Question

The gases in the cylinder of an internal combustion engine have a specific internal energy of 800 kJ/kg and a specific volume of 0.06 m3/kg at the beginning of expansion. The expansion of the gases may be assumed to take place according to pn1.5=constant, from 55 bar to 1.4 bar. The specific internal energy after expansion is 230 kJ/kg. Calculate the heat rejected to the cylinder cooling water per kilogram of gases during the expansion stroke.
Ans: 104 kJ/kg
Please explain how this answer is the result

Answer 1

Answer on Question #71021, Physics / Molecular Physics | Thermodynamics

Question. The gases in the cylinder of an internal combustion engine have a specific internal energy of $800\, \text{kJ/kg}$ and a specific volume of $0.06\, \text{m}^3/\text{kg}$ at the beginning of expansion. The expansion of the gases may be assumed to take place according to $pV^{1.5} = \text{constant}$, from 55 bar to 1.4 bar. The specific internal energy after expansion is $230\, \text{kJ/kg}$. Calculate the heat rejected to the cylinder cooling water per kilogram of gases during the expansion stroke.

Given.

$$u_0 = 800\, \frac{\text{kJ}}{\text{kg}}; \quad u_f = 230\, \frac{\text{kJ}}{\text{kg}}; \quad p_1 = 55\, \text{bar} = 5.5 \cdot 10^6\, \text{Pa}; \quad p_2 = 1.4\, \text{bar} = 0.14 \cdot 10^6\, \text{Pa}; \quad v_1 = 0.06\, \text{m}^3/\text{kg};$$ $$pv^{1.5} = \text{const}.$$

Find.

$$q - ?.$$

Solution.

According to the $1^{\text{st}}$ law of thermodynamics

$$q = \Delta u + A,$$

where $q$ is the heat added to the system per kg; $\Delta u$ is a specific internal energy change; $A$ is work done on the system per kg. So

$$\Delta u = u_f - u_0 = 230 - 800 = -570\, \text{kJ/kg}.$$

Then

$$p_1 v_1^{1.5} = p_2 v_2^{1.5} \rightarrow v_2^{1.5} = v_1^{1.5} \frac{p_1}{p_2} \rightarrow v_2 = v_1 \left(\frac{p_1}{p_2}\right)^{1/1.5} = 0.06 \left(\frac{55}{1.4}\right)^{1/1.5} = 0.69\, \text{m}^3/\text{kg}$$ $$A = \frac{p_1 \cdot v_1 - p_2 \cdot v_2}{n - 1} = \frac{5.5 \cdot 10^6 \cdot 0.06 - 0.14 \cdot 10^6 \cdot 0.69}{1.5 - 1} = 0.466 \cdot 10^6\, \frac{\text{J}}{\text{kg}} = 466\, \text{kJ/kg}$$

Finally

$$q = \Delta u + A = -570 + 466 = -104\, \text{kJ/kg}$$

Answer. $q = -104\, \text{kJ/kg}$.

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