Respuesta :
Answer:
871.8008 KJ/kg
2573.42085 KJ/kg
0.3743
Explanation:
Solution:-
- We are to analyze an ideal Rankine cycle, where the condenser and boiler operating temperatures are defined.
- We start of by evaluating the properties of water at each state before and after the process.
State 1: Condenser Exit / Pump Inlet
T1 = 40°C ---> P1,sat = 7.3851 KPa , v1 = vf = 0.001008 m^3/kg
sat liquid h1 = hf = 167.53 KJ/kg , s1 = sf = 0.5724 KJ/kg.K
P1 = condenser pressure
State 2: Pump Exit / Boiler Inlet
P2 = P3 = Psat,300°C = 8587.9 KPa
Process 1: " Isentropic Compression - constant volume "
The work done by pump in the compression process is:
wp = v1* ( P2 - P1 )
wp = ( 0.001008 ) * ( 8587.9 - 7.3851 )
wp = 8.64915 KJ /kg
Determine the enthalpy at " State 2 " by energy balance on pump ( control Volume) :
h2 = h1 + wp
h2 = 167.53 + 8.64915
h2 = 176.17915 KJ/kg
State 3: Boiler Exit / Turbine Inlet
T1 = 300°C ---> P3,sat = 8587.9 KPa
sat vapor h3 = hg = 2749.6 KJ/kg , s3 = sg = 5.7059 KJ/kg.K
P3 = Boiler pressure
Process 2: " Heat Addition - constant pressure "
The heat supplied in the boiler is:
qb = h3 - h2
qp = ( 2749.6 - 176.17915 )
qb = 2573.42085 KJ /kg .... Answer ( b )
State 4: Turbine Exit / Condenser Inlet ( Isentropic )
P4 = P1 = 7.3851 KPa ..... sfg = 7.685 KJ/kg.K
s4 = s3 = 5.7059 KJ/kg.K sf = 0.5716994 KJ/kg.K
hf = 170.32524 KJ/kg
hfg = 2406.11 KJ/kg
Process 3: Isentropic Expansion - Determine the quality of liquid-vapor mixture phase ( x ) at state (4):
x = (s4 - sf) / sfg
x = (5.7059 - 0.5716994) / 7.685
x = 0.66808
h4 = hf + x*hfg
h4 = 170.32524 + 0.66808*2406.11
h4 = 1777.79920 KJ/kg
- The work-done by the turbine in the isentropic expansion process ( wt ) is:
wt = h3 - h4
wt = 2749.6 - 1777.7992
wt = 971.8008 KJ/kg ... Answer ( a )
- To determine the thermal efficiency ( nth ) of the rankine cycle. We need to determined the net work produced by the cycle ( wn ). The net work is the energy balance between the isentropic compression ( work done - pump ) and isentropic expansion ( work produced - turbine ):
wn = wt - wp
wn = 971.8008 - 8.64915
wn = 963.15165 KJ/kg
- The thermal efficiency of a power cycle is the ratio of net work-produced ( wn ) and the heat supplied to the working fluid in the boiler ( qb ) as follows:
nth = wn / qb
nth = 963.15165 / 2573.42085
nth = 0.3743 ..... Answer ( c )
