Draw (to scale) the hydraulic grade line (HGL) and the energy grade line (EGL) of the syst...

Draw (to scale) the hydraulic grade line (HGL) and the energy grade line (EGL) of the system in Figure 4.3.6 (example 4.3.5). Take the length of each pipe as given and neglect the height of the elbows.

Figure 4.3.6 Pipe system for example 4.3.5.

EXAMPLE 4.3.5

Water flows from reservoir 1 to reservoir 2 through a 12-in diameter, 600-ft long pipeline as shown in Figure 4.3.6. The reservoir 1 surface elevation is 1000 ft and reservoir 2 surface elevation is 950 ft. Consider the minor losses due to the sharp-edged entrance, the butterfly valve (θ = 20°), the two bends (90° elbow), and the sharp-edged exit. The pipe is galvanized iron with ks = 0.0005 ft. Determine the discharge from reservoir 1 to reservoir 2. For 120°F, v = 0.609 × 10−5 ft2/s.

SOLUTION

The energy equation between 1 and 2 is

Table 4.3.1 Minor Loss Coefficients for Pipe Flow

*Loss coefficients for partially open conditions may vary widely. Individual manufacturers should be consulted for specific conditions.

Source: Adapted from Velon and Johnson (1993).

For the reservoir surface p1/γ = 0 and V2/2g = 0, so

where

The minor losses are

Table 4.3.2 Loss Coefficients for Various Transitions and Fittings

(a) ASHRAE (1977)

(b) Streeter (1961)

(c) Beij (1938)

(d) Idel’ chik (1966)

Source: After Roberson et al. (1988).

where Ke = 0.5, Kelbow = 0.25, and Kvalve = 1.5. The energy equation is now expressed as

Assuming fully turbulent flow and using ks/D = 0.0005/1 = 0.0005, we get f = 0.0165 from Figure 4.3.5, then

Compute . Referring to Figure 4.3.5 (Moody diagram), we see that the value of f is OK. Now use the continuity equation to determine Q:

Figure 4.3.5 Resistance coefficient f versus Re (from Moody (1944)).