Question

(1 a Brayton cycle with an ideal regenerator has inlet (ompressor) at 2900K, 50 kPa with the highest PT as nato kla, 17000K. Find the specific heat transfer (au), and the cycle efficiency (using two approaches) using cold air properties. (50 points) (2) At the beginning of compression in a Diesel cycle T = 30ook, P=200kPa, and after combustion Cheat addition) is complete T=1500 K, and P=7.0 MPa. Find the compression ratio (w/v), the Thermal efficiency lusing two approaches) and the mean effective pressure (map). I Draw Pav, and tos diagrams, 150 points) This is a TAKE HOME EXAM. Please work independently. GOOD LOCK.

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Answer 1

Since the regeneration process is ideal the effectiveness of the regenerator is 100%. Further, the efficiency of the Brayton cycle can be found using two different methods. One using the standard efficiency equation and the other by finding the ratio of network done to the net heat supplied. In the diesel cycle we can find the efficiency using standard equations as well as adopting normal procedures.

Q. An ideal Brayton cycle with ideal Regenerator can be drawn in a Ts diagram, as shown below, P, P, 5 . The heat required for 2-3 is extracted from 5-6 in exhaust gas. Since it is given that the regenerator is ideali, its effectiveness is 100% ie; E= 1= Torta - To Tz

ie; Tz =TE Pressure ratio; rp Since Pz = Pz (same pressure line) 1170 = 13 90 = Sp. heat ratio 'ſ=14 cair) T=290k 8 T2 = 603-48K Ts = enteng lo thing To E 816.91K .: Ty=816.91K,

. The specific heat transfer (@H), is the net heat transfer inside the combustor, ie; (WH), = cp ( Tex-T3) cpair = 1.00565/29 (9#), = 1.005(1700–816.91) (941), = 887.49 KJ/Kg. Ist approach Bragton = Whet Din Whet = Wp - Wc wqa work done by Turbine Wc in ☺ compressor, Wq = cp (ThrTs) ane = 887.50 kJ/kg We = Cp (T2-Ti) = 315.04 K55 Whet = 572.4456J/kg. Qin = ), = 887,49 KI]ky

Trayton = 572.45 887.49 2011 . brayhon 64.50% and approach 1. Using standard equation of efficiency of ideal regenerative Brayton cycle, Zbrayfon- ideal Reg = 1- Tmin (p) * Tmin=290k Tmax = 17ook, rp = 13, Ma 104 air 0:4/14 brayhon = 1 - 290 (13) 1700 = 64.5%

→ > Qrejected -- > M - = 300x, 1-200 ka = 1500 , 3-P, = 7MIA . ज ।। Vie compression ratio - Y = 1.4 v=Y = 1267

Qrej = CVC Te-to) T = rejas Ye = expansion ratio re= cut off rati (constant 'Pressure Process Soo 828.48 :: rc a = 18 recept - M - / / / .V=V4 ( constant . volume Proces) - response the Me Feet = 12.67 37 re 1.81

Pressure = constant 93 T 2 t contact - Volume = contact Rcycle = whet Qin Whet = Qin-Qrej B Meycle = (- Quej Qin Mal Pin = GP CT3 -Tz) =(0 , T2 = 300x(12:67) = 828.48K Li Qina 1.005 (1500-82848) = 674.87 kg/kg

From, | Adiabatic gas L esuation ] - TH= 73 04 (7304 cre). Ty = 688-73K · Qrej = CV CTA -T.) ; Craivz0.71&KIJEG KO 0.718 (688.73 -300) = 279-11 kislig - diesel = 1- Qred din = 58.64%. of Now using standard equation efficiency of diesel cycle.

@ST ning all values dieses = 58.62% To find Mean effective Pressure, we can use the quation Whet) cycle = Pmean (V-V2) where mean = Meam eff. Pressure V-V2= swept vollume.

Using ideal Gas equation, e Rair=0:287 kJ/kce P, 19, = RTI L, CR TI = 0.287 x 300 Pi 200 = 0.43 milks. i llego = 12-67, -r

veig = e = 0.033 m3) kg, 12.67 Whet/cycle = Qin - Qrej = 674.87 – 279.11. - 395.70 kJ/ky. o Whettycle = Pream (11-02) Pmean = 395.76 a u-u2 - Pmean = 996-87 kPa.