Fan Laws

This sounds vaguely like a HW question which isn't endorsed here. If you're more candid about it, some members will help guide you to the answer; however, pretty much no one here will give you the answer.

Anyways!

Here's whatchu need to know:

It's a little blurry. If you can't read it, let me know and I will type it out!

Law 1. With impeller diameter (D) held constant:

Law 1a. Flow is proportional to shaft speed: ^[1], ^[2]

Law 1b. Pressure or Head is proportional to the square of shaft speed:

Law 1c. Power is proportional to the cube of shaft speed:

Law 2. With shaft speed (N) held constant: ^[1]

Law 2a. Flow is proportional to the cube of impeller diameter:

Law 2b. Pressure or Head is proportional to the square of impeller diameter:

Law 2c. Power is proportional to the fifth power of impeller diameter:

Kiran,

Your helpful Wikipedia extract propagates their error in the power/diameter relationship equation. They show ^3 when they mean ^5.

Alert to casual users!

Mark Bingham
Relativity PL

Dear Mark...oh my mistake.....thanx for correcting the mistake...

Friends,

Actually there are were two errors in the Wikipedia article. I checked it against the citations and believe the text and citations are correct while the formulae were incorrect. I submitted the correction. These corrections are in Law 2a and Law 2c.

--John M.

This is almost impossible to work out as is.

If you have a fan with the current data, you must get the manufacturer's information regarding the Charted curves to give you what you need by looking at the curves.

The curves will show you what will be the power required and the flow rate at the required static pressure of 0.50 inch WG. The rpm remains the same (practically).

Otherwise, you will need to select a different fan model that will have the values you require: Same flow at 0.50 inch, and give you the power. The fan design would have changed (different diameter of impeller).

Lucke's statement is valid.

Using a ratio of N_2/N_1 = (P_2/P_1)^2 to solve for N_2 provides an RPM that will provide greater q_dot than required. If N_2 = 2051 RPM, then q_dot will be 1978 CFM.

You wrote:

"If N_2 = 2051 RPM, then q_dot will be 1978 CFM"

True. But at the old 0.25 in. wg static pressure, not 0.50 in. wg as requested

Yes, you are right and that is why I said he should try and get the chart curves from the maker so that he could extrapolate.

The Fan laws are valid, each one, separately when fixing one of the items that is not in the equation:

Q1/Q2 = N1/N2 for the same H1 (go horizontally on the chart...)

H1/H2 = (N1/N2)^2 for the same Q1, going Vertically on the chart...

The power equation is more complicated. You can have an approximate value and must take some precaution. You will not know what is the exact efficiency of the pump at the N2 rpm, with the same (or not) Q. It will be an indication of the approximate magnitude power range you will be expecting.

Therefore, he should Try a higher BHP motor, by 20% from the resulting calculation.

never mind

Why will the system static pressure increase?

hi Nurisso: You will need the fzn performance curve... it should be availoable from the fan supplier and is a must..

Superimpose the system curve on the fan curve...you can construct the system curve using tables showing duct pressure drop for given duct sizes and length.

Good luck.

Lou Bindner