Explanation Write the expression for the mass balance where the bottle acts as a system. m in − m out = Δ m system m e = m 2 (I) Here, mass of the fluid entering the system is m in , mass of the fluid leaving the system is m out , change in mass of the system is Δ m system , exit mass of the nitrogen is m e and mass of the nitrogen at state 2 is m 2 . Write the expression for the energy balance equation for the closed system. E in − E out = Δ E system − m e h e = m 2 u 2 − m 1 u 1 − m e c p T e = m 2 c v T 2 − m 1 c v T 1 (II) Here, net energy transfer in to the control volume is E in , net energy transfer exit from the control volume is E out , change in internal, kinetic and potential energies of system is Δ E system , amount of heat transferred to the cylinder is Q out , exit enthalpy of the nitrogen is h e , mass of the nitrogen at state 1 is m 1 , internal energy of the nitrogen at state 1 is u 1 , internal energy of the nitrogen at state 2 is u 2 , temperature of the nitrogen at state 1 is T 1 , temperature of the nitrogen at state 2 is T 2 , constant pressure specific heat of nitrogen is c p , constant volume specific heat of nitrogen is c v and exit temperature of nitrogen is T e . Write the formula to calculate the mass of the nitrogen at state 1 ( m 1 ) . m 1 = P V R T 1 (III) Here, pressure of the nitrogen is P , volume of the cylinder is V and gas constant of nitrogen is R . Write the formula to calculate the mass of the nitrogen at state 2 ( m 2 ) . m 2 = m 1 2 (IV) Write the formula to calculate the pressure of the nitrogen at state 2 ( T 2 ) . P 2 = m 2 R T 2 V (V) Write the formula to calculate the exit temperature of the nitrogen ( T e ) . T e = T 1 + T 2 2 (VI) Write the formula to calculate the exit pressure of the nitrogen ( P e ) . P e = P 1 + P 2 2 (VI) For the closed system, write the simplified the entropy balance equation on the extended system which includes water and its immediate surroundings. S in − S out + S gen = Δ S system − m e s e + S gen = m 2 s 2 − m 1 s 1 S gen = m 2 s 2 − m 1 s 1 + m e s e S ˙ g e n = m 2 ( c p ln T 2 − R ln P 2 ) − m 1 ( c p ln T 1 − R ln P 1 ) + m e ( c p ln T e − R ln P e ) (VII) Here, entropy generation is S gen , change of entropy of the system is Δ S system , entropy at inlet condition is S in , entropy at outlet condition is S out , temperature of the reservoir is T R , exit entropy of the nitrogen is s i , entropy of the nitrogen at state 1 is s 1 and entropy of the nitrogen at state 2 is s 2 . Write the formula to calculate exergy destruction during the process ( X destroyed ) . X destroyed = T 0 S gen (VIII) Here, surrounding temperature is T 0 . Conclusion: Refer Table A-2, “Ideal-gas specific heats of various common gases”, obtain the following properties of nitrogen. R = 0.2968 kPa ⋅ m 3 / kg ⋅ K c p = 1.039 kJ / kg ⋅ K c v = 0.743 kJ / kg ⋅ K k = 1 .4 Substitute 1000 kPa for P , 0.100 m 3 for V , 0.2968 kPa ⋅ m 3 / kg ⋅ K for R and 20 ° C for T 1 in Equation (III). m 1 = ( 1000 kPa ) ( 0.100 m 3 ) ( 0.2968 kPa ⋅ m 3 / kg ⋅ K ) ( 20 ° C ) = ( 1000 kPa ) ( 0.100 m 3 ) ( 0.2968 kPa ⋅ m 3 / kg ⋅ K ) ( 20 + 273 K ) = ( 1000 kPa ) ( 0.100 m 3 ) ( 0.2968 kPa ⋅ m 3 / kg ⋅ K ) ( 293 K ) = 1.150 kg Substitute 1

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Author: CENGEL

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Chapter 8.8, Problem 118RP

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The change in the exergy content of the tank.

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Chapter 8.8, Problem 117RP

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The change in the exergy content of the tank.

Solution Summary: The author writes the expression for the mass balance where the bottle acts as a system.

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