7.7-hydraulic and energy grade

7.7 Hydraulic and Energy Grade LinesThis section introduces the hydraulic grade line (HGL) and the energy grade line (EGL), which are graphical

representations that show head in a system. This visual approach provides insights and helps one locate and

correct trouble spots in the system (usually points of low pressure).

The EGL, shown in Fig. 7.7, is a line that indicates the total head at each location in a system. The EGL is related

to terms in the energy equation by

(7.38)

Notice that total head, which characterizes the energy that is carried by a flowing fluid, is the sum of velocity

head, the pressure head, and the elevation head.

Figure 7.7 EGL and HGL in a straight pipe.

The HGL, shown in Fig. 7.7, is a line that indicates the piezometric head at each location in a system:

(7.39)

Since the HGL gives piezometric head, the HGL will be coincident with the liquid surface in a piezometer as

shown in Fig.7.7. Similarly, the EGL will be coincident with the liquid surface in a stagnation tube.

Tips for Drawing HGLs and EGLs

1. In a lake or reservoir, the HGL and EGL will coincide with the liquid surface. Also, both the HGL and

EGL will indicate piezometric head. For example, see Fig. 7.7.

2. A pump causes an abrupt rise in the EGL and HGL by adding energy to the flow. For example, see Fig.

7.8.

Hydraulic and Energy Grade Lines http://edugen.wiley.com/edugen/courses/crs2436/crowe9771/crowe9771...

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3. For steady flow in a Pipe of constant diameter and wall roughness, the slope (∆hL/∆L) of the EGL and

the HGL will be constant. For example, see Fig. 7.7

4. Locate the HGL below the EGL by a distance of the velocity head (αV2/2g).

5. Height of the EGL decreases in the flow direction unless a pump is present.

6. A turbine causes an abrupt drop in the EGL and HGL by removing energy from the flow. For example,

see Fig. 7.9.

7. Power generated by a turbine can be increased by using a gradual expansion at the turbine outlet. As

shown in Fig. 7.9, the expansion converts kinetic energy to pressure. If the outlet to a reservoir is an

abrupt expansion, as in Fig. 7.11, this kinetic energy is lost.

8. When a pipe discharges into the atmosphere the HGL is coincident with the system because p/γ = 0 at

these points. For example, in Figures 7.10 and 7.12, the HGL in the liquid jet is drawn through the jet

itself.

9. When a flow passage changes diameter, the distance between the EGL and the HGL will change (see Fig.

7.10 and Fig. 7.11) because velocity changes. In addition, the slope on the EGL will change because the

head loss per length will be larger in the conduit with the larger velocity (see Fig. 7.11).

10. If the HGL falls below the pipe, then p/γ is negative, indicating subatmospheric pressure (see Fig. 7.12)

and a potential location of cavitation.

Figure 7.8 Rise in EGL and HGL due to Pump.

Figure 7.9 Drop in EGL and HGL due to turbine.


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Figure 7.10 Change in HGL and EGL due to flow through a nozzle.

Figure 7.11 Change in EGL and HGL due to change in diameter of pipe.

Figure 7.12 Subatmospheric pressure when pipe is above HGL.

The recommended procedure for drawing an EGL and HGL is shown in Example 7.6. Notice how the tips from

pp. 233–234 are applied.


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EXAMPLE 7.6 EGL A%D HGL FOR A SYSTEM

A pump draws water (50°F) from a reservoir, where the water-surface elevation is 520 ft, and forces

the water through a pipe 5000 ft long and 1 ft in diameter. This pipe then discharges the water into a

reservoir with water-surface elevation of 620 ft. The flow rate is 7.85 cfs, and the head loss in the

pipe is given by

Determine the head supplied by the pump, hp, and the power supplied to the flow, and draw the HGL

and EGL for the system. Assume that the pipe is horizontal and is 510 ft in elevation.

Problem Definition

Situation: Water is pumped from a lower reservoir to a higher reservoir.

Find:

1. Pump head (in ft).

2. Power (in hp) supplied to the flow.

3. Draw HGL. Draw EGL.

Properties: Water (50°F), Table A.5: γ = 62.4 lbf/ft3.

Sketch:

Plan

1. Apply the energy equation 7.29 between sections 1 and section 2.

2. Calculate terms in the energy equation.

3. Find the power by applying the power equation 7.30a.

4. Draw the HGL and EGL by using the tips given on p. 270.

Solution

1. Energy equation (general form)


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· Velocity heads are negligible because V1 ≈ 0 and V2 ≈ 0.

· Pressure heads are zero because p1 = p2 = 0 gage.

· ht = 0 because there are no turbines in the system.

Interpretation: Head supplied by the pump provides the energy to lift the fluid to a higher

elevation plus the energy to overcome head loss.

2. Calculations of terms in the energy equation

· Calculate V using the flow rate equation.

· Calculate head loss.

· Calculate hp.

3. Power

4. HGL and EGL

· From Tip 1 on p. 233, locate the HGL and EGL along the reservoir surfaces.

· From Tip 2, sketch in a head rise of 178 ft corresponding to the pump.

· From Tip 3, sketch the EGL from the pump outlet to the reservoir surface. Use the fact

that the head loss is 77.6 ft. Also, sketch EGL from the reservoir on the left to the pump

inlet. Show a small head loss.

· From Tip 4, sketch the HGL below the EGL by a distance of V2/2g ≈ 1.6 ft.

· From Tip 5, check the sketches to ensure that EGL and HGL are decreasing in the

direction of flow (except at the pump).

Sketch: HGL (dashed black line) and EGL (solid blue line)


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7.7-hydraulic and energy grade

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