Question

Process Combuster Compressor ip Compressor Turbine 10 Intercooler Figure 1 shows a typical Brayton power cycle with inter-cooling Air enters the compressor at 98 kPa and 24 °C. Compression is done in two stages with inter-cooling between the stages. The compressor has an isentropic efficiency of 87% The gas exits the combustor at 490 kPa and 850 °C. The turbine has an isentropic efficiency of 95%. A gas stream with a temperature of 950 "C, recovered from another process, is added to the cycle and increases the mass flow in the cycle by 15%. 1.1. Calculate the thermal efficiency for this cycle. Assume ideal conditions where necessary. Where ideal conditions are assumed, it must be clearly stated,

Answer 1

T7=950c Ts=8500 1223K 5 490kpa 니' 672 8 U TA 2. 98 kPa Wcz اليا 1 Ti=24c on 297K (bean come 87% C) Ison 95% sen turbo from process 1-2 그 2 19.13 kPa 490x98 P2=P2 = Per API for ideal condition 73=T1 , Tu=Tz Wc = Wcz 낼 "P2 T2 To 024 104 Tz . 219,13 98 297 Ta - 373.776

comp . (tz-Ti) Ta'-To 0.875 373.77- 297 Ta'-297 Tu' Ta'a = 385.24k W=Wcz = Cp (T2"-T.) choos8385.24-297) Wa=Wez Klug for process, 7-8 Ooy Try Pz Pg 88.68 KB) Tz To ooy 1-4 /1223 TE ugo 98 To = 772.18K Curbine =95% Tz- Tgl Tz- Tg 0.952 1223 - Tg! 1223-772-18 To = 794.7214 W Turbine = ep (T7-To!) 1,005x(1223- 794.72

Turbine = 430.42 KJ o ling Whet thermal Cadd W Turbine - 2x Wcomp = C add Z 430042 – 2X 88.68 1,005X(1223 – 385624) = 30% thermal AN
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