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Solution Manual For Gas Dynamics, 3rd Edition

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    1 C Ch ha ap pt te er r O On ne e B BA AS SI IC C E EQ QU UA AT TI IO ON NS S O OF F C CO OM MP PR RE ES SS SI IB BL LE E F FL LO OW W 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 m 3 . Atmospheric pressure is 101 kPa and the local acceleration of gravity is 9.81 m/s 2 . (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. K kg / kJ 7 28 . 0 R s / m 81 . 9 g m 1 C 300 273 27 T kpa 1 22 101 120 P P P 2 3 atm gage abs ⋅ = = = ∀ ° = + = = + = + = a) 3 m kg 5668 . 2 ) 300 )( 287 . 0 ( 221 RT P = = = ρ N 1801 . 25 ) 81 . 9 )( 1 )( 5668 . 2 ( g mg W = = ∀ ρ = = b) 3 earth moon m kg 5668 . 2 = ρ = ρ N 1967 . 4 W 6 1 W 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, V 2 /2, (m 2 /s 2 ) in order to add this term to specific enthalpy h (kJ/kg). Air

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    2 a) 2 2 2 2 3 2 3 2 V s m s N m kg 1 kg m N kg m m N p m kg , m N p ≈ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⋅ ⋅ − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ≈ ρ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ≈ ρ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ≈ b) kg kJ 1000 1 J 0 100 kJ 1 m N J 1 kg m N P = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ ⋅ ≈ ρ c) c 2 2 2 2 g 1000 1 factor kg kJ J 0 100 kJ 1 m N J 1 m kg s N 1 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. e i m m & & = s s p p s p i V A V A m m m ρ + ρ = + = & & & e e e V A m ρ = & e e s s p p V A V A V A = + ∴ So e s s p p e A V A V A V + = p s e A A A + = 2 p p D 4 A π = 2 p 2 o s D 4 D 4 A π − π = 2 o e D 4 A π = i e s p

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    3 ( ) ( ) s p 2 o 2 p s 2 o s 2 p 2 o p 2 p e s s p p e V V D D V D V D D V D A V A V A V − + = − + = + = ( ) s / m 6667 . 6 5 20 30 10 5 2 2 = − + = Problem 4 . – A slow leak develops in a storage bottle and oxygen slowly leaks out. The volume of the bottle is 0.1 m 3 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 e A T p 04248 . 0 m = & 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/m 2 , temperature K, area m 2 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. 3 m 1 . 0 = ∀ MPa 0 1 p 1 = 2 1 T K 293 T = = MPa 5 p 2 = K kg J 8219 . 259 32 3 . 314 , 8 R ⋅ = = From the continuity equation e m dt dm & − = but RT p m ∀ = so p T A 8 0424 . 0 m dt dp RT dt dm e e − = − = ∀ = & Integrating we get, O 2 m(t) d = 0.1mm e e A T p 04248 . 0 m = &

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    4 t A T R 8 0424 . 0 p p ln e 1 2 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ∀ − = hrs 59 7 9 . 12 sec 4076 . 713 , 46 2 1 ln 293 ) 8219 . 259 ( mm 0 100 m mm 1 . 0 4 ) 04248 . 0 ( 1 . 0 p p ln T R A ) 04248 . 0 ( 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. s / m 0 50 V 1 = ? V 2 = kPa 0 5 p 1 = kPa 7 13 p 2 = K 0 25 T 1 = K 3 . 343 T 2 = 2 1 A A = From the continuity equation 2 1 m m & & = So 2 2 2 1 1 1 V A V A ρ = ρ s / m 5839 . 250 ) 500 ( 250 3 . 343 137 50 V T T p p V RT / p RT / p V V 1 1 2 2 1 1 2 2 1 1 1 2 1 2 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = = = ρ ρ = 2 1

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    5 Problem 6 . – A gas flows steadily in a 2.0 cm diameter circular tube with a uniform velocity of 1.0 cm/s and a density ρ o . At a cross section farther down the tube, the velocity distribution is given by V = U o [1-( r/R) 2 ], with r in centimeters. Find U o , assuming the gas density to be ρ o [1+( r/R) 2 ]. s / cm 1 V 1 = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = 2 o 2 R r 1 U V o 1 ρ = ρ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ρ = ρ 2 o 2 R r 1 2 1 m m & & = o 2 2 1 o R o 1 1 1 R o 1 1 R R V rdr 2 V dA V m ρ π = π ρ = π ρ = ρ = ∫ ∫ & ( ) o o 2 2 o o 1 o 5 2 2 o o 2 2 o 2 2 R o R o o 2 2 2 U R 3 2 6 1 2 1 R U 2 R r where d R 2 U rdr 2 R r 1 U R r 1 dA V m ρ π = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − πρ = = ξ ξ ξ − ξ π ρ = π ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + ρ = ρ = ∫ ∫ ∫ & o o 2 o 2 U R 3 2 R ρ π = ρ π ∴ so s / cm 2 3 V o = Problem 7 . – For the rocket shown in Figure 1.6, determine the thrust. Assume that exit plane pressure is equal to ambient pressure. ( ) ( ) ( ) e e 2 o H e e o H o H e e e atm e A m m V m m m m 0 V m A p p ρ + = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ρ + + + = + − = & & & & & & & T r 1 2

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    6 Problem 8 . – Determine the force F required to push the flat plate of Figure Pl.8 against the round air jet with a velocity of 10 cm/s. The air jet velocity is 100 cm/s, with a jet diameter of 5.0 cm. Air density is 1.2 kg/m 3 . Figure P1.8 To obtain steady state add + V p to all velocities V m F & = ( ) ( ) ( ) 1 . 0 1 5 . 0 4 2 . 1 AV m 2 + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ π = ρ = & s / kg 2 00259 . 0 = ( )( ) N 1 00285 . 0 1 . 1 002592 . 0 F = = Problem 9 . – A jet engine (Figure P1.9) is traveling through the air with a forward velocity of 300 m/s. The exhaust gases leave the nozzle with an exit velocity of 800 m/s with respect to the nozzle. If the mass flow rate through the engine is 10 kg/s, determine the jet engine thrust. Exit plane static pressure is 80 kPa, inlet plane static pressure is 20 kPa, ambient pressure surrounding the engine is 20 kPa, and the exit plane area is 4.0 m 2 . F x V = -10 s cm V j = 100 s cm V j = 110 s cm x V = 0 F

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    7 Figure P1.9 ( ) ( ) ( )( ) ( )( ) kN 245 5 240 300 800 10 4 20 80 V V m A p p i e e atm e = + = − + − = − + − = & T Problem 10 . – A high-pressure oxygen cylinder, typically found in most welding shops, accidentally is knocked over and the valve on top of the cylinder breaks off. This creates a hole with a cross-sectional area of 6.5 x 10 -4 m 2 . Prior to the accident, the internal pressure of the oxygen is 14 MPa and the temperature is 27 ˚ C. Based on critical flow calculations, the velocity of the oxygen exiting the cylinder is estimated to be 300 m/s, the exit pressure 7.4 MPa and the exit temperature 250 K. How much thrust does the oxygen being expelled from the cylinder generate? What percentage is due to the pressure difference? What percentage due to the exiting momentum? Atmospheric pressure is 101 kPa. Also note that 0.2248 lb f = 1 N. Figure P1.10 s / m 0 30 V e = 2 4 e m 10 5 . 6 A − × = MPa 4 . 7 p e = MPa 1 10 . 0 kPa 1 10 p atm = = k 0 25 T e = e e e e e e e V A RT p V A m = ρ = & k kg J 82 . 259 R ⋅ = ( )( ) ( )( ) s / kg 2 . 22 300 10 5 . 6 250 82 . 259 m N 10 4 . 7 4 2 6 = × ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ × = − 300 m/s 800 m/s

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    8 ( ) ( ) ( )( ) ( ) f 2 2 4 3 e e atm e lb 3 . 548 , 2 N 0 . 336 , 11 N 6 . 6664 4 . 4671 m kg s N 1 s m kg 300 2 . 22 N 10 4 . 6 10 101 7400 V m A p p = = + = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⋅ ⋅ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⋅ + × × × − = + − = − & T The thrust due to the pressure is 41% of total and that due to momentum 59%. Problem 11 . – Air enters a hand held hair dryer with a velocity of 3 m/s at a temperature of 20°C and a pressure of 101 kPa. Internal resistance heaters warm the air and it exits through an area of 20 cm 2 with a velocity of 10 m/s at a temperature of 80°C. Assume that internal obstructions do not appreciably affect the pressure between inlet and exit and that heat transfer to the surroundings are negligible. Determine the power in kW needed to operate the hair dryer at steady state. ( ) ( )( ) ( )( ) ( ) ( ) ( ) s kg 019939 . 0 10 100 m 20 353 287 10 101 V A RT p V A m 2 2 3 2 2 2 2 2 2 2 = = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = ρ = & ( ) ( ) kg kJ 3 . 60 20 80 K kg kJ 005 . 1 T T c h h 1 2 p 1 2 = − ⋅ = − = − ( )( ) kg kJ 0455 . 0 2000 3 10 3 10 g 0 100 1 2 V V c 2 1 2 2 = − + = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − i e

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    9 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + = − 2 V h m 2 V h m W Q 2 1 1 2 2 2 & & & & ( ) ( )( ) W 2051 . 203 , 1 kg kJ 203205 . 1 0455 . 0 3 . 60 019939 . 0 2 V V h h m W 2 1 2 2 1 2 = = + = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − + − = − & & Problem 12 . – Air is expanded isentropically in a horizontal nozzle from an initial pressure of 1.0 MPa, of a temperature of 800 K, to an exhaust pressure of 101 kPa. If the air enters the nozzle with a velocity of 100 m/s, determine the air exhaust velocity. Assume the air behaves as a perfect gas, with R = 0.287 kJ/kg · K and γ = 1.4. Repeat for a vertical nozzle with exhaust plane 2.0 m above the intake plane. (a) Horizontal nozzle 2 V h 2 V h h 2 2 2 2 1 1 o + = + = ( ) ( ) 2 1 2 1 2 1 p 2 1 2 T T 1 R 2 V T T c 2 V V − − γ γ + = − + = k kg J 287 R ⋅ = 5194 . 0 1000 101 p p T T 4 . 1 4 . 1 1 2 1 2 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = γ − γ K 5 . 415 T 2 = ( ) ( )( )( ) ( ) s / m 8 56 . 884 5 . 460 , 772 000 , 10 5 . 415 800 4 . 0 287 4 . 1 2 100 V 2 2 = + = − + = (b) Vertical nozzle 1 2 p2 = 101kPa p 1 = 1MPa T 1 = 800K V 1 = 100m/s

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    10 2 2 2 2 1 2 1 1 gz 2 V h gz 2 V h + + = + + ( ) ( ) ( )( )( ) 2 81 . 9 2 5 . 460 , 782 z z g 2 T T 1 R 2 V V 2 1 2 1 2 1 2 + = − + − − γ γ + = s / m 0 59 . 884 74 . 499 , 782 = = Problem 13 . – Nitrogen is expanded isentropically in a nozzle from a pressure of 2000 kPa, at a temperature of 1000 K, to a pressure of 101 kPa. If the velocity of the nitrogen entering the nozzle is negligible, determine the exit nozzle area required for a nitrogen flow of 0.5 kg/s. Assume the nitrogen to behave as a perfect gas with constant specific heats, mean molecular mass of 28.0, and γ = 1.4. 2 V h 2 V h h 2 2 2 2 1 1 o + = + = ( ) ( ) 2 1 p 2 1 2 T T c 2 h h 2 V − = − = ( ) K 1 . 426 2000 101 1000 P P T T 4 . 1 4 . 0 1 1 2 1 2 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = γ − γ K kg J 9 . 296 28 3 . 8314 R ⋅ = = ( ) ( ) ( )( )( ) s / m 2 . 1092 1 . 426 1000 9 . 296 7 T T R 7 T T 1 R 2 V 2 1 2 1 2 = − = − = − − γ γ = ( )( ) 3 2 2 2 m kg 798 . 0 1 . 426 9 . 296 000 , 101 RT p = = = ρ p1 = 2000kPa T 1 = 1000K V 1 ~ small 1 2 p2 = 101kPa m = 0.5kg/s = ρ 2 A 2 V 2 • A 2 = ? V 2 = ?

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    11 ( )( ) 2 2 2 2 2 cm 4 73 . 5 m 4 000573 . 0 2 . 1092 798 . 0 5 . 0 V m A = = = ρ = & Problem 14 . – Air enters a compressor with a pressure of 100 kPa and a temperature 20°C; the mass flow rate is 0.25 kg/s . Compressed air is discharged from the compressor at 800 kPa and 50°C. Inlet and exit pipe diameters are 4.0 cm. Determine the exit velocity of the air at the compressor outlet and the compressor power required. Assume an adiabatic, steady, flow and that the air behaves as a perfect gas with constant specific heats; c p = 1.005 kJ/kg · K and R = 0.287kJ/kg·K. kg kJ 005 . 1 c p = k kg kJ 287 . 0 R ⋅ = s / kg 5 2 . 0 m m m 2 1 = = = & & & ( ) 2 2 2 2 1 m 6 0012 . 0 04 . 0 4 d 4 A A = π = π = = ( )( ) 3 1 1 1 m kg 189 . 1 293 0.287 kPa 0 . 10 RT p = = = ρ ( )( ) 3 1 2 2 m kg 630 . 8 323 0.287 800 RT p = = = ρ ( )( ) s m 3 . 167 00126 . 189 . 1 25 . 0 A m V 1 1 1 = = ρ = & ( )( ) s m 1 . 23 00126 . 63 . 8 25 . 0 A m V 2 2 2 = = ρ = & p 1 = 100kPa T 1 = 293K d = 4.cm 1 2 W = ? • m = 0.25 kg/s • p 2 = 800KPa T 2 = 323K V 2 = ?

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    12 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + = − 2 V h m 2 V h m W Q 2 1 1 1 2 2 2 2 & & & & ( ) ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ − + − = − 2 V V h h m W 2 1 2 2 1 2 & & ( ) ( )( ) ( )( ) ( )( ) ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − + + − = 1000 2 1673 1 . 23 1 . 23 3 . 167 293 323 005 . 1 25 . 0 ( )( ) kW 1 . 4 s kJ 1 . 4 73 . 13 15 . 30 25 . 0 = = − = Problem 15 . – Hot gases enter a jet engine turbine with a velocity of 50 m/s , a temperature of 1200 K, and a pressure of 600 kPa. The gases exit the turbine at a pressure of 250 kPa and a velocity of 75 m/s . Assume isentropic steady flow and that the hot gases behave as a perfect gas with constant specific heats (mean molecular mass 25, γ = 1.37). Find the turbine power output in kJ/(kg of mass flowing through the turbine). K kg J 6 . 332 25 3 . 8314 R ⋅ = = 37 . 1 = γ ( )( ) K kg kJ 2314 . 1 37 . 6 . 332 37 . 1 1 R c p ⋅ = = − γ γ = K 3 . 947 600 250 1200 p p T T 37 . 1 37 . 0 1 1 2 1 2 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = γ − γ 1 2 W • V 1 = 50m/s T 1 = 1200K p 1 = 600kPa p 2 = 250 kPa V 2 = 75 m/s

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    13 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + = − 2 V h m 2 V h m W Q 2 1 1 1 2 2 2 2 & & & & ( ) ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ − + − = 2 V V h h m W 2 2 2 1 2 1 & & ( ) ( )( ) 2000 V V V V T T c 2 V V h h m W W 2 1 2 1 2 1 p 2 2 2 1 2 1 − + + − = − + − = = & & ( )( ) ( )( ) kg kJ 6 . 309 563 . 1 18 . 311 2000 25 125 3 . 947 1200 2314 . 1 W = − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − + − = Problem 16 . – Hydrogen is stored in a tank at 1000 kPa and 30°C. A valve is opened, which vents the hydrogen and allows the pressure in the tank to fall to 200 kPa. Assuming that the hydrogen that remains in the tank has undergone an isentropic process, determine the amount of hydrogen left in the tank. Assume hydrogen is a perfect gas with constant specific heats; the ratio of specific heats is 1.4, and the gas constant is 4.124 kJ/kg · K. The tank volume is 2.0m 3 . K 303 T kPa 1000 p 1 1 = = kPa 200 p 2 = ( ) K 3 . 191 1000 200 303 p p T T 4 . 1 4 . 0 1 1 2 1 2 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = γ − γ ( )( ) ( )( ) kg 507 . 0 3 . 191 124 . 4 2 200 RT p m 2 2 2 = = ∀ =

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    14 Problem 17 . – Methane enters a constant-diameter, 3 cm duct at a pressure of 200 kPa, a temperature of 250 K, and a velocity of 20 m/s . At the duct exit, the velocity reaches 25 m/s . For isothermal steady flow in the duct, determine the exit pressure, mass flow rate, and rate at which heat is added to the methane. Assume methane behaves as a perfect gas; the ratio of specific heats is 1.32 (constant) and the mean molecular mass is 16.0. 2 2 2 1 1 1 V A V A ρ = ρ 2 2 2 1 1 1 V RT p V RT p ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ 2 2 1 1 V p V p = ( ) 2 2 1 1 2 m N 160 25 20 200 V V p p = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = ( ) 2 2 m 000707 . 0 03 . 0 4 A = π = K kg J 519.6 16 3 . 8314 R ⋅ = = ( ) ( ) ( ) ( ) ( ) s kg 02176 . 0 20 250 000707 . 0 6 . 519 200 V A RT p V A m 1 1 1 1 1 1 1 = = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = ρ = & ( ) ( )( ) W 448 . 2 2 5 20 25 02176 . 0 2 V V m Q 2 1 2 2 = + = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − = & & Problem 18 . – Air is adiabatically compressed from a pressure of 300 kPa and a temperature of 27 C to a pressure of 600 kPa and a temperature of 327 C. Is this compression actually possible? 1 2 T = constant Q • p 1 = 200kPa T 1 = 250K V 1 = 20 m/s p 2 = ? m = ? • V 2 = 25m/s Q = ? • d = 3cm γ = 1.32 MW = 16

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    15 kPa 300 p 1 = kPa 600 p 2 = K 300 273 27 T 1 = + = K 600 273 327 T 2 = + = 300 600 ln R 300 600 ln c p p ln R T T ln c s s p 1 2 1 2 p 1 2 − = − = − ( ) 0 2 ln c 2 ln R c v p > = − = possible ∴ Problem 19 . – Two streams of air mix in a constant-area mixing tube of a jet ejector. The primary jet enters the tube with a speed of 600 m/s, a pressure of 200 kPa and a temperature of 400 ˚ C. The secondary stream enters with a velocity of 30 m/s, a pressure of 200 kPa and a temperature of 100 ˚ C. The ratio of the area of the secondary flow to the primary jet is 5:1. The air behaves as a perfect gas with constant specific heats, c p = 1.0045 kJ/kg· K. Using the iterative numerical procedure described in Example 1.9 determine the velocity, pressure and temperature of the air leaving the mixing tube. g c 1 α 5 γ 1.4 R 287 c p 1004.5 Primary Secondary V 600.00 30.00 T 673 373 P 200,000 200,000 A 43,122.5078 B 263,528.7595 C 706,538.5693 n V e (m/s) P e (Pa) T e (K) 1 0.0000 101,000.0 293.1500 2 125.1620 244,722.8 695.5757 3 122.5671 245,112.7 695.8957 4 122.4284 245,133.6 695.9126 5 122.4210 245,134.7 695.9135 6 122.4206 245,134.7 695.9136 7 122.4206 245,134.7 695.9136

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