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

Answer:

The heat transfer $\mathbf{\mathbf{Q_{1-2}} = 35.904 \ Btu}$

$W_{2-3}= -71.312 \ Btu$

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  

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

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

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

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

$\mathbf{Q_{1-2}= W+ \Delta U}$

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

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

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

$\mathbf{\mathbf{Q_{1-2}} = 35.904 \ Btu}$

b) the work for Process 2–3, in Btu.

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

$Q_{2-3} = W_{2-3} + \Delta U$

$Q_{2-3} = W_{2-3} + m(u_3-u_2)$

$Q_{2-3} = W_{2-3} +2.2(577.86-590.4)$

$-98.9 = W_{2-3} +2.2(577.86-590.4)$

$-W_{2-3}= 2.2(577.86-590.4)+98.9$

$-W_{2-3}= -27.588+98.9$

$-W_{2-3}= 71.312$

$W_{2-3}= -71.312 \ Btu$