BE Civil Engineering (IOE, TU) Hydraulics (IOE, CE 604) Question Paper 2079 Nepal

(a) Condition for the most economical trapezoidal section

For a trapezoidal channel of bed width $b$, depth $y$ and side slope $z$ (horizontal : vertical):

• Area: $A = (b + zy)\,y$
• Wetted perimeter: $P = b + 2y\sqrt{1+z^2}$

For a fixed area $A$ and side slope $z$, the section is most economical when $P$ is minimum (this maximises $R = A/P$ and hence discharge).

From $A = (b+zy)y \Rightarrow b = \dfrac{A}{y} - zy$.

Substitute into $P$:

$$P = \frac{A}{y} - zy + 2y\sqrt{1+z^2}$$

Differentiate w.r.t. $y$ (with $A$, $z$ constant) and set $\dfrac{dP}{dy}=0$:

$$\frac{dP}{dy} = -\frac{A}{y^2} - z + 2\sqrt{1+z^2} = 0$$ $$\Rightarrow A = y^2\left(2\sqrt{1+z^2} - z\right)$$

But $A = (b+zy)y$, so:

$$b + zy = y\left(2\sqrt{1+z^2} - z\right) \Rightarrow \boxed{b = 2y\left(\sqrt{1+z^2} - z\right)}$$

This is the condition for the most economical trapezoidal section: the top width equals twice the length of one sloping side, and a semicircle drawn with centre on the water surface is tangent to the three sides.

Show $R = y/2$:

$P = b + 2y\sqrt{1+z^2} = 2y(\sqrt{1+z^2}-z) + 2y\sqrt{1+z^2} = 2y(2\sqrt{1+z^2} - z)$.

$A = y^2(2\sqrt{1+z^2}-z)$.

$$R = \frac{A}{P} = \frac{y^2(2\sqrt{1+z^2}-z)}{2y(2\sqrt{1+z^2}-z)} = \frac{y}{2}\qquad\blacksquare$$

(b) Design for $Q = 28\ \mathrm{m^3/s}$, $z=2$, $n=0.018$, $S_0=0.0009$

Geometry factor:

$$\sqrt{1+z^2} = \sqrt{1+4} = \sqrt{5} = 2.2361$$ $$2\sqrt{1+z^2} - z = 2(2.2361) - 2 = 2.4721$$

So for the economical section:

$$A = 2.4721\,y^2,\qquad R = \frac{y}{2}$$

Apply Manning's equation $Q = \dfrac{1}{n}A R^{2/3} S_0^{1/2}$:

$$28 = \frac{1}{0.018}\,(2.4721\,y^2)\left(\frac{y}{2}\right)^{2/3}(0.0009)^{1/2}$$

Compute constants:

• $(0.0009)^{1/2} = 0.03$
• $\left(\tfrac{1}{2}\right)^{2/3} = 0.6300$

$$28 = \frac{1}{0.018}\times 2.4721 \times 0.6300 \times 0.03 \times y^{2}\,y^{2/3}$$ $$28 = 55.556 \times 2.4721 \times 0.6300 \times 0.03 \times y^{8/3}$$ $$28 = 2.5953\, y^{8/3}$$ $$y^{8/3} = \frac{28}{2.5953} = 10.789$$ $$y = 10.789^{3/8} = 10.789^{0.375}$$

$\ln(10.789) = 2.3785$; $\times 0.375 = 0.8919$; $e^{0.8919} = 2.4399$.

$$\boxed{y \approx 2.44\ \mathrm{m}}$$

Bed width:

$$b = 2y(\sqrt{1+z^2}-z) = 2(2.44)(2.2361 - 2) = 2(2.44)(0.2361) = 1.152\ \mathrm{m}$$ $$\boxed{b \approx 1.15\ \mathrm{m}}$$

Check velocity:

$$A = 2.4721\,y^2 = 2.4721(2.44)^2 = 2.4721(5.954) = 14.72\ \mathrm{m^2}$$ $$V = \frac{Q}{A} = \frac{28}{14.72} = 1.90\ \mathrm{m/s}$$

This is a reasonable non-scouring, non-silting velocity for an earthen/lined channel. Verify with Manning: $R = y/2 = 1.22\ \mathrm{m}$, $R^{2/3}=1.144$, $V = \frac{1}{0.018}(1.144)(0.03) = 1.91\ \mathrm{m/s}$ ✓ (matches $Q/A$).

Design summary: depth $y \approx 2.44\ \mathrm{m}$, bed width $b \approx 1.15\ \mathrm{m}$, side slope $2H{:}1V$, velocity $\approx 1.90\ \mathrm{m/s}$.

(a) Specific energy

Specific energy $E$ is the energy per unit weight of water measured with respect to the channel bed:

$$E = y + \frac{V^2}{2g} = y + \frac{Q^2}{2gA^2}$$

For a rectangular channel with specific discharge $q = Q/b$:

$$E = y + \frac{q^2}{2gy^2}$$

Specific-energy curve (E on x-axis, y on y-axis) for constant $q$:

  y |        upper limb (subcritical, Fr<1)
    |      /
 yc |---- C  (critical, minimum E = Emin)
    |    /
    |   /  lower limb (supercritical, Fr>1)
    |  /
    +-------------------- E
           Emin

For any $E > E_{min}$ there are two alternate depths (one subcritical on the upper limb, one supercritical on the lower limb). The minimum specific energy occurs at the critical depth $y_c$, where $Fr = 1$.

(b) Computations

Given: $b = 4\ \mathrm{m}$, $Q = 12\ \mathrm{m^3/s}$, $y_1 = 1.8\ \mathrm{m}$.

$q = Q/b = 12/4 = 3\ \mathrm{m^2/s}$.

$A_1 = 4(1.8) = 7.2\ \mathrm{m^2}$, $V_1 = 12/7.2 = 1.667\ \mathrm{m/s}$.

Specific energy:

$$E_1 = y_1 + \frac{V_1^2}{2g} = 1.8 + \frac{1.667^2}{2(9.81)} = 1.8 + \frac{2.778}{19.62} = 1.8 + 0.1416 = 1.942\ \mathrm{m}$$

Critical depth:

$$y_c = \left(\frac{q^2}{g}\right)^{1/3} = \left(\frac{3^2}{9.81}\right)^{1/3} = \left(\frac{9}{9.81}\right)^{1/3} = (0.9174)^{1/3} = 0.9717\ \mathrm{m}$$ $$\boxed{y_c \approx 0.972\ \mathrm{m}}$$

Froude number at $y_1$:

$$Fr_1 = \frac{V_1}{\sqrt{g y_1}} = \frac{1.667}{\sqrt{9.81(1.8)}} = \frac{1.667}{\sqrt{17.658}} = \frac{1.667}{4.202} = 0.397$$

Since $Fr_1 < 1$, the flow is subcritical.

Alternate depth $y_2$ (same $E$, same $q$): solve $E_1 = y_2 + \dfrac{q^2}{2gy_2^2}$:

$$1.942 = y_2 + \frac{9}{2(9.81)y_2^2} = y_2 + \frac{0.4587}{y_2^2}$$

The alternate depth is supercritical (below $y_c$). Solve by iteration:

• $y_2 = 0.55$: $0.55 + 0.4587/0.3025 = 0.55 + 1.516 = 2.066$ (too high)
• $y_2 = 0.60$: $0.60 + 0.4587/0.36 = 0.60 + 1.274 = 1.874$ (too low)
• $y_2 = 0.58$: $0.58 + 0.4587/0.3364 = 0.58 + 1.364 = 1.944$ ✓

$$\boxed{y_2 \approx 0.58\ \mathrm{m}\ (\text{supercritical alternate depth})}$$

(c) Flow over a hump $\Delta z = 0.25\ \mathrm{m}$

With no energy loss, the specific energy over the hump is reduced by the hump height:

$$E_2 = E_1 - \Delta z = 1.942 - 0.25 = 1.692\ \mathrm{m}$$

Check choking: minimum specific energy at the hump $E_{min} = \frac{3}{2}y_c = 1.5(0.9717) = 1.458\ \mathrm{m}$. Since $E_2 = 1.692 > E_{min} = 1.458$, the flow is not choked; upstream depth is unaffected.

The upstream flow is subcritical, so the depth over the hump $y_h$ stays on the subcritical limb (closer to $y_c$ but $> y_c$):

$$1.692 = y_h + \frac{0.4587}{y_h^2}$$

Iterate (subcritical root, $y_h > y_c = 0.972$):

• $y_h = 1.50$: $1.50 + 0.4587/2.25 = 1.50 + 0.2039 = 1.704$ (slightly high)
• $y_h = 1.49$: $1.49 + 0.4587/2.2201 = 1.49 + 0.2066 = 1.697$
• $y_h = 1.48$: $1.48 + 0.4587/2.1904 = 1.48 + 0.2094 = 1.689$
• $y_h = 1.485$: $1.485 + 0.4587/2.2052 = 1.485 + 0.2080 = 1.693$ ✓

$$\boxed{y_h \approx 1.485\ \mathrm{m}}$$

Change in upstream water-surface level:

• Water-surface elevation upstream (above bed datum) $= y_1 = 1.80\ \mathrm{m}$.
• Water-surface elevation over hump $= \Delta z + y_h = 0.25 + 1.485 = 1.735\ \mathrm{m}$.

The water surface dips by $1.80 - 1.735 = 0.065\ \mathrm{m}$ over the hump (typical for subcritical flow). Since the flow is not choked, there is no rise in the upstream water level.

Drop in water surface over hump $\approx 0.065\ \mathrm{m}$.

(a) Sequent-depth (momentum) equation

Apply the momentum equation between sections 1 (before) and 2 (after) the jump on a horizontal bed, neglecting boundary friction. The hydrostatic pressure force minus equals the rate of change of momentum:

$$\frac{\gamma b y_1^2}{2} - \frac{\gamma b y_2^2}{2} = \rho Q (V_2 - V_1)$$

Using $Q = b q$, $V = q/y$, and dividing by $\gamma b$:

$$\frac{y_1^2}{2} - \frac{y_2^2}{2} = \frac{q}{g}\left(\frac{q}{y_2} - \frac{q}{y_1}\right) = \frac{q^2}{g}\left(\frac{1}{y_2}-\frac{1}{y_1}\right)$$ $$\frac{1}{2}(y_1^2 - y_2^2) = \frac{q^2}{g}\cdot\frac{y_1 - y_2}{y_1 y_2}$$

Divide both sides by $(y_1 - y_2)$ (noting $y_1^2 - y_2^2 = (y_1-y_2)(y_1+y_2)$):

$$\frac{1}{2}(y_1 + y_2) = \frac{q^2}{g\,y_1 y_2}$$ $$y_1 y_2 (y_1 + y_2) = \frac{2q^2}{g}$$

With $Fr_1^2 = \dfrac{q^2}{g y_1^3}$, dividing through by $y_1^3$ and solving the quadratic in $(y_2/y_1)$:

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

(b) Numerical computation

Given: $b = 5\ \mathrm{m}$, $y_1 = 0.45\ \mathrm{m}$, $Q = 20\ \mathrm{m^3/s}$.

$q = Q/b = 20/5 = 4\ \mathrm{m^2/s}$.

$V_1 = q/y_1 = 4/0.45 = 8.889\ \mathrm{m/s}$.

Upstream Froude number:

$$Fr_1 = \frac{V_1}{\sqrt{g y_1}} = \frac{8.889}{\sqrt{9.81(0.45)}} = \frac{8.889}{\sqrt{4.4145}} = \frac{8.889}{2.1011} = 4.231$$ $$\boxed{Fr_1 \approx 4.23}$$

Sequent depth:

$$\frac{y_2}{y_1} = \frac{1}{2}\left(\sqrt{1+8(4.231)^2}-1\right) = \frac{1}{2}\left(\sqrt{1+8(17.901)}-1\right)$$ $$= \frac{1}{2}\left(\sqrt{1+143.21}-1\right) = \frac{1}{2}(\sqrt{144.21}-1) = \frac{1}{2}(12.009-1) = 5.504$$ $$y_2 = 5.504(0.45) = 2.477\ \mathrm{m}$$ $$\boxed{y_2 \approx 2.48\ \mathrm{m}}$$

Energy loss:

$$\Delta E = \frac{(y_2 - y_1)^3}{4\,y_1 y_2} = \frac{(2.477 - 0.45)^3}{4(0.45)(2.477)}$$ $$= \frac{(2.027)^3}{4.4586} = \frac{8.328}{4.4586} = 1.868\ \mathrm{m}$$ $$\boxed{\Delta E \approx 1.87\ \mathrm{m}}$$

Cross-check via total energies: $V_2 = 4/2.477 = 1.615\ \mathrm{m/s}$. $E_1 = 0.45 + \frac{8.889^2}{19.62} = 0.45 + 4.027 = 4.477\ \mathrm{m}$. $E_2 = 2.477 + \frac{1.615^2}{19.62} = 2.477 + 0.133 = 2.610\ \mathrm{m}$. $\Delta E = 4.477 - 2.610 = 1.867\ \mathrm{m}$ ✓

Length of jump:

$$L_j = 6.9(y_2 - y_1) = 6.9(2.027) = 13.99\ \mathrm{m} \approx 14.0\ \mathrm{m}$$

Classification: Since $Fr_1 = 4.23$ lies in the range $4.5 > Fr_1 > 2.5$... more precisely $Fr_1 = 4.23$ is in $2.5$–$4.5$, this is an oscillating jump (transitional). For $Fr_1 > 4.5$ it would be a steady jump. With $Fr_1 = 4.23$ the jump is well-developed and effective for energy dissipation.

Energy dissipation efficiency: $\Delta E / E_1 = 1.868/4.477 = 0.417$, i.e. about 42% of the upstream energy is dissipated.

(a) GVF assumptions and dynamic equation

Assumptions:

1. The flow is steady; discharge constant.
2. The pressure distribution is hydrostatic (streamlines nearly parallel; small curvature).
3. The head-loss formula for uniform flow (Manning/Chezy) is valid for the gradually varied flow using the local depth.
4. The channel slope is small ($\cos\theta \approx 1$, $\sin\theta\approx\tan\theta = S_0$).
5. The roughness coefficient is constant and independent of depth.
6. The channel is prismatic.

Derivation: Total energy head above datum:

$$H = z + y + \frac{V^2}{2g}$$

Differentiate w.r.t. $x$ (distance along channel):

$$\frac{dH}{dx} = \frac{dz}{dx} + \frac{dy}{dx} + \frac{d}{dx}\left(\frac{V^2}{2g}\right)$$

• $\dfrac{dH}{dx} = -S_f$ (slope of energy line), $\dfrac{dz}{dx} = -S_0$.
• $\dfrac{d}{dx}\left(\dfrac{V^2}{2g}\right) = \dfrac{d}{dy}\left(\dfrac{Q^2}{2gA^2}\right)\dfrac{dy}{dx} = -\dfrac{Q^2 T}{g A^3}\dfrac{dy}{dx} = -Fr^2\dfrac{dy}{dx}$.

Thus:

$$-S_f = -S_0 + \frac{dy}{dx} - Fr^2\frac{dy}{dx}$$ $$\frac{dy}{dx}(1 - Fr^2) = S_0 - S_f$$ $$\boxed{\frac{dy}{dx} = \frac{S_0 - S_f}{1 - Fr^2}}$$

(b) Normal depth, critical depth, slope class, profile type

Critical depth (wide rectangular, per unit width $q=2.5$):

$$y_c = \left(\frac{q^2}{g}\right)^{1/3} = \left(\frac{2.5^2}{9.81}\right)^{1/3} = \left(\frac{6.25}{9.81}\right)^{1/3} = (0.6371)^{1/3} = 0.8606\ \mathrm{m}$$ $$\boxed{y_c \approx 0.861\ \mathrm{m}}$$

Normal depth (wide channel, $R \approx y$), Manning: $q = \dfrac{1}{n}y_n^{5/3}S_0^{1/2}$:

$$2.5 = \frac{1}{0.020}\,y_n^{5/3}(0.0004)^{1/2} = 50\,y_n^{5/3}(0.02) = y_n^{5/3}$$

Wait, $50 \times 0.02 = 1.0$, so $2.5 = 1.0\, y_n^{5/3}$:

$$y_n^{5/3} = 2.5 \Rightarrow y_n = 2.5^{3/5} = 2.5^{0.6}$$

$\ln 2.5 = 0.9163$; $\times 0.6 = 0.5498$; $e^{0.5498} = 1.733$.

$$\boxed{y_n \approx 1.733\ \mathrm{m}}$$

Slope classification: $y_n = 1.733\ \mathrm{m} > y_c = 0.861\ \mathrm{m}$, so the slope is mild (M).

Profile type: At the section, $y = 1.8\ \mathrm{m}$. Compare: $y = 1.8 > y_n = 1.733 > y_c = 0.861$, so the depth lies in Zone 1 of a mild slope. This is an M1 profile (backwater curve), with depth increasing downstream and asymptotically approaching $y_n$ upstream. Moving upstream from $y=1.8\ \mathrm{m}$ the depth decreases toward $y_n$.

(c) Direct-step method (single step), $1.8\ \mathrm{m} \to 1.7\ \mathrm{m}$

Direct-step formula:

$$\Delta x = \frac{E_2 - E_1}{S_0 - \bar{S_f}}$$

where $E = y + \dfrac{q^2}{2gy^2}$ and $\bar{S_f}$ is the mean friction slope.

$\dfrac{q^2}{2g} = \dfrac{6.25}{19.62} = 0.3186$.

Section 1 ($y=1.8$): $E_1 = 1.8 + 0.3186/1.8^2 = 1.8 + 0.3186/3.24 = 1.8 + 0.0983 = 1.8983\ \mathrm{m}$. $V_1 = 2.5/1.8 = 1.389\ \mathrm{m/s}$. $S_{f1} = \dfrac{n^2 V_1^2}{y_1^{4/3}} = \dfrac{(0.020)^2(1.389)^2}{1.8^{4/3}}$. $1.8^{4/3} = e^{(4/3)\ln1.8} = e^{1.333(0.5878)} = e^{0.7837} = 2.1896$. $S_{f1} = \dfrac{0.0004 \times 1.9293}{2.1896} = \dfrac{0.0007717}{2.1896} = 0.0003525$.

Section 2 ($y=1.7$): $E_2 = 1.7 + 0.3186/1.7^2 = 1.7 + 0.3186/2.89 = 1.7 + 0.1102 = 1.8102\ \mathrm{m}$. $V_2 = 2.5/1.7 = 1.471\ \mathrm{m/s}$. $1.7^{4/3} = e^{1.333(0.5306)} = e^{0.7074} = 2.0288$. $S_{f2} = \dfrac{(0.020)^2(1.471)^2}{2.0288} = \dfrac{0.0004\times 2.1638}{2.0288} = \dfrac{0.0008655}{2.0288} = 0.0004266$.

Mean friction slope: $\bar{S_f} = \tfrac12(0.0003525 + 0.0004266) = 0.00038955$.

Step length:

$$\Delta x = \frac{E_2 - E_1}{S_0 - \bar{S_f}} = \frac{1.8102 - 1.8983}{0.0004 - 0.00038955} = \frac{-0.0881}{0.00001045}$$ $$= -8431\ \mathrm{m}$$

The negative sign with $E_2<E_1$ and direction convention indicates the depth $1.7\ \mathrm{m}$ occurs upstream of the $1.8\ \mathrm{m}$ section by:

$$\boxed{|\Delta x| \approx 8.43\ \mathrm{km}}$$

Interpretation: The very small denominator ($S_0$ barely exceeds $\bar S_f$ because both depths are near normal depth) gives a long M1 profile, consistent with backwater curves that extend far upstream of a control.

(a) Water hammer; rapid vs gradual closure

Water hammer is the transient pressure surge produced in a closed conduit when the velocity of the flowing liquid is changed rapidly (e.g. by closing a valve). The kinetic energy of the moving column is converted into pressure (strain) energy, producing a pressure wave that travels back and forth along the pipe at the celerity $a$, alternately compressing and rarefying the liquid until friction damps it out.

• Sudden / rapid closure: the valve closes in a time $T < T_c = \dfrac{2L}{a}$ (the time for the pressure wave to travel to the reservoir and back). The full Joukowsky pressure rise develops: $\Delta p = \rho a \Delta V$. The pipe near the valve experiences the maximum surge.
• Gradual / slow closure: the valve closes in $T > T_c = \dfrac{2L}{a}$. The reflected negative wave returns before closure is complete and partly relieves the pressure, so the maximum pressure rise is smaller and can be estimated by $\Delta p \approx \dfrac{\rho L V_0}{T}$ (rigid-column approximation).

(b) Celerity, critical time, instantaneous pressure rise

Effective bulk modulus / celerity for an elastic pipe:

$$a = \sqrt{\dfrac{K/\rho}{1 + \dfrac{K\,D}{E\,t}}}$$

Compute the correction term:

$$\frac{K D}{E t} = \frac{(2.1\times10^{9})(0.5)}{(2.0\times10^{11})(0.008)} = \frac{1.05\times10^{9}}{1.6\times10^{9}} = 0.65625$$ $$\frac{K}{\rho} = \frac{2.1\times10^{9}}{1000} = 2.1\times10^{6}\ \mathrm{m^2/s^2}$$ $$a = \sqrt{\frac{2.1\times10^{6}}{1 + 0.65625}} = \sqrt{\frac{2.1\times10^{6}}{1.65625}} = \sqrt{1.2680\times10^{6}} = 1126\ \mathrm{m/s}$$ $$\boxed{a \approx 1126\ \mathrm{m/s}}$$

Critical (pipe-period) time:

$$T_c = \frac{2L}{a} = \frac{2(600)}{1126} = \frac{1200}{1126} = 1.066\ \mathrm{s}$$ $$\boxed{T_c \approx 1.07\ \mathrm{s}}$$

Pressure rise for instantaneous closure (Joukowsky):

$$\Delta p = \rho a V_0 = 1000 \times 1126 \times 2.5 = 2.815\times10^{6}\ \mathrm{Pa}$$ $$\boxed{\Delta p \approx 2.82\ \mathrm{MPa}\ (\approx 287\ \mathrm{m\ head})}$$

(Head: $\Delta H = \Delta p/(\rho g) = 2.815\times10^6/9810 = 287\ \mathrm{m}$.)

(c) Closure in $T = 1.5\ \mathrm{s}$

Compare with critical time: $T = 1.5\ \mathrm{s} > T_c = 1.07\ \mathrm{s}$, therefore the closure is slow (gradual).

For slow closure, use the rigid-column estimate:

$$\Delta p = \frac{\rho L V_0}{T} = \frac{1000 \times 600 \times 2.5}{1.5} = \frac{1.5\times10^{6}}{1.5} = 1.0\times10^{6}\ \mathrm{Pa}$$ $$\boxed{\Delta p \approx 1.0\ \mathrm{MPa}\ (\approx 102\ \mathrm{m\ head})}$$

This is much smaller than the instantaneous-closure surge (2.82 MPa), confirming the benefit of slow valve operation. Head: $\Delta H = 1.0\times10^6/9810 = 101.9\ \mathrm{m}$.

(a) Uniform vs non-uniform flow

(b) Discharge

Given: $b = 2.5\ \mathrm{m}$, $y = 1.2\ \mathrm{m}$, $S_0 = 1/1600 = 0.000625$, $n=0.015$.

$A = b y = 2.5(1.2) = 3.0\ \mathrm{m^2}$.

$P = b + 2y = 2.5 + 2(1.2) = 4.9\ \mathrm{m}$.

$R = A/P = 3.0/4.9 = 0.6122\ \mathrm{m}$.

$R^{2/3} = 0.6122^{2/3}$: $\ln0.6122 = -0.4908$; $\times \tfrac23 = -0.3272$; $e^{-0.3272}=0.7210$.

$S_0^{1/2} = \sqrt{0.000625} = 0.025$.

$$V = \frac{1}{n}R^{2/3}S_0^{1/2} = \frac{1}{0.015}(0.7210)(0.025) = 66.667 \times 0.018025 = 1.2017\ \mathrm{m/s}$$ $$Q = A V = 3.0 \times 1.2017 = 3.605\ \mathrm{m^3/s}$$ $$\boxed{Q \approx 3.61\ \mathrm{m^3/s}}$$

(a) Critical flow

Critical flow is the flow condition at which the specific energy is a minimum for a given discharge (equivalently, the discharge is maximum for a given specific energy). Characterising conditions:

1. Froude number $Fr = 1$, i.e. $\dfrac{Q^2 T}{g A^3} = 1$.
2. Specific energy is minimum for the given $Q$.
3. The flow velocity equals the celerity of a small gravity wave, $V = \sqrt{g\,A/T} = \sqrt{g\,D}$ (where $D = A/T$ is the hydraulic depth).

(b) Critical depth in a triangular channel

For a triangular section with side slope $z = 1.5$ (horizontal:vertical):

• Area: $A = z y^2 = 1.5 y^2$
• Top width: $T = 2 z y = 3 y$

Critical-flow condition $\dfrac{Q^2 T}{g A^3} = 1$:

$$\frac{Q^2 (2zy)}{g (zy^2)^3} = 1 \Rightarrow \frac{Q^2 \cdot 2z y}{g z^3 y^6} = 1 \Rightarrow Q^2 = \frac{g z^2 y^5}{2}$$

Solve for $y_c$:

$$y_c = \left(\frac{2Q^2}{g z^2}\right)^{1/5} = \left(\frac{2(3.0)^2}{9.81(1.5)^2}\right)^{1/5} = \left(\frac{2(9)}{9.81(2.25)}\right)^{1/5} = \left(\frac{18}{22.0725}\right)^{1/5}$$ $$= (0.8155)^{1/5}$$

$\ln 0.8155 = -0.2039$; $\times 0.2 = -0.04078$; $e^{-0.04078} = 0.9600$.

$$\boxed{y_c \approx 0.960\ \mathrm{m}}$$

Critical velocity:

$$A_c = 1.5 y_c^2 = 1.5(0.960)^2 = 1.5(0.9216) = 1.3824\ \mathrm{m^2}$$ $$V_c = \frac{Q}{A_c} = \frac{3.0}{1.3824} = 2.170\ \mathrm{m/s}$$ $$\boxed{V_c \approx 2.17\ \mathrm{m/s}}$$

Check $Fr=1$: hydraulic depth $D = A/T = 1.3824/(3\times0.960) = 1.3824/2.88 = 0.480\ \mathrm{m}$; $\sqrt{gD} = \sqrt{9.81\times0.480}=\sqrt{4.709}=2.170\ \mathrm{m/s} = V_c$ ✓

(a) M-profiles on a mild slope ($y_n > y_c$)

On a mild slope the channel has two reference lines: the normal depth line (NDL) at $y_n$ and the critical depth line (CDL) at $y_c$, with $y_n > y_c$. Three zones:

  ----------------------------- NDL (y = yn)   Zone 1: y > yn   -> M1
  ----------------------------- CDL (y = yc)   Zone 2: yc<y<yn  -> M2
  ///////////////////////////// bed            Zone 3: y < yc   -> M3

• M1 (backwater curve), Zone 1, $y > y_n > y_c$: subcritical; depth increases downstream, surface asymptotic to NDL upstream and rises toward horizontal downstream. Concave-up rising profile.
• M2 (drawdown curve), Zone 2, $y_n > y > y_c$: subcritical; depth decreases in the downstream direction from $y_n$ toward $y_c$. The surface drops down toward critical at the downstream end.
• M3 profile, Zone 3, $y < y_c$: supercritical; depth increases downstream toward $y_c$, usually terminating in a hydraulic jump. Short, steep rising profile.

In all three the slope of the water surface is governed by $\dfrac{dy}{dx}=\dfrac{S_0-S_f}{1-Fr^2}$.

(b) Practical occurrences

• M1: Backwater upstream of a dam, weir or sluice gate raising the level above normal depth in a mild channel; also flow approaching a reservoir.
• M2: Drawdown at a free overfall at the end of a mild channel, or where a mild channel meets a steeper reach (sudden enlargement / increase in slope).
• M3: Flow issuing from beneath a sluice gate onto a mild bed (high-velocity supercritical jet) before it rises to a hydraulic jump; or flow from a steep reach onto a mild reach.

(a) Surges

A surge is a moving wave front (an abrupt, rapidly varied change of depth) that travels along an open channel as a result of a sudden change in flow rate or depth, e.g. rapid gate operation. It is essentially a moving hydraulic jump.

• Positive surge: results in an increase of depth (an advancing wall of water). It has a steep, stable front. Examples: surge moving upstream when a downstream gate is suddenly closed; surge moving downstream when an upstream gate is suddenly opened (tidal bore).
• Negative surge: results in a decrease of depth; it is a gradually flattening (dispersive, unstable) wave, e.g. when a downstream gate is suddenly opened or an upstream gate is suddenly closed.

(b) Speed of an upstream positive surge

Let the absolute surge velocity be $V_w$ (positive in the downstream direction; an upstream-moving surge will give $V_w < 0$). Reduce the unsteady problem to steady by superposing $-V_w$ (i.e. view in a frame moving with the surge).

Given: $b=3\ \mathrm{m}$, $y_1 = 1.5\ \mathrm{m}$ (initial, downstream of front), $V_1 = 1.2\ \mathrm{m/s}$, $y_2 = 2.0\ \mathrm{m}$ (raised depth behind front).

Continuity (relative velocities across the front), with $V_w$ the surge absolute speed:

$$(V_1 - V_w)\,y_1 = (V_2 - V_w)\,y_2$$

Momentum across the moving surge (treated as a hydraulic jump in the moving frame):

$$(V_1 - V_w)^2 = \frac{g\,y_2}{2 y_1}(y_1 + y_2)$$

Compute the right side:

$$(V_1 - V_w)^2 = \frac{9.81 \times 2.0}{2 \times 1.5}(1.5 + 2.0) = \frac{19.62}{3.0}(3.5) = 6.54 \times 3.5 = 22.89$$ $$V_1 - V_w = \pm\sqrt{22.89} = \pm 4.784\ \mathrm{m/s}$$

For a surge travelling upstream, $V_w$ is negative and $(V_1 - V_w)$ is positive, so take the $+$ root:

$$V_1 - V_w = 4.784 \Rightarrow V_w = V_1 - 4.784 = 1.2 - 4.784 = -3.584\ \mathrm{m/s}$$

The negative sign confirms the surge moves upstream. Its absolute speed is:

$$\boxed{|V_w| \approx 3.58\ \mathrm{m/s}\ (\text{travelling upstream})}$$

Optional check (velocity behind surge): from continuity, $V_2 = V_w + \dfrac{(V_1-V_w)y_1}{y_2} = -3.584 + \dfrac{4.784(1.5)}{2.0} = -3.584 + 3.588 = 0.004\ \mathrm{m/s} \approx 0$, i.e. the water behind the surge is nearly brought to rest — consistent with a near-complete gate closure.

(a) Broad-crested weir

A broad-crested weir is a flat-topped hydraulic structure placed across a channel whose crest is long enough (in the flow direction) that the streamlines become parallel and critical flow is established over the crest. It is used to measure or regulate discharge and to raise upstream water levels. Because critical depth occurs on the crest, the discharge is fixed by the upstream head.

Discharge equation (with critical flow on the crest, $y_c = \tfrac23 H$):

$$Q = C_d\,L\,\sqrt{g}\left(\frac{2}{3}\right)^{3/2} H^{3/2}$$

or in the common simplified form

$$Q = 1.705\,C_d\,L\,H^{3/2}$$

where $1.705 = \sqrt{g}\,(2/3)^{3/2}$ (with $g=9.81$). $L$ = crest length (width across channel), $H$ = head over crest.

(b) Discharge

Given: $L = 8\ \mathrm{m}$, $H = 0.6\ \mathrm{m}$, $C_d = 0.88$.

Compute the coefficient: $\sqrt{g}\,(2/3)^{3/2} = 3.1321 \times 0.5443 = 1.7050$.

$$Q = 1.705\,C_d\,L\,H^{3/2} = 1.705 \times 0.88 \times 8 \times (0.6)^{3/2}$$

$(0.6)^{3/2} = 0.6\sqrt{0.6} = 0.6 \times 0.7746 = 0.4648$.

$$Q = 1.705 \times 0.88 \times 8 \times 0.4648 = 1.705 \times 0.88 = 1.5004;\ \times 8 = 12.003;\ \times 0.4648 = 5.579\ \mathrm{m^3/s}$$ $$\boxed{Q \approx 5.58\ \mathrm{m^3/s}}$$

Cross-check via critical-flow form: $y_c = \tfrac23 H = 0.4\ \mathrm{m}$; $q = \sqrt{g\,y_c^3}=\sqrt{9.81(0.064)}=\sqrt{0.6278}=0.7924\ \mathrm{m^2/s}$; ideal $Q = qL = 6.339\ \mathrm{m^3/s}$; with $C_d=0.88$: $Q = 0.88\times6.339 = 5.58\ \mathrm{m^3/s}$ ✓

(a) Factors governing stable channel design

1. Discharge to be carried (design $Q$).
2. Permissible (non-scouring) maximum velocity and non-silting minimum velocity of the boundary material; or the permissible tractive (shear) force on bed and banks.
3. Nature of the boundary material / lining (governs $n$, side slopes and permissible velocity).
4. Longitudinal bed slope available from topography.
5. Side slopes consistent with the angle of repose of the soil (for unlined channels) or structural stability (for lined).
6. Manning's / Chezy roughness of the surface.
7. Sediment load (silt charge) — use of Kennedy's or Lacey's regime theory for alluvial channels.
8. Freeboard, economy (most economical section) and seepage considerations.

(b) Normal depth of a lined trapezoidal canal

Given: $Q=15\ \mathrm{m^3/s}$, $b=4\ \mathrm{m}$, $z=1$, $S_0=0.0005$, $n=0.016$.

Manning: $Q = \dfrac{1}{n}A R^{2/3}S_0^{1/2}$ with $A = (b+zy)y = (4+y)y$, $P = b + 2y\sqrt{1+z^2} = 4 + 2\sqrt2\,y = 4 + 2.8284 y$.

$S_0^{1/2} = \sqrt{0.0005} = 0.022361$. $\dfrac{1}{n}S_0^{1/2} = \dfrac{0.022361}{0.016} = 1.3975$.

So we need $A R^{2/3} = \dfrac{Q}{1.3975} = \dfrac{15}{1.3975} = 10.733$.

Solve by trial for $y$:

Try $y = 1.6$: $A=(4+1.6)1.6 = 5.6(1.6)=8.96$; $P=4+2.8284(1.6)=8.5254$; $R=8.96/8.5254=1.0510$; $R^{2/3}=1.0338$; $AR^{2/3}=8.96(1.0338)=9.263$ (low).

Try $y = 1.8$: $A=(4+1.8)1.8=5.8(1.8)=10.44$; $P=4+2.8284(1.8)=9.0911$; $R=10.44/9.0911=1.1484$; $R^{2/3}=1.0969$; $AR^{2/3}=10.44(1.0969)=11.452$ (high).

Try $y = 1.72$: $A=(4+1.72)1.72=5.72(1.72)=9.838$; $P=4+2.8284(1.72)=8.8649$; $R=9.838/8.8649=1.1098$; $R^{2/3}=1.0721$; $AR^{2/3}=9.838(1.0721)=10.548$ (slightly low).

Try $y = 1.74$: $A=(4+1.74)1.74=5.74(1.74)=9.988$; $P=4+2.8284(1.74)=8.9214$; $R=9.988/8.9214=1.1195$; $R^{2/3}=1.0784$; $AR^{2/3}=9.988(1.0784)=10.771$ (slightly high).

Try $y = 1.737$: $A=(4+1.737)1.737=5.737(1.737)=9.965$; $P=4+2.8284(1.737)=8.9129$; $R=1.1181$; $R^{2/3}=1.0775$; $AR^{2/3}=9.965(1.0775)=10.737 \approx 10.733$ ✓

$$\boxed{y_n \approx 1.74\ \mathrm{m}}$$

Check: $A = 9.97\ \mathrm{m^2}$, $V = Q/A = 15/9.97 = 1.50\ \mathrm{m/s}$ — an acceptable non-scouring velocity for a lined canal.