Question

A vapor-compression refrigeration system with a capacity of 10 tons has superheated Refrigerant 134 a vapor entering the compressor at \(15^{\circ} \mathrm{C}, 4\) bar, and exiting at 12 bar. The compression process can be modeled by \(p v^{1.01}=\) constant. At the condenser exit, the pressure is \(11.6\) bar, and the temperature is \(44^{\circ} \mathrm{C}\). The condenser is water-cooled, with water entering at \(20^{\circ} \mathrm{C}\) and leaving at \(30^{\circ} \mathrm{C}\) with a negligible change in pressure. Heat transfer from the outside of the condenser can be neglected. Determine (a) the mass flow rate of the refrigerant, in \(\mathrm{kg} / \mathrm{s}\). (b) the power input and the heat transfer rate for the compressor, each in \(\mathrm{kW}\). (c) the coefficient of performance. (d) the mass flow rate of the cooling water, in \(\mathrm{kg} / \mathrm{s}\). (e) the rates of exergy destruction in the condenser and expansion valve, each expressed as a percentage of the power input. Let \(T_{0}=20^{\circ} \mathrm{C}\).

Short answer

Mass flow rate of refrigerant is obtained using the system capacity. Power input and heat transfer rates determine the COP. Rates of exergy destruction are then calculated as percentages.

Step-by-step solution

Determine Refrigerant Properties

Obtain necessary thermodynamic properties for Refrigerant 134a at the given state points using refrigerant tables or software. For the inlet to the compressor (superheated at 15°C, 4 bar) and the outlet of the compressor (unknown temperature, 12 bar), and for the state at the condenser exit (44°C, 11.6 bar).

Calculate Enthalpies

Find the specific enthalpy (h) at the various state points:- h₁: the enthalpy at the inlet of the compressor (superheated at 15°C, 4 bar)- h₂: the enthalpy at the outlet of the compressor (isentropic compression assumption)- h₃: the enthalpy at the condenser exit (liquid state at 44°C, 11.6 bar)

Calculate Mass Flow Rate of Refrigerant

Use the capacity of the refrigeration system (10 tons):1 ton of refrigeration = 3.517 kWGiven the refrigeration capacity (\text{Capacity}) = 10 * 3.517 kW = 35.17 kW.Using the relation \[ \text{Capacity} = \text{Mass Flow Rate} (\text{h}_1 - \text{h}_3) \]Solve for the mass flow rate (ṁ) of refrigerant.

Calculate Power Input and Heat Transfer Rate

For the work done by the compressor, use the mass flow rate (\text{ṁ₁}) and enthalpy difference between inlet and outlet:\[ \text{Power Input} = \text{ṁ} (\text{h}_2 - \text{h}_1) \]For the heat transfer rate in the condenser, use enthalpies before and after the condenser:\[ \text{Heat Transfer Rate} = \text{ṁ} (\text{h}_2 - \text{h}_3) \]

Determine Coefficient of Performance (COP)

Calculate COP using the formula:\[ \text{COP} = \frac{\text{Capacity}}{\text{Power Input}} \]

Calculate Mass Flow Rate of Cooling Water

Use the heat balance for the water: \[ \text{Heat Transfer Rate} = \text{Mass Flow Rate of Water} \times \text{Specific Heat of Water} \times (\text{Temperature Out} - \text{Temperature In}) \]Solve for the mass flow rate of cooling water.

Evaluate Rates of Exergy Destruction

Using the definition of exergy destruction, calculate it for both condenser and expansion valve assuming steady state and no heat transfer with surroundings:\[ \text{Exergy Destruction} = \text{Mass Flow Rate} \times (T₀) \times (\text{s}_{\text{exit}} - \text{s}_{\text{inlet}}) \]Express as a percentage of the power input.

Key concepts

thermodynamic properties of refrigerants

Thermodynamic properties of refrigerants are essential in understanding how refrigerants behave under different conditions within a system. For Refrigerant 134a, these properties include pressure, temperature, specific volume, enthalpy, and entropy.

These properties are used to determine the state points, like the conditions at the compressor entry and exit, and the condenser exit. Refrigerant tables or software like REFPROP can help obtain these properties accurately.

In our exercise, the specific states are given by: inlet to the compressor (superheated at 15°C, 4 bar), the outlet of the compressor (unknown temperature, 12 bar), and the state at the condenser exit (44°C, 11.6 bar). By knowing these properties, we can move on to performing essential calculations.

enthalpy calculations

Enthalpy is a measure of the total energy of a thermodynamic system. It's crucial in calculating the energy changes throughout the refrigeration cycle.

In our system, we need to determine the enthalpies at various states:

• h1: the enthalpy at the inlet of the compressor (superheated at 15°C, 4 bar)
• h2: the enthalpy at the outlet of the compressor (isentropic compression assumption, 12 bar)
• h3: the enthalpy at the condenser exit (liquid state at 44°C, 11.6 bar)

These enthalpies can also be derived from the thermodynamic properties obtained earlier. Accurate enthalpy calculations allow us to determine the work done by the compressor, the heat transfer rate in the condenser, and other critical aspects of the system.

compressor work and efficiency

The compressor is a key component in the vapor-compression refrigeration system, as it increases the pressure of the refrigerant. The work input required by the compressor can be calculated using the enthalpy values determined earlier.

The formula for power input by the compressor is: \[ \text{Power Input} = \dot{m} \times (h_2 - h_1) \]

Where \ dot{m} \ is the mass flow rate of the refrigerant. The efficiency of the compressor can further be evaluated, considering the real-world deviations from ideal isentropic processes. This work and efficiency directly impact the overall performance and energy consumption of the refrigeration system.

coefficient of performance (COP)

The Coefficient of Performance (COP) is a key indicator of the efficiency of a refrigeration cycle. It is the ratio of the refrigeration effect (cooling provided) to the work input required by the compressor.

The formula for COP is: \[ \text{COP} = \frac{\text{Capacity}}{\text{Power Input}} \]

In our example, the capacity is given as 10 tons, which translates to 35.17 kW. By determining the power input from the previous calculations, we can find the COP. Higher COP values indicate a more efficient refrigeration system, as more cooling is achieved per unit of work input.

mass flow rate determination

Knowing the mass flow rate of the refrigerant is crucial for a variety of calculations within the system. It connects the refrigeration effect with the energy transfers occurring through enthalpies at different states.

To determine the mass flow rate in our system, we use the relation: \[ \text{Capacity} = \dot{m} \times (h_1 - h_3) \]

Given the capacity of 35.17 kW and previously found enthalpies, solving for \dot{m} gives us the mass flow rate in \ kg/s \. This value is then used to compute other parameters, such as the power input and heat transfer rate in the condenser. An accurate determination of mass flow rate ensures precise energy balances and system performance assessments.

exergy destruction analysis

Exergy destruction analysis helps identify the losses due to irreversibility in different components of the refrigeration system. Exergy is the measure of useful work potential of a system, and destruction of exergy indicates inefficiencies.

The formula for exergy destruction is: \[ \text{Exergy Destruction} = \dot{m} \times T_0 \times (s_{\text{exit}} - s_{\text{inlet}}) \]

where \( T_0 \) is the reference (ambient) temperature. In our system, we calculate exergy destruction for both the condenser and expansion valve. By evaluating these as a percentage of the power input, we can pinpoint where the highest losses occur and improve the system's overall efficiency. Analyzing exergy destruction guides us in optimizing the design and operation of vapor-compression refrigeration systems.