GIVEN: Air enters the compressor of a Brayton Cycle power plant at 1 bar and 300
ID: 2997751 • Letter: G
Question
GIVEN:
Air enters the compressor of a Brayton Cycle power plant at 1 bar and 300K. When the air leaves, its temperature is 1400K. The compressor pressure ratio is 10. The mass flow rate is 6 kg/sec. THE TURBINE AND COMPRESSOR ARE ISENTROPIC.
REQUIRED:
Using IDEAL GAS PROPERTIES OF AIR table data ONLY, find the back work and the turbine total work.
*** Additional Information: ***
From my computations (interpolation) - T2 = 524.6 K ; h2 = 528.37 kJ/kg , also h4 = 1515.42 kJ/kg
PLEASE SHOW ALL WORK CLEARLY. I REALLY NEED TO UNDERSTAND HOW TO APPROACH THE COMBUSTOR. Thank you.
GIVEN: Air enters the compressor of a Brayton Cycle power plant at 1 bar and 300K. When the air leaves, its temperature is 1400K. The compressor pressure ratio is 10. The mass flow rate is 6 kg/sec. THE TURBINE AND COMPRESSOR ARE ISENTROPIC. REQUIRED: Using IDEAL GAS PROPERTIES OF AIR table data ONLY, find the back work and the turbine total work. *** Additional Information: *** From my computations (interpolation) - T2 = 524.6 K ; h2 = 528.37 kJ/kg , also h4 = 1515.42 kJ/kg & W Comp= m(h2-h1) = 1369.08 kWExplanation / Answer
m = 6 kg/s
STATE 1
P1 = 1 bar
T1 = 300 K
Pv = RT
1*10^5*v1 = 287*300
v1 = 0.861 m^3/kg
STATE 2
P2 = 10*P1
P2 = 10 bar
since 1 -2 is isentropic
P^(1-k)/k*T = const (k = 1.4)
P1^(1-k)/k*T1 = P2^(1-k)/k*T2
T2 = 300*(1/10)^(1-1.4)/1.4
T2 = 579.2 K
P1*v1^1.4 = P2*v2^1.4
v2 = v1*(1/10)^1/1.4
v2 = 0.166 m^3/kg
STATE 3
Combustion is constant Pressure process
P3 = P2 = 10 bar
v2/T2 = v3/T3
STATE 4
T4 = 1400 K
P4 = P1 = 1 bar
P4*v4 = R*T4
1*10^5*v4 =287*1400
v4 = 4.018 m^3/kg
and it is Isentropic process
P^(1-k)/k*T = const (k = 1.4)
P3^(1-k)/k*T3 = P4^(1-k)/k*T4
T3 = T4*(P4/P3)^(1-k)/k
T3 = 1400*(1/10)^(1-1.4)/1.4
T3 = 2703 K
Also
v3/T3 = v2/T2
v3 = 0.166*(2703/579.2)
v3 = 0.775 m^3/kg
Turbine work
Wt = m*(h3-h4)
Wt = m*Cp*(T3-T4)
Wt = 6*1.004*(2703-1400)
Wt = 7849.3 kW
Compressor work input
Wc = m*(h2-h1)
Wc = m*Cp*(T2-T1)
Wc = 6*1.004*(579.2-300)
Wc = 1682 kW
Therefore
Back work
Wb = Wc = 1682 kW
BAck work ratio = Wc/Wt = 1682/7849.3 = 0.214
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