Explanation Given : Inlet temperature is 30 ° C . Inlet pressure is 1000 kPa . mass flow rate of ammoniais 0.01 kg/s . Exit pressure is 300 kPa . Inlet velocity is 0 m/s . Final velocity is 450 m/s . Concept used: Write the expression for energy balance for nozzle. V 2 2000 = h 1 − h 2 …… (1) Here, V is the exit velocity, h 1 is the inlet enthalpyand h 2 is the exit enthalpy. Write the expression of enthalpy. h = h f + x h f g …… (2) Here, h is enthalpy, h f is enthalpy of fluid and h f g is enthalpy of mixture. Write the expression of volume. v = v f + x v f g …… (3) Here, v is specific volume, v f is specific volume of fluid and v f g is specific volume of mixture. Write the expression of mass flow rate. m ˙ = A V v 2 …… (4) Here, v 2 is specific volume, V is the exit velocity, m ˙ is mass flow rate and A is exit area. Calculation: Properties of ammonia at 30 ° C and 1000 kPa are given as follows: h 1 = 1493.5 kJ/kg . Substitute 1493.5 kJ/kg for h 1 and 450 m/s for V in equation (1). ( 450 m/s ) 2 2000 = ( 1493.5 kJ/kg ) − h 2 h 2 = 1392.25 kJ/kg Properties of ammonia at 300 kPa are given as follows: T e = − 9.24 ° C , h f = 137.42 kJ/kg , v f = 0

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Final temperature, quality and exit area.

Solution Summary: The author describes the final temperature, quality, and exit area of the nozzle.

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