BE Civil Engineering (IOE, TU) Hydraulics (IOE, CE 604) Question Paper 2078 Nepal

Given: $q = 4.2\ \text{m}^3/\text{s/m}$, $y_1 = 0.45\ \text{m}$, $g = 9.81\ \text{m/s}^2$.

(a) Upstream conditions

$$V_1 = \frac{q}{y_1} = \frac{4.2}{0.45} = 9.333\ \text{m/s}$$ $$Fr_1 = \frac{V_1}{\sqrt{g\,y_1}} = \frac{9.333}{\sqrt{9.81\times 0.45}} = \frac{9.333}{2.101} = 4.44$$

Since $Fr_1 > 1$ the upstream flow is supercritical, so a jump is possible.

Sequent depth (Belanger equation):

$$y_2 = \frac{y_1}{2}\left(\sqrt{1+8\,Fr_1^{2}}-1\right) = \frac{0.45}{2}\left(\sqrt{1+8(4.44)^2}-1\right)$$ $$= 0.225\left(\sqrt{158.78}-1\right) = 0.225(12.601-1) = \mathbf{2.61\ m}$$ $$V_2 = \frac{q}{y_2} = \frac{4.2}{2.611} = 1.609\ \text{m/s}, \qquad Fr_2 = \frac{1.609}{\sqrt{9.81\times 2.611}} = \mathbf{0.318}$$

Downstream flow is subcritical, as expected.

(b) Energy loss

$$E_L = \frac{(y_2-y_1)^3}{4\,y_1\,y_2} = \frac{(2.611-0.45)^3}{4(0.45)(2.611)} = \frac{(2.161)^3}{4.700} = \frac{10.09}{4.700} = \mathbf{2.15\ m}$$

Check by specific energy: $E_1 = y_1 + \dfrac{V_1^2}{2g} = 0.45 + \dfrac{9.333^2}{19.62} = 4.890\ \text{m}$; $E_2 = 2.611 + \dfrac{1.609^2}{19.62} = 2.743\ \text{m}$; $E_1-E_2 = 2.147\ \text{m}$ — agrees.

Power dissipated per metre width:

$$P = \rho g\, q\, E_L = 9810 \times 4.2 \times 2.147 = \mathbf{88.5\ kW/m}$$

(c) Classification: With $Fr_1 = 4.44$ (range $4.5 > Fr_1 > 2.5$ is steady; $Fr_1 > 9$ is strong), the jump lies at the upper edge of the steady/oscillating-to-steady range and behaves essentially as a well-formed steady jump with good, stable energy dissipation. Such jumps are deliberately created in stilling basins downstream of spillways and sluice gates to dissipate kinetic energy and protect the downstream bed from scour.

       supercritical              subcritical
        ----->                       ----->
  ___________  ~~~~~ roller ~~~~~  ______________
   y1=0.45 m  /\/\/\/\/\/\/\/        y2=2.61 m
  ====================================== bed

(a) Derivation

For steady uniform flow the friction slope equals the bed slope, $S_f = S_0$. Chezy's equation gives the mean velocity

$$V = C\sqrt{R\,S_0}$$

where $R$ is the hydraulic radius. Manning related Chezy's $C$ to roughness by $C = \dfrac{1}{n}R^{1/6}$ (empirical, SI units). Substituting,

$$\boxed{V = \frac{1}{n}R^{2/3}S_0^{1/2}}, \qquad Q = AV = \frac{1}{n}A\,R^{2/3}S_0^{1/2}$$

Assumptions: prismatic channel, constant discharge, hydrostatic pressure, small bed slope so depth measured vertically ≈ normal to bed, fully turbulent rough flow.

(b) Geometry at $y = 1.20\ \text{m}$ ($z = 2$):

$$A = (b + z y)y = (5 + 2\times1.20)(1.20) = (7.4)(1.20) = 8.88\ \text{m}^2$$ $$P = b + 2y\sqrt{1+z^2} = 5 + 2(1.20)\sqrt{1+4} = 5 + 2.4\times2.236 = 10.367\ \text{m}$$ $$R = \frac{A}{P} = \frac{8.88}{10.367} = 0.857\ \text{m}$$ $$V = \frac{1}{0.018}(0.857)^{2/3}(0.001)^{1/2} = 55.556 \times 0.902 \times 0.03162 = 1.585\ \text{m/s}$$ $$Q = AV = 8.88 \times 1.585 = \mathbf{14.07\ m^3/s}$$

(c) State of flow

Top width $T = b + 2zy = 5 + 2(2)(1.20) = 9.80\ \text{m}$; hydraulic (mean) depth $D = \dfrac{A}{T} = \dfrac{8.88}{9.80} = 0.906\ \text{m}$.

$$Fr = \frac{V}{\sqrt{gD}} = \frac{1.585}{\sqrt{9.81\times0.906}} = \frac{1.585}{2.982} = \mathbf{0.53}$$

Since $Fr < 1$, the flow is subcritical (tranquil). The channel slope is therefore a mild slope for this discharge.

(a) Definition. Specific energy is the energy head measured with respect to the channel bed: $E = y + \dfrac{V^2}{2g} = y + \dfrac{q^2}{2g\,y^2}$ for a rectangular section (unit discharge $q$). For a given $q$, $E$ is minimum at the critical depth; any $E > E_{min}$ has two alternate depths — one supercritical ($y < y_c$) and one subcritical ($y > y_c$).

  y |          subcritical limb (y > yc)
    |        /
    |      /
 yc |----*  <- critical point (E = Emin)
    |    \
    |      \  supercritical limb (y < yc)
    |________\__________ E
         Emin

(b) Unit discharge $q = \dfrac{Q}{b} = \dfrac{8}{3.5} = 2.286\ \text{m}^3/\text{s/m}$.

$$y_c = \left(\frac{q^2}{g}\right)^{1/3} = \left(\frac{2.286^2}{9.81}\right)^{1/3} = \left(0.5327\right)^{1/3} = \mathbf{0.811\ m}$$ $$V_c = \frac{q}{y_c} = \frac{2.286}{0.811} = \mathbf{2.82\ m/s} \quad (=\sqrt{g y_c}=2.82,\ Fr=1\ \checkmark)$$ $$E_{min} = 1.5\,y_c = 1.5 \times 0.811 = \mathbf{1.216\ m}$$

(c) At $y = 0.60\ \text{m}$ ($< y_c$, so supercritical):

$$V = \frac{q}{y} = \frac{2.286}{0.60} = 3.810\ \text{m/s}$$ $$E = y + \frac{V^2}{2g} = 0.60 + \frac{3.810^2}{19.62} = 0.60 + 0.740 = \mathbf{1.340\ m}$$

Alternate depth $y'$ satisfies $E = y' + \dfrac{q^2}{2g\,y'^2} = 1.340$ with $y' > y_c$. Solving $1.340 = y' + \dfrac{0.2663}{y'^2}$ by iteration:

• $y' = 1.10 \Rightarrow 1.10 + 0.220 = 1.320$
• $y' = 1.14 \Rightarrow 1.14 + 0.205 = 1.345$
• $y' = 1.132 \Rightarrow 1.132 + 0.208 = 1.340$ ✓

$$\boxed{y' \approx 1.13\ \text{m}}$$

The two depths $0.60\ \text{m}$ and $1.13\ \text{m}$ are alternate depths carrying the same discharge with the same specific energy $1.34\ \text{m}$.

(a) Normal and critical depths

Unit discharge $q = 10/5 = 2.0\ \text{m}^3/\text{s/m}$.

$$y_c = \left(\frac{q^2}{g}\right)^{1/3} = \left(\frac{4.0}{9.81}\right)^{1/3} = (0.4077)^{1/3} = \mathbf{0.742\ m}$$

Normal depth from Manning $Q = \dfrac{1}{n}A R^{2/3}S_0^{1/2}$ with $A = 5y$, $P = 5+2y$. By iteration:

So $y_n = \mathbf{1.225\ m}$.

Since $y_n (1.225) > y_c (0.742)$, the slope is mild (M).

(b) Profile type. Both working depths $1.30\ \text{m}$ and $1.40\ \text{m}$ exceed $y_n$ (and $y_c$): the depth lies in Zone 1 of a mild slope, so the profile is an M1 backwater curve (e.g. upstream of a weir or dam). Depth increases in the downstream direction; the surface approaches the horizontal asymptotically.

(c) Direct-step method (one step). Use $\Delta x = \dfrac{E_2 - E_1}{S_0 - \bar S_f}$ with $E = y + \dfrac{V^2}{2g}$ and $S_f = \left(\dfrac{nV}{R^{2/3}}\right)^2$.

Section 1 ($y_1 = 1.30$): $A=6.50$, $V=1.538$ m/s, $R=6.50/7.60=0.855$, $E_1 = 1.30 + \dfrac{1.538^2}{19.62} = 1.421\ \text{m}$, $S_f = \left(\dfrac{0.017\times1.538}{0.855^{2/3}}\right)^2 = 8.43\times10^{-4}$.

Section 2 ($y_2 = 1.40$): $A=7.00$, $V=1.429$ m/s, $R=7.00/7.80=0.897$, $E_2 = 1.40 + \dfrac{1.429^2}{19.62} = 1.504\ \text{m}$, $S_f = \left(\dfrac{0.017\times1.429}{0.897^{2/3}}\right)^2 = 6.81\times10^{-4}$.

$$\bar S_f = \frac{8.43+6.81}{2}\times10^{-4} = 7.62\times10^{-4}$$ $$\Delta x = \frac{E_2-E_1}{S_0-\bar S_f} = \frac{1.504-1.421}{0.001-0.000762} = \frac{0.0834}{2.38\times10^{-4}} = \mathbf{350\ m}$$

The positive sign confirms the depth grows in the downstream direction, consistent with an M1 backwater curve. The two depths are about 350 m apart.

(a) Types of closure. Let $T$ be the valve-closure time and $T_c = \dfrac{2L}{a}$ the time for the pressure wave to travel to the reservoir and return.

• Rapid / sudden closure: $T \le T_c$. The wave returns only after the valve is fully shut, so the full Joukowsky pressure rise develops at the valve.
• Gradual closure: $T > T_c$. Reflected relief waves reach the valve before closure completes, reducing the peak pressure (approximately $\Delta H \approx \dfrac{2LV_0}{gT}$).

Critical time here: $T_c = \dfrac{2\times1800}{1200} = \mathbf{3.0\ s}$. Closure faster than 3 s is treated as sudden.

(b) Maximum pressure rise (Joukowsky equation) for instantaneous closure:

$$\Delta p = \rho\, a\, V_0 = 1000 \times 1200 \times 2.0 = 2.4\times10^{6}\ \text{Pa} = \mathbf{2.40\ MPa}$$

As a head:

$$\Delta H = \frac{a\,V_0}{g} = \frac{1200\times2.0}{9.81} = \mathbf{244.6\ m\ of\ water}$$

This rise is added to the static head when sizing the penstock wall thickness.

(c) Surge tank. A surge tank is an open vertical chamber connected to the pressure conduit, placed as close as practicable to the turbine/valve (at the junction of the low-pressure headrace tunnel and the high-pressure penstock). On sudden load rejection the rising water in the tank converts the kinetic energy of the decelerating column into potential energy, reflecting the water-hammer wave and confining the high-pressure transient to the short length of penstock below the tank. It thus protects the long upstream tunnel from water hammer and provides reserve flow during sudden load demand.

For the best hydraulic trapezoidal section the cross-section is half of a regular hexagon: side slope $z = \dfrac{1}{\sqrt3}$, hydraulic radius $R = \dfrac{y}{2}$, area $A = \sqrt3\,y^2$ and bottom width $b = \dfrac{2y}{\sqrt3}$.

Apply Manning's equation with $R = y/2$:

$$Q = \frac{1}{n}A\,R^{2/3}S_0^{1/2} = \frac{1}{0.022}(\sqrt3\,y^2)\left(\frac{y}{2}\right)^{2/3}(0.0005)^{1/2}$$ $$20 = \frac{1.732}{0.022}\times 0.6300\, y^{2}\,y^{2/3}\times 0.02236 = 1.108\,y^{8/3}$$

where $\left(\tfrac12\right)^{2/3} = 0.6300$. Hence

$$y^{8/3} = \frac{20}{1.108} = 18.05 \quad\Rightarrow\quad y = 18.05^{3/8} = \mathbf{2.96\ m}$$

Bottom width:

$$b = \frac{2y}{\sqrt3} = \frac{2\times2.96}{1.732} = \mathbf{3.42\ m}$$

Mean velocity:

$$A = \sqrt3\,(2.96)^2 = 1.732\times8.76 = 15.17\ \text{m}^2, \qquad V = \frac{Q}{A} = \frac{20}{15.17} = \mathbf{1.32\ m/s}$$

This velocity (≈1.3 m/s) is non-silting and non-scouring for an ordinary earthen channel, so the design is acceptable. (Provide suitable freeboard of ≈0.3–0.5 m above $y$ in final construction.)

A positive surge (moving hydraulic bore) is analysed by superimposing the wave speed $V_w$ so the unsteady problem becomes steady relative to the surge.

Momentum across the surge (treating it as a moving jump) gives the surge speed:

$$(V_w - V_1)^2 = \frac{g\,y_2}{2\,y_1}\,(y_1 + y_2)$$ $$(V_w - 1.0)^2 = \frac{9.81\times2.2}{2\times1.5}(1.5+2.2) = \frac{21.582}{3.0}\times3.7 = 7.194\times3.7 = 26.62$$ $$V_w - 1.0 = \sqrt{26.62} = 5.159 \quad\Rightarrow\quad \boxed{V_w = 6.16\ \text{m/s (downstream)}}$$

Continuity in the moving frame $(V_w - V_1)y_1 = (V_w - V_2)y_2$:

$$V_2 = V_w - (V_w - V_1)\frac{y_1}{y_2} = 6.159 - (5.159)\frac{1.5}{2.2} = 6.159 - 3.518 = \mathbf{2.64\ m/s}$$

Check continuity: $(6.159-1.0)(1.5) = 7.74$ and $(6.159-2.642)(2.2) = 7.74$ ✓

So the surge travels downstream at about $6.16\ \text{m/s}$ and the flow velocity behind it increases to about $2.64\ \text{m/s}$ as the deeper, faster discharge propagates.

(a) Critical flow on the crest. A broad-crested weir has a crest long enough (in the flow direction) that the streamlines become parallel and hydrostatic pressure is re-established over the crest. As water passes from the upstream pool over the raised crest the specific energy is forced toward its minimum, so the flow passes through the critical state on the crest. The depth on the crest equals the critical depth:

$$y_c = \frac{2}{3}H = \frac{2}{3}\times0.45 = \mathbf{0.30\ m}$$

(neglecting approach velocity, since at critical flow $E = \tfrac32 y_c = H$).

(b) Discharge. The broad-crested weir formula is

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

where the constant $1.705 = \dfrac{2}{3}\sqrt{\dfrac{2g}{3}}$ in SI units and $L = 3\ \text{m}$ is the crest width across the flow.

$$Q = 1.705 \times 0.85 \times 3 \times (0.45)^{3/2}$$ $$(0.45)^{3/2} = 0.45\sqrt{0.45} = 0.45\times0.6708 = 0.3019$$ $$Q = 1.705 \times 0.85 \times 3 \times 0.3019 = \mathbf{1.31\ m^3/s}$$

Thus the weir discharges about $1.31\ \text{m}^3/\text{s}$ at the given head.

On a mild slope $y_n > y_c$, giving the normal depth line (NDL) above the critical depth line (CDL). Three zones and three M-profiles arise:

  ----------------------------------- NDL (y = yn)   Zone 1 above
      M1  (y > yn)  : backwater, rising toward horizontal
  ===================================  yn
      M2  (yn > y > yc) : drawdown, falling toward yc
  - - - - - - - - - - - - - - - - - -  CDL (y = yc)
      M3  (y < yc)  : supercritical, rising toward yc
  ===================================  bed

(a) Profiles

(b) Physical examples

• M1: Water surface upstream of a dam, weir or a partly closed sluice gate on a mild-slope river (the classic backwater curve).
• M2: Flow approaching a free overfall or a sudden enlargement / drop at the downstream end of a mild channel (drawdown).
• M3: Supercritical flow issuing from beneath a sluice gate onto a mild bed, before it forms a hydraulic jump back to subcritical flow.

(a) Specific force. Applying the momentum equation to a short channel reach (horizontal, frictionless) gives a quantity that is conserved across a hydraulic jump — the specific force per unit weight. For a rectangular channel of unit width:

$$F_s = \frac{q^2}{g\,y} + \frac{y^2}{2}$$

The first term is the momentum flux; the second is the hydrostatic pressure force. Plotting $F_s$ vs $y$ gives a curve with a minimum at the critical depth. A horizontal line ($F_s =$ const) cuts the curve at two depths — the conjugate (sequent) depths $y_1$ and $y_2$ — which are exactly the depths upstream and downstream of a hydraulic jump (equal specific force, but unequal specific energy because energy is lost in the jump).

   y |        \          /
     |         \        /  upper limb (subcritical)
  yc |----------*------  <- minimum Fs (critical)
     |         /        lower limb (supercritical)
     |________/______________ Fs
            Fs,min

(b) Numerical check ($q = 3.0$, $g = 9.81$, so $q^2/g = 9/9.81 = 0.9174$).

Critical depth: $y_c = \left(\dfrac{q^2}{g}\right)^{1/3} = (0.9174)^{1/3} = 0.972\ \text{m}$.

At $y_c = 0.972$:

$$F_s = \frac{0.9174}{0.972} + \frac{0.972^2}{2} = 0.944 + 0.472 = \mathbf{1.416\ m^2}$$

At $y = 0.50$ (supercritical, $y < y_c$):

$$F_s = \frac{0.9174}{0.50} + \frac{0.50^2}{2} = 1.835 + 0.125 = \mathbf{1.960\ m^2}$$

Since $F_s(0.50) = 1.960 > F_s(y_c) = 1.416$, the specific force is indeed minimum at the critical depth, confirming that critical flow corresponds to the minimum of the momentum function (just as it does for specific energy).

(a) Depth over the hump. Upstream specific energy:

$$V_1 = \frac{q}{y_1} = \frac{2.4}{1.20} = 2.0\ \text{m/s}, \qquad E_1 = y_1 + \frac{V_1^2}{2g} = 1.20 + \frac{4.0}{19.62} = 1.404\ \text{m}$$

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

$$E_2 = E_1 - \Delta z = 1.404 - 0.15 = 1.254\ \text{m}$$

Over the hump, $E_2 = y_2 + \dfrac{q^2}{2g\,y_2^2} = y_2 + \dfrac{0.2936}{y_2^2}$ (since $\dfrac{q^2}{2g} = \dfrac{5.76}{19.62} = 0.2936$). For subcritical flow take the larger root by iteration:

• $y_2 = 1.00 \Rightarrow 1.00 + 0.294 = 1.294$
• $y_2 = 1.05 \Rightarrow 1.05 + 0.266 = 1.316$ (too high → wrong branch direction)
• $y_2 = 1.02 \Rightarrow 1.02 + 0.282 = 1.302$
• $y_2 = 0.98 \Rightarrow 0.98 + 0.306 = 1.286$
• $y_2 = 1.02$ gives 1.302; we need 1.254, so try lower-energy values... recheck root:

Solve $y_2 + 0.2936/y_2^2 = 1.254$ (subcritical, $y_2 > y_c$). Critical depth $y_c = (q^2/g)^{1/3} = (0.5872)^{1/3} = 0.838\ \text{m}$.

• $y_2 = 1.05 \Rightarrow 1.05 + 0.2936/1.1025 = 1.05+0.266 = 1.316$
• $y_2 = 0.95 \Rightarrow 0.95 + 0.2936/0.9025 = 0.95+0.325 = 1.275$
• $y_2 = 0.93 \Rightarrow 0.93 + 0.2936/0.8649 = 0.93+0.339 = 1.269$
• $y_2 = 0.90 \Rightarrow 0.90 + 0.2936/0.81 = 0.90+0.363 = 1.263$
• $y_2 = 0.88 \Rightarrow 0.88 + 0.2936/0.7744 = 0.88+0.379 = 1.259$
• $y_2 = 0.87 \Rightarrow 0.87 + 0.2936/0.7569 = 0.87+0.388 = 1.258$

The subcritical root converges to about $y_2 \approx \mathbf{0.87\ m}$ (depth measured above the hump). The water surface therefore dips by $1.20 - (0.87 + 0.15) = 0.18\ \text{m}$ over the hump, as expected for subcritical flow.

(b) Maximum (critical) hump height before choking. Choking begins when the flow over the hump just reaches critical, i.e. $E_2 = E_{c,min} = 1.5\,y_c$:

$$E_{min} = 1.5 \times 0.838 = 1.257\ \text{m}$$ $$\Delta z_{max} = E_1 - E_{min} = 1.404 - 1.257 = \mathbf{0.147\ m}$$

The given hump of $0.15\ \text{m}$ is marginally above $\Delta z_{max} \approx 0.147\ \text{m}$, so it is essentially at the choking limit — the flow over the crest is on the verge of becoming critical (consistent with $y_2$ in part (a) sitting almost at $y_c = 0.838\ \text{m}$). Any larger hump would force the upstream depth to rise (back-up) to pass the same discharge.