(a) Explanation Write the expression for the mass of steam. m = V v 1 (I) Here, the mass of steam is m , the volume is V and specific volume at initial state is v 1 . Write the expression for enthalpy at final state. W in = m ( h 2 − h 1 ) W in m = h 2 − h 1 h 2 = ( W in m + h 1 ) (II) Here, the enthalpy at final state is h 2 , the input work is W in and the enthalpy at initial state is h 1 . Write the expression for dryness fraction at final state. x 2 = ( h 2 − h f h f g ) (III) Here, the dryness fraction at final state is x 2 , the enthalpy of saturated liquid vapour mixture at final state is h f g and the enthalpy of saturated liquid at final state is h f . Write the expression for entropy at final state. s 2 = s f + ( x 2 s f g ) (IV) Here, the entropy at final state is s 2 , the entropy of saturated liquid at final state is s f and the entropy of saturated liquid vapour at final state is s f g . Write the expression for internal energy at final state. u 2 = u f + ( x 2 u f g ) (V) Here, the internal energy at final state is u 2 , the internal energy of saturated liquid u f and the internal energy of saturated liquid vapour is u f g . Write the expression for specific volume at final state. v 2 = v f + ( x 2 ) ( v g − v f ) (VI) Here, the specific volume at final state is v 2 , the specific volume of saturated liquid at final state is v f and the specific volume of saturated vapour at final state is v g . Calculate the reversible work input. W rev,in = − m [ ( u 1 − u 2 ) − T 0 ( s 1 − s 2 ) + P 0 ( v 1 − v 2 ) ] (VII) Here, the reversible work input is W rev,in , the internal energy at initial state is u 1 , the surrounding temperature is T 0 , the entropy at initial state is s 1 and the pressure of surrounding is P 0 . Conclusion: Refer to table A-4 “saturated water-Temperature table” at initial state corresponding to 120 kPa to obtain internal energy of saturated liquid ( u 1 ) as 439.27 kJ/kg , the specific volume of saturated liquid ( v 1 ) as 0.001047 m 3 /kg , the enthalpy of saturated liquid ( h 1 ) as 439.36 kJ/kg and entropy of saturated liquid ( s 1 ) as 1.3609 kJ/kg ⋅ K . Substitute 0.8 L for V and 0.001047 m 3 /kg for v 1 in Equation (I). m = ( 0.8 L ) ( 0.001047 m 3 /kg ) = ( 0.8 L ) ( 1 m 3 1000 L ) ( 0.001047 m 3 /kg ) = 0.76408 kg Substitute 439.36 kJ/kg for ( h 1 ) and 1400 kJ for W in and 0.76408 kg for m in Equation (II). h 2 = ( ( 1400 kJ 0.76408 kg ) + ( 439.36 kJ/kg ) ) = ( 1832.2688 kJ/kg ) + ( 439.36 kJ/kg ) ≈ 2271.6288 kJ/kg Refer to table A-4 “saturated water-Temperature table” to obtain enthalpy of saturated liquid ( h f ) as 439.36 kJ/kg , enthalpy of liquid-vapour ( h f g ) as 2243.7 kJ/kg , entropy of saturated liquid ( s f ) as 1.3609 kJ/kg ⋅ K , entropy of liquid-vapour ( s f g ) as 5.93687 kJ/kg ⋅ K , internal energy of saturated liquid as ( u f ) as 439 (b) To determine The energy destroyed during the process.

8th Edition

ISBN: 9780073398174

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

Publisher: McGraw-Hill Education

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Chapter 8.8, Problem 33P

(a)

To determine

The minimum work by the cylinder piston device.

(b)

To determine

The energy destroyed during the process.

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Chapter 8.8, Problem 32P

Chapter 8.8, Problem 35P

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