Question

An insulated piston-cylinder device contains 4 L of saturated liquid water at a constant pressure of 175 kPa. Water is stirred by a paddle wheel while a current of 7 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 300 kJ, determine the voltage of the source.

Answer 1

Answer:

The voltage of the source is 207.5 V

Explanation:

Given:

Volume of the water V = 4 L

Pressure P = 175 KPa

Dryness fraction x2 = 0.5

The current I = 7 Amp

Time T = 45 min

The paddle-wheel work Wpw = 300 KJ

Obs: Assuming the kinetic and potential energy changes, thermal energy stored in the cylinder and cylinder is well insulated thus heat transfer are negligible

1 KJ/s = 1000 VA

Energy Balance

Ein - Eout = ΔEsys

We,in + Wpw, in - Wout = ΔU

IVΔT + Wpw, in = ΔH = m(h2 - h1)

Using the steam table (A-5) at P = 175 KPa, x1 = 0

v1 = vf = 0.001057 $m^{3}/ Kg$

h1 = h2 = 487.01 $\frac{KJ}{Kg}$

Using the steam table (A-5) at P = 175 KPa, x1 = 0.5

h2 = hf + x2 (hg - hf)  = 487.1  + 0.5 * (2700.2 - 487.1) = 1593.65 $\frac{KJ}{Kg}$

The mass of the water is

m = V/v1

m = 0.004/0.001057 = 3.784 Kg

The voltage is

V = $\frac{m(h2 - h1) - Wpw,in}{I (delta)t}$

V = $\frac{3.784 * (1593.65 - 478.1) - 300}{7 * (45 * 60)}$ = 0.207 * 1000 = 207.5 V