Question

A combined cycle gas turbine/vapor power plant uses the turbine exhaust as the energy source for the boiler. Each power system uses a single turbine. The gas power system is modeled as an ideal air-standard Brayton cycle. The vapor power system is modeled as an ideal Rankine cycle. Given specific operating conditions determine the temperature and pressure at each state, the rate of heat transfer in the boiler, the power output of each turbine, and the overall efficiency.

--Given Values--

Brayton Compressor Inlet: T1 (K) = 322

Brayton Compressor Inlet: P1 (kPa) = 190

Brayton Heat Exchanger Exit: T5 (K) = 530

Brayton: rp = 9.7

Brayton: rc = 2.45

Brayton: AV (m3/s) = 2.2

Rankine Turbine Inlet: T7 (C) = 480

Rankine Turbine Inlet: P7 (MPa) = 12

Rankine Condenser Inlet: P8 (kPa) = 50

a) Determine the specific enthalpy (kJ/kg) at the compressor inlet.

b) Determine the specific enthalpy (kJ/kg) at the combustor inlet.

c) Determine the specific enthalpy (kJ/kg) at the gas turbine inlet.

d) Determine the temperature (K) at the gas turbine inlet.

e) Determine the pressure (kPa) at the gas turbine inlet.

f) Determine the specific enthalpy (kJ/kg) at the gas turbine exit.

g) Determine the specific enthalpy (kJ/kg) of the gas at the heat exchanger exit.

h) Determine the mass flow rate of the air (kg/s) in the gas-power system.

i) Determine the net-power output (kW) of the Brayton cycle.

j) Determine the rate of heat transfer (kW) to the combustor.

k) Determine the rate of heat transfer (kW) to the water in the boiler.

l) Determine the specific enthalpy at the turbine inlet (kJ/kg).

m) Determine the specific entropy (kJ/kg-K) at the turbine inlet.

n) Determine the specific enthalpy at the condenser inlet (kJ/kg).

o) Determine the specific enthalpy at the pump inlet (kJ/kg).

p) Determine the specific enthalpy at the pump exit (kJ/kg).

q) Determine the mass flow rate of the water (kg/s) in the vapor-power system.

r) Determine the net power (kW) of the Rankine cycle.

s) Determine the thermal efficiency (%) of the combined power plant.

Answer 1

Ts=530k >Wy Heat exchanger I Brayton S cycle 4 2 190 kPa Wc A tares T1=322K 1 T1 7 To=4806 Claud [ Ranking Rankinę Cycle 12 MPa >W g Wp 50 kPa torej 8 specific enthalpy at comprens inlet hiz CP T hi 1,005 X 322 The= 323.61 KI/kg And こ B at combusion inlet Specific enthalpy hst = Cp X T5 = 1.005 X 530 Inst 532.65 KS kg Ang

specific enthply at the gas 2 turbine inlet rc Tz Ts Tz = 2045× 530 }=1298.5k 3 Cpx 1298.5 h =1.005 X 1298.5 = 13ou.gg KJ kJ/kg Any ca turbine Inlet Temperature at the gos = 1298.5k [from the past co Any P2 = 13 = Pixop 9.7X190 113 1843 kPa = Ang from procen 3-4 Or4 P3 Itu Pu C To Tu 1298.5 Tu= 1843 10 y lo 190 (Tuz 678.43K hu = Cpx Ty = 1-005X678.43= 681.82 Kthing Ang

3 heat exchanger Specific enthalpy of the exit at gas 2.4 To = oy Tix opt 322 x 9.74 T2= 616.29K hza Cpx Tz =1005X616.29 619.37 kJ Kolly And mass flow rate of air =PG = (2x 212 = 2064 g/s An Q Net power output S m ( Wf-Wc) = r [ (hs-hu) - (hu) 2,64x{ (1304. 99-681-83) -(619,37- 323,60) 드 864.36 KW Net Power output ☺ Net heat transfer to the combustor mx (ha-hs) = 2.64% (1304.99- 532.65 2038.97 kW Ary

rate of heat transfer to water = mx Chu-hu = 2.64X ( 681.82 – 323.61) 45.67 KW Am At p= el 12Mpa Tau&oc specific enthply at turbine inlet 329513 kg [ from steam table Aru 17= m) specific entropy at turbine inlet At p=12 MPa T=480c St 6, 4186 Klugh [from steam table n to determine the specific enthply at the condensor inlet S = Sg = Sf te (sg - Sf) At 50 kPa or P=005 Mia = log12 hf 340.54 Kyung Sg = 7.5930 kJ/ ng =2645.2 Sf KJ byk KJ/kg Klugh S7=Sq = 6.4186= 1-0912 + x (7.5930–100912) xz 0.819 hq = hf tax (hg-hf)

hq = 340.54 +0.819 x(2645-2- 340.54) Tho 2228.05 Ks/vg Aus Specific enthply at pump inlet Ing=ht At p=50kPa [from steam team table - 340.54 KJ/ - US AP = = (P) Specific enthely at pump exit WP ho-ng 0.00102993% (12000 -50) ho-hg= 12.30 KJ) KI/kg [of=from Steam table at hg + 12.30 340.54 +12:30 352.84 KJ Ang P= 50Kpa) hor ho= Kong q flow rate of water mass Prej] Broy 945.67 = mwx (h7 ho [from from port (k] rwx [3295.3–352.84] this = 0.3214

© Wivet Net power (kw) of the Rankine Cycle (WT-Wpxm I(haha)-(as hy) Jxmi [(32953 95:3 -2228005) - (12:30)] X0:321 - s IWnet - 338.63 kW Ary Whet (hemoro Romion = = thers Brayton 864.36 2038.97 Cadd = 0.423 & Bray. are 42.3% Erker (ther Wnet Clodd 338.63 in Chr-ho) 338.63 0.321X(3295:3-352.84) 0.3585 or 35.85% (hcombined) - be thRank - Bray Blank 0.423 +0.3585-00423 80.355 0.6298 62.98% Combor Any
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