Question

Problem 9.025 SI An air-standard dual cycle has a compression ratio of 8.6 and displacement of Vo = 2.2 L. At the beginning of compression, P1 = 95 kPa, and T1 = 290 K. The heat addition is 4.25 kJ, with one quarter added at constant volume and the rest added at constant pressure. Determine: c) each of the unknown temperatures at the various states, in K. the net work of the cycle, in kJ. the power developed at 3000 cycles per minute, in kW. the thermal efficiency. the mean effective pressure, in kPa. e)

Answer 1

Dual cycle : tas Vac 3 9 POB RT Vd = V – V = 2.8L P = GS kPa as, t Qs 2 = 4.25 5 -2.2X0.001 २90K [vi-va Qs = 1.0695 kJ = 0.0022 m²) u = v= 8.6 Vo - 86 Ose = 3.1875 kJ 7.6V7: OOOR Vi= 8.68₂ Vo = 0.000289 m = 0.002489 m) - V5 V2= V3 mars of air m= Pivi - m= 95800002489 m = 0.00284 kg 0287 x 290 @ Temp at the end of various stages : Poacens I-2 (rev acliabatic ) Asia mau (T3-Te ) 1.0625 T2 - 689.8 0.0028470718 T2 = 2900 (8.690.4 Ta = 1206.85k Ta= 685.8K Q se = mcp (Tu-T₂ ) 3. 1875 - Tu-1206.85 0.0028 4x1.005 Ty = 2323.62 K

to Net work in the cycle: 9 = gets no Whet = Qu-AR. OR: m Cv (Tyn Tp) Ts : 12967 Op = 0900284 X0718 ( 1876.7 - 290) Whet=4.25-2017 PR = 2.012 KI] Wnet = 2.238 kJ I net work. I © Thermal efficiency (1) = Whet As ^ - 7.893_ x 100 Process 3-4 (contant Pressure) Vy - T4 (p = 13) V3 T3 . (V. = Vg) V4 = 0.000189 % 2323-63 1206.85 [V4 = 0.000556 m3 I put this in a no 14 0.00748910.4 0.000556 4.25 In = 52.661. To Ts : 2323.62 @ Mean effective presure (Pon) & Pm - Whet Swepol volume Swept volume = displacement Ts = 127 6.7 K 9 Pm - Po P38 0.0023 m = 10 17.27 kpal
​​​​​​I have given the solution of all parts and also provided the p-v and t-s diagram for the processea so that you can understand it nicely. Hope you like the solution. Thank you.