Chapter 16, Problem 16.1P

Interpretation Introduction

(a)

Interpretation:

To interpret the value of ideal power required5 $\left({\dot{W}}_{ideal}\right)$ for the process.

Concept Introduction:

The initial stage contains liquid water at $21°C$ . The final stage contains ice at $0°C$ . Heat of fusion of water is $333.5\frac{kJ}{kg}$ and ice is producing at a rate of $0.5\frac{kg}{s}$ .

Answer to Problem 16.1P

The ideal power required $\left({\dot{W}}_{ideal}\right)$ for the process is $13.1\text{5 kW}$ .

Explanation of Solution

At initial stage,

  $\begin{array}{l}{H}_i=88.50\text{43 }\frac{\text{kJ}}{\text{kg}}\\ S_i=0.312\text{16 }\frac{\text{kJ}}{\text{kg-K}}\end{array}$

At final stage,

  $\begin{array}{l}{H}_f=-333.5\frac{\text{kJ}}{\text{kg}}\\ S_f=1.22\frac{\text{kJ}}{\text{kg-K}}\end{array}$

The value of $T_{\sigma }$ is $T_{\sigma }=294.15$

The flowsheet of the given process is as follows,

The various positions in the flowsheet are described as follows,

Point A: saturated vapor at $0°C$ .

Point B: Superheated vapor at $P=5.84atm$ and the entropy of Point A.

Point C: saturated liquid at $21°C.P=5.84atm$ .

Point D: Mixture of saturated liquid & saturated vapor at $0°C$ with the enthalpy of Point C.

Data for Points A, C, & D from Table 9.1. Data for Point B from Fig. G.2. For reference, Table 9.1 and Figure G.2 are below.

The equation to calculate the ideal work done $\left(W_{ideal}\right)$ is $W_{ideal}=H_f-H_i-T_{\sigma }\left(S_f-S_i\right)$ .

Upon substituting the values,

  $\begin{array}{c}{W}_{ideal}=H_f-H_i-T_{\sigma }\left(S_f-S_i\right)\\ =\left(-333.5\right)-\left(88.5043\right)-\left(294.15\left(1.22-.31216\right)\right)\\ W_{ideal}=-29.22\frac{KJ}{Kg}\\ \left|W_{ideal}\right|=29.22\frac{KJ}{Kg}\end{array}$

Let the mass flow rate be $\dot{m}=0.45\frac{Kg}{s}$

So, the ideal power required is ${\dot{W}}_{ideal}=\dot{m}\times W_{ideal}$

  $\begin{array}{c}{\dot{W}}_{ideal}=\dot{m}\times W_{ideal}\left(\frac{Kg}{s}\times \frac{KJ}{Kg}\right)\\ =0.45\times 29.22\frac{KJ}{s}\\ =13.15KW\left(\because \frac{J}{s}=W\right)\end{array}$

So, the ideal power required is ${\dot{W}}_{ideal}=13.145KW$

Interpretation Introduction

(b)

Interpretation:

To calculate the ideal power required $\left({\dot{W}}_{ideal}\right)$ for a single Carnot Heat Pump operated between $273.15K\left(0°C\right)\&294.15K\left(21°C\right)$ and its thermodynamic efficiency.

Concept Introduction:

A single Carnot heat pump is operated between sink and source at $273.15K\left(0°C\right)\&294.15K\left(21°C\right)$ respectively.

Answer to Problem 16.1P

The ideal power required $\left({\dot{W}}_{ideal}\right)$ for given single Carnot Heat Pump is $14.6\text{ kW}$ and thermodynamic efficiency ${\eta }_t=0.9$ .

Explanation of Solution

For the Carnot heat pump, heat equal to the enthalpy change of the water is extracted from a cold reservoir at $0°C$, with heat rejection to the surroundings at $21°C$ .

The given data is as follows,

  $\begin{array}{c}{T}_H=21°C\left(or{T}_{\sigma }\right)\\ T_C=0°C\\ H_i=88.5043\frac{kJ}{kg}\\ H_f=-333.5\frac{\text{kJ}}{\text{kg}}\\ {\eta }_T=ThermodynamicEfficiency\end{array}$

The heat removed is $Q_C=H_f-H_i$ .

  $\begin{array}{c}{Q}_C=H_f-H_i\\ =-333.5-88.5043\\ Q_C=422\frac{KJ}{Kg}\end{array}$

The work done can be calculated as $W=\left|Q_C\right|\left(\frac{T_H-T_C}{T_C}\right)$

  $\begin{array}{c}W=\left|Q_C\right|\left(\frac{T_H-T_C}{T_C}\right)\\ =\left|-422\right|\left(\frac{\left(21+273.15\right)-\left(0+273.15\right)}{\left(0+273.15\right)}\right)\\ =\left|-422\right|\left(\frac{21}{273.15}\right)\\ W=32.44\frac{KJ}{Kg}\end{array}$

The power required is $\dot{W}=\dot{m}\times W$ .

  $\begin{array}{c}\dot{W}=\dot{m}\times W\left(\frac{kg}{s}\times \frac{kJ}{kg}\right)\\ =0.45\times 32.44\frac{kJ}{s}\\ \dot{W}=14.6kW\left(\because \frac{J}{s}=W\right)\end{array}$

The thermodynamic efficiency is ${\eta }_T=\frac{{\dot{W}}_{ideal}}{\dot{W}}$

  $\begin{array}{c}{\eta }_T=\frac{{\dot{W}}_{ideal}}{\dot{W}}\\ =\frac{13.15}{14.6}\\ {\eta }_T=0.9\end{array}$

Interpretation Introduction

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Interpretation:

To interpret the power requirement of an ideal tetrafluoroethane vapor-compression refrigeration cycle and thermodynamic efficiency of the process.

Concept Introduction:

Vapor-compression refrigeration cycle is operated using tetrafluoroethane. Ideal condition implies Isentropic Compression, Infinite cooling water rate in the condenser and minimum heat transfer driving forces in evaporator and condenser.

Answer to Problem 16.1P

The power requirement of an ideal tetrafluoroethane vapor-compression refrigeration cycle is $16.67\text{ kW}$ . Thermodynamic efficiency of the process is ${\eta }_t=0.788$ .

Explanation of Solution

Conventional refrigeration cycle under ideal conditions of operation: Isentropic compression, infinite flow rate of cooling water, & minimum temperature difference for heat transfer = 0.

The data is as follows,

For saturated liquid and vapor at $0°C$,

  $\begin{array}{l}{H}_A=250.28\frac{kJ}{kg}\\ S_A=0.93\frac{kJ}{kg-K}\end{array}$

For saturated liquid at $18°C$,

  $\begin{array}{l}{H}_C=79.735\frac{kJ}{kg}\\ H_D=H_C\end{array}$

For superheated vapor at $5.84atm$ and $S=0.93\frac{kJ}{kg-K}$,

  $H_B=265.164\frac{kJ}{kg}$

Refrigeration Circulation Rate:

  $\begin{array}{l}\dot{m}=\frac{-\left(H_f-H_i\right)\times 0.45\frac{Kg}{s}}{H_A-H_D}\\ \dot{W}=\dot{m}\times \left(H_B-H_A\right)\\ {\eta }_A=\frac{{\dot{W}}_{ideal}}{\dot{W}}\end{array}$

Substitute the values in the above equations,

  $\begin{array}{c}\dot{m}=\frac{-\left(H_f-H_i\right)\times 0.45\frac{kg}{s}}{H_A-H_D}\\ =\frac{-\left(333.5-88.5043\right)\times 0.45}{\left(250.28-79.735\right)}\\ \dot{m}=1.12\frac{kg}{s}\\ \\ \dot{W}=\dot{m}\times \left(H_B-H_A\right)\\ =1.12\times \left(265.164-250.28\right)\\ \dot{W}=16.67\\ \\ {\eta }_t=\frac{{\dot{W}}_{ideal}}{\dot{W}}\\ =\frac{13.15}{16.67}\\ {\eta }_t=0.788\end{array}$

Interpretation Introduction

(d)

Interpretation:

To interpret the power requirement $\left({\dot{W}}_{ideal}\right)$ of a tetrafluoroethane vapor-compression cycle for which the compressor efficiency is $75\%$ .

Concept Introduction:

Vapor-compression refrigeration cycle is operated using tetrafluoroethane. The thermodynamic efficiency is $75\%$ . Ideal condition implies Isentropic Compression, Infinite cooling water rate in the condenser and minimum heat transfer driving forces in evaporator and condenser.

Answer to Problem 16.1P

The power requirement $\left({\dot{W}}_{ideal}\right)$ of a tetrafluoroethane vapor-compression cycle for which the compressor efficiency of $75\%$ is $46.8\text{ k}W$ and the thermodynamic efficiency is ${\eta }_t=0.281$

Explanation of Solution

Given, efficiency is $\eta =0.75$

The practical cycle has 4 major points,

Point A: Saturated vapor at $5°C$ .

Point B: Superheated vapor at $9.17atm$ .

Point C: Saturated Liquid at $10°C$ .

Point D: Mix of saturated liquid and saturated vapor at $5°C$ with H of point C,

For saturated liquid and vapor at $5°C$,

  $\begin{array}{l}{H}_{liq}=45.54\frac{kJ}{kg}\\ S_{liq}=0.18\frac{kJ}{kg-K}\\ H_{vap}=247.67\frac{kJ}{kg}=H_A\\ S_{vap}=0.914\frac{kJ}{kg-K}=S_A\end{array}$

For saturated liquid at $10°C$,

  $\begin{array}{l}{H}_C=102.9\frac{kJ}{kg}\\ S_C=0.38\frac{kJ}{kg-K}\end{array}$

For isentropic compression, the entropy of Point B is $S_B=0.934\frac{kJ}{kg-K}$ at $P=9.17atm$ .

  $\begin{array}{l}{H^{\prime }}_B=274.47\frac{KJ}{Kg}\left(FromFig.G-2\right)\\ H_B=H_A+\frac{{H^{\prime }}_B-H_A}{\eta }\\ H_B=247.67+\frac{274.47-247.67}{0.75}\\ H_B=283.4\frac{kJ}{kg}\end{array}$

  $\begin{array}{c}{x}_D=\frac{H_D-H_{liq}}{H_{vap}-H_{liq}}\\ =\frac{102.9-45.54}{247.67-45.54}\\ x_D=0.284\end{array}$

Entropy at point D can be calculated as

\n

  $\begin{array}{c}{S}_D=S_{liq}+x_D\left(S_{vap}-S_{liq}\right)\\ =0.18+\left(0.284\left(0.914-0.18\right)\right)\\ S_D=0.388\frac{kJ}{kg-K}\end{array}$

Refrigerant circulation rate can be calculated as

\n

  $\begin{array}{c}\dot{m}=\frac{-\left(H_f-H_i\right)\times 0.45\frac{kg}{s}}{H_A-H_D}\\ =\frac{-\left(333.5-88.5043\right)\times 0.45}{\left(247.67-102.9\right)}\\ \dot{m}=1.31\frac{kg}{s}\\ \\ \dot{W}=\dot{m}\times \left(H_B-H_A\right)\\ =1.31\times \left(283.4-247.67\right)\\ \dot{W}=46.8\\ \\ {\eta }_t=\frac{{\dot{W}}_{ideal}}{\dot{W}}\\ =\frac{13.15}{46.8}\\ {\eta }_t=0.281\end{array}$

Thermodynamic Analysis

\n

Here $T_{\sigma }=21°C$,

  $\begin{array}{c}{\dot{W}}_{ideal}=13.15KW\\ {\dot{W}}_{LostCompressor}=\dot{m}{T}_{\sigma }\left(S_B-S_A\right)\\ =1.31\times 294.15\times \left(0.934-0.914\right)\\ {\dot{W}}_{LostCompressor}=8.305kW\\ \\ {\dot{Q}}_{Condenser}=\dot{m}\left(H_C-H_B\right)\\ =1.31\times \left(102.9-283.4\right)\\ {\dot{Q}}_{Condenser}=-236.455\\ {\dot{W}}_{LostCondenser}=\dot{m}{T}_{\sigma }\left(S_C-S_B\right)-{\dot{Q}}_{Condenser}\\ =1.31\times 294.15\left(0.38-0.934\right)-\left(-236.455\right)\\ {\dot{W}}_{LostCondenser}=14.178\text{ kW}\\ \\ {\dot{W}}_{Throttle}=\dot{m}{T}_{\sigma }\left(S_D-S_C\right)\\ =1.31\times 294.15\left(0.388-0.377\right)\\ {\dot{W}}_{Throttle}=4.24\text{ kW}\end{array}$