Differential head calculation

[srisua]

Dear Seniors,

I am using HFO as media. While calculating the pump head, i calculated pressure drop in MWC and the same i converted in MLC by using the following formulae

MLC = MWC x Specific gravity of fluid.

I have used the above formulae for all my calculations. But one of my consultant commented as MLC = MWC/ specific gravity of fluid.

Now you are reqeusted to please suggest me which one is correct.

Thanks in advance

Srinivas

 

[quark]

Rho_L x g x H _L = Rho_W x g x H _W

Hope, you can do rest of the calculation.

PS: Check all your calculations.

 

[LittleInch]

If your SG is <1, then to achieve the same pressure / equivalent water column at the foot of your column, the height needs to be higher.

however calculating pressure drop using water is also not the best as the viscosity of the fluid has a distinct impact on pressure drop.

It seems you're trying to do this very simply and not understanding the basic physics behind this.

Check lots of times before you buy the pump!



My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[srisua]

Dear Mr. Quark & Little Inch,

Thanks for your explanation. The following is understood from your formulae

As G is gets cancel (HL = HW x Rho water/ Rho liquid). This means always Meter Liquid column height is more than Meter water column if specific gravity of fluid less than 1.

The above complies the statement of Mr. Little inch.

Thanks a lot

Regards
Srinivas

 

[electricpete]

This means always Meter Liquid column height is more than Meter water column if specific gravity of fluid less than 1.

Not sure what you're saying. If you have two columns which both have a height of 1 meter, then they have the same height / head (one meter). The one with lower sg has lower corresponding [differential] pressure [from bottom to top]

=====================================
(2B)+(2B)' ?