Question

Consider a regenerative gas refrigeration cycle using helium as the working fluid. Helium enters the compressor at \(100 \mathrm{kPa}\) and \(-10^{\circ} \mathrm{C}\) and is compressed to \(300 \mathrm{kPa}\). Helium is then cooled to \(20^{\circ} \mathrm{C}\) by water. It then enters the regenerator where it is cooled further before it enters the turbine. Helium leaves the refrigerated space at \(-25^{\circ} \mathrm{C}\) and enters the regenerator. Assuming both the turbine and the compressor to be isentropic, determine ( \(a\) ) the temperature of the helium at the turbine inlet, ( \(b\) ) the coefficient of performance of the cycle, and ( \(c\) ) the net power input required for a mass flow rate of \(0.45 \mathrm{kg} / \mathrm{s}\).

Short answer

Question: Determine the temperature of the helium at the turbine inlet (State 4). Answer: To find the temperature at the turbine inlet (State 4), use the isentropic relations with the temperature and pressure at State 3: \(T_4 = T_3 \times \left(\frac{P_1}{P_2}\right)^{(k-1) / k}\) where \(T_3 = 20^\circ \, \mathrm{C}\), \(P_1 = 100 \, \mathrm{kPa}\), \(P_2 = 300 \, \mathrm{kPa}\), and \(k = 1.66\). Calculate the value of \(T_4\). Question: Calculate the coefficient of performance (COP) of the cycle. Answer: To calculate the COP, use the formula: \(COP = \frac{\Delta h_{45}}{\Delta h_{12} - \Delta h_{34}}\) Where \(\Delta h_{12}\), \(\Delta h_{34}\), and \(\Delta h_{45}\) are the enthalpy differences calculated in Step 3. Calculate the COP using these values. Question: Calculate the net power input required for a mass flow rate of 0.45 kg/s. Answer: To calculate the net power input, use the formula: \(P_{net} = 0.45 \times (\Delta h_{12} - \Delta h_{34})\) Where \(\Delta h_{12}\) and \(\Delta h_{34}\) are the enthalpy differences calculated in Step 3 and the mass flow rate of helium is 0.45 kg/s. Calculate the value of \(P_{net}\).

Step-by-step solution

1. Determine the properties of helium at each state

To determine the properties of helium at each state, we can use the temperature and pressure data given in the problem: State 1: \(100 \, \mathrm{kPa} \, ,-10^\circ \mathrm{C}\) (Compressor inlet) State 2: \(300 \, \mathrm{kPa}\) (Compressor outlet, after isentropic compression) State 3: \(20^\circ \, \mathrm{C}\) (After cooling by water) State 4: Turbine inlet State 5: \(-25^\circ \, \mathrm{C}\) (After the refrigerator and before entering the regenerator) For isentropic processes, we have the relation: \(\frac{P_2}{P_1} = \left(\frac{T_2}{T_1}\right)^{\frac{k}{k-1}}\) Where \(k\) is the specific heat ratio of helium, which is 1.66. Rearrange the equation and solve for \(T_2\): \(T_{2}=T_{1} \times\left(\frac{P_{2}}{P_{1}}\right)^{(k-1) / k}\)

2. Find the temperature at the turbine inlet (State 4)

Using the temperature and pressure at state 2, we can find the temperature at state 4 using the isentropic relations: \(T_4 = T_3 \times \left(\frac{P_1}{P_2}\right)^{(k-1) / k}\)

3. Calculate the enthalpy differences

We can calculate the enthalpy differences for the isentropic process using the relation: \(\Delta h_{isentropic}=c_{p} \times \Delta T\) Where \(c_p\) for helium is \(5.18 \, \mathrm{kJ/kg . K}\). Calculate the enthalpy differences for the processes 1-2, 3-4, and 4-5: \(\Delta h_{12} = c_p \times (T_2 - T_1)\) \(\Delta h_{34} = c_p \times (T_4 - T_3)\) \(\Delta h_{45} = c_p \times (T_5 - T_4)\)

4. Calculate the coefficient of performance (COP)

The COP for the refrigeration cycle is given by the ratio of the heat absorbed by the refrigerated space (Q_L) to the net work input (W_net): \(COP = \frac{Q_L}{W_{net}}\) The heat absorbed by the refrigerated space is the enthalpy difference between state 4 and state 5: \(Q_L = \Delta h_{45}\) The net work input is the difference between the work input to the compressor and the work output of the turbine: \(W_{net} = \Delta h_{12} - \Delta h_{34}\) Now, calculate the COP using the enthalpy differences obtained in step 3: \(COP = \frac{\Delta h_{45}}{\Delta h_{12} - \Delta h_{34}}\)

5. Calculate the net power input

The net power input is given by: \(P_{net} = \dot{m} \times W_{net}\) Where \(\dot{m}\) is the mass flow rate of helium (0.45 kg/s). Calculate the net power input using the values from Step 3 and the mass flow rate: \(P_{net} = 0.45 \times (\Delta h_{12} - \Delta h_{34})\) Now, you can find all the required values for the given exercise.

Key concepts

Isentropic Process

An isentropic process is a theoretical concept in thermodynamics where entropy remains constant as a system evolves from one equilibrium state to another. This means that the process is both reversible and adiabatic—no heat is transferred into or out of the system, and no entropy is produced within the system. The term 'isentropic' is derived from the Greek words 'isos' (equal) and 'entropy'.

When analyzing an isentropic process involving an ideal gas, like the helium in our regenerative gas refrigeration cycle, the specific heats remain constant. Given that the ratio of the specific heats (\( k \)), for helium is 1.66, we can apply the isentropic relations to calculate the temperature at various points in the cycle, assuming both the compressor and the turbine operate isentropically. This assumption simplifies the analysis and allows us to use the relation \( \frac{P_2}{P_1} = \left(\frac{T_2}{T_1}\right)^{\frac{k}{k-1}} \) to calculate the temperatures needed to solve other parts of the problem.

For students aiming to thoroughly understand this concept, it's crucial to grasp the principles of reversible processes and adiabatic efficiency, along with the implications of assuming isentropic conditions in practical applications. Understanding isentropic processes also underpins the study of the second law of thermodynamics and the concept of entropy itself.

Coefficient of Performance

In thermodynamics, the coefficient of performance (COP) quantifies the efficiency of a refrigeration cycle or heat pump. Specifically, COP is the ratio of useful heating or cooling provided to the work required to produce that heating or cooling. The higher the COP, the more efficient the cycle is. For a refrigeration cycle, COP is calculated as \( COP = \frac{Q_L}{W_{net}} \), where \( Q_L \) is the heat absorbed by the refrigerated space (useful cooling), and \( W_{net} \) is the net work input to the cycle.

Intuitively, a high COP means that the system is able to absorb a lot of heat from the refrigerated space relative to the work input needed by the compressor. In the context of our exercise, knowing the enthalpy differences at critical points of the cycle enables us to calculate the COP by establishing the relationship between the heat absorbed and the net work done. Teaching these concepts effectively involves explaining how the first law of thermodynamics applies to calculating the net work input and the importance of COP in evaluating refrigeration systems.

Enthalpy Differences

Enthalpy is a measure of the total energy of a thermodynamic system, including internal energy and energy required to sustain the system's volume and pressure. In the context of thermodynamics education, students need to understand how to calculate enthalpy differences to analyze thermodynamic cycles. For an ideal gas undergoing a process, the enthalpy change \( \Delta h \) can be approximated as \( \Delta h = c_p \times \Delta T \), where \( c_p \) signifies the specific heat at constant pressure, and \( \Delta T \) represents the temperature difference.

Accordingly, in this problem with helium as the working fluid and \( c_p = 5.18 \, \mathrm{kJ/kg . K} \), the enthalpy differences calculated for the process steps (1-2, 3-4, and 4-5) help us determine the work done by the compressor and the turbine as well as the heat absorbed from the refrigerated space. Clearly comprehending enthalpy differences is vital for students in evaluating energy transfers and the performance of thermodynamic cycles.