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Refrigerant \(134 \mathrm{a}\) enters an air conditioner compressor at 4 bar, \(20^{\circ} \mathrm{C}\), and is compressed at steady state to 12 bar, \(80^{\circ} \mathrm{C}\). The volumetric flow rate of the refrigerant entering is \(4 \mathrm{~m}^{3} /\) \(\min\). The work input to the compressor is \(60 \mathrm{~kJ}\) per \(\mathrm{kg}\) of refrigerant flowing. Neglecting kinetic and potential energy effects, determine the heat transfer rate, in \(\mathrm{kW}\).

Short Answer

Expert verified
The heat transfer rate is 65.25 kW.

Step by step solution

01

- State properties and values

List the given values for the refrigerant: 1. Inlet pressure, \(P_1 = 4 \text{ bar} = 400 \text{ kPa}\)2. Inlet temperature, \(T_1 = 20^{\text{C}}\)3. Inlet volumetric flow rate, \(\text{v虈}_1 = 4 \text{ m}^3/\text{min}\)4. Outlet pressure, \(P_2 = 12 \text{ bar} = 1200 \text{ kPa}\)5. Outlet temperature, \(T_2 = 80^{\text{C}}\)6. Work input to compressor, \(w_{\text{in}} = 60 \text{ kJ/kg}\)
02

- Determine refrigerant's specific volume

Using thermodynamic tables for refrigerant \(134a\) at the given temperatures and pressures, find the specific volumes: For inlet: \( v_1 = 0.1484 \text{ m}^3/\text{kg}\) For outlet: \( v_2 = 0.05402 \text{ m}^3/\text{kg} \)
03

- Calculate mass flow rate

The mass flow rate at the inlet can be calculated using: \[ \text{m虈} = \frac{\text{v虈}_1}{v_1} \] Given \(\text{v虈}_1 = 4 \text{ m}^3/\text{min}\) and \( v_1 = 0.1484 \text{ m}^3/\text{kg} \), \[ \text{m虈} = \frac{4 \text{ m}^3/\text{min}}{0.1484 \text{ m}^3/\text{kg}} = 26.95 \text{ kg/min} \] Converting this to \text{ kg/s }: \[ \text{m虈} = \frac{26.95 \text{ kg}}{60 \text{ s}} \text{ = 0.449 \text{ kg/s} } \]
04

- Determine specific enthalpies

Using the thermodynamic tables for refrigerant \(134a\), find the specific enthalpies: For inlet: \(h_1 = 411.00 \text{ kJ/kg}\) For outlet: \(h_2 = 496.37 \text{ kJ/kg}\)
05

- Apply energy balance for the compressor

Using the first law of thermodynamics for steady-state flow processes: \[ \text{q虈} - \text{w虈} = \text{m虈} (h_2 - h_1 ) \] Given \( \text{w_in} = 60 \text{ kJ/kg} \), so \text{w虈} = \text{m虈} \text{w_in}. Solving for \text{q虈}: \[ \text{q虈} = \text{m虈} (h_2 - h_1) + \text{m虈} \text{w_in} \] Substitute the values: \[ \text{q虈} = 0.449 \text{ kg/s} (496.37 - 411.00) + 0.449 \times 60 \] So, \[ \text{q虈} = 0.449 \times 85.37 + 0.449 \times 60 \] \[ \text{q虈} = 38.31 + 26.94 = 65.25 \text{ kW} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Refrigerant Properties
Understanding the properties of the refrigerant used in any refrigeration cycle is crucial. For this exercise, we are dealing with refrigerant R-134a. This refrigerant is widely used due to its low environmental impact and favorable thermodynamic properties. The key properties we need are:
- Inlet pressure, temperature, and specific volume: These help us determine the state of the refrigerant at the entry of the compressor.
- Outlet pressure, temperature, and specific volume: These are needed to understand the changes undergone by the refrigerant during compression.
By looking up the thermodynamic tables for R-134a, we can obtain the specific enthalpies and specific volumes at the given states. This forms the basis for other calculations in our cycle analysis.
Energy Balance
The energy balance is a fundamental principle in thermodynamics, especially in steady-state processes like the refrigeration cycle. For our compressor, we apply the first law of thermodynamics, which states:
\[ \text{q虈} - \text{w虈} = \text{m虈} (h_2 - h_1 ) \] This tells us that the net heat transfer rate (q虈) and the work input rate (w虈) relate to the change in enthalpy (h) of the refrigerant multiplied by its mass flow rate (\text{m虈}). By plugging in the values we obtain from the tables for inlet and outlet state properties, and knowing the work input, we can solve for the heat transfer rate. The energy balance ensures that energy is conserved and helps us understand how much energy is added or removed at different stages of the cycle.
Mass Flow Rate Calculation
Next, determining the mass flow rate of the refrigerant is essential to understand how much refrigerant is being processed per unit time. Using the relation:
\[ \text{m虈} = \frac{\text{v虈}_1}{v_1} \] where \(\text{v虈}_1\) is the volumetric flow rate at the inlet and \(v_1\) is the specific volume at the inlet, we calculate the mass flow rate. This tells us how many kilograms of refrigerant pass through the compressor each second. For our case, converting from per minute to per second gives a clearer picture of the dynamic nature of the cycle. Calculating the mass flow rate correctly is critical for all further energy calculations in the system.
Thermodynamic Tables
Thermodynamic tables serve as an essential tool for analyzing any refrigeration cycle. These tables provide crucial data such as specific volumes, specific enthalpies, and other properties at various temperatures and pressures. For refrigerant R-134a, the tables give us:
- Specific volume at 20掳C and 400 kPa
- Specific volume at 80掳C and 1200 kPa
- Specific enthalpy at 20掳C and 400 kPa
- Specific enthalpy at 80掳C and 1200 kPa
With these values, we can accurately describe the state of the refrigerant at different points in the cycle, which is necessary for performing any energy and mass flow rate calculations. These table values ensure that our theoretical calculations align with real-world behavior.
Heat Transfer Rate
Finally, we address the heat transfer rate, which is often one of the key outputs of a refrigeration cycle analysis. Using the energy balance equation:
\[ \text{q虈} = \text{m虈} (h_2 - h_1) + \text{m虈} \text{w_in} \] we substitute the known values: specific enthalpies (h鈧, h鈧), mass flow rate (\text{m虈}), and specific work input (w_in). The calculation yields the heat transfer rate in kW, which tells us how much energy is being transferred due to heat per second. This helps in understanding the efficiency of the refrigeration cycle and how well the compressor is performing. A precise heat transfer rate is crucial for designing and operating any refrigeration system efficiently.

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Most popular questions from this chapter

Refrigerant 134 a enters a compressor operating at steady state as saturated vapor at \(0.12 \mathrm{MPa}\) and exits at \(1.2 \mathrm{MPa}\) and \(70^{\circ} \mathrm{C}\) at a mass flow rate of \(0.108 \mathrm{~kg} / \mathrm{s}\). As the refrigerant passes through the compressor, heat transfer to the surroundings occurs at a rate of \(0.32 \mathrm{~kJ} / \mathrm{s}\). Determine at steady state the power input to the compressor, in \(\mathrm{kW}\).

A \(380-\mathrm{L}\) tank contains steam, initially at \(400^{\circ} \mathrm{C}, 3\) bar. A valve is opened, and steam flows out of the tank at a constant mass flow rate of \(0.005 \mathrm{~kg} / \mathrm{s}\). During steam removal, a heater maintains the temperature within the tank constant. Determine the time, in s, at which \(75 \%\) of the initial mass remains in the tank; also determine the specific volume, in \(\mathrm{m}^{3} / \mathrm{kg}\), and pressure, in bar, in the tank at that time.

A rigid copper tank, initially containing \(1 \mathrm{~m}^{3}\) of air at \(295 \mathrm{~K}, 5\) bar, is connected by a valve to a large supply line carrying air at \(295 \mathrm{~K}, 15\) bar. The valve is opened only as long as required to fill the tank with air to a pressure of 15 bar. Finally, the air in the tank is at \(310 \mathrm{~K}\). The copper tank, which has a mass of \(20 \mathrm{~kg}\), is at the same temperature as the air in the tank, initially and finally. The specific heat of the copper is \(c=0.385 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\). Assuming ideal gas behavior for the air, determine (a) the initial and final mass of air within the tank, each in \(\mathrm{kg}\), and (b) the heat transfer to the surroundings from the tank and its contents, in kJ, ignoring kinetic and potential energy effects.

Air enters a compressor operating at steady state at \(14.7\) \(\mathrm{lbf} / \mathrm{in} .^{2}\) and \(60^{\circ} \mathrm{F}\) and is compressed to a pressure of \(150 \mathrm{lbf} / / \mathrm{in} .^{2}\) As the air passes through the compressor, it is cooled at a rate of 10 Btu per lb of air flowing by water circulated through the compressor casing. The volumetric flow rate of the air at the inlet is \(5000 \mathrm{ft}^{3} / \mathrm{min}\), and the power input to the compressor is 700 hp. The air behaves as an ideal gas, there is no stray heat transfer, and kinetic and potential effects are negligible. Determine (a) the mass flow rate of the air, lb/s, and (b) the temperature of the air at the compressor exit, in \({ }^{\circ} \mathrm{F}\).

If a kitchen-sink water tap leaks one drop per second, how many gallons of water are wasted annually? What is the mass of the wasted water, in lb? Assume that there are 46,000 drops per gallon and that the density of water is \(62.3 \mathrm{lb} / \mathrm{ft}^{3}\).

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