Energy and Momentum Principles in Open channel flow

Specific force (momentum function)

Specific force $F$ (or specific momentum $M$) is the sum of the momentum flux and the hydrostatic pressure force per unit weight of water at a section of the channel:

$$F=\frac{Q^2}{gA}+A\bar y$$

where $A$ is the flow area and $\bar y$ is the depth of the centroid of area below the free surface. Its unit is m³.

Derivation (momentum equation)

Apply the momentum equation between two sections 1 and 2 over a short horizontal reach of a prismatic channel (friction and weight component neglected, as in a hydraulic jump):

$$\gamma A_1\bar y_1-\gamma A_2\bar y_2=\rho Q(V_2-V_1)=\rho Q^2\left(\frac1{A_2}-\frac1{A_1}\right)$$

Dividing by $\gamma$ and rearranging:

$$\frac{Q^2}{gA_1}+A_1\bar y_1=\frac{Q^2}{gA_2}+A_2\bar y_2\;\Rightarrow\;F_1=F_2$$

Proof that $F$ is minimum at critical flow

For a given discharge, differentiate $F$ with respect to depth $y$:

$$\frac{dF}{dy}=-\frac{Q^2}{gA^2}\frac{dA}{dy}+\frac{d(A\bar y)}{dy}$$

With top width $T$, $dA=T\,dy$. Also $\dfrac{d(A\bar y)}{dy}=A$ (the first moment of area increases by the area when depth increases: $d(A\bar y)=A\,dy$). Therefore

$$\frac{dF}{dy}=-\frac{Q^2T}{gA^2}+A$$

For minimum $F$, $\dfrac{dF}{dy}=0$:

$$\frac{Q^2T}{gA^3}=1\;\Rightarrow\;\frac{Q^2}{g}=\frac{A^3}{T}$$

This is exactly the condition for critical flow ($Fr=V/\sqrt{gA/T}=1$). The second derivative $\dfrac{d^2F}{dy^2}=\dfrac{2Q^2T^2}{gA^3}-\dfrac{Q^2}{gA^2}\dfrac{dT}{dy}+T$ is positive for usual channel sections, so $F$ is a minimum there. Hence for a given discharge the specific force is minimum when the flow is critical.

  y ^
    |  \      /
    |   \    /  F-curve: two depths (conjugate
    | y1 \  / y2   depths) for each F > Fmin
    |     \/
    | yc  *  <- minimum F
    +----------------> F

Uses of the concept

1. Hydraulic jump: conjugate (sequent) depths $y_1$ and $y_2$ have equal specific force, so $F_1=F_2$ gives the depth after the jump.
2. Calculation of the force on structures (sluice gate, weir, baffle blocks, stilling basin) from the difference in specific force.
3. Locating the jump position in a stilling basin.
4. Analysis of flow where energy loss is not known (jump, abrupt expansion, flow below gates).
5. Check of flow type: the lower limb of the F-curve is supercritical, the upper limb subcritical.

Specific energy

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

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

It has the unit of length and, unlike total head, can increase or decrease along the flow.

Proof of minimum energy at critical flow

For constant $Q$, $E$ is a function of $y$ only. Differentiate with respect to $y$ ($dA=T\,dy$):

$$\frac{dE}{dy}=1-\frac{Q^2}{gA^3}\frac{dA}{dy}=1-\frac{Q^2T}{gA^3}$$

For minimum $E$, $dE/dy=0$:

$$\frac{Q^2T}{gA^3}=1\;\Rightarrow\;\frac{V^2}{g(A/T)}=1\;\Rightarrow\;Fr=\frac{V}{\sqrt{gD_h}}=1$$

Since $\dfrac{d^2E}{dy^2}=\dfrac{3Q^2T^2}{gA^4}-\dfrac{Q^2}{gA^3}\dfrac{dT}{dy}>0$ for usual sections, this is a minimum. The condition $Fr=1$ is critical flow. Hence the specific energy is minimum at critical flow.

For a rectangular channel: $y_c=\sqrt[3]{q^2/g}$ and $E_{min}=\dfrac32y_c$.

Specific energy curve

  y ^           /  E = y (45 deg line)
    |          /
    |         /      subcritical (Fr<1)
 y1 |--------/--*
    |       /  /
 yc |------*  /
    |     / \/
 y2 |----*   \   supercritical (Fr>1)
    |   /  \   \
    +------+----+---------> E
         Emin  E

• The curve has two branches for a given $E>E_{min}$ and a single point at $E=E_{min}$.
• Alternate depths: the two depths $y_1$ (subcritical, upper limb) and $y_2$ (supercritical, lower limb) have the same specific energy.
• Critical depth $y_c$ and $E_{min}$ are at the nose of the curve.
• Subcritical zone: $y>y_c$, $V<\sqrt{gD_h}$, upper limb (asymptote to the 45° line $E=y$).
• Supercritical zone: $y<y_c$, $V>\sqrt{gD_h}$, lower limb (asymptote to the $E$-axis).
• For a rectangular section the curve is cubic: $y^3-Ey^2+\dfrac{q^2}{2g}=0$.

Set-up

A rectangular channel with discharge $q$ per unit width, approach depth $y_1$ (subcritical) and specific energy $E_1=y_1+\dfrac{q^2}{2gy_1^2}$. A hump of height $\Delta z$ is placed on the bed. With no energy loss, specific energy over the hump is

$$E_2=E_1-\Delta z$$

On the specific energy diagram, moving over the hump shifts the state to the left by $\Delta z$. The critical hump height is $\Delta z_c=E_1-E_{min}=E_1-\tfrac32y_c$.

  y ^                      /
    |                     /
    |        y1 (upstream)*
    |                   / |
    |     yc  *--------/--|
    |         |      /    |
    |   y2 *--|     /     |
    |      <--dz-->       
    +----+--------+--------> E
       E_min   E2    E1

(i) Hump height less than critical ($\Delta z<\Delta z_c$)

• $E_2=E_1-\Delta z>E_{min}$, so flow over the hump is possible with the same discharge.
• The upstream depth $y_1$ is unchanged.
• Depth over the hump falls, on the subcritical limb (it is lower than $y_1$, but still greater than $y_c$). The water surface drops over the hump by less than $\Delta z$.
• The surface rises again after the hump to $y_1$.

(ii) Hump height equal to critical ($\Delta z=\Delta z_c$)

• $E_2=E_{min}$, so the depth over the hump is exactly critical ($y=y_c$).
• The upstream depth is still $y_1$ (unchanged), but this is the limit: the hump is acting as a control (broad-crested weir).
• Discharge is the maximum that can pass for this upstream energy.

(iii) Hump height greater than critical ($\Delta z>\Delta z_c$)

• $E_1-\Delta z<E_{min}$: the available energy over the hump is less than the minimum required for $q$. The flow cannot pass with the given upstream energy.
• The flow backs up: upstream depth rises to $y_1'$ so that upstream energy becomes $E_1'=E_{min}+\Delta z$.
• Critical depth occurs over the hump, and downstream of the hump the flow becomes supercritical (and may end in a hydraulic jump).
• Increasing the hump further raises the upstream depth further.

If the hump is combined with a constriction of constant width, the same logic applies with $q$ increased in the contracted section; the critical condition is $E_1-\Delta z=\tfrac32y_c$ with $y_c=\sqrt[3]{q_2^2/g}$ and $q_2=Q/b_2$.

Critical depth occurs at a section when the available specific energy there is equal to the minimum specific energy for the discharge per unit width at that section: $E_2=E_{min}=\tfrac32y_c$, where $y_c=\sqrt[3]{q^2/g}$.

Approach flow

$Q=2.4$ m³/s, $b=2$ m, $y_1=1.0$ m.

$$q_1=\frac{2.4}{2}=1.2\ \text{m}^2/\text{s},\quad V_1=1.2\ \text{m/s},\quad Fr_1=\frac{1.2}{\sqrt{9.81\times1}}=0.383\ (\text{subcritical})$$ $$E_1=y_1+\frac{V_1^2}{2g}=1+\frac{1.44}{19.62}=1.0734\ \text{m}$$ $$y_{c1}=\sqrt[3]{\frac{1.2^2}{9.81}}=0.5275\ \text{m},\qquad E_{min1}=1.5y_{c1}=0.7913\ \text{m}$$

(a) Hump of 0.20 m (width unchanged)

Energy over hump $E_2=E_1-\Delta z=1.0734-0.20=0.8734$ m $>E_{min1}=0.7913$ m.

So critical depth does not occur. The flow stays subcritical over the hump, with depth found from $y+\dfrac{1.44}{19.62y^2}=0.8734$:

$$y_2=0.739\ \text{m}\quad(>y_c=0.528\ \text{m})$$

The upstream depth remains 1.0 m.

(b) Side-wall constriction to 1.7 m (no hump)

$$q_2=\frac{2.4}{1.7}=1.4118\ \text{m}^2/\text{s},\quad y_{c2}=\sqrt[3]{\frac{q_2^2}{g}}=0.5879\ \text{m},\quad E_{min2}=0.8818\ \text{m}$$

$E_2=E_1=1.0734>E_{min2}=0.8818$, so critical depth does not occur. Depth in the constriction (subcritical): $y_2=0.964$ m. Upstream depth unchanged.

(c) Hump and constriction together

Available energy at the section: $E_2=E_1-0.20=0.8734$ m, while the minimum required is $E_{min2}=0.8818$ m (for $q_2=1.412$).

$E_2<E_{min2}$, so critical depth occurs (choking) over the hump in the constriction, $y_c=0.588$ m. The channel cannot pass the flow at the original upstream energy, so the upstream depth is affected: it rises until

$$E_1'=E_{min2}+\Delta z=0.8818+0.20=1.0818\ \text{m}$$ $$y_1'+\frac{1.2^2}{2g\,y_1'^2}=1.0818\;\Rightarrow\;y_1'=1.0098\ \text{m}$$

Rise in upstream depth $=1.0098-1.000=0.0098$ m (about 10 mm).

Answer: (a) not critical ($y=0.74$ m over hump); (b) not critical ($y=0.96$ m); (c) critical depth $0.588$ m occurs; upstream depth rises from 1.000 m to 1.010 m, about 10 mm.

Specific force (momentum function)

Specific force $F$ at a section is the sum of the momentum flux and the hydrostatic pressure force per unit weight of water:

$$F=\frac{Q^2}{gA}+A\bar y$$

where $\bar y$ = depth of centroid of the flow area below the water surface.

Development (from the momentum equation)

For a short horizontal reach (friction and weight component negligible), the net force equals the change in momentum flux:

$$\gamma A_1\bar y_1-\gamma A_2\bar y_2=\rho Q(V_2-V_1)=\rho Q^2\left(\frac{1}{A_2}-\frac{1}{A_1}\right)$$ $$\frac{Q^2}{gA_1}+A_1\bar y_1=\frac{Q^2}{gA_2}+A_2\bar y_2\;\Rightarrow\;F_1=F_2$$

For a rectangular channel of width $b$ ($q=Q/b$, $A\bar y=by^2/2$) the specific force per unit width is

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

Conjugate (sequent) depths

For a given $Q$ and a value of $F>F_{min}$, the equation $F(y)=$ const has two positive roots, $y_1$ (supercritical) and $y_2$ (subcritical). These are the conjugate depths: they have the same specific force. They are the depths before and after a hydraulic jump, where energy is lost but specific force is conserved. For a rectangular channel:

$$\frac{y_2}{y_1}=\frac12\left(\sqrt{1+8Fr_1^2}-1\right)$$

Specific force curve

  y ^
    |  \              /
    |   \            /   subcritical (upper limb)
 y2 |----\----------/--- conjugate depth y2
    |     \        /
 yc |------\------*  <- F_min, critical depth
    |       \    /
 y1 |--------\--/----    conjugate depth y1
    |         \/         supercritical (lower limb)
    +--------------------> F
          F1 = F2

• Lower limb ($y<y_c$): supercritical flow.
• Nose of the curve: $F=F_{min}$ at $y=y_c$, where $\dfrac{Q^2T}{gA^3}=1$ (critical flow).
• Upper limb ($y>y_c$): subcritical flow.
• The horizontal line at a given $F$ cuts the curve at the two conjugate depths. The difference in specific energy at these depths is the energy loss of the jump.

Given data

Upstream $b_1=3.5$ m, $y_1=1$ m, $Q=4$ m³/s; contracted $b_2=2.5$ m with bed raised $\Delta z=0.10$ m.

$$q_1=\frac{4}{3.5}=1.1429\ \text{m}^2/\text{s},\quad V_1=1.143\ \text{m/s},\quad E_1=1+\frac{1.143^2}{19.62}=1.0666\ \text{m}$$ $$q_2=\frac{4}{2.5}=1.6\ \text{m}^2/\text{s},\quad y_{c2}=\sqrt[3]{\frac{1.6^2}{9.81}}=0.6390\ \text{m},\quad E_{min2}=0.9586\ \text{m}$$

Depth in the contracted section

$$E_2=E_1-\Delta z=1.0666-0.10=0.9666\ \text{m}>E_{min2}=0.9586\ \text{m}$$

Flow is possible without choking; the flow remains subcritical:

$$y_2+\frac{1.6^2}{19.62y_2^2}=y_2+\frac{0.13048}{y_2^2}=0.9666\;\Rightarrow\;y_2=0.701\ \text{m}$$

Maximum rise in bed without affecting the upstream depth

The limit is when the contracted section has critical flow with $E_2=E_{min2}$:

$$\Delta z_{max}=E_1-E_{min2}=1.0666-0.9586=0.1080\ \text{m}\approx10.8\ \text{cm}$$

Answer: depth in the contracted section $=0.70$ m; maximum bed rise without affecting the upstream depth $=0.108$ m (about 10.8 cm).

Given data

$b_1=3$ m, $Q=3$ m³/s, $y_1=1$ m. Contracted width $b_2=2$ m. $q_1=1$ m²/s, $q_2=1.5$ m²/s.

$$E_1=1+\frac{1^2}{19.62}=1.0510\ \text{m},\qquad y_{c2}=\sqrt[3]{\frac{1.5^2}{9.81}}=0.6121\ \text{m},\quad E_{min2}=0.9182\ \text{m}$$

(a) Bed raised by 10 cm: depth in the contracted section

$$E_2=E_1-0.10=0.9510\ \text{m}>E_{min2}$$

so the flow is not choked (subcritical):

$$y_2+\frac{1.5^2}{19.62y_2^2}=y_2+\frac{0.11468}{y_2^2}=0.9510\;\Rightarrow\;y_2=0.744\ \text{m}$$

(b) Maximum rise without affecting the upstream depth

$$\Delta z_{max}=E_1-E_{min2}=1.0510-0.9182=0.1328\ \text{m}\approx13.3\ \text{cm}$$

(c) Rise of 30 cm ($>\Delta z_{max}$)

The flow is choked. Critical flow occurs in the contraction and the upstream depth rises to satisfy

$$E_1'=E_{min2}+0.30=0.9182+0.30=1.2182\ \text{m}$$ $$y_1'+\frac{1}{19.62y_1'^2}=1.2182\;\Rightarrow\;y_1'=1.182\ \text{m}$$

Water surface elevations (above the original upstream bed):

So the upstream water surface rises by 0.182 m (to 1.182 m), and the water surface at the contracted section is 0.912 m (critical depth 0.612 m).

Answer: (a) $y_2=0.74$ m; (b) $\Delta z_{max}=0.133$ m; (c) upstream depth rises from 1.00 m to 1.18 m, and critical flow ($y=0.61$ m) occurs in the contraction.

Specific energy

Specific energy is the energy per unit weight of water at a section measured above the channel bed:

$$E=y+\frac{V^2}{2g}\qquad(\alpha\approx1)$$

Proof: critical flow gives maximum discharge for given $E$

Consider a rectangular channel (width $b$, $q=Q/b$). With $V=q/y$:

$$E=y+\frac{q^2}{2gy^2}\;\Rightarrow\;q=y\sqrt{2g(E-y)}$$

For constant $E$, $q$ is a function of $y$ only. Differentiate $q^2=2gy^2(E-y)$:

$$\frac{d(q^2)}{dy}=2g\left(2yE-3y^2\right)=0\;\Rightarrow\;y=\frac{2E}{3}$$

At this depth the second derivative is negative, so $q$ is a maximum. Then

$$q_{max}=\frac{2E}{3}\sqrt{2g\left(E-\frac{2E}{3}\right)}=\sqrt g\left(\frac{2E}{3}\right)^{3/2}$$

Critical depth in a rectangular channel is $y_c=\sqrt[3]{q^2/g}$, i.e. $q^2=gy_c^3$. Comparing: $y=\dfrac{2E}{3}=y_c$ and $E=\dfrac32y_c=E_{min}$. Also

$$V=\frac{q}{y}=\sqrt{\frac{g\cdot y^3}{y^2}}=\sqrt{gy}\;\Rightarrow\;Fr=1$$

Hence for a given specific energy, the discharge is maximum when the flow is critical. (For any section, $dQ/dy=0$ gives $Q^2T/(gA^3)=1$, the same critical-flow condition.)

  y ^
    |            q increases
 yc |-----------*  q_max (critical)
    |         /  \
    |        /    \  q decreases towards
    |       /      \ both y=0 and y=E
    +------+--------> q (E constant)

Alternate depths $y_1$ and $y_2$ have the same specific energy and the same discharge per unit width $q$.

Express $E$ in terms of Froude number

For a rectangular channel, $Fr=\dfrac{V}{\sqrt{gy}}=\dfrac{q}{\sqrt g\,y^{3/2}}$, so

$$\frac{V^2}{2g}=\frac{Fr^2\,y}{2}$$ $$E=y+\frac{Fr^2y}{2}=\frac{y\,(2+Fr^2)}{2}$$

Equal specific energy

$$\frac{y_1(2+F_1^2)}{2}=\frac{y_2(2+F_2^2)}{2}\;\Rightarrow\;\frac{y_1}{y_2}=\frac{2+F_2^2}{2+F_1^2}\qquad(1)$$

Ratio of Froude numbers

Since $q$ is the same for both depths,

$$\frac{F_2}{F_1}=\frac{q/(\sqrt g\,y_2^{3/2})}{q/(\sqrt g\,y_1^{3/2})}=\left(\frac{y_1}{y_2}\right)^{3/2}$$ $$\left(\frac{F_2}{F_1}\right)^{2/3}=\frac{y_1}{y_2}\qquad(2)$$

Result

From (1) and (2):

$$\boxed{\left(\frac{F_2}{F_1}\right)^{2/3}=\frac{2+F_2^2}{2+F_1^2}}$$

Hence proved.

Given data

Rectangular channel $b=5$ m, upstream $y_1=5$ m, $V_1=4$ m/s, no losses.

$$q=V_1y_1=4\times5=20\ \text{m}^2/\text{s},\quad Q=100\ \text{m}^3/\text{s}$$ $$E_1=y_1+\frac{V_1^2}{2g}=5+\frac{16}{19.62}=5.8155\ \text{m}$$ $$Fr_1=\frac{4}{\sqrt{9.81\times5}}=0.571\ (\text{subcritical}),\quad y_c=\sqrt[3]{\frac{20^2}{9.81}}=3.442\ \text{m},\quad E_{min}=5.163\ \text{m}$$

(i) Step rise of 0.2 m

Energy equation (datum at the upstream bed): $E_1=\Delta z+E_2$

$$E_2=5.8155-0.2=5.6155\ \text{m}>E_{min}$$

so flow is possible, and since the approach flow is subcritical, the flow stays subcritical:

$$y_2+\frac{20^2}{19.62\,y_2^2}=y_2+\frac{20.387}{y_2^2}=5.6155\;\Rightarrow\;y_2=4.688\ \text{m}$$

Water surface elevation above the upstream bed $=0.2+4.688=4.888$ m. Change in water level $=4.888-5.000=-0.112$ m, i.e. the water surface falls by 0.112 m.

(ii) Step drop of 0.2 m

$$E_2=E_1+0.2=6.0155\ \text{m}$$

Subcritical root: $y_2+\dfrac{20.387}{y_2^2}=6.0155\Rightarrow y_2=5.286\ \text{m}$

Water surface elevation above the upstream bed $=5.286-0.2=5.086$ m; change in water level $=5.086-5=+0.086$ m, i.e. the water surface rises by 0.086 m.

 (i) rise:   y1=5 ____        (ii) drop:  y1=5 ____
                  \___ y2=4.69               ____/ y2=5.29
         step 0.2 |___|                    |___| drop 0.2

Answer: (i) $y_2=4.69$ m, water level falls by 0.11 m; (ii) $y_2=5.29$ m, water level rises by 0.09 m.

(Subcritical flow: the surface drops over a rise and goes up over a drop, the opposite of the bed movement.)

Condition for critical flow

Minimum specific energy occurs at critical flow:

$$\frac{Q^2T}{gA^3}=1\quad\text{or}\quad \frac{Q^2}{gA^2}=\frac{A}{T}\qquad(1)$$

Triangular section

Side slope $z$ (H:V), depth $y$:

$$A=zy^2,\qquad T=2zy,\qquad \frac{A}{T}=\frac{zy^2}{2zy}=\frac y2$$

 T = 2zy
 \~~~~~~~/
  \     /  y
   \   /
    \ /

At critical depth $y=y_c$

From (1):

$$\frac{Q^2}{gA^2}=\frac{A}{T}=\frac{y_c}{2}\;\Rightarrow\;\frac{V_c^2}{2g}=\frac{Q^2}{2gA^2}=\frac{y_c}{4}$$

(the velocity head at critical flow is half the hydraulic depth, $\dfrac{A}{2T}=\dfrac{y_c}{4}$).

Minimum specific energy

$$E_c=y_c+\frac{V_c^2}{2g}=y_c+\frac{y_c}{4}=\frac54\,y_c$$ $$\boxed{E_c=\frac54\,y_c}$$

Hence proved. (Compare: rectangular channel $E_c=\tfrac32y_c$; parabolic channel $E_c=\tfrac43y_c$. In general $E_c=y_c+\dfrac{A_c}{2T_c}$.)

An obstruction (hump) of height $\Delta z$ across the bed does not affect the upstream depth as long as the specific energy over it is at least the minimum, $E_{min}$. The limiting height is when the flow over the obstruction is just critical.

Given data

$b=2$ m, $Q=3$ m³/s, $y_1=1.5$ m.

$$q=\frac{3}{2}=1.5\ \text{m}^2/\text{s},\quad V_1=\frac{1.5}{1.5}=1.0\ \text{m/s}$$ $$E_1=y_1+\frac{V_1^2}{2g}=1.5+\frac{1.0^2}{19.62}=1.5510\ \text{m}$$

Critical condition at the obstruction

$$y_c=\sqrt[3]{\frac{q^2}{g}}=\sqrt[3]{\frac{1.5^2}{9.81}}=0.6121\ \text{m}$$ $$E_{min}=\frac32y_c=0.9182\ \text{m}$$

For no rise upstream, $E_1=E_{min}+\Delta z_{max}$:

$$\Delta z_{max}=E_1-E_{min}=1.5510-0.9182=0.633\ \text{m}$$

Answer: maximum height of the obstruction $=0.633$ m (about 63 cm). A higher obstruction will cause the upstream water level to rise.

Set-up

For a rectangular channel with discharge $q$ per unit width, two alternate depths $y_1$ and $y_2$ have the same specific energy $E$:

$$E=y_1+\frac{q^2}{2gy_1^2}=y_2+\frac{q^2}{2gy_2^2}\qquad(1)$$

Eliminate $q$

From the equality in (1):

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

Cancelling $(y_2-y_1)\neq0$:

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

Specific energy

Substitute (2) into $E=y_1+\dfrac{q^2}{2gy_1^2}$:

$$E=y_1+\frac{y_1^2y_2^2}{(y_1+y_2)y_1^2}=y_1+\frac{y_2^2}{y_1+y_2}$$ $$E=\frac{y_1(y_1+y_2)+y_2^2}{y_1+y_2}$$ $$\boxed{E=\frac{y_1^2+y_1y_2+y_2^2}{y_1+y_2}}$$

Hence proved.

Critical flow at the transition requires $E_2=E_{min}=\tfrac32y_c$ with $y_c=\sqrt[3]{q_2^2/g}$ at that section.

Given data

$b=50$ m, $Q=250$ m³/s, $y_1=5$ m. $q_1=\dfrac{250}{50}=5$ m²/s, $V_1=1$ m/s.

$$E_1=5+\frac{1^2}{19.62}=5.0510\ \text{m},\qquad y_{c1}=\sqrt[3]{\frac{25}{9.81}}=1.366\ \text{m},\quad E_{min1}=2.049\ \text{m}$$

The approach flow is subcritical ($Fr_1=0.143$).

(i) Step in the bed (width constant)

$$E_1=E_{min1}+\Delta z\;\Rightarrow\;\Delta z=5.0510-2.0489=3.002\ \text{m}$$

Height of step $=3.00$ m.

(ii) Width reduction (bed unchanged)

Critical flow at the contraction with $E_2=E_1$:

$$y_c=\frac23E_1=3.3673\ \text{m},\qquad q_2=\sqrt{gy_c^3}=\sqrt{9.81\times3.3673^3}=19.354\ \text{m}^2/\text{s}$$ $$b_2=\frac{Q}{q_2}=\frac{250}{19.354}=12.92\ \text{m}$$

Reduction in width $=50-12.92=37.08$ m.

(iii) Combination of step and width reduction

Many combinations are possible. Take a step of $\Delta z=1.5$ m (assumed). Then energy at the contraction:

$$E_2=E_1-1.5=3.5510\ \text{m},\qquad y_c=\frac23E_2=2.3673\ \text{m}$$ $$q_2=\sqrt{9.81\times2.3673^3}=11.408\ \text{m}^2/\text{s},\qquad b_2=\frac{250}{11.408}=21.91\ \text{m}$$

Reduction in width $=50-21.91=28.09$ m.

Answer: (i) step $=3.00$ m; (ii) width reduced by 37.08 m to 12.92 m; (iii) e.g. step 1.5 m with width reduced by 28.09 m to 21.91 m.

Given conditions (rectangular channel, width $b$, discharge $Q$)

1. Flow is critical.
2. The wetted perimeter is minimum (for the given discharge).

Critical flow in a rectangular channel

$$q=\frac Qb,\quad y_c=\sqrt[3]{\frac{q^2}{g}}=\left(\frac{Q^2}{gb^2}\right)^{1/3},\qquad E_{min}=\frac32y_c$$

Minimum wetted perimeter

$$P=b+2y_c=b+2\left(\frac{Q^2}{g}\right)^{1/3}b^{-2/3}$$

Differentiate with respect to $b$ ($Q$ constant) and equate to zero:

$$\frac{dP}{db}=1-\frac43\left(\frac{Q^2}{g}\right)^{1/3}b^{-5/3}=0$$ $$b^{5/3}=\frac43\left(\frac{Q^2}{g}\right)^{1/3}$$

Since $\left(\dfrac{Q^2}{g}\right)^{1/3}=y_c\,b^{2/3}$ (from $y_c^3=Q^2/(gb^2)$):

$$b^{5/3}=\frac43y_cb^{2/3}\;\Rightarrow\;b=\frac43y_c$$

(The second derivative is positive, so this is a minimum.)

Relation with $E_{min}$

From $E_{min}=\tfrac32y_c$, $y_c=\tfrac23E_{min}$:

$$b=\frac43\times\frac23E_{min}=\frac89E_{min}$$ $$\boxed{b=\frac89E_{min}}$$

Hence proved.

Given data

$y_1=1.6$ m, $V_1=1.10$ m/s, $b_1=4.0$ m; contracted width $b_2=3.5$ m with a hump of $\Delta z=0.35$ m; no loss.

$$Q=b_1y_1V_1=4\times1.6\times1.10=7.04\ \text{m}^3/\text{s}$$ $$E_1=y_1+\frac{V_1^2}{2g}=1.6+\frac{1.21}{19.62}=1.6617\ \text{m},\qquad Fr_1=\frac{1.10}{\sqrt{9.81\times1.6}}=0.278$$

Check for critical flow at the contraction

$$q_2=\frac{7.04}{3.5}=2.0114\ \text{m}^2/\text{s},\quad y_{c2}=\sqrt[3]{\frac{q_2^2}{g}}=0.7444\ \text{m},\quad E_{min2}=1.1165\ \text{m}$$

Energy available over the hump: $E_2=E_1-\Delta z=1.6617-0.35=1.3117$ m $>E_{min2}$. So flow passes without choking and the upstream depth is unchanged.

Depth over the hump (subcritical)

$$y_2+\frac{q_2^2}{2gy_2^2}=y_2+\frac{0.2062}{y_2^2}=1.3117\;\Rightarrow\;y_2=1.1579\ \text{m}$$

(here $q_2^2/2g=0.2062$). Velocity $V_2=q_2/y_2=1.737$ m/s.

Water surface elevations (datum: upstream bed)

• At a section upstream: $1.600$ m.
• At the contracted section: $0.35+1.1579=1.5079$ m.

Drop in water surface $=1.600-1.508=0.092$ m.

Specific energy diagram

  y ^                     /  E = y
    |                    /
 1.6|-----------------*(1)   upstream: E1 = 1.662
    |                /
y2  |------*(2)    /         over hump: E2 = 1.312
    |     /|      /
yc2 |----/-|-----/
    +----+------+--------> E
       E2      E1
       |<-0.35->|

Point (1) is on the upper (subcritical) limb of the curve for $q_1=1.76$ m²/s; point (2) is on the upper limb of the curve for $q_2=2.011$ m²/s, shifted left by $\Delta z=0.35$ m.

Answer: depth over the hump $=1.158$ m, water surface elevation there $=1.508$ m (above the upstream bed); upstream elevation $=1.600$ m (unchanged).

Given data

$b=2$ m, $Q=6$ m³/s, so $q=3$ m²/s. Hump height $\Delta z=0.5$ m, no energy loss.

Point A (upstream, subcritical): $y_A=1.63$ m, $E_A=1.63+\dfrac{3^2}{19.62\times1.63^2}=1.8026\approx1.80$ m. The alternate depth is A' ($y=0.62$ m, supercritical) with the same $E=1.80$ m.

Critical state on the curve

$$y_c=\sqrt[3]{\frac{q^2}{g}}=\sqrt[3]{\frac{9}{9.81}}=0.972\ \text{m},\qquad E_{min}=\frac32y_c=1.458\ \text{m}$$

(i) Is the specific energy adequate?

Energy available over the hump:

$$E_{hump}=E_A-\Delta z=1.80-0.50=1.30\ \text{m}$$

Since $E_{hump}=1.30<E_{min}=1.458$ m, the flow does not have adequate specific energy to pass 6 m³/s over the 0.5 m hump. (Maximum allowable hump height for the present upstream state: $E_A-E_{min}=1.8026-1.4575=0.345$ m, less than 0.5 m.)

(ii) What happens (profile upstream and downstream)

The hump acts as a control. Upstream, the water surface rises (backwater) until the specific energy upstream is just enough for critical flow over the hump:

$$E_1'=E_{min}+\Delta z=1.4575+0.50=1.9575\ \text{m}$$

Upstream depth (subcritical root of $y+\dfrac{9}{19.62y^2}=1.9575$): $y_1'=1.819$ m.

• On the hump: critical flow, $y=y_c=0.972$ m.
• Downstream of the hump: the flow passes through critical depth and becomes supercritical. After the hump, bed returns to original level, energy $=1.9575$ m, so the supercritical depth is $y_3=0.576$ m. This may be followed by a hydraulic jump further downstream when the tailwater is deep enough.

  y (m)
  1.82 |            * A1 (upstream after rise, E=1.96)
  1.63 |        * A (original upstream, E=1.80)
       |
  0.97 |    * critical on hump (E=1.46)
  0.62 |        * A' (E=1.80)
  0.58 |            * downstream supercritical (E=1.96)
       +----+----+----+----+--> E (m)
          1.3  1.46 1.80 1.96

Final results on the curve: A moves up to 1.82 m (E 1.96 m); hump point at the nose (0.97 m, 1.46 m); downstream supercritical point at 0.58 m (E 1.96 m).

Answer: (i) No, since $E=1.30$ m $<E_{min}=1.46$ m; (ii) upstream depth rises to 1.82 m, critical depth 0.97 m over the hump, then supercritical flow 0.58 m downstream.

Given data

$b=3$ m, $Q=15$ m³/s, $y_1=2$ m. $q=\dfrac{15}{3}=5$ m²/s, $V_1=2.5$ m/s.

$$E_1=2+\frac{2.5^2}{19.62}=2.3186\ \text{m}$$ $$y_c=\sqrt[3]{\frac{q^2}{g}}=\sqrt[3]{\frac{25}{9.81}}=1.3659\ \text{m},\qquad E_{min}=1.5y_c=2.0489\ \text{m}$$

(a) Minimum hump height for critical depth over the hump

$$\Delta z_{min}=E_1-E_{min}=2.3186-2.0489=0.270\ \text{m}$$

(b) Hump height for upstream depth 2.5 m

The upstream depth is raised, so the hump causes critical flow. Upstream energy:

$$E_1'=2.5+\frac{(5/2.5)^2}{19.62}=2.5+\frac{4}{19.62}=2.7039\ \text{m}$$ $$\Delta z=E_1'-E_{min}=2.7039-2.0489=0.655\ \text{m}$$

(c) Hump of 0.20 m

Since $0.20<\Delta z_{min}=0.270$ m, the flow is not choked: upstream depth remains 2.0 m. Depth over the hump (subcritical):

$$E_2=E_1-0.20=2.1186\ \text{m},\qquad y_2+\frac{25}{19.62\,y_2^2}=y_2+\frac{1.2742}{y_2^2}=2.1186$$ $$y_2=1.651\ \text{m}$$

Answer: (a) minimum hump height $=0.270$ m; (b) hump height $=0.655$ m; (c) upstream depth 2.00 m, depth over the 0.2 m hump $=1.65$ m.

Given data

$b_1=3.5$ m, $Q=4$ m³/s, $y_1=1.2$ m; contracted $b_2=2.0$ m, bed raised $\Delta z=0.15$ m.

$$q_1=\frac{4}{3.5}=1.1429\ \text{m}^2/\text{s},\quad V_1=\frac{1.1429}{1.2}=0.952\ \text{m/s},\quad E_1=1.2+\frac{0.952^2}{19.62}=1.2462\ \text{m}$$

Critical conditions at the contracted section

$$q_2=\frac{4}{2}=2\ \text{m}^2/\text{s},\quad y_{c2}=\sqrt[3]{\frac{2^2}{9.81}}=0.7415\ \text{m},\quad E_{min2}=1.5y_{c2}=1.1123\ \text{m}$$

Energy available at the contracted section if the upstream depth is unchanged:

$$E_2=E_1-\Delta z=1.2462-0.15=1.0962\ \text{m}<E_{min2}=1.1123\ \text{m}$$

The available energy is less than the minimum needed, so the flow is choked: critical flow occurs in the reduced section and the upstream depth must rise.

Depth at the reduced section

$$y_2=y_{c2}=0.742\ \text{m}$$

Upstream depth

Upstream energy must be $E_1'=E_{min2}+\Delta z=1.1123+0.15=1.2623$ m.

$$y_1'+\frac{q_1^2}{2gy_1'^2}=y_1'+\frac{0.06657}{y_1'^2}=1.2623\;\Rightarrow\;y_1'=1.217\ \text{m}$$

The upstream depth rises from 1.20 m to 1.217 m.

Answer: depth at the reduced section $=0.74$ m (critical); depth upstream $=1.22$ m.