Question

An Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 95 kPa and 27°C and 750 kJ/kg of heat is transferred to the air during the heat addition process. Accounting for the variation of specific heats with temperature, determine: a) the temperature [K] and pressure [kPa] after the compression process; b) the temperature [K] and pressure [kPa] after the heat addition process; c) the net power output [kJ/kg]; d) the thermal efficiency and; e) the mean effective pressure [kPa]. Include a sketch defining the system, with energy terms shown. Show the cycle and states on a P-v diagram. You do NOT need to interpolate. You may approximate necessary properties using closest values.

Answer 1

to solve the question first find all the properties of state ,i.e, p,v,t at states 1,2,3 by adiabatic law, ideal gas equation

after that thermal efficiency = work done / heat addition = 1 -(1/r)$\small \gamma -1$

from here we can calculate thermal efficiency and work done in the process

lastly mean effective pressure(P mean) = work done / Swept volume

Solution) Given, For air, Compression ratio (1)=8 8=1.4 195 kPa, Ti= RTC = 300k Sodd = 750k/kg = Cv (T3 -T₂ ) - 0 Process 1-2 is compression process (adiabatic). 2 3 - heat addicion (constant volume process) 344 adiabatic expansion (a) 4513 heat rejection (constant volume proces At state , from ideal gas law Values, & Cp8 cv. A PIV = MRT Tempor ce Cv 95 x 103 x V = 1 x 0.287 X 103 x 300 300k/1.4/1.005 0716 689 K 13651-022 0-785 (V = 0.906 m/kg/ Also, ea- v thus V2 = on 2 = 0.906 = 0.113 m² /kg) 8 - Now, we have ViiV₂, T, IP, now we apply adiabatic relations given by & Pi Vir= P₂ V 2 12 = SVO PEN 13 T B 0.90671-4- 300 C 0.113 T2 = 300 x [7.99]0.4 P₂ = 958814 (P2 = 1746.02 kPa) 12 = 689-122 kl Thus after compression, T2 = 689,122 8 P2 = 1746-02 kPa (b) we know that Godd - 750 kJ/kg = Cv (T3-T2) Jusing cv = 0.785 kJ/kg - 750 0.785 [T3 - 689.122] » 955.41 = T3 - 689.122 T3 = 1644.53K) Aven

As process P. process 2-3 2-3 is constant volume process thus or P= P₂ x T3 - 1746.02 x 16 44.53 0 6890 122 3 This IP₂ - 4166.74 kPa) addition process, P.-4166.74 KPQ & T3 - 162 Arowen Thus after heat addition pron | Net power output (What) we know that otto cycle efficiency Thus n = W.D - 1 - (2) HA = WD = – 0.4352 = efficiency (n) = Work done - Heat addition I - CH 0.564 This HA WD = 0. WD = 0.5647 x Qats - 0.5647X 750 (W.D= 423.54 kJ) Answer 2) Thermal efficiency (w = 0.5647 or 56.47% (from above - Mean effective pressure (pmean) we know that Vsi swept volume Vs- V. - V₂ kmean = 423.54 Vs = 0.906-0-113 0.793 (Vs = 0.793 m 1kg) kmean = 534.098 kPa Answer mean = What Thus Vs