(a) To determine Whether the operation of the refrigerator is reversible, irreversible or impossible. Explanation The following figure shows the cycle on the P − v diagram. Figure (1) Figure (I) shows the operation of the cycle, in this P − v digram process 1 − 2 shows the adiabatic expansion of the steam in the turbine, process 3 − 2 represents constant pressure heat rejection in the condenser, process 3 − 4 represents adicbatic compression in the pump and the process 4 − 1 represents constant pressure heat addition in the Boiler. Write the expression for the time rate equation. ∮ δ Q T = − σ c y c l e (I) Here, the heat interation is Q and the temperaure of the reservoir is T and the strength of inequality is σ c y c l e . Simpify the Equation (I) for the cycle. Q 4 − 1 ( T s a t ) 1 MPa − Q 2 − 3 ( T s a t ) 20 kPa = − σ c y c l e (II) Here, the heat added to the steam in the boiler is Q 4 − 1 , heat rejected in the boiler is Q 2 − 3 , saturation temperation of steam at 1 MPa is ( T s a t ) 1 MPa and the saturation temperature at 20 kPa is ( T s a t ) 20 kPa . Write the expression for the heat added in the boiler. Q 4 − 1 = ( h g ) 1 MPa − ( h f ) 1 MPa (III) Here, the enthlpy of saturated liquid at 1 MPa is ( h f ) 1 MPa and enthalpy of the saturated vapour at 1 MPa is ( h g ) 1 MPa . Write the expression for the heat rejeted in the condenser. Q 2 − 3 = h 2 − h 3 (IV) Here, the enthalpy at inlet of condenser is h 2 and enthalpy at the exit of the condenser is h 3 . Write the expression for the enthelpy at the inlet of the condenser. h 2 = ( h f ) 20 kPa + x 2 ( h f g ) 20 kPa (V) Here, the enthlpy of saturated liquid at 20 kPa is ( h f ) 20 kPa , latent heat of the liqid at 20 kPa is ( h f g ) 20 kPa and the dryness fraction or quality of steam at point 2 is x 2 . Write the expression for the enthelpy at the exit of the condenser. h 3 = ( h f ) 20 kPa + x 3 ( h f g ) 20 kPa (VI) Here, the dryness fraction or quality of steam at point 3 is x 3 . Conclusion: Refer to the Table A-3 “Properties of Saturated Water (Liquid–Vapor): Pressure Table” at 1 MPa to obtain the value of ( h g ) 1 MPa as 2778.1 kJ / kg , ( h f ) 1 MPa as 762.81 kJ / kg and ( T s a t ) 1 MPa as 179.9 °C Substitute 2778.1 kJ / kg for ( h g ) 1 MPa and 762.81 kJ / kg for ( h f ) 1 MPa in the Equation (III). Q 4 − 1 = ( 2778.1 kJ / kg ) − ( 762.81 kJ / kg ) = ( 2778.1 − 762.81 ) kJ / kg = 2015.29 kJ / kg ≈ 2015 (b) To determine The thermal efficiency of the cycle. (c) To determine The comparison of the thermal efficiency of the cycle with the Carnot efficiency.

Fundamentals of Engineering Thermodynamics

8th Edition

ISBN: 9781118832318

Author: MORAN

Publisher: WILEY

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Chapter 5.11, Problem 90P

(a)

To determine

Whether the operation of the refrigerator is reversible, irreversible or impossible.

(b)

To determine

The thermal efficiency of the cycle.

(c)

To determine

The comparison of the thermal efficiency of the cycle with the Carnot efficiency.

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Chapter 5.11, Problem 89P

Chapter 5.11, Problem 91P

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