An ideal Diesel cycle has a compression ratio of 18 and a cutoff ratio of 1.5. Determine the (1) maximum air temperature and (2) the rate of

Question

An ideal Diesel cycle has a compression ratio of 18 and a cutoff ratio of 1.5. Determine the (1) maximum air temperature and (2) the rate of heat addition to this cycle when it produces 150 hp of power, the cycle is repeated 1200 times per minute, and the state of the air at the beginning of the compression is 95 kPa and 17°C. Use constant specific heats at room temperature. The properties of air at room temperature are cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4.

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Ngọc Diệp 4 years 2021-07-27T07:53:23+00:00 1 Answers 299 views 0

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  1. kiet
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    2021-07-27T07:54:52+00:00
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    Answer:

    (1) The maximum air temperature is 1383.002 K

    (2) The rate of heat addition is 215.5 kW

    Explanation:

    T₁ = 17 + 273.15 = 290.15

    \frac{T_2}{T_1} =r_v^{k - 1} =18^{0.4} =3.17767

    T₂ = 290.15 × 3.17767 = 922.00139

    \frac{T_3}{T_2} =\frac{v_3}{v_2} = r_c = 1.5

    Therefore,

    T₃ = T₂×1.5 = 922.00139 × 1.5 = 1383.002 K

    The maximum air temperature = T₃ = 1383.002 K

    (2)

    \frac{v_4}{v_3} =\frac{v_4}{v_2} \times \frac{v_2}{v_3} = \frac{v_1}{v_2} \times \frac{v_2}{v_3} = 18 \times \frac{1}{1.5} = 12

    \frac{T_3}{T_4} =(\frac{v_4}{v_3} )^{k-1} = 12^{0.4} = 2.702

    Therefore;

    T_4 = \frac{1383.002}{2.702} =511.859 \ k

    Q_1 = c_p(T_3-T_2)

    Q₁ = 1.005(1383.002 – 922.00139) = 463.306 kJ/jg

    Heat rejected per kilogram is given by the following relation;

    c_v(T_4-T_1)  = 0.718×(511.859 – 290.15) = 159.187 kJ/kg

    The efficiency is given by the following relation;

    \eta = 1-\frac{\beta ^{k}-1}{\left (\beta -1 \right )r_{v}^{k-1}}

    Where:

    β = Cut off ratio

    Plugging in the values, we get;

    \eta = 1-\frac{1.5 ^{1.4}-1}{\left (1.5 -1 \right )18^{1.4-1}}= 0.5191

    Therefore;

    \eta = \frac{\sum Q}{Q_1}

    \therefore 0.5191 = \frac{150}{Q_1}

    Heat supplied = \frac{150}{0.5191} = 288.978 \ hp

    Therefore, heat supplied = 215491.064 W

    Heat supplied ≈ 215.5 kW

    The rate of heat addition = 215.5 kW.

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