Question

174 2 combustor , =1500 K Problem 2 (22 pts): Brayton cycle and entropy calculations An open Brayton cycle is used to generate electricity. Air enters the isentropic compressor at T1 = 300 K and P1 = 100 kPa with a mass flow rate m = 1.5 kg/s. The combustor can be modeled as a constant-pressure heat exchanger, supplying 1540 kW of thermal power to the air from a thermal reservoir (Th = 1750 K). Air enters the isentropic turbine at T3 = 1500 K and exits at P4 = 100 kPa. 0,2 = 1540 kW 7 a) (5 pts) On the combustor T-s diagram," show the isobars compressor turbine P1 and P2 as W, dash lines, and (P = 100 kPa indicate the ${T, = 300 K processes 1+2, m=1.5 kg/s {PA = 100 kPa 2+3, and 3>4 b) (5 pts) By performing an energy balance on the combustor, derive an expression for h2, and determine T2 C) (4 pts) Calculate the pressure ratio rp = P2/P1 d) (8 pts) By formulating the entropy change across the combustor, calculate the rate of entropy generation, Sgen, for the combustor • variations of specific heats with temperature will be accounted for • use the closest value from the ideal gas air property table (do not interpolate)

Answer 1

P1 4

T1 = 300 K

P1 = 100 kPa

m = 1.5 kg/s

By Energy balance equation on combustor

Q - W = m (h2 - h1 + v22 - v12 / 2 + g z2 - g z1)

In heat exchanger,

1. No work transfer, i.e. W = 0

2. No change in velocity (KE = 0)

3. No potential head (z2 z1)

Q2-3 = m (h3 - h2)

h3 @ 1500 K = 300.19 kJ/kg

1540 = 1.5 (1635.97 - h2)

h2 = 609.3033 kJ/kg

At h = 609.3033 kJ/kg, T = 600 K

Therefore, T2 = 600 K

For isentropic process,

P2/P1 = (T2/T1)k/k-1

P2/P1 = (600 / 300)1.4/1.4-1

P2/P1 = 11.3137

At state 2, 600 K

s2 = 2.40902 kJ/kg-K

At state 3, 1500 K

s3 = 3.44516 kJ/kg-K

sgen = S - Q / TH

S = m (s3 - s2) = 1.5 (3.44516 - 2.4090) = 1.55424‬ kW/K

Q / TH = 1540 / 1750 = 0.88‬ kW/K

sgen = 1.55424‬ - 0.88‬

sgen = 0.67424‬ kW/K

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