Lesson 11:
Force, Pressure and Head Calculations
Objective
In this lesson we will learn the following:
- How to calculate force of an object.
- The different types of head and how to calculate them.
- How to calculate pressure (psi).
Lecture
Introduction
Before we can perform calculations to determine force, pressure and head, we need to understand what we will be calculating.
Force - Push exerted by water on any confined surface. It can be expressed in pounds, tons, grams or kilograms.
Pressure - Force per unit area. Most common way of expression is pounds per square inch (psi).
Head - Vertical distance or height of water above a reference point. It is usually expressed in feet.
*FYI: In the case of water, head and pressure are related.
Force and Pressure
Force is an interaction between objects that causes them to change motion. We measure force in either pounds or newtons (N), which is the scientific unit for measuring weight. One newton is equal to approximately 0.22 pounds. A newton is categorized as force due to gravity.
Pressure and force are related, which means you can calculate one if you know the other by using the physics equation: pressure = force/area. This pressure can either be documented in pounds per square inch, psi, or newtons per square meter (N/m2). The unit N/m2 is also known as a pascal, which is equivalent to 1 newton per square meter, and is abbreviated as Pa.
Hydrostatic pressure, or the pressure a fluid exerts at equilibrium at a certain point in the fluid due to gravity, increases at lower depths as the fluid can exert more force from the liquid above that point. This hydrostatic pressure can be determined with the formula: pressure = force/area.
Example:
Determine the amount of force exerted, in pounds, on 144 square inches at 45 psi.
Example:
Determine the amount of force exerted, in pounds, on 144 square feet at 45 psi.
You must first convert square feet to square inches so the units will cancel, leaving only pounds for force:
Now you can determine the amound of force, in pounds, that is exerted.
A cubical container measuring 1 foot on each side can hold 1 cubic foot of water.
It is a basic fact of science that 1 cubic foot of water weighs 62.4 pounds and contains 7.48 gallons. That means that the force acting on the bottom of the container would be 62.4 pounds per square foot. The area of the bottom of the cube, in square inches, is:
1 ft2 = 12 in x 12 in = 144 in2
Therefore, the pressure, in pounds per square inch (psi) is:
Since the head is the vertical distance above the reference point, the head would be 1 foot for the cube, if we made the bottom of the cube the reference point. The calculation above shows that 1 foot of head is equivalent to 0.433 psi.
Shows the relationship between pressure and head
*An important point to remember is that force acts in a particular direction; water in a tank exerts force downward on the bottom and out the sides. Pressure, however, acts in all directions. A rock at a water depth of 1 foot would experience 0.433 psi of pressure acting inward on all sides. Because of this relationship between pressure and head you have the following equations:
Pressure, psi = 0.433, psi x head, ft
Head, ft = 2.31 ft x pressure, psi
Let's look at how force can be measured in a different type of unit, known as newtons:
Force, newtons = (Pressure, Pa)(Area, m2)
Example:
Determine the amount of force exerted, in newtons, on 144 square meters at 45 pascals.
Force, newtons = (45 Pa)(144 m2) *Remember that 1 Pa = 1 N/m2
Force, newtons = 6480 N
You can now convert the above force into pounds. We know that 1 newton is equivalent to 0.22 lb:
Force, lb = Force, newtons x 0.22 lb/newton
Force, lb = 6480 N x 0.22 lb/N
Force, lb = 1425.6 lb
Head
Total Head
As we mentioned earlier, head is the vertical distance the water must be lifted from the supply tank or unit process to the discharge.
The total head includes the vertial disance the liquid must be lifted (static head), the loss to friction (friction head), and the energy required to maintain the desired velocity (velocity head).
Total head = Static head + Friction head + Velocity head
Static Head
Static head is the vertical distance, in feet or meters, from a reference point to the water surface when the water is not moving and can be calculated by:
Static head = Discharge elevation - Supply elevation
Example:
A supply tank is located at an elevation of 108 ft. The discharge point is at an elevation of 205 ft. What is the static head, in feet?
Static head = Discharge elevation - Supply elevation
Static head = 205 ft - 108 ft
Static head = 97 ft
Friction Head
Friction head is the equivalent distance of the energy that must be supplied to overcome friction. Engineering references include tables showing the equivalent vertical distance for various sizes and types of pipes, fittings, and valves. The total friction head is the sum of the equivalent vertical distances for each component and can be determined by:
Friction head, ft = Energy losses due to friction
We will not dive into friction head, just know that it is part of the total head.
Velocity Head
Velocity head is the equivalent distance of the energy consumed in achieving and maintaining the desired velocity in the system and can be calculated by:
Velocity head, ft = Energy losses to maintain velocity
Again, we will not go into how to determine the velocity head; just now it is part of the total head calculation.
Pressure/Head
The pressure exerted by water or wastewater is directly proportional to its depth or head in the pipe, tank, or channel. If the pressure is known, the equivalent head can be calculated by:
Head, ft = Pressure, psi x 2.31 ft/psi
Example:
The pressure gauge on the discharge line from the influent pump reads 75.3 psi. What is the equivalent head in feet?
Head, ft = Pressure, psi x 2.31 ft/psi
Head, ft = 75.3 psi x 2.31 ft/psi
Head, ft = 173.9 ft
Head/Pressure
If the head is known, the equivalent pressure can be calculated by:
Example:
A tank is 15 feet deep. What is the pressure, in psi, at the bottom of the tank when it is filled with water?
Now that we know how total head is determined and the part that pressure and force play in total head, let's look at some different examples.
Example:
Convert 40 psi to feet head.
The "psi" unit cancels, leaving ft.
To solve this it would be 40 ÷ 0.433.
Example:
Convert 40 feet to psi.
The "ft" unit cancels, leaving psi.
To solve this it would be 40 x 0.433.
Notice that the relationship between psi and feet is almost two to one. It takes slightly more than 2 feet (2.31 ft) to make 1 psi; therefore, when looking at a problem where the data are in pressure, the result should be in feet, and the answer should be at least twice as large as the starting number. For instance, if the pressure is 25 psi, we intuitively know that the head is over 50 feet; therefore, we must divide by 0.433 to obtain the correct answer.
Now, let's combine math principles from above along with others from previous lessons.
Example:
A 150 ft diameter cylindrical tank contains 2.0 MG of water. (1) What is the water depth? (2) At what pressure would a gauge at the bottom of the tank read, in psi?
First, convert MG to cubic feet:
Next, using volume solve for depth:
Now that we know the head, or depth, let's determine the psi at that depth:
Example:
The pressure in a pipe is 70 psi. (1) What is the pressure in feet of water? (2) What is the pressure in pounds per square foot (psf)?
First convert pressure to feet of water:
Head, ft = Pressure, psi x 2.31 ft/psi
Head, ft = 70 psi x 2.31 ft/psi
Head, ft = 161.7 ft of water
Now, convert psi to to psf for the second part of the question:
70 psi x 144 in2/ft2= 10,080 psf
Example:
The pressure in a pipeline is 6476 psf. What is the head on the pipe?
Head on pipe = ft of pressure
Pressure = Weight x Height
6476 psf = 62.4 lb/ft3x Height
Height = 6476 psf / 62.4 lb/ft3
Height = 104 ft
Summary
Force - Push exerted by water on any confined surface. It can be expressed in pounds, tons, grams or kilograms.
Pressure - Force per unit area. Most common way of expression is pounds per square inch (psi).
Head - Vertical distance or height of water above a reference point. It is usually expressed in feet.
It is a basic fact of science that 1 cubic foot of water weighs 62.4 pounds and contains 7.48 gallons. An important point to remember is that force acts in a particular direction; water in a tank exerts force downward on the bottom and out the sides. Pressure, however, acts in all directions. A rock at a water depth of 1 foot would experience 0.433 psi of pressure acting inward on all sides. Because of this relationship between pressure and head you have the following equations:
Pressure, psi = 0.433, psi x head, ft
Head, ft = 2.31 ft x pressure, psi
The total head includes the vertial disance the liquid must be lifted (static head), the loss to friction (friction head), and the energy required to maintain the desired velocity (velocity head).
Assignment
Complete themath worksheetfor this lesson. You must be logged into Canvas to submit this assignment. Make sure you choose the appropriate semester.
