---
title: "Clark IUH, 128 km2 basin"
subject: "Engineering Hydrology"
chapter: 7
key: "semester_5/engineering_hydrology:iuh-128km2"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/iuh-128km2
site: BCT Engineering
---

# Clark IUH, 128 km2 basin

Engineering Hydrology · Chapter 7: Flow Routing

- Asked in: 2072 Kartik, 8 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/iuh-128km2
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-7

## Question

A basin having 128 km$^2$ of drainage area has 22 hours and 14 hours of concentration time and storage constant respectively. Determine the IUH for this basin if the inter-isochrones area distribution is as below:

| Travel time (hr) | 0-3 | 3-6 | 6-9 | 9-12 | 12-15 | 15-18 | 18-21 | 21-24 | 24-27 |
|---|---|---|---|---|---|---|---|---|---|
| Area (km$^2$) | 2 | 7 | 17 | 25 | 31 | 23 | 14 | 6 | 3 |

## Short answer

Time-area inflows are computed as I=2.78Aᵳ/Δ t and routed through a linear reservoir with C0=0.0968, C2=0.8065. Peak IUH ordinate is 13.70 m³/s per cm at 21 h.

## Answer

#### 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 $U_n = (O_n + O_{n-1})/2$.

| t (h) | Inflow I (m³/s) | IUH ordinate O (m³/s/cm) | 3-h UH (m³/s) |
|---|---|---|---|
| 0 | 0.00 | 0.00 | 0.00 |
| 3 | 1.85 | 0.18 | 0.09 |
| 6 | 6.49 | 0.95 | 0.57 |
| 9 | 15.75 | 2.92 | 1.94 |
| 12 | 23.17 | 6.12 | 4.52 |
| 15 | 28.73 | 9.96 | 8.04 |
| 18 | 21.31 | 12.87 | 11.42 |
| 21 | 12.97 | 13.70 | 13.29 |
| 24 | 5.56 | 12.84 | 13.27 |
| 27 | 2.78 | 11.16 | 12.00 |
| 30 | 0.00 | 9.27 | 10.22 |
| 33 | 0.00 | 7.48 | 8.37 |
| 36 | 0.00 | 6.03 | 6.75 |
| 39 | 0.00 | 4.86 | 5.45 |
| 42 | 0.00 | 3.92 | 4.39 |
| 45 | 0.00 | 3.16 | 3.54 |
| 48 | 0.00 | 2.55 | 2.86 |
| 51 | 0.00 | 2.06 | 2.30 |
| 54 | 0.00 | 1.66 | 1.86 |
| 57 | 0.00 | 1.34 | 1.50 |
| 60 | 0.00 | 1.08 | 1.21 |
| 63 | 0.00 | 0.87 | 0.97 |
| 66 | 0.00 | 0.70 | 0.79 |
| 69 | 0.00 | 0.57 | 0.63 |

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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Although t_c is 22 h, which strips are used?
   A. Only first seven (21 h)
   B. All nine 3-h strips (27 h)
   C. Only the largest
   D. None
   Answer: B. The table extends to 27 h. (Key point: Strips)

2. For K = 14 h, dt = 3 h, C0 and C2 are:
   A. 0.0968 and 0.8065
   B. 0.0909 and 0.8182
   C. 0.1 and 0.9
   D. 0.0476 and 0.9048
   Answer: A. C0 = 3/31, C2 = 25/31. (Key point: Coefficients)

3. Inflow from the 31 km^2 strip is:
   A. 31/3 = 10.3
   B. 2.78 x 31 x 3
   C. 31 x 14
   D. 2.78 x 31/3 = 28.73 m^3/s
   Answer: D. I = 2.78 A/dt. (Key point: Inflow)

4. IUH peak is:
   A. 28.73 at 15 h
   B. 12.87 at 18 h
   C. 13.70 m^3/s per cm at 21 h
   D. 13.29 at 21 h as IUH
   Answer: C. O peaks at 13.70 at t = 21 h. (Key point: Peak)

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