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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

Question Bank: v127

HGE - Hydraulics / Open Channels / HGE May 2022

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

Water flows at 18 m3/s at a depth of 0.8 m in a trapezoidal canal having bottom width 5.5 m, side slope 2.5H:1V, and Manning roughness coefficient 0.015.

v127

Obtain the specific energy.

  1. 0.983 m
  2. 1.259 m
  3. 1.431 m
  4. 1.587 m

Evaluate the canal slope for uniform flow using Manning's formula.

  1. 0.003045
  2. 0.004434
  3. 0.003899
  4. 0.004917

Obtain the boundary shearing stress on the canal surface.

  1. 18.28 Pa
  2. 26.61 Pa
  3. 29.51 Pa
  4. 23.40 Pa
For a trapezoidal channel, $$A=y(b+zy)=6,\quad P=b+2y\sqrt{1+z^2}=9.80813184571,\quad R=A/P=0.611737290484.$$ The velocity is $v=Q/A=3$ m/s. Therefore, $$E=y+\frac{v^2}{2g}=1.25871559633\text{ m}.$$ Manning's formula gives $$S=\left(\frac{vn}{R^{2/3}}\right)^2=0.00389946554481,$$ and $$\tau_0=\gamma RS=23.4012496547\text{ Pa}.$$

Computed answers:
1. 1.259 m
2. 0.003899
3. 23.40 Pa

Question Bank: v150

HGE - Hydraulics / Open Channels / HGE Refresher Series

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

A rectangular open channel carries 0.3 m3/s. Its bottom width is 1 m, flow depth is 0.75 m, and Manning coefficient is 0.014.

Determine the hydraulic radius.

  1. 0.234 m
  2. 0.341 m
  3. 0.300 m
  4. 0.378 m

Determine the mean velocity.

  1. 0.400 m/s
  2. 0.312 m/s
  3. 0.455 m/s
  4. 0.504 m/s

Determine the bed slope required for uniform flow.

  1. 0.000122
  2. 0.000178
  3. 0.000197
  4. 0.000156
Compute A = by and R = A/P. Then use V = Q/A and rearrange Manning's equation to $S$ = (Vn/R2/3)2.
Computed answers: 0.3, 0.4, 0.000156152058724.

Question Bank: v154

HGE - Hydraulics / Open Channels / HGE Refresher Series

For a fixed flow area, which open-channel section has the smallest wetted perimeter?

  1. Rectangular
  2. Trapezoidal
  3. Triangular
  4. Semicircular
A semicircle provides the minimum wetted perimeter for a specified flow area.

Question Bank: v155

HGE - Hydraulics / Open Channels / HGE Refresher Series

What term describes flow whose depth and mean velocity do not vary from one section to another along the channel?

  1. Continuous flow
  2. Laminar flow
  3. Steady flow
  4. Uniform flow
Uniform flow has no spatial change in depth or mean velocity along the reach.

Question Bank: v162

HGE - Hydraulics / Open Channels / HGE Refresher Series

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

A rectangular irrigation canal is 7 m wide, carries water 0.8 m deep, has bed slope 0.0008, and Manning coefficient 0.012.

Compute the mean velocity.

  1. 1.38 m/s
  2. 2.01 m/s
  3. 1.77 m/s
  4. 2.23 m/s

Compute the discharge.

  1. 7.74 m^3/s
  2. 11.27 m^3/s
  3. 12.50 m^3/s
  4. 9.92 m^3/s

For the same area and slope, find the depth of the most efficient rectangular section.

  1. 1.67 m
  2. 1.31 m
  3. 1.90 m
  4. 2.11 m
Apply Manning's equation. For the most efficient rectangular section, b = 2y and A = 2y2.
Computed answers: 1.77074815497, 9.91618966781, 1.67332005307.

Question Bank: v166

HGE - Hydraulics / Open Channels / HGE Refresher Series

For the hydraulically best trapezoidal channel, how does the sloping-side length compare with one-half of the top width?

  1. They are equal.
  2. It is one-half as large.
  3. It is twice as large.
  4. It is √3 times as large.
In the best trapezoidal section, one sloping side equals half the top width.

Question Bank: v168

HGE - Hydraulics / Open Channels / HGE Refresher Series

What kind of flow has a discharge at a section that is unchanged with time?

  1. Continuous
  2. Laminar
  3. Steady
  4. Uniform
Steady flow is time-invariant at a fixed location.

Question Bank: v169

HGE - Hydraulics / Open Channels / HGE Refresher Series

What term applies when mean velocity is the same at every cross-section along a reach?

  1. Continuous
  2. Laminar
  3. Steady
  4. Uniform
Uniform flow is spatially unchanged along the reach.

Question Bank: v171

HGE - Hydraulics / Open Channels / HGE Refresher Series

Which term means the same discharge passes every section of a stream at a given instant?

  1. Continuous flow
  2. Steady flow
  3. Critical flow
  4. Uniform flow
Continuity requires equal flow rate through successive sections when no storage or lateral flow occurs.

Question Bank: v172

HGE - Hydraulics / Open Channels / HGE Refresher Series

At a given discharge, which open-channel state has minimum specific energy?

  1. Continuous
  2. Steady
  3. Critical
  4. Subcritical
Specific energy reaches its minimum at critical flow.

Question Bank: v178

HGE - Hydraulics / Open Channels / HGE Refresher Series

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

A rectangular channel 3.5 m wide and 1.1 m deep and a semicircular channel carry the same uniform-flow discharge with the same slope and roughness. The semicircle is full to its diameter.

Find the semicircular channel diameter.

  1. 2.357 m
  2. 3.018 m
  3. 3.431 m
  4. 3.805 m
Equate A R2/3 for the two sections because slope and Manning coefficient are unchanged.
Computed answers: 3.01777448792.

Question Bank: v190

HGE - Hydraulics / Open Channels / HGE Refresher Series

Which statements describe the most economical trapezoidal channel?

  1. Half the top width equals a sloping side.
  2. Hydraulic radius equals half the flow depth.
  3. A semicircle tangent to the sides also touches the waterline.
  4. All of these.
All listed geometric properties are equivalent conditions for the best trapezoidal section.

Question Bank: v193

HGE - Hydraulics / Open Channels / HGE Refresher Series

For the most economical rectangular channel, what is the relation between depth and width?

  1. Depth is one-fourth the width.
  2. Depth is three times the hydraulic radius.
  3. Depth is one-half the width.
  4. None of these.
The best rectangular section has b = 2y.

Question Bank: v197

HGE - Hydraulics / Open Channels / HGE Refresher Series

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

A wide rectangular channel carries 20 m3/s with unit discharge based on width 8 m.

Find the minimum specific energy.

  1. 1.29 m
  2. 1.01 m
  3. 1.47 m
  4. 1.63 m
At critical flow in a rectangular channel, yc = (q2/g)1/3 and Emin = 3yc/2.
Computed answers: 1.29070877417.

Question Bank: v200

HGE - Hydraulics / Open Channels / HGE Refresher Series

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

A rectangular canal is 5.75 m wide and 1.2 m deep with Manning n = 0.014 and slope 0.0011.

Compute the discharge.

  1. 14.63 m^3/s
  2. 11.43 m^3/s
  3. 16.63 m^3/s
  4. 18.45 m^3/s

Estimate the wetted-perimeter saving if rebuilt as the best rectangular section of equal area.

  1. 0.563 m
  2. 0.819 m
  3. 0.908 m
  4. 0.720 m

State the excavation-area saving for equal flow area.

  1. 0.431 m^2
  2. 0.628 m^2
  3. 0.552 m^2
  4. 0.696 m^2
Apply Manning's equation and the best rectangular condition b = 2y.
Computed answers: 14.6289016099, 0.720329751597, 0.552.

Question Bank: v203

HGE - Hydraulics / Open Channels / HGE Refresher Series

Which statement about specific energy in an open channel is incorrect?

  1. It is energy relative to the channel floor.
  2. Alternate depths can have the same specific energy.
  3. Velocity is critical when specific energy is maximum.
  4. Critical flow has Froude number 1.
Critical flow occurs at minimum, not maximum, specific energy.

Question Bank: v207

HGE - Hydraulics / Open Channels / HGE Refresher Series

What is the principal driving force for ordinary open-channel flow?

  1. Gravity
  2. Atmospheric pressure
  3. Hydrostatic pressure
  4. Mechanical pressure
Gravity supplies the component of force along the channel.

Question Bank: v216

HGE - Hydraulics / Open Channels / HGE Refresher Series

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

A trapezoidal channel has depth 2.6 m, mean velocity 1 m/s, and one of three stated side-slope configurations.

Compute discharge for equal side lengths inclined 60° from horizontal.

  1. 73.87 m^3/s
  2. 107.55 m^3/s
  3. 94.59 m^3/s
  4. 119.27 m^3/s

Compute discharge when each side slope is 2V:3H.

  1. 11.44 m^3/s
  2. 16.65 m^3/s
  3. 18.47 m^3/s
  4. 14.65 m^3/s

Compute discharge for the minimum-seepage section with side angle 55°.

  1. -20.89 m^3/s
  2. -30.41 m^3/s
  3. -26.75 m^3/s
  4. -33.73 m^3/s
Determine the cross-sectional area for each geometry and use Q = AV.
Computed answers: 94.5866526685, 14.6466666667, -26.7473947302.

Question Bank: v220

HGE - Hydraulics / Open Channels / HGE Refresher Series

For the most economical triangular channel, what angle does each side make with the vertical?

  1. 90°
  2. 30°
  3. 60°
  4. 45°
The best triangular section has a 90-degree included angle, or 45 degrees per side from vertical.

Question Bank: v237

HGE - Hydraulics / Open Channels / HGE Refresher Series

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

A hydraulically efficient triangular channel has Manning n=0.022, slope 0.0015, and discharge 2 m3/s.

Find flow depth.

  1. 0.82 m
  2. 1.19 m
  3. 1.05 m
  4. 1.32 m
For the best triangular section, side slopes are 1H:1V; substitute its A and R in Manning's equation.
Computed answers: 1.04900520381.

Question Bank: v242

HGE - Hydraulics / Open Channels / HGE Refresher Series

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

A rectangular channel width 3 m has specific energy 1.3 m and actual depth 0.65 m.

Find critical depth.

  1. 0.87 m
  2. 0.68 m
  3. 0.99 m
  4. 1.09 m

Return 1 for supercritical and 2 for subcritical.

  1. Flow code 1
  2. Flow code 11
  3. Flow code 21
  4. Flow code 31

Find maximum discharge at the given specific energy.

  1. 5.92 m^3/s
  2. 7.58 m^3/s
  3. 8.62 m^3/s
  4. 9.56 m^3/s
At maximum discharge, flow is critical and yc=2E/3.
Computed answers: 0.866666666667, 1, 7.58112920349.

Question Bank: v248

HGE - Hydraulics / Open Channels / HGE Refresher Series

What is the abrupt transition from supercritical to subcritical flow called?

  1. Hydraulic jump
  2. Water hammer
  3. Turbulence
  4. Shock wave
A hydraulic jump dissipates energy as flow changes regime.

Question Bank: v249

HGE - Hydraulics / Open Channels / HGE Refresher Series

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

An efficient triangular channel carries 5 m3/s with Manning n=0.014.

Find critical depth.

  1. 0.672 m
  2. 0.978 m
  3. 0.860 m
  4. 1.085 m

Find critical velocity.

  1. 5.274 m/s
  2. 6.753 m/s
  3. 7.678 m/s
  4. 8.516 m/s

Find critical slope.

  1. 0.04368
  2. 0.03412
  3. 0.04967
  4. 0.05509
Use critical-flow geometry and Manning's equation.
Computed answers: 0.860472516116, 6.75298830648, 0.043684300607.

Question Bank: v256

HGE - Hydraulics / Open Channels / HGE Refresher Series

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

A trapezoidal canal has one vertical side and one side inclined 45°. It carries 19 m3/s at mean velocity 0.85 m/s and is proportioned as a most efficient section.

Estimate required depth.

  1. 2.38 m
  2. 3.46 m
  3. 3.04 m
  4. 3.84 m
Use A=Q/V and the efficient-section geometry.
Computed answers: 3.04284265026.

Question Bank: v269

HGE - Hydraulics / Open Channels / HGE Refresher Series

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

A hydraulically best rectangular channel carries 4.75 m3/s on slope 0.001 with Manning coefficient 0.014.

Find the optimum channel width.

  1. 2.42 m
  2. 1.89 m
  3. 2.76 m
  4. 3.06 m
For the best rectangle b = 2y and R = y/2; substitute into Manning's equation.
Computed answers: 2.42359117924.