# Estimating Exhaust Requirements

*EDITOR'S NOTE: Tables 1 and 2 were too large to reproduce clearly in this article. They can be downloaded by clicking on the link at the end of the article.*

Ovens and dryers use various methods to remove moisture and dry products, including hot circulating air, infrared, microwave and other techniques. The circulating or surrounding air absorbs the moisture as the products dry. In many cases, the air also contains products of combustion. Exhausting this hot, moisture-laden air from the oven and processing facility is essential to ensure satisfactory drying processes, indoor air quality and safety.

Establishing an exhaust flow rate is necessary for proper ventilation of an oven or dryer. The production rate for the product, inlet temperature and humidity, and exhaust temperature and humidity will determine the flow. It is useful to know the amount of water that a cubic foot of saturated air can hold.

Saturated air is air at 100% relative humidity. Relative humidity is roughly the weight of water in a volume of air divided by the maximum weight of water that volume of air could hold. As air temperature in-creases, the air is able to absorb more water. Tables presented in this article are based on standard air density at sea level and 70^{o}F (21^{o}C) with a density of 0.075 lb/ft^{3}. Use this information along with knowledge of the product to determine the flow rate.

Suppose you want to produce 500 lb/hr of medicinal herbs with average initial moisture content of 65% and final moisture content of 2.5%. The flow rate can be determined by figuring the weight of the moisture initially in the product, the residual moisture in the finished product and the production rate. The weight of the gross input product would be 500 lb/hr divided by the net product percentage of 35%, which equals 1,429 lb/hr. Moisture removed is the difference in the gross and the net, or 929 lb of water per hr.

As an example, consider a facility with 72^{o}F (22^{o}C) air at 50% relative humidity. The dryer incoming air is heated to 220^{o}F (104^{o}C) in a countercurrent flow dry-er with an exhaust temperature of 140^{o}F (60^{o}C) and relative humidity of 80% saturated. The discharge relative humidity of 80% will cool some of the exhaust gas before discharging. If the discharge cools below 130^{o}F (54^{o}C), moisture will begin to condense in the ductwork and fan. From table 1, the weight of the water in one cubic foot of 72^{o}F (22^{o}C) air at 50% relative humidity is 0.00062 lb/ft^{3} with an air density of 0.0743 lb/ft^{3}. In addition, the weight of the water in one cubic foot of 140^{o}F (60^{o}C), 80% relative humidity air is 0.00653 lb/ft^{3} with an air density of 0.0623 lb/ft^{3}. The net moisture absorbed is the difference of the two moisture values, which is 0.00591 lb/ft^{3} of air. Exhaust flow is approximately equal to the total moisture removed divided by net moisture per cubic foot removed. That is 929 lb/hr or 0.00591lb/ft^{3}, which equals 157,191 ft^{3}/hr or 2,620 cubic feet per minute (cfm). Add-ing heat throughout the drying process also is necessary to vaporize water. This is a conservative calculation. It does not take into account that the incoming air, when raised to 140^{o}F (60^{o}C), is equivalent to about 7% relative humidity and expands.

**Entering and Exiting Air Volumes**. The volume of air entering a dryer is not equal to the volume exiting a dryer. There are two reasons for this: the air expands as it is heated, and the moisture vaporized during drying increases ex-haust volume. The sum of the mass entering the dryer in the form of cool air and the product moisture is equal to the exhausted mass. The moisture of 929 lb/hr divided by 60 min equals 15.5 lb/min. The exhaust density of 140^{o}F (60^{o}C), 80% relative humidity air is 0.0623 lb/ft^{3}. The air density of 0.0623 lb/ft^{3} times the flow of 2,620 cfm is 163.2 lb/min. Subtracting the 15.5 lb/min of moisture gives the inlet air mass of 147.7 lb/min. Dividing the inlet air mass of 147.7 by the air density of 0.0743 lb/ft^{3} for 72^{o}F (22^{o}C) air at 50% relative humidity equals the inlet volume of 1,988 cfm inlet air. Add any products of combustion to the exhaust requirement and adjust the temperature and density.

## Estimating an Exhaust System

With the required exhaust flow of 2,620 cfm at 140^{o}F (60

^{o}C) and an air density of 0.0623 lb/ft

^{3}, an exhaust system can be estimated. Start by sketching out the duct layout. The duct will start with pickup at the dryer discharge and will discharge out of the building. You should include approximate lengths and elbow locations as well as any other fittings such as dampers, filters and stack caps.

Aim to maintain the temperature of the exhaust gas to avoid condensation. To maintain the exhaust gas temperature:

- Keep duct runs as short as possible.
- Keep vertical runs to a minimum.
- Insulate the ductwork if necessary.
- Avoid introduction of cool air into the exhaust air.

If the exhaust gas cools noticeably, it could fall to the saturation temperature and moisture will condense in the exhaust ductwork and fan.

Size exhaust ductwork for a velocity between 1,500 and 3,000 ft/min based on the required flow and duct diameter. Use table 2 to determine duct diameter at various velocities. In the above example, a duct diameter of 14" would result in a velocity of approximately 2,500 ft/min in the duct. Count the number of 90^{o} elbows in the layout. Use table 3 to convert the elbows to an equivalent length of straight pipe based on the diameter of the duct and radius of the elbows.

Total the length of straight duct and add the equivalent length for the elbows. Use table 2 to determine the static pressure per 100' of duct length. Multiply the value from table 2 by the length of equivalent duct divided by 100. Add static pressure for the other fittings (available from the manufacturer). The total is the static pressure necessary to move the required exhaust flow through the duct system and fittings. This loss is based on standard air with a density of 0.075 lb/ft^{3}.

## Selecting a Fan

Fan performance tables are based on standard air. Selecting a fan using standard tables will provide a fan capable of moving the required flow and developing the required pressure. As the temperature increases and humidity decreases, system resistance and the pressure developed by the fan will decrease at the same rate, and a constant flow rate is maintained. Altitude reduces the air pressure and density along with the amount of moisture the air can carry. The density and moisture values in table 1 must be divided by the correction factors in table 4 to determine the actual density and mois-ture content per cubic foot.
In the above example, 120' of straight 14" duct with three short radius elbows (three times 30 equals 90 from table 3) equal a total of 210 equivalent feet. The loss for 2,620 cfm through the 210' of duct is 0.59 times 2.1, which equals 1.24". Add to this the dryer, an indirect-fired gas heater, damper and stack cap at a total of 1.0" static pressure loss for a grand total of 2.24" water gauge. The type of fan necessary will be dependent on the flow and pressure relationship.

^{o}F (260

^{o}C).

For higher pressure ranges, a backward curved centrifugal fan often is selected. This design offers pressures up to 16" water gauge. They also offer the benefit of having the shaft and bearings outside the fan, and they can be mounted on the roof with a weather cover.

As with all fan systems, it is important to remember that only available air can be exhausted. Fans cannot operate in a vacuum. If the exhausted air cannot be replaced in the facility, fan performance will drop off and production will suffer. The production facility must have makeup air for every cubic foot that is exhausted from the facility. The makeup air can be provided through general ventilation systems or as part of a heating, ventilating and air conditioning system.

NOTE: Tables 1 and 2 were too large to reproduce clearly in this article online. They can be downloaded by clicking on the link below.