Question

5-30 Air enters an adiabatic nozzle steadily at 300 kPa, 200°C, and 30 m/s and leaves at 100 kPa and 180 m/s. The inlet area of the nozzle is 80 cm². Determine (a) the mass flow rate through the nozzle, (b) the exit temperature of the air, and (c) the exit area of the nozzle. Answers: (a) 0.5304 kg/s, (b) 184.6°C, (c) 38.7 cm P = 300 kPa T, = 200°C Vi = 30 m/s A = 80 cm AIR P2 = 100 kPa V2 = 180 m/s 5-36 Air at 80 kPa and 127°C enters an adiabatic diffuser steadily at a rate of 6000 kg/h and leaves at 100 kPa. The velocity of the airstream is decreased from 230 to 30 m/s as it passes through the diffuser. Find (a) the exit temperature of the air and (b) the exit area of the diffuser. 5-54 Argon gas enters an adiabatic turbine steadily at 900 kPa and 450°C with a velocity of 80 m/s and leaves at 150 kPa with a velocity of 150 m/s. The inlet area of the turbine is 60 cm². If the power output of the turbine is 250 kW, determine the exit temperature of the argon. 5-56 Refrigerant-134a enters an adiabatic compressor as saturated vapor at -24°C and leaves at 0.8 MPa and 60°C. The mass flow rate of the refrigerant is 1.2 kg/s. Determine (a) the power input to the compressor and (b) the volume flow rate of the refrigerant at the compressor inlet. 5-68 A well-insulated valve is used to throttle steam from 8 MPa and 500°C to 6 MPa. Determine the final temperature of the steam. Answer: 490.1°C

Answer 1

5-30 a) mass flow rate (n) = S, AVI 3 Density of air at enlet of nozzle of Dis es assumed as a perfect gas a obeys êded gas equote Hence. Pa PRT B S 3- 300X16 . 2. 20gglegim? - RT = 0.287 183 x 473 - o characteric gas constant for air= 0.287 kJ/kg k & T = (200 +273) = 473k 200 X 10 - a n = 2.2099x 80 X 10 X 188030 Kgls = 0.53038 kg/5 b) Using steady flow Energy Equation I KW Igle Eachon is en IIIg here A=0 [adiabatic nozzle] Kneto 7 het veren = het hele = p.I, & oo = I2 + ve Data about detum ir not given. 12

900 2000 1.00 32400 2000 Since, Air is assumed to be an éded gas, Hence Enthalpy hi f(t) Only Loos kykgex (442) + 300 = (1.00 S&T) + Beyond = 30 m/s 2= 180 m/s. - 1.005 (473-T) = 31500 15.75 - To=457.3284k - 184.324c no () exist are of nozzle (Q). RD (0.2878 457.3284) = 0.76188 kg/m3 uring con Fruity equoteon. I SAVI - za 4 A SAU - (0.53038 22 T0.76)888 180) = 0.00 38 674 m² = 38.674 cm?
Explanation-Following our guidelines,i solved the first question only as you had not mentioned about any one.Used steady flow energy equation to find the temperature at the exit of the nozzle as it is said that it is adiabatic and there is no useful work output so,both gets zero in SFEE.Air is assumed as an ideal gas so its enthalpy is a function of temperature only.Please feel free to ask if you have any doubt.

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