Answer:
1) 287760.4 Hp
2) 18410899.5 kPa
Explanation:
The parameters given are;
p₁ = 14.7 lbf/in² = 101325.9 Pa
v₁ = 0.0196 ft³ = 0.00055501 m³
T₁ = 80°F = 299.8167 K
k = 1.4
Assumptions;
1) Air standard conditions are appropriate
2) There are negligible potential and kinetic energy changes
3) The air behaves as an ideal gas and has constant specific heat capacities of temperature and pressure
1) Process 1 to 2
Isentropic compression
T₂/T₁ = (v₁/v₂)^(1.4 - 1) = 10^0.4
p₂/p₁ = (v₁/v₂)^(1.4)
p₂ = p₁×10^0.4 = 101325.9*10^0.4 = 254519.153 Pa
T₂ = 299.8167*10^0.4 = 753.106 K
p₃ = 1080 lbf/in² = 7,446,338 Pa
Stage 2 to 3 is a constant volume process
p₃/T₃ = p₂/T₂
7,446,338/T₃ = 254519.153/753.106
T₃ = 7,446,338/(254519.153/753.106) = 22033.24 K
T₃/T₄ = (v₁/v₂)^(1.4 - 1) = 10^0.4
T₄ = 22033.24/(10^0.4) = 8771.59 K
The heat supplied, Q₁ = cv(T₃ - T₂) = 0.718*(22033.24 -753.106) = 15279.14 kJ
The heat rejected = cv(T₄ - T₁) = 0.718*(8771.59 - 299.8167) = 6082.73 kJ
W(net) = The heat supplied - The heat rejected = (15279.14 - 6082.73) = 9196.41 kJ
The power = W(net) × RPM/2*1/60 = 9196.41 * 2800/2*1/60 = 214582.9 kW
The power by the engine = 214582.9 kW = 287760.4 Hp
2) The mean effective pressure, MEP = W(net)/(v₁ - v₂)
v₁ = 0.00055501 m³
v₁/v₂ = 10
v₂ = v₁/10 = 0.00055501/10 = 0.000055501
MEP = 9196.41/(0.00055501 - 0.000055501) = 18410899.5 kPa