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Chapter 6: Problem 44

Two \(\mathrm{kg}\) of Refrigerant \(134 \mathrm{a}\) initially at \(1.4\) bar, \(60^{\circ} \mathrm{C}\), are compressed to saturated vapor at \(60^{\circ} \mathrm{C}\). During this process, the temperature of the refrigerant departs by no more than \(0.01^{\circ} \mathrm{C}\) from \(60^{\circ} \mathrm{C}\). Determine the minimum theoretical heat transfer from the refrigerant during the process, in \(\mathrm{kJ}\).

Short Answer

Expert verified
The minimum theoretical heat transfer is -44.40 kJ.

Step by step solution

01

- Understand the initial state

The refrigerant starts at a pressure of 1.4 bar and a temperature of 60掳C. The mass of the refrigerant is 2 kg.
02

- Identify the final state

The refrigerant is compressed to a saturated vapor state at 60掳C. This means the final state is at the saturation pressure corresponding to 60掳C.
03

- Find saturation pressure and properties at 60掳C

From the refrigerant 134a tables, find the saturation pressure at 60掳C. It is approximately 14.67 bar. Also note the specific enthalpy of the vapor at 60掳C, which is approximately 271.25 kJ/kg.
04

- Calculate initial specific enthalpy

Using the initial state properties (1.4 bar and 60掳C) and the refrigerant tables, find the specific enthalpy for the initial state. At 60掳C and 1.4 bar, the specific enthalpy of the superheated vapor is approximately 293.45 kJ/kg.
05

- Calculate change in specific enthalpy

The change in specific enthalpy, \(\triangle h\), is given by:\[ \triangle h = h_{\text{final}} - h_{\text{initial}} = 271.25 - 293.45 = -22.20 \text{ kJ/kg} \]
06

- Calculate total heat transfer

Multiply the change in specific enthalpy by the mass of the refrigerant to find the total heat transfer: \[ Q = m \times \triangle h = 2 \times (-22.20) = -44.40 \text{ kJ} \]

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Key Concepts

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

Refrigerant 134a properties
Refrigerant 134a, also known as R-134a or tetrafluoroethane, is commonly used in refrigeration and air conditioning systems. It belongs to the hydrofluorocarbon (HFC) family and is favored for its low impact on the ozone layer. Understanding its properties is crucial for solving thermodynamic problems.

Key properties to consider:
  • Pressure: The pressure of a refrigerant significantly influences its state and its ability to absorb or release heat.
  • Temperature: Along with pressure, temperature helps to determine the phase (liquid or vapor) of the refrigerant.
  • Enthalpy: This is a measure of the total energy content of the refrigerant, including both internal energy and the energy required to make room for the refrigerant by displacing its environment.
Refrigerant tables are essential tools that provide these properties at various temperatures and pressures. These tables help to identify whether the refrigerant is in the superheated state (as in the initial condition of the problem) or in the saturated vapor state (as in the final condition).
Specific enthalpy change
Specific enthalpy change is an important concept in thermodynamics. It represents the difference in specific enthalpy between two states of a substance. For solving the given problem, we need to determine the initial and final specific enthalpies of the Refrigerant 134a.

Steps to find specific enthalpy change:
  • Identify the initial and final states of the refrigerant using relevant properties like pressure and temperature.
  • Use refrigerant tables to find the specific enthalpy values. For the given problem:
    • Initial state: 1.4 bar, 60掳C, specific enthalpy = 293.45 kJ/kg.
    • Final state: Saturated vapor state at 60掳C, specific enthalpy = 271.25 kJ/kg.
  • Calculate the change in specific enthalpy: \[ \triangle h = h_{\text{final}} - h_{\text{initial}} = 271.25 - 293.45 = -22.20 \text{kJ/kg} \]
The negative sign indicates that the enthalpy of the refrigerant decreases during the process, meaning it releases energy. This energy release is crucial for calculating heat transfer.
Heat transfer calculation
Heat transfer is a key concept in thermodynamic processes. It represents the energy exchanged between a system and its surroundings due to temperature differences. In the given problem, we aim to calculate the heat transfer from the refrigerant during compression.

Steps to calculate heat transfer:
  • Determine the mass of the refrigerant, which is provided as 2 kg.
  • Find the specific enthalpy change, \[ \triangle h \], which has been calculated as -22.20 kJ/kg.
  • Use the formula for heat transfer: \[ Q = m \times \triangle h = 2 \times (-22.20) = -44.40 \text{ kJ} \]
This negative value signifies that the refrigerant loses 44.40 kJ of heat to its surroundings. In thermodynamic terms, this means the system is performing work on the environment while losing heat, consistent with the compression process described.

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Most popular questions from this chapter

Two insulated tanks are connected by a valve. One tank initially contains \(0.45 \mathrm{~kg}\) of air at \(93^{\circ} \mathrm{C}, 1\) bar, and the other contains \(0.9 \mathrm{~kg}\) of air at \(38^{\circ} \mathrm{C}, 2\) bar. The valve is opened and the two quantities of air are allowed to mix until equilibrium is attained. Employing the ideal gas model with \(c_{v}=0.7 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\), determine (a) the final temperature, in \({ }^{\circ} \mathrm{C}\). (b) the final pressure, in bar. (c) the amount of entropy produced, in \(\mathrm{kJ} / \mathrm{K}\).

Saturated water vapor at \(100 \mathrm{kPa}\) enters a counterflow heat exchanger operating at steady state and exits at \(20^{\circ} \mathrm{C}\) with a negligible change in pressure. Ambient air at \(275 \mathrm{~K}\), 1 bar enters in a separate stream and exits at \(290 \mathrm{~K}, 1\) bar. The air mass flow rate is 170 times that of the water. The air can be modeled as an ideal gas with \(c_{p}=1.005 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\). Kinetic and potential energy effects can be ignored. (a) For a control volume enclosing the heat exchanger, evaluate the rate of heat transfer, in kJ per kg of water flowing. (b) For an enlarged control volume that includes the heat exchanger and enough of its immediate surroundings that heat transfer from the control volume occurs at the ambient temperature, \(275 \mathrm{~K}\), determine the rate of entropy production, in \(\mathrm{kJ} / \mathrm{K}\) per \(\mathrm{kg}\) of water flowing.

By what means can entropy be transferred across the boundary of a closed system? Across the boundary of a control volume?

If a closed system would undergo an internally reversible process and an irreversible process between the same end states, how would the changes in entropy for the two processes compare? How would the amounts of entropy produced compare?

An electric water heater having a 200 liter capacity employs an electric resistor to heat water from 23 to \(55^{\circ} \mathrm{C}\). The outer surface of the resistor remains at an average temperature of \(80^{\circ} \mathrm{C}\). Heat transfer from the outside of the water heater is negligible and the states of the resistor and the tank holding the water do not change significantly. Modeling the water as incompressible, determine the amount of entropy produced, in \(\mathrm{kJ} / \mathrm{K}\), for (a) the water as the system. (b) the overall water heater including the resistor. Compare the results of parts (a) and (b), and discuss.
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