BE Civil Engineering (IOE, TU) Hydraulics (IOE, CE 604) Question Paper 2080 Nepal

(a) Definitions and conditions

Normal depth is the depth of flow at which uniform flow occurs in a prismatic channel for a given discharge, i.e. the depth at which the energy/bed/water-surface slopes are all equal and the gravity component driving the flow exactly balances the boundary resistance, so the depth remains constant along the channel.

Most economical (best hydraulic) section: For a given discharge, slope and roughness, the most economical section is the one that conveys the discharge with the minimum cross-sectional area (hence minimum excavation/lining cost). From Manning's equation $Q=\frac{1}{n}A R^{2/3}S_0^{1/2}$, for fixed $A$ the discharge is maximum when the hydraulic radius $R=A/P$ is maximum, which means the wetted perimeter $P$ is minimum.

For a trapezoidal section with side slope $m\,\text{H}:1\,\text{V}$, the two geometric conditions of the most economical section are:

1. Top width equals twice the sloping side length: $b + 2my = 2y\sqrt{1+m^2}$, equivalently the half-top-width equals the sloping side length.
2. Hydraulic radius equals half the depth: $R = \dfrac{y}{2}$.

(Equivalently, a semicircle of radius $y$ centred at the water surface is tangent to the three sides — the section can inscribe such a circle.)

(b) Design of the most economical section

Given: $Q=25\ \text{m}^3/\text{s}$, $S_0=0.0009$, $m=1.5$, $n=0.018$.

Step 1 — Geometry of the economical section. For the most economical trapezoidal section:

$$b = 2y\left(\sqrt{1+m^2}-m\right)$$

With $m=1.5$: $\sqrt{1+1.5^2}=\sqrt{3.25}=1.80278$.

$$b = 2y(1.80278-1.5) = 2y(0.30278) = 0.60555\,y$$

Step 2 — Area and hydraulic radius. Area: $A=(b+my)\,y=(0.60555y+1.5y)\,y = 2.10555\,y^2$.

For the economical section $R=y/2 = 0.5\,y$.

(Check via perimeter: $P=b+2y\sqrt{1+m^2}=0.60555y+2y(1.80278)=0.60555y+3.60555y=4.21110y$; $R=A/P=2.10555y^2/4.21110y=0.5y$. ✓)

Step 3 — Apply Manning's equation.

$$Q=\frac{1}{n}A R^{2/3}S_0^{1/2}$$ $$25 = \frac{1}{0.018}(2.10555\,y^2)(0.5\,y)^{2/3}(0.0009)^{1/2}$$

Compute the constants:

• $(0.5)^{2/3}=0.62996$
• $(0.0009)^{1/2}=0.03$
• $\frac{1}{0.018}=55.5556$

So:

$$25 = 55.5556 \times 2.10555 \times 0.62996 \times 0.03 \times y^{2}\,y^{2/3}$$ $$25 = 2.21118\,y^{8/3}$$

Intermediate: $55.5556\times2.10555=116.975$; $\times0.62996=73.6826$; $\times0.03=2.21048$. (Using $2.2105$.)

Step 4 — Solve for $y$.

$$y^{8/3}=\frac{25}{2.2105}=11.3097$$ $$y = 11.3097^{3/8}$$ $$\ln(11.3097)=2.42565,\quad \times\tfrac{3}{8}=0.909619,\quad e^{0.909619}=2.4833$$ $$\boxed{y \approx 2.48\ \text{m}}$$

Step 5 — Bed width.

$$b = 0.60555\times2.4833 = 1.5037\ \text{m}\approx 1.50\ \text{m}$$

Verification. $A=2.10555\times2.4833^2=2.10555\times6.1668=12.984\ \text{m}^2$; $R=0.5\times2.4833=1.2417\ \text{m}$; $R^{2/3}=1.1525$.

$$Q=55.5556\times12.984\times1.1525\times0.03 = 24.94\ \text{m}^3/\text{s}\approx25\ \text{m}^3/\text{s}\ ✓$$

Result: bed width $b \approx 1.50\ \text{m}$, normal depth $y \approx 2.48\ \text{m}$ (side slope $1.5\text{H}:1\text{V}$).

(a) Derivation of sequent-depth relation

Apply the momentum equation to a control volume across the jump on a horizontal, frictionless bed. The net hydrostatic pressure force equals the rate of change of momentum:

$$\bar{p}_1 A_1 - \bar{p}_2 A_2 = \rho Q (V_2 - V_1)$$

For a rectangular channel of width $B$, take discharge per unit width $q=Q/B$, so $V=q/y$, $A=By$, and the centroid of pressure acts at $y/2$:

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

Dividing by $\rho g B$ and rearranging:

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

Factor $\frac{(y_1-y_2)(y_1+y_2)}{2}$ on the left and cancel the common factor $(y_1-y_2)$:

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

Using $F_1^2 = \dfrac{q^2}{g y_1^3}$, this rearranges to the standard form:

$$\boxed{\dfrac{y_2}{y_1}=\tfrac{1}{2}\left(\sqrt{1+8F_1^2}-1\right)}$$

(b) Numerical computation

Discharge per unit width: $q = Q/B = 30/5 = 6\ \text{m}^2/\text{s}$.

Upstream velocity & Froude number:

$$V_1 = q/y_1 = 6/0.60 = 10\ \text{m/s}$$ $$F_1 = \frac{V_1}{\sqrt{g y_1}} = \frac{10}{\sqrt{9.81\times0.60}} = \frac{10}{\sqrt{5.886}} = \frac{10}{2.4262}=4.122$$

Supercritical ($F_1>1$). ✓

(i) Sequent depth:

$$\frac{y_2}{y_1}=\tfrac12\left(\sqrt{1+8(4.122)^2}-1\right)=\tfrac12\left(\sqrt{1+8(16.991)}-1\right)=\tfrac12\left(\sqrt{136.93}-1\right)$$ $$=\tfrac12(11.702-1)=\tfrac12(10.702)=5.351$$ $$y_2 = 5.351\times0.60 = \boxed{3.21\ \text{m}}$$

(ii) Downstream Froude number:

$$V_2 = q/y_2 = 6/3.211 = 1.8686\ \text{m/s}$$ $$F_2 = \frac{1.8686}{\sqrt{9.81\times3.211}}=\frac{1.8686}{\sqrt{31.50}}=\frac{1.8686}{5.6125}=0.333$$

Subcritical ($F_2<1$). ✓ ($F_1=4.12$, $F_2=0.33$.)

(iii) Energy loss. For a rectangular-channel jump:

$$\Delta E = \frac{(y_2-y_1)^3}{4\,y_1 y_2}=\frac{(3.211-0.60)^3}{4(0.60)(3.211)}=\frac{(2.611)^3}{7.7064}=\frac{17.802}{7.7064}=2.310\ \text{m}$$ $$\boxed{\Delta E \approx 2.31\ \text{m}}$$

(Cross-check with specific energies: $E_1 = y_1+\frac{V_1^2}{2g}=0.60+\frac{100}{19.62}=0.60+5.097=5.697\ \text{m}$; $E_2 = y_2+\frac{V_2^2}{2g}=3.211+\frac{3.4917}{19.62}=3.211+0.178=3.389\ \text{m}$; $\Delta E=5.697-3.389=2.308\ \text{m}$ ✓)

(iv) Efficiency:

$$\eta=\frac{E_2}{E_1}=\frac{3.389}{5.697}=0.595=\boxed{59.5\%}$$

Head loss is $\Delta E/E_1 = 2.308/5.697 = 40.5\%$ of upstream energy.

Classification: Since $F_1 = 4.12$ lies in the range $4.5 > F_1 > 2.5$... it is just below 4.5, so $2.5<F_1<4.5$ → oscillating jump (transition toward steady). At $F_1=4.12$ the jump is an oscillating jump with appreciable energy dissipation.

(a) Derivation of the GVF equation

Assumptions: (1) steady flow; (2) gradually (slowly) varying depth so streamlines are nearly straight and pressure is hydrostatic; (3) channel slope is small ($\cos\theta\approx1$); (4) the head-loss at a section equals that of uniform flow at the same depth and velocity (Manning/Chezy applies locally, friction slope $S_f$); (5) prismatic channel, constant $\alpha\approx1$; (6) constant discharge.

Total energy head above a horizontal datum:

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

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

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

Now $\dfrac{dH}{dx}=-S_f$ (energy gradient) and $\dfrac{dz}{dx}=-S_0$ (bed drops in $+x$). Also

$$\frac{d}{dx}\!\left(\frac{V^2}{2g}\right)=\frac{d}{dy}\!\left(\frac{Q^2}{2gA^2}\right)\frac{dy}{dx}=-\frac{Q^2}{gA^3}\frac{dA}{dy}\frac{dy}{dx}=-\frac{Q^2 T}{gA^3}\frac{dy}{dx}=-F^2\frac{dy}{dx}$$

since $F^2=\dfrac{Q^2 T}{gA^3}$. Substituting:

$$-S_f=-S_0+\frac{dy}{dx}-F^2\frac{dy}{dx}\;\Rightarrow\; \boxed{\frac{dy}{dx}=\frac{S_0-S_f}{1-F^2}}$$

Slope classification: Compare normal depth $y_n$ with critical depth $y_c$.

• Mild (M): $y_n>y_c$ (subcritical normal flow)
• Steep (S): $y_n<y_c$
• Critical (C): $y_n=y_c$
• Horizontal (H): $S_0=0$ ($y_n\to\infty$)
• Adverse (A): $S_0<0$ (no real $y_n$)

Mild-slope profiles (zones: 1 above both depths, 2 between, 3 below both):

• M1 ($y>y_n>y_c$): backwater curve upstream of a dam/weir; $dy/dx>0$, asymptotic to $y_n$ upstream and horizontal downstream. Rises in flow direction.
• M2 ($y_n>y>y_c$): drawdown curve, e.g. approach to a free overfall; $dy/dx<0$, depth falls toward $y_c$.
• M3 ($y_n>y_c>y$): supercritical flow downstream of a sluice gate on a mild slope; depth rises toward $y_c$, usually terminated by a hydraulic jump.

        M1
  ~~~~~~~~~~~  <- water surface rises to y_n asymptote
 - - - - - - -  y_n (NDL)
        M2 (drawdown)
 ............. y_c (CDL)
        M3 (rises to jump)
=============== bed (mild)

(b) Direct step method

Given wide rectangular: use unit width, $R\approx y$, $q=3.0\ \text{m}^2/\text{s}$, $n=0.025$, $S_0=0.001$.

Normal depth from Manning (per unit width, $R=y$): $q=\frac{1}{n}y\,y^{2/3}S_0^{1/2}=\frac{1}{n}y^{5/3}S_0^{1/2}$.

$$y_n=\left(\frac{q\,n}{S_0^{1/2}}\right)^{3/5}=\left(\frac{3.0\times0.025}{0.0316228}\right)^{3/5}=\left(\frac{0.075}{0.0316228}\right)^{3/5}=(2.37171)^{3/5}$$ $$\ln2.37171=0.863594,\;\times0.6=0.518156,\;e^{0.518156}=1.6790\Rightarrow y_n=1.679\ \text{m}$$

Critical depth (rectangular): $y_c=\left(\frac{q^2}{g}\right)^{1/3}=\left(\frac{9}{9.81}\right)^{1/3}=(0.91743)^{1/3}=0.9717\ \text{m}$.

Since $y_n(1.68)>y_c(0.97)$ → MILD slope. The depths $1.40$–$1.60$ m lie between $y_c$ and $y_n$ (zone 2), so this is an M2 (drawdown) profile.

Direct step (single step from $y_1=1.40$ to $y_2=1.60$ m): Specific energy $E=y+\dfrac{q^2}{2g y^2}$, with $\dfrac{q^2}{2g}=\dfrac{9}{19.62}=0.458716$.

Friction slope $S_f=\left(\dfrac{qn}{y^{5/3}}\right)^2=\dfrac{q^2 n^2}{y^{10/3}}$. With $q^2n^2=9\times0.000625=0.005625$:

• At $y=1.40$: $y^{10/3}=e^{(10/3)\ln1.40}=e^{3.33333\times0.336472}=e^{1.121574}=3.06987$; $S_f=0.005625/3.06987=0.0018323$.
• At $y=1.60$: $y^{10/3}=e^{3.33333\times0.470004}=e^{1.566680}=4.79086$; $S_f=0.005625/4.79086=0.0011741$.

Mean friction slope: $\bar{S_f}=\tfrac12(0.0018323+0.0011741)=0.0015032$.

Direct-step formula:

$$\Delta x=\frac{E_2-E_1}{S_0-\bar{S_f}}=\frac{1.77919-1.63404}{0.001-0.0015032}=\frac{0.14515}{-0.0005032}=-288.5\ \text{m}$$

The negative sign indicates the depth increases in the upstream direction (consistent with an M2 drawdown computed against the flow). The magnitude of the distance is:

$$\boxed{|\Delta x|\approx 288\ \text{m}}$$

Thus the depth changes from 1.40 m to 1.60 m over about 288 m measured upstream (the deeper section, 1.60 m, lies upstream nearer $y_n$).

(a) Water hammer and closure criterion

Water hammer is the sudden rise (and subsequent oscillation) of pressure in a closed conduit caused by a rapid change in the velocity of flow, typically when a valve is closed quickly. The kinetic energy of the moving column is converted into a pressure (strain) wave that travels back and forth along the pipe at the acoustic celerity $c$, alternately compressing the water and stretching the pipe wall.

The critical (or pipe-period) time is the time for the pressure wave to travel to the reservoir and back:

$$T_c=\frac{2L}{c}$$

• Rapid / sudden closure: valve closing time $T \le T_c=2L/c$. The full Joukowsky pressure rise develops, $\Delta p=\rho c V$ (or head $\Delta H=cV/g$), since the reflected relief wave has not yet returned to the valve.
• Slow / gradual closure: $T > T_c$. The reflected wave returns before closure completes and partially relieves the pressure, giving a smaller rise approximated (rigid-column / Michaud) by $\Delta H=\dfrac{2LV}{gT}$.

(b) Computations

Step 1 — Wave celerity with pipe elasticity (thin-walled pipe):

$$c=\frac{1}{\sqrt{\rho\left(\dfrac{1}{E_w}+\dfrac{D}{t\,E_p}\right)}}$$

Compute the two compliance terms:

• $\dfrac{1}{E_w}=\dfrac{1}{2.07\times10^9}=4.8309\times10^{-10}\ \text{Pa}^{-1}$
• $\dfrac{D}{t\,E_p}=\dfrac{0.5}{0.008\times200\times10^9}=\dfrac{0.5}{1.6\times10^9}=3.1250\times10^{-10}\ \text{Pa}^{-1}$
• Sum $=7.9559\times10^{-10}\ \text{Pa}^{-1}$

$$\rho\times\text{sum}=1000\times7.9559\times10^{-10}=7.9559\times10^{-7}$$ $$c=\frac{1}{\sqrt{7.9559\times10^{-7}}}=\frac{1}{8.9196\times10^{-4}}=1121.1\ \text{m/s}$$ $$\boxed{c\approx 1121\ \text{m/s}}$$

Step 2 — Critical time:

$$T_c=\frac{2L}{c}=\frac{2\times2400}{1121.1}=\frac{4800}{1121.1}=4.282\ \text{s}$$

Step 3 — Instantaneous (sudden) closure — Joukowsky:

$$\Delta p=\rho c V=1000\times1121.1\times2.0=2.2422\times10^{6}\ \text{Pa}=\boxed{2.24\ \text{MPa}}$$

As head: $\Delta H=\dfrac{cV}{g}=\dfrac{1121.1\times2.0}{9.81}=228.6\ \text{m}$ of water.

Step 4 — Gradual closure in $T=4\ \text{s}$. Here $T=4\ \text{s} < T_c=4.282\ \text{s}$, so the closure is technically still rapid (the relief wave has not returned by the time the valve shuts at $t=4$ s). Therefore the full Joukowsky rise applies:

$$\Delta p=\rho c V = 2.24\ \text{MPa},\quad \Delta H=228.6\ \text{m}.$$

For comparison, if one (incorrectly) applied the slow-closure Michaud formula:

$$\Delta H_{slow}=\frac{2LV}{gT}=\frac{2\times2400\times2.0}{9.81\times4}=\frac{9600}{39.24}=244.6\ \text{m}$$

This exceeds the Joukowsky value, confirming the closure is in the rapid regime and the governing pressure rise is the Joukowsky value $\Delta p\approx2.24\ \text{MPa}$ ($\approx229\ \text{m}$ head).

Note: Had the valve closed in, say, $T=8\ \text{s}>T_c$, the slow-closure formula would govern: $\Delta H=\dfrac{2\times2400\times2.0}{9.81\times8}=122.3\ \text{m}$.

Step 5 — Hoop (circumferential) stress at peak pressure. The peak internal pressure (above the static line) is $\Delta p=2.2422\times10^{6}\ \text{Pa}$. For a thin-walled pipe the hoop stress is:

$$\sigma_h=\frac{\Delta p\,D}{2t}=\frac{2.2422\times10^{6}\times0.5}{2\times0.008}=\frac{1.1211\times10^{6}}{0.016}=7.007\times10^{7}\ \text{Pa}$$ $$\boxed{\sigma_h\approx 70.1\ \text{MPa}}$$

This is the additional hoop stress due to water hammer; it must be added to the stress from the steady operating pressure when checking the pipe against the allowable steel stress.

(a) Specific energy and critical depth

Specific energy $E$ is the energy head measured relative to the channel bed at a section:

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

For a rectangular channel ($q=Q/B$): $E=y+\dfrac{q^2}{2gy^2}$.

Critical depth $y_c$ is the depth at which the specific energy is minimum for a given discharge; at this depth the Froude number $F=1$. For a rectangular channel $y_c=\left(\dfrac{q^2}{g}\right)^{1/3}$ and $E_{min}=\tfrac{3}{2}y_c$.

Specific-energy curve ($E$ on x-axis, $y$ on y-axis): a C-shaped curve with two limbs.

 y |        /  upper limb (subcritical, F<1)
   |       /
 y2|------*    alternate depths y1, y2 for same E
   |     / \
 yc|----*---  (nose = E_min, F=1)
 y1|---* \
   |  /   \ lower limb (supercritical, F>1)
   +------------------ E
        E_min

For any $E>E_{min}$ there are two alternate depths: a larger subcritical depth (upper limb) and a smaller supercritical depth (lower limb). They merge at the nose where $E=E_{min}$ and $y=y_c$.

(b) Hump analysis

Data: $B=4\ \text{m}$, $Q=12\ \text{m}^3/\text{s}$, $y_1=2.0\ \text{m}$, $q=Q/B=3.0\ \text{m}^2/\text{s}$.

Upstream specific energy:

$$V_1=q/y_1=3/2=1.5\ \text{m/s},\quad E_1=y_1+\frac{V_1^2}{2g}=2.0+\frac{1.5^2}{19.62}=2.0+0.11468=2.1147\ \text{m}$$

Froude $F_1=\dfrac{1.5}{\sqrt{9.81\times2}}=\dfrac{1.5}{4.429}=0.339$ → subcritical (upper limb).

Critical depth and minimum energy over the hump (q unchanged):

$$y_c=\left(\frac{q^2}{g}\right)^{1/3}=\left(\frac{9}{9.81}\right)^{1/3}=(0.91743)^{1/3}=0.9717\ \text{m}$$ $$E_{c}=\tfrac32 y_c=\tfrac32(0.9717)=1.4576\ \text{m}$$

Maximum hump height (chokes flow at critical): Energy balance (no loss): $E_1=\Delta z+E_{over\,hump}$. The minimum possible $E$ over the hump is $E_c$, so the maximum hump that does not yet force upstream depth to rise is:

$$\Delta z_{max}=E_1-E_c=2.1147-1.4576=0.6571\ \text{m}\approx\boxed{0.657\ \text{m}}$$

Any higher hump would choke the flow and raise the upstream depth.

Depth over a $\Delta z=0.10\ \text{m}$ hump. Since $0.10\ \text{m}<0.657\ \text{m}$, the flow is not choked and the upstream depth stays at $2.0$ m. Specific energy over the hump:

$$E_2=E_1-\Delta z=2.1147-0.10=2.0147\ \text{m}$$

Solve $E_2=y_2+\dfrac{q^2}{2g y_2^2}=y_2+\dfrac{0.458716}{y_2^2}=2.0147$. The flow stays subcritical (upper limb), so iterate near $y_2\approx y_1$:

• Try $y_2=1.88$: $0.458716/1.88^2=0.458716/3.5344=0.12979$; $E=1.88+0.12979=2.00979$ (low).
• Try $y_2=1.885$: $/3.5532=0.12910$; $E=1.885+0.12910=2.01410$ (≈ target).
• Try $y_2=1.8849$: $E\approx2.01400$.

Thus $y_2\approx\boxed{1.885\ \text{m}}$ (subcritical root).

Check: depth drops from 2.000 m to 1.885 m, a fall of 0.115 m, while the bed rises 0.10 m; the water surface therefore dips slightly over the hump — characteristic of subcritical flow over a rise. ✓

(c) Hump height $0.90\ \text{m}$ — choking

Since $\Delta z=0.90\ \text{m} > \Delta z_{max}=0.657\ \text{m}$, the hump is too high: the upstream specific energy $E_1=2.1147\ \text{m}$ is insufficient to pass the flow over the crest at critical without raising the upstream water level. The flow chokes — the upstream depth must rise so that the available specific energy increases by enough to clear the hump while passing critical flow on the crest.

Over the crest the flow is now critical with $E_{crest}=E_c=1.4576\ \text{m}$ (since $q$ is unchanged). The new required upstream specific energy is:

$$E_1'=E_c+\Delta z=1.4576+0.90=2.3576\ \text{m}$$

Solve for the new (subcritical) upstream depth $y_1'$:

$$y_1'+\frac{0.458716}{y_1'^2}=2.3576$$

Iterating (subcritical root):

• $y_1'=2.265$: $2.265+0.458716/5.1302=2.265+0.08942=2.35442$ (low)
• $y_1'=2.268$: $2.268+0.458716/5.1438=2.268+0.08918=2.35718$ (≈ target)

$$\boxed{y_1'\approx 2.268\ \text{m}}$$

So the upstream depth rises from $2.000$ m to about 2.27 m (a backwater build-up of ~0.27 m), the flow passes through critical depth ($y_c=0.972$ m) on the crest, and accelerates to supercritical on the downstream face of the hump. This demonstrates choking at an excessive bed rise.

(a) Chezy–Manning relation

Chezy: $V=C\sqrt{R S}$; Manning: $V=\frac{1}{n}R^{2/3}S^{1/2}$. Equating:

$$\boxed{C=\frac{1}{n}R^{1/6}}\quad(\text{SI units})$$

(b) Discharge and velocity

Geometry: $b=3\ \text{m}$, $y=1.2\ \text{m}$, $S_0=1/1600=6.25\times10^{-4}$, $n=0.015$.

• Area $A=b\,y=3\times1.2=3.6\ \text{m}^2$
• Wetted perimeter $P=b+2y=3+2(1.2)=5.4\ \text{m}$
• Hydraulic radius $R=A/P=3.6/5.4=0.66667\ \text{m}$
• $R^{2/3}=0.66667^{2/3}$: $\ln0.66667=-0.405465$, $\times\tfrac23=-0.270310$, $e^{-0.270310}=0.76314$
• $S_0^{1/2}=\sqrt{6.25\times10^{-4}}=0.025$

Velocity:

$$V=\frac{1}{0.015}(0.76314)(0.025)=66.6667\times0.76314\times0.025=1.2719\ \text{m/s}\approx\boxed{1.27\ \text{m/s}}$$

Discharge:

$$Q=AV=3.6\times1.2719=4.579\ \text{m}^3/\text{s}\approx\boxed{4.58\ \text{m}^3/\text{s}}$$

(c) Froude number

$$F=\frac{V}{\sqrt{g y}}=\frac{1.2719}{\sqrt{9.81\times1.2}}=\frac{1.2719}{\sqrt{11.772}}=\frac{1.2719}{3.4310}=0.3707$$

Since $F=0.37<1$, the flow is subcritical.

(a) General critical-flow condition

For a channel of arbitrary cross-section, flow is critical when the Froude number equals 1, i.e.

$$\boxed{\frac{Q^2 T}{g A^3}=1}\quad\Longleftrightarrow\quad \frac{Q^2}{g}=\frac{A^3}{T}$$

where $A$ = flow area and $T$ = top width, both evaluated at the critical depth $y_c$.

(b) Trial computation of critical depth

Data: $b=2.5\ \text{m}$, side slope $m=2$, $Q=8\ \text{m}^3/\text{s}$.

• Area: $A=(b+my)y=(2.5+2y)y$
• Top width: $T=b+2my=2.5+4y$
• Required: $\dfrac{A^3}{T}=\dfrac{Q^2}{g}=\dfrac{64}{9.81}=6.5240\ \text{m}^5$

Trial 1: $y=0.70$ m

• $A=(2.5+1.4)(0.70)=3.9\times0.70=2.730\ \text{m}^2$; $A^3=20.346$
• $T=2.5+2.8=5.30\ \text{m}$
• $A^3/T=20.346/5.30=3.839$ (< 6.524 → too small, increase $y$)

Trial 2: $y=0.90$ m

• $A=(2.5+1.8)(0.90)=4.3\times0.90=3.870\ \text{m}^2$; $A^3=57.961$
• $T=2.5+3.6=6.10\ \text{m}$
• $A^3/T=57.961/6.10=9.502$ (> 6.524 → too large, decrease $y$)

Trial 3: $y=0.82$ m

• $A=(2.5+1.64)(0.82)=4.14\times0.82=3.3948\ \text{m}^2$; $A^3=39.126$
• $T=2.5+3.28=5.78\ \text{m}$
• $A^3/T=39.126/5.78=6.769$ (slightly high → decrease $y$ a little)

Trial 4: $y=0.81$ m

• $A=(2.5+1.62)(0.81)=4.12\times0.81=3.3372\ \text{m}^2$; $A^3=37.169$
• $T=2.5+3.24=5.74\ \text{m}$
• $A^3/T=37.169/5.74=6.476$ (≈ 6.524, slightly low)

Interpolating between $y=0.81$ (6.476) and $y=0.82$ (6.769) for target 6.524:

$$y_c\approx0.81+0.01\times\frac{6.524-6.476}{6.769-6.476}=0.81+0.01\times0.164=0.8116\ \text{m}$$ $$\boxed{y_c\approx 0.812\ \text{m}}$$

(a) Classification of surges

A surge is a moving wave-front (a moving hydraulic jump/bore) produced by a sudden change in flow. Types:

• Positive surge: the depth increases as the surge passes (steep, stable front). e.g. surge from gate closure.
• Negative surge: the depth decreases (flatter, dispersive front), e.g. from gate opening. Each can travel upstream (against the flow) or downstream (with the flow). Here: a positive surge moving upstream (depth rises 1.5→2.1 m).

(b) Surge celerity and velocity behind the surge

Reduce the unsteady problem to steady by superimposing a velocity equal to the absolute surge speed. Let:

• Upstream (undisturbed): $y_1=1.5\ \text{m}$, $V_1=1.0\ \text{m/s}$ (flowing toward the gate, +x downstream).
• Behind surge: $y_2=2.1\ \text{m}$, $V_2=?$
• Absolute surge velocity $V_w$ directed upstream (take magnitude $V_w$, moving in −x).

Continuity (relative to the moving front, surge speed $V_w$ upstream so relative velocities are $V_1+V_w$ and $V_2+V_w$):

$$(V_1+V_w)\,y_1=(V_2+V_w)\,y_2 \quad\quad(1)$$

Momentum across the surge (treated as a moving jump) gives the relative approach velocity:

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

Step 1 — Compute relative approach velocity from (2):

$$V_1+V_w=\sqrt{\frac{9.81\times2.1}{2\times1.5}(1.5+2.1)}=\sqrt{\frac{20.601}{3.0}\times3.6}=\sqrt{6.867\times3.6}=\sqrt{24.7212}=4.9721\ \text{m/s}$$

Step 2 — Surge celerity (absolute):

$$V_w=4.9721-V_1=4.9721-1.0=3.9721\ \text{m/s}$$ $$\boxed{V_w\approx 3.97\ \text{m/s (upstream)}}$$

Step 3 — Velocity behind surge from continuity (1):

$$V_2+V_w=\frac{(V_1+V_w)\,y_1}{y_2}=\frac{4.9721\times1.5}{2.1}=\frac{7.4582}{2.1}=3.5515\ \text{m/s}$$ $$V_2=3.5515-V_w=3.5515-3.9721=-0.4206\ \text{m/s}$$

The negative sign means the water behind the surge flows in the −x direction (upstream) at about

$$\boxed{V_2\approx 0.42\ \text{m/s (reversed, upstream)}}$$

This reversal is physically reasonable for a strong gate-closure surge that backs the flow up.

Check (continuity in absolute terms): discharge change handled consistently; relative continuity $(4.9721)(1.5)=7.458=(3.5515)(2.1)=7.458$ ✓.

(a) Critical flow and discharge formula

Over a sufficiently long (broad) crest, the streamlines become straight and parallel and the flow accelerates until it reaches the condition of minimum specific energy for the available head — i.e. critical flow establishes on the crest. The crest essentially acts as a control section where $F=1$.

Let $H$ = upstream head (specific energy) above the crest measured from the upstream pool (approach velocity neglected). On the crest, with critical depth $y_c$ over a rectangular crest of width $L$ and unit discharge $q$:

$$y_c=\frac{2}{3}H \quad(\text{since }E_{min}=\tfrac32 y_c=H)$$

For critical flow on a rectangular section $q=\sqrt{g\,y_c^3}$:

$$q=\sqrt{g}\,y_c^{3/2}=\sqrt{g}\left(\tfrac{2}{3}H\right)^{3/2}=\sqrt{g}\left(\tfrac{2}{3}\right)^{3/2}H^{3/2}$$

Total discharge over length $L$, introducing a coefficient of discharge $C_d$ for real losses:

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

Evaluate the constant: $\sqrt{9.81}=3.1321$, $(2/3)^{3/2}=0.54433$, product $=1.7049$.

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

(b) Discharge

Given $C_d=0.92$, $L=4\ \text{m}$, $H=0.45\ \text{m}$:

$$H^{3/2}=0.45^{1.5}=0.45\sqrt{0.45}=0.45\times0.67082=0.301869$$ $$Q=1.705\times0.92\times4\times0.301869$$

Step: $1.705\times0.92=1.5686$; $\times4=6.2744$; $\times0.301869=1.8941$.

$$\boxed{Q\approx 1.89\ \text{m}^3/\text{s}}$$

(a) Control section

A control section is a section in an open channel where a definite, unique relationship exists between the depth (stage) and the discharge — typically a section where the flow passes through critical depth ($F=1$). Because depth is fixed there for a given $Q$, it acts as the starting point (control) for computing GVF profiles: subcritical profiles are controlled from downstream, supercritical from upstream.

Examples: (1) the crest of a weir/spillway or a broad-crested weir (critical flow); (2) a free overfall (brink, where flow passes through critical just upstream); also a sluice-gate opening (vena contracta) and the break in grade from mild to steep slope.

(b) Profile sequence: sluice gate → mild slope → free overfall

Setup: mild slope ($y_n>y_c$); a sluice gate at the upstream end issues a thin supercritical jet (depth below $y_c$); far downstream the channel ends in a free overfall.

Sequence in the direction of flow:

1. Just downstream of the gate (vena contracta): depth $< y_c < y_n$ → flow is in zone 3 of a mild slope → M3 profile. The supercritical depth increases in the flow direction as friction decelerates the jet, rising toward $y_c$.

2. Hydraulic jump: Because the channel is mild, the flow must ultimately return to the subcritical normal depth $y_n$. The rising M3 supercritical profile meets the subcritical requirement through a hydraulic jump, located where the upstream M3 depth and the downstream subcritical sequent depth satisfy the momentum (conjugate-depth) relation.

3. After the jump (subcritical): the flow is now subcritical. If the jump brings it close to normal depth it runs near $y_n$; approaching the overfall it must draw down. Between the jump and the brink the surface is an M2 (drawdown) profile — depth $y_n>y>y_c$, decreasing toward the overfall.

4. Free overfall (brink): at the edge the flow passes through approximately critical depth ($y\approx y_c$, the control), then plunges over.

Summary sequence: gate → M3 (supercritical, rising) → hydraulic jump → subcritical flow near $y_n$ → M2 (drawdown) → critical depth at the free overfall.

 gate                         jump                overfall
  |                            ||                    |
  | M3 (rising) ............(jump)  M2 (drawdown)     v
  *___---^^^----============~~~~~~~~~~~~~~~~~~~~------.
 - - - - - - - - - - - - - - - - - - - - - - - - -  y_n
 .................................................. y_c
 =================================================== mild bed

(a) Design factors and freeboard

Factors governing design of a non-erodible (lined) channel:

1. Design discharge $Q$ and required carrying capacity.
2. Available/permissible bed slope and the ground topography.
3. Kind of lining material and its roughness (Manning's $n$).
4. Maximum permissible (and minimum non-silting) velocity.
5. Most economical (best hydraulic) section to minimise cost.
6. Side slopes (governed by lining type and bank stability).
7. Freeboard and provision for fluctuations/waves.
8. Cost of excavation and lining.

Freeboard is the vertical distance between the full-supply (design) water surface and the top of the channel lining/bank. It is a safety margin that prevents overtopping due to wave action, surges, unsteady flow, or higher-than-design discharge.

(b) Area and total depth

Required cross-sectional area from continuity $Q=AV$:

$$A=\frac{Q}{V}=\frac{15}{1.5}=\boxed{10.0\ \text{m}^2}$$

Total depth of lined section = depth of flow + freeboard:

$$D=y+\text{freeboard}=1.8+0.5=\boxed{2.3\ \text{m}}$$

Thus the lining must be carried up to a total depth of 2.3 m, providing a 0.5 m freeboard above the 1.8 m design flow depth, with the wetted section carrying $A=10.0\ \text{m}^2$ at $1.5\ \text{m/s}$.