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Refrigerant- \(134 \mathrm{a}\) enters the condenser of a steadyflow Carnot refrigerator as a saturated vapor at 90 psia, and it leaves with a quality of \(0.05 .\) The heat absorption from the refrigerated space takes place at a pressure of 30 psia. Show the cycle on a \(T\) -s diagram relative to saturation lines, and determine \((a)\) the coefficient of performance, \((b)\) the quality at the beginning of the heat-absorption process, and ( \(c\) ) the net work input.

Short Answer

Expert verified
Based on the given information and step by step solution: (a) The coefficient of performance for the steady-flow Carnot refrigerator with refrigerant-134a is 1.68. (b) The quality at the beginning of the heat-absorption process is 0.136. (c) The net work input required for the refrigerator is 5.765 Btu/lbm.

Step by step solution

01

Identify and label states on a T-s diagram

Draw a T-s diagram and show the saturation lines for the refrigerant. Label the four states of the steady flow Carnot cycle as follows: 1. Refrigerant enters the condenser as a saturated vapor at 90 psia (State 1). 2. Refrigerant leaves the condenser with a quality of 0.05 (State 2). 3. Refrigerant begins the heat-absorption process at 30 psia (State 3). 4. Refrigerant after the heat-absorption process (State 4).
02

Determine temperatures and enthalpies

Using a refrigerant-134a property table or software, find the corresponding saturation temperatures and enthalpies at State 1, State 2, and State 3. a) State 1: P1 = 90 psia, T1 = 105.511 F (saturated vapor temperature at 90 psia), h1 = h_g = 109.379 Btu/lbm (saturated vapor enthalpy at 90 psia) b) State 2: P2 = 90 psia, T2 = T1 = 105.511 F (isothermal process), h2 = (1 -0.05) * h_g + 0.05 * h_f = 0.95 * 109.379 + 0.05 * 48.542 = 103.614 Btu/lbm c) State 3: P3 = 30 psia, T3 = -16.045 F (saturated temperature at 30 psia), Since saturated liquid-vapor mixture, find the saturated enthalpy values: h_f = 47.267 Btu/lbm, h_g = 110.146 Btu/lbm
03

Find quality at the beginning of heat-absorption (x3)

Since we have a Carnot cycle (reversible and adiabatic processes between condenser and evaporator), changes in enthalpy between State 2 and State 3 equal changes in enthalpy between State 1 and State 4: Δh_23 = Δh_14 h4 - h3 = h1 - h2 Assume we have a saturated liquid-vapor mixture at State 3: h3 = (1 - x3) * h_f + x3 * h_g = (1 - x3) * 47.267 + x3 * 110.146 Now, we can solve for x3: x3 = (h1 - h2 - h_f) / (h_g - h_f) = (109.379 - 103.614 - 47.267) / (110.146 - 47.267) = 0.136 So, the quality at the beginning of the heat-absorption process is 0.136.
04

Calculate net work input and coefficient of performance

The net work input (W_net_in) can be determined as the difference between the work done in the turbine and the work done in the compressor: W_net_in = h1 - h2 The heat absorption from the refrigerated space, Q_in, can be calculated as the difference between enthalpies at State 4 and State 3, knowing that h3 has a quality of 0.136: h3 = 47.267 + 0.136 * (110.146 - 47.267) = 56.967 Btu/lbm Q_in = h4 - h3 Now, we can calculate the coefficient of performance (COP) as the ratio between Q_in and W_net_in: COP = Q_in / W_net_in = (h4 - h3) / (h1 - h2) = (56.967 - 47.267) / (109.379 - 103.614) = 9.7 / 5.765 ≈ 1.68 (a) The coefficient of performance is 1.68. (b) The quality at the beginning of the heat-absorption process is 0.136. (c) The net work input is 5.765 Btu/lbm.

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Key Concepts

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

Coefficient of Performance
The coefficient of performance (COP) is a key concept when evaluating refrigeration systems, like the Carnot refrigerator cycle. In simple terms, the COP is a measure of a refrigerator's efficiency, or more specifically, how effectively a refrigerator can transfer heat from a cooler place to a warmer place for each unit of work input. Here, it's calculated as the ratio of the heat absorbed from the refrigerated space (Q_in) to the net work input (W_net_in) required to move that heat. Using the formula to ascertain COP, it acts as an important criterion for gauging the performance of the refrigeration cycle. The higher the COP, the less work required to achieve the same amount of cooling. For instance, a COP of 1.68, as found in our exercise, suggests that for every unit of work put into the system, approximately 1.68 units of heat are moved from a cold to a warm space.

Refrigerant Quality
Refrigerant quality, often expressed as a decimal or percentage, indicates the amount of vapor in the mixture of liquid and vapor within the refrigerant. It is crucial for understanding how well the refrigerant can absorb heat during the refrigeration cycle. At a quality of 0.05, the refrigerant leaving the condenser primarily consists of liquid, with only 5% of it being in vapor form. Quality is especially significant in discerning the enthalpy of the refrigerant during phases of transition from liquid to vapor and vice versa, as the heat absorption process typically takes place when the refrigerant is partially vaporized. The quality at the beginning of the heat absorption process, calculated as 0.136 in our exercise, tells us that 13.6% of the refrigerant is vapor, which forms the basis for computing the changes in enthalpy and subsequently, other critical parameters of the cycle.

T-s Diagram
A Temperature-Entropy (T-s) diagram is a valuable visual tool in thermodynamics used to represent the changes in temperature (T) and entropy (S) throughout a thermodynamic cycle, like the Carnot refrigerator cycle in our exercise. When illustrating the Carnot cycle, the T-s diagram helps visualize the isothermal and adiabatic processes that comprise the cycle. The saturation lines for a given refrigerant are marked on this diagram, and the four states of the refrigerant during the cycle can be plotted accordingly. The area within the cycle loop on a T-s diagram corresponds to the net work done, which is intricately related to the energy efficiency of the cycle. By understanding each point and transition on the T-s diagram, students can better grasp the thermodynamic processes at play and apply this understanding to the practical optimization of refrigeration systems.

Heat Absorption
Heat absorption is the process by which the refrigerant takes in thermal energy from the space that needs to be cooled. In our exercise, this occurs at a pressure of 30 psia. During this stage, the refrigerant's capability to absorb heat is impacted by its pressure, temperature, and quality. As the refrigerant absorbs heat, it changes phase from a liquid to a vapor, which is an endothermic process; hence, its enthalpy increases. The amount of heat absorbed (Q_in) can be determined by finding the difference in enthalpy between two states during heat absorption. This value of Q_in is fundamental when calculating the COP, providing a direct measure of the machine's cooling capacity.

Net Work Input
Net work input is the total amount of work required to complete one cycle of the refrigeration process. In the Carnot cycle, it is the amount of energy needed to drive the compressor and, in some cases, other components like pumps or turbines. To compute net work input, as carried out in our example exercise, the enthalpy values at varying states of the refrigeration cycle must be known. From these values, the work is then calculated as the difference in enthalpy between specific states. The net work figure is pivotal; alongside heat absorption, it is employed in the calculation of the COP, giving us a concise metric for understanding the energy trade-offs intrinsic to the operation of the refrigerator.

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Most popular questions from this chapter

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