You are designing an industrialscale naturalgasfired power p

You are designing an industrial-scale, natural-gas-fired power plant, based on a Brayton cycle with regeneration. Evaluate the performance of an air-standard Brayton cycle operating under the following conditions: The compressor inlet (State 1) has a pressure P_1 = 90 kPa and temperature T_1 = 21degree C. Due to material limits, the max temperature is 750 degree C. The net cycle power (W_turbine + W_compressor, where W_compressor

Solution

solution:

1)for given brayton cycle without regenerator efficiency is less but iducing regenerator efficiency increases and total heat supplied reduces

A)2)for PR=Rp=2

hence for temperature

T2=T1(Rp)^(.4/1.4)

T3=T4(Rp)^(.4/1.4)

we get that

T2=358.39 k

T4=839.20 k

2)but compressor and turbine efficiency are

nc=T2-T1/T2\'-T1

nt=T3-T4\'/T3-t4

we get actual cycle temperature as

T2\'=372.52 k

T4\'=861.25 k

5) mass flow rate is obtained as

net work=Wn=Wt-Wc=m\'Cp((T3-T4\')-(T2\'-T1))

hence for net work=1100 kw and cp=1.005 kj/kg k

we get

m\'=13.15 kg/s

hence specific net work=W/m\'=83.6461 kj/kg

6)cycle efficiency without regeneration is

n=Wn/Qh

Qh=m\'Cp*(T3-T2\')=8596.58 kw

n=.1279

6)on reneration effectiveness heat extraxted suppliesd to air before combustion hence

Er=Ts-T2\'/T4\'-T2\'

Er=.85

Ts=767.07 k

7)heat saved is

Qh\'=Qh-Qs

Qs=m\'*Cp(Ts-T4\')=5401.12 kw

Qh\'=3195.46

nreg=Wn/Qh\'=.3442

8)saving in cost=Qs*86400*8.3*10^-6=3873.25 dollar

B)for PR=Rp=3

hence for temperature

T2=T1(Rp)^(.4/1.4)

T3=T4(Rp)^(.4/1.4)

we get that

T2=402.40 k

T4=747.40 k

2)but compressor and turbine efficiency are

nc=T2-T1/T2\'-T1

nt=T3-T4\'/T3-t4

we get actual cycle temperature as

T2\'=426.19 k

T4\'=780.47 k

5) mass flow rate is obtained as

net work=Wn=Wt-Wc=m\'Cp((T3-T4\')-(T2\'-T1))

hence for net work=1100 kw and cp=1.005 kj/kg k

we get

m\'=9.919 kg/s

hence specific net work=W/m\'=110.89 kj/kg

6)cycle efficiency without regeneration is

n=Wn/Qh

Qh=m\'Cp*(T3-T2\')=5949.35 kw

n=.1848

6)on reneration effectiveness heat extraxted suppliesd to air before combustion hence

Er=Ts-T2\'/T4\'-T2\'

Er=.85

Ts=727.32 k

7)heat saved is

Qh\'=Qh-Qs

Qs=m\'*Cp(Ts-T4\')=3001.92 kw

Qh\'=2947.43

nreg=Wn/Qh\'=.3732

8)saving in cost=Qs*86400*8.3*10^-6=2152.73 dollar

C)for PR=Rp=4

hence for temperature

T2=T1(Rp)^(.4/1.4)

T3=T4(Rp)^(.4/1.4)

we get that

T2=436.88 k

T4=688.42 k

2)but compressor and turbine efficiency are

nc=T2-T1/T2\'-T1

nt=T3-T4\'/T3-t4

we get actual cycle temperature as

T2\'=468.24 k

T4\'=728.56 k

5) mass flow rate is obtained as

net work=Wn=Wt-Wc=m\'Cp((T3-T4\')-(T2\'-T1))

hence for net work=1100 kw and cp=1.005 kj/kg k

we get

m\'=9.105 kg/s

hence specific net work=W/m\'=120.8 kj/kg

6)cycle efficiency without regeneration is

n=Wn/Qh

Qh=m\'Cp*(T3-T2\')=5078.47 kw

n=.2166

6)on reneration effectiveness heat extraxted suppliesd to air before combustion hence

Er=Ts-T2\'/T4\'-T2\'

Er=.85

Ts=689.51 k

7)heat saved is

Qh\'=Qh-Qs

Qs=m\'*Cp(Ts-T4\')=2024.75 kw

Qh\'=3051.58 kw

nreg=Wn/Qh\'=.3604

8)saving in cost=Qs*86400*8.3*10^-6=1451.98 dollar

d)in this way performance can be obtain for Rp=5,it is observe that efficiency increases with increase in pressure ratio and cost of saved heat is less at higher pressure ratio