Electrical work in a piston-cylinder filled with refrigerant

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 7K views
JJBladester
Gold Member
Messages
281
Reaction score
2

Homework Statement



0.8kg of saturated liquid R-134a with an initial temperature of -5 °C is contained in a well-insulated, weighted piston-cylinder device. This device contains an electrical resistor to which 10 volts are applied causing a current of 2 amperes to flow through the resistor. Determine the time required for the refrigerant to be converted to a saturated vapor, and the final temperature.

Homework Equations



Assumption: The system is at constant pressure throughout the process since both the atmospheric pressure and the weight of the piston remain constant throughout.

[tex]Q_{in}+W_{e,in}-W_{b}=\Delta U+\Delta KE+ \Delta PE[/tex]

The Attempt at a Solution



[tex]Q_{in}=\Delta KE= \Delta PE=0[/tex]

[tex]W_{e,in}=\Delta U+W_{b}=\Delta H[/tex]

[tex]VI\Delta t=m(h_{2}-h_{1})[/tex]

[tex]\Delta t=\frac{m(h_2-h_1)}{VI}[/tex]

State 1
P1=Psat @ -5 °C = 243.5kPa
Saturated Liquid
Thus, h1=hf @ 243.5kPa = 45.143 kJ/kg

State 2
P2=P1=243.5kPa
Saturated Vapor
Thus, h2=hg @ 243.5kPa = 247.49 kJ/kg

T2 = Tsat @ 243.5 kPA = -5 °C

Rearranging the energy balance equation for t yields 135 minutes for the time it takes to change the saturated liquid refrigerant into saturated vapor state.

I am having doubts about my calculation for T2. It seems like the temperature would increase as that is the nature of a resistance heater. Any thoughts?
 
Physics news on Phys.org
Hi,
I don't think your steam tables consider the circumstances that the problem gives you, the temperature would probably increase slightly (I'm just a student, not nearly as good as other people on this forum).

However, looking through my steam tables, I found that hf @ 243.3kPa = 193.32kJ/kg.

Same with hg, I found that hg @ 243.3kPa = 395.49kJ/kg.
 
TaxOnFear said:
Hi,
I don't think your steam tables consider the circumstances that the problem gives you, the temperature would probably increase slightly (I'm just a student, not nearly as good as other people on this forum).

However, looking through my steam tables, I found that hf @ 243.3kPa = 193.32kJ/kg.

Same with hg, I found that hg @ 243.3kPa = 395.49kJ/kg.

Hey there TaxOnFear. The problem deals with refrigerant R-134a, not water, so the h values I posted should be correct.

Perhaps the temperature does stay the same from state 1 to state 2, but the energy that the resistor circuit provides to the system goes into the increasing volume?