Question #290847

3. A working substance undergoes a reversible non flow process during which P

= (-3V + 100) psia, where V changes from 10 ft3/min to 30 ft3/min. The change

of internal energy for the process is – 100 BTU/min. For a non-flow process,

determine a) work (BTU/min) b) heat transferred (BTU/min).


1
Expert's answer
2022-01-30T14:15:44-0500

P=(3V+100)psiaP=(-3V+100)psia



a) Work=V1V2PdV=\int_{V_1}^{V_2}PdV


=10303V+100dV=\int_{10}^{30}-3V+100dV




3V22+100V1030\frac{-3V^2}{2}+100V\mid_{10}^{30}



3(302)2+100(30)+3(102)2100(10)      =800BTU/min\frac{-3(30^2)}{2}+100(30)+\frac{3(10^2)}{2}-100(10)\\~~~~~~=800BTU/min


b) Heat transferred=100800=700BTU/min= 100-800= -700BTU/min


Note:Eint=QWNote: E_{int}=Q-W







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