Question

Ammonia in a piston–cylinder assembly undergoes two processes in series. At the initial state, p1 = 120 lbf/in.2 and the quality is 100%. Process 1–2 occurs at constant volume until the temperature is 100°F. The second process, from state 2 to state 3, occurs at constant temperature, with Q23 = –98.9 Btu, until the quality is again 100%. Kinetic and potential energy effects are negligible. For 2.2 lb of ammonia, determinea) the heat transfer for Process 1–2, in Btu. b) the work for Process 2–3, in Btu.

Answer 1

The heat transfer
`\mathbf{\mathbf{Q_(1-2)} = 35.904 \ Btu}`

`W_(2-3)= -71.312 \ Btu`

Step-by-step explanation:

At the initial state when P₁ = 10 lbf/in :

We obtain the internal energy u₁ and specific volume v₁.

u₁ =
`u_g` = 574.08 btu/lbm

v₁ =
`v_g` = 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

`\mathbf{u_2 = 591.28 + (2.4746-2.5917)/(2.117-2.5917)*(587.68 -591.28)}`

`\mathbf{u_2 = 591.28 + 0.246682115(-3.6)}`

`\mathbf{u_2 = 591.28+ (-0.888055614)}`

`\mathbf{u_2 \approx 590.4 \ btu/lbm}`

`\mathbf{Q_(1-2)= W+ \Delta U}`

`\mathbf{Q_(1-2)= W+m( u_2 -u_1)}`

`\mathbf{\mathbf{Q_(1-2)} = 0+2.2(590.4-574.08)}`

`\mathbf{\mathbf{Q_(1-2)} = 0+2.2(16.32)}`

`\mathbf{\mathbf{Q_(1-2)} = 35.904 \ Btu}`

b) the work for Process 2–3, in Btu.

At
`T_3 = 100 ^0 \ F` ;
`u_3 = 577.86 \ Btu/lbm`

`Q_(2-3) = W_(2-3) + \Delta U`

`Q_(2-3) = W_(2-3) + m(u_3-u_2)`

`Q_(2-3) = W_(2-3) +2.2(577.86-590.4)`

`-98.9 = W_(2-3) +2.2(577.86-590.4)`

`-W_(2-3)= 2.2(577.86-590.4)+98.9`

`-W_(2-3)= -27.588+98.9`

`-W_(2-3)= 71.312`

`W_(2-3)= -71.312 \ Btu`