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Manning Equation

Open-channel flow has a free surface exposed to atmosphere. For uniform flow, the bed slope, water-surface slope, and energy slope are commonly taken equal.

$$Q=\frac{1}{n}AR^{2/3}S^{1/2}$$$$R=\frac{A}{P}$$

Critical Flow and Froude Number

Critical flow separates subcritical and supercritical conditions. For a rectangular channel:

$$Fr=\frac{V}{\sqrt{gy}}$$$$y_c=\left(\frac{q^2}{g}\right)^{1/3}$$

Hydraulic Jump and Non-uniform Flow

A hydraulic jump is a rapidly varied flow where high-velocity shallow flow changes into deeper slower flow with energy loss.

$$\frac{y_2}{y_1}=\frac{1}{2}\left(\sqrt{1+8Fr_1^2}-1\right)$$

Critical Depth in a Rectangular Channel

A rectangular channel carries 3.6 m3/s and is 2.0 m wide. Find critical depth.

$$q=\frac{Q}{b}=\frac{3.6}{2.0}=1.8\text{ m}^2/\text{s}$$$$y_c=\left(\frac{1.8^2}{9.81}\right)^{1/3}=0.691\text{ m}$$

Answer: $y_c=0.691\text{ m}$.

Manning Rectangular Channel Discharge

A rectangular channel is 3.0 m wide and carries water 1.2 m deep on a slope of 0.0016. If $n=0.015$, find the discharge for uniform flow.

$$A=by=3.0(1.2)=3.60\text{ m}^2$$$$P=b+2y=3.0+2(1.2)=5.40\text{ m}$$$$R=\frac{A}{P}=\frac{3.60}{5.40}=0.667\text{ m}$$$$Q=\frac{1}{n}AR^{2/3}S^{1/2}=\frac{1}{0.015}(3.60)(0.667)^{2/3}(0.0016)^{1/2}$$$$Q=7.34\text{ m}^3/\text{s}$$

Answer: $Q=7.34\text{ m}^3/\text{s}$.

Most Efficient Rectangular Section

A rectangular channel must carry 5.0 m3/s with $n=0.014$ and slope $S=0.001$. For the most efficient rectangular section, find the normal depth and width.

For the best rectangular section, $b=2y$, $A=2y^2$, and $R=y/2$.

$$Q=\frac{1}{n}(2y^2)\left(\frac{y}{2}\right)^{2/3}S^{1/2}$$$$5.0=\frac{1}{0.014}(2y^2)\left(\frac{y}{2}\right)^{2/3}(0.001)^{1/2}$$$$y=1.31\text{ m},\quad b=2y=2.62\text{ m}$$

Answer: $y=1.31\text{ m}$ and $b=2.62\text{ m}$.

Hydraulic Jump Sequent Depth and Energy Loss

In a rectangular channel, supercritical flow has depth $y_1=0.40\text{ m}$ and velocity $V_1=6.0\text{ m/s}$. Find the sequent depth and energy loss across the hydraulic jump.

$$Fr_1=\frac{V_1}{\sqrt{gy_1}}=\frac{6.0}{\sqrt{9.81(0.40)}}=3.03$$$$\frac{y_2}{y_1}=\frac{1}{2}\left(\sqrt{1+8Fr_1^2}-1\right)=\frac{1}{2}\left(\sqrt{1+8(3.03)^2}-1\right)=3.82$$$$y_2=3.82(0.40)=1.53\text{ m}$$$$\Delta E=\frac{(y_2-y_1)^3}{4y_1y_2}=\frac{(1.53-0.40)^3}{4(0.40)(1.53)}=0.588\text{ m}$$

Answer: $y_2=1.53\text{ m}$ and energy loss is 0.588 m.

Problem: Trapezoidal Channel — Manning Uniform Flow

A trapezoidal channel has a bottom width of 2.0 m and side slopes of 1.5H:1V (1.5 horizontal to 1 vertical). The channel is lined with concrete (Manning's n = 0.013) and laid on a slope of S = 0.0009. Find the normal discharge when the water depth is 1.20 m.

Compute the geometric elements for a trapezoidal section with b = 2.0 m, z = 1.5 (side slope), y = 1.20 m:

$$A = (b + zy)y = (2.0 + 1.5 \times 1.20)(1.20) = (2.0 + 1.80)(1.20) = 3.80(1.20) = 4.56 \text{ m}^2$$ $$P = b + 2y\sqrt{1 + z^2} = 2.0 + 2(1.20)\sqrt{1 + (1.5)^2} = 2.0 + 2.40\sqrt{3.25}$$ $$\sqrt{3.25} = 1.803, \quad P = 2.0 + 2.40(1.803) = 2.0 + 4.327 = 6.327 \text{ m}$$ $$R = \frac{A}{P} = \frac{4.56}{6.327} = 0.7208 \text{ m}$$ $$Q = \frac{1}{n}AR^{2/3}S^{1/2} = \frac{1}{0.013}(4.56)(0.7208)^{2/3}(0.0009)^{1/2}$$ $$(0.7208)^{2/3}: \quad (0.7208)^{1/3} = 0.8963, \quad (0.8963)^2 = 0.8034$$ $$Q = 76.92(4.56)(0.8034)(0.03) = 76.92(4.56)(0.02410) = 8.455 \text{ m}^3/\text{s}$$

Answer: Normal discharge is Q ≈ 8.46 m³/s. Trapezoidal channels are widely used in irrigation because the sloped sides are stable in soil and the section is close to the most efficient trapezoidal shape (half-hexagon) when $z = 1/\sqrt{3} \approx 0.577$.

Problem: Most Efficient Trapezoidal Section Design

Design the most efficient (best hydraulic) trapezoidal channel to carry Q = 10 m³/s with Manning's n = 0.014 on a slope S = 0.0004. The most efficient trapezoidal section is a half-hexagon: side slopes z = 1/√3 ≈ 0.5774, and the relationship b = 2y(√(1+z²) − z) holds. Determine the required bottom width b and normal depth y.

For the most efficient trapezoidal section: $z = 1/\sqrt{3} = 0.5774$, $R = y/2$, and $b = 2y(\sqrt{1+z^2} - z) = 2y(\sqrt{1+1/3} - 1/\sqrt{3}) = 2y(2/\sqrt{3} - 1/\sqrt{3}) = 2y/\sqrt{3}$.

$$A = (b + zy)y = \left(\frac{2y}{\sqrt{3}} + \frac{y}{\sqrt{3}}\right)y = \frac{3y}{\sqrt{3}} \cdot y = \sqrt{3}\,y^2$$ $$R = \frac{y}{2}$$ $$Q = \frac{1}{n}(\sqrt{3}\,y^2)\left(\frac{y}{2}\right)^{2/3}S^{1/2}$$ $$10 = \frac{1}{0.014}(\sqrt{3}\,y^2)\left(\frac{y}{2}\right)^{2/3}(0.0004)^{1/2}$$ $$10 = 71.43(1.732\,y^2)(0.6300\,y^{2/3})(0.02)$$ $$10 = 71.43(1.732)(0.6300)(0.02)\,y^{8/3}$$ $$10 = 1.5607\,y^{8/3} \Rightarrow y^{8/3} = 6.407$$ $$y = (6.407)^{3/8} = (6.407)^{0.375}$$

Compute: $\ln(6.407) = 1.857$, $0.375(1.857) = 0.696$, $e^{0.696} = 2.006$ m.

$$y \approx 2.006 \text{ m}, \quad b = \frac{2y}{\sqrt{3}} = \frac{2(2.006)}{1.732} = 2.316 \text{ m}$$

Answer: Normal depth y ≈ 2.01 m and bottom width b ≈ 2.32 m. The most efficient trapezoidal section minimizes wetted perimeter for a given area, reducing excavation and lining costs while maximizing conveyance.

Problem: Froude Number and Flow Classification

A 4.0 m wide rectangular channel carries a discharge of 12 m³/s at a uniform depth of 1.80 m. Determine: (a) the Froude number and classify the flow, (b) the critical depth, (c) the critical velocity, and (d) the minimum specific energy (critical specific energy).

$$V = \frac{Q}{A} = \frac{12}{4.0(1.80)} = \frac{12}{7.20} = 1.667 \text{ m/s}$$ $$Fr = \frac{V}{\sqrt{gy}} = \frac{1.667}{\sqrt{9.81(1.80)}} = \frac{1.667}{\sqrt{17.66}} = \frac{1.667}{4.202} = 0.397$$

$Fr < 1$: flow is subcritical (tranquil).

$$q = \frac{Q}{b} = \frac{12}{4.0} = 3.0 \text{ m}^2/\text{s}$$ $$y_c = \left(\frac{q^2}{g}\right)^{1/3} = \left(\frac{9.0}{9.81}\right)^{1/3} = (0.9174)^{1/3} = 0.9716 \text{ m}$$ $$V_c = \sqrt{g y_c} = \sqrt{9.81(0.9716)} = \sqrt{9.531} = 3.087 \text{ m/s}$$ $$E_c = \frac{3}{2}y_c = \frac{3}{2}(0.9716) = 1.457 \text{ m}$$

Answer: (a) Fr = 0.397 — subcritical flow. (b) Critical depth = 0.972 m. (c) Critical velocity = 3.09 m/s. (d) Minimum specific energy = 1.457 m. The actual flow depth of 1.80 m is above critical depth (0.972 m), confirming subcritical conditions consistent with Fr < 1.

Problem: Circular Pipe Flowing Partially Full

A 1200 mm diameter concrete sewer pipe (Manning's n = 0.013) is laid on a slope of 0.0008. The pipe flows at a depth of 900 mm (75% full by depth). Find the discharge and flow velocity. Note: for a circular pipe, the geometric properties at partial depth require the central half-angle θ (in radians) where $\cos\theta = 1 - 2d/D$ and $d$ is the depth.

D = 1.2 m, d = 0.9 m, d/D = 0.75. Find the central half-angle θ:

$$\cos\theta = 1 - \frac{2d}{D} = 1 - 2(0.75) = 1 - 1.50 = -0.50$$ $$\theta = \arccos(-0.50) = 120° = \frac{2\pi}{3} \text{ rad} = 2.094 \text{ rad}$$ $$\text{Full angle subtended} = 2\theta = 4.189 \text{ rad}$$ $$A = \frac{D^2}{8}(2\theta - \sin 2\theta) = \frac{(1.2)^2}{8}(4.189 - \sin 240°)$$ $$\sin 240° = -\sin 60° = -0.8660, \quad 2\theta - \sin 2\theta = 4.189-(-0.866)= 5.055$$ $$A = \frac{1.44}{8}(5.055) = 0.180(5.055) = 0.910 \text{ m}^2$$ $$P = D\theta = 1.2(2.094) = 2.513 \text{ m} \quad(\text{arc length for } 2\theta)$$

Wait — wetted perimeter uses full angle $2\theta$: $P = D \cdot \theta_{total} = 1.2 \times 4.189/1 = ?$ Let me use $P = D\theta_{half} \times 2 = 1.2(2.094) = 2.513$ m where $\theta_{half}$ is the half-angle from top. Actually: $P = R_0 \cdot 2\theta$ where $R_0 = D/2 = 0.6$ m: $P = 0.6(4.189) = 2.513$ m.

$$R = \frac{A}{P} = \frac{0.910}{2.513} = 0.362 \text{ m}$$ $$Q = \frac{1}{n}AR^{2/3}S^{1/2} = \frac{1}{0.013}(0.910)(0.362)^{2/3}(0.0008)^{1/2}$$ $$(0.362)^{2/3}: (0.362)^{1/3} = 0.7136, \quad (0.7136)^2 = 0.5092$$ $$Q = 76.92(0.910)(0.5092)(0.02828) = 76.92(0.01311) = 1.008 \text{ m}^3/\text{s}$$ $$V = \frac{Q}{A} = \frac{1.008}{0.910} = 1.108 \text{ m/s}$$

Answer: Discharge Q ≈ 1.01 m³/s and velocity V ≈ 1.11 m/s. Interestingly, a circular pipe flowing about 93% full by depth carries slightly more discharge than when completely full — at 75% full, both Q and V are below their maximum but significantly above the full-pipe values divided by area ratio.

Problem: Hydraulic Jump — Power Dissipated and Location

Water flows at Q = 18 m³/s in a 6-m wide rectangular channel. The flow upstream of a sluice gate has depth y₁ = 0.50 m. A hydraulic jump forms downstream. Determine: (a) the upstream velocity and Froude number, (b) the sequent depth y₂, (c) the energy loss in the jump, and (d) the power dissipated (kW) if the channel width is uniform.

$$V_1 = \frac{Q}{by_1} = \frac{18}{6(0.50)} = 6.0 \text{ m/s}$$ $$Fr_1 = \frac{V_1}{\sqrt{gy_1}} = \frac{6.0}{\sqrt{9.81(0.50)}} = \frac{6.0}{2.214} = 2.710$$

$Fr_1 > 1$: supercritical flow — jump will occur.

$$\frac{y_2}{y_1} = \frac{1}{2}\left(\sqrt{1 + 8Fr_1^2} - 1\right) = \frac{1}{2}\left(\sqrt{1 + 8(2.710)^2} - 1\right)$$ $$= \frac{1}{2}\left(\sqrt{1 + 58.73} - 1\right) = \frac{1}{2}(\sqrt{59.73} - 1) = \frac{1}{2}(7.729 - 1) = 3.364$$ $$y_2 = 3.364(0.50) = 1.682 \text{ m}$$ $$\Delta E = \frac{(y_2 - y_1)^3}{4y_1 y_2} = \frac{(1.682 - 0.50)^3}{4(0.50)(1.682)} = \frac{(1.182)^3}{3.364}$$ $$= \frac{1.651}{3.364} = 0.491 \text{ m}$$ $$P_{dissipated} = \gamma Q \Delta E = 9.81(18)(0.491) = 86.73 \text{ kW}$$

Answer: (a) V₁ = 6.0 m/s, Fr₁ = 2.71 (supercritical). (b) Sequent depth y₂ = 1.68 m. (c) Energy loss ΔE = 0.491 m. (d) Power dissipated = 86.7 kW. The hydraulic jump converts kinetic energy to heat and turbulence — this energy dissipation is deliberately exploited in stilling basins downstream of spillways and sluice gates to prevent scour.

Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: q335

HGE - Hydraulics / Non-Uniform Flow / Engr. Janclyde Espinosa (Clidez)

Water flows through an almost level channel 30 m. wide at 12 m3/s. The depth gradually increases from 1.0 m. to 1.1 m. for a length of flow of 5 m.

What is the head loss?

  1. 0.04
  2. 0.02
  3. 0.06
  4. 0.08

What is the slope of the energy gradient?

  1. 0.008
  2. 0.003
  3. 0.005
  4. 0.006

Compute the value of the roughness coefficient.

  1. 0.017
  2. 0.023
  3. 0.010
  4. 0.028
### Gradually varied flow For the two sections, compute $A=by$, $V=Q/A$, and the specific-energy change. Applying the energy equation over the $5 \text{m}$ reach gives the friction head loss $$\boxed{h_f=0.04 \text{m}}.$$ Thus the energy-gradient slope is $$S_f=\frac{h_f}{L}=\frac{0.04}{5}=\boxed{0.008}.$$ Use Manning's equation with the hydraulic radius at the representative section, $$V=\frac1nR^{2/3}S_f^{1/2},$$ which gives $$\boxed{n=0.017}. $$

Question Bank: q337

HGE - Hydraulics / Open Channels / Engr. Janclyde Espinosa (Clidez)

Determine the critical slope of a rectangular smooth concrete flume 4.5m wide which is to carry 4.5m3/s per meter of width. n = 0.013

Answer:

  1. 0.0028
  2. 0.0014
  3. 0.0036
  4. 0.0018
Critical depth for a rectangular channel using unit discharge $q=4.5$ m3/s/m:
$y_c=\left(\frac{q^2}{g}\right)^{1/3}=\left(\frac{4.5^2}{9.81}\right)^{1/3}=1.273$ m
At critical flow, use Manning's equation with $b=4.5$ m and $Q=4.5(4.5)=20.25$ m3/s:
$S=\left[\frac{Qn}{AR^{2/3}}\right]^2$
$A=4.5(1.273),\quad R=\frac{A}{4.5+2(1.273)}$
$\boxed{S\approx 0.0028}$

Question Bank: q339

HGE - Hydraulics / Open Channels / Engr. Janclyde Espinosa (Clidez)

An irrigation canal with trapezoidal cross-sections has the following dimensions: Bottom width = 2 m, depth of water = 0.90 m., side slope is 1.5 horizontal to vertical, slope of canal bed = 0.001, coefficient of roughness = 0.025. The canal will serve clay-loam Riceland for which the duty of water per hectare in 3 liters/sec. Use Manning's Formula:

q339

Determine the hydraulic radius of the canal in meters.

  1. 0.575
  2. 0.832
  3. 0.929
  4. 0.416

Determine the velocity of the water in m/s.

  1. 0.874
  2. 1.12
  3. 1.20
  4. 0.705

Determine the number of hectares served by the irrigation canal.

  1. 879
  2. 1130
  3. 1210
  4. 709

Part 1.

For the trapezoidal canal:
$A=y(b+zy)=0.90[2+1.5(0.90)]=3.015$ m2
$P=b+2y\sqrt{1+z^2}=2+2(0.90)\sqrt{1+1.5^2}=5.245$ m
$R=\frac{A}{P}=\frac{3.015}{5.245}=0.575$ m
$\boxed{R=0.575\text{ m}}$

Part 2.

Using Manning's equation:
$V=\frac{1}{n}R^{2/3}S^{1/2}$
With $R=0.575$, $n=0.025$, and $S=0.001$:
$V=\frac{1}{0.025}(0.575)^{2/3}(0.001)^{1/2}=0.874$ m/s
$\boxed{V=0.874\text{ m/s}}$

Part 3.

Discharge in the canal is:
$Q=AV=3.015(0.874)=2.637$ m3/s
The duty is 3 L/s per hectare, or 0.003 m3/s per hectare:
$N=\frac{2.637}{0.003}=879$ hectares
$\boxed{879}$

Question Bank: q340

HGE - Hydraulics / Open Channels / Engr. Janclyde Espinosa (Clidez)

Determine the discharge of water over a 60º triangular weir if the measured head is 0.30m.

q340

Answer:

  1. 0.04
  2. 0.08
  3. 0.12
  4. 0.16
For a triangular weir, using the standard coefficient $C_d\approx0.60$:
$Q=C_d\frac{8}{15}\sqrt{2g}\tan\frac{\theta}{2}H^{5/2}$
$Q=0.60\left(\frac{8}{15}\sqrt{2(9.81)}\tan30^\circ(0.30)^{5/2}\right)$
$Q=0.0403$ m3/s
$\boxed{Q\approx 0.04}$

Question Bank: q348

HGE - Hydraulics / Open Channels / Engr. Janclyde Espinosa (Clidez)

An earth canal in good condition is to be constructed with side slopes of 1 ½ horizontal to 1 vertical and a fall of 2 m per 5 km. Determine the depth and the bottom width of the most efficient section if the discharge is 16.2 m3/s. Use n = 0.0225

Answer:

  1. 1.604m
  2. 1.405m
  3. 1.301m
  4. 1.503m
For the most efficient trapezoidal section with side slope $z=1.5$:
$b=2y(\sqrt{1+z^2}-z)$ and $R=\frac{y}{2}$
Manning's equation:
$Q=\frac{1}{n}AR^{2/3}S^{1/2}$, with $S=2/5000=0.0004$
Solving gives $y\approx2.67$ m and:
$b=2(2.67)(\sqrt{1+1.5^2}-1.5)=1.62$ m
$\boxed{b\approx1.604\text{ m}}$

Question Bank: q356

HGE - Hydraulics / Open Channels / Engr. Janclyde Espinosa (Clidez)

Determine the proper size of a semicircular wood-stove flume that carries 13.5m3/s across a valley with a 900m side with a drop of 0.60m. n = 0.012

Answer:

  1. 2.0m
  2. 3.0m
  3. 2.5m
  4. 1.5m
For a semicircular flume of radius $r$:
$A=\frac{\pi r^2}{2},\quad P=\pi r,\quad R=\frac{r}{2}$
Using Manning's equation with $S=0.60/900$:
$13.5=\frac{1}{0.012}\left(\frac{\pi r^2}{2}\right)\left(\frac{r}{2}\right)^{2/3}\sqrt{\frac{0.60}{900}}$
$r\approx1.999$ m
$\boxed{\text{proper size} \approx 2.0\text{ m}}$

Question Bank: q361

HGE - Hydraulics / Open Channels / Engr. Janclyde Espinosa (Clidez)

In order to provide water from a nearby spring, a triangular flume of efficient cross-section was provided on a slope of 0.21 percent. Assuming the roughness coefficient of the channel to be n = 0.018. Obtain the depth of flow in meter(s) of the water in the flume if it is discharging at the rate of 2 m2/sec.

Answer:

  1. 1.18
  2. 1.22
  3. 1.26
  4. 1.14
For the most efficient triangular flume, the side slopes are symmetrical at 45°, so:
$A=y^2,\quad R=\frac{y}{2\sqrt2}$
Using Manning's equation with $Q=2$ m3/s, $S=0.21\%=0.0021$, and $n=0.018$:
$2=\frac{1}{0.018}(y^2)\left(\frac{y}{2\sqrt2}\right)^{2/3}\sqrt{0.0021}$
$y=1.1846$ m
$\boxed{y\approx1.18}$

Question Bank: q366

HGE - Hydraulics / Open Channels / Engr. Janclyde Espinosa (Clidez)

An irrigation canal with trapezoidal cross-section has the following dimensions: Bottom width = 2.50 m, depth of water = 0.90 m, side slope = 1.5 horizontal to 1 vertical, slope of the canal bed = 0.001, coefficient of roughness = 0.025. The canal will serve clay-loam Riceland for which the duty of water per hectare is 3.0 liters/sec. Using Manning’s Formula:

Determine the hydraulic radius of the canal, in meters.

  1. 0.603
  2. 0.621
  3. 0.647
  4. 0.640

Compute the velocity of water in m/s.

  1. 0.903
  2. 0.912
  3. 0.974
  4. 0.942

Determine the number of hectares served by the irrigation canal.

  1. 1043
  2. 3130
  3. 1087
  4. 3260

Part 1.

For the trapezoidal canal:
$A=y(b+zy)=0.90[2.50+1.5(0.90)]=3.465$ m2
$P=b+2y\sqrt{1+z^2}=2.50+2(0.90)\sqrt{1+1.5^2}=5.745$ m
$R=\frac{A}{P}=\frac{3.465}{5.745}=0.603$ m
$\boxed{R=0.603\text{ m}}$

Part 2.

Using Manning's equation:
$V=\frac{1}{n}R^{2/3}S^{1/2}$
With $R=0.603$, $n=0.025$, and $S=0.001$:
$V=\frac{1}{0.025}(0.603)^{2/3}(0.001)^{1/2}=0.903$ m/s
$\boxed{V=0.903\text{ m/s}}$

Part 3.

The canal discharge is:
$Q=AV=3.465(0.903)=3.129$ m3/s
Duty is 3 L/s per hectare, or 0.003 m3/s per hectare:
$N=\frac{3.129}{0.003}=1043$ hectares
$\boxed{1043}$

Question Bank: q373

HGE - Hydraulics / Discharge / Engr. Janclyde Espinosa (Clidez)

A 50mm pipe 15m long extends vertically downward from the bottom of an elevated tank and discharges into air. The entrance from tank to pipe is square-cornered. When the water in the tank is 3m deep over the entrance to the pipe, what is the discharge? Neglect head loss.

Answer:

  1. 0.037
  2. 0.045
  3. 0.026
  4. 0.058
With head loss neglected, the total head from the tank surface to the pipe outlet is $3+15=18$ m:
$V=\sqrt{2gH}=\sqrt{2(9.81)(18)}=18.79$ m/s
$Q=AV=\frac{\pi(0.05)^2}{4}(18.79)=0.0369$ m3/s
$\boxed{Q\approx0.037}$

Question Bank: v2

HGE - Hydraulics / Flow Rate / HGE May 2019

Formula-mode item rendered with fixed values for lecture/PDF export.

Water flows through a rectangular irrigation canal 500 mm deep by 1.2 m wide with a mean velocity of 1.7 m/sec. Determine the rate of flow in m3/min.

  1. 1.02 m3/min
  2. 61.20 m3/min
  3. 30.60 m3/min
  4. 73.44 m3/min

The discharge is the cross-sectional area times the mean velocity, with depth $d=\dfrac{d_{mm}}{1000}$ in metres:

$$Q=A v=(b\,d)\,v\ \text{m}^3/\text{s}.$$

Convert to per-minute by multiplying by 60:

$$Q=(b\,d\,v)\times 60\ \text{m}^3/\text{min}.$$
Computed answer: 61.20 m3/min

Question Bank: v67

HGE - Hydraulics / Flow Rate / HGE November 2019

Formula-mode item rendered with fixed values for lecture/PDF export.

Water flows at 6.5 m3/s under a total head of 11 m. Obtain the horsepower in the flow.

  1. 941 hp
  2. 837 hp
  3. 1223 hp
  4. 978 hp
Hydraulic power is $P=Q\gamma H$ in kW. Convert using $1$ hp $=0.7457$ kW: $$P_{hp}=\frac{Q\gamma H}{0.7457}=\frac{(6.5)(9.81)(11)}{0.7457}=940.612846989\text{ hp}.$$
Computed answer: 941 hp

Question Bank: v71

HGE - Hydraulics / Open Channels / HGE November 2019

Formula-mode item rendered with fixed values for lecture/PDF export.

Water flows in a rectangular channel 5.5 m wide at depth 1 m. The bed slope is 0.0012 and Manning roughness is 0.02. Obtain the discharge.

  1. 9.30 m³/s
  2. 6.66 m³/s
  3. 7.75 m³/s
  4. 5.66 m³/s
For the rectangular channel, $A=by$, $P=b+2y$, and $R=A/P$. Manning gives $$Q=\frac1nAR^{2/3}S^{1/2}=7.74683431451\text{ m}^3/\text{s}.$$
Computed answer: 7.75 m³/s

Question Bank: v106

HGE - Hydraulics / Non-Uniform Flow / HGE November 2021

Formula-mode item rendered with fixed values for lecture/PDF export.

Water flows in an almost level rectangular channel of width 2.7 m at 15 m^3/s. The depth increases from d1 = 0.8 m to d2 = d1 + 0.05 m over a length L = 6 m. Determine the head loss, energy-grade-line slope, and Manning roughness coefficient.

The head loss, in meters.

  1. 0.231 m
  2. 0.173 m
  3. 0.300 m
  4. 0.369 m

The slope of the energy grade line.

  1. 0.02883
  2. 0.04998
  3. 0.03844
  4. 0.06151

The roughness coefficient.

  1. 0.0140
  2. 0.0233
  3. 0.0280
  4. 0.0186
Continuity gives $v_1=Q/(bd_1)$ and $v_2=Q/(bd_2)$. The energy equation gives $$HL=d_1+ rac{v_1^2}{2g}-d_2- rac{v_2^2}{2g}=0.230667489425 ext{ m},$$ so $S=HL/L=0.0384445815708$. Applying Manning's formula at both sections and averaging gives $$n=0.0186401051222.$$

Computed answers:
1. 0.231 m
2. 0.03844
3. 0.0186