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3 years ago

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A closed, rigid, 0.45 m^3 tank is filled with 12 kg of water. The initial pressure is p1 = 20 bar. The water is cooled until the

pressure is p2 = 4 bar. Determine the initial quality, x1, and the heat transfer, in kJ.

1 answer:

liberstina [14]3 years ago

7 0

Answer:

initial quality = 0.3690

heat transfer = 979.63 kJ/kg

Explanation:

Given data:

volume of tank 0.45^3

weight of water 12 kg

Initial pressure 20 bar

final pressure 4 bar

Specific volume $v = \frac {0.45}{12} = 0.0375 m^3/kg$

At Pressure = 20 bar, from saturated water table

$v_f = 0.01177 m^/kg$

$v_g = 0.099587 m^3/kg$

$x = \frac{v -v_f}{v_g -v_f} = \frac{0.0375 - 0.001177}{0.099587 - 0.001177}$

inital quality is x =0.3690

Heat transfer is calculated as

$u_1 = h_f + x(h_g - h_f) = v_f + x( h_{fg})$

from saturated water table, for pressure 20 bar ,

$h_f = 908.79 kJ/kg, h_{fg} = 1890.7 kJ/kg$

=908.79 + 0.0357(1890.7)

= 979.63 kJ/kg

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a) Potential in A: -2700 V

b) Potential difference: -26,800 V

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Explanation:

a)

The electric potential at a distance r from a single-point charge is given by:

$V(r)=\frac{kq}{r}$

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$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

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r is the distance from the charge

In this problem, we have a system of two charges, so the total potential at a certain point will be given by the algebraic sum of the two potentials.

Charge 1 is

$q_1=+2.00\mu C=+2.00\cdot 10^{-6}C$

and is located at the origin (x=0, y=0)

Charge 2 is

$q_2=-3.00 \mu C=-3.00\cdot 10^{-6}C$

and is located at (x=0, y = 0.40 m)

Point A is located at (x = 0.40 m, y = 0)

The distance of point A from charge 1 is

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So the potential due to charge 2 is

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The distance of point A from charge 2 is

$r_{2A}=\sqrt{0.40^2+0.40^2}=0.566 m$

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Therefore, the net potential at point A is

$V_A=V_1+V_2=+4.50\cdot 10^4 - 4.77\cdot 10^4=-2700 V$

b)

Here we have to calculate the net potential at point B, located at

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The distance of charge 1 from point B is

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So the potential due to charge 2 at point B is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.412}=-6.55\cdot 10^4 V$

Therefore, the net potential at point B is

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So the potential difference is

$V_B-V_A=-29,500 V-(-2700 V)=-26,800 V$

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