Question

An air-standard dual cycle has a compression ratio of 12.5. At the beginning of compression, p₁=100kPa,T₁ =300 K, and V₁ =14 L. The total amount of energy added by heat transfer is 22.7 kJ. The ratio of the constant-volume heat addition to total heat addion is zero. Determine: (a) the temperatures at the end of each heat addition process, in K. (b) the net work per unit of mass of air, in kJ/kg. (c) the percent thermal efficiency. (d) the mean effective pressure, in kPa.

Answer 1

(a) The compression ratio of the air-standard dual cycle is 12.5Given p₁ = 100 kPa, T₁ = 300 K, and V₁ = 14 L.Air-standard dual cycleThe air-standard dual cycle is represented on p-V and T-s diagrams below:For air-standard dual cycle, the heat transfer processes occur at constant volume and constant pressure.The steps involved in the air-standard dual cycle are as follows:Step 1-2: Isentropic compression.Step 2-3: Constant volume heat addition.Step 3-4: Isentropic expansion.Step 4-5: Constant volume heat rejection.(a) The temperatures at the end of each heat addition process are:Process 2-3 is a constant volume process, hence the pressure at state 3 can be obtained as:P₁V₁ᵣⁿ = P₂V₂ᵣⁿP₂ = P₁V₁ᵣⁿ / V₂ᵣⁿ= 100 x (14 / 12.5)¹².⁵ = 356.95 kPaT₃ = T₂ + Qᵥ / Cᵥ= 300 + 22.7 / (0.718 x 0.014) = 2367.24 KProcess 4-5 is a constant volume process, hence the pressure at state 4 can be obtained as:P₄ = P₃ (V₃ / V₄)ᵣⁿP₄ = 356.95 x (14 / 12.5)¹².⁵ = 156.89 kPaT₅ = T₄ - Qₓ / Cᵥ= 2367.24 - 22.7 / (0.718 x 0.014) = 1039.09 K(b) The work done per cycle can be determined by considering the area enclosed by the cycle on a p-V diagram and can be written as:Wnet = Wc + Wt + WrThe work done during the constant volume heat addition process is given as:Qv = Cv(T3 - T2)Wt = QvWt = 0.718 x (2367.24 - 798.27)Wt = 1245.11 kJ/kgThe work done during the isentropic compression process can be calculated using the formula:Ws = Cp(T2 - T1)Ws = 1.005 x (1654.78 - 300)Ws = 1396.25 kJ/kgThe work done during the isentropic expansion process can be calculated using the formula:We = Cp(T4 - T5)We = 1.005 x (1328.29 - 1039.09)We = 293.41 kJ/kgThe net work per unit of mass of air is given as:Wnet = Ws + Wt - We= 1396.25 + 1245.11 - 293.41= 2348.95 kJ/kg(c) Thermal EfficiencyThe thermal efficiency of the cycle is given by:ηth = Wnet / Qhηth = 2348.95 / 22.7= 103.2%(d) Mean Effective PressureThe mean effective pressure can be calculated using the formula:MEP = Wnet / (Vd)MEP = (2348.95 x 10³) / (0.014 x 12.5)= 1.676 MPaThus, the mean effective pressure is 1.676 MPa.