Hydrograph Analysis

Unit hydrograph (UH)

A unit hydrograph (Sherman, 1932) is the direct runoff hydrograph resulting from 1 unit (1 cm or 1 mm) of effective rainfall falling uniformly over the whole basin at a uniform rate during a specified duration $D$. It is written as the $D$-hour unit hydrograph.

 Q |      /\
   |     /  \
   |    /    \___
   |___/         \____
   +------------------- t
   |<-- Tb (base) --->|

Assumptions

1. Constant intensity: effective rainfall is uniform in intensity over the duration $D$.
2. Uniform distribution: effective rainfall is uniformly spread over the whole drainage area.
3. Constant base time: the base time of the direct runoff hydrograph is the same for all storms of equal duration, whatever the intensity.
4. Linearity (proportionality): ordinates of DRH due to the same duration are directly proportional to the effective rainfall depth; for twice the rain, ordinates are doubled.
5. Superposition (time invariance): the hydrographs of successive rain periods can be added; the catchment response does not change with time.
6. All characteristics of the catchment are reflected in the shape of the UH.

Limitations

1. Applies to catchments of about 25 to 5000 km$^2$; very small (< 25 km$^2$) or very large basins (> 5000 km$^2$) violate the assumptions.
2. Not valid when rainfall is not uniform in space or time, or storms move across the basin.
3. Not suitable for snowmelt-dominated or heavily regulated basins, or basins with large storage (lakes, reservoirs).
4. Cannot handle effective rainfall of different durations without S-curve conversion.
5. Real catchment behaviour is not exactly linear; UH from small storms may differ from that of large floods.
6. Needs reliable rainfall and streamflow records; an ungauged basin needs synthetic methods.
7. Derived UH is valid only for that point on the stream.

Method: effective rainfall (ER) is rainfall minus $\phi$ loss for each hour; the DRH is found by superposition of the 1-h unit hydrographs.

Unit hydrographs (hourly ordinates, triangles)

(Before: rising limb 0 to 3.6 in 2 h; falling limb 3.6 to 0 in 4 h. After: rising 0 to 6.0 in 1 h; falling 6.0 to 0 in 3 h.)

Effective rainfall

Storm intensities: 4 cm/h in hour 1 and 3 cm/h in hour 2.

Storm hydrograph $Q(t) = P_1\,u(t) + P_2\,u(t-1)$

(a) Increase in peak

$$\text{Increase}=\frac{30.00-17.19}{17.19}\times100 = 74.5\ \%$$

(b) Volume of flood runoff

$$V = \Delta t\sum Q = 3600\times\sum Q$$

• Before: $\sum Q = 60.48$ m$^3$/s, $V = 217\,728$ m$^3$ (this equals 5.6 cm over the 3.888 km$^2$ catchment area implied by its unit hydrograph).
• After urbanization: $\sum Q = 74.40$ m$^3$/s, $V = 267\,840$ m$^3$ (6.2 cm over 4.32 km$^2$ implied by its unit hydrograph).

Answer: (a) Peak increases by about 74.5 % (17.19 to 30.00 m$^3$/s). (b) The flood runoff volume after urbanization is about 2.68 x 10$^5$ m$^3$, an increase of about 50 100 m$^3$ (23 %) over the pre-urban 2.18 x 10$^5$ m$^3$.

Unit hydrograph

A unit hydrograph is the direct runoff hydrograph produced by 1 cm of effective rainfall, falling uniformly over the basin at a constant rate during a specified duration $D$. It assumes linearity and time invariance.

Preparing a UH from a multiple (complex) storm

When a storm has several periods of effective rainfall, the observed direct runoff hydrograph (DRH) is the sum of the responses to each period. The UH is obtained by deconvolution.

Steps

1. Separate the baseflow from the storm hydrograph to get the DRH ordinates $Q_n$.
2. Compute the effective rainfall for each period of duration $D$ by deducting losses ($\phi$-index): $P_1, P_2, \dots, P_M$.
3. The DRH ordinate at time $n$ is

$$Q_n=\sum_{m=1}^{\min(n,M)} P_m\,U_{n-m+1}$$

4. Solve for the UH ordinates $U_n$ successively:

$$U_1=\frac{Q_1}{P_1},\quad U_2=\frac{Q_2-P_2U_1}{P_1},\quad U_3=\frac{Q_3-P_2U_2-P_3U_1}{P_1},\ \dots$$

5. Check that the area under the UH equals 1 cm over the basin: $\sum U\,\Delta t = 1\ \text{cm}\times A$.
6. For more accuracy, use many storms and the least-squares method, and average the UH ordinates.

Example

Two periods of 1 h with effective rainfall $P_1=2$ cm and $P_2=3$ cm. Observed DRH: $Q_1=20$, $Q_2=90$, $Q_3=130$, $Q_4=80$, $Q_5=30$ m$^3$/s.

The 1-h UH ordinates are 10, 30, 20, 10 m$^3$/s at 1, 2, 3, 4 h.

The storm lasts 3 h but the UH is for 6 h, so first derive the 3-h UH from the 6-h UH by the S-curve method, then apply the rainfall excess.

Step 1: Rainfall excess

$$ER = P - \phi t = 4 - 0.5\times 3 = 2.5 \text{ cm}$$

Step 2: S-curve and 3-h UH

The S-curve is the sum of the 6-h UH repeated every 6 h. Its values at 6-h points are 0, 8, 22, 50, 66, 76, 82, 86, 86 (running sum of the UH ordinates). A smooth S-curve is drawn through them and read at 3-h intervals, so that the curve can be lagged by 3 h:

$$S(t) = U_6(t) + S(t-6)$$

The 3-h UH is the S-curve lagged by 3 h, subtracted, and multiplied by 6/3:

$$U_3(t) = \frac{6}{3}\,[S(t)-S(t-3)]$$

Step 3: Storm hydrograph

$$Q(t) = 2.5\,U_3(t) + 2$$

The S-curve reaches a constant value of 86 m³/s (sum of the 6-h UH ordinates), as it should.

Answer: peak of the storm hydrograph ≈ 73.1 m³/s at t = 18 h; it returns to the 2 m³/s base flow after t = 42 h.

Approach. The rain falls in three 2-h periods and the UH ordinates are tabulated at 2-h intervals, so the ordinates are used as the UH response to a 2-h rainfall excess period, each period being lagged by 2 h (the usual convolution at the 2-h step). Converting the given ordinates to a 2-h UH by the S-curve method gives erratic (even negative) values for this data, so the table is used directly.

Step 1: Rainfall excess per 2-h period

Loss per period = φ × 2 h = 1.25 × 2 = 2.5 cm.

Step 2: Convolution with the UH

$$Q(t) = 0.70\,U(t) + 0.50\,U(t-2) + 2.50\,U(t-4) + 40$$

Answer: peak flood ≈ 704.0 m³/s at t = 12 h; the flow returns to the base flow of 40 m³/s at t = 28 h. The last column of the table is the flood hydrograph.

The storm lasts 3 h while the UH is for 6 h, so the 3-h UH is derived by the S-curve method and then applied to the rainfall excess.

Step 1: Rainfall excess

Initial loss = 0.5 cm, then infiltration at φ = 1 cm/h during the 3-h storm.

$$ER = 15 - 0.5 - (1\times 3) = 11.5 \text{ cm}$$

Step 2: 3-h UH from the S-curve

The S-curve at 6-h points is the running sum of the UH ordinates (0, 6, 18, 43, 53, 61, 67, 70, 70). Draw a smooth S-curve through them, read it at 3-h intervals, lag it by 3 h, and take the difference:

$$S(t) = U_6(t) + S(t-6), \qquad U_3(t) = \frac{6}{3}\,[S(t)-S(t-3)]$$

Step 3: Storm hydrograph

$$Q(t) = 11.5\,U_3(t) + 4$$

The S-curve becomes constant at 70 m³/s (sum of the 6-h UH ordinates).

Answer: peak of the storm hydrograph ≈ 297.1 m³/s at t = 15 h; it returns to the base flow of 4 m³/s after t = 42 h.

Approach. The rain falls in three 2-h periods and the UH ordinates are tabulated at 2-h intervals, so the ordinates are used as the UH response to a 2-h rainfall excess period, each period being lagged by 2 h (the usual convolution at the 2-h step). Converting the given ordinates to a 2-h UH by the S-curve method gives erratic (even negative) values for this data, so the table is used directly.

Step 1: Rainfall excess per 2-h period

Loss per period = φ × 2 h = 1.25 × 2 = 2.5 cm.

Step 2: Convolution with the UH

$$Q(t) = 1.00\,U(t) + 2.00\,U(t-4) + 50$$

Answer: peak flood ≈ 570.0 m³/s at t = 12 h; the flow returns to the base flow of 50 m³/s at t = 28 h. The last column of the table is the flood hydrograph.

Base flow separation divides the storm hydrograph into direct runoff and base flow so that the direct runoff hydrograph (DRH) can be analysed. The common methods are:

 Q |        peak
   |       /\
   |  A   /  \  N days
   | ____/    \____B___C
   |      \ ___/  ..
   |       (1)(2)(3)
   +------------------------ t

1. Straight-line method (Method 1): join the point A where the rising limb begins to the point B on the recession limb where direct runoff ends (a horizontal or straight line). It is simple and used for small catchments and perennial streams.

2. Fixed-base length method (Method 2): from the peak, find the end of direct runoff at $N$ days after the peak,

$$N = 0.83\,A^{0.2}$$

($A$ in km$^2$, $N$ in days). Extend the recession curve backward from A to a point below the peak, and join it to the point at $N$ days after the peak by a straight line.

3. Variable-slope method (Method 3): extend the base flow recession curve (from before the storm) forward to the time of the peak (A to D); extend the recession limb backward to the point of inflection C; join D to C by a straight line.

4. Master recession curve / semi-log method: plot the recession limb on semi-log paper ($\log Q$ against $t$); the point where the straight (base flow) recession line begins to depart is the end of direct runoff.

All methods are arbitrary, so the same method should be used for all storms of the catchment. Volume of direct runoff = area between hydrograph and base flow line.

Flood hydrograph

A flood (storm) hydrograph is the graph of discharge against time at a river section during a flood caused by a storm. It has a rising limb, crest (peak), and recession (falling) limb, and includes both direct runoff and base flow.

 Q |       peak
   |       /\
   | rising/  \ recession
   |   ___/    \___
   |__/  base flow  \____
   +--------------------- t

Direct runoff hydrograph (DRH)

The DRH is the hydrograph of the surface runoff (and interflow) only, obtained by subtracting the base flow from the storm hydrograph. The area under the DRH equals the volume of direct runoff, equal to the effective rainfall depth multiplied by the basin area.

Unit hydrograph

The unit hydrograph is the DRH due to 1 cm (1 mm) of effective rainfall, uniformly distributed over the catchment, at a constant rate during a specified duration (the $D$-hour UH).

Methods of base-flow separation

1. Straight-line method: a horizontal or straight line from the start of rise to the point where the recession ends.
2. Fixed-base length method: the end of direct runoff is taken $N = 0.83A^{0.2}$ days after the peak ($A$ in km$^2$); connect the start of rise to this point by a straight line (the start of rise is joined with the recession curve extended back).
3. Variable-slope (three-point) method: the pre-storm recession is extended forward to the time of peak; the recession limb is extended backward to the inflection point; the two are joined by a straight line.
4. Master recession curve (semi-log) method: the point on the semi-log recession plot where the slope changes marks the end of direct runoff.

The area between the hydrograph and the base-flow line gives the direct runoff volume.

S-curve method: the S-curve (S-hydrograph) is the response to a continuous effective rainfall of intensity 1 cm per 2 h. It is built by adding the 2-h UH ordinates lagged by 2 h each. The S-curve is then lagged by 3 h (the new duration) and subtracted; the difference is the response to 3 h of rain at 1 cm per 2 h, which is scaled by $2/3$ to get 1 cm per 3 h.

$$U_3(t) = \frac{2}{3}\left[S(t) - S(t-3)\right]$$

Step 1: S-curve

$S(t) = U_2(t) + S(t-2)$

Step 2: Lag the S-curve by 3 h

$S(t-3)$ is found by linear interpolation between the S-curve ordinates (for example, $S(1)=6$, $S(3)=36$, $S(5)=99$, $S(7)=195$, $S(9)=288$).

Answer: 3-h UH ordinates at t = 0, 2, 4, 6, 8, 10, 12 h are 0, 8, 36, 68, 102, 86, 60 m$^3$/s.

(The given 2-h UH is cut off at 12 h with a non-zero ordinate, so the 3-h UH is computed only for this range.)

Factors affecting the flood hydrograph

Climatic factors

• Rainfall intensity and duration: high intensity gives a higher, sharper peak; long duration gives larger volume and broader hydrograph.
• Areal distribution and storm movement: a storm moving downstream gives a higher peak than upstream movement.
• Antecedent moisture: wet soil increases runoff.
• Evaporation, temperature and snow-melt.

Catchment factors

• Size: larger area gives a longer time base and later, lower specific peak.
• Shape: fan or circular shape gives a high, sharp peak; elongated gives a flatter one.
• Slope: steep slopes shorten the time to peak and increase the peak.
• Land use, vegetation and soil: forest and permeable soil reduce the peak; urban and bare areas raise it.
• Drainage density: higher density gives quicker response.
• Storage: lakes, ponds, depressions and flood plains flatten the peak.

Channel factors: cross-section, roughness and slope, and channel storage.

Methods of baseflow separation

1. Straight-line method: join the start of the rising limb with the end of direct runoff by a straight line.
2. Fixed-base length method: end of direct runoff at $N=0.83A^{0.2}$ days after the peak ($A$ in km$^2$); join to the start of the rise.
3. Variable-slope method: extend the pre-storm recession forward to the peak time and the recession limb backward to an inflection point; join the two by a straight line.
4. Semi-log recession (master recession curve) method.

Method: subtract the baseflow (10 m$^3$/s) to get the direct runoff hydrograph (DRH), find the volume and the effective rainfall depth, and divide the DRH ordinates by this depth to get the 4-h UH (ordinates per 1 cm).

Direct runoff volume

$$V=\sum DRH\times\Delta t = 885\times4\times3600 = 12,744,000\ \text{m}^3 = 12.744\times10^6\ \text{m}^3$$

Effective rainfall depth over the catchment

$$d=\frac{V}{A}=\frac{12,744,000}{600\times10^6}=2.124\ \text{cm}$$

Unit hydrograph

$$U(t)=\frac{DRH(t)}{2.124}$$

Check: $\sum U\times\Delta t = 416.7\times14400 = 6,000,000$ m$^3$ = 1 cm over 600 km$^2$ (6,000,000 m$^3$).

Answer: The 4-h UH has a peak of 103.6 m$^3$/s at 8 h and ordinates as shown in the last column (0, 42.4, 103.6, 84.8, 56.5, 42.4, 28.2, 23.5, 18.8, 9.4, 4.7, 2.4, 0 m$^3$/s at 0, 4, ..., 48 h).

Unit hydrograph

A unit hydrograph (UH) is the direct runoff hydrograph resulting from one unit (1 cm or 1 mm) of effective rainfall generated uniformly over the drainage basin at a uniform rate during a specified period (the unit duration $D$, e.g. the 3-h UH).

It is based on the assumptions of linearity and time invariance, so hydrographs of other storms of the same duration are found by scaling and superposition.

 Q |      /\
   |     /  \     area = 1 cm x A
   |    /    \__
   |___/        \____
   +------------------- t
        |<--- Tb --->|

Uses of the hydrograph (unit hydrograph) in engineering

1. Estimation of flood hydrograph and peak flow for any storm on the catchment, by multiplying the UH by the effective rainfall and adding baseflow.
2. Design flood for dams, spillways, culverts, bridges and flood-control works, using the design storm.
3. Flood forecasting and warning from observed rainfall.
4. Synthetic UH (Snyder, SCS) for ungauged basins, derived from catchment properties.
5. Studying the effect of land-use change (urbanization, deforestation) on runoff.
6. Deriving UH of other durations by the S-curve method.
7. Reservoir and flow routing studies which need inflow hydrographs.
8. Water-resources planning, such as estimating storm runoff volume and time to peak.

The 4-h UH is given at 1-h intervals (the ordinate is 0 at t = 0 and 6 m$^3$/s at t = 1 h). Check: the volume under the UH is $\sum Q\times3600 = 297\times3600 = 1.069\times10^6$ m$^3$, which is 0.99 cm over 300 km$^2$, so the UH is a 1 cm UH.

Method

1. Build the S-curve: $S(t)=U_4(t)+S(t-4)$ (add the 4-h UH repeatedly, each lagged by 4 h).
2. Lag the S-curve by the new duration 3 h: $S(t-3)$.
3. The difference is the response to 3 h of rain at the 4-h rate (1 cm per 4 h); scale by $4/3$:

$$U_3(t)=\frac{4}{3}\left[S(t)-S(t-3)\right]$$

The S-curve reaches an equilibrium value of about 207 m$^3$/s, in agreement with $Q_e=2.78A/T_r = 2.78\times300/4 = 208.5$ m$^3$/s.

Answer: The 3-h UH ordinates at 1-h intervals are 0, 8.0, 48.0, 88.0, 113.3, 101.3, 84.0, 72.0, 62.7, 54.7, 44.0, 36.0, 30.7, 25.3, 20.0, 13.3, 12.0, 6.7, 5.3, 0, 2.0 m$^3$/s (t = 0 to 20 h), with a peak of 113.3 m$^3$/s at 4 h. The small ripples at the tail come from rounding of the given ordinates.

Definitions

• Storm hydrograph: the graph of total discharge against time at a river section during and after a storm. It consists of the rising limb, crest, recession limb and includes base flow.
• Direct runoff hydrograph (DRH): the hydrograph of surface runoff (plus interflow), obtained from the storm hydrograph by removing the base flow.
• Base flow: the part of the stream flow that comes from groundwater and delayed sub-surface flow; it keeps the stream flowing between storms.

Methods of base flow separation

 Q |        /\ peak
   |       /  \
   |    A /    \       B = end of direct runoff
   |  ---/------\------B----
   |   (1) straight line AB
   +--------------------------- t

1. Straight-line method: join the point A, where the rising limb starts, to the point B on the recession limb where direct runoff ends, by a straight line (horizontal when base flow is level). Simple, for perennial streams.

 Q |        /\  peak
   |       /  \
   |     A/    \
   |  ---/__    \__ B      (2) A'-B with
   |   A' \ \__________ N days after peak
   +-------------------------- t

2. Fixed-base length method: the end of direct runoff B is taken $N$ days after the peak, $N = 0.83A^{0.2}$ ($A$ in km$^2$). The base flow line joins A (extended back from the recession) with B.

 Q |         /\ peak
   |        /  \
   |       /    \
   |   .--D      C   <- inflection
   |  /  extended recession
   |_/____________________ t

3. Variable-slope method: extend the pre-storm recession forward to a point D under the peak, extend the recession limb backward to the inflection point C, and join D and C. It is considered the most realistic.

4. Master recession (semi-log) method: on a semi-log graph the base flow recession is a straight line; where the plot of the falling limb merges into this line is the end of direct runoff.

Assumptions: the 6-h hydrograph is the response to one 6-h effective-rainfall pulse. A 12-h effective storm is two successive 6-h pulses, the first giving 3.3 cm and the second 5.5 cm. The base flow of 40 m$^3$/s (the initial and final flow) is constant.

Step 1: Direct runoff and 6-h UH

DRH = Q - 40. Sum of DRH ordinates $= 1790$ m$^3$/s.

$$V=1790\times6\times3600 = 38.66\times10^6\ \text{m}^3,\qquad d=\frac{V}{A}=\frac{38.66\times10^6}{770\times10^6}=5.02\ \text{cm}$$

So the observed storm gave about 5.02 cm of effective rainfall, and the 6-h UH is $U = DRH/5.02$.

Step 2: Superposition

$$Q(t)=40+3.3\,U(t)+5.5\,U(t-6)$$

Answer: The flood hydrograph peaks at about 638 m$^3$/s at 30 h; ordinates are in the last column.

Part 1: 3-h UH from the 2-h UH by the S-curve method

$S(t)=U_2(t)+S(t-2)$, and $U_3(t)=\dfrac{2}{3}\left[S(t)-S(t-3)\right]$ with $S(t-3)$ interpolated linearly.

The S-curve reaches equilibrium (896 m$^3$/s) at 22 h.

Part 2: Flood hydrograph

Effective rainfall ($\phi$ = 0.3 cm/h, assumed consecutive storms: 3 h of 8 cm, then 2 h of 7 cm):

• 3-h storm: $8 - 0.3\times3 = 7.1$ cm
• 2-h storm: $7 - 0.3\times2 = 6.4$ cm, starting at t = 3 h.

$$Q(t)=10+7.1\,U_3(t)+6.4\,U_2(t-3)$$

$U_3$ and $U_2$ are read at hourly intervals (linear interpolation of the S-curve; $U_2(t)=S(t)-S(t-2)$).

Answer: The 3-h UH ordinates are 0, 16.7, 75, 140, 180, 176.7, 130, 83.3, 43.3, 23.3, 16.7, 9, 2 m$^3$/s at 0, 2, ..., 24 h. The flood hydrograph peaks at about 2384 m$^3$/s at 10 h and returns to the base flow of 10 m$^3$/s by about 27 h.

Part 1: 2-h UH from the 4-h UH (S-curve method)

$S(t)=U_4(t)+S(t-4)$, and $U_2(t)=\dfrac{4}{2}\left[S(t)-S(t-2)\right]$ (the S-curve lagged by the new duration, 2 h, and scaled by $4/2$).

The raw S-curve points zig-zag because the ordinates of the given 4-h UH are irregular. A smooth S-curve is drawn through them (3-point smoothing of the ordinates, ending at the mean equilibrium value 149 m$^3$/s). Differences are taken from the smooth curve.

(The sum of the $U_2$ ordinates is 298, equal to the sum of the $U_4$ ordinates, so the volume of 1 cm is conserved.)

Part 2: Effective rainfall ($\phi=0.15$ cm/h)

Part 3: Flood hydrograph

$Q(t)=40+\sum_j P_j\,U_2(t-2j)$ with $P=(2.2,3.7,3.7,2.7,2.7,2.7)$.

Answer: The 2-h UH ordinates (0, 2, ..., 24 h) are 0, 15, 20, 34, 46, 50.5, 42.5, 32.5, 24.5, 16.5, 11.5, 5, 0 m$^3$/s. The flood hydrograph, including base flow of 40 m$^3$/s, peaks at about 726 m$^3$/s at 14 h.

Step 1: Baseflow and effective rainfall

The storm hydrograph begins at 0.5 m$^3$/s and ends at 0.5 m$^3$/s, so a constant base flow of 0.5 m$^3$/s is assumed.

Losses $=0.2$ cm/h:

Total ER = 2.0 cm.

Step 2: Hourly (1-h) UH by successive deconvolution

Direct runoff $DRH(t)=Q-0.5$ and

$$DRH(t)=0.5\,U(t)+0.7\,U(t-1)+0\cdot U(t-2)+0.8\,U(t-3)$$ $$U(t)=\frac{DRH(t)-0.7U(t-1)-0.8U(t-3)}{0.5}$$

Examples: $U(1)=44/0.5=88$; $U(2)=(110-0.7\times88)/0.5=96.8$; $U(3)=(85-0.7\times96.8)/0.5=34.5$; $U(4)=(102.3-0.7\times34.5-0.8\times88)/0.5=15.5$.

Step 3: S-curve and 2-h UH

For the 1-h UH, $S(t)=\sum U$ (lag 1 h), and

$$U_2(t)=\frac{1}{2}\left[S(t)-S(t-2)\right]=\frac{U(t)+U(t-1)}{2}$$

(The tiny negative and positive values after 8 h come from rounding of the given ordinates and are taken as zero. Check: $\sum DRH = 508.4$ and total ER $=2.0$ cm, so the sum of the 1-h UH ordinates should be $508.4/2.0 = 254.2$; the computed sum is 254.4, so the UH is consistent.)

Answer: 1-h UH ordinates at t = 0 to 8 h: 0, 88, 96.8, 34.5, 15.5, 10.4, 6.1, 2.8, 0.3 m$^3$/s. 2-h UH ordinates at t = 1 to 9 h: 44, 92.4, 65.6, 25.0, 13.0, 8.3, 4.5, 1.5, 0.1 m$^3$/s.

Check of the UH: $\sum U\times3600=55.31\times3600=199\,116$ m$^3$, which is 0.996 cm over 20 km$^2$, so it is a 1 cm, 3-h UH.

The storm consists of 4 h at 2 cm/h followed by 3 h at 1 cm/h. These durations are not multiples of 3 h, so the 3-h UH is first converted to a 1-h UH by the S-curve method.

Step 1: S-curve

$S(t)=U_3(t)+S(t-3)$ (response to a continuous 1 cm per 3 h = 0.333 cm/h).

Step 2: 1-h UH

$$U_1(t)=\frac{3}{1}\left[S(t)-S(t-1)\right]$$

Step 3: Hourly effective rainfall

Hours 1 to 4: 2 cm each; hours 5 to 7: 1 cm each.

$$Q(t)=\sum_{j=0}^{6} P_j\,U_1(t-j),\qquad P=(2,2,2,2,1,1,1)\ \text{cm}$$

The $U_1$ ordinates after 12 h are rounding noise and are taken as zero. The total ER is $2\times4+1\times3=11$ cm.

Answer: Peak flow $\approx 76.5$ m$^3$/s, occurring 7 h after the commencement of rainfall (at the end of the 7-h storm).

Direct runoff = rainfall excess; the peak of the flood hydrograph = (ER depth in cm) × (UH peak) + base flow, because a 6 h storm on a 6 h UH needs no lagging.

Data

• UH duration = storm duration = 6 h, UH peak = 96 m³/s per cm
• Base flow = 25 m³/s
• Infiltration is at its minimum rate, so it is taken as the φ-index = 2.5 mm/h = 0.25 cm/h

Rainfall excess

Rainfall intensity = 18.3 / 6 = 3.05 cm/h

$$\begin{aligned} ER &= (i-\phi)\,t = (3.05 - 0.25)\times 6 \\ &= 16.8 \text{ cm} \end{aligned}$$

Peak discharge

$$\begin{aligned} Q_p &= ER \times U_p + Q_b \\ &= 16.8 \times 96 + 25 \\ &= 1637.8 \text{ m}^3/\text{s} \end{aligned}$$

Answer: peak discharge ≈ 1637.8 m³/s (direct runoff 1612.8 m³/s + base flow 25 m³/s).

A unit hydrograph (UH) derived from a complex storm uses a direct runoff hydrograph (DRH) that results from several successive periods of rainfall excess of different intensity. Because the UH ordinates are not separable as in a single burst, the DRH is deconvoluted using the convolution (discrete) equation.

Basic equation

If the net rain in successive periods of duration $D$ is $P_1, P_2, \dots, P_M$ (cm), the UH of duration $D$ has ordinates $U_1, U_2, \dots, U_N$, and the DRH has ordinates $Q_1, Q_2, \dots, Q_n$, then

$$Q_n = \sum_{m=1}^{n\le M} P_m\, U_{n-m+1}, \qquad n = M+N-1$$

Writing it out for each ordinate:

$$\begin{aligned} Q_1 &= P_1U_1 \\ Q_2 &= P_1U_2 + P_2U_1 \\ Q_3 &= P_1U_3 + P_2U_2 + P_3U_1 \\ &\ \ \vdots \end{aligned}$$

Procedure

1. Select a storm with several periods of rain of uniform duration $D$ over the whole basin, and its stream-flow record.
2. Separate base flow from the storm hydrograph to get the DRH ordinates $Q_n$ at intervals $D$.
3. Compute rainfall excess $P_m$ of each period by deducting losses (φ-index or a loss model), keeping only periods with $P_m>0$.
4. Number of UH ordinates $N = n - M + 1$ (n = number of DRH ordinates, M = number of excess-rain periods).
5. Solve the equations successively (forward substitution):

$$U_1=\frac{Q_1}{P_1},\quad U_2=\frac{Q_2-P_2U_1}{P_1},\quad U_3=\frac{Q_3-P_2U_2-P_3U_1}{P_1},\ \dots$$

6. Check: the area of the derived UH must equal 1 cm of runoff over the basin: $\sum U_i \cdot D\cdot 3600 = A\times 0.01$ m. Adjust (smooth) the ordinates if needed.
7. Plot the UH and, if necessary, re-compute DRH from it to confirm it reproduces the observed DRH.

Remarks

• Forward substitution accumulates errors, so ordinates may oscillate or become negative. Then the least-squares (matrix) form $\mathbf{Q}=[\mathbf{P}]\mathbf{U}$, with $\mathbf{U}=(\mathbf{P}^T\mathbf{P})^{-1}\mathbf{P}^T\mathbf{Q}$, or a trial-and-error / optimisation approach with a non-negative constraint is used (Collins' method: assume a UH, compute the DRH, correct the UH, and repeat).
• The UH obtained from the average of several storms is more reliable.

A unit hydrograph represents 1 cm of direct runoff from the whole basin. So the volume under the UH equals area × 1 cm.

Step 1: Volume under the UH

Ordinates are at 1 h interval, so the volume is the sum of ordinates × 1 h (3600 s).

$$\sum Q = 4+25+44+60+70+61+52+45+38+32+27+22+18+14+11+8+6+4+2+1 = 544 \text{ m}^3/\text{s}$$ $$V = 544 \times 3600 = 1,958,400 \text{ m}^3$$

Step 2: Area

$$A = \frac{V}{\text{runoff depth}} = \frac{1,958,400}{0.01} = 195,840,000 \text{ m}^2 = 195.8 \text{ km}^2$$

Answer: Area of basin ≈ 195.8 km² (using the 1 h ordinates with the trapezoidal-type sum; the first and last ordinates are small, so the error is negligible).

The hydrograph is a trapezoid produced by constant net rain of 0.20 in/hr for 5 hr. The rise stops when the whole basin contributes (equilibrium), and the recession starts when rain stops.

 Q
 Qp |        ________
    |      /|       |\
    |    /  |       |  \
    |  /    |       |    \
    |/______|_______|______\___ t (hr)
    0       3       5       8
         rain 0 -> 5 hr

(a) Excess release time

Rain stops at t = 5 hr and the direct runoff ends at t = 8 hr. The time taken for the last excess water to leave the basin is

$$t_{release} = 8 - 5 = 3 \text{ hr}$$

This equals the time of concentration, $t_c = 3$ hr (also the duration of the rising limb, 0 to 3 hr).

(b) Contributing area at t = 4 hr

Since $t = 4\text{ hr} > t_c = 3\text{ hr}$, water from every part of the basin, including the farthest point, has reached the outlet. So

$$\frac{A_{contributing}}{A} = 100\%$$

(At t < 3 hr the contributing fraction is t/3, e.g. 33% at t = 1 hr.)

(c) Peak discharge $Q_p$

With the whole basin contributing, $Q_p$ = net rain intensity × area (equilibrium discharge).

$$\begin{aligned} Q_p &= i\,A = \frac{0.20}{12}\ \text{ft/hr} \times 4250 \times 43560\ \text{ft}^2 \\ &= 3,085,500 \text{ ft}^3/\text{hr} \\ &= \frac{3,085,500}{3600} = 857 \text{ cfs} \end{aligned}$$

Answer: $Q_p$ ≈ 857 cfs (the sketch values are only approximate readings).

(d) Why the plateau between t = 3 and t = 5 hr

After t = 3 hr (= $t_c$), the entire basin already contributes. Rain continues at the same constant intensity, so the inflow to the outlet from the basin is constant: the rate of runoff leaving equals the rate of net rain falling (equilibrium). No new area can be added to increase the flow, so the hydrograph stays flat at $Q_p$ until the rain stops at t = 5 hr. It then recedes to zero in another $t_c$ = 3 hr.

A 2-hr UH gives the response to 1 inch of net rain in 2 hr (intensity 0.5 in/hr). Ordinates are directly proportional to the net rain depth, and responses of successive blocks are added with a time lag (principle of proportionality and superposition).

Read from the sketch (t, hr: Q, cfs): 0: 0; 1: 200; 2: 400; 3: 600 (peak); 4: 400; 5: 200; tail to about 0 at 9.

(a) Peak discharge for 5.00 in/hr for 2 hr

Duration is the same as the UH duration (2 hr), so just scale.

$$\begin{aligned} \text{net rain} &= 5.00 \times 2 = 10 \text{ in} \\ Q_{peak} &= 10 \times 600 = 6000 \text{ cfs} \end{aligned}$$

Answer: 6000 cfs (at t = 3 hr).

(b) Total direct surface runoff

Direct runoff depth equals net (effective) rain depth.

$$\text{Runoff} = 5.00 \text{ in/hr} \times 2 \text{ hr} = 10 \text{ in}$$

Answer: 10 inches over the basin.

(c) Storm of 0.50 in/hr for 4 hr, discharge at 8 p.m.

Net rain is 0.5 × 4 = 2 in, which is two successive 2-hr blocks of 0.5 in/hr × 2 hr = 1 in each. Each block gives one UH (multiplier 1); the second is lagged by 2 hr.

Rain starts at 4 p.m., so 8 p.m. is t = 4 hr:

$$Q(4) = 1\times U(4) + 1\times U(4-2) = 400 + 400 = 800 \text{ cfs}$$

Answer: 800 cfs at 8 p.m.

The S-hydrograph is the response to a continuous net rain of 1 cm/h that continues indefinitely. Here its ordinate is 1 (unit: cm/h of runoff) from t = 0 and stays constant, i.e. $S(t)=1$ for $t \ge 0$.

(a) 2-hour unit hydrograph

Lag the S-curve by $D = 2$ h and subtract. This gives the response to 1 cm/h for 2 h = 2 cm. Divide by $D\times 1$ = 2 to get a UH for 1 cm:

$$U_2(t) = \frac{S(t)-S(t-2)}{2}$$

So the 2-h UH is a rectangle of height 0.5 per hour (cm/h per cm of rain) from t = 0 to t = 2 h, and zero afterwards. Check: area = 0.5 × 2 = 1 cm. ✓.

 U (1/h)
 0.5 |#########
     |#########
   0 +---------+--------- t (h)
     0         2

(b) 4-hour UH from the 2-hour UH

Add two 2-h UHs, the second lagged by 2 h, and divide by 2 (4 h/2 h = 2 blocks, and the sum is for 2 cm):

$$U_4(t) = \frac{U_2(t) + U_2(t-2)}{2}$$

Answer: the 4-h UH is a rectangle of height 0.25 per hour from t = 0 to 4 h (area = 0.25 × 4 = 1 cm ✓). The same result comes directly from $[S(t)-S(t-4)]/4 = 1/4$.