Question

A closed loop Brayton cycle using Helium as the working fluid is designed as a land based power station. The plant layout and limiting operating parameters are shown in the figure below. State any assumptions. Tz=2100 [K] 3 N Heat Exchanger Heat addition Wnet Brayton Turbine 1 Compressor Helium Brayton Cycle Heat Exchanger 4 T= 300 [K] Heat Rejection Py = 2 [kg/m3 Answer the following questions for a standard Brayton cycle designed to give maximum specific work. a) What are the limiting factors in a Brayton cycle in an attempt to achieve maximum specific work. b) Determine the Carnot cycle efficiency for this cycle, c) Determine the ratio of T,/T(maximum specific work) and the corresponding cycle efficiency. Answer the following questions assuming a cycle efficiency of 70% for this Brayton cycle. d) Determine the temperature ratio T//T. e) Determine the pressure ratio P2/P. f) Determine the absolute pressures P, and P2. g) Determine the absolute temperatures T, and T4 h) Determine the specific work of the compressor from 1-2, 1W2. i) Determine the specific work of the turbine from 3-4, 3W4. j) Determine the specific heat of the heat exchanger (heat addition) from 2—3, 293.

Answer 1

2 T3 = 2100K 3 Heat addition 3 What Qin Turbine f Compress Wy T 2 4 Ti = 80081 14. werden ti . Pq = 2 kg/m Qout 1 Heat rejection S 2igim3. achieving • Given that, To = compreber inlet temperature = 300k T3 = Turbine inlet temperature, Tmore 2000K. Ij = Density at inlet of compressor = working fluid is Helium which is a mano atomic gas. for, Helicem, r=1-67. @ The two limiting factors in a Brayton cycle in an attewept of maximum specific work ů, The maximum temperature is lenited due to metallurgical conditions of the turbine blade. ciig. The minimum temperature that generally depends upon atmospheric condition and we carit go below atmospheric temperature and pressure. carnot excle efficiency is written as.. Trmin T, Tmax Tz 6 Mcarmot = 1 1- 300 as = 0.8571 2100 Marnot 85.71% CS Scanned with CamScanner

for maximum specific work output, == STITE Heue T3 2000 I! 세서 백 Ta 300 ㅋ 세에 - 2646 and we know that o efficiency of Brayton cycle ['pa pressure ratio) 건 - 밥 and also, T2 - = (op) TE N . M-1 TZ 들 7 1 - T, ㅠ 1 초크 = 1- 2. 646 7 5 0 6221 } 건 = 62.21 1. be TO% @ Atsuming the efficieney from above formula, of efficieney. 키 nei - T 도 - = 1-n 포 > امام 007=. TT 03. 3:333. TZ .Te Scanned with CamScanner CS

Pa/P we know that pressure ratio Yp (&p) T ryy Vr- - Ta Ti 1.67%.67 > B. P = (3-3333) = 20.11 P2 PL - 20.de from Ideal gas equation, P = GRT constant Ra Charalteristics gas of 300K = 2:0769 kilig k Pet SRT,= 2x2-0769 x 200 Helium at =) → ܒܘ = 12 46.14 kPa P. =1.246 MPa and P2 = 20.1] =) P2=24922.8 kPa. PE P. = 24.923 MPa we know that presseuse ratio, (rp) V T2 TE 0-67/1.67 T2 = T(&p) > T2 = 300 x(20-11) CS Scannes) yrit = 1000 K CamScanter

Since, Pg=P₂ and P = P4 Hence, P2. Ps PE Pu riyy. rp = Tz Tu CP) - Tu To Y-yr (e) => T4 2100x3 lo U 4 =.630k Now specific work of the compress ar; We = cp [(T2-T₂) = 5.1926.[1000 -300] i go for Helium = s. 1926 Istigh We = 3634.82 kJ/kg ☺ Specific work for tussine, Wy= Cp (T3-T6) = 5-192612100-630) Wy = 7633.12 ksling Specific heat addition through heat exchanger q = cp [T3-T2) = 5.1926 (2100--1000) Lin 5711-86 Kglig in CS Scanned with CamScanner