Air Flow

INTRODUCTIONSince the job of the TAB Technician is to regulate flow rates, you must understand the basics of fluid flow. Air and water are both fluids (although air is a gas and water is a liquid). This unit will explain the basic terms of flow for both air and water and also explain how they are used by the TAB Techni cian.AIRFLOWThe TAB Technician must know how to measure airflow and how to change its rate of flow.Velocity is the speed of the airflow and is ex pressed in feet per minute (FPM).Volume is the quantity of airflow and is ex pressed as cubic feet per minute (CFM). CFM is often written as Q (quantity).

• Metric Equivalents for this Unit• English• FPM (feet per minute)• CFM (cubic feet per minute)• "WG (inches water gage)• ft. H2O (feet of water)• GPM (gallons per minute)• Metric• m/s (meters per second)• 1/s (liters per second)• m3/s (cubic meters per second)• kPa (kilopascals)• kPa (kilopascals)• 1/s (liters per second)• m3/s (cubic meters per second)• Standard air: 70°F at sea level 20°C at sea level 0.075 lbs/ft3Density of standard air:1.204 kg/m3

• Fig. 1: One cubic foot• A cubic foot is the size of a box that measures one foot on all sides (Fig. 1).• The basic equation for airflow is:• Q = AV• Where:• Q = Cubic feet per minute (CFM)• A = Cross-sectional area of duct in square feet (sq. ft.)• V = Velocity of air in feet per minute (FPM)• Example• The air velocity in a duct that measures 12" x 24" is 1000 FPM. What is the volume of

air delivered?• Duct sizes are given in inches. Since one square foot equals 144 square inches, the

cross-sectional area can be converted to square feet by dividing by 144.

• A = 12” x 24” • 144 sq. in. per sq. ft.• A = 2 sq. ft. cross-sectional area• Now the air volume can be determined:• Q = AV• Q = 2 sq. ft. x 1000 FPM• Q = 2000 CFM• Note that for CFM, usually the nearest whole number or the nearest

5 is the degree of accuracy required.• Example• A duct measures 16" x 20" and has an air velocity of 1600 FPM.

What is the CFM?• Q = AV• Q= 16" x 20" • 144 sq. in. x 1600 FPM• Q = 3556 CFM

• Example• A duct that measures 14" x 30" must deliver 3500 CFM. What must

the air velocity be?• In this case, velocity is the unknown. Therefore rearrange the

equation Q = AV to solve for V:• V= Q• A• A= 14" x 30"• 144 sq. in. per sq. ft.• A= 2.917 sq. ft.• V = 3500 CFM• 2.917 sq. ft.• V = 1200 FPM

• Air Pressure• Air flows in a duct if there is a difference in pressure. It flows from a

higher pressure to a lower pressure. A higher pressure is created at the fan outlet, so the air flows to the lower pressure at the outlet of the duct (Fig. 2). The greater the difference in pressure, the faster the air moves.

• Positive pressure (Fig. 2) is a pressure that is greater than atmospheric pressure. It is created at the fan outlet where air is pushed out of the fan.

• Negative pressure (Fig. 2) is a pressure that is less than atmospheric pressure. It is created at the fan inlet where air is pulled into the fan.

• Standard atmospheric pressure is 14.7 psi (pounds per square inch). The additional pressure that the fan adds to the duct system is a very small amount compared to atmospheric pressure. There fore duct pressures are measured in inches of water column, which indicate relatively small pressures.

• Inches of water means the height that a given pressure will raise a column of water in a tube. In Fig. 3, if a hose is connected to a glass tube at point A, and the other end of the hose is inserted in the duct, the total air pressure in the duct will raise the water in the U-shaped glass tube. At point C, the water has raised 0.06" above the original water level. The difference in height between points B and C is 0.12". Therefore the total air pressure in the duct is 0.12 inches of water ("WG). Inches of water is also called water column (WC), water (H20), and water gage (WG).

• The total pressure in a duct seldom exceeds 0.25 psi (pounds per square inch), and 0.25 psi equals about 6.9" WG

• Types of Pressure• The TAB Technician determines three different

pressures in HVAC systems:• Static pressure• Velocity pressure• 3. Total pressure• Static pressure (SP) is the pressure exerted in

all directions. It is static pressure that keeps a balloon inflated (Fig. 4) or a tire filled. Static pressure in a duct exerts pressure against the sides of the duct (Fig. 5).

• Velocity pressure (VP) is the pressure that results from the air being in motion. When a balloon is opened, velocity pressure results from air rushing out (Fig. 4). Velocity pressure of wind is also the force that keeps a kite in the air. Velocity pressure in a duct is the result of the air moving in the duct (Fig. 5). The greater the air velocity in the duct, the greater the velocity pressure.

• Total Pressure (TP) in the ductwork is the sum of velocity pressure and static pressure (TP = VP + SP).

• Since total pressure equals static pressure plus velocity pressure, there is a direct relationship be tween SP and VP. If one decreases, the other must increase as long as the TP remains the same.

• To illustrate this, assume that a duct system has a constant flow of air with a constant total pressure. (Actually, because of friction and losses due to elbows and other fittings, the total pressure will decrease as the air travels down the duct. However, it is easier to assume constant total pressure in order to understand the relationship between SP and VP.) With a constant quantity of air flowing in the duct, the velocity of the air will change with every cross- sectional duct area change. If the duct becomes smaller, both the air velocity and the VP will increase.

• For example, the duct in Fig. 6 at point A is 24" x 12" with 2000 CFM flowing through it. The velocity at this point is 1000 FPM. For these con ditions the VP for this duct is 0.06" WG.

• Now assume that the constant total pressure is 2.00" WG. Use the equation TP = SP + VP and solve for SP:

• SP = TP — VP

• SP = 2.00" WG — 0.06" WG

• SP = 1.94" WG

• At point B (Fig. 6), the duct size is reduced to 20" x 10". Since the quantity is constant at 2000 CFM, when the area decreases, the velocity in creases to 1440 FPM. The corresponding VP in creases to 0.13" WG.

• To find the SP, use the equation:• SP = TP — VP• SP = 2.00" WG — 0.13" WG • SP = 1.87• There are several important principles to

remem ber about static pressure, velocity pressure, and total pressure:

• SP = 1.94" WG SP = 1.87" WG

• VP = 0.06" WG VP = 0.13" WG

• TP = 2.00" WG TP = 2.00" WG

• Fig. 6: Changes in VP and SP

• TP = VP + SP

• For the same CFM, if duct size decreases, the air velocity increases.

• If air velocity increases, velocity pressure also increases.

• If VP changes, SP changes in the opposite direction.

• VP cannot be measured directly. It can be determined by subtracting SP from TP or by using an equation with a constant:

• VP = V

• 4005

• 6. Use a version of the same equation to find

• V if VP is known:

• V = 4005 √VP