Moving boundary with electrical input

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SUMMARY

The discussion revolves around calculating the voltage of a power source in an insulated piston-cylinder device containing 5L of saturated liquid water at a constant pressure of 175kPa. An electrical current of 8A flows for 45 minutes through a resistor, while a paddle wheel does 400kJ of work and half of the liquid evaporates. The key equations used include Ohm's Law (V=IR), energy balance equations, and the work done by the boundary at constant pressure. The user successfully calculates internal energy values but struggles with converting units to find the voltage.

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  • Understanding of thermodynamic principles, specifically energy balance in closed systems
  • Familiarity with electrical concepts, particularly Ohm's Law
  • Knowledge of saturated liquid properties and property tables for water
  • Ability to perform unit conversions and calculations involving specific volume
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  • Research how to calculate mass from volume using density at specific temperatures
  • Learn about the properties of saturated water at various pressures, specifically at 175kPa
  • Study the relationship between electrical work and thermodynamic processes
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Homework Statement


Insulated piston-cylinder device contains 5L of saturated liquid water at constant pressure of 175kPa. Water is stirred by a paddle wheel. A current of 8A flows for 45 minutes through a resistor in the water.

If 1/2 of the liquid is evaporated during the constant pressure process and the paddle wheel work is 400kJ determine the voltage of the source.


Homework Equations


V=IR
v_avg = v_f + X*v_fg
Wboundary = integral(P*dV)

The Attempt at a Solution



There is no heat loss so no Q_out
I took 1/2 of the liquid evaporating to mean that quality, X is 1/2.

The energy balance would be Q - W = dU + dKE + dPE. KE and PE are both zero.

Electrical work is V*I*t

This is constant pressure so Wboundary = P*(V_1 - V_2)

So, with all this summarized I -think- I get.

V*I*t + P(V_1 - V_2) + 400kJ = U_2 - U_1

To get U_2 and U_1 I looked in my property table for saturated water at 175kPa

U_1 is 487 kJ/kg
I used U_avg = U_f + X*U_fg to find U_2
U_2 = 487 + 0.5 * 2038
U_2 = 1505 kJ/kg

So I am stuck with some unit of kJ/kg. How do I get rid of these kg^-1? I'm guessing I do something with v (specific volume) based on how many liters of water I have.

I also need the Volumes.

Could someone help? Thank you
 
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They tell you the initial volume of water. But it seems that you would have to know the initial temperature (not given) and pressure (given) in order to figure out how many kg of water there are. Perhaps you are to assume room temperature at the beginning?
 
Ah, wait. At what temperature is liquid water saturated, when P=175 kPa? That will be the initial temperature. From that you can figure out the density of water, and hence the mass.
 

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