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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.

Respuesta :

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 [tex]m^{3}/ Kg[/tex]

h1 = h2 = 487.01 [tex]\frac{KJ}{Kg}[/tex]

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 [tex]\frac{KJ}{Kg}[/tex]

The mass of the water is

m = V/v1

m = 0.004/0.001057 = 3.784 Kg

The voltage is

V = [tex]\frac{m(h2 - h1) - Wpw,in}{I (delta)t}\\[/tex]

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

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