Having a hard time completely figuring out this question out. Some step by step
ID: 1857899 • Letter: H
Question
Having a hard time completely figuring out this question out. Some step by step help or any would be nice. Thanks.
In a power plant operating on a Rankine cycle steam leaves the boiler at P3=20 MPa and T3=480 C. It leaves the turbine at P4=6 kPa and T4=80 C. It leaves the condenser as saturated liquid. Water enters the boiler at P2=20 MPa and T2=40 C.
a) Work generated by the turbine per unit mass of steam.
b) work required by the pump per unit mass of water.
c) Heat transfer per unit mass steam in the boiler.
d) The efficiency of the cycle.
Consider an ideal Otto cycle with a compression ratio of 8. Air is compressed at P1=100 KPa and T1=300 K. If the maximum temperature of the cycle is T3=1200 K. Find:
a) The air temperature after expansion (T4).
b) The net work per unit mass of air.
c) The mean effective pressure of the cycle.
d) the efficiency of the cycle.
In an ideal Diesel Cycle with a compression ratio of 15 air at P1=100kPa and T1=300 K is compressed. heat is transferred to the air after compression until its volume doubles Find:
a) The max temperature of the cycle.
b) The work required for compression per unit mass of air.
c) The work of expansion per unit mass of air?
d) The efficiency of the cycle?
Assume that the maximum allowable temperature in a brayton cycle is 1200 K. Air is compressed from P=100 Kpa and T=300 Kto P=1200 KPa. Find:
a) max allowable heat transfer per unit mass of air after compression
b) work of expansion per unit mass of air.
c) the work of compression per unit mass of air.
d) thermal efficiency of the cycle.
Explanation / Answer
work done by turbine= h3 - h4
=Cp(T3 - T4 )
=1.1(480-80)
=440 j/kg
work required by pump=h2-h1
= h1=h4
=Cp(T2- T4)
= 1.1(40-80)
= - 44 J/kg
heat transfer per unit mass= total wok transfer
=(work done by turbine- work required by pump)
=(440-(-44))
= 484 j/kg
heat in = h3 - h2
= Cp(T3-T2)
=1.1(480-40)
=484 j/kg
effiviency =heat transfer per unit mass/heat in
= 484*100/484
= 100%
2)
T3/T4=T2/T1=(compression ratio)^y-1
compression ratio=(r)^y-1
T4 =1200/(8)^1.04-1
= 1104.22 K
similarly
T2=T1*(r)^y-1
=300*(8)^1.04-1
=326.02 k
net work =Qs -Qr
Qs= mCv(T3-T2)
=1*.743(1200-326.02)
=649.36 j
Qr=mCv(T4-T1)
=1*.743(1104.22-300)
=597.54 j
so net work= Qs-Qr
=649.36 - 597.54
=51.82 j
Mean effective pressure = work output/ swept volume
swept volume= V1-V2
=V2(r-1)
so MEP= 51.82/V2(r-1)
Efficiency = 1- 1/(r)^y-1
= 1- 1/(8)^1.04-1
= 0.0798
=7.98
In diesel cycle
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