Question

14. An I.C engine operating on Limited Pressure Cycle has a cylinder bore and stroke length of 250 mm and 400 mm respectively. Its compression ratio and expansion ratio are 9 and 5 respectively. The initial pressure and temperature of the air are 1 bar and 30°C. The heat added at constant pressure is twice the heat rejected at constant volume. The compression and expansion follows the law pv 1.25 - constant. Determine: (Pressures and temperatures at all salient points. (ii) Mean effective pressure of the cycle. (iii) Efficiency of the cycle. Take: Cp = 1 kJ/kg. K and Cy - 0.71kJ/kg.K (6 Marks)

Answer 1

There is one mistake in the question. It is given like heat added at constant pressure equal to twice of heat rejection at constant volume. It can't be true. Because when I calculated like this then I was getting value of temperature at point 3 is less than at point two, which can't be possible.

Also I have done so many types of questions in IC Engine. They give like this " heat supplied at constant pressure is twice of heat supplied at constant volume".

Because it happens practically also. So I have taken like this only. Please find the attached solution.

Cinen thel Bore (D)=250mm Stroke (L) = Yoomm Compression ratio (c)=9,= ( 1 ) Expansion refio. (Me) = 5' = Uusivu) P= l bar, Ti = 30°C = 303K (Pesche en du&u-elements > m Cp fru-T3) sim cuts 73) For compression and enforsion spuls=c Cp = 1 kJ Cv=0.71 kJ kgok vg-ks P-u diagra P 4 . as(p=2) Qs =) WE 1-2 s Compression 2-3.- Heef addition (N=6) 3-4 Heat addition (Pac) 4-5 » Expansion St → heet rejection Wc V2 = V3 Vy V=V5 il 1.25-1 FOR 1-2 To I ( 0:25 A - To = 303 x (9) د) جی 524.8k 1.25 Pul2sc Pe Pi IV 11:25 V2.) ~ P2 = 1 + (9) 125 P = 15:59 bar/

Vy • Cut off ratio (3) = {of pac, tau,7 Tu va TB V3 Mc re Vilva V5lvu { V = Vs and V2 = Vail re Ys/v3 Vu (о S V.3 re بالا لا So 8 1.8 ol's los Ty. T3 =>

Gimentul m Cp (fu-ts) = d.ma fis-72) 1 x (Th="73) = 2x0:71 | To-T2) 118 T3 -T3 = 1.42 T3 - 1:42 X 524.8 Tot 120.96k so Tu =)-8X T3 ти: 2163k 1,257 FOR 4-5 Tu (vs Vs -0.25 (s jo Vu Ts Ts 2163 1.4953 175 - 1446k FOR 2-3 → v=c → To Ta PO P3 = 15:59x ([201.96 524.8 P3 = 35.70 bar FOR 3-4 → PB = P4 35.To bai For 4-5 > 1:25 Ps. ITs 1-25-1 Pu Tu 1446 Ps = 35.7 x 2163 1:25 0.25 Ps=4.7668 bar

co state Pressurelbar) itemperature (k): I 303 2 15.59 524,58 120)-96 35.7 4 > 35.7 2163 4.7668 144.6 i mean effective pressure - Whet L Pm X Vs - of where Pm = mean offe fine pressure V, = 7 x D²XL - haftad Tx(6.25)** (64) Vs= swept volume ? ->V = 7-85x1013 m3 8.728104 m3 vi V2. q 9 Vz V, -V₂ = so vs 6.978810-3 m3 " Whet = We - We = mm Co | Tu-ts)- m coltz-7) -3 Pir=mrti sm= 100x103x 7-85x10 287x303 m= 9:0278 103 kg PVL RTI ( - Whef =9-027xlo"[(2163–1946) – 1 (504.8-303) ] Whel 4.47 kJ Now in eq" © → Pm x 6.9788103 Pm = 640.58 kPa 4.47

n Whet Qs Ps = mer (73-72) tm Co (Turts) 9.027x10-2 (0.71 ( 1201.96–524-8) + 1 ( 2163–120):9) Qs as 13.0lki 4:47 son Whef Qs Bool ORIn = 34.34. n = 0.3434