Lake discharge problem

One of the classic problems in open channel flow is the “Lake Discharge Problem”. In this problem, water from a lake is discharging to a rectangular channel. The system has varying characteristics of channel width, channel slope, channel roughness, and lake level. Knowing these characteristics, one should be able to solve for the discharge coming from the lake.

Introduction

In the discharge problem, the lake discharges to a rectangular channel with characteristics of width, roughness, and slope. The discharge at the lake outlet will vary based on the normal depth calculated from the lake and channel characteristics. Channel width describes the width of the channel from bank to bank. Channel roughness is a characteristic described by the Gauckler-Manning coefficient (also called Manning's n). This can be determined using a table which describes the channel in terms of vegetation, size of bed sediment, and sinuosity. If a table is unavailable, roughness can also be determined using a Wolman pebble count. Channel slope can be found using topographic maps to determine changes in elevation along the reach. The level of the lake (y) above the outlet point is also important when determining discharge, and can be determined simply by measuring the lake level.

An important variable of interest is the normal depth of the water in the channel. The normal depth is the depth of water under conditions of uniform flow. If the reach is mild, then the normal depth must be calculated using an energy balance between the energy in the lake and the energy at the lake outlet. This normal depth is then used to find the actual discharge to the mild reach. Figure 1 below illustrates the difference between the lake level and the normal depth.

Figure 1. Illustration of lake level, specific energy of lake (E₀), critical depth (y_c) and normal depth of channel (y₀) in steep channel.

Determining the discharge to a steep reach

Suppose you are determining the discharge of a lake with the following characteristics:

Slope, S = 0.01 ft

Manning roughness, n = 0.02

Channel width, w = 10 ft

Lake level above channel outlet, y = 2 ft

Using this information, you can determine the discharge of the channel and whether the reach is steep. Assume that the reach is steep. Therefore, the flow at the lake outlet is critical and the depth of the lake is equal to the critical energy of the lake.

$y_{c}={2 \over 3}E_{0}\,\!$ (Eq 1)

Where:

$y_{c}\,$ is the critical depth in L (ft)

$E_{0}\,$ is the specific energy and critical energy in L (ft)

In this example, y_c is 1.33 ft.

Use the critical depth found to determine the discharge. Since this is a rectangular channel, the equation for specific discharge, q, can be used to find the discharge per unit width. This is then multiplied by the width to find the discharge, Q. Equation 2 below shows how to calculate specific discharge.

$q={\sqrt {y_{c}^{3}g}}$  (Eq 2)

Where:

$q\,$ is the specific discharge in L²/t (ft²/s)

$y_{c}\,$ is the critical depth in L (ft)

$g\,$ is the force due to gravity L/t² (ft/s²)

In this example, q is 8.74 ft²/s, and with a width of 10 ft, Q is 87.4 ft³/s.

Next, determine the normal depth needed to solve for a discharge of the same value (in this case, 87.4 ft³/s). To do this, use the Manning’s Roughness equation for velocity, described below in Equation 3. Multiply the velocity found by the cross sectional area of the channel to determine discharge.

$v={1.49 \over n}R^{2 \over 3}S^{1 \over 2}$   (Eq 3)

Where:

$v\,$ is the velocity of the flow in L/t (ft/s)

$R\,$ is the hydraulic radius in L (ft)

$S\,$is the slope in L/L (ft/ft)

$n\,$ is Manning’s roughness coefficient

While n can be determined from a table and slope is given, both hydraulic radius and subsequently velocity depend upon the normal depth. Similarly, the area of the channel, which is needed to calculate discharge, also depends upon the normal depth. Therefore, an iterative method is needed to solve for normal depth. One simple way to do this is using the Goal Seek tool in Microsoft Excel. Using this method, the normal depth calculated is 1.20 ft. To determine whether the reach is steep, find the Froude number of the flow. A steep reach has supercritical flow, meaning the Froude number is greater than 1. To determine the Froude number, use the following equation:

$Fr={v \over {\sqrt {gy_{0}}}}$   (Eq 4)

Where:

$Fr\,$ is the Froude Number

$v\,$ is velocity in L/t (ft/s)

$g\,$ is the force due to gravity in L²/t (ft²/s)

$y_{0}\,$ is the normal depth of water in L (ft)

If the Froude number is greater than 1, the flow is supercritical and the slope is steep. The discharge remains the same for a steep slope with a defined set of lake and reach characteristics. If the Froude number is less than 1, the flow is subcritical and the discharge needs to be recalculated, as discussed below. In this instance, the Froude number is 1.17, indicating supercritical flow and a steep reach.

Determining the discharge of a mild reach

Changing any one of the channel characteristics will affect whether the reach is steep or mild. For this example, the slope of the channel will be reduced while the other channel characteristics remain constant to understand how discharge changes with the change in slope.

As mentioned above, when the flow is subcritical, defined by a Froude number less than 1, then the reach is mild. To traverse from a steep reach to a mild reach, the flow must pass through critical conditions. Critical conditions occur when the Froude number is equal to one.

In many cases, the discharge and normal depth for a mild slope will not initially be known. The discharge can be calculated through a trial and error process. First, a discharge must be assumed so that the normal depth can be calculated. For this example, the original discharge of 87.4 ft³/s will be assumed for the mild slope. The following equation solves for discharge in terms of velocity and area. Normal depth is a function of both velocity and area and can therefore be solved using this equation.

$Q=vA$  (Eq 5)

Where

$Q\,$ is discharge in L³/s (ft³/s)

$v\,$ is velocity in L/s (ft/s)

$A\,$ is area of the channel in L (ft)

Using the initial channel characteristics, but reducing the slope from 0.01 to0.004, the resulting normal depth is found by inserting the Manning’s Roughness Equation (Eq 3) in for velocity in equation 5. A trial and error process is used to find the normal depth that corresponds to the stream characteristics and a discharge of 87.4 ft³/s . This process is made easier by using Goal seek in excel. The normal depth associated with the given discharge and stream characteristics is 1.62 ft. The variables used in the calculation of this normal depth are listed in Table 1 and include normal depth (y₀), cross sectional area of the rectangular channel (A), wetted perimeter (P), hydraulic radius (R), velocity (v), and discharge (Q). To determine if the discharge and depth are possible with this particular slope, specific energy must be determined upstream of the lake outlet and at the lake outlet. Conservation of energy must occur, so the upstream energy must equal the downstream energy. Specific energy is given as:

$E={v^{2} \over 2g}+y$ (Eq 5)

Where

$E\,$ is the energy of the flow in L (ft)

$v\,$ is velocity in L/s (ft/s)

$g\,$ is gravity in L/t² (ft/s²)

$y\,$ is water depth in L (ft)

The velocity in a lake is negligible, effectively taking the first term of the energy equation to 0, and showing that the energy in a lake equals the depth of water from the lake invert elevation to the general level of the lake surface. In this example, the lake level is 2 feet higher than the outlet invert elevation; therefore, the energy in the lake is also 2 feet. The energy at the lake outlet is calculated using the velocity calculated during the trial and error process for normal depth, shown in the first iteration in Table 1 (v=5.39 ft/s ), and the normal depth at the lake outlet, or y=1.62 ft. Using equation 6, energy at the lake outlet equals 2.07 ft. This energy is greater than the energy needed to allow the current discharge to pass through the outlet.

To determine the normal depth that is necessary to create a discharge meeting the energy requirements, the trial and error process is repeated. Below, Table 1 shows how decreasing the normal depth decreases the discharge, and as a result decreases the energy at the lake outlet. Iterations are performed until a normal depth is found that satisfies the energy balance. The Goal Seek function in Microsoft Excel is a useful tool for iterative analysis. In this case, this is when the energy at the lake outlet equals 2 ft, or the lake level. The discharge needed to maintain the energy balance is 82.9 ft³/s.

Table 1. Normal depth iterations to balance energy for a slope of 0.004

Y (ft)	A (ft2)	P (ft)	R (ft)	V (ft/s)	Q (ft3/s)	E (ft)	Fr	Comments
1.62	16.2	13.2	1.22	5.39	87.4	2.07	0.75	Shows reach is mild, because Froude number is less than 1
1.5	15	13	1.15	5.18	77.8	1.91	0.75	Need to increase depth because energy is less than 2
1.55	15.5	13.1	1.18	5.27	81.7	1.98	0.75	Need to increase depth because energy is less than 2
1.56	15.6	13.1	1.19	5.30	82.25	2	0.75	Correct discharge because energy is equal to 2

Let’s repeat this process once again to show how the discharge changes when the reach becomes even milder.

Using the original channel characteristics and discharge, but changing the slope to 0.0005, the normal depth is recalculated. The first row of Table 2 shows the initial normal depth calculated from the assumed discharge of 87.4 ft³/s. The energy related to this normal depth is 3.42 ft. Because this energy is greater than the initial energy in the lake of 2 ft, the normal depth needed to meet the conservation of energy requirements is less than 3.31 feet. Let’s try a depth of 2.5 ft for the next iteration. A normal depth of 2.5 ft returns a specific energy of 2.59 ft. This energy is again greater than the energy needed to push the water down the reach. A normal depth of 1.9 ft results in an energy of 2 ft, showing that energy is conserved from the lake to the reach.

Table 2. Normal depth iterations to balance energy for a slope of 0.0005

Y (ft)	A (ft2)	P (ft)	R (ft)	V (ft/s)	Q (ft3/s)	E (ft)	Fr	Comments
3.31	33.1	16.9	1.99	2.64	87.4	3.42	0.26	Shows reach is mild, because Froude number is less than 1
2.5	25	15	1.67	2.34	58.5	2.59	0.26	Need to decrease depth because energy is greater than 2
1.8	18	13.6	1.32	2.01	36.1	1.86	0.26	Need to increase depth because energy is less than 2
1.9	19.3	13.9	1.39	2.08	40.2	2	0.26	Correct discharge because energy is equal to 2

This iterative process can be used to assess how changing other channel characteristics, such as channel width or roughness, affects the normal depth and energy of the stream.

Figure 2. Illustration of mild reach. Normal depth is the depth of water at the discharge point as well as downstream of the discharge point.

Relationship between discharge and channel mildness

While the slope of the reach is steep, meaning the Froude number is greater than or equal to one, the discharge of the reach remains the same. Once the reach reaches subcritical condition, with a Froude number less than one, the discharge gradually decreases. As the reach gets milder, the discharge tends toward zero. Table 3 provides a list of slopes and their corresponding discharges, using the given channel characteristics on this page. The discharges in this table were calculated using Goal Seek for each of the slopes.

table 3=

Figure 3 gives a visualization of what happens as slope changes for a given set of channel characteristics. Figure 3 was created using the slopes and corresponding discharges from Table 3. Again, notice how the discharge reaches zero as the slope decreases. It is also important to see how the discharge reacts to steep slopes. Slopes greater than 0.0072 result in a Froude Number greater than 1. Any slope greater than 0.0072 is a steep slope and therefore, has a discharge of 87.4 ft³/s.

Shows the process used to determine discharge at a lake outlet