Compressible - Flow, Homework Solutions

May 6 Compressible-Flow-Homework Solutions 11.67 The stagnation pressure ratio across a normal shock in an ideal gas flow is 0.6. Determine the Mach number of the flow entering the shock if the gas is air. For air k = 1.4 and we can use Figure D-4 to get the Mach number, Max, from the stagnation pressure ratio, p0.y/p0.x. Alternatively we can solve equation 11.156 for p0.y/p0.x as a function of Max by an iterative technique to find the value of Max that gives p0.y/p0.x = 0.6. The latter approach gives Max = 2.26 for p0.y/p0.x = 0.6. 11.68 Just upstream of a normal shock in an ideal gas flow, Ma = 3.0, T = 600 R, and p = 30 psia. Determine the values of Ma, T0, T, p0, p, and V downstream of the shock if the gas is (a) air; (b) helium. This solution will use equations for both air and helium. Note that for air we can use Figure D.4 for k = 1.4 to obtain these results for air, except for the stagnation temperature. For air, k = 1.4 and R = 1716 ft·lbf/slug·R; for helium, k = 1.66 and R = 1.242x104 ft·lbf/slug·R. The downstream Mach number, May is found from equation 11.149. For air with k = 1.4: Ma y = For helium with k = 1.66: 2 2 32 + 1.4 − 1 = 0.375 k −1 = 2 2 2Ma x 23 −1 −1 1.4 − 1 k −1 Ma x2 + Ma y = ( ) 2 2 32 + 1.66 − 1 = 0.521 k −1 = 2 2Ma x 2 32 −1 −1 1.66 − 1 k −1 Ma x2 + ( ) The value of T0 is constant across a shock. The upstream value of T0 can be calculated from equation 11.56 using the data given for Tx and Max. This is the same as the downstream value. This relationship for air can be found in Figure D.1, not D.4.)  k −1   1.4 − 1 2  For air with k = 1.4: T0 = T 1 + Ma 2  = (600 R )1 + 3  2 2     −1 = 1680 R  k −1   1.66 − 1 2  For helium with k = 1.66:: T0 = T 1 + Ma 2  = (600 R )1 + 3  2 2     −1 = 2382 R The temperature ratio across a shock is found from equation 11.151; the results for air and helium are shown below. Page 2  Ty  T  x  Ty  T  x ( )     Air 2    1.4 − 1 2  2(1.4) 3 2   k −1 2  2kMa x Ma x  − 1 1 + − 1 3  1 +    2 2   k − 1  1.4 − 1   = 2.679 = = 2 2 2 2 (1.4 + 1) 3 − 1 (k + 1) Ma x −1 2(1.4 − 1) 2(k − 1)     He 2    1.66 − 1 2  2(1.66) 3 2   k −1 2  2kMa x Ma x  − 1 1 + − 1 3  1 +    2 2   k − 1  1.66 − 1   = 3.643 = = 2 2 2 2 (1.66 + 1) 3 − 1 (k + 1) Ma x −1 2(1.66 − 1) 2(k − 1) ( ) ( ) ( ) ( ) ( ) Multiplying these ratios by Tx = 600 R gives Ty = 1607 R for air; Ty = 2186 R for helium The pressure ratio across a shock is found from equation 11.151. ( ) 2kMa x2 − k + 1 2(1.4 ) 32 − 1.4 + 1 For air: p y = p x = px = (30 psia ) = 310 psia 1.4 + 1 px k +1 py ( ) 2kMa x2 − k + 1 2(1.66 ) 3 2 − 1.66 + 1 For helium: p y = p x = px = (30 psia ) = 330 psia px k +1 1.66 + 1 py We can find the stagnation pressure after the shock by using equation 11.59 for the ratio of pressure to stagnation pressure. Note that this equation requires the use of the local (downstream) Mach number. The results for air and helium are. p 0, y = p y p 0, y = p y p 0, y py p 0, y py  k −1  = p y 1 + Ma 2  2   k /( k −1)   1.4 − 1 0.375 2  = (310 psia )1 + 2   1.4 /(1.4−1)  k −1  = p y 1 + Ma 2  2   k /( k −1)  1.66 − 1  = (330 psia )1 + 0.5212  2   1.66 /(1.66−1) For air, p0,y = 362 psia; for helium, p0,y = 409 psia The speed is simply found as the product of the Mach number and the sound speed. For air:  1716 ft ⋅ lb f V = cMa = Ma kRT = 0.375 1.4  slug ⋅ R  1 slug ⋅ ft (1607 R ) =934 ft/s  2 ⋅ lb s  f  1.242 x10 4 ft ⋅ lb f For helium: V = cMa = Ma kRT = 0.521 1.66  slug ⋅ R  11.71  (2186 R )1 slug ⋅ ft =3500 ft/s  lb f ⋅ s 2  An aircraft cruises at a Mach number of 2.0 at an altitude of 15 km. Inlet air is decelerated to a Mach number of 0.4 at the engine compressor inlet. A normal shock occurs in the inlet diffuser upstream of the compressor inlet at a section where the Mach number is 1.2. Page 3 For isentropic diffusion, except across the shock, and for standard atmosphere determine the stagnation temperature and pressure of the air entering the engine compressor. This is the problem of an isentropic flow from an inlet condition of Mach 2 to a shock at Mach 1.2, followed by another isentropic flow to Mach 0.4. The entire process takes place in the diffuser of the engine; the “inlet” for this flow is the diffuser inlet and the outlet is the diffuser outlet whidch is the same as the compressor inlet. The diffuser inlet conditions are standard air at an altitude of 15 km for which we can find the following properties for the standard atmosphere in Table C.2: T = –56.50oC = 216.65 K and p = 1.211x104 N/m2 = 12.11 kPa. Since the stagnation temperature is constant for both isentropic flows and across a normal shock, the stagnation temperature at the entry to the compressor at (Ma = 0.4) is the same as the stagnation temperature at the Ma = 2 inlet where T = 216.65 K. We can find this stagnation temperature from equation 11.56 using k = 1.4 for air. (We could also use Figure D.1 for isentropic flow to find T0.)  k −1   1.4 − 1 2  T0 = T 1 + Ma 2  = (216.65 K )1 + 2  2 2     −1 = 390 K To find the stagnation pressure at the compressor inlet we have to analyze the three sections of the flow: (1) isentropic to the shock, (2) across the shock, and (3) isentropic to the compressor inlet. The isentropic flow has the same stagnation pressure so we can use equation 11.59 to relate two different pressures in the same isentropic flow. In particular, the pressure just upstream of the shock wave, p2, can be found from the inlet pressure, p1 = 12.11 kPa. k /( k −1) k /( k −1)  k −1   k −1  p0, upstream = p1 1 + Ma12  Ma22  = p 2 1 + ⇒ 2 2     k /( k −1) 1.4 /(1.4 −1) 1.4 − 1 2     k −1 = 39.07 kPa 2  Ma12   1+ 1+ 2 2   p2 = p1  = (12.11 kPa )  1 + 1.4 − 11.2 2   1 + k − 1 Ma 2     2  2 2     We can use equation 11.151 for the pressure ratio across a shock to get p3, the start of the next isentropic flow region. p3 = p 2 py px = p2 ( ) 2kMa x2 − k + 1 2(1.4) 1.2 2 − 1.4 + 1 = (39.07 kPa ) = 59.13 kPa 1.4 + 1 k +1 We can find the Mach number, Ma3, downstream from the shock by using equation 11.149. Ma3 = 2 2 1.2 2 + k −1 = 1.4 − 1 = 0.842 2 2 2Ma2 2 1.2 −1 −1 1.4 − 1 k −1 Ma22 + ( ) The stagnation pressure at this point, the start of the isentropic flow to the compressor inlet, will be the same as the stagnation pressure at the end of the isentropic flow, the compressor inlet. Thus we can use equation 11.59 that relates pressure to stagnation pressure to find the common stagnation pressure for the isentropic flow, which will be the desired stagnation pressure at the diffuser exit, which is the compressor inlet. Page 4  k −1  p 0 = p 1 + Ma 2  2   11.72 k /( k −1)  1.4 − 1  = (59.13 kPa )1 + 0.842 2  2   k /( k −1) = 94.1 kPa Determine, for the air flow through the frictionless and adiabatic converging-diverging duct of Example 11.8, the ratio of duct exit pressure to duct inlet stagnation pressure that will result in a standing normal shock at (a) x = +0.1 m; (b) x = +0.2 m; (c) x +0.4 m. How large is the stagnation pressure loss in each case? Example 11.8 computes the isentropic flow through a converging-diverging duct whose area is given by the following equation: A = 0.1 + x2, x (m) A (m2) Ma T/T0 p/p0 where x is in meters and A is in square meters. +0.1 0.11 1.37 0.73 0.33 The entrance and exit of the duct are located at x +0.2 0.14 1.76 0.62 0.18 = -0.5 m and x = +0.5 m, respectively. (The throat +0.4 0.26 2.48 0.45 0.06 is at x = 0.) The table at the left shows the results from Example 11.8 for the locations specified in this problem for a supersonic flow in the diverging region. For a standing shock at the various locations specified in the problem the upstream Mach number into the shock is the Mach number at the particular location found in the isentropic calculation in the table above. We start our calculations by using this Mach number to find the shock relations from Figure D.4. At the location x = +0.1 m, the Example 11.8 table gives a Mach number of 1.37; using this value as Max = 1.37 in Figure D.4, we find the Mach number downstream from the shock as May = 0.75 and the stagnation pressure ratio across the shock p0.y/p0.x = 0.96. Downstream from the shock, the new isentropic flow has a new value of A*, the area at which a given isentropic flow would become sonic. This new value of A* is used to characterize the isentropic flow following the shock. From Figure D.1, we see that A/A* is 1.1 for the value of May = 0.75. Since the area at this location is 0.11 m2, the new value of A* = A/(A/A*) = (0.11 m2)/1.1 = 0.1 m2. The exit area at x = 0.5 m is found from the area formula for this particular duct to be Aexit = 0.1 m2 + (0.5 m)2 = 0.35 m2. So, at the exit A/A* = (.35 m2)/(0.1 m2) = 3.5. For this exit value of A/A*, and subsonic flow, Figure D.1 gives a Mach number of 0.17 and a p/p0 value of 0.98 at the exit. To find the ratio of duct exit pressure, pexit, to inlet stagnation pressure, p0,in, we note that the stagnation pressure in an isentropic flow is constant. Thus the inlet stagnation pressure is the stagnation pressure just upstream of the shock and the exit stagnation pressure is the stagnation pressure just downstream from the shock; we previously found the ratio of these two stagnation pressures across the shock to be 0.96. So the ratio of exit pressure to inlet stagnation pressure is found as follows: p 0, y p0,exit pexit p p = exit = exit = (0.96)(0.98) = 0.94 p0,in p0,exit p0,in p0,exit p0, x To compute the loss of stagnation pressure across the shock we have to use the data from Example 11.8 that the stagnation pressure upstream of the shock was 101 kPa. We can then find the stagnation pressure loss as follows. ∆p0 = p0, x − p0, y = p0, x − p0, x p 0, y p 0, x  p 0, y = p 0, x 1 −  p 0, x    = (101 kPa )(1 − 0.96 ) = 4 kPa   We repeat the same calculations for x = +0.2 m, where the Example 11.8 table gives the Mach number as 1.76. Entering Figure D.4 with Max = 1.76 gives May = 0.62 and p0,y/p0,x = 0.83. For the isentropic flow downstream from the shock with Ma = 0.62 we find A/A* = 1.16 from Figure D.1. The area at x = +0.2 m is 0.14 m2, so the A* value for the isentropic flow to the exit is May 6 CMP homework solutions Page 5 (0.14 m2) / 1.16 = 0.1207 m2. At the exit area of 0.35 m2, A/A* = (0.35 m2) / (0.1207 m2) = 2.9. At this exit value of value of A/A*, we find (p/p0)exit = 0.97. We then find the ratio of exit pressure to inlet stagnation pressure and the stagnation pressure loss as before. p 0, y p0,exit pexit p p = exit = exit = (0.97 )(0.83) = 0.8 p0,in p0,exit p0,in p0,exit p0, x ∆p0 = p0, x − p0, y = p0, x − p0, x p 0, y p 0, x  p 0, y = p 0 , x 1 −  p 0, x    = (101 kPa )(1 − 0.83) = 17 kPa   We repeat the same calculations for x = +0.4 m, where the Example 11.8 table gives the Mach number as 2.48. Entering Figure D.4 with Max = 2.48 gives May = 0.515 and p0,y/p0,x = 0.51. For the isentropic flow downstream from the shock with Ma = 0.515 we find A/A* = 1.3 from Figure D.1. The area at x = +0.4 m is 0.26 m2, so the A* value for the isentropic flow to the exit is (0.26 m2) / 1.3 = 0.2 m2. At the exit area of 0.35 m2, A/A* = (0.35 m2) / (0.2 m2) = 1.75. At this exit value of value of A/A*, we find (p/p0)exit = 0.92. We then find the ratio of exit pressure to inlet stagnation pressure and the stagnation pressure loss as before. p 0, y p0,exit pexit p p = exit = exit = (0.92 )(0.51) = 0.47 p0,in p0,exit p0,in p0,exit p0, x ∆p0 = p0, x − p0, y = p0, x − p0, x p 0, y p 0, x  p 0, y = p 0 , x 1 −  p 0, x    = (101 kPa )(1 − 0.51) = 50 kPa   We see that lower exit pressures create the standing shock wave closer to the exit of the duct; this results in a stronger shock wave with a larger loss of stagnation pressure. 11.73 A normal shock is positioned in the diverging portion of a frictionless, adiabatic converging-diverging air flow duct where the cross section area is 0.1 ft2 and the local Mach number is 2.0. Upstream of the shock, p0 = 200 psia and T0 = 1200 R. If the duct exit area is 0.15 ft2, determine the exit area temperature and pressure and the duct mass flowrate. We have to determine ratios across the shock and for the frictionless, adiabatic (isentropic) flow following the shock. The downstream Mach number from the shock can be found from Figure D.1 or from the following equation. Ma y = 2 k −1 = 2 Ma 2y + 2 Ma y k −1 −1 22 + 2 1.4 − 1 = 0.577 2 ( ) 22 −1 1.4 − 1 The stagnation pressure ratio across the shock is found from Figure D.1 as p0,y/p0,x = 0.72. Since we are given data on area, we have to determine the new A* for the isentropic flow following the shock. At the value of May = 0.577 we can find the value of A/A* = 1.22 from Figure D.1. Since the area at the shock location is 0.10 ft2, the new value of A* = (0.1 ft2)/1.22 = 0.0822 ft2. At the exit area of 0.15 ft2, the value of A/A* = (0.15 ft2) / (0.0822 ft2) = 1.825, we can find the Mach number from Figure D.1 to be Maexit = 0.34. We know the stagnation temperature T0, which is constant for both isentropic flow and across a shock; thus the exit stagnation temperature is the value of 1200 R upstream from the shock. We can find the exit temperature from Figure D.1 or the equation below. May 6 CMP homework solutions Texit Page 6  k −1 2  Maexit = T0 1 +  2   −1  1.4 − 1 (0.34)2  = (1200 R )1 + 2   −1 = 1173 R The stagnation pressure ratio across the shock can be found from Figure D.4 to be 0.72; thus the stagnation pressure downstream from the shock is (200 psia)(0.72) = 144 psia. This stagnation pressure is constant for the isentropic flow from the shock to the exit. We can use Figure D.1 to find the ratio of pressure to stagnation pressure at the exit Mach number of 0.34: p/p0 = 0.92. Thus the exit pressure is (0.92)(144 psia). Pexit = 132 psia . The exit mass flow rate is found from the usual equation: Ma(kRT)1/2, m = ρVA , where V = cMa = and ρ = P/RT.  1716 ft ⋅ lb f V = cMa = Ma kRT = 0.34 1.4  slug ⋅ R  1 slug ⋅ ft 570 ft (1173 R ) =  s lb f ⋅ s 2  132 lb f 144 in 2 ρ= 0.00952 slug P in 2 ft 2 = = RT 1716 ft ⋅ lb f ft 3 (1173 R ) slug ⋅ R m = ρVA = ( ) 0.00952 slug 570 ft 0.15 ft 2 = 0.81 slug/s 3 s ft Using equations instead of Figure D.1 and D.4 to solve the problem, including an iterative solution of the equation to find the Mach number for a given value of A/A*, results in a final mass flow rate of 0.814 slug/s.