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Why is the throttling valve not replaced by an isentropic turbine in the ideal vapor-compression refrigeration cycle?

Short Answer

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Question: Explain the difference in efficiency and energy recovery potential between a throttling valve and an isentropic turbine in an ideal vapor-compression refrigeration cycle, and why a throttling valve is preferred over an isentropic turbine. Answer: In an ideal vapor-compression refrigeration cycle, a throttling valve is responsible for reducing pressure and temperature, converting liquid refrigerant to vapor state without any energy input. An isentropic turbine, on the other hand, recovers work from the expanding refrigerant to improve efficiency. However, achieving high efficiencies in real-world applications is challenging, and the added complexity and cost of a turbine may outweigh the potential efficiency gains. The throttling valve is preferred due to its simplicity, lower cost, ease of maintenance, and effectiveness in achieving the desired pressure difference in the system, which is sufficient for most refrigeration applications.

Step by step solution

01

Understand the ideal vapor-compression refrigeration cycle

In the ideal vapor-compression refrigeration cycle, the refrigerant goes through four main components: compressor, condenser, throttling valve, and evaporator. The cycle aims to transfer heat from a low-temperature area to a high-temperature area by circulating the refrigerant through these components.
02

Roles of a throttling valve and an isentropic turbine

The throttling valve is responsible for reducing the pressure and temperature of the refrigerant after passing through the condenser, causing a pressure drop and converting liquid refrigerant to vapor state without any energy input. It maintains the desired pressure difference between the high and low sides of the system to achieve the necessary refrigeration effect. On the other hand, an isentropic (ideal) turbine aims to recover useful work from the refrigerant by allowing it to expand through the turbine. The energy recovery potential is used to drive another component, such as a compressor, in the cycle.
03

Efficiency in an isentropic turbine compared to a throttling valve

An isentropic turbine has a higher efficiency compared to a throttling valve due to its ability to recover work from the expanding refrigerant. However, the overall efficiency of the turbine in an ideal vapor-compression refrigeration cycle may be limited due to various reasons: 1. In a vapor-compression cycle, the process of expansion through a turbine is reversible, and all the recovered work would ideally be utilized by the compressor. However, in real-world applications, achieving high efficiencies in both the expansion and compression processes is challenging. 2. The additional mechanical complexities and costs associated with the installation and maintenance of a turbine can outweigh the potential efficiency gains in most applications.
04

Why the throttling valve is preferred over an isentropic turbine

The throttling valve is preferred over an isentropic turbine in an ideal vapor-compression refrigeration cycle due to its simplicity, lower cost, and ease of maintenance. Although a turbine has the potential for energy recovery, the efficiency gains may not be significant enough to justify the added complexity and expense. Additionally, the throttling valve is effective in achieving the desired pressure drop and maintaining the necessary pressure difference between the high and low sides of the system, which is sufficient for most refrigeration applications.

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Most popular questions from this chapter

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 \(\mathrm{kJ} / \mathrm{kg}\)

A large refrigeration plant is to be maintained at \(-15^{\circ} \mathrm{C},\) and it requires refrigeration at a rate of \(100 \mathrm{kW}\) The condenser of the plant is to be cooled by liquid water, which experiences a temperature rise of \(8^{\circ} \mathrm{C}\) as it flows over the coils of the condenser. Assuming the plant operates on the ideal vapor-compression cycle using refrigerant-134a between the pressure limits of 120 and \(700 \mathrm{kPa}\), determine \((a)\) the mass flow rate of the refrigerant, \((b)\) the power input to the compressor, and (c) the mass flow rate of the cooling water.

Refrigerant- 134 a enters the compressor of a refrigerator at \(100 \mathrm{kPa}\) and \(-20^{\circ} \mathrm{C}\) at a rate of \(0.5 \mathrm{m}^{3} / \mathrm{min}\) and leaves at 0.8 MPa. The isentropic efficiency of the compressor is 78 percent. The refrigerant enters the throttling valve at \(0.75 \mathrm{MPa}\) and \(26^{\circ} \mathrm{C}\) and leaves the evaporator as saturated vapor at \(-26^{\circ} \mathrm{C}\). Show the cycle on a \(T\) -s diagram with respect to saturation lines, and determine ( \(a\) ) the power input to the compressor, \((b)\) the rate of heat removal from the refrigerated space, and ( \(c\) ) the pressure drop and rate of heat gain in the line between the evaporator and the compressor.

A thermoelectric generator receives heat from a source at \(340^{\circ} \mathrm{F}\) and rejects the waste heat to the environment at \(90^{\circ} \mathrm{F}\). What is the maximum thermal efficiency this thermoelectric generator can have?

A heat pump using refrigerant-134a heats a house by using underground water at \(8^{\circ} \mathrm{C}\) as the heat source. The house is losing heat at a rate of \(60,000 \mathrm{kJ} / \mathrm{h}\). The refrigerant enters the compressor at \(280 \mathrm{kPa}\) and \(0^{\circ} \mathrm{C}\), and it leaves at \(1 \mathrm{MPa}\) and \(60^{\circ} \mathrm{C}\). The refrigerant exits the condenser at \(30^{\circ} \mathrm{C}\). Determine \((a)\) the power input to the heat pump, (b) the rate of heat absorption from the water, and (c) the increase in electric power input if an electric resistance heater is used instead of a heat pump.

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