Question

7. The principal states in a cold air-standard Diesel cycle are given below. The mass of air is 1 kg. The Cv of air is 0.7 kJ/kg.K and k=1.4. Determine: (a) the heat addition and work done in each process, in kJ. (b) the net work output, in kJ. (c) the thermal efficiency State 1: p= 100 kPa; T = 300 K State 2: p= 2000 kPa State 3: T = 1100 K (35 points)

Answer 1

7) Cold air-standard Diesel Gyde Vio 2 3 fr P 4 = 100 kPa, Ti = 300K P2 = 2000 kPa T3 = 1100 K Tope re Qs = Heat supplied (addition) Pe=Heat rejected لیا V/1-1 T PT -Y - с 2 4 Y = 1.4 q = 1.4 Су sp=1.4)(0.7) a) Heat Addition = 0.98 kJ kg.k Pizz -0 92-3 = mc (T3-T2) r/1-1 r/1-1 = PAT2 1-4%-14 (100)(300) (2000) (T2) P, T, 1.4%-14 = T2 = 706.06 K P2-3 = (1)(0.98) (1100 – 706.06) = 386.06 kJ

=0 사 P3-4 P44 = mer (Ty-To) Y- r/1-1 På Tg = P4 T4 1-4-1 (2000) = (100) (TE 1-4-1.4 = ססון HL - = 467.38 K Pyy=(1)(0.7)(467-36 – 300 117.166 kJ b Net work done output What Ps - PR - Q2-3-Pu-l 386-06-117.166 = 268.894 KJ c) Thermal efficiency, nth, diesel = 1- PR Os Q4-1 92-3 = 117.166 -1- 386.06 - 10-6965 or 69.657