Question Bank • QB_EXP_FP_CYC_038 NAT • 2 Marks

Section 3: Farm Power

Sources of Power

Subtopic: Thermodynamic cycles of IC engines

Question Statement:

Referring to the Diesel cycle P-v diagram, the temperature at state 4 is $T_4 = 650\text{ K}$ and at state 1 is $T_1 = 300\text{ K}$. If the mass of air is $1\text{ kg}$ and $c_v = 0.718\text{ kJ/(kg}\cdot\text{K)}$, determine the heat rejected per cycle $Q_{\text{out}}$ in $\text{kJ}$ (round off to one decimal place).

Answer Key:

251.3

Step-by-Step Solution & Explanation

Heat rejected occurs at constant volume $4 \to 1$: $Q_{\text{out}} = m c_v (T_4 - T_1) = 1 \times 0.718 \times (650 - 300) = 0.718 \times 350 = 251.3\text{ kJ}$.
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