Exam (elaborations) TEST BANK FOR Gas Dynamics 3rd Edition By James John, Theo Keith (Instructor's Solution Manual)

Problem 1. – Air is stored in a pressurized tank at a pressure of 120 kPa (gage) and a temperature of 27°C. The tank volume is 1 m3. Atmospheric pressure is 101 kPa and the local acceleration of gravity is 9.81 m/s2. (a) Determine the density and weight of the air in the tank, and (b) determine the density and weight of the air if the tank was located on the Moon where the acceleration of gravity is one sixth that on the Earth. g 9.81m/ s R 0.287 kJ / kg K 1m T C P P P kpa 2 3 abs gage atm = = ⋅ ∀ = = + = ° = + = + = a) m3 2.5668 kg (0.287)(300) 221 RT ρ = P = = W = mg = ρ∀g = (2.5668)(1)(9.81) = 25.1801N b) moon earth 3 m ρ = ρ = 2.5668 kg W 4.1967N 6 W 1 g g W earth earth earth moon moon = = = Problem 2. – (a) Show that p/ρ has units of velocity squared. (b) Show that p/ρ has the same units as h (kJ/kg). (c) Determine the units conversion factor that must be applied to kinetic energy, V2/2, (m2/s2) in order to add this term to specific enthalpy h (kJ/kg). Air 2 a) 2 2 2 2 3 2 2 3 V s m N s 1 kg m kg N m kg m m p N m , kg m p N ≈ = ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ ⋅ − ⋅ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ ≈ ρ ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ ≈ ρ ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ ≈ b) kg kJ 1000 1 1000 J 1kJ N m 1 J kg P N m = ⎟⎠ ⎞ ⎜⎝ ⎛ ⎟⎠ ⎞ ⎜⎝ ⎛ ⋅ ⋅ ≈ ρ c) c 2 2 2 2 1000g factor 1 kg kJ 1000 J 1 kJ N m 1 J kg m 1 N s s m 2 V ∴ = ≈ ⎟⎠ ⎞ ⎜⎝ ⎛ ⎟⎠ ⎞ ⎜⎝ ⎛ ⋅ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⋅ ⋅ ≈ Problem 3. – Air flows steadily through a circular jet ejector, refer to Figure 1.15. The primary jet flows through a 10 cm diameter tube with a velocity of 20 m/s. The secondary flow is through the annular region that surrounds the primary jet. The outer diameter of the annular duct is 30 cm and the velocity entering the annulus is 5 m/s. If the flows at both the inlet and exit are uniform, determine the exit velocity. Assume the air speeds are small enough so that the flow may be treated as an incompressible flow, i.e., one in which the density is constant. m& i = m& e m& i = m& p + m& s = ρApVp + ρAsVs m& e = ρAeVe ∴ApVp + AsVs = AeVe So e p p s s e A A V A V V + = Ae = As + Ap 2p p D 4 A π = 2p 2 s o D 4 D 4 A π − π = 2 e Do 4 A π = i e s p 3 ( ) 2 ( p s ) o 2p 2 s o s 2p 2 p o 2p e p p s s e V V D D V D D V D D V A A V A V V = + − + − = + = (20 5) 6.6667m/ s 30 5 102 2 = + − = Problem 4. – A slow leak develops in a storage bottle and oxygen slowly leaks out. The volume of the bottle is 0.1 m3 and the diameter of the hole is 0.1 mm. The initial pressure is 10 MPa and the temperature is 20°C. The oxygen escapes through the hole according to the relation e Ae T m& = 0.04248 p where p is the tank pressure and T is the tank temperature. The constant 0.04248 is based on the gas constant and the ratio of specific heats of oxygen. The units are: pressure N/m2, temperature K, area m2 and mass flow rate kg/s. Assuming that the temperature of the oxygen in the bottle does not change with time, determine the time it takes to reduce the pressure to one half of its initial value. ∀ = 0.1 m3 p1 = 10 MPa T1 = 293K = T2 p2 = 5 MPa kg K 259.8219 J 32 R 8,314.3 ⋅ = = From the continuity equation me dt dm = − & but RT m p ∀ = so p T 0.04248 A m dt dp dt RT dm e = − e = − ∀ = & Integrating we get, O2 m(t) d = 0.1mm e Ae T m& = 0.04248 p 4 t 0.04248 R TA p p ln e 1 2 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ∀ = − 46,713.4076sec 12.9759 hrs 2 ln 1 (259.8219) 293 1000 mm 0.1 mm m 4 (0.04248) 0.1 p p ln (0.04248)A R T t 2 1 2 e = = ⎟⎠ ⎞ ⎜⎝ ⎛ ⎟⎠ ⎞ ⎜⎝ ⎛ ⎟⎠ ⎞ ⎜⎝ ⎛ π = − ∀ = − Problem 5. – A normal shock wave occurs in a nozzle in which air is steadily flowing. Because the shock has a very small thickness, changes in flow variables across the shock may be assumed to occur without change of cross-sectional area. The velocity just upstream of the shock is 500 m/s, the static pressure is 50 kPa and the static temperature is 250 K. On the downstream side of the shock the pressure is 137 kPa and the temperature is 343.3 K. Determine the velocity of the air just downstream of the shock. V1 = 500 m/ s V2 = ? p1 = 50 kPa p2 = 137 kPa T1 = 250 K T2 = 343.3 K A1 = A2 From the continuity equation m& 1 = m& 2 So ρ1A1V1 = ρ2A2V2 (500) 250.5839m/ s 250 343.3