An industrial process discharges 200,000 cubic feet/min of gaseous combustion pr
ID: 1907492 • Letter: A
Question
An industrial process discharges 200,000 cubic feet/min of gaseous combustion products at 400 F and 1 atm. As shown in the figure, a proposed system for utilizing the combustion products combines a heat-recovery steam generator with a turbine. At steady state, heat transfer from the outer surfaces of the steam generator and turbine can be ignored, as can the changes in kinetic and potential energies of the flowing streams. There is no significant pressure drop for the water flowing through the steam generator. The combustion products can be modeled as air as an ideal gas. Determine the turbine inlet temperature, in F. Determine the power developed by the turbine, in Btu/min.Explanation / Answer
1 atm = 14.7 psi
For air, gas constant R = 1716 ft-lb/slug-R
For air, Cp = 0.24 Btu/lb-R
Density of inlet combustion products = P1/(R*T1) = 14.7*12^2 / (1716*(400+260)) = 0.00187 slug/ft^3 = 0.00187*32.2 lb/ft^3
From steam properties, Inlet enthalpy of water at 40 psi and 102 F is h3 = 70.1 Btu/lb
Energy balance in generator: (2*10^5/60) *(0.00187*32.2)*0.24*(400-260) = (275/60)*(h4 - 70.1)
thus, h4 = 1541.5 Btu/lb
From steam properties at P4= 40 psi and h4 = 1541.5 Btu/lb, we get T4 = 1010 deg F
b)
From steam properties at 1 psi and quality = 93%, we get entahlpy h5 = 1030 Btu/lb
Turbine output = m*(h4 - h5)
Turbine power = 275*(1541.5 - 1030) = 140665.6 Btu/min
PLEASE RATE "BOTH: MY RESPONSES AS 5-STARS. OTHERWISE I WILL NOT GET ANY CREDIT. THANKS. LET ME KNOW IF ANY DOUBTS.