fluid_mechanics
542 Chapter 10 ■ Open-Channel Flow Chapter 3 that the energy line is located a distance z 1the elevation from some datum to the channel bottom2 plus the pressure head 1 pg2 plus the velocity head 1V 2 2g2 above the datum. Therefore, Eq. 10.6 can be written as y 1 y 2 1V 2 2 V 2 12 1S 2g f S 0 2/ (10.7) If there is no head loss, the energy line is horizontal 1S f 02, and the total energy of the flow is free to shift between kinetic energy and potential energy in a conservative fashion. In the specific instance of a horizontal channel bottom 1S 0 02 and negligible head loss 1S f 02, Eq. 10.7 simply becomes y 1 y 2 1V 2 2 V 2 12 2g E The specific energy is the sum of potential energy and kinetic energy (per unit weight). q = constant y 10.3.1 Specific Energy The concept of the specific energy or specific head, E, defined as E y V 2 2g (10.8) is often useful in open-channel flow considerations. The energy equation, Eq. 10.7, can be written in terms of E as E 1 E 2 1S f S 0 2/ (10.9) If head losses are negligible, then S f 0 so that 1S f S 0 2/ S 0 / z 2 z 1 and the sum of the specific energy and the elevation of the channel bottom remains constant 1i.e., E 1 z 1 E 2 z 2 , a statement of the Bernoulli equation2. If we consider a simple channel whose cross-sectional shape is a rectangle of width b, the specific energy can be written in terms of the flowrate per unit width, q Qb Vybb Vy, as E y 2gy 2 (10.10) which is illustrated by the figure in the margin. For a given channel of constant width, the value of q remains constant along the channel, although the depth, y, may vary. To gain insight into the flow processes involved, we consider the specific energy diagram, a graph of E E1y2, with q fixed, as shown in Fig. 10.7. The relationship between the flow depth, y, and the velocity head, V 2 2g, as given by Eq. 10.8 is indicated in the figure. q2 y V 2 2g y = E y sub y q = q" > q' y c E min E y sup q = q' y neg F I G U R E 10.7 diagram. Specific energy
10.3 Energy Considerations 543 For a given value of specific energy, a flow may have alternate depths. For given q and E, Eq. 10.10 is a cubic equation 3y 3 Ey 2 1q 2 2g2 04 with three solutions, y sup , y sub , and y neg . If the specific energy is large enough 1i.e., E 7 E min , where E min is a function of q2, two of the solutions are positive and the other, y neg , is negative. The negative root, represented by the curved dashed line in Fig. 10.7, has no physical meaning and can be ignored. Thus, for a given flowrate and specific energy there are two possible depths, unless the vertical line from the E axis does not intersect the specific energy curve corresponding to the value of q given 1i.e., E 6 E min 2. These two depths are termed alternate depths. For large values of E the upper and lower branches of the specific energy diagram 1y sub and y sup 2 approach y E and y 0, respectively. These limits correspond to a very deep channel flowing very slowly 1E y V 2 2g S y as y S with q Vy fixed2, or a very high-speed flow in a shallow channel 1E y V 2 2g S V 2 2g as y S 02. As is indicated in Fig. 10.7, y sup 6 y sub . Thus, since q Vy is constant along the curve, it follows that V sup 7 V sub , where the subscripts “sub” and “sup” on the velocities correspond to the depths so labeled. The specific energy diagram consists of two portions divided by the E min “nose” of the curve. We will show that the flow conditions at this location correspond to critical conditions 1Fr 12, those on the upper portion of the curve correspond to subcritical conditions 1hence, the “sub” subscript2, and those on the lower portion of the curve correspond to supercritical conditions 1hence, the “sup” subscript2. To determine the value of E min , we use Eq. 10.10 and set dEdy 0 to obtain dE dy 1 q2 gy 0 3 or y c a q2 g b 13 (10.11) where the subscript “c” denotes conditions at E min . By substituting this back into Eq. 10.10 we obtain E min 3y c 2 By combining Eq. 10.11 and V c qy c , we obtain V c q 1y3 2 c g 12 2 1gy y c y c c y Fr < 1 Fr > 1 E Fr = 1 or Fr c V c1gy c 2 12 1. Thus, critical conditions 1Fr 12 occur at the location of E min . Since the layer is deeper and the velocity smaller for the upper part of the specific energy diagram 1compared with the conditions at E min 2, such flows are subcritical 1Fr 6 12. Conversely, flows for the lower part of the diagram are supercritical. This is shown by the figure in the margin. Thus, for a given flowrate, q, if E 7 E min there are two possible depths of flow, one subcritical and the other supercritical. It is often possible to determine various characteristics of a flow by considering the specific energy diagram. Example 10.2 illustrates this for a situation in which the channel bottom elevation is not constant. E XAMPLE 10.2 Specific Energy Diagram—Quantitative GIVEN Water flows up a 0.5-ft-tall ramp in a constant width rectangular channel at a rate q 5.75 ft 2 s as is shown in Fig. E10.2a. 1For now disregard the “bump.”2 The upstream depth is 2.3 ft and viscous effects are negligible. FIND Determine the elevation of the water surface downstream of the ramp, y 2 z 2 .
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10.3 Energy Considerations 543<br />
For a given value<br />
of specific energy, a<br />
flow may have alternate<br />
depths.<br />
For given q and E, Eq. 10.10 is a cubic equation 3y 3 Ey 2 1q 2 2g2 04 with three<br />
solutions, y sup , y sub , and y neg . If the specific energy is large enough 1i.e., E 7 E min , where E min is a<br />
function of q2, two of the solutions are positive and the other, y neg , is negative. The negative root,<br />
represented by the curved dashed line in Fig. 10.7, has no physical meaning and can be ignored.<br />
Thus, for a given flowrate and specific energy there are two possible depths, unless the vertical<br />
line from the E axis does not intersect the specific energy curve corresponding to the value of q<br />
given 1i.e., E 6 E min 2. These two depths are termed alternate depths.<br />
For large values of E the upper and lower branches of the specific energy diagram 1y sub and y sup 2<br />
approach y E and y 0, respectively. These limits correspond to a very deep channel flowing<br />
very slowly 1E y V 2 2g S y as y S with q Vy fixed2, or a very high-speed flow in a<br />
shallow channel 1E y V 2 2g S V 2 2g as y S 02.<br />
As is indicated in Fig. 10.7, y sup 6 y sub . Thus, since q Vy is constant along the curve, it<br />
follows that V sup 7 V sub , where the subscripts “sub” and “sup” on the velocities correspond to the<br />
depths so labeled. The specific energy diagram consists of two portions divided by the E min “nose”<br />
of the curve. We will show that the flow conditions at this location correspond to critical conditions<br />
1Fr 12, those on the upper portion of the curve correspond to subcritical conditions 1hence, the<br />
“sub” subscript2, and those on the lower portion of the curve correspond to supercritical conditions<br />
1hence, the “sup” subscript2.<br />
To determine the value of E min , we use Eq. 10.10 and set dEdy 0 to obtain<br />
dE<br />
dy 1 q2<br />
gy 0 3<br />
or<br />
y c a q2<br />
g b 13<br />
(10.11)<br />
where the subscript “c” denotes conditions at E min . By substituting this back into Eq. 10.10 we<br />
obtain<br />
E min 3y c<br />
2<br />
By combining Eq. 10.11 and V c qy c , we obtain<br />
V c q 1y3 2<br />
c g 12 2<br />
1gy<br />
y c y c<br />
c<br />
y<br />
Fr < 1<br />
Fr > 1<br />
E<br />
Fr = 1<br />
or Fr c V c1gy c 2 12 1. Thus, critical conditions 1Fr 12 occur at the location of E min . Since the<br />
layer is deeper and the velocity smaller for the upper part of the specific energy diagram 1compared<br />
with the conditions at E min 2, such flows are subcritical 1Fr 6 12. Conversely, flows for the lower<br />
part of the diagram are supercritical. This is shown by the figure in the margin. Thus, for a given<br />
flowrate, q, if E 7 E min there are two possible depths of flow, one subcritical and the other<br />
supercritical.<br />
It is often possible to determine various characteristics of a flow by considering the specific<br />
energy diagram. Example 10.2 illustrates this for a situation in which the channel bottom elevation<br />
is not constant.<br />
E XAMPLE 10.2<br />
Specific Energy Diagram—Quantitative<br />
GIVEN Water flows up a 0.5-ft-tall ramp in a constant width<br />
rectangular channel at a rate q 5.75 ft 2 s as is shown in Fig.<br />
E10.2a. 1For now disregard the “bump.”2 The upstream depth is<br />
2.3 ft and viscous effects are negligible.<br />
FIND Determine the elevation of the water surface downstream<br />
of the ramp, y 2 z 2 .