Hey guys im having a lot of trouble with past exam question hopefully someone wi
ID: 1819501 • Letter: H
Question
Hey guys im having a lot of trouble with past exam question hopefully someone will be able to help me!!
Air flows under steady conditions through a compressor from 100kPa and 17*C to a pressure of 700kPa at a mass flow rate of 0.0333 kg/s. Evaluate the temperature at the exit from the compress and the compressor power inputs if the process is accomplished under..
(a) Reversible and adiabatic conditions.
(b) Reversible and isothermal conditions.
Treat the air as an ideal gas with variable specific heats and neglect changes in kinetic and potential energies. The specific gas constant for air, R, is 0.287 kJ/kg/K
Explanation / Answer
a)Reversible and adiabatic conditions( air assumed to be diatomic because mostly on N@ and o2)
p1^(1-k) * T1^k=p2^(1-k) * T2^k condition for revesible adiabatic process where k = 1.4 for air
P1= 100 kpa T1=290 k ,P2= 700 k pa substituing in above equation
T2=1.7436*T1= 505.655k
WORkDONE(Reversible and adiabatic)
rate of W by gas= -power input =k*mass flow rate* R*(T2- T1)/1-k mass flow rate= 0.0333 kg/s.
R= 0.287 kJ/kg/K
substituting values in above equation=k*0.0333 * 0.287*(505.655-290 )/(1-1.4)= - 7.21kj/s so P0WER INPUT= 7.21KJ/S
b) Reversible and isothermal conditions.
T2= T1= 290 k=T
WORKDONE
W BY GAS= mass flow rate* R*T*ln(p1/p2)=0.0333 * 0.287*290* ln( 100/700)=-5.393kj/s
POWER INPUT=5.39KJ/S