Manning's Equation: Formula, n Values and Worked Examples

Manning’s equation gives the mean velocity of steady, uniform flow in an open channel from the roughness of its boundary, the shape of the water’s cross-section and the slope it runs on. In SI units it is $V = \frac{1}{n} R^{2/3} S^{1/2}$ , and the discharge follows as $Q = VA$ . It is the equation used to size ditches, canals, culverts and storm sewers, and to find the depth at which a known flow will run in them.

Further down, two worked examples on one earth channel, the flow at a given depth and the depth for a given flow, are solved by hand and checked against Rischio’s engine.

The formula and its terms

In SI units, with lengths in metres and time in seconds:

V = \frac{1}{n} R^{2/3} S^{1/2}, \qquad Q = V A = \frac{1}{n} A R^{2/3} S^{1/2}

$V$ is the mean velocity over the cross-section (m/s).
$Q$ is the discharge (m³/s).
$n$ is Manning’s roughness coefficient of the channel’s boundary.
$A$ is the flow area, the cross-section of the water (m²).
$P$ is the wetted perimeter, the length of boundary in contact with the water (m).
$R = A/P$ is the hydraulic radius (m).
$S$ is the slope of the energy line (m/m), which in uniform flow equals the slope of the bed.

The product $A R^{2/3}$ depends only on the shape of the section and the depth of the water. It is called the section factor for uniform flow, and $K = \frac{1}{n} A R^{2/3}$ is the conveyance of the section, so that $Q = K S^{1/2}$ . The section factor is what the second worked example solves for.

In much of Europe the equation is written with Strickler’s coefficient, $k_{St} = 1/n$ , as $V = k_{St} R^{2/3} S^{1/2}$ . A trowelled concrete channel with $n = 0.013$ has $k_{St} \approx 77$ .

Manning’s equation in US units

With $V$ in ft/s and $R$ in feet:

V = \frac{1.49}{n} R^{2/3} S^{1/2}

The equation was fitted to measurements, and its units don’t balance: $n$ formally carries $\text{s/m}^{1/3}$ . The 1.49 (1.486 to four figures) is the cube root of 3.2808, the number of feet in a metre. It lets the same $n$ serve in both systems, which is why tables of $n$ give one number for both.

What the equation assumes

Manning’s equation describes uniform flow: the state a channel reaches when the pull of gravity down the slope balances the friction of the boundary. It assumes:

Steady flow. The discharge at a section doesn’t change with time.
Uniform flow. Depth, area and velocity are the same from one section to the next, which takes a prismatic channel (the same shape, slope and lining along the reach) long enough for the flow to settle. The bed, the water surface and the energy line then run parallel, so the $S$ of the equation is the bed slope, the water-surface slope and the friction slope at once.
Fully rough turbulent flow. $n$ is a constant of the boundary only when the flow is turbulent and the roughness of the boundary controls the friction. That holds in practically every channel and sewer, though not in very shallow sheet flow over smooth surfaces.
A free surface. The water surface is at atmospheric pressure; a pipe flowing full under pressure is a different problem.

Near a gate, a weir, a culvert inlet or a change of slope, the flow is no longer uniform and the equation stops giving the depth. It still gives the friction slope inside the calculations that do.

The hydraulic radius and the geometry of common sections

The hydraulic radius is the flow area divided by the wetted perimeter, $R = A/P$ . Only the solid boundary under the water counts in $P$ ; the free surface doesn’t, because the air above it offers almost no resistance. $R$ measures how much water there is per metre of boundary slowing it down, so a deep, compact section carries more than a wide, shallow one with the same area.

Two shortcuts help check a result. In a wide, shallow channel, $P$ is close to the top width and $R$ is close to the depth. In a circular pipe flowing full or half full, $R = D/4$ , a quarter of the diameter: the hydraulic radius of a pipe isn’t its radius.

For the usual prismatic sections, with the depth $y$ measured from the bed and side slopes of $z$ horizontal to 1 vertical:

Section	Flow area $A$	Wetted perimeter $P$	Top width $T$
Rectangle, width $b$	$b\,y$	$b + 2y$	$b$
Trapezoid, bottom width $b$	$(b + z y)\,y$	$b + 2y\sqrt{1 + z^2}$	$b + 2 z y$
Triangle (V-shaped)	$z\,y^2$	$2y\sqrt{1 + z^2}$	$2 z y$
Circle, diameter $D$	$\frac{D^2}{8}(\theta - \sin\theta)$	$\frac{D\,\theta}{2}$	$D \sin\frac{\theta}{2}$

The rectangle is the trapezoid with $z = 0$ and the triangle is the trapezoid with $b = 0$ . $T$ is the width of the water surface. Manning’s equation doesn’t use it, but the Froude number and the critical depth do. For the circle, $\theta$ is defined in the next section.

The channel of Example 1 at a depth of 1.00 m. The wetted perimeter is the bed plus both wet sides, 3.00 + 2 × 1.80 = 6.61 m, and R = 4.50 / 6.61 = 0.681 m.

A circular pipe flowing partly full

A storm sewer, a sanitary sewer or a culvert that isn’t full has a free surface, so it is an open channel and Manning’s equation applies. Its geometry goes through the angle $\theta$ , in radians, that the water surface subtends at the centre of the pipe:

\theta = 2 \arccos\left(1 - \frac{2y}{D}\right), \qquad A = \frac{D^2}{8}(\theta - \sin\theta), \qquad P = \frac{D\,\theta}{2}, \qquad R = \frac{D}{4}\left(1 - \frac{\sin\theta}{\theta}\right)

Take a 600 mm concrete storm sewer with $n = 0.013$ , laid at 0.3% ( $S = 0.003$ ) and running with 0.45 m of water, three quarters full:

\theta = 2 \arccos(1 - 1.5) = 2 \arccos(-0.5) = 240° = 4.189 \text{ rad}

A = \frac{0.60^2}{8}(4.189 - \sin 240°) = 0.045 \times (4.189 + 0.866) = 0.2275 \text{ m}^2, \qquad P = \frac{0.60 \times 4.189}{2} = 1.257 \text{ m}

R = \frac{0.2275}{1.257} = 0.1810 \text{ m}, \qquad V = \frac{1}{0.013} \times 0.1810^{2/3} \times 0.003^{1/2} = 76.92 \times 0.3200 \times 0.05477 = 1.348 \text{ m/s}

Q = V A = 1.348 \times 0.2275 = 0.307 \text{ m}^3/\text{s}

Three quarters full, the 600 mm sewer carries 0.307 m³/s at 1.35 m/s, 91% of its full-flow capacity.

Flowing full, the same pipe has $A = 0.2827$ m² and $R = D/4 = 0.150$ m, so $V = 1.19$ m/s and $Q = 0.336$ m³/s. Three quarters full, it carries 91% of that discharge at a velocity 13% higher. With $n$ taken as constant, the discharge peaks at about $y/D = 0.94$ , near 1.08 times the full flow, and the velocity at about $y/D = 0.81$ , near 1.14 times the full velocity: close to the crown, the wetted perimeter grows faster than the area. Design normally takes the full-flow capacity as the limit and leaves that small gain aside.

Sewer codes add their own rules. In the United States, the Recommended Standards for Wastewater Facilities (the Ten States Standards) set the minimum slopes of sanitary sewers with $n = 0.013$ , whatever the pipe material, so that they reach at least 0.6 m/s (2 ft/s) flowing full and solids don’t settle.

Typical values of Manning’s n

Manning’s $n$ gathers everything at the boundary that resists the flow: the roughness of the surface, vegetation, irregularities of the section, changes along the reach, obstructions and meanders. It can’t be calculated from the material alone; the published values come from channels whose discharge was measured. These are typical values with their usual range:

Channel or lining	Typical $n$	Usual range
PVC or other smooth plastic pipe	0.010	0.009 to 0.011
Concrete culvert, straight and free of debris	0.011	0.010 to 0.013
Concrete, trowel finish	0.013	0.011 to 0.015
Vitrified clay sewer	0.014	0.011 to 0.017
Concrete, float finish	0.015	0.013 to 0.016
Concrete, unfinished	0.017	0.014 to 0.020
Shotcrete (gunite), good section	0.019	0.016 to 0.023
Corrugated metal storm drain	0.024	0.021 to 0.030
Cemented rubble masonry	0.025	0.017 to 0.030
Earth, straight and clean, recently completed	0.018	0.016 to 0.020
Earth, straight and clean, weathered	0.022	0.018 to 0.025
Earth with gravel, uniform section, clean	0.025	0.022 to 0.030
Earth with short grass and few weeds	0.027	0.022 to 0.033
Earth, winding, with grass and some weeds	0.030	0.025 to 0.033
Rock cut, smooth and uniform	0.035	0.025 to 0.040
Unmaintained channel, dense weeds as high as the flow	0.080	0.050 to 0.120
Natural stream, clean and straight, no deep pools	0.030	0.025 to 0.033
Natural stream, clean and winding, some pools and shoals	0.040	0.033 to 0.045
Mountain stream, gravel and cobbles with few boulders	0.040	0.030 to 0.050
Mountain stream, cobbles with large boulders	0.050	0.040 to 0.070
Floodplain, pasture with short grass	0.030	0.025 to 0.035
Floodplain, light brush and trees, in summer	0.060	0.040 to 0.080

Pick within the range for the condition the channel will be in during its life. Concrete roughens with age, and earth channels grow weeds between cleanings.

A grass lining doesn’t keep a constant $n$ . As the flow gets deeper and faster, the grass bends and lies flat and its $n$ drops, so grassed waterways are designed with curves of $n$ against the product $VR$ for each class of grass. For a natural channel, Cowan’s method builds $n$ from a base value for the bed material, adds amounts for irregularity, changes of section, obstructions and vegetation, and multiplies the sum by a factor for meandering.

The discharge at a given depth is inversely proportional to $n$ , so an $n$ that is 20% too low overstates the capacity by 25%. The depth is less sensitive: the 8.00 m³/s of the second worked example runs 1.22 m deep with $n = 0.020$ , 1.37 m deep with $n = 0.025$ and 1.51 m deep with $n = 0.030$ . Rischio’s engine flags any $n$ outside 0.010 to 0.150, the usual span from smooth concrete to dense vegetation, and still computes.

Normal depth, critical depth and the flow regime

The normal depth $y_n$ is the depth at which a channel carries a given discharge in uniform flow, the depth Manning’s equation gives. It depends on the section, $n$ and $S$ , and for most shapes it has no closed form, so it is found by trial or numerically. A wide rectangular channel is the exception: with $R \approx y$ , $y_n = (n q / S^{1/2})^{3/5}$ , where $q$ is the discharge per metre of width.

The critical depth $y_c$ is the depth at which the same discharge flows with the least specific energy, $E = y + V^2/2g$ . There

\frac{Q^2 T}{g A^3} = 1

which depends on the discharge and the section only, and not on $n$ or the slope. In a rectangle it reduces to $y_c = (q^2/g)^{1/3}$ .

The Froude number compares the velocity with the speed of a small surface wave:

Fr = \frac{V}{\sqrt{g\,A/T}}

$A/T$ is the hydraulic depth, the mean depth of the section. It equals $y$ in a rectangle and is smaller in a section that widens upward, such as a trapezoid. Uniform flow deeper than critical, $y_n > y_c$ , has $Fr < 1$ and is subcritical: slow and deep, so a disturbance can travel upstream and the depth is controlled from downstream. Uniform flow shallower than critical has $Fr > 1$ and is supercritical: fast and shallow, and controlled from upstream. A slope on which $y_n > y_c$ is called mild, and one on which $y_n < y_c$ is steep; at the critical slope $S_c$ the two depths are equal.

Close to $Fr = 1$ the water surface turns wavy and unstable, because a small change of energy makes a large change of depth there. Rischio’s engine warns of near-critical flow when $0.8 < Fr < 1.2$ . Where supercritical flow meets subcritical flow, the water rises through a hydraulic jump. Where subcritical flow speeds up into supercritical, at a free overfall or a sharp increase of slope, it passes through the critical depth. Both are rapidly varied flow, outside Manning’s equation.

Composite roughness: when n changes across the section

Real sections rarely have one lining. A canal can have a concrete bed and grassed sides, and a river has a main channel between floodplains under brush. Rischio’s open channel tool takes a section drawn vertex by vertex, with its own $n$ on every segment, and handles it this way at every water level it computes:

The wet segments form the section at that level. A segment that crosses the water surface is cut there, and only its wet part counts.
The section is divided into parts at every vertex where $n$ changes, and at any dry stretch between two pools.
A part made only of segments steeper than 1 vertical to 5 horizontal joins the part it runs down to. The sides of a channel, with an $n$ of their own, stay with its bed, as HEC-RAS keeps a main channel with steep banks whole.
A part with more than one $n$ takes the Horton-Einstein equivalent, which assumes the whole part moves at one mean velocity: $n = \left( \frac{\Sigma P_i n_i^{1.5}}{P} \right)^{2/3}$
Each part carries its share of the discharge, worked out from its own area, wetted perimeter and $n$ , and the section carries their sum. The vertical lines between parts are not wetted perimeter: $Q = \sum_j \frac{1}{n_j} A_j R_j^{2/3} S^{1/2}$

Take the channel of the worked examples, at a depth of 1.00 m, with its bed lined in concrete ( $n = 0.015$ ) and its sides left in grass ( $n = 0.027$ ). The sides, at 1.5 horizontal to 1 vertical, are steeper than 1 to 5, so the section stays one part. The bed wets 3.000 m and the two sides 3.606 m:

n = \left( \frac{3.000 \times 0.015^{1.5} + 3.606 \times 0.027^{1.5}}{6.606} \right)^{2/3} = \left( \frac{0.00551 + 0.01600}{6.606} \right)^{2/3} = 0.02197

At the same depth, the discharge is inversely proportional to $n$ : $Q = 4.407 \times 0.025 / 0.02197 = 5.015$ m³/s, where 4.407 m³/s is what the all-earth channel carries in Example 1 below. Lining the bed adds 14%, and $V = 5.015 / 4.50 = 1.114$ m/s. Rischio’s engine gives the same: for 5.015 m³/s it finds the normal depth at 1.000 m, with $n = 0.0220$ by Horton-Einstein.

Steeper than 1 to 5, the grassed sides stay one part with the concrete bed, and their n is weighted by wetted perimeter: n = 0.0220.

A floodplain flatter than 1 to 5 whose $n$ differs from the channel’s is a separate part. The water over a floodplain is shallow and slow next to the deep, fast water of the main channel. Lumped into one section, the floodplain’s long wetted perimeter would cut the hydraulic radius of the whole and understate what the channel carries. When the wet section is one part, Rischio’s calculation report gives its equivalent $n$ ; when the section is divided, it gives $V = Q/A$ instead, since a single $n$ for the whole would come out lower than that of any of its parts.

A floodplain with the same $n$ as its channel isn’t divided from it, because its $n$ doesn’t change at the bank. As the water first spreads over that floodplain, the wetted perimeter grows much faster than the area, and the discharge can drop while the depth rises, so that more than one depth carries the same $Q$ . Rischio warns when this happens; the fix is to give the floodplains their own $n$ . The method is written out in the docs, in the section and its roughness and normal and critical flow.

The editor below is the studio’s, on the same engine. Its Compound tab is a lined channel between rougher floodplains. Select the cross-section to see the roughness of its segments on the drawing, then change the discharge and press Calculate to see how far the water spreads over the floodplains.

Try it: change a number, calculate

Worked example 1: the discharge at a given depth

A trapezoidal channel dug in earth, clean with some gravel ( $n = 0.025$ ), has a bottom width of 3.00 m, sides at 1.5 horizontal to 1 vertical ( $z = 1.5$ ) and banks 2.00 m above the bed. It runs on a slope of 0.1% ( $S = 0.001$ ). What discharge does it carry in uniform flow with the water 1.00 m deep? The figure in the geometry section above draws this case.

Step 1: the geometry

A = (b + z y)\,y = (3.00 + 1.5 \times 1.00) \times 1.00 = 4.50 \text{ m}^2

P = b + 2y\sqrt{1 + z^2} = 3.00 + 2 \times 1.00 \times \sqrt{1 + 1.5^2} = 3.00 + 2 \times 1.803 = 6.606 \text{ m}

R = \frac{A}{P} = \frac{4.50}{6.606} = 0.6812 \text{ m}, \qquad T = b + 2 z y = 3.00 + 2 \times 1.5 \times 1.00 = 6.00 \text{ m}

Step 2: the velocity

V = \frac{1}{n} R^{2/3} S^{1/2} = \frac{1}{0.025} \times 0.6812^{2/3} \times 0.001^{1/2} = 40 \times 0.7742 \times 0.03162 = 0.979 \text{ m/s}

Step 3: the discharge

Q = V A = 0.979 \times 4.50 = 4.41 \text{ m}^3/\text{s}

Step 4: the flow regime

\frac{A}{T} = \frac{4.50}{6.00} = 0.750 \text{ m}, \qquad Fr = \frac{0.979}{\sqrt{9.81 \times 0.750}} = \frac{0.979}{2.712} = 0.361

With $Fr < 1$ , the flow is subcritical.

Rischio’s engine gives the same. For a discharge of 4.407 m³/s in this section, it finds the normal depth at 1.000 m, with $V = 0.979$ m/s and $Fr = 0.361$ , and adds the critical depth, 0.548 m, well below the water.

Worked example 2: the normal depth for a given discharge

The same channel has to carry 8.00 m³/s. How deep will it run?

In a trapezoid the depth can’t be isolated in Manning’s equation, so it is found by trial. Moving everything that depends on the depth to one side gives the section factor the channel needs:

A R^{2/3} = \frac{n Q}{S^{1/2}} = \frac{0.025 \times 8.00}{0.03162} = 6.325 \text{ m}^{8/3}

Each trial depth gives $A$ , $P$ and $R$ , and from them $A R^{2/3}$ , to compare with 6.325:

Trial	$y$ (m)	$A$ (m²)	$P$ (m)	$R$ (m)	$A R^{2/3}$	$Q$ (m³/s)
1	1.00	4.500	6.606	0.681	3.484	4.41
2	1.50	7.875	8.408	0.937	7.538	9.54
3	1.35	6.784	7.868	0.862	6.146	7.77
4	1.37	6.925	7.940	0.872	6.322	8.00

Trials 1 and 2 bracket the answer. A straight line between them gives the third trial,

y_3 = 1.00 + 0.50 \times \frac{6.325 - 3.484}{7.538 - 3.484} = 1.35 \text{ m}

which falls short, and a line between trials 3 and 2 gives the fourth:

y_4 = 1.35 + 0.15 \times \frac{6.325 - 6.146}{7.538 - 6.146} = 1.369 \approx 1.37 \text{ m}

At 1.37 m the section factor is 6.322, within 0.05% of the target, and one more interpolation moves the depth by a third of a millimetre. The normal depth is $y_n = 1.370$ m. There $A = 6.927$ m², $P = 7.941$ m, $R = 0.872$ m and $T = 3.00 + 2 \times 1.5 \times 1.370 = 7.11$ m, and the mean velocity is

V = \frac{Q}{A} = \frac{8.00}{6.927} = 1.155 \text{ m/s}

Critical depth and flow regime

The critical depth solves $Q^2 T / (g A^3) = 1$ , by trial again. At $y = 0.783$ m, $A = (3.00 + 1.5 \times 0.783) \times 0.783 = 3.269$ m² and $T = 3.00 + 2 \times 1.5 \times 0.783 = 5.349$ m:

\frac{Q^2 T}{g A^3} = \frac{8.00^2 \times 5.349}{9.81 \times 3.269^3} = \frac{342.3}{342.6} = 1.00

so $y_c = 0.783$ m. At the normal depth the hydraulic depth is $A/T = 6.927 / 7.111 = 0.974$ m, and

Fr = \frac{1.155}{\sqrt{9.81 \times 0.974}} = 0.374

The flow is subcritical and the slope is mild: the normal depth, 1.37 m, sits well above the critical depth, 0.78 m. For this discharge the critical slope is about 0.008, so the channel would have to be eight times steeper to carry 8.00 m³/s at the critical depth.

Capacity and freeboard

Full to its banks, at $y = 2.00$ m, the channel has $A = 12.00$ m², $P = 10.21$ m and $R = 1.175$ m, and carries

Q_{bank} = \frac{1}{0.025} \times 12.00 \times 1.175^{2/3} \times 0.001^{1/2} = 16.90 \text{ m}^3/\text{s}

The design flow uses 47% of that and leaves 0.63 m of freeboard.

Rischio’s engine gives the same: a normal depth of 1.370 m, a velocity of 1.155 m/s, a Froude number of 0.374, a critical depth of 0.783 m and a bank-full capacity of 16.90 m³/s. It finds the normal depth with Brent’s method, which brackets the root as trials 1 and 2 did and then closes in on it, to about $10^{-12}$ m.

Example 2: the normal depth sits above the critical depth, so the flow is subcritical (Fr = 0.374) and the slope is mild.

The calculator below opens on this channel carrying 8 m³/s. Set Solve for to Flow and it takes a depth of 1 m, which is Example 1. Change $n$ or the slope to see how far the depth moves. Its page, the Manning equation calculator, explains how it works.

Calculator · Rischio's engine · Manning's equation

Solve for

Section

Bottom width · b

Side slope · z

H:1V

Horizontal for every 1 vertical.

Channel depth

From the bed to the top of the banks.

Flow

Manning's n

Trowelled concrete, about 0.013; clean earth, 0.022; earth with short grass, 0.027.

Slope · S

m/m

0.001 is 0.1%.

Flow · Q

m³/s

These data give no result: check the values.

Rischio's engine calculated this. In the studio you keep the model, change it on the drawing and get the calculation report, every step shown.

Start free

This calculator is offered for information and technical support. It doesn't replace professional judgment or the reviews due, and its use is the user's responsibility.

Where Manning’s equation is used, and its limits

It is the working equation for:

roadside ditches, drainage channels and irrigation canals, to size the section and its lining for a design flow, check the freeboard and check the velocity against what the lining tolerates;
culverts and storm sewers flowing partly full, for the depth and velocity in the barrel; when a culvert flows full, Manning’s $n$ also gives the friction loss along the barrel;
sanitary sewers, for their capacity and their self-cleansing velocity;
natural streams, for the normal depth of a flood flow or, working back from high-water marks, the peak of a past flood (the slope-area method).

Its limits follow from its assumptions:

Near a control, such as a gate, a weir, a culvert inlet, a drop or a change of slope or section, the depth changes along the channel. The water-surface profile then needs a gradually varied flow calculation, such as the standard step method, in which Manning’s equation supplies the friction slope of each step.
Rapidly varied flow, such as a hydraulic jump or the flow over a weir, needs the momentum and energy equations.
Unsteady flow, such as a flood wave moving down a river, needs the Saint-Venant equations, with Manning’s equation as their friction term.
Pipes flowing full under pressure use Darcy-Weisbach or Hazen-Williams.
In a sand-bed channel the bed forms ripples and dunes that change with the flow, and $n$ changes with them.
The answer is only as good as $n$ , the least certain input in most designs.

Rischio’s open channel tool works inside those limits: steady, uniform flow in a prismatic channel of any cross-section. Backwater profiles, unsteady flow, culvert hydraulics, weirs and sediment transport are outside it. Inside them, it takes any section drawn vertex by vertex with an $n$ on every segment, floodplains included, and gives the normal and critical depths, the Froude number, the specific energy, the rating curve from the bed to the banks and the bank-full capacity check. Its calculation report writes out every step, and the docs explain the method and its assumptions. For a rectangle, a trapezoid or a triangle with one $n$ , the free Manning equation calculator needs no account.

Frequently asked questions

What is the formula for the Manning equation?

In SI units, $V = \frac{1}{n} R^{2/3} S^{1/2}$ and $Q = VA$ , with $V$ in m/s, $R$ in metres and $S$ in m/m. In US units the 1 becomes 1.49, with $V$ in ft/s and $R$ in feet; $n$ is the same number in both.

What is Manning’s formula used for?

To relate the discharge, the depth and the velocity of uniform flow in open channels and in pipes flowing partly full. With it engineers size ditches, canals and storm sewers, find the depth a flow will run at, check the capacity of a channel and estimate the flow of a stream.

How do you use Manning’s equation?

For the discharge at a known depth, work out $A$ and $P$ from the section, then $R = A/P$ , then $V$ from the equation and $Q = VA$ . For the depth that carries a known discharge, work out the section factor $nQ/S^{1/2}$ and try depths until $A R^{2/3}$ matches it, as in the second worked example.

What does R stand for in Manning’s equation?

The hydraulic radius, the flow area divided by the wetted perimeter: $R = A/P$ . In a wide channel it is close to the depth, and in a pipe flowing full it is a quarter of the diameter.

Can you use Manning’s equation for pipe flow?

Yes, for pipes flowing partly full under gravity, such as sewers, storm drains and culverts: they have a free surface and behave as open channels. For a pipe flowing full under pressure, Darcy-Weisbach or Hazen-Williams are the usual equations.

What is Manning’s n for concrete?

Typically 0.013 for a trowelled finish, 0.015 for a float finish and 0.017 for unfinished concrete. Concrete sewers and culverts are commonly designed with $n = 0.013$ .