Answer to Question #175229 in Mechanical Engineering for Von Nikko

Question

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

Answer 1

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 C_v(T₁-T₂)

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

=> 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