(a) Explanation The below figure shows the working of cascade vapor compression refrigeration system. Figure-(1) Write the expression for temperature at state 5. T 5 = T 3 − T b (I) Here, the condensing temperature of R-134a is T 3 and the temperature below the condensing temperature is T b . Write the expression for enthalpy at point 8. h 8 = h f 8 + x 8 h f g 8 (II) Here, the enthalpy of liquid at point 8 is h f 8 , the enthalpy of vapor liquid mixture at point 8 is h f g 8 and the dryness fraction is x 8 . Write the expression for specific entropy at point 8. s 8 = s f 8 + x 8 ( s g 8 − s f 8 ) (III). Write the expression for mass flow rate of R-134a refrigerant. m ˙ = Q ˙ i n ( h 1 − h 4 ) (IV) Here, the refrigeration capacity is Q ˙ i n , the enthalpy at point 1 is h 1 , the enthalpy at point 4 is h 4 . Write the expression for power input to the first compressor. W c 1 = m ˙ ( h 2 − h 1 ) (V) Here, the enthalpy at point 2 is h 2 and the enthalpy at point 1 is h 1 . Write the expression for energy balance in heat exchanger. m ˙ ( h 2 − h 3 ) = m ˙ 1 ( h 5 − h 8 ) (VII) Here, the enthalpy at point 5 is h 5 , the enthalpy at point 8 is h 8 . Write the expression for power input to the compressor 2. W ˙ c 2 = m ˙ 1 ( h 6 − h 5 ) (VIII) Conclusion: Refer table A-10E “Properties of saturated refrigerant R-134a temperature table” to obtain the specific enthalpy and the specific entropy at temperature − 30 ° F as h 1 = 97.32 Btu / lb and s 1 = 0.2266 Btu / lb ° R Since the process 1-2 is isentropic process, so s 1 = s 2 Refer table A-12E “Properties of superheated refrigerant 134a vapor table” to obtain the enthalpy at pressure 50 lbf / in 2 and specific entropy 0.2266 Btu / lb ° R by using interpolation method of two variable. Write the formula of interpolation method of two variables. y 2 = ( x 2 − x 1 ) ( y 3 − y 1 ) ( x 3 − x 1 ) + y 1 (V) Here, the variables denote by x and y are specific entropy and specific enthalpy. Show the specific enthalpy at entropy 0.2189 Btu / lb ° R and 0.2276 Btu / lb ° R as in below table. Specific entropy, Btu / lb ° R specific enthalpy, Btu / lb 0.2189 Btu / lb ° R 107.43 Btu / lb 0.2266 Btu / lb ° R ? 0.2276 Btu / lb ° R 111.85 Btu / lb Substitute 0.2189 Btu / lb ° R for x 1 , 0.2266 Btu / lb ° R for x 2 , 0.2276 Btu / lb ° R for x 3 , 107.43 Btu / lb for y 1 , and 111.85 Btu / lb for y 3 in Equation (V). y 2 = ( 0.2266 Btu / lb ° R − 0.2189 Btu / lb ° R ) ( 111.85 Btu / lb − 107.43 Btu / lb ) ( 0.2276 Btu / lb ° R − 0.2189 Btu / lb ° R ) + 107.43 Btu / lb = ( 7.7 × 10 − 3 Btu / lb ° R ) ( 4.42 Btu / lb ) 8.7 × 10 − 3 Btu / lb ° R + 107.43 Btu / lb = 3.91195 Btu / lb + 107.43 Btu / lb = 111.3419 Btu / lb So, the specific enthalpy is 111.3419 Btu / lb . Refer table A-14E “Properties of saturated refrigerant 134a (liquid-vapor) Pressure table” to obtain the enthalpy, specific entropy and saturation temperature at Pressure 50 lbf / in 2 as h 3 = 24.14 Btu / lb , s 3 = 0.0523 Btu / lb ° R and T 3 = 40.27 ° F Since the process 3-4 is throttling process, so h 4 = h 3 Substitute 40.27 ° F for T 3 and 5 ° F for T b in Equation (I). T 5 = 40.27 ° F − 5 ° F = 35.27 ° F Refer table A-7E “Properties of saturated refrigerant R-22 (liquid-vapor) temperature table” to obtain the enthalpy at temperature 35.27 ° F by using interpolation method of two variable. Write the formula of interpolation method of two variables. y 2 = ( x 2 − x 1 ) ( y 3 − y 1 ) ( x 3 − x 1 ) + y 1 (V) Here, the variables denote by x and y are temperature and specific enthalpy. Show the specific enthalpy at temperature 35 ° F and 40 ° F as in below table. Temperature, ° R specific enthalpy, Btu / lb 35 ° F 107.78 Btu / lb 35.27 ° F ? 40 ° F 108.20 Btu / lb Substitute 35 ° F for x 1 , 35.27 ° F for x 2 , 40 ° F for x 3 , 107.78 Btu / lb for y 1 , and 108.20 Btu / lb for y 3 in Equation (V). y 2 = ( 35.27 ° F − 35 ° F ) ( 108.20 Btu / lb − 107.78 Btu / lb ) ( 40 ° F − 35 ° F ) + 107.78 Btu / lb = ( 0.27 ° F ) ( 0.42 Btu / lb ) 5 ° F + 107.78 Btu / lb = 0.02268 Btu / lb + 107.78 Btu / lb = 107.80 Btu / lb So, the specific enthalpy is 107.80 Btu / lb . Refer table A-7E “Properties of saturated refrigerant R-22 (liquid-vapor) temperature table” to obtain the entropy at temperature 35.27 ° F by using interpolation method of two variable. Write the formula of interpolation method of two variables. y 2 = ( x 2 − x 1 ) ( y 3 − y 1 ) ( x 3 − x 1 ) + y 1 (V) Here, the variables denote by x and y are temperature and specific enthalpy. Show the specific entropy at temperature 35 ° F and 40 ° F as in below table. Temperature, ° F Specific entropy, Btu / lb ° R 35 ° F 0.221 Btu / lb ° R 35.27 ° F ? 40 ° F 0 (b) To determine The coefficient of performance. (c) To determine The rate of exergy destruction in heat exchanger.

Fundamentals of Engineering Thermodynamics

8th Edition

ISBN: 9781118832301

Author: SHAPIRO

Publisher: JOHN WILEY+SONS,INC.-CONSIGNMENT

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Chapter 10.7, Problem 28P

(a)

To determine

The power input to each compressor.

(b)

To determine

The coefficient of performance.

(c)

To determine

The rate of exergy destruction in heat exchanger.

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Fundamentals of Engineering Thermodynamics

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