Explanation Given: The temperature is 80 F . The inlet temperature is 140 F . The temperature is 80 F . The pressure is 20 lbf / in 2 The volume of tank is 20 ft 3 . The initial pressure is 180 lbf / in 2 . Concept used: Write the expression for energy balance equation. Q T = m 2 ⋅ u 2 − m 1 ⋅ u 1 − m i ⋅ h i …… (1) Here, u i is specific internal energy at entry, m 1 is initial mass, h i specific enthalpy at inlet u 2 specific internal energy at exit, m 2 final mass, Q T is heat transfer through tank and m i is mass at inlet. From superheated ammonia table F.8.2 corresponding to 80 F and 20 psia the exiting specific volume is 16.765 ft 3 / lbm From superheated ammonia table F.8.1 corresponding to 80 F the specific volume are : The specific volume for liquid is 0 .026677 ft 3 / lbm . The specific volume for gas is 1 .95260 ft 3 / lbm . Write the expression for initial mass. m i = ( V v f + V v g ) − V v 1 …… (2) Here, m 2 is mass, V is volume of tank, v 1 for specific volume, v f is specific volume of liquid, v g specific volume of gas. Write the expression for mass at stage 1. m 1 = V v 1 …… (3) Write the expression for mass at stage 2. m 2 = V v f + V v g …… (4) Write the expression for quality factor. x 2 = v f v g …… (5) Here, x 2 is quality factor. From superheated ammonia table F.8.1 corresponding to 80 F the specific internal energy are: The specific internal energy for liquid is 130 .92 Btu / lbm . The specific internal energy for vapor and mixture is 443 .37 Btu / lbm . Write the expression for specific internal energy. u 2 = u f + x 2 u f g …… (6) Here, x 2 is quality factor, u f is specific internal energy for liquid and u 2 is specific internal energy at stage 2 and u f g is specific internal energy at vapor and liquid. From superheated ammonia table F.8.2 corresponding to 140 F and 175 psia the specific enthalpy is 667.67 Btu / lbm : From superheated ammonia table F.8.2 corresponding to 140 F and 200 psia the specific enthalpy is 663.90 Btu / lbm : Write the expression for inlet specific enthalpy. h i = 667.67 + P − 175 200 − 175 ( 663.90 − 667.67 ) …… (7) Here, P is pressure. Calculation: Substitute 20 ft 3 for V , 16.765 ft 3 / lbm for v 1 , 0

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ISBN: 9781119321453

Author: Sonntag

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Chapter 4, Problem 4.160EP

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