Explanation Given: Volume of tank is 1 m 3 . Mass of R-410A in tank is 100 kg . Initial temperature of R-410A is 15 ° C . Pressure of saturated liquid is 100 kPa . Final temperature of tank is 15 ° C . Concept used: Write the expression for specific volume of R-410A in tank. v 1 = V m 1 ..... (1) Here, v 1 is thespecific volume of R-410A in tank, V is the initial volume of tank and m 1 is the mass of R-410A in tank. Write the expression for Quality of vapor in tank. x = v 1 − v f v f g ..... (2) Here, x is the Quality of vapor in tank, v f is the specific enthalpy of saturated liquid and v f g is the specific volume of evaporation at 15 ° C . Write the expression for internal energy of R-410A. u 1 = u f + x u f g ..... (3) Here, u 1 is the internal energy of R-410A, u f is the specific internal energy of liquid R-410A and u f g is the specific internal energy of evaporation of R-410A corresponding to 15 ° C . Write the expression for final mass of R-410A. m 2 = V v 2 ..... (4) Here, m 2 is the final mass of R-410A and v 2 is the final specific volume. Write the expression for mass entering the tank. m e = m 1 − m 2 ..... (5) Here, m e is the mass entering the tank. Write the expression for heat transfer by tank. Q = m 2 u 2 − m 1 u 1 + m e h e ..... (6) Here, Q is the heat transfer by tank. Calculation: Substitute 1 m 3 for V and 100 kg for m 1 in equation (1). v 1 = 1 100 = 0.01 m 3 /kg Refer to table B.4.1 "Properties of saturated R-410A " to obtain the specific volume of saturated liquid R-410A as 0 .000904 m 3 /kg and specific volume of evaporation as 0 .01955 m 3 /kg corresponding to 15 ° C . Substitute 0 .01955 m 3 /kg for v f g , 0

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ISBN: 9781119803645

Author: Sonntag

Publisher: WILEY

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Chapter 4, Problem 4.88P

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Chapter 4 Solutions

FUNDAMENTALS OF THERMODYNAMICS

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