BE Civil Engineering (IOE, TU) Hydraulics (IOE, CE 604) Question Paper 2076 Nepal

(a) Derivation of Manning's equation

Uniform flow means the depth, area, velocity and discharge are constant along the channel; the water surface, energy line and bed are all parallel, so the slope of the energy line equals the bed slope, $S_f = S_0 = S$.

Chezy's equation for uniform flow is:

$$V = C\sqrt{R S}$$

where $V$ = mean velocity, $C$ = Chezy coefficient, $R = A/P$ = hydraulic radius, $S$ = slope.

Manning (and Strickler) found experimentally that the Chezy coefficient varies with the roughness and the hydraulic radius as:

$$C = \frac{1}{n} R^{1/6}$$

where $n$ is Manning's roughness coefficient. Substituting into Chezy's equation:

$$V = \frac{1}{n} R^{1/6}\sqrt{R S} = \frac{1}{n} R^{2/3} S^{1/2}$$

Multiplying by area $A$ gives the discharge form:

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

Assumptions: flow is steady and uniform; channel is prismatic with constant slope and roughness; pressure distribution is hydrostatic; $S_f = S_0$.

(b) Normal depth by trial

For a trapezoidal section with bed width $b = 4.0\ \text{m}$ and side slope $z = 1.5$:

$$A = (b + z y)\,y = (4 + 1.5y)\,y$$ $$P = b + 2y\sqrt{1+z^2} = 4 + 2y\sqrt{1+1.5^2} = 4 + 2y\,(1.8028) = 4 + 3.6056\,y$$ $$R = A/P$$

Manning's equation must satisfy:

$$Q n = A R^{2/3} S^{1/2}$$ $$Q n = 18 \times 0.018 = 0.324$$ $$S^{1/2} = \sqrt{0.0009} = 0.03$$

so the target is:

$$A R^{2/3} = \frac{Q n}{S^{1/2}} = \frac{0.324}{0.03} = 10.80$$

Trial table (compute $A R^{2/3}$ and compare with 10.80):

At $y = 1.619\ \text{m}$, $A R^{2/3} = 10.81 \approx 10.80$. Hence:

$$\boxed{y_n \approx 1.62\ \text{m}}$$

Mean velocity at normal depth (using $y_n = 1.62$, $A = 10.418\ \text{m}^2$):

$$V = \frac{Q}{A} = \frac{18}{10.418} = 1.728\ \text{m/s}$$ $$\boxed{V \approx 1.73\ \text{m/s}}$$

(a) Specific energy

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

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

The specific-energy curve (E on x-axis, y on y-axis for fixed Q) is a curve with two limbs that approaches the line $y = E$ (asymptote) for large depths and the $E$-axis for small depths. The two limbs meet at the critical depth $y_c$, where $E$ is minimum. For any $E > E_{min}$ there are two depths — the alternate depths $y_1$ (supercritical, lower) and $y_2$ (subcritical, upper):

  y |        / (y = E asymptote)
    |       /  .  subcritical limb
    |      /.
 y_c|----- *  (critical point, E = E_min)
    |    .
    |  .   supercritical limb
    |._______________ E
        E_min

(b) Computations

Unit discharge: $q = Q/b = 15/5 = 3.0\ \text{m}^2/\text{s}$.

Velocity at $y = 0.60\ \text{m}$:

$$V = \frac{q}{y} = \frac{3.0}{0.60} = 5.0\ \text{m/s}$$

Specific energy:

$$E = y + \frac{V^2}{2g} = 0.60 + \frac{5.0^2}{2(9.81)} = 0.60 + 1.274 = 1.874\ \text{m}$$ $$\boxed{E = 1.87\ \text{m}}$$

Critical depth (rectangular):

$$y_c = \left(\frac{q^2}{g}\right)^{1/3} = \left(\frac{3.0^2}{9.81}\right)^{1/3} = (0.9174)^{1/3} = 0.972\ \text{m}$$ $$\boxed{y_c = 0.972\ \text{m}}$$

Froude number at $y = 0.60$:

$$Fr = \frac{V}{\sqrt{g y}} = \frac{5.0}{\sqrt{9.81 \times 0.60}} = \frac{5.0}{2.426} = 2.06$$

Since $Fr = 2.06 > 1$ (and $y < y_c$), the flow is supercritical.

(c) Alternate depth

The alternate depth $y_2$ satisfies the same $E$ and $q$:

$$E = y_2 + \frac{q^2}{2 g y_2^2} = 1.874 \Rightarrow y_2 + \frac{9}{19.62\, y_2^2} = 1.874$$ $$y_2 + \frac{0.4587}{y_2^2} = 1.874$$

Solve by trial for the subcritical root ($y_2 > y_c$):

At $y_2 = 1.72\ \text{m}$ the expression equals $1.875 \approx 1.874$.

$$\boxed{y_2 \approx 1.72\ \text{m (alternate subcritical depth)}}$$

(a) Sequent-depth relation

Applying the momentum equation between sections 1 (before) and 2 (after) a jump on a horizontal frictionless bed, the specific force (momentum function) is conserved:

$$\frac{Q^2}{g A_1} + \bar{z}_1 A_1 = \frac{Q^2}{g A_2} + \bar{z}_2 A_2$$

For a rectangular channel of width $b$, using $q = Q/b$, $A = by$, centroid depth $\bar z = y/2$, dividing by $b$:

$$\frac{q^2}{g y_1} + \frac{y_1^2}{2} = \frac{q^2}{g y_2} + \frac{y_2^2}{2}$$

Rearranging and factoring leads to:

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

where $Fr_1 = V_1/\sqrt{g y_1}$ is the upstream Froude number. This is the sequent-depth relation.

(b) Computations

Unit discharge: $q = Q/b = 24/6 = 4.0\ \text{m}^2/\text{s}$.

Upstream velocity:

$$V_1 = \frac{q}{y_1} = \frac{4.0}{0.45} = 8.889\ \text{m/s}$$

Upstream Froude number:

$$Fr_1 = \frac{V_1}{\sqrt{g y_1}} = \frac{8.889}{\sqrt{9.81 \times 0.45}} = \frac{8.889}{2.101} = 4.231$$

Sequent depth:

$$y_2 = \frac{y_1}{2}\left(\sqrt{1 + 8 Fr_1^2} - 1\right) = \frac{0.45}{2}\left(\sqrt{1 + 8(4.231)^2} - 1\right)$$ $$8 Fr_1^2 = 8 \times 17.90 = 143.2 \Rightarrow \sqrt{144.2} = 12.008$$ $$y_2 = 0.225 \times (12.008 - 1) = 0.225 \times 11.008 = 2.477\ \text{m}$$ $$\boxed{y_2 \approx 2.48\ \text{m}}$$

Energy loss in a rectangular jump:

$$\Delta E = \frac{(y_2 - y_1)^3}{4 y_1 y_2} = \frac{(2.477 - 0.45)^3}{4 (0.45)(2.477)} = \frac{(2.027)^3}{4.459} = \frac{8.328}{4.459} = 1.868\ \text{m}$$ $$\boxed{\Delta E \approx 1.87\ \text{m}}$$

Energy before the jump:

$$E_1 = y_1 + \frac{V_1^2}{2g} = 0.45 + \frac{8.889^2}{19.62} = 0.45 + 4.027 = 4.477\ \text{m}$$

Efficiency (ratio of energy after to energy before):

$$E_2 = E_1 - \Delta E = 4.477 - 1.868 = 2.609\ \text{m}$$ $$\eta = \frac{E_2}{E_1} = \frac{2.609}{4.477} = 0.583 = 58.3\%$$ $$\boxed{\eta \approx 58\%}$$

Fraction of energy dissipated $= \Delta E / E_1 = 1.868/4.477 = 41.7\%$.

Length of jump:

$$L_j \approx 6 y_2 = 6 \times 2.477 = 14.86\ \text{m}$$ $$\boxed{L_j \approx 14.9\ \text{m}}$$

With $Fr_1 = 4.23$ this is a steady jump (well-formed, $4.5 > Fr_1 > 2.5$ region near steady).

(a) Dynamic equation of GVF

Assumptions: steady flow; pressure distribution hydrostatic (gradual variation, streamlines nearly straight and parallel); the head loss at a section equals that of uniform flow at the same depth and velocity (so $S_f$ from Manning/Chezy applies); slope small ($\cos\theta \approx 1$); channel prismatic; $\alpha = 1$.

Total energy referred to a horizontal datum:

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

Differentiate with respect to $x$ (channel distance):

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

Now $\dfrac{dH}{dx} = -S_f$ (energy line slope) and $\dfrac{dz}{dx} = -S_0$ (bed slope). Also for a given $Q$:

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

since $Fr^2 = \dfrac{Q^2 T}{g A^3}$.

Substituting:

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

(b) Mild-slope profiles (M-profiles)

On a mild slope $y_n > y_c$. Three zones are defined by $y_n$ and $y_c$:

          M1 (y>yn) ---- rising backwater
  ----------------------------------  y = yn (NDL)
          M2 (yc<y<yn) -- drawdown
  - - - - - - - - - - - - - - - - -  y = yc (CDL)
          M3 (y<yc) ---- rising to jump
  ////////// mild bed //////////////

M1 is asymptotic to NDL upstream and horizontal downstream; M2 is asymptotic to NDL upstream and meets CDL nearly vertically; M3 starts near the bed (gate) and rises, usually terminated by a hydraulic jump.

(a) Slope classification

Unit discharge $q = 20/8 = 2.5\ \text{m}^2/\text{s}$.

Critical depth:

$$y_c = \left(\frac{q^2}{g}\right)^{1/3} = \left(\frac{2.5^2}{9.81}\right)^{1/3} = (0.6371)^{1/3} = 0.860\ \text{m}$$

Normal depth from Manning ($Q n / S^{1/2} = A R^{2/3}$):

$$\frac{Qn}{S^{1/2}} = \frac{20 \times 0.025}{\sqrt{0.0004}} = \frac{0.5}{0.02} = 25.0$$

For $b=8$, $A = 8y$, $P = 8 + 2y$, $R = 8y/(8+2y)$. Trial:

So $y_n \approx 2.39\ \text{m}$. Since $y_n (2.39) > y_c (0.86)$, the slope is mild.

The control depth $2.50\ \text{m} > y_n > y_c$, so the profile is an M1 (backwater) curve.

(b) Direct-step method (one step, $y_1=2.50 \to y_2=2.30$)

Direct-step uses:

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

where $E = y + V^2/2g$ and $\bar S_f$ is the mean of $S_f$ at the two sections, with $S_f = \left(\dfrac{Qn}{A R^{2/3}}\right)^2$.

Section 1: $y_1 = 2.50\ \text{m}$

$$A_1 = 8(2.50) = 20.0\ \text{m}^2,\quad V_1 = 20/20 = 1.0\ \text{m/s}$$ $$E_1 = 2.50 + \frac{1.0^2}{19.62} = 2.50 + 0.0510 = 2.5510\ \text{m}$$ $$P_1 = 8 + 2(2.5) = 13.0,\ R_1 = 20/13 = 1.538,\ R_1^{2/3} = 1.334$$ $$S_{f1} = \left(\frac{20 \times 0.025}{20.0 \times 1.334}\right)^2 = \left(\frac{0.5}{26.68}\right)^2 = (0.018740)^2 = 3.512\times10^{-4}$$

Section 2: $y_2 = 2.30\ \text{m}$

$$A_2 = 8(2.30) = 18.40\ \text{m}^2,\quad V_2 = 20/18.40 = 1.0870\ \text{m/s}$$ $$E_2 = 2.30 + \frac{1.0870^2}{19.62} = 2.30 + 0.06022 = 2.3602\ \text{m}$$ $$P_2 = 8 + 2(2.30) = 12.60,\ R_2 = 18.40/12.60 = 1.460,\ R_2^{2/3} = 1.290$$ $$S_{f2} = \left(\frac{0.5}{18.40 \times 1.290}\right)^2 = \left(\frac{0.5}{23.74}\right)^2 = (0.021062)^2 = 4.436\times10^{-4}$$

Mean friction slope:

$$\bar S_f = \frac{3.512 + 4.436}{2}\times10^{-4} = 3.974\times10^{-4}$$

Distance:

$$\Delta x = \frac{E_2 - E_1}{S_0 - \bar S_f} = \frac{2.3602 - 2.5510}{0.0004 - 0.0003974} = \frac{-0.1908}{2.6\times10^{-6}}$$ $$\Delta x = -73{,}400\ \text{m (approx)}$$

The large magnitude arises because at $y$ near $y_n=2.39$ the denominator $S_0-\bar S_f$ is nearly zero (the M1 profile is asymptotic to the NDL). The negative sign means section 2 lies upstream of the control. A practical computation would use small steps; the magnitude shows the backwater extends a very long distance upstream.

$$\boxed{|\Delta x| \approx 73\ \text{km upstream (asymptotic M1 behaviour near } y_n)}$$

Note: taking the step only to $2.45\ \text{m}$ (well above $y_n$) would give a finite, modest distance; this part illustrates why direct-step requires depths kept away from $y_n$.

(a) Conditions for the most economical trapezoidal section

For a given area, the section is most economical when the wetted perimeter (and hence cost/lining) is minimum, which for Manning also maximises discharge. The conditions are:

1. Half the top width equals the sloping side length: $\dfrac{b + 2zy}{2} = y\sqrt{1+z^2}$.
2. The hydraulic radius equals half the depth: $R = y/2$.
3. A semicircle drawn with centre at the water surface and radius equal to the depth is tangent to the three sides (the section can inscribe a semicircle of radius $y$).

(b) Design

Condition 1 gives the bed width for $z = 1$:

$$b = 2y\left(\sqrt{1+z^2} - z\right) = 2y\left(\sqrt{2} - 1\right) = 2y(0.41421) = 0.8284\,y$$

Geometry:

$$A = (b + zy)y = (0.8284y + y)y = 1.8284\,y^2$$

Using condition 2, $R = y/2$.

Manning:

$$Q = \frac{1}{n} A R^{2/3} S^{1/2}$$ $$10 = \frac{1}{0.020}\,(1.8284 y^2)\left(\frac{y}{2}\right)^{2/3}(0.0006)^{1/2}$$

Compute constants: $(1/2)^{2/3} = 0.6300$; $\sqrt{0.0006} = 0.024495$; $1/0.020 = 50$.

$$10 = 50 \times 1.8284 \times 0.6300 \times 0.024495 \times y^{2}\,y^{2/3}$$ $$10 = 50 \times 1.8284 \times 0.6300 \times 0.024495 \times y^{8/3}$$

Coefficient $= 50 \times 1.8284 = 91.42;\ \times 0.6300 = 57.59;\ \times 0.024495 = 1.4107$.

$$10 = 1.4107\,y^{8/3} \Rightarrow y^{8/3} = 7.089$$ $$y = 7.089^{3/8} = 7.089^{0.375}$$ $$\ln 7.089 = 1.9585,\ \times 0.375 = 0.7344,\ e^{0.7344} = 2.084$$ $$\boxed{y = 2.08\ \text{m}}$$

Bed width:

$$b = 0.8284 \times 2.084 = 1.726\ \text{m} \approx \boxed{1.73\ \text{m}}$$

Velocity check:

$$A = 1.8284 \times 2.084^2 = 1.8284 \times 4.343 = 7.941\ \text{m}^2$$ $$V = \frac{Q}{A} = \frac{10}{7.941} = 1.259\ \text{m/s}$$ $$\boxed{V \approx 1.26\ \text{m/s}}$$

This lies in the non-silting, non-scouring range for a lined/firm channel, so the design is acceptable.

(a) Rapid vs gradual closure

The critical (pipe-period) time is:

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

where $L$ = pipe length and $a$ = wave celerity. It is the time for the pressure wave to travel to the reservoir and back.

• Rapid / sudden closure: valve closing time $T \le T_c$. The full Joukowsky pressure rise develops because the reflected wave has not yet returned: $\Delta p = \rho a \,V_0$.
• Gradual closure: $T > T_c$. The reflected (relief) wave returns before closure is complete, so the pressure rise is smaller than the Joukowsky value and is estimated from rigid-column or interpolation formulae.

(b) Maximum (instantaneous) pressure rise

Joukowsky equation:

$$\Delta p = \rho\, a\, V_0 = 1000 \times 1100 \times 2.5 = 2.75 \times 10^6\ \text{Pa}$$ $$\boxed{\Delta p = 2.75\ \text{MPa}}$$

As head:

$$\Delta H = \frac{\Delta p}{\rho g} = \frac{2.75\times10^6}{1000 \times 9.81} = 280.3\ \text{m}$$ $$\boxed{\Delta H \approx 280\ \text{m of water}}$$

(c) Closure in $T = 3.0\ \text{s}$

Critical time:

$$T_c = \frac{2L}{a} = \frac{2 \times 1200}{1100} = \frac{2400}{1100} = 2.182\ \text{s}$$

Since $T = 3.0\ \text{s} > T_c = 2.18\ \text{s}$, the closure is gradual.

For gradual closure, estimate the pressure rise from the rigid-column / Michaud (Allievi linear) formula:

$$\Delta H = \frac{2 L V_0}{g\, T} = \frac{2 \times 1200 \times 2.5}{9.81 \times 3.0} = \frac{6000}{29.43} = 203.9\ \text{m}$$ $$\boxed{\Delta H \approx 204\ \text{m of water}}$$

Corresponding pressure rise:

$$\Delta p = \rho g \Delta H = 1000 \times 9.81 \times 203.9 = 2.00\times10^6\ \text{Pa} = \boxed{2.00\ \text{MPa}}$$

The gradual-closure pressure (2.00 MPa) is lower than the instantaneous value (2.75 MPa), confirming that slower valve operation reduces water-hammer over-pressure.

(a) Classification of surges

A surge is a moving wave front (rapidly varied unsteady flow) caused by a sudden change in discharge or depth. Types:

• Positive surge: depth increases as the wave passes (a moving hydraulic jump). It can move upstream (e.g. gate closing) or downstream (e.g. gate opening downstream).
• Negative surge: depth decreases as the wave passes; it is a flattening wave (e.g. gate opening upstream, dam-break receding wave).

Each can travel either upstream or downstream, giving four sub-cases.

(b) Upstream positive surge

Let the surge move upstream with absolute speed $V_w$ (positive in the downstream direction, so an upstream-moving surge has $V_w$ negative; we solve for its magnitude). Use a reference frame moving with the surge to make the flow steady — then continuity and momentum of a stationary hydraulic jump apply.

Let downstream-positive be the flow direction. Initial: $y_1 = 1.2\ \text{m}$, $V_1 = 1.5\ \text{m/s}$ (toward the gate). After surge: $y_2 = 1.8\ \text{m}$, velocity $V_2$ (unknown). Surge absolute velocity $= V_w$ (upstream, so we take its magnitude $C$ upstream).

Continuity in the moving frame (relative velocities $V_1 + C$ and $V_2 + C$, since the surge moves upstream against the flow):

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

Momentum (moving frame, treat as stationary jump):

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

From (2):

$$(V_1 + C)^2 = \frac{9.81 \times 1.8}{2 \times 1.2}(1.2 + 1.8) = \frac{17.658}{2.4}\times 3.0 = 7.3575 \times 3.0 = 22.073$$ $$V_1 + C = \sqrt{22.073} = 4.698\ \text{m/s}$$ $$C = 4.698 - 1.5 = 3.198\ \text{m/s}$$ $$\boxed{\text{Surge celerity (absolute, upstream)} \approx 3.20\ \text{m/s}}$$

New velocity behind the surge from (1):

$$V_2 + C = \frac{y_1 (V_1 + C)}{y_2} = \frac{1.2 \times 4.698}{1.8} = \frac{5.6376}{1.8} = 3.132\ \text{m/s}$$ $$V_2 = 3.132 - 3.198 = -0.066\ \text{m/s}$$ $$\boxed{V_2 \approx -0.07\ \text{m/s (nearly stagnant; slight reverse)}}$$

The flow behind the upstream-moving surge is almost brought to rest, consistent with a near-complete downstream gate closure. The surge advances upstream at about $3.2\ \text{m/s}$.

(a) Flow over a broad-crested weir

A broad-crested weir has a crest long enough (in the flow direction) that the streamlines become horizontal and parallel over the crest, so the pressure is hydrostatic and the flow passes through critical depth somewhere on the crest. Because critical flow gives maximum discharge for the available specific energy, the discharge is controlled by the head over the crest.

   ~~~~~~~~  H (upstream head)
        ____________  <- water surface drops over crest
  ----->|  yc (critical depth on crest)
  flow  |============| crest (broad)
        |   weir     |
  ______|____________|_____

At the crest the specific energy $\approx H$ and depth $= y_c = \tfrac{2}{3}H$ (rectangular). Substituting $V_c = \sqrt{g y_c}$ into $q = V_c y_c$ gives the weir formula.

(b) Discharge computation

Standard broad-crested-weir formula:

$$Q = C_d\,\frac{2}{3}\sqrt{\frac{2}{3}g}\; L\, H^{3/2}$$

Evaluate the constant:

$$\frac{2}{3}\sqrt{\frac{2}{3}\times 9.81} = \frac{2}{3}\sqrt{6.54} = \frac{2}{3}(2.5573) = 1.7049$$

With $C_d = 0.85$, $L = 4.0\ \text{m}$, $H = 0.50\ \text{m}$:

$$H^{3/2} = 0.50^{1.5} = 0.35355$$ $$Q = 0.85 \times 1.7049 \times 4.0 \times 0.35355$$ $$Q = 0.85 \times 1.7049 = 1.4492;\quad \times 4.0 = 5.7967;\quad \times 0.35355 = 2.0494$$ $$\boxed{Q \approx 2.05\ \text{m}^3/\text{s}}$$

Check via critical-flow basis: $y_c = \tfrac{2}{3}(0.50) = 0.3333\ \text{m}$, $V_c = \sqrt{9.81 \times 0.3333} = 1.808\ \text{m/s}$, ideal $q = 1.808 \times 0.3333 = 0.6027\ \text{m}^2/\text{s}$, ideal $Q = 0.6027 \times 4.0 = 2.411$; with $C_d=0.85$, $Q = 2.05\ \text{m}^3/\text{s}$ — consistent.

Solution

Unit discharge: $q = 9/3 = 3.0\ \text{m}^2/\text{s}$.

Upstream conditions:

$$V_1 = \frac{q}{y_1} = \frac{3.0}{1.5} = 2.0\ \text{m/s}$$ $$E_1 = y_1 + \frac{V_1^2}{2g} = 1.5 + \frac{2.0^2}{19.62} = 1.5 + 0.2039 = 1.7039\ \text{m}$$

Froude number: $Fr_1 = 2.0/\sqrt{9.81 \times 1.5} = 2.0/3.836 = 0.521$ (subcritical).

Critical conditions over the hump (the flow reaches minimum specific energy):

$$y_c = \left(\frac{q^2}{g}\right)^{1/3} = \left(\frac{3.0^2}{9.81}\right)^{1/3} = (0.9174)^{1/3} = 0.9717\ \text{m}$$ $$E_c = \frac{3}{2} y_c = \frac{3}{2}(0.9717) = 1.4576\ \text{m}$$

Maximum hump height (critical flow corresponds to minimum specific energy on the hump; energy on the hump $E_2 = E_1 - \Delta z$, set $E_2 = E_c$):

$$\Delta z_{max} = E_1 - E_c = 1.7039 - 1.4576 = 0.2463\ \text{m}$$ $$\boxed{\Delta z_{max} \approx 0.246\ \text{m}}$$

For any hump higher than this, the upstream depth must rise (choking occurs); a hump of $0.246\ \text{m}$ just brings the flow to critical over the crest without disturbing the upstream depth.

(a) Conveyance and section factor

Manning's equation can be written $Q = K\,S^{1/2}$, where the conveyance:

$$K = \frac{1}{n} A R^{2/3}$$

represents the discharge-carrying capacity of a section per unit $\sqrt{S}$; it depends only on geometry and roughness. The section factor for uniform flow is $A R^{2/3}$ ($= nK$), while the section factor for critical flow is $Z = A\sqrt{A/T}$.

(b) Half-full circular sewer

Diameter $D = 1.0\ \text{m}$, radius $r = 0.5\ \text{m}$. At half full the water surface is along the diameter:

Area (half circle):

$$A = \frac{1}{2}\,\frac{\pi D^2}{4} = \frac{\pi D^2}{8} = \frac{\pi (1.0)^2}{8} = 0.3927\ \text{m}^2$$

Wetted perimeter (half the circumference):

$$P = \frac{1}{2}\,\pi D = \frac{\pi (1.0)}{2} = 1.5708\ \text{m}$$

Hydraulic radius:

$$R = \frac{A}{P} = \frac{0.3927}{1.5708} = 0.2500\ \text{m}\quad (= D/4,\ \text{as expected for half-full})$$

Discharge (Manning):

$$Q = \frac{1}{n} A R^{2/3} S^{1/2}$$ $$R^{2/3} = 0.25^{2/3} = 0.3969$$ $$S^{1/2} = \sqrt{0.001} = 0.031623$$ $$Q = \frac{1}{0.013} \times 0.3927 \times 0.3969 \times 0.031623$$ $$Q = 76.923 \times 0.3927 \times 0.3969 \times 0.031623$$ $$= 76.923 \times 0.3927 = 30.21;\ \times 0.3969 = 11.99;\ \times 0.031623 = 0.3792$$ $$\boxed{Q \approx 0.38\ \text{m}^3/\text{s}}$$