(a) Explanation Sketch the T − s diagram for the two stage cascade refrigeration system as in Figure (1). Write the energy balance equation for the heat exchanger. E ˙ in = E ˙ out ∑ m ˙ e h e = ∑ m ˙ i h i m ˙ A ( h 5 − h 8 ) = m ˙ B ( h 2 − h 3 ) (I) Here, energy entering the system is E ˙ in , energy leaving the system is E ˙ out , mass flowrate of the refrigerant at the exit of heat exchanger is m ˙ e , specific enthalpy of the refrigerant at the exit of heat exchanger is h e , mass flow rate of refrigerant at the inlet is m ˙ i , specific enthalpy of the refrigerant at the inlet is h i , the mass flow rate of refrigerant through the upper cycle is m ˙ A , the mass flow rate of refrigerant through the lower cycle is m ˙ B , specific enthalpy of refrigerant at the inlet of second stage compressor is h 5 , specific enthalpy of refrigerant at the exit of second stage compression is h 6 , specific enthalpy of refrigerant at the inlet of first stage compression is h 1 , and the specific enthalpy of the refrigerant at the exit of first stage compression is h 2 . Write the relation between the specific enthalpies at the inlet and exit of throttling process. h in = h out (II) Here, specific enthalpy at the inlet of throttling is h in , and the specific enthalpy at the exit of throttling is h out . Conclusion: From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at compressor inlet pressure ( P 1 ) of 0.14 MPa as follows: h 1 = h g = 239.19 kJ / kg s 1 = s g = 0.94467 kJ / kg ⋅ K Here, specific enthalpy of the saturated vapor is h g and the specific entropy of the saturated vapor is s g . The specific entropy at the end of isentropic compression ( s 2 ) is calculated as s 2 = s 1 = 0.94467 kJ / kg ⋅ K Refer to Table A-13, “Superheated R-134a”, and obtain the values of R-134a at pressure of 0 .4 MPa and specific entropy of 0.94467 kJ / kg ⋅ K by using interpolation method. 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 (III) Here, the variables denoted by x and y are specific entropy and specific enthalpy. Show temperature and enthalpy values from the Table A-13E. Entropy ( s 2 ) , in kJ / kg ⋅ K Enthalpy ( h 2 ) , in kJ / kg 0.9306 256.59 0.94467 ? 0.9628 265.88 Substitute 0.9306 for x 1 , 0.94467 for x 2 , 0.9628 for x 3 , 256.59 for y 1 , and 265.88 for y 3 in Equation (III) to calculate the specific enthalpy at the isentropic exit of compressor ( h 2 ) . y 2 = ( 0.94467 − 0.9306 ) ( 265.88 − 256.59 ) ( 0.9628 − 0.9306 ) + 256.59 = 260.65 kJ / kg From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at second stage compressor inlet pressure ( P 2 ) of 0 (b) To determine The rate at which heat removed from the refrigerated space. The power required to the compressor. (c) To determine The coefficient of refrigeration for the two-stage cascade refrigeration system.

8th Edition

ISBN: 9780100257054

Author: CENGEL

Publisher: YUZU

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Chapter 11.10, Problem 56P

(a)

To determine

The mass flow rate of the refrigerant through the lower compression cycle.

(b)

To determine

The rate at which heat removed from the refrigerated space.

The power required to the compressor.

(c)

To determine

The coefficient of refrigeration for the two-stage cascade refrigeration system.

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Chapter 11.10, Problem 54P

Chapter 11.10, Problem 57P

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