ME 322 Worksheet Winter 2007 Introduction to Compressible Flow

Transcription

1 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 Sound Speed m/

2 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 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 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 m m Subtitute numerical value into Equation 4. From Table 1.6 in Munon, Young and Okiihi inide book cover, E v N/m W N π m m m 675, 000 N 0.5 m If a mall car weigh 1.5 ton 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 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 N π m 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 5 3. Evaluate c krt. Ue abolute temperature and g c to get correct unit. Air: R R u M k 1.4 ft lbf mol R mol ft lb f R c krt ft lb f ft lbf lb m R R c 1130 ft H : R R u M k 1.4 ft lbf mol R mol ft lb f R coincidentally ame a air c krt ft lb f ft lbf lb m R R c 475 ft

6 6 kg mol K kg kg mol 4. Ue c krt with k 1.4, R 8315 J At 1000 m: J kg K c krt J kg 336 J kg K 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/

7 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 inch Hg 14.7 pia pia 9.9 inch Hg Compute tagnation temperature. k 1.4 for air. T o T 1 + k 1 Ma 457 R R10.8 T o 4940 R Hot! Compute tagnation preure. p o p k/k 1 To pia /0.4 T p o 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 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 pia T 0 V 0 p 14.7 pia m / Check for choked flow. p e 14.7 pia p pia 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 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 T R 833 R Recall that R 100 ft lb f / / R for the ga not air V e ft lb f 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 9 known from olution to part b. 100 ft lb f A e R 833 R 5.8 lb f 144 in 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 ft 144 in ft ft in ft A e 1.63 in

10 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 k 1 Ma k/k / p 0 p p pia Since ṁ i given the formula for the maximum flow rate i ueful ṁ max,air p 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 [ ].4/ A A and A 1.5 in 1.6 in 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 T p 0A [ p0 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 lb f 1.6 in in 0.5 lbm lbmft lb f ft lb f R Ma 1 A A * A factor of g c i T 0 51 R

11 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 ft lb f R 16 percent 51 R ft lbf V 101 ft p p p pia p 13.5 pia