# Electrical work in a piston-cylinder filled with refrigerant

1. Feb 12, 2012

1. The problem statement, all variables and given/known data

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.

2. Relevant 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.

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

3. 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=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?

2. Feb 12, 2012

### 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 hf @ 243.3kPa = 193.32kJ/kg.

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

3. Feb 13, 2012