Stage Discharge Tabulation for Only Orifice Flow

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1 Stage Discharge Tabulation for Only Orifice Flow DEPTH STAGE DISCHARGE (meters) (feet) (meters) (feet) (m 3 /s) (ft 3 /s) Weirs Relationships for sharp-crested, broad-crested, V-notch, and proportional weirs are provided in the following sections: Sharp Crested Weirs Typical sharp crested weirs are illustrated in Figure Equation 8-19 provides the discharge relationship for sharp crested weirs with no end contractions (illustrated in Figure 8-13a). Q = C scw L H 1.5 (8-19) Q = Discharge, m 3 /s (ft 3 /s) L = Horizontal weir length, m (ft) H = Head above weir crest excluding velocity head, m (ft) C scw = (H/H c ) [ (H/H c ) in English units] As indicated above, the value of the coefficient C SCW is known to vary with the ratio H/H c (see Figure 8-13c for definition of terms). For values of the ratio H/H c less than 0.3, a constant C SCW of 1.84 (3.33 in English units) is often used. Equation 8-20 provides the discharge equation for sharp-crested weirs with end contractions (illustrated in Figure 8-13(b)). As indicated above, the value of the coefficient C scw is known to vary with the ratio H/H c (see Figure 8-13c for definition of terms). For values of the ratio H/H c less than 0.3, a constant C scw of 1.84 (3.33 in English units) is often used. Q = C scw (L 0.2 H) H 1.5 (8-20) 8-23

2 Figure Sharp crested weirs. Sharp crested weirs will be effected by submergence when the tailwater rises above the weir crest elevation, as shown in Figure 8-13(d). The result will be that the discharge over the weir will be reduced. The discharge equation for a submerged sharp-crested weir is: (49) Q s = Q r (1 (H 2 / H 1 ) 1.5 ) (8-21) Q s = Submerged flow, m 3 /s (ft 3 /s) Q r = Unsubmerged weir flow from Equation 8-19 or 8-20, m 3 /s (ft 3 /s) H 1 = Upstream head above crest, m (ft) H 2 = Downstream head above crest, m (ft) Flow over the top edge of a riser pipe is typically treated as flow over a sharp crested weir with no end constrictions. Equation 8-19 should be used for this case. Example 8-6 Given: A riser pipe as shown in Figure 8-14 with the following characteristics: diameter (D) = 0.53 m (1.74 ft) crest elevation = 10.8 m (35.4 ft) weir height (H c ) = 0.8 m (2.6 ft) Find: Stage - discharge rating for the riser pipe between 10 m (32.8 ft) and 12.0 m (39.4 ft). 8-24

3 Figure Riser pipe. Solution: Since the riser pipe functions as both a weir and an orifice (depending on stage), the rating is developed by comparing the stage - discharge produced by both weir and orifice flow as follows: SI Units Using Equation 8-18 for orifices with D = 0.53 m yields the following relationship between the effective head on the orifice (H o ) and the resulting discharge: Q = K or D 2 H o 0.50 Q = (2.09)(0.53) 2 H o 0.50 Q = H o 0.50 Using Equation 8-19 for sharp crested weirs with C SCW = 1.84 (H/H c assumed less than 0.3), and L = pipe circumference = 1.67 m yields the following relationship between the effective head on the riser (H) and the resulting discharge: Q = C SCW L H 1.5 Q = (1.84)(1.67) H 1.5 Q = H 1.5 English Units Using Equation 8-18 for orifices with D = 1.74 ft yields the following relationship between the effective head on the orifice (H o ) and the resulting discharge: Q = K or D 2 H o 0.50 Q = (3.78)(1.74) 2 H o 0.50 Q = H o 0.50 Using Equation 8-19 for sharp crested weirs with C SCW = 3.33 (H/H c assumed less than 0.3), and L = pipe circumference = 5.5 ft yields the following relationship between the effective head on the riser (H) and the resulting discharge: Q = C SCW L H 1.5 Q = (3.33)(5.5) H 1.5 Q = (18.32 H

4 The resulting stage - discharge relationship is summarized in the following table: STAGE EFFECTIVE HEAD ORIFICE FLOW WEIR FLOW (m) (ft) (m) (ft) (m 3 /s) (ft 3 /s) (m 3 /s) (ft 3 /s) * 0.37* 0.45* 0.53* 0.59* 0.64* * 13.1* 15.9* 18.7* 20.8* 22.6* * * *Designates controlling flow. The flow condition, orifice or weir, producing the lowest discharge for a given stage defines the controlling relationship. As illustrated in the above table, at a stage of 10.9 m (35.76 ft) weir flow controls the discharge through the riser. However, at and above a stage of 11.0 m (36.09 ft), orifice flow controls the discharge through the riser. Broad-Crested Weir The equation typically used for a broad-crested weir is: (49) Q = C BCW L H 1.5 (8-22) Q = Discharge, m 3 /s (ft 3 /s) C BCW = Broad-crested weir coefficient, (2.34 to 3.32) L = Broad-Crested weir length, m (ft) H = Head above weir crest, m (ft) If the upstream edge of a broad-crested weir is so rounded as to prevent contraction and if the slope of the crest is as great as the loss of head due to friction, flow will pass through critical depth at the weir crest; this gives the maximum C value of 1.70 (3.09 in English units). For sharp corners on the broad crested weir, a minimum value of 1.44 (2.62 in English units) should be used. Additional information on C values as a function of weir crest breadth and head is given in Table

5 Table 8-1. SI Units - Broad-Crested Weir Coefficient C Values as a Function of Weir Crest. Broad-Crested Weir Coefficient C Values as a Function of Weir Crest Breadth and Head (coefficient has units of m 0.5 /sec) (1) Head (2) Breadth of Crest of Weir (m) (m) (1) Modified from Reference 49 (2) Measured at least 2.5 H c upstream of the weir Table 8-1. English Units - Broad-Crested Weir Coefficient C Values as a Function of Weir Crest. Broad-Crested Weir Coefficient C Values as a Function of Weir Crest Breadth and Head (coefficient has units of ft 0.5 /sec) (1) Head (2) Breadth of Crest of Weir (ft) (ft) (1) Table is taken from Reference

6 V-Notch Weir The discharge through a v-notch weir is shown in Figure 8-15 and can be calculated from the following equation: (49) Q = K u [tan (θ / 2)] H 2.5 (8-23) Q = Discharge, m 3 /s (ft 3 /s) θ = Angle of v-notch, degrees H = Head on apex of v-notch, m (ft) K u = 1.38, (2.5 in English units) Proportional Weir Figure V-notch weir. Although more complex to design and construct, a proportional weir may significantly reduce the required storage volume for a given site. The proportional weir is distinguished from other control devices by having a linear head-discharge relationship. This relationship is achieved by allowing the discharge area to vary nonlinearly with head. Design equations for proportional weirs are as follows: (50) Q = K u a 0.5 b (H a/3) (8-24) x / b = 1 (0.315) [arctan (y / a) 0.5 ] (8-25) K U = 2.74, (4.96 English units) Q = Discharge, m 3 /s (ft 3 /s) H = Head above horizontal sill, m (ft) Dimensions a, b, x, and y are shown in Figure

7 Figure Proportional weir dimensions Discharge Pipes Discharge pipes are often used as outlet structures for detention facilities. The design of these pipes can be for either single or multistage discharges. A single step discharge system would consist of a single culvert entrance system and would not be designed to carry emergency flows. A multistage inlet would involve the placement of a control structure at the inlet end of the pipe. The inlet structure would be designed in such a way that the design discharge would pass through a weir or orifice in the lower levels of the structure and the emergency flows would pass over the top of the structure. The pipe would need to be designed to carry the full range of flows from a drainage area including the emergency flows. For single stage systems, the facility would be designed as if it were a simple culvert. Appropriate design procedures are outlined in Hydraulic Design of Highway Culverts (HDS- 5). (2) For multistage control structures, the inlet control structure would be designed considering both the design flow and the emergency flows. A stage-discharge curve would be developed for the full range of flows that the structure would experience. The design flows will typically be orifice flow through whatever shape the designer has chosen while the higher flows will typically be weir flow over the top of the control structure. Orifices can be designed using the equations in Section and weirs can be designed using the equations in Section The pipe must be designed to carry all flows considered in the design of the control structure. In designing a multistage structure, the designer would first develop peak discharges that must be passed through the facility. The second step would be to select a pipe that will pass the peak flow within the allowable headwater and develop a performance curve for the pipe. Thirdly, the designer would develop a stage-discharge curve for the inlet control structure, recognizing that the headwater for the discharge pipe will be the tailwater that needs to be considered in designing the inlet structure. Last, the designer would use the stage-discharge curve in the basin routing procedure. 8-29

8 Example 8-7 Given: A corrugated steel discharge pipe as shown in Figure 8-14 with the following characteristics: maximum head on pipe = 0.75 m (2.3 ft) (conservative value of 5 m less than the riser height specified in Example 3-8. inlet invert = 1 m (32.8 ft) length (L) = 50 m (164 ft) slope = 4 m/m (ft/ft) roughness = 24 square edge entrance (K e = 0.5) discharge pipe outfall is free (not submerged) Runoff characteristics as defined in Example 3-8. Find: The size pipe needed to carry the maximum allowable flow rate from the detention basin. Solution: From Example 3-8, the maximum predeveloped discharge from the watershed is 0.55 m 3 /s (19.4 ft 3 /s). Since the discharge pipe can function under inlet or barrel control, the pipe size will be evaluated for both conditions. The larger pipe size will be selected for the final design. Using Chart 2 from HDS-5 (2) yields the relationship between head on the pipe and the resulting discharge for inlet control. From the chart, the pipe diameter necessary to carry the flow is 750 mm (30 in). Using Chart 6 from HDS-5 (2) yields the relationship between head on the pipe and discharge for barrel control. From the chart, the pipe diameter necessary to carry the flow is 675 mm (27 in). For the design, select pipe diameter = 750 mm (30 in) Emergency Spillway The purpose of an emergency spillway is to provide a controlled overflow relief for storm flows in excess of the design discharge for the storage facility. The inlet control structure discussed in Section is commonly used to release emergency flows. Another suitable emergency spillway for detention storage facilities for highway applications is a broad-crested overflow weir cut through the original ground next to the embankment. The transverse cross-section of the weir cut is typically trapezoidal in shape for ease of construction. Such an excavated emergency spillway is illustrated in Figure The invert of the spillway at the outfall should be at an elevation 0.3 m (1 ft) to 0.6 m (2 ft) above the maximum design storage elevation. It is preferable to have a freeboard of 0.6 m (2 ft) minimum. However, for very small impoundments (less than 0.4 to 0.8 hectare surface area) an absolute minimum of 0.3 meter of freeboard may be acceptable. (40) 8-30

9 Figure Emergency spillway design schematic. 8-31

10 Equation 8-26 presents a relationship for computing the flow through a broad-crested emergency spillway. The dimensional terms in the equation are illustrated in Figure Q = C sp b H p 1.5 (8-26) Q = Emergency spillway discharge, m 3 /s (ft 3 /s) C SP = Discharge coefficient b = Width of the emergency spillway, m (ft) H p = Effective head on the emergency spillway, m (ft) The discharge coefficient, C SP, in Equation 8-26 varies as a function of spillway bottom width and effective head. Figure 8-18 illustrates this relationship. Table 8-2, modified from Reference 51, provides a tabulation of emergency spillway design parameters. The critical slopes of Table 8-2 are based upon an assumed n = 40 for turf cover of the spillway. For a paved spillway, the n should be assumed as 15. Equations 8-27 and 8-28 can be used to compute the critical velocity and slope for spillway materials having other roughness values. V c = K SP (Q / b) 0.33 (8-27) V c = Critical velocity at emergency spillway control section, m/s (ft/s) Q = Emergency spillway discharge, m 3 /s (ft 3 /s) b = Width of the emergency spillway, m (ft) K SP = 2.14 (3.18 in English units) S c = K' SP n 2 [(V c b) / Q)] 0.33 (8-28) S c = Critical slope, m/m (ft/ft) n = Manning's coefficient V c = Critical velocity at emergency spillway control section, m/s (ft/s) Q = Emergency spillway discharge, m 3 /s (ft 3 /s) b = Width of the emergency spillway, m (ft) K SP ' = 9.84 (14.6 in English units) 8-32

11 Figure 8-18a. Discharge coefficients for emergency spillways, SI units. Figure 8-18b. Discharge coefficients for emergency spillways, English units. 8-33

12 Table 8-2. Emergency Spillway Design Parameters (SI units). H p Spillway Bottom Width, b, meters m Q V c S c 3.4% 3.3% 3.2% 3.2% 3.3% 3.3% 3.2% 3.2% 3.2% 3.2% Q V c S c 3.0% 3.0% 3.0% 3.0% 3.0% 3.0% 3.0% 3.0% 3.0% 3.0% Q V c S c 2.7% 2.7% 2.8% 2.8% 2.8% 2.8% 2.8% 2.8% 2.8% 2.8% 2.8% 2.8% 2.8% 2.8% 2.8% Q V c S c 2.5% 2.6% 2.6% 2.6% 2.6% 2.6% 2.6% 2.6% 2.6% 2.6% 2.6% 2.6% 2.6% 2.6% 2.6% Q V c S c 2.4% 2.4% 2.5% 2.5% 2.5% 2.5% 2.5% 2.5% 2.5% 2.5% 2.5% 2.5% 2.5% 2.5% 2.5% Q V c S c 2.3% 2.3% 2.4% 2.4% 2.4% 2.4% 2.4% 2.4% 2.4% 2.4% 2.4% 2.4% 2.4% 2.4% 2.4% Q V c S c 2.2% 2.2% 2.3% 2.3% 2.3% 2.3% 2.3% 2.3% 2.3% 2.3% 2.3% 2.3% 2.3% 2.3% 2.3% Q V c S c 2.1% 2.2% 2.2% 2.2% 2.2% 2.2% 2.2% 2.2% 2.2% 2.2% 2.2% 2.2% 2.2% 2.2% 2.2% Q V c S c 2. 1% 2.1% 2.1% 2.1% 2.1% 2.1% 2.1% 2.1% 2.1% 2.2% 2.2% 2.2% 2.2% 2.2% 2.2% NOTE 1. For a given H p, decreasing exit slope from S c decreases spillway discharge, but increasing exit slope from S c does not increase discharge. 2. If a slope S e steeper than S c is used, velocity V e in the exit channel will increase according to the following relationship: V e = V c (S e /S c ) After Maryland SCS 8-34

13 Table 8-2. Emergency Spillway Design Parameters (English units). H p Spillway Bottom Width, b, feet ft Q V c S c Q V c S c Q V c S c Q V c S c Q V c S c Q V c S c Q V c S c NOTE: 1. For a given H p, decreasing exit slope from S c decreases spillway discharge, but increasing exit slope from Sc does not increase discharge. 2. If a slope S e steeper than S c is used, velocity V e in the exit channel will increase according to the following relationship: V e = V c (S e /S c ) After Maryland SCS 8-35

14 Example 8-8 Given: An emergency spillway with the following characteristics: invert elev. = 11.6 m (38.0 ft) width (b) = 5 m (16.4 ft) discharge coeff. (C SP ) = See Figures 8-18 SI & English Find: The stage - discharge rating for the spillway up to an elevation of 12.0 m (39.4 ft). Solution: Using Equation 8-26 with the given parameters and for Hp of 0.3 m (0.98 ft) with C SP varies with head on spillway. Using Figure 8-18 SI units and English units yields: SI Units English Units Q = C SP b Hp 1.5 Q = C SP b Hp 1.5 Q = 1.43 (5) (0.3) 1.5 Q = 2.55 (16.4) (0.98) 1.5 Q = 1.17 m 3 /s Q = 40.6 ft 3 /s The following table provides the stage-discharge tabulation: STAGE EFFECTIVE HEAD ON SPILLWAY C SP SI C SP Eng. SPILLWAY DISCHARGE (m) (ft) (m) (ft) (m 3 /s) (ft 3 /s) ~1.35* ~1.35* ~2.4* ~2.4* *Use the value on lowest curve since no other data is available Infiltration Analysis of discharges from retention facilities requires knowledge of soil permeabilities and hydrogeologic conditions in the vicinity of the basin. Although infiltration rates are published in many county soils reports, it is advised that good field measurements be made to provide better estimates for these parameters. This is particularly important in karst areas where the hydrogeologic phenomenon controlling infiltration rates may be complex. Discharges controlled by infiltration processes are typically several orders of magnitude smaller than design inflow rates. If a retention facility includes an emergency overflow structure, it is often reasonable to ignore infiltration as a component of the discharge performance curve for the structure. However, if the retention facility is land-locked and has no outlet, it may be important to evaluate infiltration rates as they relate to the overall storage volume required (see Section 8.7 for discussion of land-locked storage). 8-36

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