Answer to Question #175229 in Mechanical Engineering for Von Nikko

Question #175229

Consider that 2lb of air(R=53.342 ft-lb/lb-R and k=1.4) has a decrease of internal energy of 41.16 btu while its Fahrenheit temperature is reduced to one third of its inital temperature during a reversible nonflow constant pressure process. Determine a inital and final fahrenheit temperature b heat added c work d change in entropy


1
Expert's answer
2021-04-01T01:09:41-0400

Given as ,

Mass(m)= 2lb

Decrease in internal energy

(U1-U2)= 41.16 btu =41.16×778.17=32029.47 lb-ft

T2 ="\\frac{T_1}{3}"

By first law of thermodynamics,

Q=U2-U1+W

The work done is given by

W=p(V2-V1)=mR(T2-T1)

U1-U2=m Cv(T1-T2)

"m \u00d7\\frac{R}{k-1}\u00d7\\frac{2T_1}{3}=2\u00d7\\frac{53.342}{1.4-1}\u00d7\\frac{2T_1}{3}=32029.47"

=> T1 = 180.13 F

T2 =T1/3=60.04 F

Work done:

W=mR(T2-T1) =2×53.342×(-120.09)=-12811.68 lb ft

Therefore heat transfer (Q) Can be found by adding work done and change in internal energy.

Q=-12811.68-32029.47=-44841.10 lbft




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