kk Air enters a nozzle with a velocity of 30m/s and a specific enthalpy of 334 k
ID: 1858291 • Letter: K
Question
kk
Air enters a nozzle with a velocity of 30m/s and a specific enthalpy of 334 kJ/kg. The temperature of the air as it enters the nozzle is 110 degree C and the absolute pressure on entry is 2 bar The inlet area of the nozzle is 80 cm2. The air exits the nozzle with a velocity of 180 m/s and an absolute pressure of 1.2 bar The specific internal energy of the air at the exit from the nozzle is 273.2 kJ/kg. It the nozzle may be assumed to be adiabatic. calculate the specific volume of the air at the nozzle entry,[10%] mass flow rate of air in the nozzle. [10%] the specific enthalpy of the air at the exit from the nozzle [20%] the density of the air at the exit from the nozzle.[20%] the temperature of the air at the exit. |10%]Explanation / Answer
Specific volume = RT/P these at entry
Specific volume(Vi) = 287*(273+110)/2*10^5 = 0.5496 m^3/Kg
Mass flow = area at entry * Velocity at entry/ specific vol = 80*10^-4 *30/0.5496 = 0.4367 Kg/s
PV^gamma = Constant
Pi*Vi^1.4 = Pe*Ve^1.4
Ve = 0.7916 m^3/kg
Specific Enthalpy = Specific Internal energy + PVe
Specific Enthalpy = 273.2 + 1.2*10^5*0.7916*10^-3 (10^-3 for converting it into KJ/Kg)
Specific Enthalpy = 368.192 KJ/Kg
Density = 1/Ve = 1.263 Kg/m^3
P^(1-gamma) * T^gamma = constant
Pi^-0.4 * (273+110)^1.4 = Â Pe^-0.4 *Te^1.4
Te = 330.98 K
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