Culverts beneath roads #

A culvert carries water beneath a road or embankment. In Figure 1, water enters from the upstream drain, passes through the covered barrel and leaves through the outlet. Its inside bottom is the invert and its inside roof is the soffit.

We need the upstream water level, the headwater, required to pass the design discharge. Two parts of the culvert can set it: the entrance, and the barrel together with its downstream end. So we calculate two candidate headwaters, both measured above the inlet invert, and design for the larger. The inlet-control headwater HW_i is what the entrance needs; the outlet-control headwater HW_o is what the barrel and downstream conditions need.

We will consider three conditions: free surface throughout, partial submergence at one end, and submergence at both ends. The first follows familiar open-channel hydraulics. For the other two, we derive both headwaters from energy and continuity, compare them with MSMA and work through numerical calculations before drawing the design conclusion. Two results organise the article:

• The outlet-control headwater has one equation for every submerged case. Whether the inlet, the outlet or both are submerged, our derivation leads to the same MSMA headwater equation. Only its downstream reference level changes with the outlet condition, and the derivation shows how exact the equation is in each case.
• The inlet-control headwater changes with the inlet condition. A free inlet passes water through a critical section; a submerged inlet behaves as an orifice. We derive the form of each relationship, but the entrance losses need coefficients fitted to experiments.

We will derive the culvert equations from physics and compare them with MSMA 2 and the supplied MSMA 3 DFR2-Rev2 draft; references to MSMA 3 mean that draft. We also redo the manual examples to verify all the calculation steps.

Figure 2 identifies the inlet and outlet submergence used to organise the article. The internal profile is part of the calculation: a submerged opening may lead to a partly full barrel, and ventilation can allow free-surface sections even when both ends are submerged. MSMA 2 §18.2.1 calls for a profile calculation to establish the filled length.

Free surface throughout #

When water stays below the roof of the barrel throughout, the culvert follows open-channel hydraulics. Our readers are very familiar with this, so we will not devote much time to it.

Interested readers can consult the Newton-to-Bernoulli article, which derives the energy balance. The tailwater and backwater article derives friction and calculates a free-surface profile.

The inlet and outlet comparison still applies. The inlet is unsubmerged, so the entrance requirement HW_i uses the free-inlet relationship derived later under partial submergence. The outlet requirement HW_o comes from a backwater calculation: integrate the free-surface profile upstream from the outlet, starting at the tailwater or critical depth, then add the inlet velocity head and entrance loss. MSMA’s closed-form outlet equation is not used here, because it counts full-barrel losses; MSMA itself limits it to headwater above three-quarters of the barrel height. As in every case, design for the larger of the two headwaters.

Partial submergence #

There are two possible condition with regards to partial submergence: a submerged inlet with a free outlet, and a free inlet with a submerged outlet. We first derive the relationships for each arrangement, then determine whether the entrance or downstream flow governs the required headwater.

Levels, pressure and section energy #

In Figure 3, headwater HW is upstream pool depth above the inlet invert; tailwater TW is receiving-pool depth above the outlet invert. The invert elevations z_i and z_o use one zero level. With barrel length L and dimensionless slope S_0 positive downward,

\displaystyle z_i-z_o=LS_0,\qquad (z_i+HW)-(z_o+TW)=HW+LS_0-TW.

All elevations, depths and heads are in metres. An inlet is submerged when its pool covers the top of the opening; a submerged outlet has TW>D , where D is the internal height or diameter. For a free inlet the upstream pool is below the inlet roof.

In Figure 4, water depth y is measured from the invert. Gauge pressure p measures pressure above atmosphere. Under hydrostatic pressure, invert pressure head is h_p=p/\gamma , where \gamma=\rho g is the specific weight of water, \rho is water density and g is gravitational acceleration. In a ventilated free-surface section, p=\rho gy and h_p=y . In a full section, y=D while h_p can vary independently. The open tube shows the pressure-supported level, or hydraulic grade line, HGL, at elevation z_b+h_p , where z_b is the local invert elevation.

Let Q be the steady discharge through one barrel in cubic metres per second. At depth y , wetted area is A(y) in square metres, mean velocity is V(y)=Q/A(y) in metres per second, wetted perimeter is P(y) in metres and hydraulic radius is R(y)=A(y)/P(y) . The water-surface width is T(y) . Full-section values carry subscript F : A_F,P_F,R_F,V_F .

Take hydrostatic pressure, a straight constant section, small bed slope, constant roughness and a dimensionless kinetic-energy correction \alpha=1 . The pools move slowly. We derive smooth free-surface and filled reaches; a hydraulic jump requires an additional momentum balance.

Geometry for the filled and free-surface reaches #

A partially submerged culvert can contain both filled and free-surface reaches. To calculate friction and velocity in each reach, we need its wetted area and perimeter. The next two cross-sections supply those quantities: use the full geometry where water touches every wall, and the depth-dependent geometry where an air space remains.

In Figure 5, a circular pipe has A_F=\pi D^2/4 and R_F=D/4 . A box of width B has A_F=BD and R_F=BD/[2(B+D)] . The wetted full-box perimeter includes its roof.

For the free-surface sections in Figure 6, a box has A(y)=By , P(y)=B+2y and T=B . For a pipe, with wetted central angle \phi in radians,

\displaystyle \begin{aligned}y&=\frac D2[1-\cos(\phi/2)],\\ A(y)&=\frac{D^2}{8}(\phi-\sin\phi),\\ P(y)&=\frac{D\phi}{2},\qquad T(y)=D\sin(\phi/2).\end{aligned}

The critical-depth calculation uses this same geometry: Q^2T(h_c)=gA(h_c)^3 , where h_c is critical depth. This is the critical-flow condition from the linked open-channel derivation. For a box, h_c=[Q^2/(gB^2)]^{1/3} ; for a pipe, solve the segment-area condition numerically.

Conservation of energy along a free surface #

Let x measure distance downstream from the inlet. Invert elevation is z(x)=z_i-S_0x . We start with Bernoulli’s section energy head: elevation plus pressure head plus velocity head. The derivation in Newton to Bernoulli gives this energy per unit weight of water.

In Figure 7, the selected section is inside a ventilated reach. Its invert is at elevation z(x) , and the water surface is y(x) above that invert. Hydrostatic pressure gives invert pressure head p/\gamma=y(x) . With \alpha=1 and continuity V(x)=Q/A(y(x)) , the three heads add to the section energy head \mathcal H(x) :

\displaystyle \begin{aligned}\mathcal H(x)&=z(x)+\frac{p(x)}{\gamma}+\frac{V(x)^2}{2g},\\ &=z(x)+y(x)+\frac{Q^2}{2gA(y(x))^2}.\end{aligned}

The HGL is the free-surface elevation z+y . The energy grade line, EGL, plots \mathcal H and is higher by V^2/(2g) . Friction lowers the EGL as water travels downstream; in Figure 7 the depth also grows downstream, so the velocity head between the two lines shrinks.

Write S_f(y) for friction head lost per unit length. Conservation of mechanical energy with wall dissipation gives d\mathcal H/dx=-S_f(y) . Since dA/dy=T , the derivative of velocity head with respect to depth is

\displaystyle \frac{d}{dy}\left(\frac{Q^2}{2gA(y)^2}\right)=-\frac{Q^2}{gA(y)^3}\frac{dA}{dy}=-\frac{Q^2T}{gA^3}.

Differentiate the section head along x , using dz/dx=-S_0 and the chain rule for y(x) :

\displaystyle -S_0+\left(1-\frac{Q^2T}{gA^3}\right)\frac{dy}{dx}=-S_f(y).

Define Fr^2=Q^2T/(gA^3) , the squared Froude number. Rearranging gives the free-surface profile equation:

\displaystyle \frac{dy}{dx}=\frac{S_0-S_f(y)}{1-Fr^2(y)}.

Manning supplies the empirical resistance input. For roughness n , in \text{s}/\text{m}^{1/3} under the SI convention,

\displaystyle V=\frac1nR^{2/3}\sqrt{S_f}\quad\Longrightarrow\quad S_f(y)=\frac{n^2Q^2}{A(y)^2R(y)^{4/3}}.

Thus continuity and energy determine how depth changes once the section geometry and calibrated resistance are supplied.

Conservation of energy along a filled reach #

In a filled reach, water touches the roof everywhere, as in Figure 8. We start from the same two statements used for the free surface. The section energy head is still

\displaystyle \mathcal H(x)=z(x)+\frac{p(x)}{\gamma}+\frac{V(x)^2}{2g},

and friction still removes head along the barrel, d\mathcal H/dx=-S_f .

Two things change once the barrel is full. First, the depth is fixed at y=D , so pressure is no longer set by depth. The invert pressure head becomes its own unknown, h_p(x) , shown by the water level in the open tube. Second, the flow area is fixed at A_F , so continuity gives a constant velocity V_F=Q/A_F and the velocity head does not change along the barrel. The section head becomes

\displaystyle \mathcal H(x)=z(x)+h_p(x)+\frac{V_F^2}{2g}.

Now differentiate along x . The invert term gives dz/dx=-S_0 and the velocity-head term gives zero, so

\displaystyle \frac{d\mathcal H}{dx}=-S_0+\frac{dh_p}{dx}=-S_{f,F}\quad\Longrightarrow\quad\frac{dh_p}{dx}=S_0-S_{f,F}.

Here S_{f,F} is the friction slope of the full section. Manning’s equation with the full area and hydraulic radius gives

\displaystyle S_{f,F}=\frac{n^2V_F^2}{R_F^{4/3}},

which is constant along a uniform barrel. So in Figure 8 the EGL and HGL are parallel straight lines falling at S_{f,F} , while the invert falls at S_0 . The pressure head h_p is the gap between the HGL and the invert, so it changes linearly at the difference of the two slopes. When friction is steeper than the bed, as drawn, h_p falls downstream.

At a ventilated transition where water just touches the roof, top gauge pressure is zero and invert pressure head is h_p=D . This provides the matching condition between the filled and free-surface calculations.

Submerged inlet/free outlet #

In Figure 9, the upstream barrel is full and a free surface forms farther downstream. Let \ell_o be the free-surface length and \ell_F=L-\ell_o the filled length. Assume low tailwater and critical depth at the outlet, y_o=h_c<D . The selected subcritical drawdown has S_f(y)>S_0 between h_c and D .

Start with the free-surface equation just derived. Solve it for a small downstream distance dx :

\displaystyle \frac{dy}{dx}=\frac{S_0-S_f(y)}{1-Fr^2(y)}\quad\Longrightarrow\quad dx=\frac{1-Fr^2(y)}{S_0-S_f(y)}\,dy.

In Figure 9, roof separation is at x=L-\ell_o , where y=D . The outlet is at x=L , where y=h_c . Integrate between those two sections:

\displaystyle \begin{aligned}\int_{L-\ell_o}^{L}dx&=\int_D^{h_c}\frac{1-Fr^2(y)}{S_0-S_f(y)}\,dy,\\ \ell_o&=\int_{h_c}^{D}\frac{1-Fr^2(y)}{S_f(y)-S_0}\,dy.\end{aligned}

The second line reverses the depth limits and changes the denominator’s sign. On this subcritical drawdown branch, both 1-Fr^2 and S_f-S_0 are positive, so the resulting length is positive.

Now use the filled-reach equation dh_p/dx=S_0-S_{f,F} . At the inlet x=0 , pressure head is h_{p,i} . At roof separation x=L-\ell_o , pressure head is D . Because the right-hand side is constant,

\displaystyle \begin{aligned}\int_{h_{p,i}}^D dh_p&=\int_0^{L-\ell_o}(S_0-S_{f,F})\,dx,\\ D-h_{p,i}&=(S_0-S_{f,F})(L-\ell_o),\\ h_{p,i}&=D+(S_{f,F}-S_0)(L-\ell_o).\end{aligned}

The inlet pressure head therefore exceeds D by the friction slope minus bed slope, multiplied by the filled length.

Let entrance loss be H_e=K_eV_F^2/(2g) , with a calibrated dimensionless coefficient referenced to full-barrel velocity. The upstream pool supplies inlet pressure head, velocity head and entrance loss:

\displaystyle HW=h_{p,i}+\frac{V_F^2}{2g}+H_e.

These equations calculate the filled length and headwater within the stated steady profile model. A contracted inlet jet can instead leave the barrel partly full immediately after the entrance; the inlet-control calculation below treats that possibility.

Reproducing MSMA’s headwater equation #

Headwater balance from the water-profile calculation #

We first rearrange our own calculation for the filled upstream reach and free-surface outlet. Define the full-section reference head H_F=(1+K_e)V_F^2/(2g)+S_{f,F}L . Substituting the inlet pressure head calculated above gives

\displaystyle \begin{aligned}HW&=D+(S_{f,F}-S_0)(L-\ell_o)+(1+K_e)\frac{V_F^2}{2g},\\ &=D+(S_0-S_{f,F})\ell_o+\left[S_{f,F}L+(1+K_e)\frac{V_F^2}{2g}\right]-LS_0,\\ &=\underbrace{D+(S_0-S_{f,F})\ell_o}_{h_{0,p}}+H_F-LS_0.\end{aligned}

Here h_{0,p}=D+(S_0-S_{f,F})\ell_o is the height of the filled-reach HGL extended to the outlet plane in Figure 9. We obtain it from the calculated free-surface length \ell_o . The actual outlet water depth is h_c . This rearrangement prepares the comparison with MSMA below.

MSMA’s outlet reference and headwater equation #

In Figure 10, MSMA uses an outlet reference height h_0 to calculate the outlet-control headwater HW_o . Despite its subscript, HW_o is not a level at the outlet. It is the upstream pool depth above the inlet invert that the barrel and downstream conditions require; the o stands for outlet control. Tailwater enters it only through h_0 . By contrast, y_o and V_o are depth and velocity at the outlet itself, and the subscript in h_0 is a zero. The two quantities have different reference bottoms: h_0 is above the outlet invert, while HW_o is above the inlet invert. MSMA 2 Eq. 18.9 / MSMA 3 Eq. 5.26 gives

\displaystyle HW_o=h_0+H-LS.

The required head H raises the upstream water level above the outlet reference; subtracting invert fall LS converts that upstream level to a depth above the higher inlet invert. In our notation, H=H_F and S=S_0 .

There is one h_0 , selected according to the downstream condition. A submerged outlet uses the actual tailwater, h_0=TW (MSMA 2 Eq. 18.10 / MSMA 3 Eq. 5.27). With a free outlet, MSMA compares tailwater with an approximate extended-HGL height halfway between critical depth and the roof, (h_c+D)/2 (MSMA 2 Eq. 18.11 / MSMA 3 Eq. 5.28), and takes the higher level. With critical depth capped at barrel height, h_c^*=\min(h_c,D) , the complete selection rule is

\displaystyle h_0\approx\begin{cases}TW,&TW\geq (h_c^*+D)/2,\\ (h_c^*+D)/2,&TW<(h_c^*+D)/2.\end{cases}

Choose one branch and insert that single height into the headwater equation. The midpoint is a vertical height at the outlet plane. The culvert length L remains the known total length; the approximation estimates headwater without calculating where water leaves the roof. MSMA 2, p. 18-7 and design procedure.

Comparison of the two headwater calculations #

Our water-profile calculation gives HW=h_{0,p}+H_F-LS_0 , with h_{0,p}=D+(S_0-S_{f,F})\ell_o . We calculate \ell_o by integrating the depth equation, then calculate the outlet reference height and upstream headwater. Roof separation is at x=L-\ell_o .

MSMA’s design approximation gives HW_o\approx h_0+H_F-LS_0 , using the outlet height selected above. The headwater balance and full-section reference head are the same; MSMA replaces the calculated h_{0,p} by its selected approximate h_0 . For the same inputs, the headwater difference is therefore HW_o-HW\approx h_0-h_{0,p} .

So the headwater equation follows from our derivation, but the midpoint itself does not. Setting h_{0,p}=(h_c+D)/2 in h_{0,p}=D+(S_0-S_{f,F})\ell_o shows when the two agree exactly:

\displaystyle \ell_o=\frac{D-h_c}{2(S_{f,F}-S_0)}.

Nothing in the profile equation forces the drawdown to have this length. The midpoint is an empirical design rule that FHWA HDS-5 also uses. Taking the higher of TW and the midpoint is likewise a design convention: when tailwater exceeds the midpoint, MSMA lets tailwater, not the drawdown, set the outlet reference. For the pipe example below, the exact-agreement length is 25.18\;\text{m} against a calculated \ell_o=25.81\;\text{m} , which is why the two headwaters differ by only millimetres.

In the pipe example below, MSMA’s midpoint gives h_0=1.0761\;\text{m} and upstream headwater HW_o=2.0569\;\text{m} . Integrating the water profile gives h_{0,p}=1.0730\;\text{m} and HW=2.0538\;\text{m} : a 0.0031\;\text{m} headwater difference. MSMA regards its shortcut as satisfactory for calculated headwater above 0.75D ; below that level it directs the designer to profile calculations or capacity charts. MSMA 2, p. 18-7.

Free inlet/submerged outlet #

In Figure 11, a free surface enters the culvert and rises toward the roof as the invert falls. The downstream reach is filled and discharges into tailwater above the outlet roof. We consider a smooth subcritical profile, with no hydraulic jump.

Let x_s be the distance from the inlet to roof contact in Figure 11. The filled length is \ell_F=L-x_s . We start at the submerged outlet, where water discharges into a slow pool. Let h_{p,o} be the pressure head at the invert just inside the outlet. Section energy there equals pool energy plus an exit loss of one barrel velocity head:

\displaystyle z_o+h_{p,o}+\frac{V_F^2}{2g}=z_o+TW+\frac{V_F^2}{2g}\quad\Longrightarrow\quad h_{p,o}=TW.

For the filled reach, start again with dh_p/dx=S_0-S_{f,F} . Its upstream end is roof contact, x=x_s , h_p=D . Its downstream end is the outlet, x=L , h_p=TW . Integrate between those limits:

\displaystyle \begin{aligned}\int_D^{TW}dh_p&=\int_{x_s}^{L}(S_0-S_{f,F})\,dx,\\ TW-D&=(S_0-S_{f,F})(L-x_s),\\ \ell_F=L-x_s&=\frac{TW-D}{S_0-S_{f,F}}.\end{aligned}

This smooth arrangement requires S_0>S_{f,F} and 0<\ell_F<L . Let y_i be depth just inside the free inlet, distinct from pool headwater HW . For the upstream free surface, separate the same depth equation:

\displaystyle dx=\frac{1-Fr^2(y)}{S_0-S_f(y)}\,dy.

At the inlet x=0 , y=y_i ; at roof contact x=x_s , y=D . Thus

\displaystyle \begin{aligned}\int_0^{x_s}dx&=\int_{y_i}^D\frac{1-Fr^2(y)}{S_0-S_f(y)}\,dy,\\ x_s=L-\ell_F&=\int_{y_i}^D\frac{1-Fr^2(y)}{S_0-S_f(y)}\,dy.\end{aligned}

Solve this expression for y_i . For the selected smooth profile, Fr<1 and S_0>S_f(y) throughout that reach. The pool-to-inlet energy balance then gives

\displaystyle HW=y_i+\frac{[Q/A(y_i)]^2}{2g}+H_e.

These relationships calculate this free-inlet/submerged-outlet profile from its downstream boundary. The entry loss H_e must retain its stated velocity reference. If an entrance produces supercritical flow and a hydraulic jump forms downstream, match the jump with momentum as well as the energy and pressure relationships; the smooth-profile equations above describe a different branch.

Comparison with the MSMA balance #

To compare this calculation with MSMA, first write the pool-to-pool energy balance for the actual two reaches. Friction loss is the integral over the free-surface length plus the constant-slope loss over the filled length:

\displaystyle H_f=\int_0^{x_s}S_f(y(x))\,dx+S_{f,F}(L-x_s).

The slow upstream pool supplies that friction loss, entrance loss and the outlet velocity-head loss:

\displaystyle \begin{aligned}(z_i+HW)-(z_o+TW)&=H_e+H_f+\frac{V_F^2}{2g},\\ HW&=TW+H_e+H_f+\frac{V_F^2}{2g}-LS_0.\end{aligned}

This is the same energy balance used to calculate y_i and headwater above. MSMA approximates the friction integral using full-section resistance over the whole length, H_f\approx S_{f,F}L . With the same entrance loss H_e=K_eV_F^2/(2g) , substitution gives

\displaystyle \begin{aligned}HW_o&\approx TW+(1+K_e)\frac{V_F^2}{2g}+S_{f,F}L-LS_0,\\ &=TW+H_F-LS_0.\end{aligned}

This is MSMA 2 Eq. 18.9 / MSMA 3 Eq. 5.26 with submerged-outlet reference h_0=TW . The balance follows conservation of energy; replacing the free-surface friction integral by full-section friction is the approximation we compare numerically. MSMA 2, §§18.2.1–18.2.2 and design procedure.

Inlet and outlet control under partial submergence #

The two arrangements above tell us which opening is under water. Control is a separate question: for the chosen discharge, which part of the culvert needs the higher upstream pool? We calculate two candidate headwaters, both measured from the inlet invert to the upstream pool surface.

With inlet control, the entrance alone limits the flow. Water passes through a critical section just inside a free inlet, or through a contracted jet at a submerged inlet, and the barrel downstream runs partly full, usually supercritical. Downstream conditions cannot travel upstream through that section, so the barrel and tailwater do not affect headwater. The inlet-control headwater HW_i depends only on the entrance shape, its size and Q .

With outlet control, the barrel and the downstream end limit the flow. The profile is subcritical or full, so tailwater and barrel friction influence reach back to the inlet. The profile calculations for the two arrangements above are this calculation: they integrate upstream from critical depth at a free outlet, or from tailwater at a submerged outlet. MSMA’s HW_o=h_0+H-LS_0 is their design shortcut.

Which opening is submerged does not decide control. A submerged inlet with a free outlet is outlet-controlled in the pipe example below but inlet-controlled in MSMA Example 18.B1. A free inlet with a submerged outlet is usually outlet-controlled; on a steep barrel, however, the entrance can still produce supercritical flow that jumps back to subcritical before the filled reach, and then the entrance governs. The same comparison applies when both ends are submerged. A jump or a change in tailwater can change which branch persists. HEC-RAS culvert-flow analysis.

The outlet-control relationships are those derived above. The next subsection supplies the inlet-control relationships.

Physical inlet relationships and MSMA calibration #

Let HW_i be the upstream headwater, above the inlet invert, that the entrance alone requires to pass Q ; it ignores any restriction from the barrel or tailwater. For a free inlet with a critical section, as in Figure 12, let V_c=Q/A(h_c) be the critical velocity and E_c=h_c+V_c^2/(2g) the specific energy at that section. The pool-to-control energy balance, with negligible approach velocity and local invert change, gives HW_i=E_c+h_{e,i} , where h_{e,i} is the entrance loss to that critical section. Dividing by height,

\displaystyle \frac{HW_i}{D}=\frac{E_c}{D}+\frac{h_{e,i}}{D}.

For a submerged inlet, let C_{d,i} be its dimensionless calibrated discharge coefficient and H_{eff} the head across the contracted jet. The orifice relationship gives

\displaystyle Q=C_{d,i}A_F\sqrt{2gH_{eff}}\quad\Longrightarrow\quad H_{eff}=\frac{Q^2}{2gC_{d,i}^2A_F^2}.

In Figure 13, the pool head HW_i supplies H_{eff} plus jet piezometric head H_j above the inlet invert. Therefore

\displaystyle \frac{HW_i}{D}=\frac{1}{2gC_{d,i}^2}\left(\frac{Q}{A_F\sqrt D}\right)^2+\frac{H_j}{D}.

The energy balances contain entrance effects that we have yet to specify: loss to a free-inlet critical section, or contraction and jet pressure at a submerged inlet. Experiments measure the headwater needed to pass known discharges through particular entrance shapes. FHWA fits coefficients to those measurements. The fitted relationships supply those entrance effects for the tested configurations and flow ranges. FHWA inlet-control experiments and equations.

Write X=K_uQ/(A_F\sqrt D) . FHWA fitted its coefficients in US customary units; K_u=1.811=\sqrt{1/0.3048} converts Q/(A_F\sqrt D) from SI to those units, so the published coefficients can be used with SI inputs. For a free inlet, start with our energy balance and substitute the fitted entrance loss, a power law in X with fitted coefficients K and M :

\displaystyle \begin{aligned}HW_i&=E_c+h_{e,i},\\ h_{e,i}&\approx DKX^M,\\ HW_i&\approx E_c+DKX^M.\end{aligned}

The published fit also applies a slope correction DK_sS_0 to headwater, with fitted coefficient K_s ; it is an empirical adjustment with no counterpart in our energy balance. Including that correction and dividing by D gives FHWA Form 1:

\displaystyle \frac{HW_i}{D}\approx\frac{E_c}{D}+KX^M+K_sS_0.

The slope correction adjusts the inlet headwater relationship; the entrance-loss term is DKX^M . Once the section geometry gives E_c and the selected entrance supplies K,M,K_s , this equation calculates HW_i directly.

For a submerged inlet, substitute Q/(A_F\sqrt D)=X/K_u into the physical relationship derived above:

\displaystyle \frac{HW_i}{D}=\frac{X^2}{2gC_{d,i}^2K_u^2}+\frac{H_j}{D}.

Now assume that, for one entrance shape, the discharge coefficient C_{d,i} and the relative jet head H_j/D stay constant over the flow range. The physical relationship then has exactly the form cX^2+Y , with

\displaystyle c=\frac{1}{2gC_{d,i}^2K_u^2},\qquad Y=\frac{H_j}{D}.

The published submerged fit has this form, with fitted coefficients c and Y plus the same empirical slope correction:

\displaystyle \frac{HW_i}{D}=\frac{H_{eff}+H_j}{D}\approx cX^2+Y+K_sS_0.

For example, the square-edged concrete pipe row used below has c=0.0398 , which corresponds to C_{d,i}=[2(9.81)(1.811^2)(0.0398)]^{-1/2}\approx0.62 , a typical orifice value.

Therefore the usable inlet equation is

\displaystyle HW_i\approx D(cX^2+Y+K_sS_0).

The fitted c,Y,K_s describe the combined measured head requirement. Separately calculating the contraction coefficient and jet pressure would require a more detailed entrance model. Calibration represents those observed entrance effects; its accuracy depends on the selected configuration and applicable flow range. The coefficients differ from K_e , which specifies the entrance loss in the barrel/outlet calculation. FHWA HDS-5, Appendix A.

MSMA 2 provides inlet-control nomographs in Charts 18.A2–18.A4. Select the culvert shape and entrance scale, use the per-barrel discharge and size to read HW_i/D , then multiply by D to obtain inlet headwater. The nomograph is a graphical route to the inlet headwater relationship; the calibrated equations above provide a numerical route. FHWA HDS-5 Appendix A explains how its inlet equations generate its nomographs. In MSMA 2 Example 18.B1 below, we compare the chart reading with the calibrated equation for the same pipe and entrance. MSMA 2 design procedure, p. 18-10.

For outlet control, MSMA 2 Charts 18.A9–18.A11 instead supply the required head H , used in HW_o=H+h_0-LS_0 . These are a separate calculation from the inlet relationship. Calculate both candidate headwaters and compare them as follows.

Selecting the governing result #

For the same discharge, calculate the upstream headwater required by the entrance, HW_i , and by the barrel and downstream conditions, HW_o . Both requirements must hold at once. If the pool is lower than HW_i , the entrance cannot pass Q , whatever the barrel could carry. If it is lower than HW_o , the barrel and tailwater cannot pass it, however freely the entrance flows. The larger requirement is therefore the bottleneck:

\displaystyle HW_{req}=\max(HW_i,HW_o).

Inlet control governs when HW_i>HW_o ; outlet control governs otherwise. The flow then follows the profile of the governing branch. In MSMA Example 18.B1, for instance, the barrel runs partly full behind the entrance, not with the filled upstream reach that the outlet calculation assumed.

Check that the resulting profile is compatible with the assumed free or submerged opening. For example, a calculated headwater above the inlet roof changes a proposed free-inlet case. The selected profile must also retain any assumed critical section, filled reach or jump.

Illustrative pipe example: submerged inlet/free outlet #

Take a square-edged concrete pipe with D=1.20\;\text{m} , L=90.0\;\text{m} , S_0=0.001 , n=0.013\;\text{s}/\text{m}^{1/3} , K_e=0.50 , Q=3.00\;\text{m}^3/\text{s} and TW=0.30\;\text{m} . Use g=9.81\;\text{m/s}^2 . This is our own example using MSMA’s headwater equations. It represents a filled upstream barrel followed by a free surface near the outlet; the calculation below checks that condition.

Figure 14 labels the inputs and MSMA estimates. The full section gives A_F=1.131\;\text{m}^2 , R_F=0.300\;\text{m} and V_F=2.653\;\text{m/s} . The circular critical-depth solve gives h_c=0.9522\;\text{m} .

Calculate headwater using the MSMA equations #

MSMA calculates two candidate upstream headwaters, both measured from the inlet invert to the upstream pool surface. HW_i is the headwater the entrance needs to pass Q (inlet control). HW_o is the headwater the barrel and downstream conditions need (outlet control); tailwater affects it only through h_0 . Neither is the tailwater itself. We use the equations established above:

\displaystyle \begin{aligned}HW_i&=D(cX^2+Y+K_sS_0),\qquad X=\frac{1.811Q}{A_F\sqrt D},\\ H_F&=(1+K_e)\frac{V_F^2}{2g}+\frac{n^2LV_F^2}{R_F^{4/3}},\\ h_0&=\max\left[TW,\frac{h_c^*+D}{2}\right],\\ HW_o&=h_0+H_F-LS_0.\end{aligned}

First calculate the inlet requirement. For the square-edged concrete headwall, the calibrated row uses c=0.0398 , Y=0.67 and K_s=-0.5 . Here Q/(A_F\sqrt D)=2.42 lies in its submerged range and X=4.385 ; use unrounded values in the arithmetic. EPA SWMM circular-concrete row 1.

\displaystyle HW_i=1.20[0.0398X^2+0.67-0.5(0.001)]=1.7219\;\text{m}.

Next calculate the outlet reference. Critical depth h_c=0.9522\;\text{m} is below D , so h_c^*=h_c :

\displaystyle h_0=\max\left[0.30,\frac{0.9522+1.20}{2}\right]=1.0761\;\text{m}.

Write H_v=V_F^2/(2g) for the full-barrel velocity head, which MSMA counts as the exit loss. The velocity, entrance and friction heads are

\displaystyle \begin{aligned}H_v&=\frac{2.6526^2}{2(9.81)}=0.3586\;\text{m},\\ H_e&=0.50H_v=0.1793\;\text{m},\\ H_f&=\frac{0.013^2(90)(2.6526)^2}{0.300^{4/3}}=0.5329\;\text{m},\\ H_F&=H_v+H_e+H_f=1.0708\;\text{m}.\end{aligned}

Substitute into the MSMA outlet-control headwater balance:

\displaystyle HW_o=h_0+H_F-LS_0=1.0761+1.0708-90(0.001)=2.0569\;\text{m}.

Compare the two upstream headwater requirements:

\displaystyle HW_{req}=\max(1.7219,2.0569)=2.0569\;\text{m}.

Outlet control governs. Headwater covers the 1.20\;\text{m} inlet and exceeds the 0.75D criterion for the shortcut. Tailwater is below critical depth, so MSMA’s outlet-velocity check uses area at critical depth:

\displaystyle V_o=\frac{Q}{A(h_c)}=\frac{3.00}{0.9624}=3.117\;\text{m/s}.

The approximate outlet HGL reference h_0=1.0761\;\text{m} and actual outlet water depth h_c=0.9522\;\text{m} are different quantities. MSMA 2 design procedure, pp. 18-10–18-11.

Calculate the water profile and compare headwater #

For the same pipe and discharge, calculate the free-surface length with the integral derived above:

\displaystyle \ell_o=\int_{0.9521656}^{1.20}\frac{1-Fr^2(y)}{S_f(y)-0.001}\,dy=25.8112\;\text{m}.

Here Fr(y) and S_f(y) use the circular segment geometry. The filled length is L-\ell_o=90.0-25.8112=64.1888\;\text{m} . The filled-section friction slope is S_{f,F}=H_f/L=0.0059210 . First calculate inlet pressure head, then upstream pool headwater:

\displaystyle \begin{aligned}h_{p,i}&=D+(S_{f,F}-S_0)(L-\ell_o),\\ &=1.20+(0.0059210-0.001)(64.1888)=1.5159\;\text{m},\\ HW&=h_{p,i}+(1+K_e)\frac{V_F^2}{2g},\\ &=1.5159+(1+0.50)\frac{2.6526^2}{2(9.81)}=2.0538\;\text{m}.\end{aligned}

The MSMA midpoint approximation gives upstream headwater HW_o=2.0569\;\text{m} ; integrating the water profile gives upstream headwater HW=2.0538\;\text{m} . Their difference is 2.0569-2.0538=0.0031\;\text{m} , or 3.1\;\text{mm} for these inputs. Inlet pressure head is the separate result h_{p,i}=1.5159\;\text{m} . Both headwater results exceed the inlet requirement and give outlet control. The water-profile calculation also locates roof separation, 64.1888\;\text{m} from the inlet.

MSMA 2 Example 18.B1: submerged inlet/free outlet #

MSMA 2 Example 18.B1 selects two 1.05\;\text{m} diameter concrete pipes for total discharge 4.80\;\text{m}^3/\text{s} . Each barrel carries 2.40\;\text{m}^3/\text{s} over L=36\;\text{m} . The invert reduced levels (RL, elevations above the survey datum) fall from RL 57.98\;\text{m} to RL 57.50\;\text{m} , so LS_0=0.48\;\text{m} . Receiving water at RL 58.30\;\text{m} gives TW=0.80\;\text{m} . The headwater limit is 1.75\;\text{m} . MSMA 2, Example 18.B1, pp. 18-27–18-29.

Figure 15 identifies the physical condition: the selected upstream headwater covers the inlet roof, while receiving water is below the outlet roof. The manual selects inlet control and uses its partly full velocity charts for the barrel downstream of the entrance.

Calculate inlet headwater #

For the selected twin pipes, per-barrel area and flow are

\displaystyle A_F=\frac{\pi D^2}{4}=\frac{\pi(1.05)^2}{4}=0.86590\;\text{m}^2,\qquad Q=\frac{4.80}{2}=2.40\;\text{m}^3/\text{s}.

The manual reads HW_i/D=1.62 from Chart 18.A2, giving HW_i\approx1.70\;\text{m} . To reproduce that reading with the calibrated submerged-inlet equation, start with

\displaystyle HW_i=D(cX^2+Y+K_sS_0),\qquad X=\frac{1.811Q}{A_F\sqrt D}.

Use c=0.0398 , Y=0.67 , K_s=-0.5 and S_0=0.48/36 . Then

\displaystyle \begin{aligned}X&=\frac{1.811(2.40)}{0.86590\sqrt{1.05}}=4.89854,\\ HW_i&=1.05[0.0398X^2+0.67-0.5(0.48/36)]\\ &=1.6993\;\text{m}.\end{aligned}

The equation result rounds to the manual’s 1.70\;\text{m} upstream headwater. This entrance equation uses calibrated coefficients; the preceding orifice energy balance explains its squared-flow dependence.

Calculate outlet headwater #

The outlet check uses the headwater balance derived above and MSMA’s midpoint approximation:

\displaystyle HW_o=h_0+H_F-LS_0,\qquad h_0=\max\left[TW,\frac{h_c^*+D}{2}\right].

The manual reads critical depth h_c=0.80\;\text{m} and reports h_0=0.93\;\text{m} . Its loss chart gives H_F=0.85\;\text{m} . Substituting the reported readings gives

\displaystyle HW_o=0.93+0.85-0.48=1.30\;\text{m}.

For the direct-equation comparison, solve Q^2T(h_c)=gA(h_c)^3 using circular segment geometry. It gives h_c=0.8755\;\text{m} , compared with the manual’s chart value 0.80\;\text{m} . Both are below D , so h_c^*=h_c . Keeping that difference visible,

\displaystyle h_0=\max\left[0.80,\frac{0.8755+1.05}{2}\right]=0.9627\;\text{m}.

Use the loss chart’s roughness n=0.012 , R_F=D/4=0.2625\;\text{m} , V_F=Q/A_F=2.77168\;\text{m/s} and K_e=0.50 . The full-section loss equation gives

\displaystyle \begin{aligned}H_F&=(1+K_e)\frac{V_F^2}{2g}+\frac{n^2LV_F^2}{R_F^{4/3}},\\ &=(1+0.50)\frac{2.77168^2}{2(9.81)}+\frac{0.012^2(36)(2.77168)^2}{0.2625^{4/3}}\\ &=0.8243\;\text{m},\\ HW_o&=0.9627+0.8243-0.48=1.3070\;\text{m}.\end{aligned}

Thus the direct equations give outlet headwater 1.3070\;\text{m} , compared with the manual’s 1.30\;\text{m} . The critical-depth and loss intermediates differ from the chart readings; the inlet result matches the reported precision.

Select headwater and check outlet velocity #

Choose the larger of the inlet and outlet requirements:

\displaystyle \begin{aligned}\text{Manual:}\quad HW_{req}&=\max(1.70,1.30)=1.70\;\text{m},\\ \text{Equations:}\quad HW_{req}&=\max(1.6993,1.3070)=1.6993\;\text{m}.\end{aligned}

Both select inlet control and satisfy the 1.75\;\text{m} headwater limit. The manual’s upstream water level is RL 57.98+1.70=59.68\;\text{m} .

For its outlet-velocity check, the manual uses Colebrook and partly full charts. The Colebrook chart gives the pipe’s full-flow capacity Q_{chart}=3.40\;\text{m}^3/\text{s} and full-flow velocity V_{chart}=3.80\;\text{m/s} . The partly full chart converts the ratio Q/Q_{chart}=2.40/3.40\approx0.71 into V_o/V_{chart}=0.96 . Hence V_o=0.96(3.80)=3.65\;\text{m/s} . This is the manual’s chart calculation. Its selected depth is about 0.71\;\text{m} ; the manual calls for an outlet scour or hydraulic-jump check. The preceding outlet-controlled pipe example instead has a filled upstream reach followed by drawdown; both examples have submerged inlet/free outlet openings.

Illustrative box example: free inlet/submerged outlet #

This is our own example, using MSMA’s headwater procedure for the reverse opening arrangement; it is not a numerical example from the manual. Take a B=D=1.00\;\text{m} box with L=50.0\;\text{m} , S_0=0.020 , Q=0.40\;\text{m}^3/\text{s} , n=0.013\;\text{s}/\text{m}^{1/3} and TW=1.80\;\text{m} . Choose a square-edged entrance with flared wingwalls, and K_e=0.40 referenced to full-barrel velocity. Neglect pool velocities. The two profile calculations use this same entrance-loss input.

Figure 16 shows the calculated profile. The outlet is submerged; the upstream pool and free-surface section remain below the inlet roof.

Calculate the water depth and filled length #

We first use the relationships derived for this condition:

\displaystyle \begin{aligned}\ell_F&=\frac{TW-D}{S_0-S_{f,F}},\qquad x_s=L-\ell_F,\\ x_s&=\int_{y_i}^D\frac{1-Fr^2(y)}{S_0-S_f(y)}\,dy,\\ HW&=y_i+\frac{[Q/A(y_i)]^2}{2g}+K_e\frac{V_F^2}{2g}.\end{aligned}

Full area is A_F=1.00\;\text{m}^2 , full hydraulic radius is R_F=0.250\;\text{m} and V_F=0.400\;\text{m/s} . Manning gives S_{f,F}=0.000171693 . The filled downstream length is

\displaystyle \ell_F=\frac{1.80-1.00}{0.020-0.000171693}=40.3464\;\text{m}.

Hence x_s=50.0-40.3464=9.6536\;\text{m} . Solve the free-surface length integral for inlet depth, using A(y)=y , R(y)=y/(1+2y) and Fr^2=0.40^2/(9.81y^3) . The result is y_i=0.803946\;\text{m} and V(y_i)=0.497546\;\text{m/s} . The pool headwater follows from the actual inlet velocity head plus the stated entrance loss:

\displaystyle HW=0.803946+\frac{0.497546^2}{2(9.81)}+0.40\frac{0.400^2}{2(9.81)}=0.8198\;\text{m}.

This is below the 1.00\;\text{m} inlet roof. The inlet Froude number is 0.177 , and the selected free-surface reach stays subcritical up to roof contact. The calculation therefore gives the proposed free-inlet/submerged-outlet condition without a hydraulic jump.

MSMA comparison and governing control #

For the selected rectangular box with 30°–75° flared wingwalls, the FHWA Form 1 row uses K=0.026 , M=1.0 and K_s=-0.5 . Critical depth is h_c=0.253601\;\text{m} , so E_c=3h_c/2=0.380401\;\text{m} ; X=1.811(0.40)=0.7244 . Here Q/(A_F\sqrt D)=0.40 lies well within the unsubmerged range of Form 1. The calibrated inlet requirement is

\displaystyle HW_i=0.380401+1.00[0.026(0.7244)-0.5(0.020)]=0.3892\;\text{m}.

The constants are from the EPA SWMM flared-wingwall box row; the entrance-loss choice is consistent with the square-edged 30°–75° wingwall entry in MoDOT’s entrance-loss table.

Next apply MSMA’s outlet headwater equation with the submerged-outlet reference:

\displaystyle HW_o=TW+H_F-LS_0,\qquad H_F=(1+K_e)\frac{V_F^2}{2g}+\frac{n^2LV_F^2}{R_F^{4/3}}.

Substituting the full-section velocity and radius gives

\displaystyle \begin{aligned}H_F&=(1+0.40)\frac{0.400^2}{2(9.81)}+\frac{0.013^2(50)(0.400)^2}{0.250^{4/3}}\\ &=0.008155+0.003262+0.008585=0.020002\;\text{m},\\ HW_o&=1.80+0.020002-50(0.020)=0.8200\;\text{m}.\end{aligned}

Compare inlet and outlet requirements:

\displaystyle HW_{req}=\max(0.3892,0.8200)=0.8200\;\text{m}.

Outlet control governs. MSMA’s approximation gives upstream headwater HW_o=0.8200\;\text{m} ; the calculation with separate free-surface and filled reaches gives HW=0.8198\;\text{m} . Their difference is 0.0002\;\text{m} , or 0.2\;\text{mm} for these inputs. Both headwater results remain below the inlet roof, while TW=1.80\;\text{m} covers the outlet. The submerged outlet velocity is V_o=Q/A_F=0.400\;\text{m/s} . This example checks the approximation against the water-profile calculation for the stated inputs; its numbers are our illustrative calculation.

Both ends submerged #

Both pools cover their respective opening roofs. First solve for a barrel filled throughout, then check that the pressure profile permits that condition. A ventilated barrel containing an internal free surface or jump requires the corresponding profile calculation.

Derivation of the MSMA full-flow equation #

In Figure 17, the purple EGL falls through entrance, friction and exit losses. Its separation from the green HGL inside a full barrel is velocity head. Write A=A_F , R=R_F and V=V_F=Q/A for this full-flow calculation.

The calibrated entrance loss is H_e=K_eV^2/(2g) . Integrating the constant filled-reach friction slope over L gives

\displaystyle H_f=LS_{f,F}=\frac{n^2LV^2}{R^{4/3}}.

At the outlet, water mixes into the slow receiving pool. Approximate that dissipation by its barrel velocity head, H_v=V^2/(2g) . The total required head is

\displaystyle H=H_e+H_f+H_v=K_e\frac{V^2}{2g}+\frac{n^2LV^2}{R^{4/3}}+\frac{V^2}{2g}.

MSMA 2 Eq. 18.5 / MSMA 3 Eq. 5.22 #

Factor out V^2/(2g) :

\displaystyle H=\left(1+K_e+\frac{2gn^2L}{R^{4/3}}\right)\frac{V^2}{2g}.

The leading 1 supplies the exit allowance, K_e the entrance loss, and 2gn^2L/R^{4/3} the friction contribution. This reproduces the MSMA full-flow loss equation using 4/3 in place of its rounded exponent 1.33 and 2g=19.62\;\text{m/s}^2 in place of a rounded 19.6. MSMA 2, pp. 18-4–18-6.

Headwater and discharge equations #

Both slow pool surfaces have zero gauge pressure. Their surface-elevation difference supplies H :

\displaystyle (z_i+HW_o)-(z_o+TW)=H.

Substitute z_i-z_o=LS_0 to obtain MSMA’s submerged-outlet balance:

\displaystyle HW_o=TW+H-LS_0.

Here h_0=TW is the actual downstream pool level, as in MSMA 2 Eq. 18.10 and MSMA 3 Eq. 5.27. Substituting V=Q/A and solving the same balance for forward discharge gives

\displaystyle Q=A\sqrt{\frac{2g(HW_o+LS_0-TW)}{1+K_e+2gn^2L/R^{4/3}}}.

This result assumed that the barrel is filled throughout. To check it, integrate the filled-reach equation dh_p/dx=S_0-S_{f,F} upstream from the outlet, where h_p=TW :

\displaystyle h_p(x)=TW+(S_{f,F}-S_0)(L-x),\qquad h_{p,i}=TW+(S_{f,F}-S_0)L.

Gauge pressure at the soffit is \gamma(h_p-D) . A ventilated barrel stays full only where that pressure is not negative, so h_p\geq D is required along the whole barrel. Because h_p(x) is linear, checking the two ends is enough: TW\geq D at the outlet and h_{p,i}\geq D at the inlet. The inlet pressure head is also consistent with the headwater balance, since h_{p,i}+(1+K_e)V^2/(2g)=TW+H-LS_0=HW_o . If h_{p,i}<D , a free surface forms inside the barrel, and the partial-submergence profile calculations apply instead.

The discharge expression is our rearrangement of MSMA’s loss and headwater equations. It requires positive driving head, steady incompressible flow, a filled constant-section barrel, the specified resistance inputs and the slow-pool exit assumption. Add any material bend, screen, blockage or sediment effects to geometry and losses.

Inlet and outlet control with both ends submerged #

The comparison is the same as under partial submergence; only the outlet calculation changes. Here the full-barrel balance gives the outlet-control headwater HW_o=TW+H-LS_0 , with no profile integral. The submerged-inlet relationship, or MSMA’s inlet chart, gives HW_i . Again take HW_{req}=\max(HW_i,HW_o) .

Outlet control is usual here, because the submerged outlet keeps the barrel full and tailwater acts on it directly. If HW_i is larger, the entrance contraction is the bottleneck. The jet may then leave the barrel partly full behind the entrance, so the filled-barrel assumption no longer holds. Check it with the pressure-head condition h_p\geq D derived above, and use a profile calculation to locate any contracted jet or internal transition. HEC-RAS flow checks.

MSMA 2 Example 18.B4: both ends submerged #

MSMA 2 Example 18.B4 uses a box B=0.75\;\text{m} wide and D=0.60\;\text{m} high, with Q=0.90\;\text{m}^3/\text{s} , L=25\;\text{m} , n=0.012\;\text{s}/\text{m}^{1/3} and K_e=0.5 . Inlet and outlet invert RLs are 35.30\;\text{m} and 35.09\;\text{m} ; receiving-water RL is 36.59\;\text{m} . MSMA 2, Example 18.B4, pp. 18-36–18-38.

Figure 18 shows the published dimensions and levels. We repeat its design sequence and compare the direct loss equation with the manual’s nomograph reading.

Calculate inlet headwater and tailwater #

The manual’s inlet chart gives HW_i=0.90\;\text{m} . At the outlet, TW=36.59-35.09=1.50\;\text{m}>D , so h_0=TW=1.50\;\text{m} .

Calculate outlet headwater #

Use the full-flow equation derived above, followed by the headwater balance:

\displaystyle H=\left(1+K_e+\frac{2gn^2L}{R^{4/3}}\right)\frac{V^2}{2g},\qquad HW_o=TW+H-LS_0.

The full section has A=BD=0.450\;\text{m}^2 , P=2(B+D)=2.70\;\text{m} , R=A/P=0.1667\;\text{m} and V=Q/A=2.00\;\text{m/s} . The components are

\displaystyle \begin{aligned}H_v&=\frac{V^2}{2g}=\frac{2.00^2}{2(9.81)}=0.2039\;\text{m},\\ H_e&=K_eH_v=0.50(0.2039)=0.1019\;\text{m},\\ H_f&=\frac{n^2LV^2}{R^{4/3}}=\frac{0.012^2(25)(2.00)^2}{(1/6)^{4/3}}=0.1570\;\text{m}.\end{aligned}

Thus H=0.4628\;\text{m} and LS_0=35.30-35.09=0.210\;\text{m} , giving

\displaystyle HW_o=1.50+0.4628-0.210=1.7528\;\text{m}.

The manual reads H=0.47\;\text{m} from its nomograph and reports HW_o=1.76\;\text{m} . Direct calculation is 0.0072\;\text{m} lower for both quantities.

Now check the filled-barrel assumption. Friction slope is S_{f,F}=H_f/L=0.1570/25=0.00628 and bed slope is S_0=0.210/25=0.0084 . The inlet pressure head is

\displaystyle h_{p,i}=TW+(S_{f,F}-S_0)L=1.50+0.1570-0.210=1.447\;\text{m}.

Both TW=1.50\;\text{m} and h_{p,i}=1.447\;\text{m} exceed D=0.60\;\text{m} , so the barrel stays full. Headwater HW_o=1.7528\;\text{m} also covers the 0.60\;\text{m} inlet, so both ends are submerged as assumed.

Select headwater and check outlet velocity #

The outlet requirement exceeds HW_i=0.90\;\text{m} , so outlet control governs. Both calculations satisfy the example’s 2.20\;\text{m} headwater limit. The submerged outlet velocity is V_o=2.00\;\text{m/s} , requiring the erosion check stated in the manual.

Conclusion #

In this article we derived the inlet- and outlet-control headwaters from first principles for culverts with one end submerged, in both arrangements, and with both ends submerged. The outlet-control headwater follows fully from energy, continuity and Manning friction; the inlet-control headwater takes its form from the same physics and its entrance coefficients from experiments. We compared both with the MSMA equations and redid the manual’s worked examples. Our derivations provide the missing link between the laws of energy and continuity and the MSMA culvert equations.