101325 = 0.754 kg/m3 = RTe 287 × 468 201 Therefore, ṁ = = ρe Ae Ve = 0.754 × π × 0.052 × 65.062 4 0.096 kg/s 11.23 Let subscripts 1 and 2 refer to conditions at inlet and exit of the tube, respectively. Given, pe = 0 Pa, T01 = 300 K and p01 = 6 × 101325 Pa, since 1 atm = 101325 Pa. 2 is much lower than the choking limit of 0.48667(for γ = (a) L = 0 m and pp01 1.67), therefore, the flow is choked. Thus, the mass flow rate is 22 1 0.7266 × 6 × 101325 π 30 × 10−2 √ ṁ = 4 RT01 For argon, molecular weight is 39.944, gas constant R is 208 J/(kg K) and γ = 1.67. Therefore, the mass flow rate becomes ṁ = 125 kg/s (b) L = 2.22 m. Therefore, 4f Lmax D = 4 × 0.005 × 2.22 30 × 10−2 = 0.148 From Fanno flow table, for γ = 1.67 and 4f Lmax D = 0.148, we have M1 = 0.71. Now, from isentropic table, for M1 = 0.71(γ = 1.67), we have p1 = 0.67778 p01 T1 T01 = 0.85552 p1 = 6 × 101325 × 0.67778 = 412056.35 Pa T1 = 300 × 0.85552 = 256.66 K The density at the inlet is ρ1 = p1 412056.35 = RT1 208 × 256.66 = 7.72 kg/m3 202 Flow with Friction and Heat Transfer Thus, the mass flow rate is ṁ = ρ1 A1 V1 = ρ1 A1 M1 a1 = 7.72 × = √ π0.32 × 0.71 1.67 × 208 × 256.66 4 115.68 kg/s 11.24 Given, p1 /p2 = 15 and p1 = 150 atm. Therefore, p2 = 150 = 10 atm 15 where p1 is the storage tank static pressure and p2 is the settling chamber static pressure. Let p2 be the pressure at which the flow chokes, i.e. p2 = p∗ . Thus, p1 = 15 p∗ From Fanno flow table, for p1 /p∗ = 15, we have 4f L∗ D = 140.66 Thus, L∗ = 140.66 × 0.1 4 × 0.005 = 703.3 m That is, the pipe will choke at a length of 703.3 m. 11.25 For the given piping system, p01 600 = 13.33 = p4 45 This pressure ratio is much more than the pressure ratio required for the flow to choke. Therefore, at the exit, the Mach number M4 can be taken as unity. Also, ! " 4f L 4 × 0.013 × 1.1 = D B 3.10 × 10−2 = 1.845 203 For this value, from Fanno Table, we get M3 ≈ 0.43. From isentropic table, for M3 = 0.43, we have A3 A∗ = 1.5 Also, A2 A3 = ! A2 A2 ∗ = A2 A3 , A3 A3 ∗ "2 6 3.1 = 3.75 since A2 ∗ = A3 ∗ Thus, A2 A2 ∗ 3.75 × 1.5 = 5.625 = From isentropic table, for AA22∗ = 5.625, we have M2 ≈ 0.1 and from from Fanno flow table, for M2 = 0.1, we have 4f L = 66.922 D For tube A, ! 4f L D " = 4 × 0.015 × 0.9 6 × 10−2 = 0.9 A Also, ! 4f L D " = 1−2 ! 4f L D " 1 − ! 4f L D " 2 Therefore, ! 4f L D " = 1 ! 4f L D " + 1−2 ! 4f L D = " = 0.9 + 66.922 2 67.822 % & From Fanno flow table, for this 4fDL = 67.822, we have M1 ≈ 0.1. From isentropic table, for M1 = 0.1, we get ρ1 ρ01 = 0.99502 T1 T01 = 0.998 204 Flow with Friction and Heat Transfer The stagnation density is ρ0 = p0 600 × 103 = RT0 287 × 390 = 5.36 kg/m3 Thus, ρ1 = 0.99502 × 5.36 = 5.33 kg/m3 T1 = 0.998 × 390 = 389.22 K a1 = / V1 = M1 a1 = 39.55 m/s A1 = π × 0.062 = 28.27 × 10−4 m2 4 γRT = 395.5 m/s Thus, the mass flow rate is ṁ = = ρ1 A1 V1 = 5.33 × 28.27 × 10−4 × 39.55 0.596 kg/s Chapter 12 MOC 205 206 MOC Chapter 13 Measurements in Compressible Flow 13.1 (a) p = (760 − 500) = 260 mm of Hg p0 = (760 − 350) = 410 mm of Hg p p0 = 260 = 0.634 410 From isentropic table the corresponding Mach number is M = 0.835 (b) p0 = 760 + 275 = 1035 mm of Hg p p0 = 260 = 0.251 1035 From isentropic table the corresponding Mach number is M = 1.56 13.2 pgauge = 3.6 × 104 Pa patm = 0.756 × 9.81 × 13.6 × 103 = 1.0086 × 105 Pa 207 208 Measurements in Compressible Flow = 50 cm of Hg = 0.5 × 9.81 × 13.6 × 103 = 6.67 × 104 Pa T0 = 300 K p0,gauge = 1.027 × 105 Pa p0abs = (1.027 + 1.0086) × 105 Pa p0 − p Therefore, ρ0 = p0 RT0 = 2.0356 × 105 287 × 300 = 2.36 kg/m3 (a) By compressible Bernoulli equation, we have γ p u2 + 2 γ−1ρ = γ p0 γ − 1 ρ0 = ! p p0 = ! 1.3686 × 105 2.0356 × 105 ρ = 0.753 × 2.36 = 1.78 kg/m3 u2 2 = 3.5 u = ρ ρ0 " γ1 ! 1 " 1.4 = 0.753 2.0356 1.3686 − 2.36 1.78 " × 105 256.06 m/s (b) For incompressible flow, ρ = ρ0 . Therefore, u = = 4 2 (p0 − p) = ρ 237.55 m/s 3 2 × 6.67 × 104 2.364 209 Note: The error committed in assuming the flow to be incompressible in this problem is 7.23%. 13.3 From isentropic table, for M1 = 0.9, we have p1 = 0.5913 p0 For M2 = 0.2, again from isentropic table, we have p2 = 0.9725 p0 Therefore, p2 p1 = 0.9725 = 1.6448 0.5913 p2 = 1.6447 × 4.15 × 105 = 6.826 × 105 Pa Therefore, p2 − p1 = (6.826 − 4.15) × 105 = 2.676 × 105 Pa 13.4 (a) T = 500◦ C = 773 K. Therefore, √ a = 1.4 × 287 × 773 = 557.3 m/s M = 400 V = = 0.718 a 557.3 From isentropic table, for M = 0.718, we have p0 = = p0 p = 1.4098. Therefore, 1.4098 × 1.01325 × 105 1.428 × 105 Pa (b) T = −50◦ C = 273 − 50 = 223 K a = √ M = p0 = p0 = 1.4 × 287 × 223 = 299.33 m/s 400 = 1.336 299.33 <3.5 ; 2 p 1 + 0.2 (1.336) 2.949 × 105 Pa 210 Measurements in Compressible Flow Note: Note the time saving in using tables instead of actual relations. 13.5 (a) From Standard atmospheric table, at 10, 000 m altitude, T∞ = 223.15 K. The speed of sound is given by, √ a∞ = 1.4 × 287 × 223.15 = 299.44 m/s The aeroplane velocity is V∞ = 900 = 250 m/s 3.6 The flight Mach number is M∞ = V∞ = 0.835 a∞ From isentropic table, for M∞ = 0.835, we get T0 = 1.139 T∞ Thus, the temperature at the stagnation region is T0 = 1.139 × 223.15 = 254.17 K (b) The temperature caused by impact is given by, ∆T = T0 − T = 254.17 − 223.15 = 31.02 K 13.6 The test section Mach number is M = 4. At 1650 m, from standard atmospheric table, ρ0 = 1.0425 kg/m3 . From isentropic table, for M = 4, we have ρ = 0.0277 ρ0 Therefore, the test-section density is ρts = 0.0277 × 1.0425 = 0.0289 kg/m3 13.7 The probe measures the stagnation temperature T0 as 100◦ C. That is, T0 = 100 + 273.15 = 373.15 K By energy equation, we have h0 = h+ V2 2 211 c p T0 = cp T + T0 = T+ T = T0 − V2 2 V2 2 cp V2 2 cp where T is the actual temperature(static temperature) of the air. For air cp = 1004.5 J/(kg K). Therefore, T 2502 = 373.15 − 31.11 2 × 1004.5 = 373.15 − = 342.04 K = 68.9◦ C 13.8 Let the subscripts ‘TS’, ‘i’ and ‘0’ refer to the test-section, nozzle inlet and stagnation state, respectively. Given MTS = 2.5 and pTS = 100 kPa. From isentropic table, for MTS = 2.5, pTS ATS = 0.0585, = 2.637 p0 Ath where Ath is the nozzle throat area. Therefore, Ai Ath p0 ATS Ath = 2 = 2 × 2.637 = 5.274 = 100 0.0585 = 1.71 MPa For Ai /Ath = 5.274, from isentropic table (subsonic solution), Mi ≈ 0.11 , pi = 0.9916 p0 212 Measurements in Compressible Flow Thus, the pressure at the nozzle inlet is pi = 0.9916 × 1.71 = 1.696 MPa