Homework hints, second problem set: 9.2,9,17,19,33,35,42,46,53,62,74,85,96
**9.1 An ideal gas flows adiabatically through a duct. At section 1 the pressure, temperature and velocity are given. Further downstream at section 2 only the pressure and temperature are given. Calculate the velocity at 2 in m/s and the change in entropy s_2 - s_1 if the gas is air or argon.
Use the energy equation for steady flow:
(h + 1/2 V^2 )_1 = (h + 1/2 V^2 )_2 + w_v- q
with h = c_p T. For air V_2 is 326 m/s. For argon V_2 is 266 m/s. s_2 - s_1 values are 337 and 266 J/kg K.
9.2 Solve 9.1 if the gas is steam using a. an ideal gas from A.4 and b. real steam fro the steam tables.
For steam, k = 1.33, R = 461 J/kg K, c_p = 1858 J/kg K. Subst in energy equation to give V_2 = 318 m/s.
s_2 - s_1 = 1858 ln (207 + 273 / 260 + 273) - 461 ln (30/140) = -195 + 710 = 515 J/kg K.
From the steam tables h_1 = 2.993E6 J/kg ; h_2 = 2.893E6 J/kg giving V_2 = 321 m/s. s_2 - s_1 = 8427 - 7915 = 512 J/kg K from the steam tables.
9.9 Steam enters a duct at p_1 = 50 psia, T_1 = 360 F, and V_1 = 200 ft/s, and leaves at p_2 = 100 pasia, t_2 = 800 F, V_2 = 1200 ft/s. How much heat was added in BTU/lbm?
Use the energy equation with no shaft work, so that q = h_2 - h_1 + 1/2 (V_2^2 - V_1^2). From steam tables h_1 = 1214.9 and h_2 = 1429.7 BTU/lbm, giving q = 243 BTU/lbm
**9.12 Assume that water follows Eq. 1.19 with n=7 and B=300. Compute the bulk modulus and the speed of sound at 1 atm and 1100 atm. Compute the speed of sound at 20 C and 9000 atm to compare with 2650 m/s measured.
Eq. (1.19) p/p_a = (B+1)(\rho/\rho_a)^n - B ; Bulk modulus K = \rho dp/d\rho = n(B+1)p_a(\rho/\rho_a)^n ; a=(K/\rho)^{1/2}
Substitute values for water at 1 atm to get K=2.129E9 Pa and a=1460 m/s, versus K=2.91E9 Pa and 1670 m/s at 1100 atm (deepest trench). At 9000 atm the density is 1217 kg/m^3, giving 2645 m/s (close to measured value).
9.17 A submarine at 800 m sends a sonar signal and receives the reflected wave back from a similar submerged object in 15 s. Estimate the distance.
Substitute 80.4 atm into (1.19) to get \rho = 1029 kg/m^3. Use the equation derived in 9.12 to give a = 1457 m/s (close to the value at 1 atm). The distance (one way) is therefore 1457x15/2 = 10900 m.
9.19 The Concorde aircraft flies at Ma = 2.3 at 11 km standard altitude. Estimate the temperature in C at the front stagnation point. At what Mach number would it have a front stagnation point temperature of 45 0 C.
At 11 km altitude T = 216.66 K, a = (kRT)^{1/2} = 295 m/s.
Then T_nose = T_0 = T[1 + (k-1)/2 Ma^2] = 446 K = 173 C.
For 450 C, Ma = 3.42.
9.33 Air flows isentropically from a reservoir, where p = 300 kPa and T = 500 K, to section 1 in a duct, where A_1 = 0.2 m^2 and V_1 = 550 m/s. Compute Ma_1 , T_1 , p_1 , mdot and A*. Is the flow choked?
Use the energy equation to get the temperature and then the Mach number at 1.
T_1 = T_0 - V_1^2/2 c_p = 500 - 550^2 / 2(1005) = 350 K.
a_1 = (1.4 x 287 x 350 )^1/2 = 375 m/s. Ma = 550/375 = 1.47. The flow must be choked to produce supersonic flow in the duct.
p_1 = p_0 / (1 + 0.2 Ma_1^2)^3.5 = 86 kPa.
\rho_1 = p_1/RT_1 = 0.854 kg/m^3 , so mdot = \rho A V = 0.854x0.2x550 =94 kg/s.
A/A* = 1/Ma (1+0.2Ma^2)^3 / 1.728 = 1.155 if Ma = 1.47. So A* = 0.2/1.155 = 0.173 m^2.
9.35 Air in a tank at 700 kPa and 20 C exhausts through a converging nozzle of throat area 0.65 cm^2 to a 1 atm environment. Compute the initial mass flow in kg/s.
The flow is choked because p_e / p_0 is less than 0.528. Eq. 9.46 holds. mdot = mdot_max = 0.6847 p_0 A* / (RT_0)^1/2 = 0.107 kg/s.
**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.
9.62 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.
9.74 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.
9.96 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)