Homework hints, seventh problem set: 9.1,7,12,19,33,35

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.

a. For air, take k=1.4, R=287 J/kg K, c_p 105 J/kg K. The adiabatic steady-flow energy equation 9.23 is used to compute the downstream velocity:

constant = (h + 1/2 V^2 )_1 = (h + 1/2 V^2 )_2 + w_v- q = 1005(260) + (1/2)75^2 = 1005(207) + (1/2)V_2^2 .

with h = c_p T. For air V_2 is 326 m/s. s_2 - s_1 = c_p ln(T_2/T_1) - R ln(p_2/p_1) = 1005 ln [(207 + 273)/(260+273)] - 287 ln (30/140) = -105 + 442 = 337 J/kg K.

b. For argon, take k=1.67, R=208 J/kg K, c_p=518 J/kg K. Repeat part a. 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.

9.7 Carbon dioxide (k=1.28) enters a constant area duct at 400 deg F, 100 lbf/in^2 absolute, and 500 ft/s. Farther down stream the properties are V_2 = 1000 ft/s and T_2 = 900 deg F. Compute a. p_2, b. the heat added between sections, c. the entropy change between sections, and d. the mass flow per unit area. Hint This problem requires the continuity equation.

Solution: For carbon dioxide, take k = 1.28, R = 1130 ft-lbf/slug deg R, and c_p = 5167 ft-lbf/slug deg R. a. The downstream pressure is found from continuity:

\rho_1 A_1 V_1 = \rho_2 A_2 V_2, cancel A.

p_1 V_1 / R T_1 = p_2 V_2 / R T_2, cancel R.

so: p_2 = p_1 (T_2/T_1)(V_1/V_2) = 100 (900+460)/(400+460) (500/100) = 79 psia.

b. The steady flow energy equation, with no shaft work, yields the heat transfer per mass:

q = c_p(T_2-T_1) + (1/2)(v_2^2 - V_1^2) = 5167(900-400) +(1/2)[1000^2 - 500^2] = 2.96E6 + 32.2 + 778.2 ft-lbf/slug = 118 Btu/lbm.

c. s_2 - s_1 = 5167 ln[(900+460)/(400+460)] - 1130 ln (79/100) = 2368 + 266 = 2630 ft-lbm/slug deg R.

d. m^dot / A = \rho_1 V_1 = [100x144/1130(400+460)](500) = 7.4 slug/s ft^2.

**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.

D9.1 Design Project

Select a rectangular wing for a fighter aircraft. Please present your design as a technical memo.