The Brayton cycle below takes in 02 kgs of air at 300 K and

The Brayton cycle below takes in 0.2 kg/s of air at 300 K and 100 kPa. The pressure ratio of the compressor is 12. Air leaving the compressor mixes with methane at standard temperature and allowed to react. You must account for the changes in specific heat capacity with temperature and composition, so it is highly recommended that you attempt this in EES. Determine the mass flow rate of methane such that the temperature at 3 does not exceed 1300 K. Find the net power and thermal efficiency of the system. Calculate the total mass (kg) of CO_2 emitted in one year from this plant.

Solution

1) given that ma\'=.2 kg/s ,T1=300 k,p1=100 Kpa,Rp=12 for bryton cycle temperature T1,T2,T3 AND T4 are obtain as follows, where k=1.4

(T2/T1)=(Rp)^(k-1/k) on putting value we get

T2=T1*(Rp)^(k-1/k)=300*(12)^(.4/1.4)=610.18 K AND

for temperature at exit of combustion chamber to be 1300 k ,temperature at exit of turbine to be T4

T4=T3/((Rp)^(k-1/k))=1300/(12^(.4/1.4))=639.15 k

2) heat supplied in combustion chambershould produce only temperature of 1300 k and hence mass flow rate of fuel methane at NET CALORIFIC VALUE=50000 KJ/KG AT cpg=1.191 kj/kg k at 1300 k

mf\'*NCV=(ma\'+mf\')*cpg*T3-ma\'*cpa*T2

mf\'*50000=(.2+mf\')*1.191*1300-.2*1.005*610.18

mf\'=2.834*10^-3 kg

3)mass of exhaust gas

mg\'=ma\'+mf\'

mg\'=.2+2.834*10^(-3)

mg\'=.202834 kg/s

4) work of compression wc=ma\'*cpa*(T2-T1)=.2*1.005*(610.18-300)=62.34 KW

WORK OF TURBINE=wt=mg\'*cpg*(T3-T4)=.202834*1.191*(1300-639.15)=159.64 KW

NET WORK=WT-WC=159.64-62.34=97.30 KW

5) THERMAL EFFICIENCY

n= net work/mf\'*NCV=97300/(2.834*10^-3*50000000)=68.66%

6) equation of combustion reaction

CH4+2O2=CO2+2H2O

HENCE OUT OF TOTAL EXHAUST (1/3) is CO2, HENCE MASS FLOW RATE OF CO2 IS

mco2\'=mg\'/3=.202834/3=.06760 kg/s

for year time=365*24*3600=31536000 sec

mco2\' for year=mco2\'*time=.06760*31536000=2131833.6 kg