Question

An aircraft engine operates on a simple ideal Brayton cycle with a pressure ratio rp of 9. Heat is added to the cycle at a rate of 490 kW; air passes through the engine at a rate of 1.1 kg/s; and the air at the beginning of the compression is at P1 = 71 kPa and T1 = 0 oC. Use constant specific heats at room temperature. The properties of air at room temperature are cp =1.005 kJ/kg.K and k = 1.4.

Determine the temperature after the compression that is T2.

Determine temperature as it enters the turbine that is T3.

Determine the temperature as it exits the turbine that is T4.

Determine the power produced by this engine and its thermal efficiency (please see the formulas below).

Question 1 (35) An aircraft engine operates on a simple ideal Brayton cycle with a pressure ratio rp of 9. Heat is added to the cycle at a rate of 490 kW; air passes through the engine at a rate of 1.1 kg/s; and the air at the beginning of the compression is at P1 = 71 kPa and Tj = 0 °C. Use constant specific heats at room temperature. The properties of air at room temperature are cp =1.005 kJ/kg.K and k=1.4. Determine the temperature after the compression that is T2. Determine temperature as it enters the turbine that is T3. Determine the temperature as it exits the turbine that is T4. Determine the power produced by this engine and its thermal efficiency (please see the formulas below). The dashes between 1 and 2 and between 3 and 4 in the T-s diagram below imply that there is no heat transfer in compression or expansion, since they are isentropic compression si = s2 and s3 = S4 isentropic expansion. in 273 K Yout Formulas T2 = Tır (k-1) because si entropy at state 1) = s2 (entropy at state 2) and T3 = T4r, (k-1) because s3 (entropy at state 3) = 54 (entropy at state 4) We =h2 - hi = Cp (T2 - T1), WT=h3 – h4 = Cp (T3 – T4), Wnet = WT-WC Power produced = Wnet = mWnet Jin=h3 - h2 = Cp (T3 – T2), qout=h4 – hi = Cp (T4 – T1) n = Wnet/qin, n = (WT-Wc)/qin, n = (qin - qout) qin Any formula can be used

Answer 1

Solution - Simple ideal Brayton cycle process, qin 1-2:- Reversible adjabatic comporession (pukac) 2-3: Constant pressure out 1 T 3 273k fi heat addition (p=c) 3-4: Reversible adiabatic expansion Cours) 4-1: constant pressure heat rejection (p=c) so Given - pressure ratio P2 P2 - P3 rp=g Pi P4 Temperatuwe at inlet of compressor T: 273 K pressure at inlet of compressor = P = 71 kPa Rate of heat addition 223 = 490 kw=lin mass flow rate m = 1.1 kg/s Air property: Specific heat at constant pressure Cp=1.005 KJ/kgk Adiabatic index K=1.4 • Temperature After compression (T2):- for isentropic (Reversible adiabatic) process: (K-1)/K Ti (1.4-1)/1.4 Te 273 238.45 °C - ( 13 T2=511-45k

• Temperature 98 it enters the turbine (T3) for process 2-3 (heat addition) Qin - micha-he)= mi Cp (T3-T2) => 490 = (1-1) (1005) ( T3 -511.45) T3 = 954.688k co 681.688 °C Temperature as it exits the turbine (T6) for process 3-4( isentropic process) T3 (VD) Ta (1:4-117.4 954.688 (9) (K-1/K - Tq P12-509. S9 K m 236.589 °C = m (Wq-We) power produced by this engine (Whet) Whet = im West Wq: Turbine work Wes compresser weste = [(y-he-he-hy] wi Cp [(T3 - T4.)- (Tz-T)] = (1.1)(1.005) [(954.688-509.589) (511.45-273)] Why= 228,45 kW -

Theomal efficiency (9) n = What 2-28.45 +90 0:4662 qin or 46.62% ni Result: Temperature After compression T2=511:45K 238.45°C • Temperature as it enters the turbine T3 = 954.680 K [email protected] °C Temperature as it exits the turbine. T4 = 509.589 k 236.589°C power produced by engine = 228.45 KW co 0 Whet •Thermal efficiency n = = 46.62% 0