How Do Different Equations for K Affect Cooling Efficiency in Ducted Systems?

[gomerpyle]

I'm working on a research project dealing with cooling an electrical component in an air duct with a fan. I'm experimenting with different shrouds and directional vanes around the component to see which offers the best cooling. Obviously this has to be judged against the pressure drop caused by any constriction in the duct.

I've seen the equation K = ΔPt/0.5ρv^2 used to compute minor losses. How does this differ from all the equations that use ratios of the cross-sectional areas of sudden expansions/contractions to calculate K? sudden contraction would be K=0.5*(1-A1/A2)^0.75
If I have a sudden contraction in my air duct caused by some type of shroud, would it be wrong to use K = ΔPt/0.5ρv^2 where the v is the velocity through the constriction? How would this differ from using the ratio of cross sectional areas?

 

[Navin A S]

The equation K = ΔPt/0.5ρv^2 is used to calculate minor losses from a sudden contraction or expansion in the airflow, and is typically used in cases where there is no change in the cross-sectional area. This equation is applicable in most cases that involve a sudden change in the flow area of an air duct, such as when a fan is installed in the duct. The equation takes into account the change in pressure due to the constriction, as well as the velocity of the air through the constriction. On the other hand, the equation K=0.5*(1-A1/A2)^0.75 uses the ratio of the cross sectional areas to calculate K, and is usually used in cases where there is a change in the cross-sectional area of the air duct. This equation is more applicable when you are dealing with a more complex design such as a shroud or directional vane, as it takes into account the pressure drop caused by the constriction as well as the ratio of the cross sectional areas. In conclusion, both equations can be used in the case of cooling an electrical component in an air duct with a fan, but the equation K=0.5*(1-A1/A2)^0.75 should be used if you are using a more complex design such as a shroud or directional vane.