# How to Calculate Flow Rate From a Tank

Written by william hirsch
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The force of gravity pulls down on liquid within a tank which then exits through a smaller pipe. Fluid flow through a closed tank and pipe system is constant. This follows from the continuity equation in physics. This equation states that the product of the cross-sectional area of the tank or pipe times the velocity of the fluid does at that place does not change. Cross-sectional area describes area of the circular side view of the pipe or tank.

Skill level:
Moderately Challenging

• Marker
• Tape measure
• Stopwatch
• Calculator

## Instructions

1. 1

Place two marks on the side of the tank that are a vertical distance of 20 inches apart. Make sure the top mark is below the current liquid level.

2. 2

Start the stopwatch when the liquid level reaches the top mark on the tank. Stop the timer when the liquid reaches the bottom mark on the tank. Note the time taken for the liquid level to fall. For example, 35 seconds.

3. 3

Divide the distant between the marks on the tank by the time taken to get the liquid velocity in the tank in inches per second. Completing this exercise leads to 20 inches divided by 35 seconds, or 0.57 inches per second.

4. 4

Determine the cross-sectional area of the tank in square inches. If the tank is rectangular in shape the cross-section is a rectangle with area of length times width. Measure on the sides of the tank. If the tank is cylindrical in shape the cross-section is a circle. First measure the circumference, or distance around the tank, in inches. Divide the circumference by 6.28 to get the radius in inches. The radius is the distance from the circle's centre to its edge. The formula for the area then equals 3.14 times the radius squared. As an illustration assume the area comes out to be 1,728 square inches.

5. 5

Measure the circumference of the exit pipe from the tank in inches, then calculate its cross-sectional area in square inches. Assuming you get a circumference of 10 inches, then the radius of the pipe is 10 inches divided by 6.28, or 1.6 inches. Now the cross-sectional area of the pipe is 3.14 times 1.6 inches times 1.6 inches which equals 8 square inches.

6. 6

Multiply the velocity of the liquid in the tank by the tank cross-sectional area. Call this number "x." Performing this step you get 0.57 inches per second times 1,728 square inches which is 985 cubic inches per second.

7. 7

Divide "x" by the cross-sectional area of the exit pipe to obtain the velocity of the liquid out of the tank in inches per second. Completing the exercise leads you to 985 cubic inches per second divided by 8 square inches, or a velocity of 123.1 inches per second. In feet per second the velocity is 10.3.

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