Fill an empty bottle with R-134a

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stoky
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Homework Statement


We want to fill a bottle with R-134a. The fluid comes from a fluid line.
  1. The bottle is initially empty, the bottle volume is 12 litres, the bottle is insulated (adiabatic filling).
  2. The fluid line contains R-134a at 20 bar, saturated steam.

Homework Equations


Slide titled Conservation of energy for control volumes, with work and energy balance equations and a diagram of a control volume with inlet i and exit e, heat and work transfers, and the dashed control volume boundary.

Using the second equation of energy balance.

The Attempt at a Solution


From the bottle perspective:
Qcv=0
Wcv=0
dEcv/dt=m•ui
me=0
hi=428.2 kJ/Kg

Then:
mi•ui=mi•hi
Then
ui=428.2 kJ/Kg
Which reading R134a tables results in superheated steam: P2=2000 kPa, u2=428.2 kJ/Kg.

Results in m2 (bottle)=1.11 Kg
Specific volume=0.01081 m^3/Kg
Temperature=83.18 celsius

My concern is if the calculations are correct.
 
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  • Slide titled Conservation of energy for control volumes, with work and energy balance equations and a diagram of a control volume with inlet i and exit e, heat and work transfers, and the dashed control volume boundary.
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stoky said:

Homework Statement


We want to fill a bottle with R-134a. The fluid comes from a fluid line.
  1. The bottle is initially empty, the bottle volume is 12 litres, the bottle is insulated (adiabatic filling).
  2. The fluid line contains R-134a at 20 bar, saturated steam.

Homework Equations


View attachment 235368
Using the second equation of energy balance.

The Attempt at a Solution


From the bottle perspective:
Qcv=0
Wcv=0
dEcv/dt=m•ui
me=0
hi=428.2 kJ/Kg

Then:
mi•ui=mi•hi
Then
ui=428.2 kJ/Kg
Which reading R134a tables results in superheated steam: P2=2000 kPa, u2=428.2 kJ/Kg.

Results in m2 (bottle)=1.11 Kg
Specific volume=0.01081 m^3/Kg
Temperature=83.18 celsius

My concern is if the calculations are correct.
I'm not sure whether you did it right or not because your notation is all screwy. The differential equation should be
$$\frac{dE_{dv}}{dt}=\dot{m}_i h_i$$
so that $$u_{final}=h_i$$
If that's what you did, then it is correct.