Explanation Write the expression for the energy balance equation. E in − E out = Δ E system (I) Here, the total energy entering the system is E in , the total energy leaving the system is E out , and the change in the total energy of the system is Δ E system . Simplify Equation (I) and write energy balance relation of an insulated cylinder. W b,in = Δ U W b,in − Δ U = 0 (II) Here, the boundary work to be done into the system is W b,in and the change in the internal energy is Δ U . Substitute Δ H for W b,in − Δ U in Equation (II). Δ H = 0 Δ H ice + Δ H water = 0 [ m c ( 0 ° C − T 1 ) solid + m h i f + m c ( T 2 − 0 ° C ) liquid ] ice + [ m ( h 2 − h 1 ) ] water = 0 (III) Here, the mass of the ice is m , the heat of fusion of ice is h i f , the specific heat of ice at liquid state is c , the initial temperature of ice is T 1 , the final temperature of ice is T 2 , the initial specific enthalpy of ice is h 1 , and the final specific enthalpy of ice is h 2 . Write the expression of initial specific volume of a two-phase system for insulated cylinder device. v 1 = v f + x v f g = v f + x ( v g − v f ) (IV) Here, the specific volume of ice for insulated cylinder device is v f , the specific volume of ice for insulated cylinder device is v g , the specific volume change upon vaporization for insulated cylinder device is v f g , and the quality of final state for insulated cylinder device is x . Write the expression of initial specific internal energy of a two-phase system for piston-cylinder device. u 1 = u f + x u f g (V) Here, the specific internal energy of ice for insulated cylinder device is u f , the specific internal energy change upon vaporization for insulated cylinder device is u f g , and the quality of final state for insulated cylinder device is x . Determine the initial specific volume of the refrigerant. v 1 = V 1 m m = V 1 v 1 (VI) Here, the mass of a ice is m and the initial volume of the saturated liquid-vapour mixture is V 1 . Conclusion: From the Table A-3 “Properties of common liquids, solids and foods”, obtain the value of specific heat of ice at 0 C and 120 C temperature as 2.11 kJ / kg ° C and 4.18 kJ / kg ° C . From the Table A-4 “Saturated water”, obtain the value of specific volume and enthalpy of saturated liquid, vapour and change upon vaporization for insulated cylinder device. At final temperature of water as 120 ° C . v f = 0.001060 v g = 0.89133 m 3 / kg h f = 503.81 h f g = 2202.1 kJ / kg Substitute 0.001060 for v f , 0.2 for x , and 0.89133 m 3 / kg for v g in Equation (I) v 1 = ( 0.001060 ) + ( 0.2 ) × ( 0.89133 m 3 / kg − 0.001060 ) = ( 0.001060 ) + ( 0

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Chapter 4.5, Problem 126RP

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The amount of ice that needs to be added in an insulated cylinder.

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Chapter 4.5, Problem 125RP

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The amount of ice that needs to be added in an insulated cylinder.

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