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14 November, 03:11

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)

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  1. 14 November, 03:15
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    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
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