Explanation Write the expression for the energy balance equation for a closed system where the steam acts as the system. E in − E out = Δ E system W b,in − Q out = Δ U W b,in − Q out = m ( u 2 − u 1 ) (I) 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, potential energies of the system is Δ E system , boundary work in the system is W b,in , amount of heat transferred from the system is Q out , change in internal energy of the system is Δ U , initial internal energy of the steam is u 1 and final internal energy of the steam is u 2 . Write the formula to calculate boundary work during the process ( w b,in ) . w b,in = P ( v 1 − v 2 ) (II) Here, pressure of the steam is P , initial specific volume of the steam is v 1 and final specific volume of the steam is v 2 . Write the formula to calculate heat transfer during the process ( q out ) using the energy balance equation. q out = w b,in − ( u 2 − u 1 ) (III) Write the expression to calculate the exergy change between the initial and final states ( Δ ϕ ) . Δ ϕ = ( ϕ 1 − ϕ 2 ) = u 2 − u 1 + P 0 ( v 1 − v 2 ) − T 0 ( s 1 − s 2 ) − q out ( 1 − T 0 T R ) (IV) Here, initial and final exergies of the steam are ϕ 1 and ϕ 2 respectively, surrounding temperature is T 0 , surrounding pressure is P 0 , initial and final specific entropies of the steam are s 1 and s 2 respectively and temperature of the thermal energy reservoir is T R . Write the formula to second-law efficiency of the process ( η II ) . η II = w b,in Δ ϕ (V) Conclusion: Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties of steam at the pressure of 75 kPa for initial state. v 1 = v g = 2 .2172 m 3 / kg u 1 = u g = 2496 .1 kJ / kg s 1 = s g = 7 .4558 kJ / kg ⋅ K Here, specific volume of saturated vapor is v g , internal energy of saturated vapor is u g and specific entropy of saturated vapor is s g . Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties of steam at the pressure of 75 kPa for final state. v 2 = v f = 0 .001037 m 3 / kg u 2 = u f = 384 .36 kJ / kg s 2 = s f = 1 .2132 kJ / kg ⋅ K Here, specific volume of saturated liquid is v f , internal energy of saturated liquid is u f and specific entropy of saturated liquid is s f . Substitute 75 kPa for P , 2 .2172 m 3 / kg for v 1 and 0 .001037 m 3 / kg for v 2 in Equation (II). w b,in = ( 75 kPa ) ( 2 .2172 m 3 / kg − 0 .001037 m 3 / kg ) = ( 75 kPa ) ( 2 .2172 m 3 / kg − 0 .001037 m 3 / kg ) ( 1 kJ 1 kPa ⋅ m 3 ) = 166.2 kJ / kg Substitute 166.2 kJ / kg for w b,in , 384 .36 kJ / kg for u 2 and 2496 .1 kJ / kg for u 1 in Equation (III)

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ISBN: 9780073398174

Author: Yunus A. Cengel Dr., Michael A. Boles

Publisher: McGraw-Hill Education

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Chapter 8.8, Problem 89RP

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