Force, Pressure, Head, and Velocity Calculations - Basic Mathematics for Water and Wastewater Operators - page 185

Distance traveled (feet) Distance traveled (feet)

Float

FIGURE 12.3 Illustration for Example 12.25.

■ Example 12.25
Problem : A float travels 320 ft in a channel (see Figure 12.3) in 2 minutes and 20
seconds. What is the velocity in the channel in feet per second?

Solution:

2 min, 20 sec = (2 × 60) + 20 = 140 sec

Velocity = 320 ft ÷ 140 sec = 2.29 ft/sec (fps)

■ Example 12.26
Problem : The distance between manhole #1 and manhole #2 is 110 ft. A fishing bob-
ber is dropped into manhole #1 and enters manhole #2 in 30 seconds. What is the
velocity of the wastewater in the sewer in feet per minute?

Solution:

30 sec ÷ 60 sec/min = 0.5 min

Velocity = 110 ft ÷ 0.5 min = 220 ft/min

■ Example 12.27
Problem : A float is placed in a channel. It takes 2.5 minutes to travel 400 ft. What is
the flow velocity in feet per minute in the channel? (Assume that the float is traveling
at the average velocity of the water.)

Solution:

Distance
Time

400 ft
2.5 min

Velocity = = = 160 ft/ min

■ Example 12.28
Problem : A cork placed in a channel travels 40 feet in 20 seconds. What is the veloc-
ity of the cork in feet per second?

Solution:

Velocity = 40 ft ÷ 20 sec = 2 ft/sec (fps)