Compressible Flow Homework Solutions

May 6 Compressible-Flow-Homework Solutions 11.5 Air flows steadily between two sections in a duct. At section (1), the temperature and pressure are 80oC and 301 kPa(abs), and at section (2), the temperature and pressure are T2 = 180oC and p2 = 181 kPa(abs). Calculate the (a) change in internal energy between sections (1) and (2), (b) change in enthalpy between sections (1) and (2), (c) change in density between sections (1) and (2), and (d) change in entropy between sections (1) and (2), How would you estimate the loss of available energy between the two sections of the flow? If we assume that the heat capacity is constant, the changes in internal energy, enthalpy and entropy can be computed by the following equations.   u 2 − u1 = cv (T2 − T1 )   h2 − h1 = c p (T2 − T1 ) T  P    s 2 − s1 = c p ln 2  − R ln 2   T1   P1  The heat capacity terms in these equations may be found from the Table 1.8 data that k = 1.4 and R = 286.9 N·m/kg·K = 286.9 J/kg·K, using the definition that k = cp/cv and ideal-gas heatcapacity relation cp = cv – R. k= cp cv = cv + R R = 1+ cv cv 1 286.9 kJ 1 717.2 kJ = = k −1 kg ⋅ K 1.40 − 1 kg ⋅ K 1004 kJ kR 286.9 kJ 1.40 = = c p = kcv = k −1 kg ⋅ K 1.40 − 1 kg ⋅ K ⇒ k −1 = R cv ⇒ cv = R Substituting these values into the equations above along with the given temperature and pressure data for stations one and two gives the following results. Note that in taking a temperature difference the result in the same in kelvins or degrees Celsius. However, in the entropy calculation where temperature ratios are used, absolute temperatures are required. Absolute pressures are also required in the entropy calculation. ( ) ( ) 717.2 J 71.72 kJ kJ   180 o C − 80 o C = u 2 − u1 = cv (T2 − T1 ) = 1000 J kg kg ⋅ K   1004 J 100.4 kJ kJ 180 o C − 80 o C = h2 − h1 = c p (T2 − T1 ) = 1000 J kg kg ⋅ K T   P  1004 J  453.15 K  286.9 J  181 kPa  396 J    −  = ln ln s 2 − s1 = c p ln 2  − R ln 2  =  T1   P1  kg ⋅ K  353.15 K  kg ⋅ K  301 kPa  kg ⋅ K The change in density is simply found by applying the ideal gas equation of state to both points and taking the difference. Note that the SI units for the gas constant can be expressed as R = 286.9 N·m/kg·K = 286.9 J/kg·K = 286.9 Pa·m3/kg·K = 0.2869 kPa·m3/kg·K. The final set of units is consistent with computing the density in kg/m3 using the pressure in kPa and the temperature in kelvins. Page 2 ρ 2 − ρ1 = p2 p (301 kPa ) (181 kPa ) 1.58 kg − 1 = − = − 3 RT2 RT1 0.2869 kPa ⋅ m 3 m3 (453.15 K ) 0.2869 kPa ⋅ m (353.15 K ) kg ⋅ K kg ⋅ K The change in density has a significant effect from pressure and the flow is compressible. We would compute the loss in energy from equation 5.108 for compressible flow. 2 1   Loss = u 2 − u1 + pd   + q net  ρ  in 1 ∫ We do not have the necessary information to perform the integral or to compute the net heat addition. 11.16 If a high-performance aircraft is able to cruise at a Mach number of 3.0 at an altitude of 80,000 ft, how fast is this in (a) mph; (b) ft/s; (c) m/s? The speed is simply the Mach number times the speed of sound. We can compute the sound speed at 80,000 ft, using the temperature data for standard air at 80,000 ft in Table C.1, page 764: T = –61.98oF = 397.69 R. At the minimum temperature of –40oF in Table B.3, the value of k for air is given as 1.401. Using these values for k and T along with the gas constant for air from Table 1.7, R = 1716 ft·lbf/slug·R gives the sound speed.  1716 ft ⋅ lb f c = kRT = 1.401  slug ⋅ R  1 slug ⋅ ft 977.8 ft (397.69 R ) =  s lb f ⋅ s 2  For Ma = 3, V = 3(977.8 ft/s) or V = 2933 ft/s Multiplying this answer by .3048 m/ft gives V = 894.1 m/s Multiplying the original answer by (30 mph) / (44 ft/s) gives V = 2000 mph 11.31 Air flows steadily and isentropically from standard atmospheric conditions to a receiver pipe through a converging duct. The cross section area of the throat of the converging duct is 0.05 ft2. Determine the mass flowrate through the duct if the receiver pressure is (a) 10 psia; (b) 5 psia. Sketch temperature-entropy diagrams for situations (a) and (b). Verify results obtained with values from the appropriate graph in Appendix D with calculations involving ideal gas equations. This problem is similar to example 11.5 in the text. We compute the mass flow rate as the product of density times velocity times area at the throat. The density is related to the Mach number and the velocity is the Mach number times the speed of sound. This problem is for a duct that has a converging section only. It cannot become supersonic because there is no increase in area beyond the minimum throat area. We first have to determine if the flow is choked or not. To do this we compute the critical pressure ratio, p* from the equation below, using the value of k = 1.4 for air and a stagnation pressure equal to the standard atmospheric pressure of 14.696 psia.  2  p* = p0    k +1 k ( k −1) 1.4 (1.4 −1)  2  = (14.696 psia )   1.4 + 1  = 7.763 psia If the receiver pressure is less than p* = 7.763 psia, the flow is choked and we know that the throat Mach number is one; if the receiver pressure is greater than p* = 7.763 psia, the flow is Page 3 not choked; the throat pressure is the receiver pressure and we have to compute the throat Mach number from this pressure. The initial receiver pressure of 10 psia is greater than p* = 7.763 psia so the flow is not choked; the throat pressure is 10 psia and we can compute the throat Mach number by rearranging equation 11.59 in the text to solve for the Mach number. (1.4−1) 1.4 ( k −1) k    2   p0   2   14.696 psia   Ma =  − 1 =  − 1 = 0.7625       k − 1   p   1.4 − 1   10 psia    Equation 11.60 gives the density at this Mach number. The density of standard air, ρ0 = 0.00238 slug/ft3 is taken from Table 1.7. 1 (1− k )  k −1  ρ = ρ 0 1 + Ma 2  2   1 (1−1.4 ) 0.00238 slug  1.4 − 1  = 0.7625 2  1 + 3 2   ft = 0.001808 slug ft 3 In order to find the velocity at the throat as V = c(Ma) we have to find c which requires the temperature. This is found from equation 11.56, using the standard temperature from table 1.7, T0 = 59oF = 518.67 R.  k −1  T = T0 1 + Ma 2  2   −1  1.4 − 1  = (518.67 R )1 + 0.7625 2  2   −1 = 464.6 R We can now compute the sound speed using the gas constant for air from Table 1.7: R = 1716 ft·lbf/slug·R.  1716 ft ⋅ lb f c = kRT = 1.40  slug ⋅ R  1 slug ⋅ ft 1057 ft (464.6 R ) =  s lb f ⋅ s 2  So our speed, V = c(Ma) = (1057 ft/s)(0.7625) = 805.6 ft/s. We now have all the terms necessary to compute the mass flow rate at the throat. m = ρVA = ( ) 0.001808 slug 805.6 ft 0.05 ft 2 = 0.0728 slug/s 3 s ft Using Figure D-1 with the pressure ratio p/p0 = (10 psia) / (14.696 psia) = 0.68, I read the following values: Ma = 0.66, T/T0 = 0.90, ρ/ρ0 = 0.77. These ratios give T = 0.90(518.67 R) = 467 R and ρ = 0.77(0.00238 slug/ft3) = 0.0018 slug/ft3. From the temperature and Mach number we can compute the velocity; we then can find the mass flow rate.  1716 ft ⋅ lb f V = cMa = Ma kRT = (0.66 ) 1.40  slug ⋅ R m = ρVA = (  1 slug ⋅ ft 805 ft (467 R ) =  2 s ⋅ lb s  f ) 0.0018 slug 805 ft 0.05 ft 2 = 0.0737 slug/s 3 s ft In the second case, the receiver pressure of 5 psia is less than p* so the throat pressure is p* = 7.763 psia and the throat Mach number is 1. We use the same equations as above to find the density and temperature at the throat. Page 4 1 (1− k )  k −1  ρ = ρ 0 1 + Ma 2  2   1 (1−1.4 ) 0.00238 slug  1.4 − 1 2  = 1  1 + 2   ft 3  k −1  T = T0 1 + Ma 2  2   −1  1.4 − 1 2  1  = (518.67 R )1 + 2   −1 = 0.001509 slug ft 3 = 432.2 R Since the Mach number is one, the velocity at the throat is the sound speed at the temperature just found. Once we find the velocity we can compute the mass flow rate at the throat.  1716 ft ⋅ lb f V = c = 1.40  slug ⋅ R m = ρVA =  1 slug ⋅ ft 1019 ft (432.2 R ) =  s lb f ⋅ s 2  ( ) 0.001509 slug 1019 ft 0.05 ft 2 = 0.0769 slug/s 3 s ft Using Figure D-1 with the Mach number of one, I read the following values: T/T0 = 0.84, ρ/ρ0 = 0.635. These ratios give T = 0.84(518.67 R) = 436 R and ρ = 0.635(0.00238 slug/ft3) = 0.0015 slug/ft3. From the temperature we can compute the sound speed which equals the velocity for the Mach number of one; we then can find the mass flow rate.  1716 ft ⋅ lb f V = c = 1.40  slug ⋅ R m = ρVA = 11.33  1 slug ⋅ ft 1023 ft (436 R ) =  s lb f ⋅ s 2  ( ) 0.0015 slug 1023 ft 0.05 ft 2 = 0.0773 slug/s 3 s ft Determine the static pressure to stagnation pressure ratio associated with the following motion in standard air: (a) a runner moving at the rate of 10 mph, (b) a cyclist moving at the rate of 40 mph, (c) a car moving at the rate of 65 mph, (d) an airplane moving at the rate of 500 mph. Equation 11.59 gives the ratio of static pressure to stagnation pressure as a function of Mach number squared = Ma2 = V2/c2 = V2/(kRT). p  k −1  = 1 + Ma 2  p0  2  k (1− k )  k −1 V 2   = 1 +   kRT 2   k (1− k ) For the first three cases we assume sea level conditions so that T = 59oF = 518.67 R for standard air. At this temperature the sound speed is  1716 ft ⋅ lb f c = 1.40  slug ⋅ R  1 slug ⋅ ft 30 mph (518.67 R ) = 761.1 mph  2 44 ft s ⋅ lb s  f We obtain the following results for the first three cases: 10 mph = 0.01314 Ma = 761.1 mph p  k −1  = 1 + Ma 2  p0  2  k (1− k ) 1.4 (1−1.4)  1.4 − 1 (0.01314)2  = 1 + 2   0.9999 Page 5 40 mph = 0.05256 Ma = 761.1 mph p  k −1  = 1 + Ma 2  2 p0   k (1− k ) 65 mph = 0.08540 Ma = 761.1 mph p  k −1  = 1 + Ma 2  2 p0   k (1− k ) 1.4 (1−1.4)  1.4 − 1 (0.05256)2  = 1 + 2   0.9981 1.4 (1−1.4)  1.4 − 1 (0.08540)2  = 1 + 2   0.9949 For the airplane we assume a typical elevation of 35,000 ft where the standard temperature is -65.61oF = 394.06 R; the sound speed at this temperature is  1716 ft ⋅ lb f c = 1.40  slug ⋅ R  1 slug ⋅ ft 30 mph (394.06 R ) = 663.4 mph  lb f ⋅ s 2 44 ft s  We can now find the stagnation pressure ratio for an airplane flying at 500 mph at an elevation of 35,000 ft. 500 mph Ma = = 0.7537 663.4 mph p  k −1  Ma 2  = 1 + p0  2  k (1− k ) 1.4 (1−1.4)  1.4 − 1 (0.7537 )2  = 1 + 2   0.686 11.45a An ideal gas is to flow isentropically from a large tank where the air is maintained at a temperature of 59oF and 80 psia to standard atmospheric discharge conditions. Describe in general terms the kind of duct involved and determine the exit Mach number and exit velocity if the gas is air. We can compute the exit Mach number where the pressure is the standard atmospheric pressure of 14.696 psia by rearranging equation 11.59 in the text to solve for the Mach number. ( k −1)  2   p0  Ma =      k − 1   p  k (1.4 −1) 1.4    2   80 psia    − 1 = 1.765 −1 =       1.4 − 1   14.696 psia   We can find the exit velocity as the product of the Mach number and the sound speed. In order to compute the sound speed we have to compute the exit temperature from the Mach number using equation 11.56 with the inlet temperature of 59oF = 518.67 R.  k −1  T = T0 1 + Ma 2  2   −1  1.4 − 1  1.765 2  = (518.67 R ) 1 + 2   −1 = 319.6 R We can now compute the sound speed using the gas constant for air from Table 1.7: R = 1716 ft·lbf/slug·R.  1716 ft ⋅ lb f c = kRT = 1.40  slug ⋅ R  1 slug ⋅ ft 875.3 ft (319.6 R ) =  s lb f ⋅ s 2  So our speed, V = c(Ma) = (876.3 ft/s)(1.765); V = 1546 ft/s . Because the exit flow is supersonic a converging-diverging nozzle is required . 11.47c Upstream of the throat of an isentropic converging-diverging nozzle at section (a), V1 = 150 m/s, p1 = 100 kPa(abs), and T1 = 20oC. If the discharge flow is supersonic and the throat area is 0.1 m2, determine the mass flowrate for the flow of helium. Page 6 Since we know the throat area we can find the mass flow rate if we know the density and velocity at the throat. We do not know the stagnation conditions for this problem, but we can compute them from the data that we are given at station (1). For helium, we find k = 1.66 and R = 2077 J/kg·K = 2077 N·m/kg·K from Table 1.8. We use these data to find the sound speed of helium at station (1) where T = 20oC = 293.15 K.  2077 N ⋅ m  1 kg ⋅ m 1005 m (293.15 K ) = c1 = kRT1 = 1.66 s N ⋅ s2  kg ⋅ K  The Mach number at station 1 is 150 m V s Ma1 = 1 = = 0.1492 m 1005 c1 s Since we know that the exit conditions are supersonic, this subsonic location must be before the throat in a converging-diverging nozzle. We can use this Mach number and the given values of T1 and P1 to find the stagnation conditions. Equation 11.56 gives the stagnation temperature.  k −1   1.66 − 1  T0 = T 1 + Ma 2  = (293.15 K )1 + 0.1492 2  = 295.3 K 2 2     Equation 11.59 gives the stagnation pressure.  k −1  p0 = p 1 + Ma 2  2   k ( k −1) 1.66 (1.66 −1)   1.66 − 1 0.1492 2  = (100 kPa )1 + 2   = 101.9 kPa At the throat the Mach number is one and we can find the throat pressure and temperature from the stagnation pressure and temperature and the equations just used that relate the stagnation conditions to the Mach number.   k −1 T = T0 1 + Ma 2  2    k −1  p = p 0 1 + Ma 2  2   k (1− k ) −1  1.66 − 1 2  1  = (295.3 K )1 + 2   −1 = 222.0 K 1.66 (1−1.66)  1.66 − 1 2  = (101.9 kPa )1 + 1  2   = 49.72 kPa The density at the throat is found from the ideal gas law. Here we use R = 2.077 kPa·m3/kg·K. ρ= (49.72 kPa ) p 0.1078 kg = = 3 RT 2.077 kPa ⋅ m m3 (222.0 K ) kg ⋅ K Since Ma = 1 at the throat, V = c there. Once we find V = c, we can compute the mass flow rate at the throat.  2077 N ⋅ m  1 kg ⋅ m 874.9 m (222.0 K ) = V = c = kRT = 1.66 s N ⋅ s2  kg ⋅ K  Page 7 m = ρVA = ( ) 0.1078 kg 874.9 m 0.1 m 2 = 9.43 kg/s 3 s m