Answer:
6.8 kJ
Explanation:
p1 = 95 kPa
T1 = 300 K
T3 = 2100 K
m = 12 g
Ideal gas equation:
p * v = R * T
v = R * T / p
R for air is 0.287 kJ/(kg K)
v1 = 0.287 * 300 / 95 = 0.9 m^3/kg
v2 = v1 / cr
v2 = 0.9 / 18 = 0.05 m^3/kg
Assuming an adiabatic compression
p*v^k = constant
k is 1.4 for air
p1 * v1 ^ k = p2 * v2 ^ k
p2 = p1 * (v1 / v2) ^k
p2 = p1 * cr^k
p2 = 95 * 18^1.4 = 5.43 MPa
p1*v1/T1 = p2*v2/T2
T2 = p2*v2*T1/(p1*v1)
T2 = 5430 * 0.05 * 300 / (95 * 0.9) = 952 K
The first principle of thermodynamics
Q = W + ΔU
Since this is an adiabatic process Q = 0
W = -ΔU
W1-2 = -m * Cv * (T2 - T1)
The Cv of air is 0.72 kJ/kg
W1-2 = -0.012 * 0.72 * (952 - 300) = -5.63 kJ
Next the combustion happens and temperature increases suddenly.
v3 = v2 = 0.05 m^3/kg
T2 * p2^((1-k)/k) = T3 * p3^((1-k)/k)
p3 = p2 * (T2/T3)^(k/(1-k)
p3 = 5430 * (952/2100)^(1.4/(1-1.4) = 86.5 MPa
The work is zero because the piston doesn't move.
Next it expands adiabatically:
v4 = v1 = 0.9 m^3/kg
T * v^(k-1) = constant (adiabatic process)
T3 * v3^(k-1) = T4 * v4^(k-1)
T4 = T3 * (v3 / v4)^(k-1)
T4 = 2100 * (0.05 / 0.9)^(1.4-1) = 661 K
p3*v3/T3 = p4*v4/T4
p4 = p3*v3*T4/(v4*T3)
p4 = 86500*0.05*661/(0.9*2100) = 1512 kPa
L3-4 = -m * Cv * (T4 - T3)
L3-4 = -0.012 * 0.72 * (661 - 2100) = 12.43 kJ
Net work:
L1-2 + L3-4 = -5.63 + 12.43 = 6.8 kJ
