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Chapter 6: Problem 12

The R 134 a in a Carnot refrigerator operates between saturated liquid and saturated vapor during the heat rejection process. If the cycle has a high temperature of \(52^{\circ} \mathrm{C}\) and a low temperature of \(-20^{\circ} \mathrm{C}\), calculate the heat transfer from the refrigerated space and the quality at the beginning of the heat addition process. (A sketch would be helpful.) The cycle COP is given as $$ \mathrm{COP}=\frac{T_{L}}{T_{H}-T_{L}}=\frac{253}{325-253}=3.51 $$ The COP is also given by \(\mathrm{COP}=q_{L} / w\), where $$ w=\Delta T \Delta s=[52-(-20)](0.9004-0.4432)=32.92 \mathrm{~kJ} / \mathrm{kg} $$ Hence, the heat transfer that cools is \(q_{L}=(\mathrm{COP})(w)=(3.51)(32.92)=115.5 \mathrm{~kJ} / \mathrm{kg}\). The quality at the beginning of the heat addition process is found by equating the entropy at the end of the heat rejection process to the entropy at the beginning of the heat addition process: $$ 0.4432=0.0996+(0.9332-0.0996) x \quad \therefore x=0.412 $$

Short Answer

Expert verified
The heat transfer from the refrigerated space is 115.5 kJ/kg. The quality at the beginning of the heat addition process is 0.412.

Step by step solution

01

Convert Temperatures to Kelvin

First, convert the given temperatures into Kelvin. For the high temperature, \[ T_H = 52^{\text{掳C}} + 273 = 325 \text{K}\] For the low temperature, \[ T_L = -20^{\text{掳C}} + 273 = 253 \text{K}\]
02

Calculate COP

Next, use the formula for the coefficient of performance (COP) of a Carnot refrigerator: \[ \text{COP} = \frac{T_L}{T_H - T_L} \] Substitute the values: \[ \text{COP} = \frac{253}{325 - 253} = 3.51 \]
03

Express COP in terms of Heat and Work

The COP can also be expressed in terms of the heat transferred from the refrigerated space (\(q_L\)) and the work input (\(w\)): \[ \text{COP} = \frac{q_L}{w} \]
04

Calculate the Work Input

Determine the work input using the temperature difference and entropy change: \[ w = \Delta T \cdot \Delta s \] Substitute the values: \[ w = [52 - (-20)](0.9004 - 0.4432) = 32.92 \text{ kJ/kg} \]
05

Calculate Heat Transfer from Refrigerated Space

Use the relationship between COP, heat, and work to find \(q_L\): \[ q_L = \text{COP} \times w \] Substitute the values: \[ q_L = 3.51 \times 32.92 = 115.5 \text{ kJ/kg} \]
06

Calculate Quality at Beginning of Heat Addition

Determine the quality (\(x\)) at the beginning of the heat addition process by using entropy values. Set the entropy at the end of heat rejection equal to entropy at the beginning of heat addition: \[ 0.4432 = 0.0996 + (0.9332 - 0.0996) x \] Solving for \(x\), we get: \[ x = \frac{0.4432 - 0.0996}{0.9332 - 0.0996} = 0.412 \]

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Key Concepts

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

Coefficient of Performance (COP)
The coefficient of performance (COP) of a Carnot refrigerator is a measure of how efficiently the system operates. It's the ratio of the heat removed from the refrigerated space (q_L ) to the work input (W ). For a Carnot refrigerator, the COP can be calculated using the temperatures of the high temperature reservoir (T_H ) and the low temperature reservoir (T_L ). The formula for COP is:

\[ \text{COP} = \frac{T_L}{T_H - T_L} \]

The COP of a Carnot refrigerator is always greater than one, meaning it transfers more heat from the cold space than the amount of work it requires. In our example, we have T_H = 325 K and T_L = 253 K, so the COP is approximately 3.51. This high COP indicates that the Carnot refrigerator is very efficient at transferring heat.
Heat Transfer Calculation
Heat transfer from the refrigerated space is a key metric in understanding the performance of a refrigerator. It tells you how much heat energy (q_L ) the system is able to remove per unit of work it performs. This is directly related to the COP of the refrigerator through the formula:

\[ q_L = \text{COP} \times w \]

Here, q_L is the heat transfer, COP is the coefficient of performance, and w is the work input. The work input can be calculated using the temperature difference (\Delta T ) and the change in entropy (\Delta s ):

\[ w = \Delta T \Delta s \]

In this example, the work input (w ) is 32.92 kJ/kg, and the COP is 3.51. Plugging these into the formula gives a heat transfer from the refrigerated space of approximately 115.5 kJ/kg.
Refrigeration Cycle
The refrigeration cycle of a Carnot refrigerator is a sequence of thermodynamic processes designed to remove heat from an enclosed space. This cycle includes four main steps:
  • Isothermal Expansion: The refrigerant absorbs heat from the refrigerated space while expanding at a low temperature.
  • Adiabatic Expansion: The refrigerant expands further without exchanging heat, resulting in a temperature drop.
  • Isothermal Compression: The refrigerant releases absorbed heat to the surroundings at a higher temperature.
  • Adiabatic Compression: The refrigerant is compressed, raising its temperature without exchanging heat.
Understanding these steps helps in grasping how the refrigerant transfers heat from the refrigerated space to the surroundings, making the refrigeration cycle a vital concept in thermodynamics.
Entropy and Quality in Thermodynamics
Entropy is a measure of the disorder or randomness in a thermodynamic system, and it plays a crucial role in determining the quality of the refrigerant. Quality (x ) refers to the ratio of the mass of vapor to the total mass of the mixture (vapor + liquid). The quality is found by equating the entropy at the end of the heat rejection process to the entropy at the beginning of the heat addition process:

\[ s_1 = s_f + (s_g - s_f) x \]

Here, s_1 is the entropy at the end of heat rejection, s_f is the entropy of the saturated liquid, and s_g is the entropy of the saturated vapor. Solving for x gives the quality of the refrigerant. In our example, the quality at the beginning of the heat addition process is 0.412, meaning 41.2% of the refrigerant is in the vapor phase and the rest is liquid.
Temperature Conversion to Kelvin
Thermodynamic calculations often require temperatures to be in Kelvin (K ) rather than Celsius (掳C ) because Kelvin is the absolute temperature scale. Conversion from Celsius to Kelvin is straightforward:

\[ T(K) = T(掳C) + 273 \]

This formula ensures that all thermodynamic equations use the same unit for temperature, leading to consistent and accurate results. For example, in the given problem, we convert 52掳C and -20掳C to Kelvin as follows:

\[ 52^{\text{掳C}} + 273 = 325 \text{K} \]
\[ -20^{\text{掳C}} + 273 = 253 \text{K} \]
Using Kelvin allows us to apply thermodynamic principles accurately, such as calculating the COP and the entropy changes, which are fundamental for analyzing refrigeration cycles.

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Most popular questions from this chapter

Air is compressed in an automobile cylinder from \(14.7\) to 2000 psia. If the initial temperature is \(60^{\circ} \mathrm{F}\), estimate the final temperature, assuming constant specific heats. SOLUTION Compression occurs very rapidly in an automobile cylinder \((Q \cong 0)\); hence, we approximate the process with an adiabatic reversible process. Using (6.17), we find the final temperature to be $$ T_{2}=T_{1}\left(\frac{P_{2}}{P_{1}}\right)^{(k-1) / k}=(520)\left(\frac{2000}{14.7}\right)^{0.4 / 1.4}=2117^{\circ} \mathrm{R} \quad \text { or } \quad 1657^{\circ} \mathrm{F} $$

A steam turbine produces \(3000 \mathrm{hp}\) from a mass flux of \(20,000 \mathrm{lbm} / \mathrm{hr}\). The steam enters at \(1000^{\circ} \mathrm{F}\) and 800 psia and exits at 2 psia. Calculate the efficiency of the turbine. The maximum possible work output is calculated first. For an isentropic process, state 2 is located as follows: $$ s_{2}=s_{1}=1.6807=0.1750+1.7448 x_{2} \quad \therefore x_{2}=0.8630 $$ The exit enthalpy is then \(h_{2}=h_{f}+x_{2} h_{f g}=94.02+(0.8630)(1022.1)=976.1 \mathrm{Btu} / \mathrm{lbm}\). The work output \(w_{s}\) associated with the isentropic process is $$ w_{s}=-\left(h_{2}-h_{1}\right)=-(976.1-1511.9)=535.8 \mathrm{Btu} / \mathrm{lbm} $$ The actual work output \(w_{a}\) is calculated from the given information: $$ w_{a}=\frac{\dot{W}_{T}}{\dot{m}}=\frac{(3000)(550) / 778}{20000 / 3600}=381.7 \mathrm{Btu} / \mathrm{lbm} $$ The efficiency is found, using (6.48), to be $$ \eta_{T}=\frac{w_{a}}{w_{s}}=\frac{381.7}{535.8}=0.712 \quad \text { or } \quad 71.2 \% $$

A 5 -lb block of copper at \(200^{\circ} \mathrm{F}\) is submerged in \(10 \mathrm{lbm}\) of water at \(50^{\circ} \mathrm{F}\), and after a period of time, equilibrium is established. If the container is insulated, calculate the entropy change of the universe. First, we find the final equilibrium temperature. Since no energy leaves the container, we have, using specific heat values from Table B. \(4 \mathrm{E}\), $$ m_{c}\left(C_{p}\right)_{c}(\Delta T)_{c}=m_{w}\left(C_{p}\right)_{w}(\Delta T)_{w} 5 \times 0.093\left(200-T_{2}\right)=(10)(1.00)\left(T_{2}-50\right) T_{2}=56.66^{\circ} \mathrm{F} $$ The entropy changes are found to be $$ \begin{aligned} &(\Delta S)_{c}=m_{c}\left(C_{p}\right)_{c} \ln \frac{T_{2}}{\left(T_{1}\right)_{c}}=(5)(0.093) \ln \frac{516.7}{660}=-0.1138 \mathrm{Btu} /{ }^{\circ} \mathrm{R} \\ &(\Delta S)_{w}=m_{w}\left(C_{p}\right)_{w} \ln \frac{T_{2}}{\left(T_{1}\right)_{w}}=(10)(1.00) \ln \frac{516.7}{510}=0.1305 \mathrm{Btu} /{ }^{\circ} \mathrm{R} \end{aligned} $$ Since no heat leaves the container, there is no entropy change of the surroundings. Hence $$ \Delta S_{\text {universe }}=(\Delta S)_{c}+(\Delta S)_{w}=-0.1138+0.1305=0.0167 \mathrm{Btu} /{ }^{\circ} \mathrm{R} $$

Two kilograms of steam are contained in a 6-liter rigid tank at \(60^{\circ} \mathrm{C}\). If \(1 \mathrm{MJ}\) of heat is added, calculate the final entropy. The initial quality is found as follows: $$ v_{1}=\frac{V_{1}}{m}=\frac{6 \times 10^{-3}}{2}=0.001017+x_{1}(7.671-0.001) \quad \therefore x_{1}=0.0002585 $$ The initial specific internal energy is then $$ u_{1}=u_{f}+x_{1}\left(u_{g}-u_{f}\right)=251.1+(0.0002585)(2456.6-251.1)=251.7 \mathrm{~kJ} / \mathrm{kg} $$ The first law, with \(W=0\), gives $$ Q=m\left(u_{2}-u_{1}\right) \quad \text { or } \quad u_{2}=u_{1}+\frac{\dot{Q}}{m}=251.7+\frac{1000}{2}=751.7 \mathrm{~kJ} / \mathrm{kg} $$ Using \(v_{2}=v_{1}=0.003 \mathrm{~m}^{3} / \mathrm{kg}\) and \(u_{2}=751.7 \mathrm{~kJ} / \mathrm{kg}\), we locate state 2 by trial and error. The quality must be the same for the temperature selected: $$ \begin{aligned} T_{2}=170^{\circ} \mathrm{C}: 0.003 &=0.0011+x_{2}(0.2428-0.0011) & & \therefore x_{2}=0.00786 \\ 751.7 &=718.3+x_{2}(2576.5-718.3) & & \therefore x_{2}=0.01797 \\ T_{2}=177^{\circ} \mathrm{C}: 0.003 &=0.0011+x_{2}(0.2087-0.0011) & & \therefore x_{2}=0.00915 \\ 751.7 &=750.0+x_{2}(2581.5-750.0) & & \therefore x_{2}=0.00093 \end{aligned} $$ A temperature of \(176^{\circ} \mathrm{C}\) is chosen. The quality from \(v_{2}\) is used since it is less sensitive to temperature change. At \(176^{\circ} \mathrm{C}\), we interpolate to find $$ 0.003=0.0011+x_{2}(0.2136-0.0011) \quad \therefore x_{2}=0.00894 $$ Hence \(S_{2}=m\left(s_{f}+x_{2} s_{f g}\right)=(2)[2.101+(0.00894)(4.518)]=4.28 \mathrm{~kJ} / \mathrm{K}\).

A steam turbine accepts \(2 \mathrm{~kg} / \mathrm{s}\) of steam at \(6 \mathrm{MPa}\) and \(600^{\circ} \mathrm{C}\) and exhausts saturated steam at \(20 \mathrm{kPa}\) while producing \(2000 \mathrm{~kW}\) of work. If the surroundings are at \(30^{\circ} \mathrm{C}\) and the flow is steady, calculate the rate of entropy production. SOLUTION The first law for a control volume allows us to calculate the heat transfer from the turbine to the surroundings: $$ \dot{Q}_{T}=\dot{m}\left(h_{2}-h_{1}\right)+\dot{W}_{T}=(2)(2609.7-3658.4)+2000=-97.4 \mathrm{~kW} $$ Hence, \(\dot{Q}_{\text {surr }}=-\dot{Q}_{T}=+97.4 \mathrm{~kW}\). The rate of entropy production is then found from (6.47) to be $$ \dot{S}_{\text {prod }}=\dot{S}_{\text {c..v. }}+\dot{m}\left(s_{2}-s_{1}\right)+\frac{\dot{Q}_{\text {surr }}}{T_{\text {surr }}}=0+(2)(7.9093-7.1685)+\frac{97.4}{303}=1.80 \mathrm{~kW} / \mathrm{K} $$
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