Question

2. An ideal Diesel cycle has a compression ratio of 18 and a cut-off ratio of 2. At the beginning of the compression process P = 95 kPa and T = 20°C. (a) Find the temper- ature, pressure and specific volume at states 1, 2, 3, and 4. (b) Find the heat supplied, heat rejected, work done by the system, work done on the system, net work done, and thermal efficiency. (c) Calculate the thermal efficiency of a Carnot cycle operating be- tween the same temperature limits. (d) Calculate the change in entropy of the working fluid for (i) the combustion process, and (ii) the exhaust process. (e) Sketch the P-v and T – s diagrams.

Answer 1

Sol: Given data compression ratio 1 = 18 = 4 calgf nako =2= 3 P = 95kPa Ti = 206 = 293k Pac 2 - we know u = R P . - 9 s = 1.1294 milig. 12 x 293 - = 95 RT RT 0.287 X 293 specific volume. volume. 2. = 11297m8/xy process 12 i sentropic benu 1 = ( 2 ) = (2) {y-14 for air ) To = (1801273 | 7 = 93% R= F, G 95401)1.4 P = 5.43 MPa q=14 Ių = 2.06276 m /kg

Process 2-3 Isobaric heat [R = R = 5.43 MPa. addition. 9 = T Tinc - 1-4-1 - =2 0 = 0-125507/19 for isobaril process 13 = 3 = r. B = 931x2 [13= 1862k / process 3-4 isentropic hente fos e [V = V = 1.12973/ky. also " 1862=0" [from ©] Tu = 773-18k] alice ) - P = 250kPa R = 250kPa]

& Heat for ais (p=100515/12 supplies - Cp (T3-T2) q=1.005 (1862–931) 19=935.65 kJ/kg C = 0.718 kJ/kg work done by system way = Cp (B3-T4) W34 = 1.005 (1862 - 773./). [Ngh=1094.26 kJ/kg. I work done on system W 12 = CP (TL-TI) =1005 (931-293) We = 641.19 kJ/ks. Net work don What = W34-212 I wnet = 453.07kJ Hemd Yuena no 453.07 925-15 n = 0.484 or 48.2.1, Carnot efficiency between Tmin=293k & Tomax = 1862 k. ne = 1-Thin -1- 293 [0=84.261 1 Tmar 1862

20 & lombustion process is Isobaric hena entropy Charge us= cp In I3 las = 0.6966 kg/kjk Exhuast is constant volune peren- hence as- coln Tu as = 0.717 In 773.18 1862 as= -0.63 kg 1 kjk heat is rejectes hen usis ~ve Already Drawn Please If Opvote. you the understood solution