Homework hints, eighth problem set: 41,43,54,63,74,86
9.41 Air, with a stagnation pressure of 100 kPa, flows through the nozzle in fig 9.41, 2 m long, with area A = 20 - 20 x + 10 x^2 with A in cm^2 and x in m. Plot the complete family of isentropic pressures p(x) in the nozzle for the range of pressures 1 < p(0) < 100 kPa.
There is a subsonic entrance region of high pressure and a supersonic entrance region of low pressure, both of which are bounded by a sonic (critical) throat, and both of which have a ratio Ax=0 / A* = 2.0. From table B-1 or Eq. 9.44 we find these two conditions to be bounded by:
a. subsonic entrance: A/A* = 2.0, Ma_e = 0.306, p_e = 0.9371 p_0 = 93.71 kPa
b. supersonic entrance: A/A* = 2.0, Ma_e = 2.197, p_e = 0.09396 p_0 = 9.396 kPa.
Thus no isentropic flow can exist between entrance pressures 9.396 kPa and 93.71 kPa. The complete family of isentropic pressure curves can be found, but require implicit inversion of 9.44 from area ratio to Ma.
For air, with k=1.4, 9.44 becomes:
A/A* = 1/Ma (1+0.2 Ma^2)^3 / 1.728
The exponent 3 is 1/2 (k+1)/(k-1) for k=1.4.
**9.42 Suppose that the duct of 9.41 has T_0 = 300 C and p(0) = 125 kPa. If the flow is isentropic and the exit is supersonic, compute a. mdot and b. pexit.
a. If the exit is supersonic, then the throat is choked, A* = 10 cm^2, thus x=0 corresponds to A/A* = 2.0, or, as discussed in 9.41;
x=0, A/A* = 2 (subsonic), so Ma(0) = ).306, p/p_0 = 0.9371, p_0 = 125/0.9371 = 133 kPa. Then mdot = mdot_max = 0.6847 p_0 A* / (RT_0)^1/2 = 0.225 kg/s.
The exit pressure follows from a supersonic interpretation of A/A* = 2.0:
a. A/A* = 2.0 (supersonic): Ma_exit = 2.197, p_exit = 0.09396 (133) = 12.5 kPa.
9.43 Air flows isentropically through a duct with T_0 = 300 C. At two sections with identical areas of 25 cm^2, the pressures are p_1 = 120 kPa and p_2 = 60 kPa. Determine a. the mass flow mdot; b. the throat area ; c. Ma_2.
If the areas are the same and the pressures different, then section 1 must be subsonic and section 2 supersonic. In other words, we need to find where:
p_1/p_0 / p_2/p_0 = 120/60 = 2.0 for the same A_1/A* = A_2/A* (search table B1 , isentropic). After interation we find Ma_1 = 0.729, Ma_2 = 1.32.
A/A* = 1.075 for both sections, A* = 25/1.075 = 23.3 cm^2.
With critical area and stagnation conditions known, we may compute the mass flow mdot:
p_0 = 120[1+0.2(0.729)^2]^3.5 = 171 kPa and t_0 = 300 + 273 = 573 K.
mdot = 0.684 p_0 A* /[RT_0]^1/2 = 0.671 kg/s .
**9.46 Consider a control volume which encloses the nozzle between two 25 cm^2 cross sections. If the pressure outside the duct is 1 atm, determine the total force acting on this section of nozzle. a. mdot = 0.671 kg/s b. throat area = 23.3 cm^2, c. Ma_2 = 1.32 from 9.43.
Enclose the sections with a control volume. To find the force we need a momentum balance, and the velocities at the sections.
Ma_1 = 0.729, t_0 = 573 K, t_1 = t_0 / (1 + 0.2 Ma_1^2) = 518 K, a_1 = 456 m/s.
V_1 = Ma_1 a_1 = 333 m/s. Similarly, Ma_2 = 1.32 gives V_2 = 545 m/s.
x-momentum: sum F_x = R_x +(p_1 - p_2)_gage A_1 = mdot (V_2 - V_1), so R_x = - 8 N
**9.53 Air flows steadily from a reservoir at 20 C through a nozzle of exit area 20 cm^2 and strikes a vertical plate. The flow is subsonic throughout. A force of 135 N is required tohold the plate stationary. Compute a. V_e ; b. Ma_e ; c. p_0 if p_a = 101 kPa.
F = 135 N = \rho_e V_e^2 A_e = k p_e Ma_e^2 A_e ; solve for Ma_e = 0.69
T_e = 293/[1+0.2(0.69)^2] = 268 K, a_e = (1.4(287)(268))^1/2 = 328 m/s
Thus, V_e = a_e Ma_e = 328x0.69 = 226 m/s.
p_tank = p_0 = 101350[1+0.2(0.69)^2]^3.5 = 139000 Pa.
9.54 For flow of air through a normal shock, the upstream conditions are V_1 = 600 m/s, T_o1 = 500 K, and p_o1 = 700 kPa. Compute the downstream conditions Ma_2, V_2, T_2, p_2, and p_o2.
Solution: First compute teh upstream Mach number:
T_1 = T_o - V_1^2/2 c_p = 500 - 600^2/2(1005) = 321 K. a_1 = 1.4(287)(321) = 359 m/s, Ma_1 = 600/359 = 1.67.
Hence, Ma_2 = [(0.4(1.67)^2 + 2)/(2.8(1.67)^2 - 0.4)]^1/2 = 0.648.
V_2/V_1 = (2 + 0.4(1.67)^2 )/(2.4(1.67)^2) = 0.465, so V_2 = 0.465(600) = 279 m/s.
T_2/T_1 = [2 + 0.4(1.67)^2 ](2.8(1.67)^2 - 0.4)/(2.4(1.67)^2) = 1.44, T_2 = 1.44(321)= 461 K.
p_1 = 700/[1 + 0.2(1.67)^2]^3.5 = 148 kPa, p_2/p_1 = [2.8(1.67)^2 - 0.4]/2.4 = 3.09, or p_2 = 458 kPa.
p_o2 = p_2 (1 + 0.2 Ma_2^2)^3.5 = 458[1 + 0.2(0.658)^2]^3/5 = 607 kPa
or Table B-2, at Ma_1 = 1.67, p_o2/p_o1 = 0.868(700) = 607 kPa.
**9.62 An atomic explosion propagates into still air at 14.7 psia and 520 R. The pressure just inside the shock is 5000 psia. Assuming k=1.4, what are the speed C of the shock and the velocity V just inside the shock?
The pressure rtio tells us the Mach number of the shock motion:
p_2 / p_1 = 5000/14.7 = 340 = 2.8 Ma_1^2 - 0.4 / 2.4 , solve for Ma_1 = 17.1.
a_1 = (1.4(1717)(520))^1/2 = 1118 ft/s so V_1 = C = 17.1(1118) = 19100 ft/s
The volocity ratio across the shock
V_2 / V_1 = [0.4 (17.1)^2 +2 ]/[2.4(17.1)^2] = 0.1695, so V_2 = 0.1695 (19100)= 3240 ft/s
Then V_inside = C-V_2 = 19100 -3240 = 15900 ft/s.
9.63 Sea level standard air is sucked into a vacuum tank through a nozzle, as in Fig P9.63. A normal shock stands where the nozzle area is 2 cm^2, as shown. Estimate a. the pressure in the tank, and b. the mass flow.
Solution: The flow at the exit section with 3 cm^2 is subsonic after a shock, and therefore must equal the tank pressure.
p_o1 = 101350 Pa, A_1/A^* = 2.0, thus Ma_1 = 2.20, p_1 = 101350/[1 + 0.2(2.2)^2]^3.5 = 9520 Pa.
p_2/p_1 = 2.8(2.2)^2 - 0.4 / 2.4 = 5.47, so p_2 = 5.47(9520) = 52030 Pa.
Also, A^*_2/A^*_1 = 1.59, or A^*_2 = 1.59(1) cm^2.
p_o2 = 101350/1.59 = 63800 Pa. A_3/A^*_2 = 3/1.59 = 1.89, read Ma_3 = 0.327, whence p_3 = 63800/[1+0.2(0.327)^2]^3.5 = 59200 Pa. With T_o = 288 K, the critical mass flow = 0.6847 p_o A^*/(RT_o)^1/2 = 0.241 kg/s.
**9.74 The perfect gas assumption leads smoothly to Mach number relations which are very convenient and tabulated. This is not so for a real gas such as steam. Let stam at t_0 = 500 C and p_0 = 2 Mpa expand isentropically through a converging nozzle whose exit area is 10 cm^2. Using the steam tables, find a. the exit pressure and b. the mass flow when the flow is sonic, or choked. What complicates the analysis?
For "ideal steam" k = 1.33, R = 461 J/kg K:
Approx: p_e = p_0 / [1 = (k-1)/2]^k/k-1 = 1.08 MPa.
mdot_max = 0.6726p_0A*/(RT_0)^1/2 = 2.25kg/s.
This shows the exit flow has pressure about 1.1 MPa. Iterate in steam tables to about 1.096 MPa for the sonic flow isentropic exit pressure: This gives mdot = 2.24 kg/s, very close to 2.25 found assuming steam is an ideal gas.
**9.85 A large tank delivers air through a nozzle of 1 cm^2 throat area and 2.7 cm^2 exit area. When the receiver pressure is 125 kPa, a normal shock stands in the exit plane. Estimate a. the throat pressure; and b. the temperature in the upsteam supply tank.
If there is a shock, the throat is choked, and the Mach number just upstream of the exit shock corresponds to an isentropic area ratio of 2.7:1.
at A_1/A* = 2.7, Ma_1 = 2.525, for which p_2 / p_1 |shock = 7.27, or p_1 = 125/7.27 = 17.2 kPa
p_01 = 17.2[1+0.2(2.525)^2]^3.5 = 305 kPa, so p_throat = p* = 0.5283(305) = 161 kPa
Part b. is impossible. There is not enough information to find T_0.
9.86 Air enters a 3 cm diameter pipe 15 m long at V_1 = 73 m/s, p_1 = 550 kPa, and T_1 = 60 C. The friction factor is 0.018. Compute V_2, p_2, T_2, and p_o2 at the end of the pipe. How much additional pipe length would cause the exit flow to be sonic?
Solution: First compute the inlet Mach number and then get (fL/D)_1:
a_1 = (1.4(287)(60+273))^1/2 = 366 m/s, Ma_1 = 73/366 = 0.2, read 9fL/D) = 14.53, for which p/p^* = 5.4554, T/T^* = 1.1905, V/V^* = 0.2182, and p_o/p^*_o = 2.9635. Then fL/D_2 = 14.53 - (0.018)(15)/(0.03) = 5.53, read Ma_2 = 0.295.
At this new Ma_2, read p/p^* = 3.682, T/T^* = 1.179, V/V^* = 0.320, and p_o/p^*_o = 2.067. Then V_2 = V_1 (V_2/V^*)/(V_1/V^*) = 73(1.179)/(0.218) = 107 m/s, p_2 = 550(3.682/5.455) = 371 kPa
T_2 = 333(1.179/1.190) = 330 K. Now we need p_o1 to get p_o2: p_o1 = 550[1 + 0.2(0.2)^2]^3.5 = 566 kPa, so p_o2 = 566(2.067/2.964) = 394 kPa.
The extra distance we need to choke the exit to sonic speed is (fL/D)_2 = 5.53. That is, delta L = 5.53 D/f = 5.53 (0.03/0.018) = 9.2 m.
**9.96 Derive and verify the adiabatic pipe flow velocity relation of eq 9.74, which is written as
f L/D + k+1 / k ln V_2/V_1 = a_0^2/k (1/V_1^2- 1/V_2^2)
This is left as a student exercise. Good luck! (do not spend more than one hour on this problem...anyone that solves it gets a gold star)