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A refrigerant-134a refrigerator is to maintain the refrigerated space at \(-10^{\circ} \mathrm{C}\). Would you recommend an evaporator pressure of 0.12 or 0.14 MPa for this system? Why?

Short Answer

Expert verified
Answer: The recommended evaporator pressure is 0.14 MPa.

Step by step solution

01

Determine desired temperature in Kelvin

To determine the desired temperature in Kelvin, we will convert the given temperature from Celsius to Kelvin: Desired Temperature (K) = \(-10^{\circ}\mathrm{C}\) + 273.15 Desired Temperature (K) = 263.15 K
02

Check saturation temperatures at given pressures

For refrigerant-134a, we can use the saturation pressure-temperature table or an online calculator to find the saturation temperature corresponding to different pressures. We will look up saturation temperatures for the given pressures 0.12 MPa and 0.14 MPa, respectively: Saturation temperature at 0.12 MPa (T1): 248.61 K Saturation temperature at 0.14 MPa (T2): 255.34 K
03

Compare saturation temperatures with desired temperature

Now we'll compare the saturation temperatures we found in step 2 with the desired temperature (263.15 K). Difference temperature at 0.12 MPa (ΔT1) = | Desired Temperature - T1 | ΔT1 = |263.15 - 248.61| = 14.54 K Difference temperature at 0.14 MPa (ΔT2) = | Desired Temperature - T2 | ΔT2 = |263.15 - 255.34| = 7.81 K
04

Choose the best pressure based on saturation temperature comparison

Based on our comparison in step 3, the saturation temperature at 0.14 MPa (ΔT2) is closer to the desired temperature of 263.15 K than the saturation temperature at 0.12 MPa (ΔT1). Therefore, to maintain the refrigerated space at \(-10^{\circ}\mathrm{C}\) (263.15 K), it is recommended to use an evaporator pressure of 0.14 MPa for the refrigerant-134a refrigerator.

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Key Concepts

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

Evaporator Pressure Selection
Choosing the right evaporator pressure in a refrigeration system is pivotal to its efficiency and performance. When selecting the evaporator pressure for a refrigerant-134a system, we must consider the desired temperature of the refrigerated space. The pressure directly influences the saturation temperature of the refrigerant; thus, selecting a pressure that corresponds to a saturation temperature close to the desired temperature is essential. As seen in the example, comparing the saturation temperatures at various pressures to the desired temperature helps identify the more suitable pressure. In the given problem, an evaporator pressure of 0.14 MPa is recommended because its corresponding saturation temperature is closer to the desired temperature, reducing the temperature difference and hence likely improving system efficiency as well as ensuring the space is maintained at the targeted temperature.
Saturation Temperature Calculation
Saturation temperature is a key parameter in the analysis and optimization of refrigeration cycles. It is the temperature at which a refrigerant changes phase at a given pressure. Understanding how to calculate or look up the saturation temperature is essential for determining the correct operating conditions of your system. In our exercise, we used a table or an online calculator to pinpoint the saturation temperatures at two different pressures for refrigerant-134a. This process highlighted the importance of having reliable resources or tools to accurately assess the physical properties of refrigerants. Given the critical role of saturation temperature in the design and analysis of refrigeration systems, ensuring precise calculations can make a significant difference in performance outcomes.
Refrigerant-134a Properties
Refrigerant-134a is a hydrofluorocarbon (HFC) widely used in refrigeration and air conditioning systems due to its favorable thermodynamic properties and lower environmental impact compared to older refrigerants like R-12. Understanding the physical and thermal properties of refrigerant-134a is crucial for designing and optimizing refrigeration systems. Key properties such as pressure, temperature, enthalpy, and entropy are needed for various calculations involving heat transfer and thermodynamic cycles. The properties can be significantly different from other refrigerants. In the context of our example, knowing these properties at both the evaporator and condenser pressure levels is necessary to determine the correct operating conditions for maintaining the desired temperature within the refrigerated space.
Temperature Conversion Celsius to Kelvin
Temperature conversions between Celsius and Kelvin scales are fundamental in refrigeration calculations because many scientific equations, including those for the refrigeration cycle, require temperature in Kelvin. The Kelvin scale is an absolute temperature scale starting at absolute zero, the theoretical temperature where particles cease to vibrate. To convert Celsius to Kelvin, we add 273.15 to the Celsius temperature. This conversion was demonstrated in the example when we determined the desired temperature for the refrigerated space in Kelvin. This precise conversion from the commonly used Celsius scale to Kelvin ensures accurate and effective communication of temperature conditions in scientific and engineering contexts.

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Most popular questions from this chapter

Does the area enclosed by the cycle on a \(T-s\) diagram represent the net work input for the reversed Carnot cycle? How about for the ideal vapor-compression refrigeration cycle?

In absorption refrigeration cycles, why is the fluid in the absorber cooled and the fluid in the generator heated?

A refrigerator operates on the ideal vapor compression refrigeration cycle with \(\mathrm{R}-134 \mathrm{a}\) as the working fluid between the pressure limits of 120 and 800 kPa. If the rate of heat removal from the refrigerated space is \(32 \mathrm{kJ} / \mathrm{s}\), the mass flow rate of the refrigerant is \((a) 0.19 \mathrm{kg} / \mathrm{s}\) \((b) 0.15 \mathrm{kg} / \mathrm{s}\) \((c) 0.23 \mathrm{kg} / \mathrm{s}\) \((d) 0.28 \mathrm{kg} / \mathrm{s}\) \((e) 0.81 \mathrm{kg} / \mathrm{s}\)

A refrigerator uses refrigerant-134a as the working fluid and operates on the ideal vapor-compression refrigeration cycle except for the compression process. The refrigerant enters the evaporator at \(120 \mathrm{kPa}\) with a quality of 34 percent and leaves the compressor at \(70^{\circ} \mathrm{C}\). If the compressor consumes \(450 \mathrm{W}\) of power, determine \((a)\) the mass flow rate of the refrigerant, ( \(b\) ) the condenser pressure, and ( \(c\) ) the COP of the refrigerator.

The manufacturer of an air conditioner claims a seasonal energy efficiency ratio (SEER) of \(16(\mathrm{Btu} / \mathrm{h}) / \mathrm{W}\) for one of its units. This unit operates on the normal vapor compression refrigeration cycle and uses refrigerant- 22 as the working fluid. This SEER is for the operating conditions when the evaporator saturation temperature is \(-5^{\circ} \mathrm{C}\) and the condenser saturation temperature is \(45^{\circ} \mathrm{C}\). Selected data for refrigerant- 22 are provided in the table below. $$\begin{array}{ccccc}\hline T,^{\circ} \mathrm{C} & P_{\text {sat }}, \mathrm{kPa} & h_{f}, \mathrm{kJ} / \mathrm{kg} & h_{g}, \mathrm{kJ} / \mathrm{kg} & s_{g}, \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K} \\\\\hline-5 & 421.2 & 38.76 & 248.1 & 0.9344 \\\45 & 1728 & 101 & 261.9 & 0.8682 \\ \hline\end{array}$$ (a) Sketch the hardware and the \(T\) -s diagram for this air conditioner. (b) Determine the heat absorbed by the refrigerant in the evaporator per unit mass of refrigerant- \(22,\) in \(\mathrm{kJ} / \mathrm{kg}\) (c) Determine the work input to the compressor and the heat rejected in the condenser per unit mass of refrigerant-22, in \(\mathrm{kJ} / \mathrm{kg}\)

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