Question

An air-standard dual cycle has a compression ratio of 14 and a cutoff ratio of \(1.2 .\) The pressure ratio during the constant-volume heat addition process is \(1.5 .\) Determine the thermal efficiency, amount of heat added, the maximum gas pressure and temperature when this cycle is operated at \(80 \mathrm{kPa}\) and \(20^{\circ} \mathrm{C}\) at the beginning of the compression. Use constant specific heats at room temperature.

Short answer

Answer: The thermal efficiency of the air-standard dual cycle is 63.5%, the maximum gas pressure is 6665.69 kPa, and the maximum gas temperature is 1667.35 K.

Step-by-step solution

Calculate temperatures and pressures at each stage of the cycle

To calculate the temperatures and pressures at each stage of the cycle, first we need to define the given parameters: - Compression ratio (r): \(r=14\) - Cutoff ratio (r_c): \(r_c=1.2\) - Pressure ratio during constant-volume heat addition (P_r): \(P_r=1.5\) - Initial pressure (P_1): \(80\,\text{kPa}\) - Initial temperature (T_1): \(20\,^{\circ}\text{C} = 293\,\text{K}\) (converted to Kelvin) We will also use the following constants for air at room temperature: - Specific heat ratio, \(\gamma = 1.4\) - Specific heat at constant volume, \(c_v = 718\,\text{J/(kg·K)}\) - Specific heat at constant pressure: \(c_p = 1.005\,\text{kJ/(kg·K)} = 1005\,\text{J/(kg·K)}\) - Gas constant \(R = 287 \text{J/(kg·K)}\) Now calculate the temperatures and pressures at each stage of the cycle: (a) End of Compression process (state 2) Using the isentropic relation for an ideal gas: \(T_2=T_1\cdot (r)^{\gamma-1}\) \(T_2=T_1\cdot (14)^{0.4}\) \(T_2=926.69\,\text{K}\) Now, calculate the pressure at state 2 (P_2) using the pressure-volume relation for isentropic processes: \(P_2=P_1\cdot (r)^{\gamma}\) \(P_2=80\cdot (14)^{1.4}\) \(P_2=4443.79\,\text{kPa}\) (b) End of the constant-volume heat addition process (state 3) Using the given pressure ratio (P_r): \(P_3=P_r\cdot P_2\) \(P_3=1.5\cdot 4443.79\) \(P_3=6665.69\,\text{kPa}\) Now, calculate the temperature at state 3 (T_3) using the ideal gas law: \(T_3=T_2\cdot\frac{P_3}{P_2}\) \(T_3=926.69\cdot\frac{6665.69}{4443.79}\) \(T_3=1389.46\,\text{K}\) (c) End of the constant-pressure heat addition process (state 4) Using the given cutoff ratio (r_c) and temperature calculated at state 2 (T_2): \(T_4=r_c\cdot T_3\) \(T_4=1.2\cdot 1389.46\) \(T_4=1667.35\,\text{K}\) (d) End of the isentropic expansion process (state 5) Using the isentropic relation, calculate the temperature at state 5 (T_5) by considering volume ratio at state 3 (V_3) and state 4 (V_4): \(T_5=T_4(\frac{V_3}{V_4})^{\gamma-1}\) \(T_5=T_4(\frac{1}{r})^{\gamma-1}\) \(T_5=1667.35(\frac{1}{14})^{0.4}\) \(T_5=603.83\,\text{K}\)

Calculate the heat added during the constant-volume and constant-pressure processes

(a) Heat added during constant-volume process (Q_23) Using the temperature difference between state 3 and state 2, and the specific heat at constant volume (c_v): \(Q_{23}=c_v(T_3-T_2)\) \(Q_{23}=718(1389.46-926.69)\) \(Q_{23}=332312.02\,\text{J/kg}\) (b) Heat added during constant-pressure process (Q_34) Using the temperature difference between state 4 and state 3, and the specific heat at constant pressure (c_p): \(Q_{34}=c_p(T_4-T_3)\) \(Q_{34}=1005(1667.35-1389.46)\) \(Q_{34}=278566.95\,\text{J/kg}\)

Calculate the heat rejected during the constant-volume process

Heat rejected during constant-volume process (Q_51) Using the temperature difference between state 5 and state 1, and the specific heat at constant volume (c_v): \(Q_{51}=c_v(T_5-T_1)\) \(Q_{51}=718(603.83-293)\) \(Q_{51}=222954.54\,\text{J/kg}\)

Calculate the thermal efficiency of the cycle

Thermal efficiency, \(\eta\), is calculated using the following formula: \(\eta=1-\frac{\text{Heat Rejected}}{\text{Heat Added}}=1-\frac{Q_{51}}{Q_{23}+Q_{34}}\) Calculate the total heat added: \(Q_{added}=Q_{23}+Q_{34}=332312.02+278566.95=610878.97\,\text{J/kg}\) Now calculate the thermal efficiency: \(\eta=1-\frac{222954.54}{610878.97}=0.635\) The thermal efficiency is \(63.5\%\)

Determine the maximum gas pressure and temperature

The maximum gas pressure occurs at state 3, and the maximum gas temperature occurs at state 4. From our calculations in Step 1: - Maximum gas pressure: \(P_{max}=6665.69\,\text{kPa}\) - Maximum gas temperature: \(T_{max}=1667.35\,\text{K}\) In conclusion, the thermal efficiency of the air-standard dual cycle is \(63.5\%\), the amount of heat added is \(610878.97\,\text{J/kg}\), the maximum gas pressure is \(6665.69\,\text{kPa}\) and the maximum gas temperature is \(1667.35\,\text{K}\).