Question

An Otto cycle with a compression ratio of 10.5 begins its compression at \(90 \mathrm{kPa}\) and \(35^{\circ} \mathrm{C}\). The maximum cycle temperature is \(1000^{\circ} \mathrm{C}\). Utilizing air-standard assumptions, determine the thermal efficiency of this cycle using (a) constant specific heats at room temperature and (b) variable specific heats.

Short answer

Based on the given solution, please provide numerical values for the following variables: a) Initial specific volume, \(v_1\) b) Specific volume after compression, \(v_2\) c) Adiabatic compression temperature, \(T_2\) d) Adiabatic expansion temperature, \(T_4\) e) Total work done throughout the cycle, \(W\) f) Heat exchange during combustion, \(Q_{in}\) g) Heat exchange during exhaust, \(Q_{out}\) h) Thermal efficiency, \(\eta_{th}\), for both constant specific heats at room temperature and variable specific heats.

Step-by-step solution

Calculate initial conditions

Calculate the initial volume of the air mixture using the ideal gas law: \(PV = mRT\) where \(P_1 = 90\:|\:mathrm{kPa}\) (Initial pressure) \(T_1 = 35^{\circ}C + 273.15 = 308.15\:\mathrm{K}\) (Initial temperature, converted to Kelvin) \(R = 0.287\: \mathrm{kPa\: m^3/(kg\: K)}\) (Specific gas constant for air) and \(m\) and \(V_1\) are the mass and initial volume of the air mixture, respectively. From the ideal gas law, we can calculate the initial specific volume: \(v_1 = RT_1/P_1\).

Apply the ideal gas law and cycle relationships

For the Otto cycle, we have a compression ratio \(r = V_1 / V_2 = 10.5\), where \(V_2\) is the volume after compression. For the compression process (1-2) from the initial state to the state after compression, we have: \(v_2 = v_1 / r\) For the expansion process (3-4), from the end of the constant-volume heat addition to the exhaust process, we have the same compression ratio: \(v_4 = v_3 / r\) To find temperatures at points 2 and 4, we can use the adiabatic relationships for air: \(T_2 = T_1( v_1/v_2)^{(\gamma-1)}\) \(T_4 = T_3( v_3/v_4)^{(\gamma-1)}\) In this problem, \(T_3\) is given, so we know that \(T_3 = 1000^{\circ}C + 273.15 = 1273.15\:\mathrm{K}\).

Specific heat capacities

For (a) constant specific heats at room temperature: \(c_v = 0.718\: \mathrm{kJ/(kg\: K)}\) (Specific heat capacity at constant volume) \(c_p = 1.005\: \mathrm{kJ/(kg\: K)}\) (Specific heat capacity at constant pressure) \(\gamma = c_p/c_v = 1.4\) For (b) variable specific heats, we will need to use the specific heat tables for air to calculate the properties at the given temperatures.

Calculate the work done throughout the cycle

For an ideal Otto cycle, the work done can be calculated as follows: \(W = W_{12} + W_{23} + W_{34} + W_{41}\) \(W_{12} = \int_{1}^{2} PdV = m c_v (T_1 - T_2)\) (adiabatic compression) \(W_{23} = 0\) (constant volume heat addition) \(W_{34} = \int_{3}^{4} PdV = m c_v (T_3 - T_4)\) (adiabatic expansion) \(W_{41} = 0\) (constant volume heat rejection) Total work done: \(W = mc_v (T_1 - T_2) + mc_v (T_3 - T_4)\)

Calculate the heat exchange

For an ideal Otto cycle, the heat exchange can be calculated as: \(Q_{in} = Q_{23} = m c_v (T_3 - T_2)\) (heat addition during combustion) \(Q_{out} = Q_{41} = m c_v (T_1 - T_4)\) (heat rejection during exhaust)

Determine the thermal efficiency

The thermal efficiency is given by: \(\eta_{th} = \frac {W_{net}}{Q_{in}} = \frac{Q_{in} - Q_{out}}{Q_{in}}\) For constant specific heats at room temperature (a), substitute the specific heat capacities and temperatures found in Steps 3 and 4. For variable specific heats (b), use the values obtained from the specific heat tables for air. Finally, compare the two efficiency values obtained and note any differences.