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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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$T_s = 285 degree celcius$

T_∞ = 35 degree celcius

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$\dot m C \Delta T = h A_s \Delta T$

solving for h

$h = \frac{14764.85*0.115*(285-35)}{2.513*(285-35)}$

h = 675.6 kw/m^2K

$Q = h A_s\Delta T$

Q = 675.6*2.513*(285-35)

Q = 424523.22 kw

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