Answer:
The heat transfer [tex]\mathbf{\mathbf{Q_{1-2}} = 35.904 \ Btu}[/tex]
[tex]W_{2-3}= -71.312 \ Btu[/tex]
Explanation:
At the initial state when P₁ = 10 lbf/in :
We obtain the internal energy u₁ and specific volume v₁.
u₁ = [tex]u_g[/tex] = 574.08 btu/lbm
v₁ = [tex]v_g[/tex] = 2.4746 ft³/lbm
Process 1–2 occurs at constant volume until the temperature is 100°F.
i.e T₂ = 100⁰ F
At T₂ = 100⁰ F : v₁ = v₂ = 2.4746 ft³/lbm
[tex]\mathbf{u_2 = 591.28 + \dfrac{2.4746-2.5917}{2.117-2.5917}*(587.68 -591.28)}[/tex]
[tex]\mathbf{u_2 = 591.28 + 0.246682115(-3.6)}[/tex]
[tex]\mathbf{u_2 = 591.28+ (-0.888055614)}[/tex]
[tex]\mathbf{u_2 \approx 590.4 \ btu/lbm}[/tex]
[tex]\mathbf{Q_{1-2}= W+ \Delta U}[/tex]
[tex]\mathbf{Q_{1-2}= W+m( u_2 -u_1)}[/tex]
[tex]\mathbf{\mathbf{Q_{1-2}} = 0+2.2(590.4-574.08)}[/tex]
[tex]\mathbf{\mathbf{Q_{1-2}} = 0+2.2(16.32)}[/tex]
[tex]\mathbf{\mathbf{Q_{1-2}} = 35.904 \ Btu}[/tex]
b) the work for Process 2–3, in Btu.
At [tex]T_3 = 100 ^0 \ F[/tex] ; [tex]u_3 = 577.86 \ Btu/lbm[/tex]
[tex]Q_{2-3} = W_{2-3} + \Delta U[/tex]
[tex]Q_{2-3} = W_{2-3} + m(u_3-u_2)[/tex]
[tex]Q_{2-3} = W_{2-3} +2.2(577.86-590.4)[/tex]
[tex]-98.9 = W_{2-3} +2.2(577.86-590.4)[/tex]
[tex]-W_{2-3}= 2.2(577.86-590.4)+98.9[/tex]
[tex]-W_{2-3}= -27.588+98.9[/tex]
[tex]-W_{2-3}= 71.312[/tex]
[tex]W_{2-3}= -71.312 \ Btu[/tex]
