NPSHa Calc Revisited

[BB3242]

Hey,
I have been searching through many of the threads that have arised and have become a bit confused. I understand about the NPSHr on the curve. I am looking into the NPSHa of the curve goign for VGO. I have a tank that is 40 Ft tall. Open up to Atmospheric press. I have all of the Line losses calculated with respect to the Kinematic Viscosity of the fluid.

Am I correct in using this equation?

NPSHa= Static Head + Head Due to Atmospheric pressure at top of the tank - Head due to line losses - Head due to Vaporization Pressure

H due to Patm= (14.7psi*2.31)/(sg of VGO liquid)
H due to Line losses = K* (Vel of VGO liq ^2)/ (2 * 32.214) For minor and = f* (Vel of VGO liq ^2)/ (2 * 32.214) For major losses
H due to Vap Pressure= (2.31*.49Psi)/(sg of VGO liquid)
Static Head = Height of the fluid above centerline of the pump

This is correct right I feel as though I should have to multiply the static height by the SG of the VGO fluid. Also even though the NPSHr is read as a guage pressure I should still calculate the Atmospheric pressure right?
Thanks for any help. I know there are alot of threads out there but I just became confused.

 

[JJPellin]

You do not need to adjust the static height of the fluid level for product specific gravity. All of the other values are in units of feet of product. The height of the fluid in the tank is already in those same units.

NPSHr is in absolute units, not gauge units. That is why you have to add in the atmospheric pressure on top of the fluid.

Even the bottom of a vacuum tower under hard vacuum has a positive NPSHa and NPSHr. To put it another way, there is no such thing as a negative value for either of these parameters. They are absolute pressures.


Johnny Pellin