Question

An 8 MW air-standard regenerative Brayton cycle operating at steady state with intercooling and reheat, with a mass flow rate of 12 kg/s is shown in the figure. Regenerator 1,444.8K P, 100 kr ২৫ T, -1111.OK Combustor P. - 100 kPa 7,-4448K P, 1200 kPa 2 T-111LOK - 1200 kP. T, -1450 K P. - 1200 kPa Compressor stage 1 Compressor stage 2 Turbine Turbine stage 2 stage 1 cycle - 10 MW 7 T. - 410.1K IP. - 300 KN Intercooler 7-300K A 300 PM - 1034.3 KL, T, -1450 K TP. - 300 kPa P-300 kPa Reheat Combustor 71 - 300K P. - 100 kPa a Operating data at the principal states in the cycle are given in the table below. State p (kPa) T(K) h (kJ/kg) 100 300 300.19 300 410.1 411.22 b 300 300 300.19 2 1200 444.8 446.50 х 1200 1111.0 1173.84 3 1200 1450 1575.57 300 1034.3 1085.31 d 300 1450 1575.57 4 100 1111.0 1173.84 y 100 444.8 446.50 с

Answer 1

Given data: power = 8 MW Mass flow rate = 12 kg/s Since all devices are steady flow devices, so that we can use Steady flow energy equation Total Energy In Total Energy Out For compressor Q=0 (As device is adiabatic). So, hi hz +W and work is given in the compressox So, We = hz- For Turbine Wy = hi hz ( work is produced in turbine.). For combustion chamber W=0 (No work is done ). From steady flow energy equation hit Q=h2 --Qadded = ha-hi From eq" 0, 2 and 3 also the values given in the table we can determine all the work & heat added. (a) Work developed for turbine stage 1 - For using eqr 6 for process ① and ② Wto hg-hea (15as. 57-1085-31) KI/kg = 490.26 kJ/kg (b) Wook required for compressor stage 1 - Using en D for process and O. We = ha-hi ( 411-22-300-19) KJ/kg = 11.03 kJ/kg

CO Here we have to find the thermal efficiency of the cycle in percent (7) Ans - We know that thermal efficienly Net work done Heat added So, n = (WT, + Wtz) - (Wet Wer Qadd Net compressor Work = -03 kJ/kg +(446-50-30019) KI, (11-03 +14631) KT/K9 = 251.34 KI/kg Net work done in plant - Winet - Wenet = (153s 57-19384) k/kg +490-265/kg -257-34 kg/k- = (891-99-254.34) KJ/kg 634.65 kJ/kg Head added in Refrigerator (QR) - hd-hc or, Q R = 1575.57-1085.3t 1173.84 = 490.26 kJ/kg Now Net heat added - Q add in t Q add in Refrigerator Combustion (40173 +490-26) kJ/kg = 891.99 KJ/kg 634.65 kJ/kg 891.99 kJ/kg Hence n= Net work done total heat added 0%, n=07114 = 71.14./