Question

An ice-making machine operates on the ideal vapor-compression cycle, using refrigerant-134a. The refrigerant enters the compressor as saturated vapor at 20 psia and leaves the condenser as saturated liquid at 80 psia. Water enters the ice machine at \(55^{\circ} \mathrm{F}\) and leaves as ice at \(25^{\circ} \mathrm{F}\). For an ice production rate of \(15 \mathrm{lbm} / \mathrm{h}\), determine the power input to the ice machine \((169 \mathrm{Btu}\) of heat needs to be removed from each \(1 \mathrm{bm}\) of water at \(55^{\circ} \mathrm{F}\) to turn it into ice at \(25^{\circ} \mathrm{F}\) ).

Short answer

Answer: To determine the power input to the ice machine, follow these steps: 1. Find the properties of refrigerant at different states (compressor inlet, compressor outlet, and condenser outlet) using the given pressure values and thermodynamic tables for refrigerant-134a. 2. Calculate the work input to the compressor using the enthalpy differences between state 2 (compressor outlet) and state 1 (compressor inlet). 3. Calculate the heat removal rate in Btu/h using the given ice production rate and the required heat removal per unit mass of water. 4. Determine the mass flow rate of refrigerant using the heat removal rate and the enthalpy difference between state 1 (compressor inlet) and state 3 (condenser outlet). 5. Calculate the power input to the ice machine using the mass flow rate of refrigerant and the work input to the compressor. By following these steps and using the provided information, you can determine the power input to the ice machine operating on the ideal vapor-compression cycle using refrigerant-134a.

Step-by-step solution

Find the properties of refrigerant at different states

Using the given pressure values for the compressor and condenser, we can look up the properties of the refrigerant at different states in the vapor-compression cycle. These properties will be used in the calculations of work input and power input. At the compressor inlet (state 1): pressure \(P_1\) = 20 psia (saturated vapor) At the compressor outlet (state 2): pressure \(P_2\) = 80 psia At the condenser outlet (state 3): pressure \(P_3\) = 80 psia (saturated liquid) Using thermodynamic tables for refrigerant-134a, we can find the following properties: State 1 (compressor inlet): \(h_1 = h_{g}(P_1)\), \(s_1 = s_{g}(P_1)\) State 3 (condenser outlet): \(h_3 = h_{f}(P_3)\) Now, we need to find the properties of the refrigerant at state 2 (compressor outlet). Since the process is isentropic between state 1 and state 2 (due to the ideal vapor-compression cycle), we have \(s_2 = s_1\). Using this information and \(P_2\), we can determine \(h_2\) by interpolating the properties in the tables.

Calculate the work input to the compressor

The work input to the compressor can be calculated using the enthalpy differences between state 2 and state 1: \(W_{in} = h_2 - h_1\)

Calculate the heat removal rate in Btu/h

We are given an ice production rate of 15 lbm/h and the required heat removal per unit mass of water, which is 169 Btu/lbm. Hence, the heat removal rate from the water can be calculated as: \(q_{removed} = (15 \,\mathrm{lbm}\, / \mathrm{h})(169\, \mathrm{Btu}\, / \mathrm{lbm})\)

Calculate the mass flow rate of refrigerant

The mass flow rate of refrigerant can be determined using the heat removal rate and the enthalpy difference between state 1 and state 3: \(\dot{m} = \frac{q_{removed}}{h_1 - h_3}\)

Calculate the power input to the ice machine

The power input to the ice machine can be calculated using the mass flow rate of refrigerant and the work input to the compressor: \(P_{input} = \dot{m} W_{in}\) With these steps, you can determine the power input to the ice machine using the given properties and the ideal vapor-compression cycle.

Key concepts

Refrigerant Properties

Refrigerants are vital in the vapor-compression refrigeration cycle, as they absorb and release heat while changing their state from vapor to liquid and back again. One important property of a refrigerant is its pressure-enthalpy relationship, which dictates the phase changes it undergoes at specific pressures and temperatures. For example, refrigerant-134a entering the compressor as saturated vapor at a certain pressure has a unique enthalpy value associated with it.

Understanding these properties, such as saturation temperature and pressure, specific volume, enthalpy, and entropy is critical. They can be found in the thermodynamic tables for the specific refrigerant in use, which in our exercise is refrigerant-134a. Selection of the right refrigerant depends on factors like operating temperatures, compatibility with materials in the refrigeration system, and environmental considerations.

Isentropic Process

An isentropic process is one that is both adiabatic and reversible, meaning there is no heat transfer with the surroundings and no entropy change within the system. In refrigeration cycles, the compression of the refrigerant is often assumed to be isentropic for simplification.

Isentropic efficiency is a measure of the actual performance of the compressor compared to the ideal isentropic process. The ideal scenario helps us with the calculations because, under isentropic compression, the entropy at the compressor inlet and outlet remains the same (\( s_2 = s_1 \) ). The exercise improvement advice would be to ensure the understanding that in real systems, slight deviations from isentropic behavior are expected due to factors like friction and heat loss.

Enthalpy

Enthalpy is a thermodynamic property that represents the total heat content of a system. It is especially important in refrigeration cycles because the amount of heat absorbed or released during phase changes of the refrigerant corresponds to changes in enthalpy.

The exercise implies the use of enthalpy values at different points in the cycle to calculate the work input to the compressor and the heat extracted from the water to be frozen. For the refrigeration cycle, we primarily look at the enthalpy of vaporization at the evaporator and the enthalpy of the liquid refrigerant leaving the condenser.

Compressor Work Calculation

In a refrigeration cycle, the work required by the compressor is a critical factor as it defines the power input needed for the system to operate. To calculate this work, we use the enthalpy values at the compressor inlet and outlet. The change in enthalpy characterizes the work input (\( W_{in} = h_2 - h_1 \) ), considering an isentropic process.

For a more accurate calculation, it's key to have correct enthalpy values, which come from reliable thermodynamic tables or software. The work directly affects the system's coefficient of performance (COP), which is a measure of its efficiency. In real-world applications, additional factors like mechanical efficiency of the compressor must also be taken into account.

Heat Transfer

Heat transfer, in the context of the vapor-compression refrigeration cycle, involves the removal of heat from one area and its release into another. This is how refrigeration systems cool a space or substance.

In our exercise, the system removes heat from the water to produce ice, quantified as a certain value in Btu/lbm. This heat must be accurately calculated to understand the refrigeration load and design an effective system. The transfer of heat in the cycle occurs at the evaporator (where the refrigerant absorbs heat) and the condenser (where it dissipates heat), governed by the principles of thermodynamics and relying on the correct refrigerant properties and cycle conditions.