Air at 70 kPa and 30degree C and 230m/s enters a diffuser at a rate of 2.0kg/s a
ID: 2326767 • Letter: A
Question
Air at 70 kPa and 30degree C and 230m/s enters a diffuser at a rate of 2.0kg/s and leaves at 40 degree C. The exit area of the diffuser is 370cm^2. Consider air as an ideal gas with R = 0.287 kJ/(kgK) and C = 1.007kJ/(kgK). Determine Exit velocity Exit pressure Air enters an adiabatic gas turbine at 1310kPa and 400 degree C and exits the turbine at 100 kPa and 125 degree C. The mass flow rate of air is 10kg/s. Determine power output from the turbine. Air is compressed from 100kPa and 30 degree C to 80 degree C in an adiabatic compressor. Air mass flow rate is 7kg/s. Determine power required for this compressor. Consider C = 1.013kJ/(kgK). Saturated liquid of water at 50 kPa is throttled to 25 kPa. Determine initial and final temperature. Also specific enthalpy at initial and final state.Explanation / Answer
solution:
1) here for diffuser it is given that process occure adiabitically then whatever kinetic energy loss is converted in temperature rise and pressure rise then steady flow energy eqution for diffuser
h1-h2=c2^2-c1^2/2
c2^2=c1^2-2*Cp(T1-T2)=230^2-2*1.007(40-30)
c2=229.95 m/s
where pressure is given by
T2/T1=(P2/P1)^(n-1/n)
n=1.4
on solving we get
P2=78424.90Pa
2)steady flow energy equation for turbine is
-W=m(h2-h1)=mCp(T2-T1)=mCp((P2/P1)^(n-1/n))=-10.93 Kw
Wt=10.93 kw
3)for centirfugal compressor wark done is given by SFEE as
W=m(h2-h1)=mCp(T2-T1)=mCp((P2/P1)^(n-1/n))
here to calculate P2 we have
((P2/P1)^(n-1/n))=T2/T1
P2=170.6719 KPa
on putting work equation we get
Wc=1.17 kw
4) for throttling process no heat transfer,work done occure on sytsem as well kinetic and potential energy also remain unaltered,hence SFEE is
0=h2-h1
h1=h2=340.5 kj/kg k
u1+p1v1=u2+p2v2
but as v1=v2
u1=u2
T1=T2=81.33 degree celsius
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