# 3.5 Drafting Guidelines

It is important to know the difference in elevation between the pump and the water source when drafting water from a pond or stream. When drafting water, the air at atmospheric pressure is removed from the hose line, creating a vacuum (negative pressure) within the pump chamber. The atmospheric pressure (weight of air) on the water's surface forces the water up through the suction hose to the pump.

The maximum height to which an engine or pump can lift water is determined by the atmospheric pressure. At sea level, the atmosphere exerts an average pressure of 14.7 pounds per square inch (psi). Atmospheric pressure will vary due to changes in the weather. However, these changes tend to moderate themselves so that the average pressure will tend to go back toward 14.7 pounds per square inch. That is why it is safe to use this value of 14.7 pounds per square inch as a constant for calculations.

Example 1 - What would be the maximum height of water that a pressure of 14.7 pounds per square inch would be capable of sustaining?

Step 1. Find the appropriate conversion in Table 3.1.
1 psi = 2.304 ft

Step 2. Set up the cancellation table so all units will cancel, except the desired unit, feet, to calculate the lift created by 14.7 pounds per square inch.

The atmospheric pressure would be capable of sustaining a column of water 33.9 feet in height.

If a pump could produce a perfect vacuum, the maximum height to which it could lift water at sea level would be 33.9 feet, as shown in Example 1. This number is the maximum theoretical lift, but in practice no pump built can produce a perfect vacuum. A fire engine in fairly good condition can lift water two-thirds of the theoretical lift, 2/3 × 33.9 = 22.5 feet. This height is called the maximum attainable lift. With an increase in elevation above sea level, atmospheric pressure decreases, thus reducing the vertical distance from the water source where drafting can be done effectively.

### ELEVATION EFFECTS

For every 1,000 feet of change in elevation, there is a loss of 1 foot in suction or lift and a 0.5 pounds per square inch decrease in atmospheric pressure.

Example 2 - An engine can lift water 22.5 feet at sea level. The same engine is driven to a fire at an elevation of 2,000 feet above sea level. What lift can the engine produce at this elevation?

Step 1. Use the conversion given for elevation change. A 1-foot loss = 1,000-foot elevation change

Step 2. Set up the cancellation table so all units will cancel, except the desired unit, feet, to calculate the loss in lift for a 2,000-foot elevation. (See Section 2.1 to review unit cancellations if desired.)

Step 3. Subtract the resulting value from the number of feet that can be lifted at sea level. 22.5 ft - 2 ft = 20.5 ft

This pump can lift 20.5 feet of water at a 2,000-foot elevation.

Example 3 - Larry is 16 feet above his water source, at an elevation of 4,000 feet. Will Larry still be able to draft water?

Step 1. Find the appropriate conversion/estimation in Table 3.1 to calculate the decrease in possible lift. At sea level, attainable lift is 22.5 feet.

Step 2. Set up the cancellation table so all units will cancel, except the desired unit, feet (loss), to calculate the loss in lift. Due to the elevation, the sustainable lift decreases by:

A 1,000-foot increase in elevation = 1-foot loss

Step 3. Calculate the adjusted attainable lift. The maximum attainable lift would now be: attainable lift - decrease due to elevation = adjusted attainable lift 22.5 ft - 4 ft = 18.5 ft

Step 4. Determine whether drafting is still possible. attainable lift = 18.5 ft so Larry would still be able to draft water up to a vertical distance of 18.5 feet. He desires to lift at least 16 ft.

18.5 feet - 16 feet = 2.5 feet above Larry's current location.

Yes, Larry is able to draft 16 feet above his water source.