Consider a gas turbine power plant operating at steady state

Consider a gas turbine power plant operating at steady state with air flowing through the components. Relevant data at key locations are given in the table below and shown on figure P4.103. Kinetic and potential energy are negligible. The thermal efficiency of the power plant is 68%. Show how you determined the [underlined] information in the table below. You must completely detail your answers to receive full credit. Figure P4.103 cannot be used as a resource to determine a missing pressure or temperature state value. You must completely detail your answers to receive full credit. state m[lb/sec] P[psi] T [degree R] h[Btu/lb] w/q [Btu/lb] W/O [Hp] 1 38.165 14.7 520 124.27 -31.24 -1686.3 2 38.165 650 155.51 349.2 18852 3 38.165 2000 504.71 268.7 14506 4 38.165 14.7 980 236.02 -111.8 -6033 5 38.165 SCHEMATIC & GIVEN DATA

Solution

solution:

1)here for compressor work isgiven by

-Wc=mCp(T2-T1)

here 1 Hp=.7067 BTu/s,Cp=.24

1191.8549=38.165*.24*(T2-520)

T2=650.12 R

here work of compression is

wc=-1191.8549/m=-31.22 BTU/lb

where enthalphy are

h1=Cp*T1=.24*520=124.8 BTu/lb

h2=Cp*T2=.24*650.12=156 BTU/lb

3)for combustor heat supplied is given by

n=net work/heat supplied

heat supplied=Wt-Wc=14506-1686.3/.68=18852.5 Hp or 13323.7020 BTU/s

heat supplied per unit mass =13323.7020/38.165=349.13 BTu/lb

Qs=MCp(T3-T2)

T3=2104.72 R

here enthlpy are

h3=Cp*T3=.24*2104.72=505.13 BTU/lb

4)for turbine work is given by

Wt=m*Cp*(T3-T4)

T4=985.42 R

work done is given per pound is

Wt=Wt/m=10252.6523/38.165=268.64 BTU/lb

5)for rejected heat we have that

Qr=mCp(T4-T1)

Qr=4263.90 BTU/s or 6032.79 HP

per lb

qr=111.72 BTU/lb

7)in this way all underlined parameter are found