Thermodynamics piston-cylinder closed system

[Andrew Pierce]

Homework Statement

An insulated piston-cylinder device contains 5 L of saturated liquid water at a constant pressure of 175 kPa. Water is stirred by a paddle wheel while a current of 8 A flows for 45 min through a resistor placed in the water, If one-half of the liquid is evaporated during this constant pressure process and the paddle -wheel work amounts to 400 kJ, determine the voltage of the source. Also, show the process on a P-v diagram with respect to saturation lines.

Homework Equations

ΔE = E_in - E_out
W_electric = VIΔt
if ΔP = 0, ΔH = m(h₂ - h₁)
ΔH = ΔU + W_boundary

The Attempt at a Solution

V = 5 L
m = 5 kg (by conversion)
P = 175 kPa
I = 8 A
Δt = 45*60 sec = 2700

Assuming that KE, and PE = 0,
ΔE = ΔU

Also since it is insulated we are assuming that the heat lost is 0

W_e + W_sh - W_boundary = ΔU
W_e + W_sh = ΔU + W_boundary
W_e + W_sh = ΔH
W_e + W_sh = m(h₂ - h₁)

Here is where I get lost. I know that we can find h₁ from the saturated water tables
h₁ = 487.01 kj/kg (at P_sat = 175 kPa)
I'm confused as to where the second enthalpy value comes from since our state has constant P.

Also the P-v should look something like this I believe:

 

[Chestermiller]

Homework Statement

An insulated piston-cylinder device contains 5 L of saturated liquid water at a constant pressure of 175 kPa. Water is stirred by a paddle wheel while a current of 8 A flows for 45 min through a resistor placed in the water, If one-half of the liquid is evaporated during this constant pressure process and the paddle -wheel work amounts to 400 kJ, determine the voltage of the source. Also, show the process on a P-v diagram with respect to saturation lines.

Homework Equations

ΔE = E_in - E_out
W_electric = VIΔt
if ΔP = 0, ΔH = m(h₂ - h₁)
ΔH = ΔU + W_boundary

The Attempt at a Solution

V = 5 L
m = 5 kg (by conversion)
P = 175 kPa
I = 8 A
Δt = 45*60 sec = 2700

Assuming that KE, and PE = 0,
ΔE = ΔU

Also since it is insulated we are assuming that the heat lost is 0

W_e + W_sh - W_boundary = ΔU
W_e + W_sh = ΔU + W_boundary
W_e + W_sh = ΔH
W_e + W_sh = m(h₂ - h₁)

Here is where I get lost. I know that we can find h₁ from the saturated water tables
h₁ = 487.01 kj/kg (at P_sat = 175 kPa)
I'm confused as to where the second enthalpy value comes from since our state has constant P.

Also the P-v should look something like this I believe:

If you are using the saturation tables, you can get $h_2$ using the lever rule, or equivalently, as $h_2=h_1+\Delta h_{vap}/2$.

Regarding the P-V diagram, your result is correct.