Disclaimer: The author of this article is not a licensed plumber. No content in this article should be put into practice unless researched and confirmed with the local appropriate plumbing codes.
Plumbers should use mathematics to ensure the pipes, appliances, etc. that they install or service are up to code. Arithmetic, geometry, and even some algebra can be used in the field of plumbing.
1. The "grade" of a drain pipe is sometimes referred to as the "pitch" or "fall". The grade describes how steep that a drain pipe should be. A drain pipe needs to be steep in order to ensure gravity creates enough flow to prevent clogging. The grade of a drain pipe is also specified in order to reduce the risk of siphoning nearby traps. If siphoning occurs, gasses can enter the building that are harmful.
A local plumber code states that a drain up to 4-inches in diameter needs a grade of 0.25 inch per liner foot. That means for every horizontal foot that the drain runs, it must fall 0.25 inches.
If a drain (that is under 4 inches in diameter) needs to run 3 feet horizontally to a branch drain, how many inches down should the branch drain be located?
We are given a conversion factor of 0.25 inches per linear foot.
\[\frac{0.25 \textbf{ inches down}}{1 \textbf{ horizontal foot}}\]
If we multiply the 3 horizontal feet by this conversion factor, the horizontal feet units will cancel, leaving us with "inches down," which is what we wanted to calculate.
\[\frac{0.25 \textbf{ inches down}}{1 \textbf{ horizontal foot}} \cdot 3 \textbf{ horizontal feet} =\]
\[\frac{0.25 \textbf{ inches down}}{1 \cancel{\textbf{ horizontal foot}}} \cdot 3 \cancel{\textbf{ horizontal feet}} =\]
\[0.25 \textbf{ inches down} \cdot 3 =\]
\[0.75 \textbf{ inches}\]
The branch drain must be \(0.75\) inches below the sink.
2. A drain runs 10 feet horizontally and falls 3.5 feet in order to reach the main drain line. Does this drain meet the "0.25 inches per linear foot" rule mentioned above? Is it more or less steep than the requirement?
We can calculate the grade of a line by calculating the slope (the rise divided by the run, or, in other words, the vertical change divided by the horizontal change). We divide the 3.5 fall by the 10 foot "run" (horizontal change).
\[\frac{3.5 \textbf{ feet}}{10 \textbf{ feet}} \approx 0.333 \textbf{ feet}\]
This slope is larger than 0.25 feet. Therefore, this drain is steeper than the requirement. Drains that are too steep might inhibit a p-trap’s (or other trap’s) ability to function as drainage can be siphoned out of the trap. This could lead to harmful gasses entering the home. Some solids might also not be washed away if the velocity of the water is too large. As nothing can be made to perfection, it is up to the discretion of the plumber whether the difference between 0.33 inches per linear foot and 0.25 inches per linear foot is significant enough to warrant a rebuild.
3. Sometimes the grade is expressed as a percentage. If a sewer ditch is 160 feet long with a 2 percent grade, what would be the total amount of fall?
The grade is just a percentage of the horizontal distance. We multiply by the decimal form of the percentage.
\[160 \textbf{ feet} \cdot 0.02 = 3.2 \textbf{ feet}\]
We expect the vertical drop of the sewer ditch to be 3.2 feet.
4. In plumbing, a "bend" is a type of pipe fitting used to change the direction of a sewer or drain line. There are bends with different degrees of rotation. A bend is typically expressed in terms of the fraction of a circle that represents the angle of the bend.
What angle is made when you offset a sewer line with a 1/6 bend?
Multiply the degrees of a circle, 360, by the fraction of the circle.
\[\frac{1}{6} \cdot 360 \textbf{ degrees} = 60 \textbf{degrees}\]
A 1/6 bend is a 60 degree angle.
5. Plumbers often also need an understanding of physics in order to calculate pressure. A test plug is a device used in plumbing and construction to seal or block a pipe or other opening temporarily during testing or inspection. Pressure on a test plug should be calculated so that the pressure does not exceed the limits of the test plug.
What is the pounds per square inch (psi) on a 6 inch diameter test plug that has a water head of 10 feet (a column of water of 10 feet is above the test plug)?
If pressure is measured in pounds per square inch, then our final units should be
\[\frac{\textbf{pounds}}{\textbf{in}^2}\]
The pressure exerted on the test plug is found by multiplying the density of water by the height of the column of water. The density of water is approximately 62.4 pounds per cubic foot.
\[\textbf{pressure} = \frac{62.4 \textbf{pounds}}{\textbf{ft}^3} \cdot 10 \textbf{ ft}\]
We have cubed feet in the denominator and feet in the numerator, which means the new units are squared feet in the denominator. You may have learned this rule in algebra:
\[\frac{1}{x^3} \cdot x = \frac{1}{x^2}\]
\[\textbf{pressure} = \frac{624 \textbf{pounds}}{\textbf{ft}^2}\]
The units of this pressure is pounds per square foot, but pounds per square inch was requested. We use the fact that 1 foot is 12 inches. As we need square feet, we’ll also square the number of inches in this conversion factor.
\[\textbf{pressure} = \frac{624 \textbf{ pounds}}{\textbf{ft}^2} \cdot \frac{1 \textbf{ ft}^2}{(12 \textbf{ in.})^2}\]
\[\textbf{pressure} = \frac{624 \textbf{ pounds}}{\textbf{ft}^2} \cdot \frac{1 \textbf{ ft}^2}{144 \textbf{ in.}^2}\]
\[\textbf{pressure} = \frac{ 624 \textbf{ pounds}}{\cancel{\textbf{ft}^2}} \cdot \frac{1 \cancel{\textbf{ ft}^2}}{144 \textbf{ in.}^2}\]
\[\textbf{pressure} = \frac{624 \textbf{ pounds}}{144 \textbf{ in}^2}\]
\[\textbf{pressure} = \frac{624 \textbf{ pounds}}{144 \textbf{ in}^2}\]
\[\textbf{pressure} = \frac{4.333 \textbf{ pounds}}{\textbf{ in}^2}\]
so the pressure is approximately 4.33 pounds per square inch.
6. A toilet tank hols 1.28 gallons of water. The toiler refills in 45 seconds. What is the flow rate, in gallons per minute, for the tank refill valve?
We can find the rate in seconds
\[\frac{1.28 \textbf{ gallons}}{45 \textbf{ seconds}} \approx\]
\[\frac{0.02844 \textbf{ gallons}}{ \textbf{ second}}\]
To convert to gallons per minute, we multiply by the conversion factor of 60 seconds for 1 minute.
\[\frac{0.02844 \textbf{ gallons}}{ \textbf{ second}} \cdot \frac{60 \textbf{ seconds}}{1 \textbf{ minute}} =\]
\[\frac{0.02844 \textbf{ gallons}}{\cancel{ \textbf{ second}}} \cdot \frac{60 \cancel{\textbf{ seconds}}}{1 \textbf{ minute}} =\]
\[\frac{0.02844 \cdot 60 \textbf{ gallons}}{\textbf{ minute}} \approx\]
\[\frac{1.71 \textbf{ gallons}}{\textbf{ minute}}\]
The flow rate is 1.71 gallons per minute.
7. A plumber has been asked to choose and install a water heater for a family. The dimensions of the water heater will be constrained based on the location of the new water heater in the home. The cylindrical tank that fits this location has a height of 61 inches and a diameter of 20 inches. What is the volume of the water tank?
The volume of an object can be found by finding the area of the base of the object and multiplying by the height. The base of a cylinder is a circle. The area of this circle is \(\pi (10)^2 \textbf{ in.}^2 \approx 314.16 \textbf{ in.}^2\). The height is 61 inches. The total volume is therefore
\[314.16 \textbf{ in.}^2 \cdot 61 \textbf{ in.} = 19,163.76 \textbf{ in.}^3\]
Note the dimensions of the units. If we multiply \(x \cdot x^3\), the result is \(x^{1 + 3} = x^4\). Similarly, when we multiply \(\textbf{ in.}^2\) and \(\textbf{ in.}\), we get \(\textbf{ in.}^3\).
Water tank capacity is usually measured in gallons, though, and not cubic inches. To convert from cubic inches to gallons, we will use the fact that 1 gallon is 231 cubic inches. We will multiply our result by \(\frac{1 \textbf{ gallon}}{231 \textbf{ in.}^3}\) so that the inches units cross cancel.
\[19,163.76 \textbf{ in.}^3 \cdot \frac{1 \textbf{ gallon}}{231 \textbf{ in.}^3} =\]
\[19,163.76 \cancel{\textbf{ in.}^3} \cdot \frac{1 \textbf{ gallon}}{231 \cancel{\textbf{ in.}^3}} =\]
\[\frac{19,163.76}{231} \textbf{ gallons} =\]
\[82.96 \textbf{ gallons}\]
1. A local plumber code states that a drain up to 4-inches in diameter needs a grade of 0.25 inch per liner foot. That means for every horizontal foot that the drain runs, it must fall 0.25 inches.
If a drain (that is under 4 inches in diameter) needs to run 2 feet horizontally to a branch drain, how many inches down should the branch drain be located?
2. A drain runs 14 feet horizontally and falls 3.5 feet in order to reach the main drain line. Does this drain meet the "0.25 inches per linear foot" rule mentioned above? Is it more or less steep than the requirement?
3. If a sewer ditch is 180 feet long with a 1.5 percent grade, what would be the total amount of fall?
4. What angle is made when you offset a sewer line with a 1/8 bend?
5. What is the pounds per square inch (psi) on a 6 inch diameter test plug that has a water head of 12 feet (a column of water of 12 feet is above the test plug)?
6. A toilet tank hols 1.1 gallons of water. The toiler refills in 40 seconds. What is the flow rate, in gallons per minute, for the tank refill valve?
7. A plumber has been asked to choose and install a water heater for a family. The dimensions of the water heater will be constrained based on the location of the new water heater in the home. The cylindrical tank that fits this location has a height of 58 inches and a diameter of 18 inches. What is the volume of the water tank?