Tailwater and upstream flooding #

Suppose we have a drain that discharges into a river. The river’s water level at the outfall is the drain’s tailwater. When that level rises during a flood, it can raise the water level in the drain and cause overflow over the drain bank. We need to account for this tailwater effect in hydraulic modelling. A drain that appears adequate for the upstream inflow can still overflow because a high downstream water level restricts its discharge. If the downstream head becomes high enough to reverse the flow, river water can also enter the drain. Both restricted discharge and backflow can increase the volume of water stored in the drain, leaving less storage available for incoming runoff.

The below diagram gives an illustration of the above scenario:

The solid blue line shows the high-river surface continuing upstream into the drain; it exceeds the grey bank line where the red arrow marks overflow. The dashed cyan line shows the low-river surface over the same brown bed. We have changed the receiving-water level, which the drain must discharge against.

The usual uniform-flow capacity check uses Manning’s equation:

\displaystyle Q=\frac{1}{n}AR^{2/3}\sqrt{S_0}.

Here Q is discharge in cubic metres per second, A the flowing section area in square metres, P the wetted perimeter in metres, R=A/P the hydraulic radius in metres, n Manning’s roughness coefficient in SI units, and S_0 the dimensionless bed slope, positive downward.

For a given drain section and slope, this calculation gives the uniform-flow discharge, or we solve it for the depth needed to carry a specified discharge. But the water depth at the downstream end of the drain is not an input to this capacity check. It gives the same result whether the river is low or high. On its own, it therefore cannot account for a river that holds water back in the drain and raises its upstream water level.

The effect of a high river level #

Return to the first diagram. When the river rises high enough to control the water level at the drain outlet, the water there becomes deeper. If the drain can no longer discharge the incoming runoff as quickly as it arrives, water accumulates and the upstream level rises. Backflow from the river can add more water, but the drain can back up even while water continues to flow towards the river.

To calculate how high the upstream water rises, we start at the known downstream water level and work upstream. Water loses energy through friction against the drain bed and walls, and through turning and mixing at junctions. The upstream water level must provide enough energy for water to pass through the drain against the downstream level while overcoming these losses.

In this article, we will use Bernoulli’s equation, with these energy losses included, to calculate the upstream water level for a given discharge and tailwater level. The Newton-to-Bernoulli article develops the energy equation we start from.

The downstream boundary and upstream unknown #

We take the case where the river is high enough to control the drain outlet, and water still flows from the drain into the river. The downstream water-surface elevation is our starting value. We want to find the water-surface elevation at a section farther upstream, then subtract the bed elevation there to obtain the water depth.

In the diagram, d marks the downstream section and u the upstream section. We write the water-surface elevations as z_{WS,d} and z_{WS,u} , in metres above the same survey reference level, or datum. The suffix WS means water surface.

The black arrow shows the direction of water flow. The brown arrow shows the direction of calculation: from the known tailwater back towards the upstream section. Bernoulli’s equation will connect these two water levels through the flow speeds and the energy lost between them.

For this first calculation, we take steady, subcritical flow: surface disturbances can travel upstream, allowing the river level to influence the drain. With no inflow added between the two sections, the same discharge passes through both. We choose sections where water flows mainly along the drain. A falling jet or the turbulent, rolling water in a hydraulic jump has strong vertical motion; we place our sections away from those regions. This lets us calculate pressure from depth below the water surface, as we will use below.

From ideal Bernoulli to the section energy equation #

The upper green line stays level: ideal flow has no loss of total head along the selected streamline. The lower green line falls because the real drain dissipates energy. We first write the ideal balance, then introduce the loss shown in the lower row.

Let z be a point’s elevation, p gauge pressure relative to atmosphere, \rho water density, g gravitational acceleration, and v local streamline speed. For steady, incompressible, inviscid flow along a streamline under gravity, Bernoulli gives

\displaystyle z+\frac{p}{\rho g}+\frac{v^2}{2g}=\text{constant}.

Each term is energy per unit weight, expressed as a head in metres. In the numerical examples we take g=9.81\;\text{m/s}^2 .

In a real drain, friction and turbulent mixing convert some of the energy associated with the water’s elevation, pressure and motion into heat. That energy is no longer available to drive the flow. We express this loss as a height of water, called the head loss, h_L , in metres. The upstream water must supply this extra head to keep flowing towards the downstream section.

We also need the flow speed at each section. The mean speed is V=Q/A : discharge divided by the flowing area. The same discharge can therefore have different speeds at the two sections if their flowing areas differ. Water also moves faster in some parts of each section than in others. The dimensionless factor \alpha corrects the velocity-energy term when we use the mean speed. A uniform speed across the section gives \alpha=1 ; using that value in a real drain is an approximation.

Under a steady section energy balance, with no pump supplying energy and no turbine extracting it, we obtain

\displaystyle \left(z+\frac{p}{\rho g}\right)_u+\alpha_u\frac{V_u^2}{2g}=\left(z+\frac{p}{\rho g}\right)_d+\alpha_d\frac{V_d^2}{2g}+h_L.

This is the engineering extension of the Bernoulli balance. The dissipation term requires a loss model; lossless Bernoulli alone cannot determine roughness or junction losses. Section averaging likewise introduces information about the velocity distribution.

Water-surface elevation and pressure head #

At the downstream end, the river holds a depth of water above the drain bed. That water presses on the bed: the deeper the water, the greater the pressure. We know the bed elevation from the survey, so the tailwater level gives us the depth above it. In the equation above, the bed elevation is z and that depth supplies the pressure term p/(\rho g) . Adding them returns the known tailwater elevation.

Let z_b be the bed elevation and y the depth of water above it, both in metres. At the bed we set z=z_b . Under our hydrostatic pressure assumption,

\displaystyle p=\rho g y,\qquad \frac{p}{\rho g}=y.

The term p/(\rho g) is the pressure head. Here it equals the water depth y . In the diagram, adding that depth to the bed elevation reaches the water surface, whose elevation is z_{WS} .

At the bed, the elevation and pressure terms in Bernoulli thus become

\displaystyle z_b+\frac{p}{\rho g}=z_b+y=z_{WS}.

Apply this at both drain sections. We can now put the known downstream water level z_{WS,d} directly into the energy equation:

\displaystyle z_{WS,u}+\alpha_u\frac{V_u^2}{2g}=z_{WS,d}+\alpha_d\frac{V_d^2}{2g}+h_L.

The upstream level z_{WS,u} is the quantity we want to calculate. The blue line in the next diagram follows the water surface. It is also called the hydraulic grade line (HGL): the elevation plus pressure head that we have just written as z_{WS} . Adding the velocity term \alpha V^2/(2g) gives the green energy grade line (EGL). Its drop from upstream to downstream is the head loss h_L .

The vertical green arrows show the velocity contribution at each section. To calculate the upstream water surface from the known tailwater, rearrange the energy equation: \displaystyle \boxed{z_{WS,u}=z_{WS,d}+h_L+\frac{\alpha_dV_d^2-\alpha_uV_u^2}{2g}}.

Starting from the known tailwater z_{WS,d} , we add the loss h_L and the difference between the two velocity contributions. We will first take the case where those velocity contributions cancel, so we can calculate the upstream level by adding the losses alone.

The equal-velocity-head approximation #

Now take the case shown above, with the same vertical gap between EGL and water surface at both sections. Subtracting those equal gaps leaves the difference unchanged. This is the cancellation we need before adding losses directly to the downstream surface level. If the two corrected velocity heads are equal,

\displaystyle \alpha_uV_u^2=\alpha_dV_d^2,

their difference cancels exactly. If their difference is sufficiently small for the design decision, we neglect it as an approximation. We then obtain

\displaystyle \boxed{z_{WS,u}\approx z_{WS,d}+h_L}.

For a reach where friction and one junction account for the relevant dissipation, define h_f as friction loss and h_j as junction loss, both in metres. Taking h_L=h_f+h_j gives the familiar result:

\displaystyle \boxed{z_{WS,u}\approx z_{WS,d}+h_f+h_j}.

Equal mean velocities and equal correction factors are one sufficient condition. Equal velocities alone are insufficient if the velocity distributions, and hence correction factors, differ materially. Two slowly moving pools can also have small velocity heads, although the connecting structure still needs its own loss or discharge model.

The condition concerns the difference in velocity heads. We do not have to assume both velocities are zero. Nor does the phrase “segmented uniform flow” establish cancellation across every junction: adjacent segments can have different sections and velocities.

Upstream depth from the simplified formula #

We can now work backwards from the river into the drain. Start with the tailwater elevation z_{WS,d} . Add the friction and junction losses to obtain the upstream water-surface elevation, provided the velocity-head difference is negligible as described above. To find how deep that water is in the upstream drain, subtract the upstream bed elevation z_{b,u} .

The diagram illustrates these two operations with an assigned total loss of 0.10\;\text{m} and equal corrected velocity heads. Starting from tailwater at 10.00\;\text{m} , we obtain an upstream surface at 10.10\;\text{m} . The upstream bed is at 9.00\;\text{m} , so the water there is 1.10\;\text{m} deep.

In symbols, these are the same two operations:

\displaystyle z_{WS,u}\approx z_{WS,d}+h_f+h_j,

\displaystyle y_u=z_{WS,u}-z_{b,u}\approx z_{WS,d}+h_f+h_j-z_{b,u}.

Here y_u is upstream water depth. If the downstream input is a depth y_d rather than a survey elevation, first add the downstream bed elevation z_{b,d} to obtain the tailwater elevation:

\displaystyle z_{WS,d}=z_{b,d}+y_d.

Substituting gives the calculation directly in terms of downstream depth:

\displaystyle \boxed{y_u\approx y_d+h_f+h_j-(z_{b,u}-z_{b,d})}.

The bed difference matters because depth is measured from the local bed at each end. In the diagram, downstream depth is 10.00-8.80=1.20\;\text{m} . Working from that depth gives y_u\approx1.20+0.10-(9.00-8.80)=1.10\;\text{m} , the same upstream depth as before. All bed and water-surface elevations must use the same survey datum.

The preceding diagram uses an assigned loss to show how elevation becomes depth. We will now calculate that loss from the drain dimensions and discharge, then work through the trial depths needed to balance the energy equation.

Iterative depth calculation with velocity and losses #

The unknown upstream depth affects the velocity and friction loss, which in turn affect the depth calculated from the energy equation. We therefore try a depth, calculate the velocity and losses, and adjust the depth until the energy equation balances. This can still require iteration when the velocity heads cancel, because the friction loss depends on depth.

Area and velocity at a trial depth #

Take a rectangular drain of width b . At either section, a water depth y gives a flowing area A=by . The water touches the bed and two walls, so the wetted perimeter is P=b+2y . Therefore

\displaystyle R=\frac{A}{P}=\frac{by}{b+2y},\qquad V=\frac{Q}{A}=\frac{Q}{by}.

Here R is hydraulic radius and V is mean velocity. We use the known downstream depth to calculate R_d and V_d , and the trial upstream depth to calculate R_u and V_u . For a very wide drain, the wetted bed dominates the perimeter: b+2y\approx b , giving R\approx y . We will retain both walls in our example.

Friction loss at a trial depth #

The deeper trial section has a larger area, so the same discharge moves more slowly there. Both the speed and the wetted geometry enter Manning’s friction calculation.

In the opening capacity check, we used the bed slope in Manning’s equation. For the loss calculation we instead use the friction slope, S_f : the head lost to friction per metre of drain. Manning’s empirical resistance relation is

\displaystyle V=\frac{1}{n}R^{2/3}\sqrt{S_f}.

Multiply by n/R^{2/3} and square:

\displaystyle \sqrt{S_f}=\frac{nV}{R^{2/3}},\qquad \boxed{S_f=\left(\frac{nV}{R^{2/3}}\right)^2=\frac{n^2V^2}{R^{4/3}}}.

For a reach of length L , a constant friction slope gives h_f=LS_f . If the section changes gradually between the two ends, we can approximate its average friction slope by the mean of the endpoint values:

\displaystyle \overline{S_f}\approx\frac{S_{f,u}+S_{f,d}}{2},\qquad \boxed{h_f\approx\frac{L}{2}\left(S_{f,u}+S_{f,d}\right)}.

This becomes h_f=L(nV/R^{2/3})^2 when we use one representative velocity and hydraulic radius for the reach. Manning’s n represents the resistance of the wetted surface; Bernoulli does not determine its value.

The alternative for a full circular pipe #

For a full pipe, we can use Darcy–Weisbach instead of Manning. Let D be pipe diameter in metres, \tau_w the wall shear stress in pascals and f the dimensionless Darcy friction factor. In steady, fully developed flow, balancing wall drag against the driving pressure and gravity forces gives

\displaystyle \rho g h_f A=\tau_w P L.

For the circular pipe, A=\pi D^2/4 and P=\pi D , so P/A=4/D . The Darcy factor is defined by \tau_w=f\rho V^2/8 . Substitute these into the force balance:

\displaystyle \begin{aligned} h_f&=\frac{\tau_w PL}{\rho g A}\\ &=\frac{f\rho V^2}{8}\frac{4L}{\rho gD}\\ &=\boxed{f\frac{L}{D}\frac{V^2}{2g}}. \end{aligned}

The factor f comes from a pipe-resistance relationship. Manning and Darcy–Weisbach are alternative ways to estimate the same friction loss; we use one of them for a reach.

Junction loss #

A sump or junction can dissipate additional energy as the water turns and mixes. Let V_{ref} be the velocity in the section to which its loss coefficient applies. The dimensionless loss coefficient K expresses the loss as a multiple of that section’s velocity head:

\displaystyle K=\frac{h_j}{V_{ref}^2/(2g)}.

Rearranging gives

\displaystyle \boxed{h_j=K\frac{V_{ref}^2}{2g}}.

The coefficient is an empirical input for the junction geometry. For several junctions we add their individual losses. For the one-junction example below, h_L=h_f+h_j .

The upstream-depth update equation #

We know the downstream water level and want to find the upstream depth. For a given discharge, the upstream depth determines the flowing area, velocity and friction loss, so we cannot calculate these quantities independently. We will start with a trial depth and use the energy equation to check it. The diagram below shows the bed elevations z_{b,u} and z_{b,d} , water depths y_u and y_d , mean velocities V_u and V_d , reach length L , and losses h_f and h_j . We take \alpha_u=\alpha_d=1 and retain both velocity heads in the calculation.

Substituting z_{WS,u}=z_{b,u}+y_u and z_{WS,d}=z_{b,d}+y_d into the energy equation gives

\displaystyle z_{b,u}+y_u+\frac{V_u^2}{2g} =z_{b,d}+y_d+\frac{V_d^2}{2g}+h_f+h_j.

Move the upstream bed elevation and velocity term to the right:

\displaystyle \boxed{y_u=y_d-(z_{b,u}-z_{b,d})+h_f+h_j+\frac{V_d^2-V_u^2}{2g}}.

The right-hand side still depends on y_u , through V_u and h_f . To calculate an updated trial, evaluate those quantities using the current trial depth, then substitute them into the right-hand side. We label the current trial y_u^{(k)} and the updated trial y_u^{(k+1)} , where k counts the trials.

The two slopes are already present in this equation. The bed slope S_0 is the fall in invert level per metre, so

\displaystyle z_{b,u}-z_{b,d}=S_0L.

The friction slope \overline{S_f} is the average friction head lost per metre, so

\displaystyle h_f^{(k)}=L\overline{S_f}^{(k)} \approx\frac{L}{2}\left(S_{f,u}^{(k)}+S_{f,d}\right).

Replace the bed difference and friction loss by these expressions. The update becomes

\displaystyle \begin{aligned} y_u^{(k+1)} &=y_d+L\left(\overline{S_f}^{(k)}-S_0\right)+h_j^{(k)}\\ &\quad+\frac{V_d^2-\left(V_u^{(k)}\right)^2}{2g}. \end{aligned}

Thus S_0 enters through the surveyed bed levels, while Manning supplies S_f through the flow depth, velocity and roughness. They are equal in uniform flow; they need not be equal in a tailwater-affected drain.

We have rearranged the section energy equation into an update for the upstream depth, starting from the known downstream depth.

Use each trial depth to recalculate the area, velocity and losses. Check the energy balance before deciding whether to update the depth.

Energy imbalance and convergence #

For each trial, compare the upstream total head with the downstream total head plus losses. Their difference is

\displaystyle \varepsilon =\left(z_{b,u}+y_u+\frac{V_u^2}{2g}\right) -\left(z_{b,d}+y_d+\frac{V_d^2}{2g}+h_f+h_j\right).

A balanced calculation has \varepsilon=0 . A negative value means the trial upstream head is too low to supply the downstream head and the calculated losses. In practice, we stop when the magnitude of this imbalance is below a chosen head tolerance, for example 0.001\;\text{m} for these worked calculations, and successive depths also differ by less than the chosen depth tolerance.

If repeated substitution oscillates or fails to converge, use a more stable numerical method, such as bisection or safeguarded Newton–Raphson, to solve the same energy-balance equation \varepsilon(y_u)=0 . The accepted depth must still satisfy the energy tolerance and the assumed subcritical flow condition.

Worked calculation: two trial rounds #

Take a rectangular drain with the following inputs. The river controls the downstream level, and the flow is steady and subcritical.

The downstream depth is fixed by the river:

\displaystyle y_d=10.00-9.00=1.00\;\text{m}.

Calculate the downstream section once:

\displaystyle \begin{aligned} A_d&=2(1)=2.00\;\text{m}^2,\\ P_d&=2+2(1)=4.00\;\text{m},\\ R_d&=\frac{2}{4}=0.5000\;\text{m},\\ V_d&=\frac{2}{2}=1.0000\;\text{m/s}. \end{aligned}

Thus

\displaystyle S_{f,d}=\left(\frac{0.015(1.0000)}{0.5000^{2/3}}\right)^2=0.00056696.

The junction uses this fixed downstream velocity, so its loss is the same in each round:

\displaystyle h_j=0.50\frac{1.0000^2}{2(9.81)}=0.02548\;\text{m}.

The fixed downstream total head is

\displaystyle z_{b,d}+y_d+\frac{V_d^2}{2g} =9.00+1.00+\frac{1.00^2}{2(9.81)} \approx10.050968\;\text{m}.

First round: upstream trial depth of 1.0000 m #

The upstream depth initially equals the downstream depth, so A_u=2.00\;\text{m}^2 , R_u=0.5000\;\text{m} and V_u=1.0000\;\text{m/s} . The upstream friction slope is also 0.00056696 . Hence

\displaystyle h_f^{(1)} =\frac{100}{2}(0.00056696+0.00056696) =0.05670\;\text{m}.

Check the energy balance at this trial depth:

\displaystyle \begin{aligned} \varepsilon^{(1)} &\approx\left(9.02+1.00000+\frac{1.00000^2}{2(9.81)}\right)\\ &\quad-\left(10.050968+0.056696+0.025484\right)\\ &\approx10.070968-10.133149\\ &\approx-0.06218\;\text{m}. \end{aligned}

The negative imbalance means the first trial supplies too little upstream head, so calculate an updated depth:

\displaystyle \begin{aligned} y_u^{(2)}&=1.00-(9.02-9.00)+0.05670+0.02548\\ &\quad+\frac{1.0000^2-1.0000^2}{2(9.81)}\\ &=1.06218\;\text{m}. \end{aligned}

Here L(\overline{S_f}-S_0)=100(0.00056696-0.0002)=0.03670\;\text{m} ; adding the junction loss gives the same 0.06218\;\text{m} depth increase. The first trial is too shallow: it does not provide enough upstream head to carry this discharge against the tailwater and losses. We raise the trial depth to 1.06218\;\text{m} .

Second round: upstream trial depth of 1.06218 m #

The larger depth changes the upstream area, hydraulic radius and speed:

\displaystyle A_u=2(1.06218)=2.12436\;\text{m}^2,\qquad P_u=2+2(1.06218)=4.12436\;\text{m},

\displaystyle R_u=\frac{2.12436}{4.12436}=0.51508\;\text{m},\qquad V_u=\frac{2.00}{2.12436}=0.94146\;\text{m/s}.

Using unrounded values,

\displaystyle S_{f,u} =\left(\frac{0.015(0.94146)}{0.51508^{2/3}}\right)^2 \approx0.00048301,

\displaystyle h_f^{(2)} =\frac{100}{2}(0.00048301+0.00056696) =0.05250\;\text{m}.

The friction loss has fallen because the deeper upstream flow is slower. The velocity terms no longer cancel. Check the energy balance again, carrying unrounded values through the calculation:

\displaystyle \begin{aligned} \varepsilon^{(2)} &\approx\left(9.02+1.06218+\frac{0.94146^2}{2(9.81)}\right)\\ &\quad-\left(10.050968+0.052499+0.025484\right)\\ &\approx10.127356-10.128951\\ &\approx-0.00160\;\text{m}. \end{aligned}

The second trial is much closer to balance but still slightly low. Update the depth again:

\displaystyle \begin{aligned} y_u^{(3)}&=1.00-0.02+0.05250+0.02548\\ &\quad+\frac{1.0000^2-0.94146^2}{2(9.81)}\\ &=1.06378\;\text{m}. \end{aligned}

Result after two rounds #

The two checks and depth updates are:

After the second update, substituting 1.06378\;\text{m} back into the energy equation leaves an imbalance of about -0.00004\;\text{m} . Continuing the same calculation gives y_u\approx1.06382\;\text{m} and therefore

\displaystyle z_{WS,u}=9.02+1.06382=10.08382\;\text{m}.

At this depth, h_f\approx0.05240\;\text{m} , h_j\approx0.02548\;\text{m} , and the velocity-head difference is about 0.00593\;\text{m} . Their sum, before rounding, is the 0.08382\;\text{m} rise above the known tailwater. The upstream depth and losses have now been calculated together, rather than assigning a loss independently of the depth.

The example uses the stated rectangular geometry and endpoint-average friction model. In a longer drain, repeat this calculation for successive reaches, carrying each calculated upstream water level into the next reach as its known downstream level.

MSMA 3 worked example (Appendix 6.5) #

We will now repeat Appendix 6.5 using its stated dimensions and discharge, recalculating the arithmetic and retaining the bed-elevation difference throughout. As in the manual, section 1 is upstream and section 2 is downstream. Their depths are y_1,y_2 , bed elevations z_1,z_2 , and velocities V_1,V_2 . The source uses \Delta x for reach length and h_L for friction head loss.

Appendix 6.5 uses a 10\;\text{m} wide rectangular channel carrying Q=20\;\text{m}^3/\text{s} , with n=0.015\;\text{s}/\text{m}^{1/3} , bed slope 1 in 2500 and reach length \Delta x=500\;\text{m} . The given downstream depth is y_2=2.5\;\text{m} , and the initial upstream trial is y_1=2.7\;\text{m} . Its friction calculation uses the wide-channel approximation R\approx y .

The example considers friction only, with \alpha_1=\alpha_2=1 and no added junction loss. Under those assumptions, our section energy equation becomes MSMA 3’s standard-step energy equation:

\displaystyle \boxed{\frac{V_1^2}{2g}+y_1+z_1 =\frac{V_2^2}{2g}+y_2+z_2+h_L}.

This is Eq. 6.21 in the supplied MSMA 3 draft. Its friction equations, Eq. 6.22, are the same Manning reduction we derived:

\displaystyle \boxed{h_L=S_f\Delta x},\qquad \boxed{S_f=\frac{n^2V^2}{R^{4/3}}}.

Here S_f is the representative friction slope over the reach. For the two-section calculation, MSMA’s example takes the mean of the endpoint slopes:

\displaystyle S_f=\frac{S_{f,1}+S_{f,2}}{2},\qquad h_L=\frac{\Delta x}{2}\left(S_{f,1}+S_{f,2}\right).

The bed difference is z_1-z_2=500/2500=0.20\;\text{m} . Choosing the downstream bed as datum gives z_2=0 and z_1=0.20\;\text{m} . Calculate the area and velocity at the known downstream section and at the first upstream trial:

\displaystyle \begin{aligned} A_2&=b y_2=10(2.5)=25\;\text{m}^2,\\ V_2&=\frac{Q}{A_2}=\frac{20}{25}=0.80\;\text{m/s},\\ A_1&=b y_1=10(2.7)=27\;\text{m}^2,\\ V_1&=\frac{Q}{A_1}=\frac{20}{27}\approx0.740741\;\text{m/s}. \end{aligned}

For a trial depth y_1 , the actual equations to evaluate are therefore

\displaystyle \boxed{\frac{(20/(10y_1))^2}{2g}+y_1+0.20 =\frac{0.80^2}{2g}+2.5+h_L},

\displaystyle S_{f,1} =\left(\frac{0.015\, [20/(10y_1)]}{y_1^{2/3}}\right)^2,\qquad S_{f,2}=\left(\frac{0.015(0.80)}{2.5^{2/3}}\right)^2,

\displaystyle \boxed{h_L=\frac{500}{2}\left(S_{f,1}+S_{f,2}\right)}.

First round: upstream trial depth of 2.7 m #

Using the source’s R\approx y assumption, the downstream friction slope is

\displaystyle S_{f,2} =\left(\frac{0.015(0.800000)}{2.5^{2/3}}\right)^2 =4.244004\times10^{-5}.

At the initial upstream depth y_1^{(1)}=2.7\;\text{m} ,

\displaystyle S_{f,1} =\left(\frac{0.015(0.740741)}{2.7^{2/3}}\right)^2 \approx3.283699\times10^{-5},

\displaystyle h_L^{(1)} =\frac{500}{2}\left(3.283699+4.244004\right)10^{-5} \approx0.018819\;\text{m}.

The trial’s upstream total head is 2.927966\;\text{m} , while the downstream head plus friction loss is 2.551439\;\text{m} . The imbalance is therefore +0.376527\;\text{m} : our trial depth is too large. The updated depth is

\displaystyle \begin{aligned} y_1^{(2)} &=2.5-0.20+0.018819\\ &\quad+\frac{0.800000^2-0.740741^2}{2(9.81)}\\ &\approx2.323473\;\text{m}. \end{aligned}

Second round: upstream trial depth of 2.323473 m #

Recalculate the upstream section at this new depth:

\displaystyle A_1\approx23.234728\;\text{m}^2,\qquad V_1\approx0.860780\;\text{m/s},

\displaystyle S_{f,1}\approx5.417312\times10^{-5},\qquad h_L^{(2)}=\frac{500}{2}\left(S_{f,1}+S_{f,2}\right) \approx0.024153\;\text{m}.

The upstream head is now 2.561238\;\text{m} and the required head is 2.556773\;\text{m} , leaving an imbalance of about +0.004464\;\text{m} . Update again:

\displaystyle \begin{aligned} y_1^{(3)} &=2.5-0.20+0.024153\\ &\quad+\frac{0.800000^2-0.860780^2}{2(9.81)}\\ &\approx2.319008\;\text{m}. \end{aligned}

These calculations use unrounded values internally. Continuing the same trials gives:

Corrected final depth and water level #

The balanced upstream depth is

\displaystyle \boxed{y_1\approx2.318949\;\text{m}}.

At that depth, V_1\approx0.862460\;\text{m/s} and h_L\approx0.024242\;\text{m} . Check the two sides of MSMA’s energy equation:

\displaystyle \begin{aligned} \frac{V_1^2}{2g}+y_1+z_1 &\approx0.037912+2.318949+0.20\\ &\approx2.556861\;\text{m},\\ \frac{V_2^2}{2g}+y_2+z_2+h_L &\approx0.032620+2.5+0+0.024242\\ &\approx2.556861\;\text{m}. \end{aligned}

The upstream water surface is z_1+y_1=2.518949\;\text{m} above our chosen datum, compared with z_2+y_2=2.500000\;\text{m} downstream. Thus the water surface is about 0.018949\;\text{m} higher upstream even though the water is shallower there: the upstream bed itself is 0.20\;\text{m} higher.

The result differs from the MSMA 3 draft’s printed depth of 2.73\;\text{m} because its worked example contains arithmetic and substitution errors; using its stated inputs gives y_1\approx2.31895\;\text{m} .

Velocity-head corrections to the simplified formula #

The simple formula z_{WS,u}\approx z_{WS,d}+h_L assumes the corrected velocity heads cancel. When they do not, the amount it leaves out is

\displaystyle \delta h_V=\frac{\alpha_dV_d^2-\alpha_uV_u^2}{2g},\qquad z_{WS,u}=z_{WS,d}+h_L+\delta h_V.

Since V=Q/A , a wider section or a greater water depth can slow the same discharge by increasing its flowing area. We will first compare a wider and a narrower downstream section, then return to the depth change in our worked calculation.

For the width comparisons, prescribe Q=2.00\;\text{m}^3/\text{s} , \alpha_u=\alpha_d=1 , tailwater z_{WS,d}=10.000\;\text{m} , and a total loss h_L=0.300\;\text{m} in each case. The endpoint areas and loss are assigned so we can isolate the velocity correction; they are not a solved drain profile.

Larger downstream area: overestimated upstream level #

The wider downstream drain doubles area from A_u=1.00\;\text{m}^2 to A_d=2.00\;\text{m}^2 . Velocity falls from 2.00 to 1.00\;\text{m/s} . The downstream velocity head is smaller, so its difference from the upstream velocity head is negative:

\displaystyle \delta h_V=\frac{1^2-2^2}{2(9.81)}=-0.1529\;\text{m}.

Thus the general equation gives

\displaystyle z_{WS,u}=10.000+0.300-0.1529=10.1471\;\text{m}.

The simple calculation gives 10.000+0.300=10.300\;\text{m} . It ignores the 0.1529\;\text{m} supplied by slowing the water, so it overestimates the required upstream water level by that same 0.1529\;\text{m} .

Smaller downstream area: underestimated upstream level #

The narrower downstream drain reverses the areas. Velocity rises from 1.00 to 2.00\;\text{m/s} , so the upstream water must supply both the loss and the increase in kinetic head:

\displaystyle \delta h_V=\frac{2^2-1^2}{2(9.81)}=+0.1529\;\text{m},

\displaystyle z_{WS,u}=10.000+0.300+0.1529=10.4529\;\text{m}.

The simple calculation again gives 10.000+0.300=10.300\;\text{m} , now an underestimate of about 0.153\;\text{m} . Omitting velocity-head change can err in either direction.

Velocity differences at equal width and different depths #

The same effect occurred in our worked drain of constant width b=2.00\;\text{m} . The known downstream depth was 1.00\;\text{m} , while the calculated upstream depth was about 1.06382\;\text{m} . The upstream flowing area was therefore larger:

\displaystyle A_u\approx2.12763\;\text{m}^2,\qquad A_d=2.00000\;\text{m}^2.

At Q=2.00\;\text{m}^3/\text{s} , this gives V_u\approx0.94001\;\text{m/s} and V_d=1.00000\;\text{m/s} . Water accelerates as it moves into the shallower downstream section, giving

\displaystyle \delta h_V=\frac{1.00000^2-0.94001^2}{2(9.81)} \approx0.00593\;\text{m}.

Using the converged losses, the simple formula gives 10.00+0.05240+0.02548\approx10.07788\;\text{m} . Including the velocity term restores the calculated 10.08382\;\text{m} upstream level. Omitting the depth-related velocity change would underestimate the upstream level by about 6\;\text{mm} for these same losses.

Discharge and velocity changes at a side-drain junction #

When a side drain adds flow, the downstream discharge also changes. In the plan view below, the main drain brings Q_u from the left and the side drain brings Q_s into the junction from above. After they meet, steady continuity gives Q_d=Q_u+Q_s . The dashed lines mark the sections with flowing areas A_u and A_d ; the arrows labelled V_u and V_d show the mean velocities there.

Dividing each discharge by its flowing area gives

\displaystyle V_u=\frac{Q_u}{A_u},\qquad V_d=\frac{Q_u+Q_s}{A_d}.

Even equal areas then need not give equal velocities. Unlike the constant-discharge reaches above, this case also involves the energy and mixing of the incoming side flow. We need a junction balance accounting for that inflow before applying the energy equation to the next reach.

For the constant-discharge comparisons, the omitted velocity term gives the difference between the simple and complete estimates when the other inputs are held fixed. In a solved profile, changing depths also changes friction, so recalculate the areas, speeds and losses together as in the worked iteration.

Conclusion #

We have derived the section energy equation used in MSMA 3’s backwater calculation from Bernoulli’s equation, with friction and junction losses included. Starting from a known tailwater level, we use it to calculate the upstream water depth and water-surface elevation. Each trial depth changes the area, velocity and losses, so we recalculate them together until the energy imbalance is within the chosen tolerance. Repeated substitution is one way to find that depth; bisection or a safeguarded Newton–Raphson method can solve the same equation if the simple iteration fails to converge.

The simplified formula z_{WS,u}\approx z_{WS,d}+h_L is adequate when the difference between the corrected velocity heads is small enough for the required accuracy. Changes in width, depth or discharge can make that difference significant, and omitting it can either overestimate or underestimate the upstream level. In that case, retain the velocity terms and solve the full steady energy equation. An unsteady simulation is needed when changing inflow, tailwater and water storage materially affect the result; a large velocity correction alone does not establish that need.