Surface Runoff

Most rivers in Nepal are ungauged, so the Department of Hydrology and Meteorology (DHM), WECS and project agencies have developed regional methods to estimate monthly flow from rainfall and catchment data.

1. MIP (Medium Irrigation Project) method

Developed in 1990 under the Medium Irrigation Project for catchments without flow records.

1. Delineate the catchment above the site and measure its area ($A$, km$^2$) from topographic maps.
2. Find the monsoon wetness index (MWI) of the catchment from the Nepal-wide MWI map (isolines based on monsoon rainfall).
3. Using the regression curves or equations, relating monthly specific discharge to MWI and area for each month of the year, read the flow of each month.
4. Multiply by the area to get the monthly discharge, and plot the 12 values as a flow series (hydrograph) for the site. It is simple, but gives only mean monthly flows and is suited to small and medium catchments below about 3000 m elevation.

2. WECS/DHM method (1990)

A regional approach using relationships between mean monthly flow and catchment characteristics (area, mean elevation, rainfall) for the various physiographic regions of Nepal, to be used in medium-sized basins.

3. Hydest (DHM, 2004)

Computer-based regional model. From the catchment area, mean elevation and the rainfall index, it gives mean monthly flows, flood estimates and low-flow estimates for ungauged rivers.

4. Catchment-area ratio (transposition) method

If a gauged station exists on a nearby similar river:

$$Q_{ungauged} = Q_{gauged}\left(\frac{A_{ungauged}}{A_{gauged}}\right)^{n}$$

with $n$ close to 1 (0.8 to 1.0), modified by the rainfall ratio of the basins.

5. Water-balance and rainfall-runoff coefficient method

Monthly runoff is found from rainfall and losses:

$$Q = P - E - \Delta S$$

with runoff coefficients for the catchment type, soil, land use and snow/glacier melt contribution in the high Himalaya.

The accuracy of these methods depends on the quality of the rainfall network; estimates should be checked with a few spot discharge measurements at the site.

A stream gauging station should be placed where the stage-discharge relation is stable and flow measurements are easy and accurate. Factors considered:

1. Straight reach: the channel should be straight and uniform for about 100 m upstream and downstream of the site, free from bends and sudden changes in cross-section.
2. Stable channel: the bed and banks should be stable (rock or hard material), with no scour, silting or shifting, so that the rating curve remains constant.
3. Control: a good natural control (rock bar, rapid or riffle) just downstream gives a sensitive and permanent stage-discharge relation. The site should not be affected by backwater from a confluence, dam, bridge or tide.
4. Flow conditions: velocity should be moderate, with no excessive turbulence, eddies, dead water or cross currents, and parallel flow lines.
5. Flood level: banks should be high and well defined, so that the gauge records the whole range of stages and flood discharge can be measured without overtopping.
6. Accessibility: the site should be reachable in all seasons for observers and equipment, near roads or villages, and with easy access to measure both low and high flows.
7. Location in the basin: near the point needed for the project, and not close to a tributary junction, so that the gauged flow represents the drainage area.
8. Suitable measuring section: a bridge, cableway or boat facility available for current meter work; gauge installation should be easy, and the cable/bridge should be safe.
9. Safety and security: protected from vandalism, floating debris, ice and erosion.
10. Permanency: the site should remain usable for long-term records, with no future development or change planned in the reach.

The slope-area method is an indirect method used to estimate peak flood discharge in a reach of a natural channel when a current meter cannot be used (high flood, no gauge). It uses the high-water marks left by the flood and the channel geometry, and applies Manning's equation.

Principle

For steady non-uniform flow, the discharge is

$$Q = K\sqrt{S_f},\qquad K = \frac{1}{n}AR^{2/3}$$

where $K$ is the conveyance, $n$ is Manning's roughness, $A$ the flow area, $R$ the hydraulic radius and $S_f$ the friction slope. The friction slope is not equal to the water surface slope when the channel is non-uniform, so it is found from the energy equation between sections 1 (upstream) and 2 (downstream).

  energy line ------\
                     \___ h_f
  water surface ---\   \____
                    \_ fall h
  bed  ------------------------
     [1]<----- L ----->[2]

$$h_f = (h_1 - h_2) + \left(\frac{\alpha_1 v_1^2}{2g} - \frac{\alpha_2 v_2^2}{2g}\right) - h_e$$ $$h_e = k_e\left|\frac{\alpha_1 v_1^2}{2g} - \frac{\alpha_2 v_2^2}{2g}\right|$$

where $h_1 - h_2$ is the fall of water surface, $\alpha$ the velocity-head coefficient and $k_e$ the eddy loss coefficient (0.3 for a gradual expansion, 0.1 for a gradual contraction).

Procedure

1. Select a reach that is straight, uniform, free of large boulders, bends, and with well-defined high-water marks on both banks; take 2 or more cross-sections normal to the flow.
2. Survey the cross-sections and the high-water mark levels, and measure the reach length $L$.
3. Compute $A$, $P$, $R$ for each section and estimate $n$ from the bed material and vegetation.
4. Compute $K_1, K_2$ and the mean conveyance $K_m=\sqrt{K_1K_2}$.
5. Assume $S_f = (h_1-h_2)/L$ and find a first estimate $Q = K_m\sqrt{S_f}$.
6. Find $v_1=Q/A_1$, $v_2=Q/A_2$, the velocity heads and $h_e$. Compute $h_f$, then a new $Q=K_m\sqrt{h_f/L}$.
7. Repeat until $Q$ no longer changes.

Merits and limits

• Needs no flow measurement during the flood; cheap and quick.
• Accuracy depends mostly on $n$ and the quality of high-water marks; error can be 10 to 25 %.
• Applies only to steady flow in a stable reach. It is used to extend the rating curve at peak stage.

Data: $A = 140$ km$^2$, $L = 21$ km, $h = 1090$ m, total channel length $L_s = 654$ km.

Drainage density

$$D_d = \frac{L_s}{A} = \frac{654}{140} = 4.67\ \text{km/km}^2$$

Average length of overland flow

$$L_o = \frac{1}{2D_d} = \frac{1}{2\times4.671} = 0.107\ \text{km} = 107\ \text{m}$$

Form factor

$$R_f = \frac{A}{L^2} = \frac{140}{21^2} = \frac{140}{441} = 0.317$$

Channel slope

$$S = \frac{h}{L} = \frac{1090}{21\,000} = 0.0519 \;(5.19\ \%) \approx 51.9\ \text{m/km}$$

Answer: $D_d = 4.67$ km/km$^2$; $L_o = 0.107$ km; form factor $= 0.317$; channel slope $= 0.052$ (5.2 %).

A high drainage density and low form factor show a well-drained, elongated, steep basin with fast runoff and lower, flatter peak flows than a circular basin.

Definition

A rating curve (stage-discharge curve) is the graph or equation that gives the relation between the stage (water level) and the discharge at a gauging section of a river. It is prepared from simultaneous measurements of stage and discharge, and is usually of the form

$$Q = C(G - a)^b$$

where $G$ is the gauge height, $a$ is the gauge reading for zero discharge, and $C$, $b$ are constants.

  Stage
   |             .'
   |          .'
   |       .'     <- measured points
   |    .'
   |__.'________________ Discharge
   a (zero flow stage)

Uses in hydrology

1. Conversion of continuous stage records into discharge: stage is easy to record continuously, while discharge cannot be measured every day. The rating curve converts the daily stage record to a flow series (hydrograph).
2. Flow-duration and flood-frequency studies: long discharge series are needed for design of hydropower, irrigation and water supply.
3. Flood forecasting and warning: the stage at a gauge can be converted quickly into discharge.
4. Estimating discharge for high flows by extrapolation beyond the measured range.
5. Checking changes in the channel: a shift of the curve shows scour, deposition or vegetation change at the section.
6. Design of structures such as weirs, bridges and intakes needing discharge-stage information.
7. Water-resources management and allocation, where daily flows are required.

The hydro-geomorphological characteristics of a river describe its flow regime and the form of its channel and valley, which result from the interaction of flow, sediment and geology.

1. Longitudinal profile (course of the river)

 elev.
  |\ upper course (steep)
  | \
  |  \__ middle course
  |      \____
  |           \______ lower course (flat)
  +------------------------- distance

• Upper (youth) stage: steep slope, V-shaped valley, rapids and waterfalls, high velocity, vertical erosion, coarse bed material.
• Middle (mature) stage: moderate slope, valley widens, lateral erosion, meanders begin; transport of sediment is the main process.
• Lower (old) stage: very flat, wide floodplain, meanders and oxbow lakes, deposition of fine sediment, delta at the mouth.

2. Channel pattern

 Straight        Meandering        Braided
 ========        ~~~~~~~~~~        =\\/=\\/=
                                    /\\ /\\

• Straight: rare and short; low sinuosity (< 1.1).
• Meandering: sinuosity > 1.5; erosion on the outer bank, deposition on the inner bank (point bars); forms cut-offs and oxbow lakes.
• Braided: many channels separated by bars, with steep slope, coarse and abundant sediment and wide fluctuating flow; typical of Himalayan rivers in the Terai (Koshi, Narayani).
• Anastomosing: several stable channels with vegetated islands.

3. Cross-section

 floodplain   bank          bank    floodplain
 ____       __/              \__       ____
     \_____/   \____channel___/   \_____/

Parts: main channel (bank-full discharge), floodplain, levees and terraces. The width-depth ratio is large in braided rivers and small in meandering rivers.

4. Other characteristics

• Stream order and drainage pattern (dendritic, trellis, radial).
• Slope and sinuosity: $S = \Delta h/L$; sinuosity = channel length / valley length.
• Bed material and sediment load: bed load (boulders, gravel), suspended load (silt, clay) and dissolved load.
• Flow regime: perennial (snow- and glacier-fed, e.g. Koshi, Gandaki, Karnali), seasonal (rain-fed, Mahabharat-origin) and flashy rivers (Siwalik-origin) with high floods in monsoon and very low flow in dry season.
• Dynamic equilibrium: rivers adjust width, depth, slope and pattern to the discharge and sediment supplied. Changes in these (e.g. by dams or land-use change) cause aggradation or degradation.

Given: $A_1=3500$ m$^2$, $P_1=650$ m, $\alpha_1=1.17$; $A_2=3250$ m$^2$, $P_2=621$ m, $\alpha_2=1.21$; fall $=0.4$ m, $L=1300$ m, $n=0.03$. No eddy-loss coefficient is given, so $h_e=0$ is assumed.

Method: Manning's equation gives conveyance $K=\dfrac{1}{n}AR^{2/3}$ and $Q = K\sqrt{S_f}$. The friction slope is found from the energy equation between the two sections, so the discharge is found by trial.

Step 1: Conveyance

$$R_1=\frac{3500}{650}=5.3846\ \text{m},\qquad R_2=\frac{3250}{621}=5.2335\ \text{m}$$ $$K_1=\frac{1}{0.03}\times3500\times5.3846^{2/3}=358412.6,\qquad K_2=\frac{1}{0.03}\times3250\times5.2335^{2/3}=326555.2$$ $$K_m=\sqrt{K_1K_2}=342113.3$$

Step 2: First trial (no velocity head)

$$S_f=\frac{0.4}{1300},\qquad Q_1=K_m\sqrt{S_f}=6001.064\ \text{m}^3/\text{s}$$

Step 3: Correct for velocity head and eddy loss

$$h_f=(h_1-h_2)+\left(\frac{\alpha_1v_1^2}{2g}-\frac{\alpha_2v_2^2}{2g}\right)-h_e,\qquad h_e=k_e\left|\frac{\alpha_1v_1^2}{2g}-\frac{\alpha_2v_2^2}{2g}\right|$$

With $h_e=0$: $h_f = 0.4 + (h_{v1}-h_{v2})$.

The trials converge to $h_f$ and $Q$ shown in the last row.

Answer: Discharge $Q \approx 5754.8\ \text{m}^3/\text{s}$.

(If a contraction loss of $k_e=0.1$ were included, $Q$ would be about 5732 m$^3$/s, a change of 0.4 %.)

Use the three-point method on the smooth curve with $Q_2 = \sqrt{Q_1Q_3}$.

Take the lowest and highest points: $(G_1,Q_1)=(20.80,100)$ and $(G_3,Q_3)=(24.00,1000)$.

$$Q_2=\sqrt{100\times1000}=316.2\ \text{m}^3/\text{s}$$

Reading $G_2$ from the curve by interpolation between $(21.95, 300)$ and $(22.37, 400)$:

$$G_2 = 21.95 + \frac{316.2-300}{400-300}\times(22.37-21.95) = 22.018\ \text{m}$$ $$a=\frac{G_1G_3-G_2^2}{G_1+G_3-2G_2}=\frac{20.80\times24.00-22.018^2}{20.80+24.00-2\times22.018}=\frac{499.20-484.79}{0.764}=18.86\ \text{m}$$

Check by trial on the log-log plot (least-squares fit of $\log Q$ against $\log(G-a)$ for trial $a$):

The straightest line occurs at $a \approx 18.9$ m, which agrees with the three-point result.

Answer: Stage for zero discharge $\approx 18.9$ m.

(The result is sensitive to the point selection, so the trial plot is preferred as a check.)

The mid-section method is used with $v = 0.035 + 0.74N$, where $N$ = revolutions / time (rps). The depths are small (max 1.7 m), so the meter reading is taken as the mean velocity of the vertical.

$$q_i = v_i\,d_i\,\frac{b_{i+1}-b_{i-1}}{2}$$

Total area $= 6.81$ m$^2$; mean velocity $= 0.882$ m/s.

Answer: Stream flow $Q \approx 6.01\ \text{m}^3/\text{s}$.

The stage column shows 23.37 m between 21.95 m and 23.00 m, which breaks the increasing order. It is taken as a misprint for 22.37 m (Q = 400 m$^3$/s). The last stage is 23.90 m.

Three-point method: choose $Q_1$ and $Q_3$ at the ends, with $Q_2=\sqrt{Q_1Q_3}$:

$$Q_1=100,\; G_1=20.80;\qquad Q_3=1000,\; G_3=23.90;\qquad Q_2=\sqrt{100\times1000}=316.2$$

Interpolating for $G_2$ between $(21.95,300)$ and $(22.37,400)$:

$$G_2=21.95+\frac{316.2-300}{100}(0.42)=22.018\ \text{m}$$ $$a=\frac{G_1G_3-G_2^2}{G_1+G_3-2G_2}=\frac{20.80\times23.90-22.018^2}{20.80+23.90-2\times22.018}=\frac{497.12-484.79}{0.664}=18.57\ \text{m}$$

Check by trial on log-log paper: for the straightest line of $\log Q$ against $\log(G-a)$ the best-fitting trial value is $a\approx18.7$ m (correlation 0.9997).

Answer: Stage for zero discharge $\approx 18.6$ m.

Mid-section method with $v = 0.045 + 0.76N$, $N$ = revolutions per second. The meter reading is taken as the mean velocity of each vertical.

$$q_i = v_i\,d_i\,\frac{b_{i+1}-b_{i-1}}{2}$$

Total area $= 6.81$ m$^2$; mean velocity $= 0.915$ m/s.

Answer: Stream flow $Q \approx 6.23\ \text{m}^3/\text{s}$.

Given: $A_1=45$ m$^2$, $P_1=26$ m, $\alpha_1=1.18$; $A_2=50$ m$^2$, $P_2=28$ m, $\alpha_2=1.23$; fall $=1.15$ m; $L=7000$ m; $n=0.03$. The area increases downstream (expansion), so $k_e=0.3$.

Method: Manning's equation gives conveyance $K=\dfrac{1}{n}AR^{2/3}$ and $Q = K\sqrt{S_f}$. The friction slope is found from the energy equation between the two sections, so the discharge is found by trial.

Step 1: Conveyance

$$R_1=\frac{45}{26}=1.7308\ \text{m},\qquad R_2=\frac{50}{28}=1.7857\ \text{m}$$ $$K_1=\frac{1}{0.03}\times45\times1.7308^{2/3}=2162.3,\qquad K_2=\frac{1}{0.03}\times50\times1.7857^{2/3}=2453.1$$ $$K_m=\sqrt{K_1K_2}=2303.1$$

Step 2: First trial (no velocity head)

$$S_f=\frac{1.15}{7000},\qquad Q_1=K_m\sqrt{S_f}=29.520\ \text{m}^3/\text{s}$$

Step 3: Correct for velocity head and eddy loss

$$h_f=(h_1-h_2)+\left(\frac{\alpha_1v_1^2}{2g}-\frac{\alpha_2v_2^2}{2g}\right)-h_e,\qquad h_e=k_e\left|\frac{\alpha_1v_1^2}{2g}-\frac{\alpha_2v_2^2}{2g}\right|$$

Here $h_{v1}>h_{v2}$ (expansion), so $h_e = 0.3(h_{v1}-h_{v2})$.

The trials converge to $h_f$ and $Q$ shown in the last row.

Answer: Discharge $Q \approx 29.6\ \text{m}^3/\text{s}$.

Development of a rating curve

1. Select a gauging section with a stable control and install a gauge.
2. Measure discharge by current meter (or other method) at different stages covering low to high flow, and record the stage at the same time.
3. Plot stage $G$ (y-axis) against discharge $Q$ (x-axis) on arithmetic scale, and also on log-log paper after subtracting the stage for zero flow $a$.
4. Fit a smooth curve $Q = C(G-a)^b$, with $a$, $C$ and $b$ found by trial or regression.
5. Check by correlation, and extend the curve to flood stage by extrapolation (log-log straight line or conveyance method).
6. Check the rating regularly with new measurements, and revise when the points depart from the curve.

Control

The control is a physical feature of the channel at or downstream of the gauge that governs the stage-discharge relation.

Permanent control: the relation does not change with time. Examples: a rock ledge, a weir or a narrow rock gorge. The same stage always gives the same discharge, and a single rating curve is valid for years.

 water   ~~~~~~~~~\
 level            \___ <- rock ledge / weir (stable control)
 bed  ____________/ \_______

Shifting control: the bed or banks are made of sand or silt, so scour, deposition, vegetation growth or debris change the section and the curve shifts. A given stage then gives different discharges at different times.

  Stage
   |      /  /  /   <- curves shift with time
   |     /  /  /       (scour: right; silting: left)
   |    /  /  /
   +------------------- Q

For shifting control, frequent discharge measurements are needed and the curve is adjusted with time-varying shift corrections (the stage-fall-discharge or a shift curve).

The mid-section method is used. Each vertical represents a strip from the mid-point of the previous vertical to the mid-point of the next vertical:

$$q_i = v_i\,d_i\,\frac{b_{i+1}-b_{i-1}}{2},\qquad Q=\sum q_i$$

The rating equation $v = 0.53N + 0.05$ gives the velocity at 0.6 depth, which is taken as the mean velocity of the vertical ($N=$ revolutions / time). The end points (banks) have zero depth.

Sample calculation at 5 m: $N = 112/150 = 0.747$ rps; $v = 0.53\times0.747+0.05 = 0.446$ m/s; width $=(7-3)/2 = 2$ m; area $=2\times2.6 = 5.2$ m$^2$; $q = 0.446\times5.2 = 2.318$ m$^3$/s.

Total area $= 20.25$ m$^2$ and mean velocity $= 0.381$ m/s.

Answer: Discharge $Q \approx 7.71\ \text{m}^3/\text{s}$.

Given: $A_1=73.3$ m$^2$, $P_1=26.80$ m, $A_2=93.4$ m$^2$, $P_2=30.23$ m, $n=0.02$, $L=10$ km, fall $h_1-h_2 = 306.920-306.650 = 0.27$ m. Velocity-head coefficients are taken as $\alpha_1=\alpha_2=1$. The area increases downstream, so the reach is an expansion and $k_e = 0.3$.

Method: Manning's equation gives conveyance $K=\dfrac{1}{n}AR^{2/3}$ and $Q = K\sqrt{S_f}$. The friction slope is found from the energy equation between the two sections, so the discharge is found by trial.

Step 1: Conveyance

$$R_1=\frac{73.3}{26.8}=2.7351\ \text{m},\qquad R_2=\frac{93.4}{30.23}=3.0896\ \text{m}$$ $$K_1=\frac{1}{0.02}\times73.3\times2.7351^{2/3}=7167.8,\qquad K_2=\frac{1}{0.02}\times93.4\times3.0896^{2/3}=9906.6$$ $$K_m=\sqrt{K_1K_2}=8426.6$$

Step 2: First trial (no velocity head)

$$S_f=\frac{0.27}{10000},\qquad Q_1=K_m\sqrt{S_f}=43.786\ \text{m}^3/\text{s}$$

Step 3: Correct for velocity head and eddy loss

$$h_f=(h_1-h_2)+\left(\frac{\alpha_1v_1^2}{2g}-\frac{\alpha_2v_2^2}{2g}\right)-h_e,\qquad h_e=k_e\left|\frac{\alpha_1v_1^2}{2g}-\frac{\alpha_2v_2^2}{2g}\right|$$

Here $v_2<v_1$ (expansion), so $h_e = 0.3(h_{v1}-h_{v2})$.

The trials converge to $h_f$ and $Q$ shown in the last row.

Answer: Discharge $Q \approx 44.2\ \text{m}^3/\text{s}$.

Given: $A_1=60$ m$^2$, $P_1=18$ m, $A_2=45$ m$^2$, $P_2=14$ m, $L=6$ km, $n=0.025$, $k_e=0.15$. The bed levels differ by 0.45 m; assuming the flow depth is the same at both sections (uniform depth reach), the water-surface fall equals the bed fall $=0.45$ m. $\alpha=1$.

Method: Manning's equation gives conveyance $K=\dfrac{1}{n}AR^{2/3}$ and $Q = K\sqrt{S_f}$. The friction slope is found from the energy equation between the two sections, so the discharge is found by trial.

Step 1: Conveyance

$$R_1=\frac{60}{18}=3.3333\ \text{m},\qquad R_2=\frac{45}{14}=3.2143\ \text{m}$$ $$K_1=\frac{1}{0.025}\times60\times3.3333^{2/3}=5355.5,\qquad K_2=\frac{1}{0.025}\times45\times3.2143^{2/3}=3920.4$$ $$K_m=\sqrt{K_1K_2}=4582.1$$

Step 2: First trial (no velocity head)

$$S_f=\frac{0.45}{6000},\qquad Q_1=K_m\sqrt{S_f}=39.682\ \text{m}^3/\text{s}$$

Step 3: Correct for velocity head and eddy loss

$$h_f=(h_1-h_2)+\left(\frac{\alpha_1v_1^2}{2g}-\frac{\alpha_2v_2^2}{2g}\right)-h_e,\qquad h_e=k_e\left|\frac{\alpha_1v_1^2}{2g}-\frac{\alpha_2v_2^2}{2g}\right|$$

The area decreases downstream (contraction), so $h_e = 0.15(h_{v2}-h_{v1})$.

The trials converge to $h_f$ and $Q$ shown in the last row.

Answer: Discharge $Q \approx 38.8\ \text{m}^3/\text{s}$.

Catchment

A catchment (drainage basin, watershed) is the area of land which drains the surface runoff from precipitation to a common outlet (a river or a point on it) through a connected system of streams. Its boundary is the water divide (ridge line).

Factors affecting runoff from a catchment

1. Climatic factors

• Type of precipitation (rain, snow, hail) and its form.
• Rainfall intensity: high intensity above infiltration capacity gives more runoff.
• Duration and areal distribution of the storm; direction of storm movement relative to the drainage network.
• Antecedent precipitation (soil moisture): wet soil gives higher runoff.
• Evaporation, temperature, wind, and snow-melt.

2. Physiographic (catchment) factors

• Size: larger area gives larger total runoff but lower specific peak, and longer lag time.
• Shape: fan-shaped or circular basins produce high, sharp peaks; elongated basins produce lower, flatter peaks.
• Slope: steep slopes give quick runoff, higher peaks and less infiltration.
• Land use and vegetation: forest and crops intercept and infiltrate water; urbanisation and bare land increase runoff.
• Soil type and geology: permeability of the soil and rock governs infiltration; sandy soil gives low runoff, clay gives high runoff.
• Drainage density and stream pattern: a high density gives quick drainage.
• Topography and elevation, orientation: affect rainfall, snow and temperature.
• Storage: lakes, ponds, depressions, wetlands and reservoirs reduce and delay peaks.

3. Channel factors

• Cross-section, roughness and slope of the channels; presence of check dams and bank storage.

4. Human factors

• Dams, diversions, irrigation, drainage works, and land-use change.

Given: reach length $L=260$ m between the u/s and d/s sections; fall $=315.50-315.15=0.35$ m; contraction coefficient $k_c=0.1$ (area decreasing downstream); $\alpha=1$. Roughness differs at each section ($n = 0.025, 0.027, 0.029$).

Method: Manning's equation gives conveyance $K=\dfrac{1}{n}AR^{2/3}$ and $Q = K\sqrt{S_f}$. The friction slope is found from the energy equation between the two sections, so the discharge is found by trial.

Step 1: Conveyance

$$K_m = 5310.8$$

With three sections of different roughness the conveyance of each section is computed with its own $n$:

The mean conveyance of the reach is taken as the geometric mean of the three sections: $K_m=(K_1K_2K_3)^{1/3}=5310.8$. The middle section has no gauge reading, so the fall is that between the end sections, and the eddy loss is applied between the end sections.

Step 2: First trial (no velocity head)

$$S_f=\frac{0.35}{260},\qquad Q_1=K_m\sqrt{S_f}=194.853\ \text{m}^3/\text{s}$$

Step 3: Correct for velocity head and eddy loss

$$h_f=(h_1-h_2)+\left(\frac{\alpha_1v_1^2}{2g}-\frac{\alpha_2v_2^2}{2g}\right)-h_e,\qquad h_e=k_e\left|\frac{\alpha_1v_1^2}{2g}-\frac{\alpha_2v_2^2}{2g}\right|$$

Velocity heads are taken at the u/s and d/s sections and $h_e=k_c|h_{v1}-h_{v2}|$ with $k_c = 0.1$.

The trials converge to $h_f$ and $Q$ shown in the last row.

Answer: Discharge $Q \approx 185.2\ \text{m}^3/\text{s}$.

Definition

A shifting control is a control (the channel feature that governs the stage-discharge relation) that changes with time, so the stage-discharge relation at the gauging section is not constant. The same stage gives different discharges at different times.

Causes of shifting control

1. Scour and deposition of the alluvial (sand, silt) bed by the flow, the commonest cause: aggradation lowers the discharge for a given stage, degradation raises it.
2. Growth and decay of aquatic vegetation or crops in the channel, which change roughness.
3. Moving bed forms (dunes, ripples) in sandy rivers.
4. Debris, boulders or ice jams at the control.
5. Changes in bank shape due to erosion or collapse.
6. Backwater effects from tributaries, downstream structures or tides.
7. Variable slope (hysteresis) between rising and falling floods; the rising stage gives more discharge than the falling stage for the same stage.
8. Human activity such as sand mining, bridge or weir construction, and dredging.

For such sections frequent discharge measurements are made and the shift is corrected with time.

(i) Rating equation

Use $Q = C(G-a)^b$ with $a = 7.5$ m. Taking logs: $\log Q = \log C + b\log(G-a)$, a straight line $y = \log C + b\,x$ with $x=\log(G-a)$, $y=\log Q$.

With $n=14$: $\sum x=-1.2617$, $\sum y=32.3255$, $\sum x^2=3.2388$, $\sum xy=1.6364$, $\sum y^2=81.3787$.

$$b=\frac{n\sum xy-\sum x\sum y}{n\sum x^2-(\sum x)^2}=1.4558$$ $$\log C=\frac{\sum y-b\sum x}{n}=2.4402\;\Rightarrow\; C=275.5$$

Rating equation: $Q = 275.5\,(G-7.5)^{1.456}$ (m$^3$/s)

Coefficient of correlation

$$r=\frac{n\sum xy-\sum x\sum y}{\sqrt{[n\sum x^2-(\sum x)^2][n\sum y^2-(\sum y)^2]}}=0.9913$$

The high value shows a very good fit to the log-log straight line.

(ii) Discharge at G = 10.5 m

$$Q = 275.5\times(10.5-7.5)^{1.456} = 275.5\times3^{1.456} = 1364\ \text{m}^3/\text{s}$$

Answer: $Q=275.5(G-7.5)^{1.456}$, $r=0.991$; $Q(10.5\ \text{m})\approx 1364$ m$^3$/s (the measured value is 1500 m$^3$/s, so the fit is about 9 % low at this stage).

Mid-section method: the cross-section is divided into strips, each centred on a measuring vertical, the strip extending from the mid-point of the preceding interval to the mid-point of the next.

$$q_i = v_i\,d_i\,\frac{b_{i+1}-b_{i-1}}{2},\qquad Q=\sum q_i$$

Velocity at 0.6 depth is the mean velocity of the vertical: $v = 0.3N + 0.004$ with $N$ = revolutions/second.

Total area $= 25.00$ m$^2$; mean velocity $= 0.335$ m/s.

Answer: Discharge $Q \approx 8.38\ \text{m}^3/\text{s}$.

A current meter (cup type Price meter, or propeller type) measures velocity from the rate of rotation of its rotor: $v = aN + b$. The way it is used depends on the depth and flow condition of the channel.

According to depth (points of velocity measurement in a vertical)

According to flow characteristics

• Shallow, small streams: the engineer wades with a rod-mounted meter (pigmy or Price AA) held at 0.6 d.
• Wide and deep rivers: the meter is suspended from a bridge, cableway or boat with a weight (sounding weight) on a cable; the weight is selected to keep the cable nearly vertical.
• Very low velocity (< 0.15 m/s) or very high turbulent flood: a current meter is unreliable, so floats or other methods are used.
• Sediment-laden or weedy flow: propeller type is cleaner and the rotor is checked before each use.
• Uniform flow: velocity measured at 20 to 30 verticals spaced so that no strip carries more than 5 to 10 % of the discharge.
• Time of observation: at least 40 to 60 s at each point, repeated when velocity fluctuates.
• Meter must be rated (calibrated) in a towing tank to give $v = aN+b$, and held pointing into the flow, away from the observer's body.

A stage-discharge relationship (rating curve) at a gauging site is established as follows.

1. Select the site with a stable, permanent control and a straight reach.
2. Install a gauge (staff gauge or automatic recorder) and note the datum.
3. Measure discharge at various stages, from low to flood, by the area-velocity (current meter) method, and simultaneously record the stage. A minimum of about 15 to 20 measurements covering the whole range is desirable.
4. Plot the stage $G$ against discharge $Q$ on arithmetic paper; draw a smooth curve through the points.
5. Fit the equation $Q = C(G-a)^b$:
   • Find $a$, the stage at zero discharge (by trial, or the three-point method $a=\dfrac{G_1G_3-G_2^2}{G_1+G_3-2G_2}$ with $Q_2=\sqrt{Q_1Q_3}$).
   • Plot $\log Q$ against $\log(G-a)$; a straight line is obtained when $a$ is correct; its slope is $b$ and intercept is $\log C$.
6. Check the correlation and test the curve by new measurements.
7. Extrapolate the curve above the highest measured stage by the log-log straight line or the conveyance (Manning) method.
8. Review periodically: if the points depart from the curve (shifting control), apply shift corrections or draw a new curve.

Area-velocity method: procedure

The discharge at a river section is $Q = \int v\,dA$, found by measuring velocity at many points of the cross-section.

1. Select a site on a straight reach with stable bed, parallel flow and a gauge nearby.
2. Mark the section with a tagged cable, rope or bridge markings. Record the stage.
3. Divide the width into 15 to 30 verticals so that no strip carries more than 10 % of the flow (closer verticals where depth varies strongly).
4. At each vertical, measure the depth (sounding rod or weight) and the velocity with a current meter held at 0.6 d (shallow) or at 0.2 d and 0.8 d (deeper) for 40 to 60 s. The velocity $v = aN + b$ from the revolutions $N$ per second.
5. Compute the mean velocity of each vertical, the area of each strip, and sum the partial discharges.

Mid-section method

Each measured vertical is assumed to represent a strip extending half-way to the verticals on either side.

   water surface
 ___|_____|_____|_____|___
 \  |  d1 |  d2 |  d3 | /
  \ |     |     |     |/
   b(i-1)  b(i)  b(i+1)
        <--w_i-->

Width of strip $i$:

$$w_i = \frac{b_{i+1}-b_{i-1}}{2}$$

Partial discharge:

$$q_i = v_i\,d_i\,w_i = v_i\,d_i\,\frac{b_{i+1}-b_{i-1}}{2}$$

Total discharge:

$$Q = \sum_{i=1}^{n} q_i$$

where $v_i$ is the mean velocity in the vertical and $d_i$ the depth at that vertical. The end verticals at the banks have zero depth. This is the method normally used because it is simple and does not need a separate area calculation.

The stage of zero discharge ($a$) is the gauge height at which the flow in the river becomes zero (the level of the lowest point of the control). It is needed in the rating equation $Q = C(G-a)^b$.

Methods

1. Field survey: survey the level of the lowest point of the channel control (the bed of the riffle or the lowest point of the section) directly. For sand-bed rivers this may change from time to time.

2. Trial and error (log-log plot): assume trial values of $a$, plot $\log Q$ against $\log(G-a)$. The correct $a$ gives a straight line. If the plot curves upward, $a$ is too high; if it curves downward, $a$ is too low.

 log Q
  |     a too low   .
  |             . '  ____ correct a (straight)
  |         .' _ -'
  |     . ' -'
  |   .'   a too high (curves up)
  +------------------- log(G-a)

3. Extrapolation of the arithmetic curve: extend the smooth stage-discharge curve down until it meets the stage axis at $Q=0$. This is the first rough estimate.

4. Three-point (analytical) method: Select three points $(G_1,Q_1)$, $(G_2,Q_2)$, $(G_3,Q_3)$ on the smooth curve with $Q_2^2 = Q_1Q_3$. Then

$$a = \frac{G_1G_3 - G_2^2}{G_1+G_3-2G_2}$$

Procedure: choose the lowest and highest points $Q_1$, $Q_3$; compute $Q_2=\sqrt{Q_1Q_3}$; read $G_2$ on the curve; substitute.

   G ^
 G3 -|         . (Q3)
 G2 -|      .   (Q2 = sqrt(Q1 Q3))
 G1 -|   .       (Q1)
  a -|.___________> Q