Dick Bennett continues his series on how to figure the effect of changes in air and gas flow.



In my last column, I showed how, by boiling it down to its essential variables, you can use the basic equation for noncompressible flow to figure the effect of changes in pressure drops, opening areas and gas gravities on the flow of air and gas in piping systems. I showed you how to derive the Square Root Law of flow and pressure drop and use it to figure flows from pressure drops, and vice versa.

Just to get us all on the same page, I'll review the equation and its terms.

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 (delta 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 (70^oF dry air at sea level barometric pressure).

Now, let's move on to some other, more complex, situations. Suppose you have a 1 million BTU/hr burner that will be operated on propane (2,500 BTU/ft³, 1.5 specific gravity). According to the setup figures in the manufacturer's data sheet, the burner requires a 6" w.c. pressure drop at high fire when burning natural gas (1,000 BTU/ft³, 0.62 specific gravity). What pressure drop should you use for propane?

This involves two variables -- flow and specific gravity. Because of its higher heating value, you need a flow of only 1,000,000 BTU/hr ÷ 2,500 BTU/ft³, or 400 ft³/hr of propane. Natural gas would have required a flow of 1,000 ft³/hr. This time, you're solving for pressure drop, so you have to shuffle the equation around, like this:

Next, you set up the before-and-after ratio. This time, I'll use the subscripts p and n (for propane and natural) instead of 1 and 2.

A, the area of the gas opening in the burner, doesn't change, and Cd, its flow coefficient, changes so little you can assume it stays the same. So these two factors cancel out, leaving us with this simplified equation:

Plugging in the numbers, you get:

which figures out to 2.3" w.c.

One other scenario is worth mentioning -- the effect of temperature on flow or pressure drop through a piping system. This has to be taken into account when combustion air is preheated or when oven recirculating or exhaust streams change in temperature.

Say you're sizing an exhaust fan for some hot air ductwork. With 10,000 scfh ambient air flowing through it, its pressure drop is 1.2" w.c. What pressure drop will you have to contend with when the air is 400^oF (204^oC)?

The density of air is inversely proportional to its absolute temperature; that is, ^oF + 460 (^oC + 273). At ambient conditions, air's gravity is 1. At 400^oF, it will be:

I'll save you the intermediate equation-bashing, but you'll find that:

One word of caution -- this is the pressure drop for 10,000 actual cubic feet per hour (acfh) of air. That's reduced density air. If you want to maintain the same standard cubic feet of air (same weight flow), you have to increase the flow rate to 10,000 ÷ 0.62 = 16,129 acfh. When you factor in this increase in volume flow, you'll find the pressure drop will actually increase to 1.94" w.c. at 400^oF. PH

Links

• Janus Technology Group
• Energy Notes: How Did They Figure That?
• Energy Notes: Archives