At the beginning of the compression process of an air-standard Diesel cycle operating with a compression ratio of 18, the temperature is 300K and the pressure is 0.1MPa, the cutoff ratio is 2. Determine: a. Mass flow through the nozzle; b. Throat area; c. Mach number at exit.

$$\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{\gamma-1}{\gamma}}$$ $$T_2 = T_1 \left(\frac{P_2}{P_1}\right)^{\frac{\gamma-1}{\gamma}} = 673.15 \text{ K} \left(\frac{0.5 \times 10^6 \text{ Pa}}{3 \times 10^6 \text{ Pa}}\right)^{\frac{1.4-1}{1.4}}$$ $$T_2 = 673.15 \left(\frac{1}{6}\right)^{\frac{0.4}{1.4}} = 673.15 \times (0.16667)^{0.2857} = 673.15 \times 0.6218 = 418.6 \text{ K}$$

$$c_p T_1 = c_p T_2 + \frac{V_2^2}{2}$$ $$V_2 = \sqrt{2 c_p (T_1 - T_2)} = \sqrt{2 \times 1004.5 \text{ J/(kg K)} \times (673.15 - 418.6) \text{ K}}$$ $$V_2 = \sqrt{2 \times 1004.5 \times 254.55} = \sqrt{511400.95} = 715.12 \text{ m/s}$$

$$\dot{m} = \rho_2 A_2 V_2 = 4.166 \text{ kg/m}^3 \times 0.005 \text{ m}^2 \times 715.12 \text{ m/s}$$ $$\dot{m} = \boxed{14.90 \text{ kg/s}}$$

b. Throat area.

The critical pressure ratio for choked flow is:

$$\frac{P_t}{P_1} = \left(\frac{2}{\gamma+1}\right)^{\frac{\gamma}{\gamma-1}} = \left(\frac{2}{1.4+1}\right)^{\frac{1.4}{1.4-1}} = \left(\frac{2}{2.4}\right)^{3.5} = (0.8333)^{3.5} = 0.5283$$ $$P_t = 0.5283 \times P_1 = 0.5283 \times 3 \times 10^6 \text{ Pa} = 1.5849 \times 10^6 \text{ Pa}$$

Since $P_2 = 0.5 \times 10^6 \text{ Pa} < P_t$, the flow is choked, and the throat pressure is $P_t$.

$$V_t = \sqrt{\gamma R T_t} = \sqrt{1.4 \times 287 \text{ J/(kg K)} \times 560.96 \text{ K}}$$ $$V_t = \sqrt{225588.32} = 474.96 \text{ m/s}$$

$$\dot{m} = \rho_t A_t V_t$$ $$A_t = \frac{\dot{m}}{\rho_t V_t} = \frac{14.90 \text{ kg/s}}{9.843 \text{ kg/m}^3 \times 474.96 \text{ m/s}} = \frac{14.90}{4670.9} = 0.00319 \text{ m}^2$$ $$A_t = \boxed{3190 \text{ mm}^2}$$

c. Mach number at exit.

$$a_2 = \sqrt{\gamma R T_2} = \sqrt{1.4 \times 287 \text{ J/(kg K)} \times 418.6 \text{ K}}$$ $$a_2 = \sqrt{168488.8} = 410.47 \text{ m/s}$$

$$M_2 = \frac{V_2}{a_2} = \frac{715.12 \text{ m/s}}{410.47 \text{ m/s}}$$ $$M_2 = \boxed{1.742}$$

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Q7. At the beginning of the compression process of an air-standard Diesel cycle operating with compression ratio of 18, the temperature is $300 \text{ K}$ and the pressure is $0.1 \text{ MPa}$. The cutoff ratio for the cycle is 2. Determine:

Given:
Air-standard Diesel cycle.
Compression ratio $r = V_1/V_2 = 18$.
Inlet temperature $T_1 = 300 \text{ K}$.
Inlet pressure $P_1 = 0.1 \text{ MPa} = 100 \text{ kPa}$.
Cutoff ratio $r_c = V_3/V_2 = 2$.
For air, assume $\gamma = 1.4$ and $R = 0.287 \text{ kJ/(kg K)}$.

a. The temperature and pressure at the end of each process of the cycle.

Process 1-2: Isentropic Compression
$P_1 = 0.1 \text{ MPa}$
$T_1 = 300 \text{ K}$

$$P_2 = P_1 \left(\frac{V_1}{V_2}\right)^\gamma = P_1 r^\gamma = 0.1 \text{ MPa} \times (18)^{1.4}$$ $$P_2 = 0.1 \times 57.19 = \boxed{5.719 \text{ MPa}}$$

$$T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma-1} = T_1 r^{\gamma-1} = 300 \text{ K} \times (18)^{1.4-1}$$ $$T_2 = 300 \times (18)^{0.4} = 300 \times 3.176 = \boxed{952.8 \text{ K}}$$

Process 2-3: Constant Pressure Heat Addition
$P_3 = P_2 = \boxed{5.719 \text{ MPa}}$
$V_3/V_2 = r_c = 2$

Since $P_2 = P_3$, for an ideal gas, $\frac{V_2}{T_2} = \frac{V_3}{T_3}$.

$$T_3 = T_2 \left(\frac{V_3}{V_2}\right) = T_2 r_c = 952.8 \text{ K} \times 2 = \boxed{1905.6 \text{ K}}$$

Process 3-4: Isentropic Expansion
$V_4 = V_1$ (constant volume heat rejection from 4 to 1)
So, $\frac{V_4}{V_3} = \frac{V_1}{V_3} = \frac{V_1}{V_2} \times \frac{V_2}{V_3} = r \times \frac{1}{r_c} = \frac{18}{2} = 9$.

$$P_4 = P_3 \left(\frac{V_3}{V_4}\right)^\gamma = P_3 \left(\frac{r_c}{r}\right)^\gamma = 5.719 \text{ MPa} \times \left(\frac{2}{18}\right)^{1.4}$$ $$P_4 = 5.719 \times \left(\frac{1}{9}\right)^{1.4} = 5.719 \times 0.0692 = \boxed{0.3956 \text{ MPa}}$$

$$T_4 = T_3 \left(\frac{V_3}{V_4}\right)^{\gamma-1} = T_3 \left(\frac{r_c}{r}\right)^{\gamma-1} = 1905.6 \text{ K} \times \left(\frac{2}{18}\right)^{1.4-1}$$ $$T_4 = 1905.6 \times \left(\frac{1}{9}\right)^{0.4} = 1905.6 \times 0.4105 = \boxed{782.3 \text{ K}}$$

b. The thermal efficiency.

$$\eta_{th} = 1 - \frac{1}{r^{\gamma-1}} \left[\frac{r_c^\gamma - 1}{\gamma(r_c - 1)}\right]$$ $$\eta_{th} = 1 - \frac{1}{(18)^{1.4-1}} \left[\frac{(2)^{1.4} - 1}{1.4(2 - 1)}\right]$$ $$\eta_{th} = 1 - \frac{1}{(18)^{0.4}} \left[\frac{2.63901 - 1}{1.4(1)}\right]$$ $$\eta_{th} = 1 - \frac{1}{3.17609} \left[\frac{1.63901}{1.4}\right]$$ $$\eta_{th} = 1 - 0.31484 \times 1.17072$$ $$\eta_{th} = 1 - 0.36859 = 0.63141$$ $$\eta_{th} = \boxed{63.14\%}$$

c. The mean effective pressure, in MPa.

$$v_1 = \frac{R T_1}{P_1} = \frac{0.287 \text{ kJ/(kg K)} \times 300 \text{ K}}{0.1 \text{ MPa}} = \frac{86.1 \text{ kJ/kg}}{0.1 \text{ MPa}} = 861 \frac{\text{kJ}}{\text{MPa kg}} = 0.861 \text{ m}^3/\text{kg}$$ $$v_2 = \frac{v_1}{r} = \frac{0.861 \text{ m}^3/\text{kg}}{18} = 0.04783 \text{ m}^3/\text{kg}$$

The heat added during constant pressure process 2-3 is $q_{in} = c_p (T_3 - T_2)$.
$c_p = \frac{\gamma R}{\gamma-1} = \frac{1.4 \times 0.287}{0.4} = 1.0045 \text{ kJ/(kg K)}$.

$$q_{in} = 1.0045 \text{ kJ/(kg K)} \times (1905.6 - 952.8) \text{ K} = 1.0045 \times 952.8 = 957.1 \text{ kJ/kg}$$ $$W_{net} = \eta_{th} q_{in} = 0.63141 \times 957.1 \text{ kJ/kg} = 604.3 \text{ kJ/kg}$$

$$MEP = \frac{W_{net}}{v_1 - v_2} = \frac{604.3 \text{ kJ/kg}}{(0.861 - 0.04783) \text{ m}^3/\text{kg}}$$ $$MEP = \frac{604.3 \text{ kJ/kg}}{0.81317 \text{ m}^3/\text{kg}} = 743.1 \text{ kPa}$$ $$MEP = \boxed{0.743 \text{ MPa}}$$