In this business, we deal with pressures, pressure drops, flows and velocities all the time. For all that familiarity, you'd think we could visualize what one of these measurements means in terms of the others. Unfortunately, most of us can't. Mentally, we have a hard time relating the velocity of a moving airstream to its velocity pressure, and we usually underestimate, by a huge margin, the force exerted on a surface by a static pressure.
Maybe we just don't use the right terms to fix the ideas in our heads. Let's try out some non-standard ways of looking at the subject.
Remember when you were a kid, riding in the front seat of the car? It was a nice summer day, and you'd stick your hand out the window. With your palm facing the road, your hand would slice through the air. If you angled it up or down, the moving air would try to make your hand climb or dive, just like an airplane wing. If you turned your palm into the wind, its force would push your arm backward. What was going on there?
You were feeling what happens when velocity is turned into pressure, and pressure generates force. The velocity of a moving airstream is directly convertible into pressure. All you have to do is put something in its way. Once the pressure meets that object, it exerts a force on it. Working backward, the force is proportional to the velocity of the moving air.
Suppose Mom or Pop was driving along at 45 mph (72 km/h). Assuming no winds outside, the air is moving past the car at that speed. You put your hand out the window. The moving air exerts a velocity pressure of about 1" w.c. (2.5 mbar) on your hand -- that's equivalent to 5.2 lb/ft 2(25.4 kg/m2). For an average kid's hand, the force trying to push his or her hand backward works out to about 0.5 lb (0.26 kg). It feels the same as if someone placed an object weighing that much in your hand.
Now, let's take this insight over to a process oven. If it's like many ovens, our typical example has an array of slots and louvers where the heated air enters the product chamber. Pressure in the air plenum forces the air through the slots, accelerating it toward the load, which it heats on contact. Suppose you read a static pressure of 0.2" w.c. (74 mbar) in the plenum. The pressure drop from the plenum to the product chamber, which is usually at around atmospheric pressure, will generate a velocity of 30 ft/sec (9 m/sec). In hand-out-the-window terms, that's about 20 mph (33 km/h) -- a fairly stiff breeze.
Obviously, ovens operate at elevated air temperatures. What happens then? You can calculate it with this equation:
V is the air velocity in feet per second.
ΔP is the pressure drop (or velocity pressure), in inches water column.
T is the air temperature inoF.
Taking our example oven one step further, let's say the plenum-to-chamber pressure drop is 0.2" w.c. (0.5 mbar) at 400oF (204oC) air temperature. Run the numbers, and you get 38 ft/sec (11.6 m/s) -- or in hand-out-of-the-window terms, 26 mph (42 km/h).
This makes sense when you think about it -- heating the air causes it to expand, so for a given pressure drop across a nozzle or louver, it has to move faster. The flip side of this comes when the hot air makes contact with the product or any other fixed surface. For the same arrival velocity, hot air will exert less pressure on that surface. Why? Because it's less dense than ambient air.
Let's return to the oven plenum. The air is at 400oF and 0.2" w.c. static pressure inside the plenum. Say the plenum wall is 40' long by 10' high (12.2 x 3.04 m). How much total force is the air exerting on it? A pressure of 0.2" w.c. equals 1.04 lb/ft2(5.08 kg/m2), so the total force on the wall is 416 lb (188 kg). Hard to believe that such a measly little pressure differential can do that, but now you may understand why oven walls often bang and thump when the fan is turned on and may continue to creak as the system heats up.
Oven designers need to maintain a certain hot-air velocity at the product in order to get the desired heat transfer rate. Because the moving air is spreading throughout the oven and mixing with air already there, it loses speed on its way to the product. Consequently, the velocity at the nozzles or louvers has to be higher than at the product surface, and it's that required velocity that dictates the circulating fan pressure.
By the way, if you have the air velocity and want to figure its velocity pressure or the nozzle drop required to generate that velocity, the equation is:
One word of caution when using this to figure nozzle or louver pressure drops: The actual air supply pressure will have to be higher than this because of friction and turbulence losses at the nozzles. PH