At the beginning of the compression process of an air-standard Diesel cycle, p1 = 95 kPa and T1 = 300 K. The maximum temperature is 2100 K and the mass of air is 12 g. For a compression ratio of 18, determine the net work developed in kJ (enter a number only)

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