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Page 1 Chapter 12 Compressible Flow Review Problems 12131 A leak develops in an automobile tire as a result of an accident. The initial mass flow rate of air through the leak is to be determined. Assumptions 1 Air is an ideal gas with constant specific heats. 2 Flow of air through the hole is isentropic. Properties The gas constant of air is R = 0.287 kPa ·m 3 /kg ·K. The specific heat ratio of air at room temperature is k = 1.4. Analysis The absolute pressure in the tire is kPa 314 94 220 atm gage = + = + = P P P The critical pressure is, from Table 122, kPa 94 > kPa 166 = kPa) 314 )( 5283 . 0 ( 5283 . 0 * 0 = = P P Therefore, the flow is choked, and the velocity at the exit of the hole is the sonic speed. Then the flow properties at the exit becomes 3 ) 1 4 . 1 /( 1 3 ) 1 /( 1 0 * 3 3 0 0 0 kg/m 327 . 2 1 4 . 1 2 ) kg/m 671 . 3 ( 1 2 kg/m 671 . 3 K) 298 )( K kg / m kPa 287 . 0 ( kPa 314 = ? ? ? ? ? ? + = ? ? ? ? ? ? + = = · · = =   k k RT P ? ? ? K 3 . 248 K) (298 1 4 . 1 2 1 2 0 * = + = + = T k T m/s 9 . 315 K) 3 . 248 ( kJ/kg 1 s / m 1000 K) kJ/kg 287 . 0 )( 4 . 1 ( 2 2 * = ? ? ? ? ? ? ? ? · = = = kRT c V Then the initial mass flow rate through the hole becomes kg/min 0.554 = kg/s 0.00924 = m/s) /4](315.9 m) (0.004 )[ kg/m 327 . 2 ( 2 3 p ? = = AV m & Discussion The mass flow rate will decrease with time as the pressure inside the tire drops. PROPRIETARY MATERIAL. © 2006 The McGrawHill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1283 Page 2 Chapter 12 Compressible Flow Review Problems 12131 A leak develops in an automobile tire as a result of an accident. The initial mass flow rate of air through the leak is to be determined. Assumptions 1 Air is an ideal gas with constant specific heats. 2 Flow of air through the hole is isentropic. Properties The gas constant of air is R = 0.287 kPa ·m 3 /kg ·K. The specific heat ratio of air at room temperature is k = 1.4. Analysis The absolute pressure in the tire is kPa 314 94 220 atm gage = + = + = P P P The critical pressure is, from Table 122, kPa 94 > kPa 166 = kPa) 314 )( 5283 . 0 ( 5283 . 0 * 0 = = P P Therefore, the flow is choked, and the velocity at the exit of the hole is the sonic speed. Then the flow properties at the exit becomes 3 ) 1 4 . 1 /( 1 3 ) 1 /( 1 0 * 3 3 0 0 0 kg/m 327 . 2 1 4 . 1 2 ) kg/m 671 . 3 ( 1 2 kg/m 671 . 3 K) 298 )( K kg / m kPa 287 . 0 ( kPa 314 = ? ? ? ? ? ? + = ? ? ? ? ? ? + = = · · = =   k k RT P ? ? ? K 3 . 248 K) (298 1 4 . 1 2 1 2 0 * = + = + = T k T m/s 9 . 315 K) 3 . 248 ( kJ/kg 1 s / m 1000 K) kJ/kg 287 . 0 )( 4 . 1 ( 2 2 * = ? ? ? ? ? ? ? ? · = = = kRT c V Then the initial mass flow rate through the hole becomes kg/min 0.554 = kg/s 0.00924 = m/s) /4](315.9 m) (0.004 )[ kg/m 327 . 2 ( 2 3 p ? = = AV m & Discussion The mass flow rate will decrease with time as the pressure inside the tire drops. PROPRIETARY MATERIAL. © 2006 The McGrawHill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1283 Chapter 12 Compressible Flow 12132 The thrust developed by the engine of a Boeing 777 is about 380 kN. The mass flow rate of gases through the nozzle is to be determined. Assumptions 1 Air is an ideal gas with constant specific heats. 2 Flow of combustion gases through the nozzle is isentropic. 3 Choked flow conditions exist at the nozzle exit. 4 The velocity of gases at the nozzle inlet is negligible. Properties The gas constant of air is R = 0.287 kPa ·m 3 /kg ·K, and it can also be used for combustion gases. The specific heat ratio of combustion gases is k = 1.33. Analysis The velocity at the nozzle exit is the sonic speed, which is determined to be m/s 6 . 335 K) 295 ( kJ/kg 1 s / m 1000 K) kJ/kg 287 . 0 )( 33 . 1 ( 2 2 = ? ? ? ? ? ? ? ? · = = = kRT c V Noting that thrust F is related to velocity by V m F & = , the mass flow rate of combustion gases is determined to be kg/s 1132 = N 1 kg.m/s 1 m/s 335.6 N 000 , 380 2 ? ? ? ? ? ? ? ? = = V F m & Discussion The combustion gases are mostly nitrogen (due to the 78% of N 2 in air), and thus they can be treated as air with a good degree of approximation. 12133 A stationary temperature probe is inserted into an air duct reads 85 °C. The actual temperature of air is to be determined. Assumptions 1 Air is an ideal gas with constant specific heats at room temperature. 2 The stagnation process is isentropic. Properties The specific heat of air at room temperature is c p = 1.005 kJ/kg ·K. Analysis The air that strikes the probe will be brought to a complete stop, and thus it will undergo a stagnation process. The thermometer will sense the temperature of this stagnated air, which is the stagnation temperature. The actual air temperature is determined from T 250 m/s C 53.9 ° = ? ? ? ? ? ? · ×  ° =  = 2 2 2 2 0 s / m 1000 kJ/kg 1 K kJ/kg 005 . 1 2 m/s) (250 C 85 2 p c V T T Discussion Temperature rise due to stagnation is very significant in highspeed flows, and should always be considered when compressibility effects are not negligible. PROPRIETARY MATERIAL. © 2006 The McGrawHill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1284Read More
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