How Did They Figure That?
Sizing burners and heating systems and figuring what they'll do at different flow rates require a number of different calculations. To the uninitiated, it can appear pretty complex and intimidating -- after all, you have to know the effect of flow rates on pressure differentials and vice versa, and sometimes the effect of things like temperatures, fluid densities and orifice sizes on flows and differentials. It must take a whole textbook full of equations to figure all this stuff.
Nope -- one equation will do. You just have to know how to tweak it for different situations. Plenty of times, I've had to do a flow or pressure drop calculation on the fly with nothing more than a pencil, paper and pocket calculator. As long as I can remember that one handy-dandy equation, I'm in business. Here it is:
Q represents the flow of air or a gas in cubic feet per hour.
Cd is the Coefficient of Discharge or Flow Coefficient. It's a dimensionless number between 0 and 1 that describes how efficiently an orifice or other flow restriction allows a fluid to pass through it.
A is the cross-sectional area of the opening in square inches.
∆P is the pressure differential across the orifice or restriction in inches water column.
g is the gravity of the gas, another dimensionless number figured by dividing the gas density by the density of "standard" air (70oF dry air at sea-level barometric pressure).
This is the basic relationship used to determine the flow of non-compressible gases across a resistance like a metering orifice or through a burner or valve. There are other equations that work equally well, but this serves our purpose nicely. By the way, non-compressible flow describes situations where gas velocities are relatively low, such as you'll normally find in air and gas piping systems and ductwork.
This equation also works with different units of measurement, but you'll have to use a coefficient different than 1,660.
Putting It to UseMost of the time, you'll be confronted with before-and-after situations; for example, if the gas flow to a burner is such-and-so at a certain pressure drop across a metering orifice, what will it be at a different reading? Or, you could be dealing with the converse of this: knowing the pressure drop required for a certain flow, what pressure drop reading is required to get a specified different flow?
Start this way -- call the "before," or known condition, Case 1, and the "after," or unknown condition, Case 2. Then calculate a ratio by dividing the flow equation by itself, like this:
Drawing on that long-ago high school algebra, you simplify this fraction by crossing out all of the terms that are the same from numerator to denominator. 1660, Cd, A and g are unchanged, or change so little, that you can drop them out of the fraction. That leaves you with a greatly simplified relationship: