PLUMBING MATH FORMULAS.docx

PLUMBING MATH FORMULAS Finding the volume of a Rectangular Tank V = volume L = length H = height W = width Formula V=LxWxH To find the volume in Liters: Measure the tank in centimeters and divide the result by 1000. (Example: 13cm x 11cm x 28cm = 4004cm3 4004cm3 /1000 = 4.004 L) Measure the tank in meters and multiply by 1000. To find the volume in US gallons: Measure the tank in inches and divide by 231. Measure the tank in feet and multiply by 7.48.

Finding the volume of a Cylindrical Tank V = volume D = diameter L = length Formula V = D x D x 0.7854 x L or V = ∏R2L To find the volume in Liters: Measure the tank in centimeters and divide the result by 1000. (Example: 10cm x 10cm x 0.7854 x 30cm = 2356.2cm3) (2356.2cm3 / 1000 = 2.36 L) Measure the tank in meters and multiply by 1000.

To find the volume in US gallons: Measure the tank in inches and divide by 231. Measure the tank in feet and multiply by 7.48.

Finding the volume of an Elliptical Tank V = volume L = length W = width H = height Formula V = H x W x 0.7854 x L To find the volume in Liters: Measure the tank in centimeters and divide the result by 1000. (EX: 20cm x 10cm x 0.7854 x 30cm = 4712.4cm3) (4712.4cm3 / 1000 = 4.71 L) Measure the tank in meters and multiply by 1000. To find the volume in US gallons: Measure the tank in inches and divide by 231. Measure the tank in feet and multiply by 7.48.

Finding the volume of a Spherical Tank V = volume D3 = diameter cubed Formula To find the volume in Liters: Measure the tank in centimeters and divide the result by 1000. (EX: 20cm x 20cm x 20cm x 0.5236 = 4188.8cm3) (4188.8cm3 / 1000 = 4.19 L)

Measure the tank in meters and multiply by 1000. To find the volume in US gallons: Measure the tank in inches and divide by 231. Measure the tank in feet and multiply by 7.48.

Finding the volume of a Frustum Tank A frustum can be multi-sided or conical, the formula below will work for both.

V = volume H = height between bases B1 = area of upper base B2 = area of lower base Formula

To find the volume in Liters: Measure the tank in centimeters and divide the result by 1000. Measure the tank in meters and multiply by 1000. To find the volume in US gallons: Measure the tank in inches and divide by 231. Measure the tank in feet and multiply by 7.48. How to Convert Decimals to Fractions

STEP 1 Since the 8 is a whole number, no conversion is necessary. The answer will start with 8 feet.

STEP 2 Now to change the 0.62 of a foot that is left over into inches. Since there are 12 inches in a foot we will multiply 0.62 x 12 = 7.44. Now the whole number that is left (7) represents the inches. STEP 3 In construction we are usually only concerned with fractions of an inch to the 1/16th but you can use this for any other fraction. To start, divide the fraction you want to find by its denominator to get a decimal equivalent. In this example we'll use 1/16th . In the last step we were left with 0.44 inches. We now have to divide that number by the decimal equivalent of the fraction. 1 / 16 = 0.0625 (find any decimal equivalent by dividing the top number by the bottom) 0.44 / 0.0625 = 7.04 or 7/16" (always reduce the fraction if possible i.e. 8/16" = 1/2") Round the decimal to the nearest whole number. So the answer for our problem is 8' 7" 7/16". PRACTICE QUESTIONS Convert 5.92 feet Convert 12.39 feet Convert 63.88 inches (scroll down for the answers) Answers - 1) 5'11"1/16" 2) 12'4"11/16" 3)5'3"7/8" Ratio of Pipe Capacity Formula An important plumbing concept is understanding the ratio between pipe size and volume output. A common question that can be found on plumbing exams is: "How many one inch pipes would it take to provide the same volume of water as a two inch pipe?". Although this seems like a simple question the answer is actually four. A simple formula is all that's required to find the capacity of larger pipes in relation to smaller ones but this will not take into account the friction loss and assumes the same pressure. SIMPLE PIPE SIZE RATIO D2 - diameter of larger pipe squared d2 - diameter of smaller pipe squared N - number of smaller pipes

N = D2 / d2 Example: How many 1-1/2" pipes would be required to provide the volume of one 3" pipe? (3x3) / (1.5x1.5) = Four 1-1/2" pipes are needed The above formula does not account for friction loss however as smaller pipes have more surface in contact with the water flowing through them. To more accurately find the ratio including friction loss use this formula:

Example: How many 2" pipes would be required to provide the volume of one 3" pipe? 3 / 2 = 1.5 1.5 x 1.5 x 1.5 x 1.5 x 1.5 = 7.59375 square root of 7.59375 = 2.756 2" pipes are needed Area of a square / rectangle Area = length x width Area of a circle Area = diameter x diameter x .7854 Surface area of a cone Area = half of circumference of base x slant height + area of base Area of ellipse Area = short diameter x long diameter x 0.7854 Surface area of a cylinder Area = diameter x 3.1416 x length x area of two bases Area of a hexagon Area = width of side x 2.598 x width of side Area of a parallelogram Area = length of base x distance between base and top

Surface area of a pyramid Area = 1/2 the base perimeter x slant height + area of base Surface area of a sphere Area = diameter x diameter x 3.1416 Pipe Offsets – Plumbing Math To calculate a pipe offset using 45 degree and 22 1/2 degree elbows use the following chart.

To use this chart simply multiply the known side by the corresponding number to find the missing value. Side to find

Known side

45 degree offsets

22 1/2 degree offsets

travel

offset

1.414

2.613

offset

travel

.707

.383

run

offset

1.0

2.414

offset

run

1.0

.414

travel

run

1.414

1.082

run

travel

.707

.924

Rolling Offset Formula

A rolling offset in a plumbing system is when a pipe changes in both the horizontal and vertical planes. To visualize the travel of the pipe, imagine a three dimensional box with the pipe entering at one corner and exiting at the farthest diagonal corner.

STEP 1 - CALCULATING THE TRUE OFFSET

The first number you need to find when calculating a rolling offset is the "true offset" which is found using Pythagoras' theorem. This simply means that the offset squared plus the rise squared will equal the true offset squared. You then need to take the square root of the result to get the true offset. STEP 2 - FINDING THE SETBACK AND DIAGONAL Once you know the true offset you can use a table to determine the setback and diagonal center to center measurements. See the table below for the most common fitting constants. If your studying for a plumbing exam you will need to remember the 45˚ constants. Fitting angle

60˚

45˚

22.5˚

Diagonal = true offset X

1.155

1.414

2.613

Setback = true offset X

0.577

1.000

2.414

Practice Question Solve for pipes A, B, and C in the image below using 45 degree fittings and again using 22.5 degree fittings. The measurements are shown in imperial but all the constants are the same for metric. Scroll down for answers.

Solve for...

22.5 Degree fitting

45 Degree fitting

True offset

14.42"

14.42"

Setback

34.81"

14.42"

Pipe A 42" - setback =

7.19"

27.58"

Pipe B

37.68"

20.39"

Pipe C 61" - setback =

26.19"

46.58"

Jumper Offset

A jumper offset in a plumbing system uses a combination of fittings to go around an obstacle, such as a cylindrical tank. If your planning on writing a plumbing exam, rest assured that a jumper offset will be on the exam.

Using simple math and various constants, it's possible to find the required center-tocenter lengths for each pipe in the offset, which are outlined below. To find the missing values in a jumper offset you will need to remember the constants for right angle triangles used in a 45 degree offset; which are: To find the hypotenuse (the long side) multiple a short side by 1.414 To find a short side when the hypotenuse is known, multiply it by 0.707

Jumper offset using 45-90-45 fittings

Use the image above to sketch out the jumper offset using the variables on the right. After a couple practice runs, try changing the numbers. R - 4" A = 3" E = 2" C - 20"

Step 1 - The first thing you need to find is the side of the secondary square (G). So add (R) + (A) = (G) 4 + 3 = 7" This number is known as the secondary square (G). Step 2 - Multiplying the secondary square by 2 will give you the length of pipe (B), and the length of pipe (F) to the centerline of the tank.

(G) x 2 = (B) so to find the end to end length of pipe (B) which ends on the center line it would be 7" x 2 = 14". Step 3 - To find the length of pipe (F) will require two steps, first you have to multiply dimension (E) by 1.414 to find the length of pipe (F) below the centerline of the tank, then add that number to result you got in step two above. 2" x 1.414 = 2.828 then 2.828" + 14" = 16.828" or roughly 16 7/8"

Step 4 - To find (H) also known as the offset, multiply the diagonal (B) by the constant 0.707 (H) = 14 x 0.707 = 9.898" Step 5 - Find dimension (J) by adding the offset (H) with (E) (H) +(E) = (J) / 9.898 + 2 = 11.898" Step 6 - Finding the length of pipe (D) is simple subtraction. (C) - (j) = (D) 20 - 11.898 = 8.102" Static Pressure Head of Water How to find the pressure head of water from various heights in pounds per square inch or kilopascals. Simply put, the pressure from static head is the force exerted in all directions onto its container from the weight of the water above. Knowing the required static head is useful when selecting pumps to lift water to specific heights. Assumes water temperature of 60F / 16C Head formula - Imperial 27.72 inches of water = 1 psi

1 foot of water = 0.433 Psi Example 3 feet X 0.433 = 1.299 Psi

Head Pressure - Metric 10 Kpa = 1.45 Psi

1 Meter of water = 9.797 Kpa Example 18 M X 9.797 = 176.346 Kpa

Pipe Expansion Formula

Finding the expansion of pipe is relatively easy using the formula for "linear expansion". There are many reasons beside trade school to know these calculations and apply the knowledge gained into your building plans. Here are the factors you will need to know in order find the change in length (ΔL - change of length)of a section of pipe for various temperatures. Please note all of the units will be metric in the following examples. ΔT -(change of temperature) This is the difference between the initial temperature and the final temperature of the pipe. L - Initial length of the pipe a - Coefficient of linear expansion for the material. The formula to find the thermal expansion of pipe would look like this: ΔL = a L ΔT

Using the formula To keep this simple (I know it can look confusing), lets suppose we need to find how much a 10m length of copper pipe will change given a 40C change in temperature. The coefficient of linear expansion of copper is 0.000017, so a=0.000017 for this example. ΔL = 0.000017 x 10 x 40 ΔL = 0.0068m or 7mm Material type

Coefficient of thermal expansion (a)

ABS

73.8 x 10-6

Aluminum

22.2 x 10-6

Brass

18.7 x 10-6

Cast iron

10.4 x 10-6

Cement

10.0 x 10-6

CPVC

66.6 x 10-6

Concrete

14.5 x 10-6

Glass pipe

4.0 x 10-6

Copper

16.6 x 10-6

Iron

12.0 x 10-6

Lead

28.0 x 10-6

PEX 02 barrier

140 x 10-6

PEX-AL-PEX

24 x 10-6

Polyethylene (PE)

200 x 10-6

Polyethylene terephthalate (PET)

59.4 x 10-6

Polypropylene (PP)

90.5 x 10-6

PVC

110 x 10-6

Rubber

77 x 10-6

Steel

13 x 10-6

Wood Framing

3.7 x 10-6

Specific Gravity Formula The specific gravity of a substance is a comparison between the mass or density of a substance as compared to an equal volume of water. The base line for specific gravity is taken from the properties of water and is given the value of 1.

Specific gravity = Mass of a substance Mass of an equal volume of water For example: A cubic meter of iron has a mass of 7901 Kg, while a cubic meter of water has a mass of 1000 Kg. Therefore if you divide the mass of the iron by the mass of an equal volume of water you will find the specific gravity of iron is 7.901 In this case, the specific gravity (the ratio of the density of iron to the density of water) indicates that a cubic meter of iron weighs about 8 times as much as a cubic meter of water. Note that there are no units of measurement. Specific gravity simply states how many times heavier a solid or liquid sample is, from water. Example 2 : The specific gravity of a certain oil is 0.8, find the mass of 1 litre of oil.

The mass of 1 litre of water is 1 Kg, so the mass of the oil would be 0.8 Kg. Formula for Pressure in Depths of Liquid Finding pressure in depths of water In the operation of pressurized systems you will be primarily concerned with the pressures exerted by water. Water pressures are directly related to both the height (depth) and density of water. Pressure is defined as the amount of force acting (pushing) on a unit area. Consider a cubic meter of water to have a mass of 1000 kg and, the force acting downward to be 1000 x 9.8 or 9800 Newton. As this force is acting on 1.0 M2 the pressure on the base of the cube will be 9800 N or 9.8 Kpa per 1.0 m2. It follows that at a depth of 2.0 m the pressure will be 2 x 9.8 or 19.6 Kpa and 3.0 m it will be 3 x 9.8 or 29.4 Kpa. Therefore, to find the pressure in water simply multiplies 9.8 by the depth in meters. Remember that the result of this calculation will give you kilopascals. Formula Pressure (P) = 9.8 x depth (m) = Kpa P = 9.8 x depth (m) x SG = Kpa (If working with substances other than water their specific gravity(SG) must be factored in) Example: Find the pressure in water at a depth of 150m. P = 9.8 x 150 P= 1470 Kpa Example 2: If a pressure gauge on a non pressurized tank reads 24.3 Kpa, how many meters of water are there in the tank? Depth= 24.3 / 9.8 depth= 2.48 m Roof Drain Calculations In order to size rain water leaders and gutters there are some simple calculations to follow. First, you must know the maximum 15 minute rainfall intensity in millimetres (mm) for your area; which can be found either through the national building code or your local environmental weather office, and the size of the area in square meters (m2) being

served by the drain. Any adjoining vertical walls extending from the roof will be accounted for as well by using half of the square area. The product of these two figures will produce a hydraulic load; in litres, that can be used to find the required pipe size according to your local codes.

Example calculation

A flat roof measuring 3m x 4m with one vertical wall extending from the roof measuring 3m x 4m has a 30mm rainfall intensity. Area = (3m x 4m= 12m2)+(3m x 4m= 12m2 /2 = 6m2)= 18m2 30mm x 18m2 = 540 litres The maximum hydraulic load for this example would be 540 litres and would require 2" leader or 3" gutter (1 in 100 slope) as sized according to the NPC of Canada 2005. Finding Grade using a Philadelphia Rod Using Survey Equipment to find Pipe Grade Although most residential plumbers will only need to know how to read a Philadelphia rod in order to pass the plumbing exam, those interested in working on industrial and commercial sites may have to find the grade of long pipe runs using a Philadelphia rod and transit (Survey Equipment).

The image below shows an imperial Philly rod that is being read to a height of 5.76 feet (the 5 foot marker is out of view). The top and bottom of each black mark is one hundredth of a foot; the top of each is even, while the bottom of each marking is odd.

Using a metric Philadelphia rod is very similar to an imperial philly rod, except each mark is one centimeter and has the even numbers starting from the bottom of each black mark. The large numbers in the image below show tenths of a meter and are marked by the line on which the numbers rest.

Grade the Pipe To properly grade a piping run using a transit and Philly rod you will need to do some simple math, take proper distance measurements, and have a note pad to record the measurements. Know the grade you need on the pipe. 1:50= 1/4" per foot, 1:100 = 1/8" per foot Measure the distance of the pipe you are finding the grade, and calculate the drop (grade) it should have. For example 64 feet of pipe at 1/8" per foot would need to drop 8 inches.

To find the actual grade of the pipe, have the rod man place the philly rod on the pipe at the high end and take a reading with the transit, then have him move to the low end and take another reading. Subtract the high side reading from the low side reading, to calculate the actual drop of the pipe. To find the grade of a pipe when you know the drop and the distance, simply divide the distance by the drop using the same units of measurement. Example: the pipe run is 38 feet and the drop is 5.5 inches. 38 x 12 = 456 inches. 456 / 5.5 = 83 so the grade would be 1:83. Knowing how to read a Philadelphia rod and calculate grade will help you with a whole section on the plumbing exam, as well as experience in the field.