ME 3 Workheet Winter 007 Introduction to Compreible Flow 1. A two-liter cylindrical tank, 10 cm in diameter, ha a piton that fit perfectly. The piton doe not leak, and there i no friction between the piton and wall of the tank. W x L d Suppoe the tank i filled with water that i initially at 1 atm of preure. How much additional weight mut be placed on the piton to move the piton 1 cm?. Suppoe the tank in the preceding exerciei filled with air that i initially at 1 atm of preure. How much additional weight mut be placed on the piton to move the piton 1 cm? 3. Calculate the peed of ound in air, and in Hydrogen at 70 F. 4. Calculate the peed of ound in air at the altitude in the following table. Aume the pecific heat ratio i k 1.4. Altitude m Temperature C 0 15 1000 8.5 5000 17.5 10000 49.9 0000 56.5 Sound Speed m/
5. What i the tagnation temperature and tagnation preure on the noe of a reentry vehicle moving at a Mach number of 7 at an altitude of 00,000 ft.? At 00,000 ft. the ambient temperaure i 457 Rand the preure i 0.58 10 inch Hg. 6. An ideal ga k 1.4, R 100 ft lb f / / R i upplied to a converging nozzle at low velocity at at 100 pia and 540 F. The nozzle dicharge to atmopheric preure, 14.7 pia. Auming frictionle, adiabatic flow, and a ma flow rate of, /, calculate a. The preure in the exit plane in pia. b. The velocity in the exit plane in ft/. c. The cro-ectional area of the exit plan in in. 7. Air flowing in a duct ha a preure of 0 pia, a Mach number of 0.6, and a flow rate of 0.5 /. The cro ectional area of the duct i 1.5 in. a. Compute the tagnation temperature of the tream. b. What i the maximum percent reduction in area that could be introduced without reducing the flow rate of the tream? c. For the maximum reduction from part b, find the velocity and preure at the minimum area. Aume that the flow i adiabatic and friction i negligible.
3 Solution 1. Apply the definition of bulk modulu to finite change in p and V E v V dp dv V p V p E v V V where V mean that a poitive preure change occur when V i negative. Ue imple geometry of the cylinder 1 V LA and V L x V V where A π 4 D i the area of the cylinder. Combine Equation 1 and Equation p E v x L + x L 3 From the definition of preure, p W mg/a, where W i the added weight and m i the ma of the piton. Define W W mg a the change in weight neceary to compre the liguid, then p W A W A p E v A x L 4 From the given geometric data L V l A 1000 cm3 l π 4 0.10 0.5 m m Subtitute numerical value into Equation 4. From Table 1.6 in Munon, Young and Okiihi inide book cover, E v.15 10 9 N/m W.15 10 9 N π m 4 0.10 m 0.010 m 675, 000 N 0.5 m If a mall car weigh 1.5 ton 1.5 000 lb f 1 lb f /4.448 N 674 N, then the weight of 1000 car i required to compre the liquid by one cm!
4. Ue the ideal ga law to find the relationhip between change in preure and change in volume. pv mrt p mrt V p p 1 m RT /V m 1 RT 1 /V 1 T T 1 V 1 V Aume that the compreion proce i quaitatic and thermal equilibrium with the environment i maintained. Then T 1 T and From the geometry p p 1 V 1 V p p 1 V 1 V V 1 LA and V L xa p p 1 L L x From definition of preure, W p A. Combining the preceding equation give p W A p L 1 L x W p 1 A L L x Subtitute numerical value W 10135 N π m 4 0.10 m 0.5 m 0.5 m 0.01 m 88 N 88 N i the weight of m W g which i the ma of a large adult male. 88 N 85 kg. 9.8 m/
5 3. Evaluate c krt. Ue abolute temperature and g c to get correct unit. Air: R R u M 1545 8.97 k 1.4 ft lbf mol R mol 53.33 ft lb f R c krt 1.4 53.33 ft lb f 3.174 ft lbf lb 70 + 460 m R R c 1130 ft H : R R u M 1545.018 k 1.4 ft lbf mol R mol 765.6 ft lb f R coincidentally ame a air c krt 1.4 765.6 ft lb f 3.174 ft lbf lb 70 + 460 m R R c 475 ft
6 kg mol K kg kg mol 4. Ue c krt with k 1.4, R 8315 J At 1000 m: 8.97 87 J kg K c krt 336 1.4 87 J kg 336 J kg K 8.5 + 73.15 K N m 336 kg kg m m kg c 336 m Repeat calculation to fill in the table. Altitude m Temperature C Sound Speed m/ 0 15 340 1000 8.5 336 5000 17.5 30 10000 49.9 300 0000 56.5 95
7 5. Apply formula for tagnation preure and temperature. Ue abolute preure and temperature. Given preure and temperature are ambient condition and are the tatic preure and temperature in the formula for tagnation temperature and preure. Given: Ma 7, T 457 R, p 0.58 10 inch Hg 14.7 pia 0.0085 pia 9.9 inch Hg Compute tagnation temperature. k 1.4 for air. T o T 1 + k 1 Ma 457 R 1 + 0.4 7 457 R10.8 T o 4940 R Hot! Compute tagnation preure. p o p k/k 1 To 0.0085 pia 10.8 1.4/0.4 T p o 11.83 pia Note: The given preure could not be a gage preure becaue that would imply that p 14.7 pia at 00,000 ft.
8 6. Aume preure at the exit plane i equal to the preure of the urrounding. Ue ubcript e to deignate condition at the exit plane. p 0 100 pia T 0 V 0 p 14.7 pia m / Check for choked flow. p e 14.7 pia p 0 100 pia 0.147 Since p e /p 0 < 0.58 the flow i choked. Therefore, Ma 1 at the mallet area along the flow path, which i the exit plane a. Ma 1 p p at the exit. p e p 0.58p 0 for k 1.4 0.58100 pia p e 5.8 pia b. Ma 1 at the exit mean V e c krt and T T. Compute T and then V. T T 0 0.833 T 0.833540 + 460 R 833 R Recall that R 100 ft lb f / / R for the ga not air V e 1.4 100 ft lb f 3.174 ft lbf lb 833 m R R V e 1937 ft c. ṁ ρ e V e A e /. Ue ideal ga law to compute ρ e and then olve for A e. A e ṁ ρ e A e, ρ e p e RT e A e ṁrt e p e V e p e i known from the olution to part a. T e and V e value are
9 known from olution to part b. 100 ft lb f A e R 833 R 5.8 lb f 144 in 1937 0.01131 ft 144 in ft ft in ft A e 1.63 in d. ṁ ρ e V e A e /. Ue ideal ga law to compute ρ e and then olve for A e. A e ṁ ρ e A e, ρ e p e RT e A e ṁrt e p e V e p e i known from the olution to part a. T e and V e value are known from olution to part b. 100 ft lb f A e R 833 R 5.8 lb f 144 in 1937 0.01131 ft 144 in ft ft in ft A e 1.63 in
10 7. Known: k 1.4, Ma 0.6, ṁ 0.5 /, A 1.5 in, p 0 pia. a. We need to find T before we can compute T 0. Start by calculating p 0. p p 0 1 1 + k 1 Ma k/k 1 1 1 + 0.4 0.6 1.4/0.4 0.784 p 0 p 0.784 0 p 0 5.51 pia 0.784 Since ṁ i given the formula for the maximum flow rate i ueful ṁ max,air 0.6847p 0A RT0 Both ṁ max,air and A are unknown. However, ince A and Ma are known we can compute A from A A 1 [ 1 + k 1 ] k+1/[k 1] Ma k + 1 Ma 1 [ 1 + 0.4 ].4/0.8 0.6.4 0.6 A A 1.188 and A 1.5 in 1.6 in 1.188 But ince ṁ doe not vary along the duct ṁ max,air ṁ 0.5 m m Ma 0.6 A 1.5 in Solve Equation for T 0 T0 0.6847p 0A [ 0.6847p0 A ] 1 T 0 ṁ max,air R ṁ max,air R Everything on the right hand ide i known. needed for the unit to work out T 0 0.6847 5.51 lb f 1.6 in in 0.5 lbm 3.174 lbmft lb f 53.33 ft lb f R Ma 1 A A * A factor of g c i T 0 51 R
11 b. A i the minimum poible area if the ma flow rate i to remain unchanged. The actual area i A. Therefore the larget percent reduction in area i 100 A A A 100 c. Find V and p at A A V krt T krt 0 T 0 1.4 1.5 1.6 1.5 53.33 ft lb f R 16 percent 51 R 0.8333 3.174 ft lbf V 101 ft p p 0 0.583 p 0.5835.51pia p 13.5 pia