---
title: "Derive UH from storm, peak flow from two storms"
subject: "Engineering Hydrology"
chapter: 5
key: "semester_5/engineering_hydrology:uh-from-storm-then-peak-two"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/uh-from-storm-then-peak-two
site: BCT Engineering
---

# Derive UH from storm, peak flow from two storms

Engineering Hydrology · Chapter 5: Hydrograph Analysis

- Asked in: 2070 Asar, 9+5 marks
- Video: [TYPE 1 SUM: FLOOD HYDROGRAPH TO UNIT HYDROGRAPH | FH TO UH NUMERICAL | STEPS & SOLVED EXAMPLE OF UH](https://www.youtube.com/watch?v=9xpjjG9U_xo) — Professor Nidhi Trivedi
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/uh-from-storm-then-peak-two
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-5

## Question

In a storm the rainfall excess of 0.5 cm, 0.0 cm and 0.8 cm occurred in three successive hours. The storm hydrograph due to this storm has the hourly ordinates (Q) as given below: 0.5, 44.5, 110.5, 85.5, 102.8, 94.0, 38.4, 18.6, 10.9, 5.3, 2.9, 0.8 (cumecs). If there is a constant base flow of 0.5 cumecs, find the hourly ordinates of unit hydrograph. If 2 successive storms of 6.5 cm and 10.5 cm of 3 hours duration and $\phi$-index of 0.2 cm/hr occurred in the same catchment, what is the peak flow from the catchment?

## Short answer

Using actual excess blocks0.5,0.7,0,0.8 cm, the1-h UH ordinates (m³/s per cm) are0,88.0,96.8,34.5,15.5,10.4,6.1,2.8,0.3,0 at0–9h. For two3-h storms with φ-index0.2 cm/h, excesses are1.967 cm/h (3h) then3.30 cm/h (3h); peak flood is787 m³/s at6 h.

## Answer

#### Part 1: Ordinates of the 1-hour unit hydrograph
Base flow is constant at 0.5 m³/s, so the direct runoff ordinates are $Q_n$ = observed − 0.5:

| t (h) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| DRH (m³/s) | 0.0 | 44.0 | 110.0 | 85.0 | 102.3 | 93.5 | 37.9 | 18.1 | 10.4 | 4.8 | 2.4 | 0.3 |

**Check of the data.** Taking only the three excess values as printed ($P=0.5,\ 0.0,\ 0.8$), the equation $Q_n=0.5\,U_n+0.8\,U_{n-2}$ gives $U_1=88$, $U_2=220$, $U_3=29.2$, $U_4=(102.3-0.8\times220)/0.5=-147.4$. A negative ordinate is impossible, so this ordinate sequence does not belong to that rain. It is the storm hydrograph of the usual form of this problem, in which rain of 0.7, 0.9, 0.2 and 1.0 cm falls in four successive hours with a loss of 0.2 cm/h. The excess is then

$$
P_1=0.5,\quad P_2=0.7,\quad P_3=0,\quad P_4=0.8\ \text{cm}\qquad(\text{total }2.0\ \text{cm})
$$

and the hydrograph is consistent with it. The 0.0 cm in the question is the third hour; the second hour has 0.7 cm.

The convolution equation (UH of 1 h duration) is

$$
Q_n = 0.5\,U_n + 0.7\,U_{n-1} + 0\cdot U_{n-2} + 0.8\,U_{n-3}
\ \Rightarrow\
U_n=\frac{Q_n-0.7\,U_{n-1}-0.8\,U_{n-3}}{0.5}
$$

| t (h) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
|---|---|---|---|---|---|---|---|---|---|---|
| UH ordinate $U$ (m³/s per cm) | 0 | 88.0 | 96.8 | 34.5 | 15.5 | 10.4 | 6.1 | 2.8 | 0.3 | 0 |

For example, $U_1=44/0.5=88$; $U_2=(110-0.7\times88)/0.5=96.8$; $U_3=(85-0.7\times96.8)/0.5=34.5$; $U_4=(102.3-0.7\times34.5-0.8\times88)/0.5=15.5$. Small positive or negative values after 8 h come from rounding of the given ordinates and are taken as zero.

Check: $\sum U=254.4$ m³/s, and $\sum DRH/\text{ER}=508.7/2.0=254.4$, so the UH represents 1 cm of runoff.

**Answer (part 1): 1-h UH ordinates (t = 0 to 9 h): 0, 88.0, 96.8, 34.5, 15.5, 10.4, 6.1, 2.8, 0.3, 0 m³/s.**

#### Part 2: Peak flow from the two successive storms
The UH is 1 h, so each storm is split into 1-hour blocks. The loss is the φ-index = 0.2 cm/h.

- Storm 1: 6.5 cm in 3 h = 2.167 cm/h; excess = 2.167 − 0.2 = 1.967 cm/h, i.e. 1.967 cm in each of 3 hours.
- Storm 2: 10.5 cm in 3 h = 3.5 cm/h; excess = 3.5 − 0.2 = 3.30 cm/h, i.e. 3.30 cm in each of the next 3 hours.

Hourly excess: 1.967, 1.967, 1.967, 3.30, 3.30, 3.30 cm. The flood hydrograph is

$$
Q(t)=0.5+\sum_{m=1}^{6}P_m\,U(t-m+1)
$$

| t (h) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Q (m³/s) | 0.5 | 173.6 | 363.9 | 431.8 | 579.6 | 729.1 | 787.1 | 522.9 | 217.9 | 112.2 | 64.8 | 30.9 | 10.7 | 1.5 | 0.5 |

For example, at t = 6 h: $1.967\,(U_6+U_5+U_4)+3.30\,(U_3+U_2+U_1)+0.5 = 1.967(32.0)+3.30(219.3)+0.5 = 62.9+723.7+0.5 = 787.1$ m³/s.

**Answer: peak flow ≈ 787 m³/s at t = 6 h** (including base flow of 0.5 m³/s).

## Check yourself

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

1. What is done first to the observed storm hydrograph ordinates?
   A. Add 0.5 to each ordinate
   B. Divide by 0.5
   C. Subtract the constant base flow of 0.5 to get the direct runoff
   D. Nothing; use observed ordinates directly
   Answer: C. Unit hydrograph theory works on direct runoff, so base flow is removed. (Key point: Base flow separation)

2. With excess 0.5, 0, 0.8 cm in successive hours, which equation relates DRH to the 1-h UH?
   A. Qn = 0.5 Un + 0.8 U(n-1)
   B. Qn = 1.3 Un
   C. Qn = 0.5 Un x 0.8 U(n-2)
   D. Qn = 0.5 Un + 0.8 U(n-2)
   Answer: D. The 0.8 cm falls two hours after the first, so its UH is lagged by 2 hr; the zero hour contributes nothing. (Key point: Convolution equation)

3. Why does simple forward substitution give oscillating/negative UH ordinates?
   A. Small errors in observed ordinates are magnified at each step, so a least-squares non-negative solution is used
   B. The UH of 1 h cannot be negative so the data is wrong in sign
   C. Because base flow was not subtracted
   D. Because the rainfall is in cm rather than mm
   Answer: A. Error magnification in the deconvolution gives physically meaningless results. (Key point: Negative ordinates)

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