Flow Routing

Flow routing is the mathematical procedure for finding the time and magnitude of a flood wave (the hydrograph) at a downstream point of a river or at the outlet of a reservoir, when the hydrograph at an upstream point (or the inflow to the reservoir) is known. As the flood moves through a reach or reservoir, storage changes its shape, so the outflow hydrograph differs from the inflow hydrograph.

Basis

Routing combines two relations:

• Continuity equation: $I - O = \dfrac{dS}{dt}$
• A storage relation: $S = f(I, O)$, for example $S = KO$ for a reservoir, or $S = K[xI + (1-x)O]$ for a river reach (Muskingum).

Effect of routing

 Q
 |      /\
 |     /  \     inflow
 |    /    \  __
 |   /   __/\/  \__   outflow
 |  /  _/         \__
 | /__/               \___
 +-------------------------> t
   lower, later, flatter peak

• Attenuation: the peak outflow is smaller than the peak inflow.
• Lag (translation): the peak outflow occurs later than the peak inflow.

Types

1. Reservoir (hydrologic storage) routing: outflow depends on storage only, e.g. Modified Puls (storage-indication) method.
2. Channel routing: storage depends on both inflow and outflow, e.g. Muskingum method.
3. Hydraulic routing: solves the Saint-Venant equations.

Uses

Flood forecasting, design of spillways and reservoir capacity, flood-control works, and flood-warning systems.

Data and method

Area of basin $A = 123\ \text{km}^2$ (sum of the inter-isochrone areas), $K = 10$ h, $\Delta t = 2$ h. The table has 9 strips of 2 h each (total 18 h), and all strips are used. Although $t_c$ is given as 14 h, the table extends to 18 h, so every strip of the table is used.

Step 1: Translation (time-area) inflow. For 1 cm of excess rain over each strip, the inflow to the imaginary linear reservoir is

$$I = \frac{2.78\,A_i}{\Delta t}\ \text{m}^3/\text{s per cm}$$

Step 2: Storage (linear reservoir) routing. With $S = KQ$ and the continuity equation, the outflow is

$$O_2 = C_0 (I_1 + I_2) + C_2 O_1,\qquad C_0 = \frac{\Delta t}{2K+\Delta t},\quad C_2 = \frac{2K-\Delta t}{2K+\Delta t}$$ $$\begin{aligned} C_0 &= \frac{2}{2(10)+2} = 0.0909\\ C_2 &= \frac{2(10)-2}{2(10)+2} = 0.8182\end{aligned}$$

Check: $C_0 + C_0 + C_2 = 1.0000$ (must equal 1).

Computation

Inflow ordinates are taken at the end of each strip (time $= i\Delta t$); inflow at $t=0$ and outflow at $t=0$ are zero. Column $O$ is the IUH ordinate (routed outflow per cm), and the $2$-h unit hydrograph ordinate is $U_n = (O_n + O_{n-1})/2$.

Beyond the last row the hydrograph is a recession, with each ordinate equal to $C_2 = 0.8182$ times the previous one (the inflow is zero after $t = 18$ h).

Flood hydrograph

The tabulated outflow is the flood hydrograph produced by 1 cm of excess rain: the IUH (column $O$) and the 2-h unit hydrograph (last column). The IUH peaks at 17.20 m³/s per cm at t = 14 h, and the 2-h unit hydrograph peaks at 16.96 m³/s at t = 16 h. For any storm, multiply the ordinates by the excess rainfall depth in cm (and superpose for successive blocks).

Data and method

Area of basin $A = 120\ \text{km}^2$ (sum of the inter-isochrone areas), $K = 10$ h, $\Delta t = 2$ h. The table has 9 strips of 2 h each (total 18 h) and all are used, although $t_c$ is stated as 14 h.

Step 1: Translation (time-area) inflow. For 1 cm of excess rain over each strip, the inflow to the imaginary linear reservoir is

$$I = \frac{2.78\,A_i}{\Delta t}\ \text{m}^3/\text{s per cm}$$

Step 2: Storage (linear reservoir) routing. With $S = KQ$ and the continuity equation, the outflow is

$$O_2 = C_0 (I_1 + I_2) + C_2 O_1,\qquad C_0 = \frac{\Delta t}{2K+\Delta t},\quad C_2 = \frac{2K-\Delta t}{2K+\Delta t}$$ $$\begin{aligned} C_0 &= \frac{2}{2(10)+2} = 0.0909\\ C_2 &= \frac{2(10)-2}{2(10)+2} = 0.8182\end{aligned}$$

Check: $C_0 + C_0 + C_2 = 1.0000$ (must equal 1).

Computation

Inflow ordinates are taken at the end of each strip (time $= i\Delta t$); inflow at $t=0$ and outflow at $t=0$ are zero. Column $O$ is the IUH ordinate (routed outflow per cm), and the $2$-h unit hydrograph ordinate is $U_n = (O_n + O_{n-1})/2$.

Beyond the last row the hydrograph is a recession, with each ordinate equal to $C_2 = 0.8182$ times the previous one (the inflow is zero after $t = 18$ h).

Flood hydrograph

The tabulated outflow is the flood hydrograph produced by 1 cm of excess rain: the IUH (column $O$) and the 2-h unit hydrograph (last column). The IUH peaks at 16.96 m³/s per cm at t = 14 h, and the 2-h unit hydrograph peaks at 16.74 m³/s at t = 16 h. Multiply by the excess rainfall (cm) for an actual storm.

When a flood wave passes through a reach or reservoir, its shape changes. The hydrograph is flattened and delayed.

Concept

• Attenuation of peak: the peak outflow is lower than the peak inflow, $\Delta Q_p = I_{max} - O_{max}$. During the rising limb, part of the inflow fills the storage ($I > O$), so the outflow is smaller than the inflow. The stored water drains later, so the falling limb is stretched out.
• Lag of peak: the peak outflow occurs later than the peak inflow, $t_{lag} = t_{p,out} - t_{p,in}$.
• The peak outflow always occurs where the outflow curve crosses the inflow curve, because there $dS/dt = I - O = 0$ (storage is maximum).

Sketch

 Q
 |      inflow
 |       /\
 |      /  \      outflow
 |     /   /‾‾\
 |    /   /    \
 |   /   /      \__
 |  /___/__________\___
 +--------------------------> t
     t_pI  t_pO
     |<--lag-->|
 Attenuation = I_peak - O_peak

Why it occurs

• In a reservoir, the water surface is level, so storage is a function of outflow alone and the peak is cut by surcharge storage.
• In a river reach, the wedge and prism storage (Muskingum) delays the flood, and friction and floodplain storage flatten it. The wave peak decreases with distance travelled.

Attenuation and lag increase with the storage constant $K$ and are important in design of spillways, flood-control reservoirs, and in forecasting arrival times of a flood downstream.

Clark's method derives a unit hydrograph by passing the excess rainfall through two effects separately: translation (movement of water to the outlet, represented by a time-area curve) and attenuation/storage (represented by a linear reservoir at the outlet).

Data needed

Catchment area $A$, time of concentration $t_c$, storage constant $K$, and the isochrone map (areas between lines of equal travel time to the outlet).

Procedure

1. Delineate isochrones. Divide the basin into zones by lines of equal travel time to the outlet at interval $\Delta t$ (up to $t_c$). Find the area $A_i$ between successive isochrones by planimeter.
2. Time-area curve. Plot the cumulative area against travel time (or the area histogram).
3. Translation hydrograph. For 1 cm of excess rain, the inflow from each strip to the outlet reservoir is

$$I_i = \frac{2.78\,A_i}{\Delta t}\ \text{m}^3/\text{s}\quad (A_i \text{ in km}^2,\ \Delta t \text{ in h})$$

4. Find $K$. Estimate the storage constant $K$ from the recession limb of an observed hydrograph, $K = -Q/(dQ/dt)$ at the inflexion point, or from regional relations.
5. Route through a linear reservoir $S = KO$:

$$O_2 = C_0 (I_1 + I_2) + C_2 O_1,\quad C_0 = \frac{\Delta t}{2K + \Delta t},\quad C_2 = \frac{2K - \Delta t}{2K + \Delta t}$$

The outflow ordinates form the instantaneous unit hydrograph (IUH) of the catchment. 6. Unit hydrograph of duration $\Delta t$. Take the average of two successive IUH ordinates:

$$U_n = \frac{O_n + O_{n-1}}{2}$$

This is the $\Delta t$-hour unit hydrograph, i.e. the response to 1 cm of excess rain in $\Delta t$ hours. 7. Use. Multiply by excess rainfall depths and superpose to get the flood hydrograph for a storm.

Flow chart

 Isochrone map -> area per strip
        |
        v
 Inflow I = 2.78 A / dt
        |
        v
 Route through linear reservoir (K)
        |
        v
 IUH ordinates O
        |
        v
 dt-hour UH = average of two IUH ordinates

Data and method

Area of basin $A = 182\ \text{km}^2$ (sum of the inter-isochrone areas), $K = 10$ h, $\Delta t = 1$ h. The strips add up to 182 km² (the stated area is 180 km²); the tabulated strip areas are used as given, the 1 % difference being negligible.

Step 1: Translation (time-area) inflow. For 1 cm of excess rain over each strip, the inflow to the imaginary linear reservoir is

$$I = \frac{2.78\,A_i}{\Delta t}\ \text{m}^3/\text{s per cm}$$

Step 2: Storage (linear reservoir) routing. With $S = KQ$ and the continuity equation, the outflow is

$$O_2 = C_0 (I_1 + I_2) + C_2 O_1,\qquad C_0 = \frac{\Delta t}{2K+\Delta t},\quad C_2 = \frac{2K-\Delta t}{2K+\Delta t}$$ $$\begin{aligned} C_0 &= \frac{1}{2(10)+1} = 0.0476\\ C_2 &= \frac{2(10)-1}{2(10)+1} = 0.9048\end{aligned}$$

Check: $C_0 + C_0 + C_2 = 1.0000$ (must equal 1).

Computation

Inflow ordinates are taken at the end of each strip (time $= i\Delta t$); inflow at $t=0$ and outflow at $t=0$ are zero. Column $O$ is the IUH ordinate (routed outflow per cm), and the $1$-h unit hydrograph ordinate is $U_n = (O_n + O_{n-1})/2$.

Beyond the last row the hydrograph is a recession, with each ordinate equal to $C_2 = 0.9048$ times the previous one (the inflow is zero after $t = 8$ h).

Result

The IUH of the catchment is given by the column $O$. Peak IUH ordinate = 34.43 m³/s per cm at t = 7 h. The 1-h unit hydrograph (last column) peaks at 34.05 m³/s at t = 7 h.

Method

Muskingum routing: $O_2 = C_0 I_2 + C_1 I_1 + C_2 O_1$, where

$$C_0 = \frac{-Kx + 0.5\Delta t}{K - Kx + 0.5\Delta t},\quad C_1 = \frac{Kx + 0.5\Delta t}{K - Kx + 0.5\Delta t},\quad C_2 = \frac{K - Kx - 0.5\Delta t}{K - Kx + 0.5\Delta t}$$

Here $K = 10$ h, $x = 0.2$, $\Delta t = 5$ h, so $K - Kx + 0.5\Delta t = 10.5$ h.

$$\begin{aligned} C_0 &= \frac{-(10)(0.2) + 2.5}{10.5} = 0.0476\\ C_1 &= \frac{(10)(0.2) + 2.5}{10.5} = 0.4286\\ C_2 &= \frac{8 - 2.5}{10.5} = 0.5238\end{aligned}$$

Check: $C_0 + C_1 + C_2 = 1.0000$.

Routing table

Peak attenuation and lag

• Peak inflow = 45 m³/s at t = 15 h
• Peak outflow = 35.00 m³/s at t = 25 h
• Attenuation of peak = 45 − 35.00 = 10.00 m³/s (22.2 % of the inflow peak)
• Lag of peak = 25 − 15 = 10 h (to the nearest time step)

Data and method

Area of basin $A = 110\ \text{km}^2$ (sum of the inter-isochrone areas), $K = 12$ h, $\Delta t = 2$ h. The table has 9 strips of 2 h each (total 18 h = $t_c$).

Step 1: Translation (time-area) inflow. For 1 cm of excess rain over each strip, the inflow to the imaginary linear reservoir is

$$I = \frac{2.78\,A_i}{\Delta t}\ \text{m}^3/\text{s per cm}$$

Step 2: Storage (linear reservoir) routing. With $S = KQ$ and the continuity equation, the outflow is

$$O_2 = C_0 (I_1 + I_2) + C_2 O_1,\qquad C_0 = \frac{\Delta t}{2K+\Delta t},\quad C_2 = \frac{2K-\Delta t}{2K+\Delta t}$$ $$\begin{aligned} C_0 &= \frac{2}{2(12)+2} = 0.0769\\ C_2 &= \frac{2(12)-2}{2(12)+2} = 0.8462\end{aligned}$$

Check: $C_0 + C_0 + C_2 = 1.0000$ (must equal 1).

Computation

Inflow ordinates are taken at the end of each strip (time $= i\Delta t$); inflow at $t=0$ and outflow at $t=0$ are zero. Column $O$ is the IUH ordinate (routed outflow per cm), and the $2$-h unit hydrograph ordinate is $U_n = (O_n + O_{n-1})/2$.

Beyond the last row the hydrograph is a recession, with each ordinate equal to $C_2 = 0.8462$ times the previous one (the inflow is zero after $t = 18$ h).

Clark's 2-h IUH (2-h unit hydrograph)

The 2-h unit hydrograph is the last column of the table, obtained by averaging two successive IUH ordinates. Its peak is 13.85 m³/s at t = 16 h; the IUH peak is 13.96 m³/s per cm at t = 14 h.

Method

Muskingum routing: $O_2 = C_0 I_2 + C_1 I_1 + C_2 O_1$, where

$$C_0 = \frac{-Kx + 0.5\Delta t}{K - Kx + 0.5\Delta t},\quad C_1 = \frac{Kx + 0.5\Delta t}{K - Kx + 0.5\Delta t},\quad C_2 = \frac{K - Kx - 0.5\Delta t}{K - Kx + 0.5\Delta t}$$

Here $K = 12$ h, $x = 0.2$, $\Delta t = 6$ h, so $K - Kx + 0.5\Delta t = 12.6$ h.

$$\begin{aligned} C_0 &= \frac{-(12)(0.2) + 3}{12.6} = 0.0476\\ C_1 &= \frac{(12)(0.2) + 3}{12.6} = 0.4286\\ C_2 &= \frac{9.6 - 3}{12.6} = 0.5238\end{aligned}$$

Check: $C_0 + C_1 + C_2 = 1.0000$.

Routing table

Peak attenuation and lag

• Peak inflow = 60 m³/s at t = 18 h
• Peak outflow = 49.58 m³/s at t = 30 h
• Attenuation of peak = 60 − 49.58 = 10.42 m³/s (17.4 % of the inflow peak)
• Lag of peak = 30 − 18 = 12 h (to the nearest time step)

Data and method

Area of basin $A = 172\ \text{km}^2$ (sum of the inter-isochrone areas), $K = 10$ h, $\Delta t = 1$ h. The strips add up to 172 km², equal to the given area.

Step 1: Translation (time-area) inflow. For 1 cm of excess rain over each strip, the inflow to the imaginary linear reservoir is

$$I = \frac{2.78\,A_i}{\Delta t}\ \text{m}^3/\text{s per cm}$$

Step 2: Storage (linear reservoir) routing. With $S = KQ$ and the continuity equation, the outflow is

$$O_2 = C_0 (I_1 + I_2) + C_2 O_1,\qquad C_0 = \frac{\Delta t}{2K+\Delta t},\quad C_2 = \frac{2K-\Delta t}{2K+\Delta t}$$ $$\begin{aligned} C_0 &= \frac{1}{2(10)+1} = 0.0476\\ C_2 &= \frac{2(10)-1}{2(10)+1} = 0.9048\end{aligned}$$

Check: $C_0 + C_0 + C_2 = 1.0000$ (must equal 1).

Computation

Inflow ordinates are taken at the end of each strip (time $= i\Delta t$); inflow at $t=0$ and outflow at $t=0$ are zero. Column $O$ is the IUH ordinate (routed outflow per cm), and the $1$-h unit hydrograph ordinate is $U_n = (O_n + O_{n-1})/2$.

Beyond the last row the hydrograph is a recession, with each ordinate equal to $C_2 = 0.9048$ times the previous one (the inflow is zero after $t = 8$ h).

Result

The IUH of the catchment is given by the column $O$. Peak IUH ordinate = 32.87 m³/s per cm at t = 7 h. The 1-h unit hydrograph (last column) peaks at 32.69 m³/s at t = 7 h.

Clark's method considers two processes in transforming excess precipitation into direct runoff.

1. Translation (movement)

Water from each part of the catchment travels to the outlet without change of shape, only delayed by the travel time. This is represented by the time-area diagram (isochrones). The area between two isochrones contributes a flow $I = 2.78\,A_i/\Delta t$ for 1 cm of excess rain.

How it is achieved: the basin is divided by isochrones at interval $\Delta t$ up to the time of concentration; the strip areas give the translation hydrograph (inflow ordinates).

2. Attenuation (storage)

Storage in the channels, soil surface, and depressions reduces the peaks and spreads the flow in time. This is represented by a single linear reservoir at the outlet, $S = KQ$.

How it is achieved: the translation hydrograph is routed through the linear reservoir using $O_2 = C_0(I_1 + I_2) + C_2 O_1$, with $C_0 = \Delta t/(2K + \Delta t)$ and $C_2 = (2K - \Delta t)/(2K + \Delta t)$. The result is the IUH, and averaging two consecutive ordinates gives the unit hydrograph of duration $\Delta t$.

Use the storage-indication (modified Puls) method. The reservoir is assumed to start full to the spillway crest (elevation 140 m), so $O = 0$ and $S = 0$ at $t = 0$.

Method

Continuity over one interval $\Delta t$: $\dfrac{I_1 + I_2}{2} - \dfrac{O_1 + O_2}{2} = \dfrac{S_2 - S_1}{\Delta t}$, which rearranges to

$$\left(\frac{S_2}{\Delta t} + \frac{O_2}{2}\right) = \frac{I_1 + I_2}{2} + \left(\frac{S_1}{\Delta t} - \frac{O_1}{2}\right)$$

$\Delta t = 6 \times 3600 = 21600$ s.

Step 1: Storage-indication curve

Step 2: Routing

At each step, $\psi_2$ is computed from the right-hand side, then $O_2$ and $S_2$ are found by linear interpolation in the table above ($S_1/\Delta t - O_1/2 = \psi_1 - O_1$).

Results

• Peak inflow = 2130 m³/s at 36 h.
• Peak outflow occurs at t = 54 h (where the outflow curve meets the inflow curve, $O \approx I$).
• Storage at that time $\approx$ 101.8 × 10⁶ m³, which lies between 144 m (95) and 145 m (140).

$$\text{Max. level} = 144 + \frac{101.8 - 95}{140 - 95} = 144.15\ \text{m}$$

Answer:

• i) Maximum reservoir level $\approx$ 144.15 m
• ii) Maximum outflow rate $\approx$ 1442 m³/s
• iii) Reduction in peak $= 2130 - 1442 =$ 688 m³/s (32.3 % of the peak inflow)

Assumption: the inflow table is cut off in the question. The inflow ordinates are assumed to be the standard set for this problem (10, 20, 50, 60, 55, 45, 35, 27, 20, 15 m³/s at 6 h intervals from 0 to 54 h). If your table differs, use the same coefficients and the same steps with your values.

Method

Muskingum routing: $O_2 = C_0 I_2 + C_1 I_1 + C_2 O_1$, where

$$C_0 = \frac{-Kx + 0.5\Delta t}{K - Kx + 0.5\Delta t},\quad C_1 = \frac{Kx + 0.5\Delta t}{K - Kx + 0.5\Delta t},\quad C_2 = \frac{K - Kx - 0.5\Delta t}{K - Kx + 0.5\Delta t}$$

Here $K = 10$ h, $x = 0.2$, $\Delta t = 6$ h, so $K - Kx + 0.5\Delta t = 11$ h.

$$\begin{aligned} C_0 &= \frac{-(10)(0.2) + 3}{11} = 0.0909\\ C_1 &= \frac{(10)(0.2) + 3}{11} = 0.4545\\ C_2 &= \frac{8 - 3}{11} = 0.4545\end{aligned}$$

Check: $C_0 + C_1 + C_2 = 1.0000$.

Routing table

Peak attenuation and lag

• Peak inflow = 60 m³/s at t = 18 h
• Peak outflow = 51.33 m³/s at t = 30 h
• Attenuation of peak = 60 − 51.33 = 8.67 m³/s (14.5 % of the inflow peak)
• Lag of peak = 30 − 18 = 12 h (to the nearest time step)

Data and method

Area of basin $A = 128\ \text{km}^2$ (sum of the inter-isochrone areas), $K = 14$ h, $\Delta t = 3$ h. The table has 9 strips of 3 h (27 h), and all are used although $t_c$ is stated as 22 h.

Step 1: Translation (time-area) inflow. For 1 cm of excess rain over each strip, the inflow to the imaginary linear reservoir is

$$I = \frac{2.78\,A_i}{\Delta t}\ \text{m}^3/\text{s per cm}$$

Step 2: Storage (linear reservoir) routing. With $S = KQ$ and the continuity equation, the outflow is

$$O_2 = C_0 (I_1 + I_2) + C_2 O_1,\qquad C_0 = \frac{\Delta t}{2K+\Delta t},\quad C_2 = \frac{2K-\Delta t}{2K+\Delta t}$$ $$\begin{aligned} C_0 &= \frac{3}{2(14)+3} = 0.0968\\ C_2 &= \frac{2(14)-3}{2(14)+3} = 0.8065\end{aligned}$$

Check: $C_0 + C_0 + C_2 = 1.0000$ (must equal 1).

Computation

Inflow ordinates are taken at the end of each strip (time $= i\Delta t$); inflow at $t=0$ and outflow at $t=0$ are zero. Column $O$ is the IUH ordinate (routed outflow per cm), and the $3$-h unit hydrograph ordinate is $U_n = (O_n + O_{n-1})/2$.

Beyond the last row the hydrograph is a recession, with each ordinate equal to $C_2 = 0.8065$ times the previous one (the inflow is zero after $t = 27$ h).

Result

The IUH of the basin is given by the column $O$. Peak IUH ordinate = 13.70 m³/s per cm at t = 21 h. The 3-h unit hydrograph (last column) peaks at 13.29 m³/s at t = 21 h.

The time-area method estimates the direct runoff hydrograph at the outlet of a catchment from effective rainfall, taking into account only the translation of runoff to the outlet and ignoring storage effects.

Basic idea

• Lines of equal travel time to the outlet (isochrones) are drawn at interval $\Delta t$: $t_1, t_2, \dots, t_c$.
• The area between two consecutive isochrones is $A_1, A_2, \dots, A_N$. Rain falling in $A_1$ reaches the outlet in the first interval, in $A_2$ in the second interval, and so on. The last strip takes time $t_c$ (time of concentration).
• The plot of cumulative area against time (or the histogram of strip areas) is the time-area diagram.

Time-area histogram

 Area
  A3 |      ____
  A2 |  ___|    |___
  A1 | |    |    |   |
     | |    |    |   |___
     +-+----+----+----+---> t
       0   Δt  2Δt 3Δt  tc

Procedure

1. Prepare the isochrone map and find the strip areas $A_1, \dots, A_N$.
2. Convert the effective rainfall hyetograph into blocks $r_1, r_2, \dots, r_M$ (cm or mm) of duration $\Delta t$.
3. The discharge at the end of interval $n$ is the sum of contributions of every rainfall block over its corresponding strip:

$$Q_n = \frac{10^4}{3600\,\Delta t}\sum_{i=1}^{n} r_i\,A_{n-i+1}\quad (A \text{ in km}^2, r \text{ in cm})$$

i.e. $Q_n = 2.78 \sum r_i A_{n-i+1}/\Delta t$ for $r$ in cm and $\Delta t$ in h. For example, $Q_1 = 2.78\,r_1A_1/\Delta t$, $Q_2 = 2.78\,(r_1A_2 + r_2A_1)/\Delta t$, $Q_3 = 2.78\,(r_1A_3 + r_2A_2 + r_3A_1)/\Delta t$. 4. Plot $Q_n$ against time to get the direct runoff hydrograph.

The calculation is a convolution of the rainfall blocks with the area strips (reversed order), and the base flow is added afterwards.

Limitations

• Storage (channel and catchment) is ignored, so peaks are overestimated and the hydrograph is too steep.
• Isochrones are hard to draw accurately and the velocity is assumed constant, independent of flow depth and rainfall intensity.
• Uniform rainfall over each strip is assumed.

Clark improved the method by routing the result through a linear reservoir to include storage.

Consider a reservoir with inflow $I$, outflow $O$ and storage $S$.

Continuity equation

$$I - O = \frac{dS}{dt}$$

Over a short interval $\Delta t$ from time 1 to time 2, use average inflow and outflow:

$$\frac{I_1 + I_2}{2}\Delta t - \frac{O_1 + O_2}{2}\Delta t = S_2 - S_1$$

Divide by $\Delta t$ and multiply by 2:

$$(I_1 + I_2) - (O_1 + O_2) = \frac{2(S_2 - S_1)}{\Delta t}$$

Collect the unknowns ($S_2$, $O_2$) on one side and the knowns on the other:

$$\left(\frac{2S_2}{\Delta t} + O_2\right) = (I_1 + I_2) + \left(\frac{2S_1}{\Delta t} - O_1\right)$$

This is the reservoir (modified Puls) routing equation. Equivalently, with $\psi = S/\Delta t + O/2$:

$$\psi_2 = \frac{I_1 + I_2}{2} + (\psi_1 - O_1)$$

Use

• $I_1, I_2$ are known from the inflow hydrograph, and $S_1, O_1$ are known from the previous step.
• A curve of $O$ against $(2S/\Delta t + O)$ is prepared from the elevation-storage and elevation-outflow data.
• For each step, the right-hand side gives $(2S_2/\Delta t + O_2)$, from which $O_2$ is read from the curve; $S_2$ then follows, and the process is repeated.

Purpose

The time-area method is used to compute the direct runoff hydrograph at a catchment outlet for a known effective (excess) rainfall hyetograph, especially for ungauged catchments (it is also the base of Clark's IUH).

The method

Draw isochrones (lines of equal travel time to the outlet) at interval $\Delta t$. Let the areas between them be $A_1, A_2, \dots, A_N$ (the last at $t_c$). The time-area histogram:

 Area (km2)
  A4 |        ____
  A3 |    ___|    |
  A2 | __|    |    |___
  A1 ||  |    |    |   |
     +-+--+---+----+---+--> t
       0  Δt  2Δt 3Δt 4Δt

Let the effective rainfall be in blocks $r_1, r_2, \dots, r_M$ (cm) of duration $\Delta t$.

• Rain $r_1$ falling on $A_1$ reaches the outlet in the first interval, on $A_2$ in the second interval, etc.
• The discharge at the end of each interval is the sum of the flows from all blocks:

$$Q_n = \frac{2.78}{\Delta t}\sum_{i=1}^{n} r_i\,A_{n-i+1}$$

So:

• $Q_1 = 2.78\,r_1A_1/\Delta t$
• $Q_2 = 2.78\,(r_1A_2 + r_2A_1)/\Delta t$
• $Q_3 = 2.78\,(r_1A_3 + r_2A_2 + r_3A_1)/\Delta t$
• and so on, until all rainfall blocks and strips have been used.

Here $A$ is in km², $r$ in cm and $\Delta t$ in h. Plotting $Q_n$ against $t$ gives the direct runoff hydrograph (base flow is added to get the total flood).

Drawbacks

• It considers only translation and ignores storage in channels and on the catchment, so the computed peak is too high and the hydrograph too sharp.
• Constant velocity is assumed, so travel times do not change with discharge or rainfall intensity.
• Drawing isochrones accurately needs a good topographic map and field knowledge.
• Rainfall and losses are taken to be uniform over each strip.

Linear reservoir

A linear reservoir is a reservoir (or storage) in which the storage is directly proportional to the outflow:

$$S = K\,O$$

where $K$ is the storage constant (units of time). With continuity $I - O = dS/dt$, the response is $K\,dO/dt + O = I$, and its IUH is $u(t) = \dfrac{1}{K}e^{-t/K}$.

Procedure for Clark UH from the time-area method

1. Isochrones and areas. Divide the catchment by isochrones at interval $\Delta t$ up to $t_c$ and find the strip areas $A_1, A_2, \dots, A_N$.
2. Time-area (translation) inflow. For 1 cm excess rain:

$$I_i = \frac{2.78\,A_i}{\Delta t}\ \text{m}^3/\text{s}$$

This is the inflow hydrograph to an imaginary linear reservoir at the outlet. 3. Storage constant. Get $K$ from the recession of an observed hydrograph, or by a regional formula. 4. Route through the linear reservoir:

$$O_2 = C_0 (I_1 + I_2) + C_2 O_1,\quad C_0 = \frac{\Delta t}{2K + \Delta t},\quad C_2 = \frac{2K - \Delta t}{2K + \Delta t}$$

starting with $O_1 = 0$. The outflow ordinates are the IUH. 5. $\Delta t$-hour unit hydrograph. The ordinates are the average of successive IUH ordinates, $U_n = (O_n + O_{n-1})/2$. 6. Longer-duration UH. For a unit hydrograph of duration $D$ (a multiple of $\Delta t$), use the S-curve or average successive ordinates.

 Time-area        Linear reservoir      IUH / UH
 inflow I(t) ---> S = K O  ----------> O(t), U(t)
 (translation)    (storage effect)

Data and method

Area of basin $A = 157\ \text{km}^2$ (sum of the inter-isochrone areas), $K = 9.5$ h, $\Delta t = 1$ h. The strips add up to 157 km², equal to the given area.

Step 1: Translation (time-area) inflow. For 1 cm of excess rain over each strip, the inflow to the imaginary linear reservoir is

$$I = \frac{2.78\,A_i}{\Delta t}\ \text{m}^3/\text{s per cm}$$

Step 2: Storage (linear reservoir) routing. With $S = KQ$ and the continuity equation, the outflow is

$$O_2 = C_0 (I_1 + I_2) + C_2 O_1,\qquad C_0 = \frac{\Delta t}{2K+\Delta t},\quad C_2 = \frac{2K-\Delta t}{2K+\Delta t}$$ $$\begin{aligned} C_0 &= \frac{1}{2(9.5)+1} = 0.0500\\ C_2 &= \frac{2(9.5)-1}{2(9.5)+1} = 0.9000\end{aligned}$$

Check: $C_0 + C_0 + C_2 = 1.0000$ (must equal 1).

Computation

Inflow ordinates are taken at the end of each strip (time $= i\Delta t$); inflow at $t=0$ and outflow at $t=0$ are zero. Column $O$ is the IUH ordinate (routed outflow per cm), and the $1$-h unit hydrograph ordinate is $U_n = (O_n + O_{n-1})/2$.

Beyond the last row the hydrograph is a recession, with each ordinate equal to $C_2 = 0.9000$ times the previous one (the inflow is zero after $t = 7$ h).

Result

The IUH of the catchment is given by the column $O$. Peak IUH ordinate = 33.86 m³/s per cm at t = 6 h. The 1-h unit hydrograph (last column) peaks at 33.01 m³/s at t = 7 h.

Flood routing is the procedure for determining how the hydrograph of a flood changes in shape, peak, and timing as it passes through a river reach or reservoir. It uses continuity ($I - O = dS/dt$) and a storage-discharge relation.

Clark IUH method

Clark's method transforms excess rainfall into a catchment outlet hydrograph, using two steps.

1. Translation: isochrones divide the catchment into strips of area $A_i$ between lines of equal travel time. For 1 cm of excess rain, the inflow to the outlet is $I_i = 2.78\,A_i/\Delta t$ (m³/s).
2. Storage: this inflow is routed through a single linear reservoir at the outlet ($S = KO$):

$$O_2 = C_0 (I_1 + I_2) + C_2 O_1,\quad C_0 = \frac{\Delta t}{2K + \Delta t},\quad C_2 = \frac{2K - \Delta t}{2K + \Delta t}$$

The outflow ordinates give the instantaneous unit hydrograph (IUH), and the average of two consecutive ordinates gives the $\Delta t$-hour unit hydrograph. Using the excess rainfall depths and superposition, the flood hydrograph is obtained. The parameters needed are $t_c$, $K$ and the isochrone areas.