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It is assumed that the direct impact covers half of the water surface and horizontal<br />

surface flow over the other half, the forced convection Nusselt number for the total<br />

water surface may be expressed as:<br />

Nu 0.5( Nu Nu )<br />

(3.37)<br />

wsf dir Cp<br />

Where Nuwsf is total water surface forced convection Nusselt number.<br />

The mixed convection on such surface may be calculated as follow [50, 53]:<br />

Nu ( Nu Nu ) [ Nu (0.5Nu 0.5 Nu ) ] (3.38)<br />

3 3 1/3 3 3 1/3<br />

wsm wsn wsf wsn dir Cp<br />

So thermal resistance on the chamber water surface is:<br />

R<br />

ws<br />

1 1<br />

<br />

A k A k<br />

Nu [ Nu (0.5Nu 0.5 Nu ) ]<br />

ws ma ws ma 3 3 1/3<br />

ws wsm<br />

ws<br />

wsn dir Cp<br />

53<br />

(3.39)<br />

Where Aws is the water surface area. Thus the heat transfer from water to chamber air<br />

can be determined when the water temperature is known.<br />

3.3.4 Heat lost from the chamber water into chamber air by evaporationQev <br />

Evaporation rate on a horizontal surface may be calculated by [54]:<br />

m k A ( C C )<br />

(3.40)<br />

ev ev ws sv Caiv<br />

When the flow is from the ADU to the chamber, this inlet has the same specific<br />

humidity as ambient air but may have different level of absolute humidity due to<br />

temperature and pressure differences between the chamber and the ambient. When the<br />

flow is reversed, then the inlet is the air re-entering chamber from HADT and the<br />

absolute humidity is that of the re-entering air from the HADT. The absolute humidity<br />

can be calculated from the specific humidity as:<br />

C<br />

v<br />

Where P is pressure in kPa and T is air temperature in K.<br />

[28.966(1 d) 18.01 d] d P<br />

(3.41)<br />

8.31T

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