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Chapter 4: Problem 32

Refrigerant \(134 \mathrm{a}\) enters an air conditioner compressor at \(3.2\) bar, \(10^{\circ} \mathrm{C}\), and is compressed at steady state to \(10 \mathrm{bar}, 70^{\circ} \mathrm{C}\). The volumetric flow rate of refrigerant entering is \(3.0 \mathrm{~m}^{3} / \mathrm{min}\). The power input to the compressor is \(55.2 \mathrm{~kJ}\) per \(\mathrm{kg}\) of refrigerant flowing. Neglecting kinetic and potential energy effects, determine the heat transfer rate, in \(\mathrm{kW}\).

Short Answer

Expert verified
Heat transfer rate in \text{kW} is \text{Total Power Input} - \text{Mass flow rate} \times (h_2 - h_1).

Step by step solution

01

- Identify the Given Data

Given:1. Initial pressure, \( P_1 = 3.2 \text{ bar} \) 2. Initial temperature, \( T_1 = 10^{\text{掳C}} \) 3. Final pressure, \( P_2 = 10 \text{ bar} \) 4. Final temperature, \( T_2 = 70^{\text{掳C}} \) 5. Volumetric flow rate, \( \text{Volumetric Flow Rate} = 3.0 \text{ } \text{m}^3 \text{/min} \) 6. Power input, \( \text{Power input} = 55.2 \text{ kJ/kg} \)
02

- Look up Refrigerant Properties

Using refrigerant tables or property software, find the specific enthalpies of the refrigerant at the given initial and final conditions: Initial state: \( h_1 \) at \( 3.2 \text{ bar} \) and \( 10^{\text{掳C}} \).Final state: \( h_2 \) at \( 10 \text{ bar} \) and \( 70^{\text{掳C}} \).
03

- Determine Mass Flow Rate

First, we need the specific volume \( v_1 \) of the refrigerant at the initial state (use refrigerant tables):\( v_1 = \text{specific volume at} \) \( 3.2 \text{ bar} \text{ and } 10^{\text{掳C}} \).Then, calculate the mass flow rate: \[ \text{Mass flow rate} (\frac{\text{kg}}{\text{s}}) = \frac{\text{Volumetric Flow Rate} (\frac{\text{m}^3}{\text{min}})} {v_1 \text{ }(\frac{\text{m}^3}{\text{kg}})} \times \frac{1}{60} \]
04

- Calculate Work Done and Heat Transfer

The power input to the compressor is given per unit mass, so we multiply by the mass flow rate to get the total power input:\[ \text{Total Power Input} (\text{kW}) = \text{Power input per unit mass} (\frac{\text{kJ}}{\text{kg}}) \times \text{Mass flow rate} (\frac{\text{kg}}{\text{s}}) \]Using the First Law of Thermodynamics for a steady-state control volume:\[ \text{Total Power Input} (\text{kW}) = \text{Mass flow rate} \times (h_2 - h_1) + \text{Heat transfer rate} (\text{kW}) \]Rearrange to solve for the heat transfer rate:\[ \text{Heat transfer rate} (\text{kW}) = \text{Total Power Input} - \text{Mass flow rate} \times (h_2 - h_1) \]

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Key Concepts

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

Thermodynamics
Thermodynamics is the study of energy and its transformations. It focuses on how energy moves between systems and the work they perform. In this exercise, an air conditioner compressor handles refrigerant 134a, transforming it from one state to another. By studying thermodynamics, we understand the energy changes, pressures, and temperatures during the compression process. This system obeys laws like the First Law of Thermodynamics.
Refrigerant Properties
Refrigerants are special substances used in cooling systems like air conditioners to absorb and release heat. Refrigerant 134a is common for its efficiency and environmental friendliness. Understanding its properties, such as pressure and temperature at various states (3.2 bar, 10掳C initially; 10 bar, 70掳C finally), is vital. These properties are found using refrigerant tables and are crucial for calculating specific enthalpy and volume, which are needed for further thermodynamic calculations.
First Law of Thermodynamics
The First Law of Thermodynamics, also known as the conservation of energy principle, is a key concept here. It states that energy cannot be created or destroyed, only transformed. In our compressor, the power input is transformed into work and heat. The equation governing this is\[\text{Total Power Input} = \dot{m}(h_2 - h_1) + \text{Heat transfer rate}\]here, \( \dot{m} \) stands for mass flow rate, and \( h_1 \) and \( h_2 \) are the specific enthalpies at initial and final states respectively. We use this equation to determine how much heat is added to or removed from the system during compression.
Mass Flow Rate Calculation
Calculating mass flow rate is a crucial step. It defines how much refrigerant enters the system per unit time. Given the volumetric flow rate of 3.0 m鲁/min, we find the specific volume (\( v_1 \) ) from refrigerant tables, then use it in the formula:\[\text{Mass flow rate} = \frac{\text{Volumetric Flow Rate}} {v_1} \times \frac{1}{60}\]This conversion ensures we have a consistent unit, kilograms per second (kg/s), to use in subsequent calculations, like determining the total power input.
Heat Transfer Rate
Finally, we reach the heat transfer rate, describing how much heat is removed or added to the air conditioner system during compression. By rearranging the First Law of Thermodynamics equation, we solve for heat transfer rate as:\[\text{Heat transfer rate} = \text{Total Power Input} - \dot{m}(h_2 - h_1)\]This calculation tells us the efficiency and effectiveness of the air conditioner's compressor in transferring heat, ensuring optimal performance and energy use.

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Most popular questions from this chapter

The electronic components of a computer consume \(0.1 \mathrm{~kW}\), of electrical power. To prevent overheating, cooling air is supplied by a 25-W fan mounted at the inlet of the electronics enclosure. At steady state, air enters the fan at \(20^{\circ} \mathrm{C}, 1\) bar and exits the electronics enclosure at \(35^{\circ} \mathrm{C}\). There is no significant energy transfer by heat from the outer surface of the enclosure to the surroundings and the effects of kinetic and potential energy can be ignored. Determine the volumetric flow rate of the entering air, in \(\mathrm{m}^{3} / \mathrm{s}\).

Propane vapor enters a valve at \(1.6 \mathrm{MPa}, 70^{\circ} \mathrm{C}\), and leaves at \(0.5 \mathrm{MPa}\). If the propane undergoes a throttling process, what is the temperature of the propane leaving the valve, in \({ }^{\circ} \mathrm{C}\) ?

Steam at 160 bar, \(480^{\circ} \mathrm{C}\), enters a turbine operating at steady state with a volumetric flow rate of \(800 \mathrm{~m}^{3} / \mathrm{min}\). Eighteen percent of the entering mass flow exits at 5 bar, \(240^{\circ} \mathrm{C}\), with a velocity of \(25 \mathrm{~m} / \mathrm{s}\). The rest exits at another location with a pressure of \(0.06\) bar, a quality of \(94 \%\), and a velocity of \(400 \mathrm{~m} / \mathrm{s}\). Determine the diameters of each exit duct, in \(\mathrm{m}\).

A tiny hole develops in the wall of a rigid tank whose volume is \(0.75 \mathrm{~m}^{3}\), and air from the surroundings at 1 bar, \(25^{\circ} \mathrm{C}\) leaks in. Eventually, the pressure in the tank reaches 1 bar. The process occurs slowly enough that heat transfer between the tank and the surroundings keeps the temperature of the air inside the tank constant at \(25^{\circ} \mathrm{C}\). Determine the amount of heat transfer, in \(\mathrm{kJ}\), if initially the tank (a) is evacuated. (b) contains air at \(0.7\) bar, \(25^{\circ} \mathrm{C}\).

A feedwater heater operates at steady state with liquid water entering at inlet 1 at 7 bar, \(42^{\circ} \mathrm{C}\), and a mass flow rate of \(70 \mathrm{~kg} / \mathrm{s}\). A separate stream of water enters at inlet 2 as a two-phase liquid-vapor mixture at 7 bar with a quality of \(98 \%\). Saturated liquid at 7 bar exits the feedwater heater at 3 . Ignoring heat transfer with the surroundings and neglecting kinetic and potential energy effects, determine the mass flow rate, in \(\mathrm{kg} / \mathrm{s}\), at inlet 2 .
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