Respuesta :
Solution :
For R-134a, we are given :
[tex]$T_i = 25^\circ C$[/tex]
[tex]$P_i=750 \ kPa$[/tex]
[tex]$P_e=165 \ kPa$[/tex]
Now we have one inlet and one exit flow, no work and no heat transfer. The energy equation is :
[tex]$h_e+\frac{1}{2}.v_e^2= h_i+\frac{1}{2}.v_i^2 $[/tex]
We also know that the gas is throttled and there is no change in the kinetic energy.
So, [tex]$v_e=v_i$[/tex]
Now from the energy equation above, we can see that the inlet and the exit enthalpies are also the same. Therefore,
[tex]$h_i=h_e$[/tex]
From the saturated R-134a table, corresponding to [tex]P_e = 165 \ kPa[/tex], we can find the exit saturation temperature.
[tex]$T_e=-15^\circ C$[/tex]
From the saturated R-134a table, corresponding to [tex]P_e = 165 \ kPa[/tex], we can find the specific enthalpies :
[tex]$h_f = 180.19 \ kJ/kg$[/tex]
[tex]$h_{fg} = 209 \ kJ/kg$[/tex]
Calculating the exit flow quality factor,
[tex]$x_e=\frac{h_e-h_f}{h_{fg}}$[/tex]
  [tex]$=\frac{234.59-180.19}{209}$[/tex]
  = 0.26
From the saturated R-134a table, corresponding to [tex]P_e = 165 \ kPa[/tex], we can find the specific volumes :
[tex]$v_f = 0.00746 \ m^3/kg$[/tex]
[tex]$v_{fg} = 0.11932 \ m^3/kg$[/tex]
Calculating the exit specific volume :
[tex]$v_e=v_f+x_e(v_{fg})$[/tex]
  = 0.000746 + 0.26 (0.11932)
  = 0.0318 [tex]m^3/kg[/tex]
The mass flow is equal to :
[tex]$\dot{m} = A_i . \frac{v}{v_i}$[/tex]
 [tex]$=A_e . \frac{v}{v_e}$[/tex]
So, [tex]$\frac{A_e}{A_i}=\frac{v_e}{v_i}$[/tex]
Therefore, the ratio of the exit pipe and the inlet pipe diameter is equal to
[tex]$\frac{D_e}{D_i}=\sqrt{\frac{A_e}{A_i}}$[/tex]
[tex]$\frac{D_e}{D_i}=\sqrt{\frac{v_e}{v_i}}$[/tex]
[tex]$\frac{D_e}{D_i}=\sqrt{\frac{0.0318}{0.000829}}$[/tex]
[tex]$\frac{D_e}{D_i}=6.19$[/tex]
  Â
