8.2: Basic Hydraulics

There are two important physical laws that apply to hydraulics, conservation of energy and conservation of mass (continuity). The energy in the water will be in any of the following forms:

• kinetic energy due to the velocity of the water,
• potential energy due to the elevation of the water relative to an arbitrary reference elevation, and
• potential energy due to water pressure.

In this book, the energy in water is expressed either as energy per unit of weight of water (head) or energy per unit of volume of water (pressure). Since energy has the dimensions FL (force x length) and weight has the dimension F, energy per unit of weight has the dimension of length (L). Energy expressed in this manner is referred to as head. Energy per unit of volume has the dimension of FL/L³ or F/L² . A common unit is pounds per square inch (psi). The energy of water in an irrigation system includes velocity head, elevation head, and pressure head.

Kinetic energy is a result of the movement of the fluid. Velocity head (h_v) is given by:

\(h_v=\dfrac{{V_m}^2}{2g} \) (8.1)

where: V_m = average velocity at a point in the pipe or channel, ft/s, and

g = gravitational constant, 32.2 ft/s² .

In general, the maximum recommended average velocity of flow in an enclosed or pressurized pipeline is 5 feet per second (ft/s). When the velocity in a mainline exceeds 5 ft/s, there is potential to develop relatively high-pressure surges which may damage pipelines. Pressure surges are due to flow being stopped suddenly while the upstream water has a large amount of momentum. When the flow is stopped too quickly, the rapid change of momentum results in an impulsive force called water hammer. The allowable maximum velocity may be higher than 5 ft/s if special precautions (pressure relief valves, surge tanks, etc.) are used to relieve possible pressure surges. The potential energy due to elevation is a result of the location of the water relative to an arbitrary reference plane. Water at a higher elevation has more potential energy than water at a lower elevation. Consider water flowing downhill. Energy is the ability to do work, and work can be described as a force acting over a distance. As water flows downhill, the force is gravity, and the distance is the length over which the water flows. The water has the ability to do work as it flows downhill such as eroding the surface, generating power, etc. The potential energy of the water decreases as it flows downhill. The letter Z will be used to represent elevation head or gravitational head. The potential energy due to the pressurization of water can be a very large component in an irrigation system. Pressure is the force per unit area exerted on the walls of a container. The pressure may be expressed as:

P = γh or h = P/γ

where: P = pressure,

γ = weight of a unit volume of fluid (specific weight), and

h = pressure head.

For water, γ = 62.4 lb/ft³ . Figure 8.1 illustrates how the pressure is related to the depth (head) of water in a container.

Figure 8.1. Pressure head for water in a vessel.

The shape and volume of the container are not important when applying Equation 8.2.

In USCS units, the following conversions are convenient:

\(\gamma = 0.433 \dfrac{\text{psi}}{\text{ft}} \) (8.3)

or

\(\dfrac{1}{\gamma}=2.31\dfrac{\text{ft}}{\text{psi}} \)

Because different fluids have different weights per unit volume (γ), Equation 8.3 is only valid for water.

In Example 8.1 the pressure is independent of the surface area of the columns, but realize of course that the forces on the container bottoms are different, one having 10 times the force as the other.

Example 8.1

Two columns of water are filled to a height of 10 feet with water. One column has a cross-sectional area of 1 in² , the other 10 in² . Find the pressure due to the fluid at the bottom of each column.

Given: h = 10 ft

γ = 0.433 psi/ft

Find: P (Figure 8.1)

Solution

\(P=\gamma h \) (Eq. 8.2)

\(P=0.4333 \dfrac{\text{psi}}{\text{ft}}(10\text{ ft})=4.33 \text{ psi} \)

The sum of the three energy forms is the total energy per unit weight of water called total head (H). Total head is:

H = velocity head + elevation head + pressure head

or

\(H=\dfrac{{V_m}^2}{2g}+Z+h \)

The sum of elevation head and pressure head is called hydraulic head. Figure 8.2 illustrates the components of hydraulic head for a pipeline with various orientations. Another important concept of water flow is continuity. In a hydraulic system mass must be conserved. Therefore, for an incompressible fluid such as water, the volumetric flow rate (Q) must be the same for all points in a system with only one inlet and one outlet. The continuity equation for an incompressible fluid, such as water, may be expressed as:

Q = V_m A_f (8.5)

where: Q = volumetric flow rate or discharge,

V_m = average flow velocity, and

A_f = cross-sectional area of flow.

The laws of conservation of mass and energy are applied in Example 8.2. The conservation of mass states that the volumetric flow rate (Q) must be the same for all points in the system. Thus, the flow rate everywhere in the system shown in Example 8.2 must be 400 gallons per minute (gpm). By combining the continuity equation and the concept of mass flow, problems other than just calculating the total head at a point may be solved. An important law of fluid mechanics is conservation of energy. Conservation of energy for irrigation systems is described by the Bernoulli Equation, which is expressed as:

H₁ = H₂ + h_L (8.6)

or

\(\dfrac{{V_1}^2}{2g}+Z_1+h_1=\dfrac{{V_2}^2}{2g}+Z_2+h_2+h_L\) (8.7)

where: H₁ = total head at point 1 in a system,

H₂ = total head at point 2 in a system, and

h_L = head loss during flow from point 1 to point 2.

Velocity head (h_v) can be determined graphically using Figure 8.3. The head loss from point 1 to point 2 is due to friction loss (h_f) from the resistance to flow along a pipeline and to minor losses (h_m) of energy through pipe fittings, etc. Thus,

h_L = h_f + h_m (8.8)

Expressed as pressure loss,

P_L = P_f + P_m (8.9)

where: P_L = pressure loss,

P_f = pressure loss due to friction, and

P_m = pressure loss due to minor losses.

Example 8.2

In the pipeline system shown, find the total head at the inlet into the 4-inch diameter (d = 4 in) pipeline.

Given: Z = 15 ft, P = 60 psi

Q = 400 gpm, d = 4 in

Find: h, H

\(V_m=\dfrac{Q}{A_f}\)

\(\text{Velocity head }= \dfrac{{V_m}^2}{2g}\)

Solution

\(h = 2.31 \dfrac{\text{ft}}{\text{psi}} (60\text{ psi})=139 \text{ ft}\)

\(H=15\text{ ft}+139\text{ ft}+1.6\text{ ft}=156\text{ ft} \)

\(A=\dfrac{\pi (4\text{ in})^2}{4}=12.57\text{ in}^2 \left(\dfrac{1\text{ ft}^2}{144\text{ in}^2}\right)=0.087 \text{ ft}^2\)

\(Q= \dfrac{400 \text{ gpm}}{450\text{ gpm/cfs}}=0.89\text{ cfs} \)

\(V= \dfrac{0.89\text{ cfs}}{0.087\text{ ft}^2}=10.23\text{ ft/s} \)

\(\text{Since }g=32.2\text{ ft}/\text{s}^2, 2g=64.4\text{ ft}/\text{s}^2 \)

\(\text{Velocity Head}=\dfrac{(10.23\text{ ft/s})^2}{2g}=\dfrac{(10.23\text{ ft/s})^2}{(64.4\text{ ft}/\text{s}^2)}=1.6\text{ ft}\)

\(\text{Pressure is the primary component of total head in this example.}\)

In many pressurized irrigation systems, such as sprinkler and micro-irrigation systems, velocity head is a minor component of the total head and thus it can be ignored. In this case, it is more convenient to express the Bernoulli equation in terms of pressure:

0.433z₁ + P₁ = 0.433Z₂ + P₂ + P_L

Application of this equation for level and sloping pipelines is shown in Figure 8.4. After studying Figure 8.4 you might ask yourself the question, “How do you apply Equation 8.10 when the pipeline goes up and over a hill?”

Example 8.3

What is the velocity head at point 2 in Example 8.2?

Given: Q₂ = Q₁ = 400 gpm

d₂ = 10 in

Find: \(A_2=\dfrac{\pi {d_2}^2}{4} \)

\(\text{V_m} \)

\(\text{Velocity head at point 2} \)

Solution

\(A_2 = \dfrac{\pi (10 \text{in})^2}{4}= 78.5 \text{ in}^2 = 0.545 \text{ ft}^2 \)

\(V_2 = \dfrac{(0.89 \text{ ft}^3/ \text{s})}{(0.545 \text{ft}^2)}=1.63 \text{ ft/s} \)

\(\text{Velocity head}=\dfrac{(1.69 \text{ ft/s})^2}{(64.4 \text{ ft}/\text{s}^2)}=0.04 \text{ ft} \)

This means that the velocity head in the 10-in pipe is 0.025 times the velocity head in the 4-in pipe.

Figure 8.3. Graph for determining velocity head in pipelines.

Figure 8.4. Application of pressure form of Bernoulli equation for level and sloping pipelines (velocity head changes assumed insignificant).