Explanation Given : Inlet pressure is 150 kPa . Inlet temperature is 25 ° C . Volume of tankis 1 m 3 . Temperature of mixing ammonia is 60 ° C . Pressure of mixing ammonia is 1200 kPa . Final temperature of air is 25 ° C . Concept used: Write the expression for mass. m = V v …… (1) Here, m is the mass, V is the volumeand v is specific volume. Write the expression forinlet mass. m i = m 2 − m 1 …… (2) Here, m i is the inlet mass, m 2 is mass after filling and m 1 is mass before filling. Write the expression of internal energy. u = u f + x u f g …… (3) Here, u isinternal energy, u f is internal energy of fluid and u f g is internal energy of mixture. Write the expression for energy balance for tank. q + m i h i = m 2 u 2 − m 1 u 1 …… (4) Here, m is the mass, q is the heat transfer, h is enthalpy and u is internal energy. i, 1and 2 are subscript for different state. Calculation: Properties of ammonia at 150 kPa and 25 ° C are given as follows: v 1 = 0.9552 m 3 /kg and u 1 = 1380.6 kJ/kg . Properties of ammonia at 25 ° C are given as follows: v f = 0.001658 m 3 /kg , v g = 0.12813 m 3 /kg , u f = 296.6 kJ/kg and u f g = 1038.4 kJ/kg . For inlet ammonia, Substitute 1 m 3 for V and 0.9552 m 3 /kg for v in equation (1). m 1 = 1 m 3 0.9552 m 3 /kg m 1 = 1.047 kg Finally, half of the fluid is liquid ammonia and half is vapor. Quality of ammonia at state 2 is calculated as follows: x = m v m v + m l x = V v g V v g + V v f x = 0

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Chapter 4, Problem 4.100P

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