What can we calculate? #

Water enters a sump, turns through 90 degrees, and leaves through another drain. We can see the turn on plan. How much does it raise the upstream water level, and how hard does the water push against the sump?

A one-dimensional drainage model can calculate the effect of an assumed sump loss on the connected network. From its flow and velocity results, we can also calculate energy loss and momentum change. However, the model gives the sump one hydraulic head; it cannot tell us how pressure or water level varies from one wall to another.

In this article we will work through these calculations, then use two examples to show what an engineer can decide from them. Local water-level rise is covered separately in The Superelevation Equation for Flow Through Curved Conduits.

Water level, head and the flow section #

Head expresses energy per unit weight of water as a height, in metres. At a pipe or channel section, let z be the elevation of the point where gauge pressure p is measured, \rho the water density, and g gravitational acceleration. Gauge pressure is measured relative to atmospheric pressure. The hydraulic grade line (HGL) is the elevation

\displaystyle H=z+\frac{p}{\rho g}.

For open-channel flow with approximately hydrostatic pressure, this is the water-surface elevation. In a full pipe, it is the height water would reach in a piezometer connected to the pipe. Thus a gravity drainage network can still have pressurised sections when it surcharges.

The energy grade line (EGL) also includes the kinetic energy of the moving water. If V is the mean speed at the section, its elevation is

\displaystyle H+\frac{V^2}{2g}.

We use a velocity-head correction factor of one in these bulk calculations. This treats the section velocity as uniform; it does not describe the fast jet and slower recirculating water that may coexist inside a sump.

For the calculations below, Q is discharge in m³/s, A is wetted flow area in m², and V=Q/A is mean speed in m/s. Subscripts i and o mean inlet and outlet. We use positive flow magnitudes along the actual flow direction, with g=9.81\ \text{m/s}^2 and \rho=1000\ \text{kg/m}^3 .

MiTS stores signed link flows. Those signs must be checked before identifying which link carries water into or out of the sump. The equations below assume one active inlet and one active outlet. Flow reversal or several simultaneous inflows need separate treatment; total node inflow alone does not tell us the direction or momentum of each stream.

How the sump loss enters the network calculation #

From the plan angle to the loss coefficient #

The deflection angle, \delta , is the change in flow direction on plan: zero for straight-through flow and 90 degrees for a right-angle turn.

Let \boldsymbol t_i and \boldsymbol t_o be unit vectors along the adjacent drain segments, pointing towards and away from the sump respectively. Their dot product gives

\displaystyle \boldsymbol t_i\cdot\boldsymbol t_o=\cos\delta, \qquad \delta=\cos^{-1}(\boldsymbol t_i\cdot\boldsymbol t_o).

The vectors must have length one. Otherwise their dot product includes the segment lengths as well as the angle. Software also limits a rounded dot product to the interval from −1 to 1 before taking the inverse cosine.

MiTS uses the following access-hole values from HEC-22 for its automatic preliminary loss calculation. HEC-22 measures the interior angle between the pipes, which is 180^\circ-\delta under this convention.

The dimensionless coefficient K expresses the loss as a multiple of the outlet velocity head. These are preliminary access-hole estimates; benching, depth, plunging inflow and unequal pipe sizes can change the actual loss. HEC-22, Fourth Edition, Sections 9.1.6.6–9.1.6.7 and Table 9.4.

Between two table entries, MiTS interpolates linearly. If \delta_1 and \delta_2 bracket the angle, and their coefficients are K_1 and K_2 , the fraction of the interval travelled is (\delta-\delta_1)/(\delta_2-\delta_1) . Therefore,

\displaystyle K=K_1+\frac{\delta-\delta_1}{\delta_2-\delta_1}(K_2-K_1).

For a 30-degree turn,

\displaystyle K=0.45+\frac{30-22.5}{45-22.5}(0.75-0.45)=0.55.

The automatic method does not extrapolate beyond 90 degrees. Also notice that straight-through flow still has a coefficient of 0.15: expansion, contraction and mixing in the chamber can consume energy even without a turn.

From the coefficient to head loss #

The kinetic energy per unit mass is V_o^2/2 . Dividing by g converts it to energy per unit weight, or velocity head, h_{V,o} . The head loss, h_L , is then

\displaystyle h_{V,o}=\frac{V_o^2}{2g}, \qquad h_L=K h_{V,o}=K\frac{V_o^2}{2g}.

To see what this means for water levels, consider steady flow between inlet and outlet sections, with no pump or turbine. If h_L accounts for the loss between those sections, the energy balance is

\displaystyle H_i+\frac{V_i^2}{2g}=H_o+\frac{V_o^2}{2g}+h_L.

Rearranging gives

\displaystyle H_i-H_o=h_L+\frac{V_o^2-V_i^2}{2g}.

Only when the two speeds are equal does the HGL difference equal the head loss. In a network, adding a sump loss can change both levels and flows, so we rerun the network to find the result.

We include the sump loss in the network’s energy calculation and solve the connected flows and water levels together. The calculation must account for both local losses and downstream backwater. Each physical loss should be included once.

The resulting HGL already includes the applied loss. Calculating h_L afterwards tells us how much loss that coefficient represents at the reported speed; adding it to the HGL again would double-count it. MiTS’s automatic angle calculation uses the connected plan geometry. The sump width does not enter this angle-only coefficient.

Checking the rim margin #

Let z_r be the rim elevation and t time. The vertical clearance between the rim and the node HGL is

\displaystyle m(t)=z_r-H(t), \qquad m_{\min}=z_r-\max_t H(t).

We compare m_{\min} with the project’s required clearance. A positive margin only establishes that the modelled head stays below the rim; if the project requires 150 mm freeboard, a 70 mm margin still fails that requirement. Surcharge means water has risen above the relevant pipe crown and may occur while the sump remains below its rim. Flooding is a separate model output and depends on the node’s overflow and ponding settings.

What else can we calculate from the flow results? #

The loss coefficient K enters the hydraulic calculation and can change the node HGL. The loss power P_L and momentum-flux change M below are calculated afterwards from the flow results. Neither is needed for the HGL or rim-margin check, and neither is fed back as an additional loss.

We can use them to compare layouts. To turn loss power into an erosion check, we would also need a method relating dissipation to the material and the area exposed. Momentum-flux change enters a force balance, but a structural check also needs the other forces and their distribution. If the question is simply whether the network has enough capacity and freeboard, these two calculations are optional.

Power dissipated by the sump loss #

Head loss is energy lost per unit weight of water. A discharge Q_o carries mass at a rate \rho Q_o , and weight at a rate \rho gQ_o . Multiplying this by the head loss gives the loss power, P_L :

\displaystyle P_L=(\rho gQ_o)h_L=\rho gQ_oK\frac{V_o^2}{2g}.

The units are N/s multiplied by m, giving watts. At the same flow and speed, a larger K means more power dissipated. However, total power alone does not tell us where that energy is dissipated or over what area, so it cannot establish whether a wall or lining will erode.

Momentum change through the turn #

Water can change momentum by changing direction even if its speed stays the same. We therefore need velocity vectors, \boldsymbol V_i and \boldsymbol V_o , as well as discharge. We take the momentum correction factor as one, consistent with the mean-velocity approximation above.

Momentum carried through a section per second is mass flow rate multiplied by velocity. The outgoing amount minus the incoming amount is

\displaystyle \boldsymbol M=\rho Q_o\boldsymbol V_o-\rho Q_i\boldsymbol V_i.

Here \boldsymbol M is the momentum-flux change and M=|\boldsymbol M| its magnitude. For approximately equal discharges Q_i=Q_o=Q and speeds |\boldsymbol V_i|=|\boldsymbol V_o|=V ,

\displaystyle M=\rho Q|\boldsymbol V_o-\boldsymbol V_i|.

For a horizontal turn through \delta , expand the squared vector difference:

\displaystyle |\boldsymbol V_o-\boldsymbol V_i|^2=V^2+V^2-2V^2\cos\delta=2V^2(1-\cos\delta).

Using 1-\cos\delta=2\sin^2(\delta/2) and taking the square root gives

\displaystyle M=2\rho QV\sin\left(\frac{\delta}{2}\right).

Thus a straight path gives zero momentum change under these equal-flow, equal-speed assumptions, while a 90-degree turn gives M=\sqrt{2}\rho QV . If the inlet and outlet speeds differ, use the vector equation instead. A plunging inlet also requires its vertical velocity component; a plan angle cannot describe that impact.

But can we use M as the force on the wall? To see what is missing, take a fixed control volume enclosing the water inside the sump. The momentum balance is

\displaystyle \boldsymbol F_w+\boldsymbol F_p+\boldsymbol W=\frac{d}{dt}\int_{\mathrm{CV}}\rho\boldsymbol v\,d\mathcal V+\boldsymbol M.

Here \boldsymbol F_w is the force of the walls and floor on the water, \boldsymbol F_p the pressure force at the inlet and outlet openings, and \boldsymbol W the water’s weight. The integral is momentum stored inside the control volume: \boldsymbol v is local velocity and d\mathcal V a small volume of water. Its time derivative accounts for acceleration of the water held inside the sump.

Even in steady flow, when that derivative is zero, pressure and weight remain in the balance. The water’s force on the structure is opposite to \boldsymbol F_w , and distributing it between individual walls needs more information again. The calculated M helps compare how much momentum two layouts redirect, but does not supply a wall design load.

Hydraulic depth and Froude number #

For a section with a free surface, let y be water depth above the invert and T the width of the water surface. The hydraulic depth, D_h , is

\displaystyle D_h=\frac{A}{T}.

It equals y in a rectangular channel because A=Ty . In a partly full circular pipe, the width changes with depth, so hydraulic depth and water depth generally differ.

For small, long surface waves under the shallow-water approximation, wave speed relative to the water is c=\sqrt{gA/T}=\sqrt{gD_h} . The Froude number compares flow speed with that wave speed:

\displaystyle Fr=\frac{V}{c}=\frac{V}{\sqrt{gD_h}}.

Take downstream as positive. Relative to the moving water, small surface disturbances can travel in either direction at speed c . To find their velocities relative to the ground, we add the flow velocity V :

\displaystyle \text{Upstream-directed wave: }V-c, \qquad \text{Downstream-directed wave: }V+c.

Here “upstream-directed” describes the wave’s motion relative to the water. Whether it reaches upstream relative to the ground depends on the flow regime:

• Subcritical, Fr<1 : V<c , so V-c<0 and V+c>0 . Disturbances can travel both upstream and downstream.
• Critical, Fr=1 : V=c , so V-c=0 . The upstream-directed wave is stationary relative to the ground, while the other travels downstream at V+c=2c .
• Supercritical, Fr>1 : V>c , so both V-c and V+c are positive. Both waves move downstream relative to the ground.

These statements concern small, long surface waves; a moving hydraulic jump requires a separate analysis. See USACE, River Hydraulics, Section 2-2.

These quantities belong to the chosen conduit section. They do not describe the whole sump. A full, pressurised pipe has no internal free surface, so this hydraulic-depth and surface-wave calculation does not apply there.

Why Fr = 1 marks critical flow and minimum specific energy #

The specific energy, E , is energy head measured above the local invert. For hydrostatic free-surface flow,

\displaystyle E=y+\frac{V^2}{2g}=y+\frac{Q^2}{2gA(y)^2}.

Why does critical flow occur at Fr=1 ? At fixed discharge and section geometry, increasing depth increases the first term but reduces velocity and hence the second. Critical energy is the minimum of their sum. Here A(y) reminds us that the wetted area changes with depth. A small depth increase dy adds an area approximately equal to T\,dy , so dA/dy=T .

Keeping Q constant and differentiating the energy equation gives

\displaystyle \frac{dE}{dy}=1+\frac{Q^2}{2g}\frac{d}{dy}\left(A^{-2}\right).

By the chain rule, d(A^{-2})/dy=-2A^{-3}(dA/dy) . Substituting this and cancelling the factor of two gives

\displaystyle \frac{dE}{dy}=1-\frac{Q^2}{gA^3}\frac{dA}{dy}=1-\frac{Q^2T}{gA^3}.

The fraction on the right looks different from the Froude-number formula above. However, it is the same quantity squared, written in terms of discharge and section geometry. Start with the original definition and substitute D_h=A/T :

\displaystyle Fr=\frac{V}{\sqrt{gD_h}}=\frac{V}{\sqrt{g(A/T)}}.

We square both sides to remove the square root and compare this expression with the term in the energy derivative. Dividing by A/T is the same as multiplying by T/A :

\displaystyle Fr^2=\frac{V^2}{gD_h}=\frac{V^2}{g(A/T)}=\frac{V^2T}{gA}.

Finally, V=Q/A , so V^2=Q^2/A^2 . Substituting gives

\displaystyle Fr^2=\frac{Q^2}{A^2}\frac{T}{gA}=\frac{Q^2T}{gA^3}.

These are equivalent forms of the same definition. We use the velocity form to compare flow speed with wave speed, and the discharge form to recognise the term in the energy derivative:

\displaystyle \frac{dE}{dy}=1-\frac{Q^2T}{gA^3}=1-Fr^2.

At the critical depth, specific energy reaches its minimum, so

\displaystyle \frac{dE}{dy}=0 \quad\Longrightarrow\quad 1-Fr^2=0 \quad\Longrightarrow\quad Fr=1.

We take the positive root because Fr is defined here using speed. Thus the point where flow speed equals small-wave speed is also the point of minimum specific energy for the given discharge and section. This is the physical reason for the critical-flow condition. We can compare that minimum with the available specific energy when checking a channel control or possible choking.

A Froude number close to one deserves attention because a small change in the flow or geometry can change the regime. It also affects whether a regime-dependent formula can be used. It is not, on its own, a sump acceptance limit.

A consistent circular-pipe example #

Consider a 400 mm diameter pipe flowing 300 mm deep at 1.50 m/s. To calculate its Froude number we first need the wetted area and surface width. Let the radius be r=0.20 m and the wetted central angle be \theta , measured in radians:

\displaystyle \theta=2\cos^{-1}\left(\frac{r-y}{r}\right)=2\cos^{-1}(-0.5)=\frac{4\pi}{3}.

The wetted area is the sector area r^2\theta/2 minus the signed triangle area r^2\sin\theta/2 . For the top width, half the surface chord forms a right triangle with sides r-y and r , so (T/2)^2=r^2-(r-y)^2 . Therefore,

\displaystyle A=\frac{r^2}{2}(\theta-\sin\theta)=0.10110\ \text{m}^2, \qquad T=2\sqrt{2ry-y^2}=0.34641\ \text{m}.

Thus

\displaystyle Q=AV=0.15164\ \text{m}^3/\text{s}, \qquad D_h=\frac{0.10110}{0.34641}=0.29184\ \text{m},

\displaystyle Fr=\frac{1.50}{\sqrt{9.81(0.29184)}}=0.887, \qquad E=0.30+\frac{1.50^2}{2(9.81)}=0.415\ \text{m}.

This section is subcritical. Notice that the water-surface width is 346 mm, although the pipe diameter is 400 mm. Substituting the diameter for the top width would change the answer.

Worked examples #

We will look at three cases: a sump that fails the rim check once its loss is included, two layouts that both pass, and a detention-drain network where adding sump losses raises the peak HGL at one node but lowers it at another. The first two use assumed values to illustrate the checks. The third uses the MiTS results from the linked case study. All elevations within each example use the same datum.

When including the loss changes the decision #

Suppose a sump has a 90-degree turn and a rim elevation of 13.10 m. For this example, the acceptance limit is the rim itself, with no additional freeboard requirement.

Including the sump loss changes an apparent 70 mm clearance into a 40 mm exceedance of the limit. We already have a reason to revise the design, without calculating a local water-level rise. The engineer could change the alignment, capacity or levels, then rerun the model.

The straightened alternative is a different network layout. Its HGL must come from its own run; we cannot obtain it by subtracting the difference in K from the original water level. Likewise, comparing losses fairly requires consistent rainfall, downstream levels and other design inputs.

If an open node spills at its rim, the solver may limit the head and report flooding instead of the assumed 13.14 m head. A real project must use its actual node and ponding settings when checking this condition.

When both layouts stay below the rim #

Now consider a separate example with the same 13.10 m rim limit. The 90-degree layout and a straight-through alternative give the following assumed results:

Both pass the stated rim check. Straightening gives another 90 mm of margin. Whether that is worth the cost depends on the project.

To compare the loss and momentum calculations, suppose both layouts have Q=0.12\ \text{m}^3/\text{s} and V=1.50\ \text{m/s} at the selected reporting time, with equal inlet and outlet speeds. This implies a flow area of A=Q/V=0.080\ \text{m}^2 ; it is a separate example from the circular-pipe section above. In a real comparison, use each layout’s own simultaneous flow and velocity results.

First calculate velocity head:

\displaystyle h_V=\frac{1.50^2}{2(9.81)}=0.11468\ \text{m}.

Then the losses are

\displaystyle h_{L,90}=1.00(0.11468)=0.11468\ \text{m}, \qquad h_{L,0}=0.15(0.11468)=0.01720\ \text{m}.

Multiplying each loss by the weight flow rate gives

\displaystyle P_{L,90}=1000(9.81)(0.12)(0.11468)\approx135\ \text{W},

\displaystyle P_{L,0}=1000(9.81)(0.12)(0.01720)\approx20.3\ \text{W}.

For the turn, the momentum-flux change is

\displaystyle M_{90}=2(1000)(0.12)(1.50)\sin45^\circ=254.6\ \text{N}.

For the straight path, \sin0^\circ=0 , so M_0=0 under the equal-flow, equal-speed assumptions. The chamber can still dissipate energy, as its nonzero K shows.

The turn therefore redirects more momentum and dissipates more energy in this comparison. If wall impact or lining durability is a concern, we would examine that layout first. Neither 135 W nor 255 N supplies an erosion or structural acceptance limit.

We can make a separate loss check if the project specifies an allowance. Suppose it allows at most 0.10 m of local head loss: the turned layout exceeds it by about 15 mm, while the straight layout passes. This allowance is an example, not a HEC-22 limit. Without such a requirement, a calculated loss of 0.115 m does not establish failure.

Does including K raise the HGL at every node? #

In the first example, including the sump loss raised the assumed HGL enough to fail the rim check. How does this work out when we calculate the levels through a connected network? We ran a detention-drain model in MiTS with and without the angle-derived sump losses, keeping the pipe layout and storm inputs unchanged. The model setup and full comparison are in Local Bend Losses and Water Superelevation in a Detention Drain.

For the 5-year ARI, 20-minute storm, two of the nodes give:

At A03, the peak rises as we might expect. At B03, it falls. The added losses alter flow and its timing through the network; the change at a node includes the effects of losses elsewhere. We therefore cannot obtain the new peak HGL by adding that node’s local head loss to its old peak.

The rim-margin calculation is still the same. With rim elevations unchanged, A03 loses 13 mm of minimum clearance and B03 gains 17 mm. We would compare the resulting margins with the project’s requirement, just as in the first two examples. The routing run supplies the HGL values that those simplified examples assumed.

What about the water rising at the outside wall? #

Head loss measures energy dissipated as water passes through the sump. Superelevation is the water-surface difference across a turn. The pressure difference across the flow helps turn the water; even an ideal curved flow with no energy loss can have a higher water surface on the outside.

Although both effects depend on velocity squared, we cannot obtain the outside-wall rise from K . Changing K can change the routed velocity and depth, and therefore change the inputs to a local-rise calculation, but it does not provide the shape of the water surface inside the sump.

The detention-drain case study also compares chamber sizes. Changing the chamber dimensions leaves the angle-only K and the routed centreline HGL unchanged in that comparison, while changing the separate local-rise estimate. Thus a chamber-size change needs to be assessed against the quantity we want to improve.

The companion superelevation article explains how the sump footprint supplies a reference radius and when its preliminary rise estimate can be used. Its geometry and free-surface flow conditions must be checked before applying it. The estimate is not a guaranteed upper limit on the local level.

Where the design depends on actual wall pressures, jets or overtopping, we need a method that resolves those effects. A suitable two-dimensional model can examine predominantly horizontal, depth-averaged flow. Vertical jets and detailed impact pressures may require validated three-dimensional computational fluid dynamics (CFD), measurements or physical modelling. Decide which level, pressure or force the design needs before commissioning the study.

Do we need a full storm simulation? #

We need to distinguish the routing method from the inflow supplied to it. In the EPA Storm Water Management Model (SWMM) used by MiTS, Steady Flow routing passes each inflow hydrograph through the conduits without travel-time delay or a change in shape; it omits backwater, entrance/exit losses and pressurised flow. Dynamic Wave routing includes these effects. EPA SWMM 5.2 User’s Manual, Section 3.4.5.

We can run Dynamic Wave with constant design inflows and a fixed downstream level, allowing the model to settle to a steady condition. Storage is still represented during that settling period. The final steady state, however, cannot show how a storm hydrograph is delayed or attenuated, or whether tributary peaks coincide with high downstream water levels. Some imposed inflows may produce continuing accumulation or flooding instead of a settled state.

For preliminary sizing, a Rational Method peak flow compared with Manning’s normal-flow capacity is useful. That isolated pipe check does not solve the upstream water level caused by a sump loss or downstream backwater. A steady energy-profile calculation that explicitly includes local losses can do more; steady conditions do not, by themselves, exclude K .

Likewise, adding every branch’s individual peak assumes that they occur together. That assumption may overstate the coincident inflow, while an unfavourable downstream level may govern the HGL at another time. Use a full event when timing or storage affects the decision, and find the maximum HGL, velocity, loss power and momentum change separately. They need not occur at the same reporting time.