Respuesta :
Answer:
The answers the question are
Final pressure p₂ is 18.166 kPa
The work done, W is -0.192 kJ/kg
The heat transfer, Q is 48.052 kJ/kg
Explanation:
A force balance reveals that the gas is at equilibrium with the spring piston. That is when the temperature increases results in the increase in volume and the force exerted on the piston is constant. Therefore the system is a constant pressure system
T₁ = 15 °C, V₁ = 0.02 m³,
u₁ = [tex]u_{f1} + x*u_{fg1}[/tex], From R-134a tables
u₁ = 71.93 + 0.4×238.39 = 238.39 kJ/kg
u₁ = 238.39 kJ/kg
The mass of R-134a in the piston cylinder arrangement
m = [tex]\frac{V_1}{v_1}[/tex]
Where
V₁ = Volume of R-134a and
v₁ = Specific volume at state 1
v₁ is given by v₁ = [tex]v_{f1} + x*v_{fg1}[/tex] = 0.804/(10³) + (0.04211 - 0.8041/10³)
= 0.01732646 m³ = 1.733 × 10⁻² m³
Therefore m = [tex]\frac{0.02}{0.01732646}[/tex] = 1.1543 kg
h₁ = [tex]h_{f1} + x*h_{fg1}[/tex] =72.32 + 0.4×(258.97 - 72.32) = 146.98 kJ/kg
Similarly s₁ =[tex]s_{f1} + x*s_{fg1}[/tex] = 0.2768 + 0.4×(0.9245-0.2768)
= 0.53588 kJ/kg-K
u₂ = 137.79 +0.4×(259.25 - 137.79) = 189.374 kJ/kg
h₂ = 139.38 + 0.4×(278.51 - 139.38) = 195.032 kJ/kg
v₂ = (0.948)/10³ + 0.4 × (0.01144 - (0.948)/10³) = 0.0051448 m³
v₂ = 0.030 m³ then V₂ = m×v₂ = 3.465×10³ m³
For a constant pressure process
Q + W = u₂ - u₁ = 186.374 - 138.514 = 47.86 kJ/kg
Q = h₂ - h₁ = 195.032 - 146.98 = 48.052 kJ/kg
W = Δu - Q = 47.86 -48.052 = -0.192 kJ/kg
From W = p·dv we have W = p×(5.145×10⁻³-1.733×10⁻²) =-1.22×10⁻²×p
That is -0.192 kJ/kg×1.1543 kg = -1.22×10⁻²×p or p = 18.166 kPa
