Explanation Assume that the rigid vessel is stationary and thus, there is no change in kinetic and potential energy. Also assume that the thermal energy stored in the rigid vessel is negligible. Here, water is taken as the system. Write the energy balance for the given closed system. E in − E out = Δ E system Q in = Δ U Q in = m ( u 2 − u 1 ) (I) Here, energy added to the system is E in , energy out of the system is E out , energy accumulated into the system is Δ E system , heat added to the system is Q in , change in internal energy of the system is Δ U , mass of the system is m , specific internal energy of system at initial point is u 1 and specific internal energy of the system at final point is u 2 . Write the initial specific volume of the refrigerant. v 1 = v f + x 1 ⋅ v f g = v f + x 1 ⋅ ( v g − v f ) (II) Here, specific volume at saturated liquid state is v f , specific volume at saturated vapor state is v g and dryness fraction is x 1 . Write the specific internal energy of the system. u 1 = u f + x 1 ⋅ u f g = u f + x 1 ⋅ ( u g − u f ) (III) Here, specific internal energy at saturated liquid state is u f and specific internal energy at saturated vapor state is u g . Write the dryness fraction for final state. x 2 = v 2 − v f v g − v f (IV) Here, the specific volume at the final state is v 2 . Write the specific internal energy of the system at the final state. u 2 = u f + x 2 ⋅ u f g = u f + x 2 ⋅ ( u g − u f ) (V) Write the mass of the refrigerant. m = V 1 v 1 (VI) Here, the initial volume is V 1 . Calculate the heat added into the system. Q in = m ( u 2 − u 1 ) (VII) Conclusion: Refer to Table A-4 “Saturated water-Temperature table” to obtain specific volume at saturated liquid state as 0.001043 m 3 / kg , specific volume at saturated vapor state as 1.6720 m 3 / kg , specific internal energy at saturated liquid state as 419.06 kJ / kg and specific internal energy at saturated vapor state as 2506 kJ / kg with respect to 100 ° C and obtain specific volume at saturated liquid state as 0.001091 m 3 / kg , specific volume at saturated vapor state as 0.39248 m 3 / kg , specific internal energy at saturated liquid state as 631.66 kJ / kg and specific internal energy at saturated vapor state as 1927.4 kJ / kg . These values were taken corresponding to the temperature 150 ° C . Substitute 0.001043 m 3 / kg for v f , 1.6720 m 3 / kg for v g and 0.123 for x in Equation (II). v 1 = 0.001043 m 3 / kg + 0.123 [ ( 1.6720 m 3 / kg ) − ( 0.001043 m 3 / kg ) ] = 0

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ISBN: 9780100257054

Author: CENGEL

Publisher: YUZU

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Chapter 4.5, Problem 31P

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The heat added into the system.

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