Question

Consider a steady-flow Carnot refrigeration cycle that uses refrigerant-134a as the working fluid. The maximum and minimum temperatures in the cycle are 30 and \(-20^{\circ} \mathrm{C}\) respectively. The quality of the refrigerant is 0.15 at the beginning of the heat absorption process and 0.80 at the end. Show the cycle on a \(T\) -s diagram relative to saturation lines, and determine (a) the coefficient of performance, ( \(b\) ) the condenser and evaporator pressures, and ( \(c\) ) the net work input.

Short answer

Question: Determine the coefficient of performance, condenser and evaporator pressures, and the net work input for a steady-flow Carnot refrigeration cycle using refrigerant-134a as the working fluid. The maximum temperature is 30°C, the minimum temperature is -20°C, the quality at the beginning of the heat absorption process is 0.15, and the quality at the end is 0.8. Answer: To determine the desired values, first, label the points on the T-s diagram and then calculate the enthalpy values at each point using refrigerant-134a property tables. Next, calculate the coefficient of performance using the formula COP = |(h1 - h4)| / |(h2 - h1)|. Then, find the condenser and evaporator pressures corresponding to the saturated vapor and liquid phases at the given temperatures. Finally, calculate the net work input using the formula |W| = |m (h2 - h1)|.

Step-by-step solution

Label the Points on the T-s Diagram

Since this is a Carnot refrigeration cycle with refrigerant-134a, we have 4 points on the T-s diagram, as follows: 1. At the beginning of the heat absorption process, the quality is 0.15 (point 1) 2. At the end of the heat absorption process, the quality is 0.8 (point 2) 3. At the maximum temperature of 30°C, the fluid is saturated vapor (point 3) 4. At the minimum temperature of \(-20^{\circ}\)C, the fluid is saturated liquid (point 4)

Determine the Enthalpy Values at Each Point in the Cycle

Using refrigerant-134a property tables, we can find the enthalpy values at each point as follows: 1. At point 1: quality=0.15 \(\Longrightarrow h_1 = h_f + x_1 * (h_fg) = h_f + 0.15 * (h_fg)\) 2. At point 2: quality=0.8 \(\Longrightarrow h_2 = h_f + x_2 * (h_fg) = h_f + 0.8 * (h_fg)\) 3. At point 3: Since we are given the maximum temperature of 30°C and the fluid is saturated vapor, we can find the corresponding enthalpy value from the property table as \(h_3 = h_g\) 4. At point 4: Since we are given the minimum temperature of \(-20^{\circ}\)C and the fluid is saturated liquid, we can find the corresponding enthalpy value from the property table as \(h_4 = h_f\)

Calculate the Coefficient of Performance

The coefficient of performance (COP) can be calculated using the formula: \(\text{COP} = \frac{|Q_L|}{|W|}\) where \(Q_L\) is the heat absorbed from the low-temperature reservoir and \(W\) is the net work input. From the enthalpy values calculated, we can determine \(Q_L\) and \(W\) as follows: \(Q_L = |m (h_1 - h_4)|\) \(W = |m (h_2 - h_1)|\) Substituting values for enthalpies from Step 2: \(\text{COP} = \frac{|h_1 - h_4|}{|h_2 - h_1|}\)

Calculate the Condenser and Evaporator Pressures

We can find the condenser and evaporator pressures corresponding to the saturated vapor and liquid phases at the given temperatures. These can be found using refrigerant-134a property tables. 1. Condenser Pressure at 30°C: \(P_C\) 2. Evaporator Pressure at \(-20^{\circ}\)C: \(P_E\).

Calculate the Net Work Input

We can calculate the net work input using the formula: \(|W| = |m (h_2 - h_1)|\) Substitute the values from Step 2 and solve for |W|.

Key concepts

Understanding Thermodynamic Cycles

Thermodynamic cycles are the cornerstone of heat engines and refrigeration systems, representing a series of processes that a working fluid undergoes and then returns to its initial state. In the context of refrigeration, such as the Carnot cycle, these cycles aim to remove heat from a low-temperature environment and discharge it at a higher temperature, inversely applying the concept of a heat engine.

The Carnot cycle is a theoretical model that provides an upper limit on the efficiency achievable by any heat engine or refrigeration cycle operating between two temperature reservoirs. This cycle consists of two isothermal processes (heat transfer at constant temperature) and two adiabatic processes (no heat transfer). For refrigeration, the focus is on maximizing the amount of heat absorbed from the cooled space (the evaporator) while minimizing the work input required to achieve this cooling.

Refrigerant-134a Properties

Refrigerants are substances used in a refrigeration cycle to absorb and release heat at different temperatures and pressures. Refrigerant-134a is a commonly used refrigerant due to its desirable thermodynamic properties and relatively low environmental impact.

Crucial properties of Refrigerant-134a, such as enthalpy and pressure, vary with its state, which could be a saturated mixture, saturated vapor, or saturated liquid. Understanding how these properties change with temperature (such as the given 30°C and -20°C in this exercise) is essential to analyze and design refrigeration systems.

Quality is an important term referring to the ratio of the mass of vapor to the total mass of a saturated mixture. Knowing the quality at different points in the cycle allows for the calculation of specific enthalpies using the property tables and subsequently the work and heat interactions required for performance evaluation.

Coefficient of Performance (COP)

The coefficient of performance (COP) is a measure of a refrigeration system's effectiveness. It is defined as the ratio of the useful refrigeration (the amount of heat removed from the cold reservoir) to the work input required to transfer that heat to the higher temperature reservoir.

For a Carnot refrigeration cycle, the COP provides an idealized benchmark of efficiency, which helps in gauging the performance of real refrigeration systems by comparison. High COP values indicate a more efficient refrigeration cycle, consuming less energy for a given cooling effect. The COP is determined by using the enthalpy values at specific points in the cycle and is greatly influenced by the difference in enthalpy during the heat absorption and work input stages.

Interpreting the T-s (Temperature-Entropy) Diagram

A Temperature-Entropy (T-s) diagram is a graphical representation that charts the changes in temperature and entropy of a thermodynamic cycle. It's particularly useful for visualizing the various stages of refrigeration cycles, including the Carnot cycle.

In a T-s diagram, the isothermal and adiabatic processes are represented by horizontal and vertical lines, respectively. The area under a process curve represents the heat transfer. For the Carnot refrigeration cycle, the diagram will generally exhibit a rectangle, highlighting the ideal nature of isothermal heat absorption and rejection, along with adiabatic compression and expansion.

The exercise's given quality values and the saturation lines for refrigerant-134a help in locating the specific states on the T-s diagram, which is essential for analyzing the cycle and determining the necessary energy exchanges between the refrigerant and its environment.