Explanation Given: Temperature of line is 300 K . Pressure of line is 0.5 MPa . Volume of tank is 50 m 3 . Pressure of tank is 0.5 MPa . The final temperature is 250 K . Concept used: Write the expression for energy balance for tank. m 2 u 2 − m 1 u 1 = m i n h i n + Q − W Here, m 2 is the final mass in tank, u 2 is the final internal energy of tank, m 1 is the initial mass in tank, u 1 is the initial internal energy of tank, m i n is the mass enter the tank, h i n is the enthalpy of Nitrogen inside the tank, Q is the heat transfer and W is the work done by turbine. Substitute 0 for m 1 and 0 for Q in above expression. m 2 u 2 − 0 u 1 = m i n h i n + 0 − W The mass enter inside the tank will be equal to the final mass of Nitrogen inside the tank. Substitute m 2 for m i n and simplify above expression. m 2 u 2 = m 2 h i n − W .......... (1) Write the expression for final mass of Nitrogen in the tank. m 2 = V v 2 .......... (2) Here, V is the volume of tank and v 2 is the specific volume of Nitrogen corresponding to 0.5 MPa . Calculation: Refer to Table B.6.2 “Properties of superheated Nitrogen” to obtain the specific enthalpy of Nitrogen at 0.5 MPa and 300 K by interpolation. h i n − 310.50 310.06 − 310.50 = 500 − 400 600 − 400 h i n = 310.28 kJ/kg Refer to Table B.6.2 “Properties of superheated Nitrogen” to obtain the specific internal energy of Nitrogen corresponding to 400 kPa and 250 K by interpolation. u 0 .4 MPa − 176.67 191.64 − 176.67 = 250 − 240 260 − 240 u 0. 4Mpa = 184.155 kJ/kg Refer to Table B.6.2 “Properties of superheated Nitrogen” to obtain the specific internal energy of Nitrogen corresponding to 600 kPa and 250 K by interpolation. u 0.6 Mpa − 176.11 191.13 − 176.11 = 250 − 240 260 − 240 u 0

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ISBN: 9781119321453

Author: Sonntag

Publisher: JOHN WILEY+SONS,INC.-CONSIGNMENT

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Chapter 4, Problem 4.105P

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The point at which temperature is 250 K and turbine work.

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