Time-area method with histogram, drawbacks

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.