Additional Deviations

07/07/2035 Glory to to God tuesday Additional derivation. The net work done per kg in the Otto cycle can also bet expressed in terms of P.H.T. P is expressed in bar i.e; 105 N/m', then work done. 5-1 - 5-1 Also, P3 = h = P2 r P4 P1 (1-pr P3 = PA = rp , where rp stands for pressure ratio. P2 P, and W= I (Ry - Pre ( P2L22 P.U. i) r-1-1 rus = F-1 (Ryv, Pize 1/1/1 1)) 24 r-1 (+FL1)) us = -5-1 W - P,24 6-1 (rp-i)) II PM )=s) = Mean effective pressure (Pm) is given r-1 - Part-AM 5-1 (G-Vs) bar. also Pm= [ Pire r-1 Pm Stroke 24-V2 Qs V.V. Qr, Pm Pm= vero 24-41 iii Pm= (8-1)(R-1)

Constant Pressure or Dierel cycle This cycle was introduced by Dr. P. Diesel in / 897. It differs from Otto cycle in that heat is supplied at Constant pressure instead of at constant volume. This cycle Comprises of the following operations: (i) 1-2 - Adiabatic Compression (ii) 2-3 - Addition of heat at constant pressure (ii) 3-4 - Adiabatic expansion (iv) 4-1 - Rejection of heat at constant volume. P T 2 3 p= const. Adiabatic 4 V Const. Adiabatic 4 1 V s Point 1 represents that the cylinder is full of air. Let P, V, and T, be the Corresponding pressure, volume and absolute temp. The piston then compresses the air adiabatically (ix, Pub c) till the values become P2 V2 and T2 respectively at point 2. Heat is then added from a not body at a constant pressure. During this addition of heat let Volume increases from V2 V3 to and temperature T2 to T3 corresponding to point 3. Point 3 is called the point of cut-off. The air then expands adiabatically to the conditions PA, V4 and T4 respectively corresponding to point 4- Finally, the air rejects the heat to the Cold body at constant volume till the point 1 where it returns to its orginal state Consider 1 kg of air. Heat supplied at constant pressure- - Cp (T3-T2 Heat rejected at Constant Volume = Cv (T1-7 Ti)

Work done = Heat supplied - Heat rejected. = Cp(T3-T2) - (T4-II) Ndiesel Heat supplied. Fs Work done - cp(Ts-T2) ext Cv T4-II) Cv CP = r) CP 73-T2) 1 r(T-T) Let Compression ratio 24, = Re, and cut-off ratio, p=V3 ( su volume at volume cut-off ) clearance Now, during adiabatic Compression, 1-2, FI 51 T2 THE = - 2 During Constant pressure process 2-3, T3 V Tz = V2 T3-ST2 = Pin = T3= Ti(Rc) 5-1 - 3 During adiabatic expansion 5-1 3-4, U2 ) & V4 v, u, x T3 V4 (Ric) = v3 is V3 U3 - V3 to T4 triog T3 T4 = r-1) T,(R) R ,74 = Rc) T4 = Ti PT 4 ReL.PRe Giving (2) ,and 4 in , todger toold n diesel = 1- (T,PT-71)

Ndiesel - 1-T/P - 1) X(R) Naiesel = 1- (PT-1) 1- I pr-17 or TCR) it P-1 5(Re)5-1(P-1) We Can see the efficiency of diesel cycle is different factor from that is of the Otto cycle only in the (bracketed factor) Henu This for a always greator than unity, because p>1. efficient. given Compression ratio, the otto cycle is move The net work for diesel cycle can be expressed in terms of Pv as follows: W = P2(2s-V2). + P3V3-P414 - P2 V2-P,V, 5-1 5-1 The last expression will be W = 1-r (1-1)] (r-1) Mean effective pressure Pn is given by: done P, U (iii)) Pm - = Work stroke volume (r=1) vy R-1 Pm - (R) (PP-1)] (5-1) (RCH1R)