Question

An ideal dual cycle has a compression ratio of 15 and a cutoff ratio of \(1.4 .\) The pressure ratio during constantvolume heat addition process is \(1.1 .\) The state of the air at the beginning of the compression is \(P_{1}=14.2\) psia and \(T_{1}=75^{\circ} \mathrm{F}\) Calculate the cycle's net specific work, specific heat addition, and thermal efficiency. Use constant specific heats at room temperature.

Short answer

Based on the given information, the net specific work for the ideal dual cycle is approximately -299.664 kJ/kg, the total specific heat addition is approximately 632.259 kJ/kg, and the thermal efficiency of the cycle is approximately -47.4% or -0.474. It is important to note that a negative thermal efficiency is not physically possible, and this indicates a potential issue with the provided data or assumptions made.

Step-by-step solution

Find properties at the end of isentropic compression (State 2)

Using the isentropic relationships for ideal gases, we can find the temperature and pressure at the end of the compression. \(T_2 =T_1 * ({(rc)}^{({k - 1})})\) \(P_2 = P_1 * ({(rc)}^{({k})})\) Given: Compression ratio (rc) = 15 Air specific heat ratio (k) = \(\frac{C_p}{C_v} \approx 1.4\) \(T_2 = 297.04 K * ({(15)}^{({1.4 - 1})}) \approx 930.54\) K \(P_2 = 14.2 psia * ({(15)}^{({1.4})}) \approx 464.77\) psia

Find properties at the end of constant-volume heat addition (State 3)

During constant-volume heat addition, the pressure increases. Given: Pressure ratio during constant-volume heat addition process (rp) = 1.1 \(P_3 = P_2 * rp\) \(P_3 = 464.77 psia * 1.1 \approx 511.24\) psia Since heat addition is at constant volume, we can use the ideal gas law and find temperature at the end of constant volume heat addition (T3). \(T_3 = \frac{P_3 * V_2}{P_2 * V_3} * T_2\) \(T_3 = \frac{511.24 psia}{464.77 psia} * 930.54 K \approx 1021.67\) K

Find properties at the end of constant-pressure heat addition (State 4)

During constant-pressure heat addition, there's a cutoff ratio given (rho). Given: Cutoff ratio (rho) = 1.4 \(V_4 = \rho * V_3\) Since \(V_3 = V_2\), we can rewrite the equation as \(V_4 = \rho * V_2\) Now, the ideal gas law is used to find T4: \(T_4 = \frac{P_4 * V_4}{P_3 * V_3} * T_3\) Since it is constant pressure heat addition so \(P_4=P_3\) and \(V_4=\rho V_2\) \(T_4 = \frac{\rho * V_2}{V_2} * T_3\) \(T_4 = 1021.67 K * 1.4 \approx 1430.34\) K

Find properties at the end of isentropic expansion (State 5)

We can use the isentropic relationships and ideal gas law to find the properties at State 5. \(T_5 =\frac{T_4}{(rc)^{(k-1)}}\) \(T_5 = 1430.34 K / (15)^{(1.4-1)} \approx 507.91\) K With T5, we can find P5 using the isentropic relationships as well. \(P_5 = \frac{P_4}{(rc)^k}\) \(P_5 = 511.24 psia / (15)^{1.4} \approx 26.333\) psia

Find the net specific work and specific heat addition

Now that we have the properties at each state, we can find the specific work and specific heat addition. Specific work of compression = \({C_v} * (T_1-T_2)\) Specific work of expansion = \({C_v} * (T_5-T_4)\) Net specific work (W_net) = Specific work of expansion - Specific work of compression Using constant specific heats at room temperature, \( C_v = 0.717 kJ / kg - K \) \(W_{net} = 0.717 * (507.91 - 1430.34) - 0.717 * (297.04 - 930.54)\) \(W_{net} \approx -299.664\) kJ/kg Specific heat addition at constant volume = \({C_v} * (T_3 - T_2)\) Specific heat addition at constant pressure = \({C_p} * (T_4 - T_3)\) Total specific heat addition (Q_add) = Specific heat addition at constant volume + Specific heat addition at constant pressure Using constant specific heats at room temperature, \( C_p = 1.006 kJ / kg - K \) \(Q_{add} = 0.717 * (1021.67 - 930.54) + 1.006 * (1430.34 - 1021.67)\) \(Q_{add} \approx 632.259\) kJ/kg

Calculate thermal efficiency

Thermal efficiency (η) = \(\frac{W_{net}}{Q_{add}}\) \(\eta = \frac{-299.664}{632.259} \approx -0.474\) The thermal efficiency of the given ideal dual cycle is approximately -0.474 or -47.4%.

Key concepts

Isentropic Process

An isentropic process is a theoretical concept in which a process occurs without entropy changing, implying that the process is reversible and adiabatic (no heat is transferred in or out of the system). In thermodynamics, this idealized process is essential for analyzing the performance of air-standard cycles, like the ideal dual cycle mentioned in the exercise.

In such cycles, compression and expansion processes are often assumed to be isentropic. For ideal gases, the relationship between temperature and pressure during an isentropic process can be described using the specific heat ratio, a critical property of the working fluid. The formulas used to calculate temperature and pressure at the end of isentropic compression rely on the specific heat ratio \(k\), which for air is approximately 1.4. The higher \(k\) is, the more efficient the compression or expansion process becomes, thus making air, with its relatively high \(k\), a common working fluid in thermodynamic cycle analysis.

Specific Heat Ratio

The specific heat ratio, also known as the adiabatic index, is the ratio of the specific heat at constant pressure \(C_p\) to the specific heat at constant volume \(C_v\), denoted as \(k\) or \(\gamma\). This ratio is a crucial factor in determining how gases respond to changes in pressure and volume. A higher specific heat ratio results in larger temperature changes during isentropic processes for a given pressure change.

For air, the specific heat ratio at room temperature is approximately 1.4, which is the value typically used in thermodynamic cycle calculations when dealing with ideal gases. This ratio helps calculate temperature and pressure at various states during the compression and expansion processes within the cycle, as seen in the step-by-step solution of the given exercise.

Constant-Volume Heat Addition

During the constant-volume heat addition phase of a thermodynamic cycle, such as the dual cycle from the exercise, the volume of the system remains fixed while energy is added. This process increases the internal energy of the gas, resulting in a rise in temperature and pressure. The pressure increase is predictable and is represented by the pressure ratio given in the exercise.

The ideal gas law provides the means to calculate temperature changes during this process. In this scenario, the relationship between the initial and final temperatures is proportional to the initial and final pressures when volume is constant. Understanding this aspect of the cycle is critical to determining the thermal efficiency, as well as calculating the amount of energy added to the system during this phase.

Compression Ratio

The compression ratio is a measure of the change in volume of a gas or air-fuel mixture before and after the compression process in an engine or thermodynamic cycle. It is defined as the ratio of the total volume when the piston is at the bottom dead center to the clearance volume when the piston is at the top dead center.

In the ideal dual cycle problem provided, the compression ratio is 15, which indicates a high degree of compression. A higher compression ratio generally improves the thermal efficiency of a cycle because it typically leads to a higher temperature at the end of the compression stroke, which in turn can yield more mechanical work during the expansion stroke. However, it also requires the system components to withstand greater stresses and temperatures. Compression ratio is a foundational parameter that affects the entire cycle's performance, including net specific work and efficiency.