Answer:
Energy Transfer = 350 kJ
Explanation:
The net work can be determined from an energy balance. That is, with assumption
∆KE + ∆PE + ∆U = Q − W
Where
∆KE = ∆PE = 0 (Since There is no significant change in the kinetic or potential energy of the steam)
The net work is the sum of the work associated with the paddlewheel Wpw
and the work done on the piston Wpiston:
W = Wpw + Wpiston
From the given information, Wpw= −18.5 kJ,
Collecting results:
Wpw + Wpiston = Q − ∆U
Wpiston = Q − ∆U − Wpw= Q − m (u2− u1) − Wpw
Where Q=80kJ, m=5kg, u2 = 2659.6 kJ/kg, u1 = 2709.9 kJ/kg
= 80 kJ − 5 kg (2659.6 − 2709.9)kJ/kg − ( −18. 5 kJ)
= 350 kJ
The energy transfer by work from steam to piston is ; [tex]W_{p}[/tex] = 350 kJ
Given that ;
ΔU = ( 2659.6 - 2709.9 ) = - 50.3 kJ/Kg
Q ( heat magnitude ) = 80 kJ
m ( mass of steam ) = 5 kg
Energy transferred by paddle wheel ( [tex]W_{pp}[/tex] ) = - 18.5 kJ
Energy transferred to piston ( [tex]W_{p}[/tex] ) = ?
Total work done given that there is no change on K.E. and P.E.
Total work done = m ( ∆U ) = Q - W ----- ( 1 )
where W = [tex]W_{pp} + W_{p}[/tex]
Equation ( 1 ) becomes
ΔU = Q - [tex]( W_{PP} + W_{P} )[/tex] ------ ( 2 )
Therefore the energy transferred to piston by work from steam ( [tex]W_{p}[/tex] )
[tex]W_{p}[/tex] = Q - m( ΔU ) - [tex]W_{pp}[/tex]
= 80 - 5(- 50.3 ) - ( -18.5 )
= 350 kJ
Hence the The energy transfer by work from steam to piston is ; [tex]W_{p}[/tex] = 350 kJ
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