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Rectangular Weir Calculations
Geometric notions
To perform the calculations necessary to describe and manage the flow in a weir, it is necessary to examine the geometry of the storage area created by the weir. This, in particular, allows us to determine the configuration of the weir when occupied by a specified volume of water, or conversely, when the volume of water in the weir pool is known, to determine the depth and weir extent.
For the purpose of the Weir model with rectangular geometry it is assumed that:
- The reach is of constant width in the neighborhood of the weir
- The river bed slopes downward at a constant angleÂ Î¸Â to the horizontal
- The weir gates are lowered after the reach has been flowing with the gates raised
- Some of the water previously assumed to have been routed downstream will now be stored behind the weir, occupying a wedge-shaped region.
Letâ€™s investigate the dimensions of this wedge-shaped volume and how they change as the water level rises and falls.Â Figure 34Â represents a cross-section of the weir, with the areas and distances labelled. Because the weir storage relationship is assumed to be constant for the length of the weir, the ratios of the areas outlined in the figure are equal to the ratios of the corresponding volumes of water. The wedge-shaped volume corresponds to the lighter-shaded triangle.
Figure 34. Configuration of water in weir
The water in the weir is regarded as made up of the water in the wedge region and the water occupying the region below the wedge, which is essentially the channel in which the water runs when the weir is not activated, and has a constant depthÂ hr. The total depthÂ hÂ at the weir wall can be expressed as:
Â
Equation 142 |
where
h_{w}Â is the height of the wedge.
Figure 35. Weir pool
The lighter-shaded area inÂ Figure 35Â is the weir pool and its lengthÂ bÂ can also be expressed in the form:
Equation 143 |
where
b_{w}Â is the distance from the weir wall to the pointÂ WÂ where the surface of the weir pool meets the top edge of the channel of depthÂ hrÂ andÂ brÂ is the length of the lighter-shaded triangle in Figure 36Â
Figure 36. Another similar triangle
If the inflow into the weir is greater than the outflow from the weir, the surface level will rise. The points labelledÂ WÂ andÂ BÂ inÂ Figure 34, where the surface meets the top of the routing area and where it meets the bed, respectively, will then move rightwards and upwards parallel to the river bed, causing the triangular cross-section to expand but leaving its proportions and internal angles unchanged. Also the lighter-shaded weir pool area inÂ Figure 35Â has the same interior angles, so its sides are in the same proportions too, and so are those of the small lighter-shaded triangle inÂ Figure 36.
Hence ifÂ hÂ andÂ h_{w}Â are the total depth of the weir pool and the depth of the wedge-shaped region at the weir, andÂ bÂ andÂ b_{w}Â are the length of the weir extent and the length of the top edge of the wedge, then as the wedge expands (or contracts) the ratiosÂ h/b,Â h_{wÂ }/Â b_{w}Â andÂ h_{rÂ }/Â b_{r}Â all maintain a common valueÂ DÂ (which in fact is equal to tanÂ Î¸). In particular, ifÂ h_{max}Â andÂ b_{max}Â are the values ofÂ hÂ andÂ bÂ when the weir pool reaches its maximal value, then:
Equation 144 |
IfÂ V_{max}Â is the volume of the weir pool when this maximal level in the weir is attained, then:
Equation 145 |
where
WÂ is the width of the reach. Similarly, if is the volume of the weir pool, then
Equation 146 |
so that:
Equation 147 |
UsingÂ Equation 144, replaceÂ hÂ andÂ h_{max}Â inÂ Equation 147Â byÂ b Ã— DÂ andÂ b_{max}Â Ã— D, respectively, and then cancel the common terms Â½ andÂ WÂ andÂ D. This gives:
Equation 148 |
or, equivalently:
Equation 149 |
In words, this says that when the volume of the weir pool reaches a certain proportion of its maximal value, the length of the backflow reaches the square root of the same proportion of its maximal value.
NowÂ Equation 144Â implies that:
Equation 150 |
soÂ Equation 149Â can also be expanded to:
Equation 151 |
In exactly the same way, the volumeÂ VwÂ of the wedge is given by:
Equation 152 |
Divide byÂ Equation 145Â again and note thatÂ b_{wÂ }/ b_{max}Â = h_{wÂ }/ h_{max}; this yields
Equation 153 |
Finally compute in the same way the volumeÂ V_{0}Â of the wedge corresponding to the small triangleÂ BWWâ€™Â and make the substitutionÂ hrÂ = b_{r}Â Ã— D, again usingÂ Equation 144:
Equation 154 |
DividingÂ Equation 154Â byÂ Equation 145Â gives:
Equation 155 |
and this leads to an expression similar toÂ Equation 149Â andÂ Equation 153Â for the ratioÂ b_{rÂ }/ b_{max}:
Equation 156 |
In applying these equations, it is convenient to choose units to makeÂ b_{max}Â = 1, so thatÂ Equation 149,Â Equation 153Â andÂ Equation 156Â reduce to:
Equation 157 |
Equation 158 |
Equation 159 |
These equations express the three distanceÂ b, b_{w}Â andÂ b_{r}Â in terms of corresponding volumesÂ V, V_{w}Â andÂ V_{0}.
It is also useful to be able to calculate the routing depth, expressed as a proportionÂ h_{rÂ }/ hÂ of the weir pool depth, in terms of the routing volumeÂ V_{routing}Â and the weir pool volumeÂ V. This is provided by the following equation, which will be derived below:
Equation 160 |
With the assumptionÂ b_{max}Â = 1,Â Equation 160Â takes the form:
Equation 161 |
To establishÂ Equation 160Â , and henceÂ Equation 162, note first thatÂ V_{routing}Â represents the volume of routing flow when the weir pool is filled to its maximal extentÂ -Â that is, when the length of the weir pool equalsÂ b_{max}. Since the routing depthÂ h_{r}Â is constant, the part of the routing volume situated downstream of the pointÂ BÂ is obtained by scaling downÂ V_{routing}Â by the factorÂ bÂ / b_{max}Â (seeÂ Figure 37). That is, it is equal to:
Figure 162 |
Next observe that, inÂ Figure 37,Â the parallelogram area formed by the channel bed, a line parallel to the channel bed, the weir wall and a vertical line that intersects the channel bed at point BÂ isÂ twice the size of the weir pool.
Figure 37. Proportional routing height
Therefore the volume contained by this parallelogram is equal toÂ 2 Ã— V. The smaller lighter-shaded parallelogram representing the routing volume differs from the large parallelogram only along the vertical dimension, so the areas of these two parallelograms (and the volumes of water they represent) are in the same proportion as their respective depthsÂ -Â that is, in the ratioÂ h_{r}Â : h. Consequently, the value ofÂ h_{r}Â / hÂ must be the same as the corresponding fraction involving the two volumes. That is:
Equation 163 |
which simplifies immediately toÂ Equation 160.
Finally, to obtain an equation expressingÂ b_{w}Â in terms ofÂ h_{r}Â andÂ b, recall thatÂ Equation 144Â implies thatÂ h_{rÂ }/ h = b_{rÂ }/Â b, whileÂ Equation 143Â implies thatÂ b_{w}Â = b - b_{r}, so that:
Figure 164 |
and hence:
Equation 165 |
Modelling the flow
As usual the reach is broken down into divisions and the study of the flow involves treating the flows into and out of each division and the change in storage volume in that division, computed for each time step. This section will give a general picture of the steps involved in building the model, without going into any details.
Step 1
Routing flow
The first step again involves studying the flow in a situation where the weir gates are raised. Then the flows and storage can be computed as for a reach, where the outflow from each division is the inflow to the next. This gives us, for each division and each time stepÂ t, an inflow , an outflow and a storage volume. The outflow from the most downstream division will be the maximum outflow that can occur from the weir in the current time step.
Step 2
Calculating proportional height and back flow
Next assume that the weir gates are in place and examine the flow into the weir over a time step, assuming that at the start of the time step the water in the weir pool occupies volume and the routing volume is. To carry out further calculations we need to determine the dimensions of the weir pool at this stage.
Equation 149Â shows that the proportional backflow bÂ / b_{max}Â (to the pointÂ B) and the proportional depthÂ hÂ /Â h_{max}Â of the weir pool are given by:
Equation 166 |
Now use equationÂ Equation 164, withÂ V_{pool}Â andÂ V_{routing}Â replaced byÂ V^{t-1}_{pool}Â andÂ V^{t-1}_{routing}, to obtain:
Â
Equation 167 |
Next equationÂ Equation 166Â shows that the lengthÂ b_{w}Â of the wedge is given by:
Equation 168 |
Knowledge of the value ofÂ bÂ / b_{max}Â tells us the location of the pointÂ B. The division containingÂ BÂ is the division in which the upstream flow enters the weir pool. If the divisions are numbered from the most upstream (number 1) to the most downstream (numberÂ ndiv, say), andÂ BÂ lies in the division number div_{inflow}, then the value ofÂ div_{inflow}Â is given by:
Equation 169 |
where
Â denotes the integral part ofÂ xÂ (that is, the largest integer â‰¤Â x).
Step 3
Calculating inflow from upstream flows (Part 1)
The inflow volume into the weir for the current time step is calculated using the routing inflow from upstream calculated inÂ Step 1, including any calculated additional fluxes (groundwater, loss and time series), and subtracting the increment in the upstream storage between the current and previous time step, which represents the part of the inflow volume that needs to be apportioned to routing (seeÂ Figure 38). Since the upstream water enters the weir pool in division div_{inflow}, the routing inflow from which this increment is to be subtracted is.
Figure 38. Change of weir configuration during one time step
The upstream volume for the current time step is the lighter-shaded area inÂ Figure 39.
Figure 39. Upstream flow calculation
The volumeÂ V_{0}Â in the small regionÂ WWâ€™BÂ is also related toÂ V_{max}Â by equationÂ Equation 159, from which it follows that:
Equation 170 |
FromÂ Figure 39Â it is clear that the volume of water upstream of the weir can be obtained by subtracting the volumeÂ V_{0}, just calculated, from the total water volume in the routing channel beyond the vertical line throughÂ W. However, if the maximal length of the weir extent is taken to be 1, then the distance ofÂ WÂ from the wall isÂ b_{w}, the horizontal extent of the part of the routing channel beyondÂ WÂ to the farthest point isÂ 1 - b_{w}. As the channel is of uniform depth, it follows that the volume filling this region isÂ (1 - b_{w})Â Ã—Â V_{routing}. It now follows that the upstream volumeÂ upstreamVol_{current}Â for the current time step is given by:
Equation 171 |
We also need to calculateÂ upstreamVol_{previous}, the upstream volume at the end of the previous time step. Once we have this figure, then, as stated above, we can derive the inflow into the weir from the formula:
Equation 172 |
However, the calculation of the upstream volume for the end of a time step (which enables us to computeÂ upstreamVol_{previous}) is deferred for now, as it depends on results to be covered in the following sections.
Step 4
Updating the weir storage table
The Source storage table displays the storage volume and surface area for a number of values of the water depth.
The program adjusts the values of storage volume input by the user to account for the presence of routing volume, which is excluded from the storage calculation. The input storage volume is adjusted for every value in the table by removing the proportion of volume that needs to be reckoned as routing volume. To determine this proportion, calculations like those carried out in the previous paragraphs are performed for each row in the table. The calculation for a particular row starts with the volume figure, sayÂ V, for that row and uses it in place of the volume in equationÂ Equation 167Â to derive corresponding values forÂ h_{r}Â andÂ b, and then calculates a value forÂ b_{w}Â from equationÂ Equation 169. These values describe how the water is distributed in the weir pool for the particular value ofÂ VÂ and the dimensions of the wedge and so can be used to determine the volume of the routing channel (the lighter-shaded area inÂ Figure 40). This quantity is now subtracted fromÂ VÂ and the resulting figure is taken as the volume for the current row.
Figure 40. Updating storage volumes
Step 5
Calculating the maximum and minimum release rates at the default outlet
The minimum and maximum operating levels are now used to adjust the minimum and maximum release curves on regulated outlets from the weir.
Assuming that the operating levels set by the user are within the regulated range of the weir, set the minimum and maximum release curves values to zero whenever their values are less than or equal to the minimum operating level on all regulated outlets. Then adjust the minimum release curve on the default outlet according the following equation, which is designed to maintain the weir operating level at the maximum operating level during periods of high flow:
Equation 173 |
This means that any inflow that will cause the weir to go above its maximum operating level will be released down the default outlet path.
Step 6
Calculating the weir pool volume and the resulting release for the current time step
The calculation of the weir pool volume and the release at the end of the current time step is carried out by the same general method used for general storage nodes. It involves calculating fluxes (inflow, evaporation, groundwater, etc), as described in the storage node section. These calculations are integrated across the time step, meaning that there is no need to recalculate volumes and level for any intermediate times during the time step.
Once the volume figures have been updated, Source calculates values for the various quantities associated with the model, including net evaporation fluxes.
Step 7
Calculating the wedge dimensions based on the calculated weir pool volume and routing volume at the end of the time step
At this stage we know the weir pool volume and we know that any changes to the volume must produce a corresponding change in the size of the wedge, but we need to determine the actual dimensions of the wedge at the end of the step. For this we simply repeat the calculations ofÂ Step 2, but substitute the volumeÂ V^{t}_{pool}Â in place ofÂ V^{t-1}_{pool}.Â Equation 167Â now becomes:
Equation 174 |
Next useÂ Equation 164, replacingÂ VÂ andÂ V_{routing}Â byÂ V^{t}_{poolÂ }andÂ V^{t}_{routing}, to derive an expression forÂ h_{rÂ }/ h:
Â
Equation 175 |
which can be used to give an expression forÂ b_{w}:
Equation 176 |
Step 8
Determining the impact of the changes to the wedge on each division and adjust the net evaporation calculations as necessary
After carrying outÂ Step 4, we can determineÂ O^{t}_{weir}, the total weir outflow volume from all outlets, at the end of the time step. To determine the wedge volumeÂ V^{t}_{wedge}Â we subtractÂ O^{t}_{weirÂ }from the overall estimated routing volumeÂ O_{trouting}Â fromÂ Step 1:
Equation 177 |
The calculation ofÂ O^{t}_{routing}Â described inÂ Step 1Â includes providing an estimate of the net evaporation fluxÂ Ev^{t}_{routing}Â (div)Â for each division. In addition it was mentioned inÂ Step 6Â that a net evaporation fluxÂ Ev^{t}_{storage}Â is also calculated for the weir pool after the storage table is updated. However, these two kinds of estimate are calculated independently and need to be combined into a single figure for each division, reflecting the relative proportions of weir storage volume and pure routing volume that make up the total for that division. This is done is by a procedure that loops through the divisions, starting from the most downstream one and working up to the most upstream one, and derives a suitable evaporation figure for each division in turn. The following paragraphs outline the procedure, suppressing some details.
For each division the relative contributions ofÂ Ev^{t}_{routingÂ }(div)Â andÂ Ev^{t}_{storage}Â are initially assigned as follows, depending on where the pointÂ WÂ (where the wedge meets the routing channel) is situated in relation to the division. There are three cases:
- WÂ is upstream of the current division. In this case, the division lies entirely within the weir pool and the net evaporation flux is taken to be a fraction ofÂ Ev^{t}_{storage}, calculated according to the relative extent of the weir pool occupied by the divisionÂ -Â that is,Â Ev^{t}_{storage}Â Ã— (length of division)/(length of weir pool).
- WÂ is downstream of the current division. In this case, the division is entirely upstream of the weir pool and the net evaporation flux is taken to be justÂ Ev^{t}_{routing}(div).
- WÂ lies inside the current division. This is the case shown inÂ Figure 42. In this case the division overlaps the edge of the wedge and the net evaporation flux is taken to be a weighted combination ofÂ Ev^{t}_{storage}Â andÂ Ev^{t}_{routing}(div), the weights being assigned to reflect the relative volumes associated with each of the two shaded areasÂ -Â that is, according to the relative distances ofÂ WÂ from the downstream end of the division and from the upstream end of the division, respectively.
The rules just described assign an initial value to the net evaporation flux. However, before processing the next division certain adjustments are made to the wedge volumeÂ V^{t}_{wedge}Â to ensure that the net evaporation flux just calculated does not deviate too much from the initial routing estimate. These are then carried forward into the next step of the processing and the corresponding total volumeÂ V_{t}(div)Â in the division is stored.
Figure 42. Allocating evaporation components according to volumes
At the end of the iterative procedure, the figureÂ upstreamVol_{previous}Â can be calculated. First add the stored valuesÂ V_{t}(div)Â from each division:
Equation 178 |
where
ndivÂ is the number of divisions.
Then, using the weir pool volumeÂ V_{pool}Â calculated inÂ Step 6, we set:
Equation 179 |
Step 9
Calculating inflow from upstream flows (Part 2)
Now that the values ofÂ upstreamVol_{current}Â andÂ upstreamVol_{previous}Â have been calculated, inÂ Step 3Â andÂ Step 8Â respectively, we can now substitute these values in equationÂ Equation 173Â and derive an expression for the inflowÂ I^{t}_{weirÂ }into the weir in the current time step.