7.7-Hydraulic and Energy Grade ..

Hydr aulic and Ener gy Grade Lines

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7.7 Hydraulic and Energy Grade Lines This 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 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 wh ich characterizes characterizes the energy that is carried by b y a flowing fluid, is the sum of velocit v elocity y head, the pressure head, and the elevation head.

Figuree 7.7 Figur 7 .7  EGL and HGL in a straight pipe.

The HGL, 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 reser voir,, the HGL HGL and EGL will coincide with the liquid surface. Also, both bo th the HGL HGL and EGL will indicate piezometric head. For example, see Fig. 7.7.

2.

A pump causes caus es an abrupt abru pt rise in the EGL EGL and HGL HGL by adding energy to the flow. For example, see Fig. 7.8.

Hydraulic and Energy Grade Lines

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3.

For steady flow in a Pipe of constant diameter and wall roughness, the slope (∆h L/∆ 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 (αV 2/2 g ).

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.

Hydraulic and Energy Grade Lines

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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 pr ocedure for drawing an EGL and HGL is shown in Example 7.6. Notice how the tips from  pp. 233–234 are applied.

Hydraulic and Energy Grade Lines

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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 p ipe 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 h ead loss in the  pipe is given by

Determine the head supplied by the pump, h p, 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/ft 3. 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)

Hydraulic and Energy Grade Lines

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· Velocity heads are negligible because V 1 ≈ 0 and V 2 ≈ 0. · Pressure heads are zero because p1 = p2 = 0 gage. · h t  = 0 because there are no turbines in the system.

Interpretation: Head supplied by the pu mp 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 h p.

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  V 2/2 g ≈ 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)

Hydraulic and Energy Grade Lines

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