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JJBladester
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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?