Electrical work in a piston-cylinder filled with refrigerant

[JJBladester]

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.

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

The Attempt at a Solution

Q_{in}=\Delta KE= \Delta PE=0

W_{e,in}=\Delta U+W_{b}=\Delta H

VI\Delta t=m(h_{2}-h_{1})

\Delta t=\frac{m(h_2-h_1)}{VI}

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

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

T₂ = T_sat @ 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?

 

[TaxOnFear]

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 h_f @ 243.3kPa = 193.32kJ/kg.

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

 

[JJBladester]

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 h_f @ 243.3kPa = 193.32kJ/kg.

Same with h_g, I found that h_g @ 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?