STEM Problem Detail

Problem #60

Civil Engineering Hard

Problem Statement

Prompt

provide me the stepwise solution for question 7.5 .

Solution Rendering...

## Solution to Hydrograph Prediction Problem 7.5

## Problem

A drought is ended over a catchment area of $$100$$ km$$^2$$ by uniform rain of $$36$$ mm falling for $$6$$ h. The relevant hydrograph of the river draining the area is given below, the rain period having been between hours $$3$$ and $$9$$. Use these data to predict the maximum discharge that might be expected following a $$50$$ mm fall in $$3$$ h on the catchment. Qualify the forecast appropriately.

| Hours | Discharge (m$$^3$$/s) | Hours | Discharge (m$$^3$$/s) |
|---|---|---|---|
| 0 | 3 | 24 | 25 |
| 3 | 3 | 27 | 21 |
| 6 | 10 | 30 | 17 |
| 9 | 25 | 33 | 13.5 |
| 12 | 39 | 36 | 10.5 |
| 15 | 43 | 39 | 8 |
| 18 | 37 | 42 | 5.5 |
| 21 | 30.5 | 45 | 4 |
| | | 48 | 3.9 |

## Solution

### Step 1: Understand the given data

- Catchment area: $$A = 100$$ km$$^2 = 100 \times 10^6$$ m$$^2$$
- Current storm: $$P_1 = 36$$ mm $$= 0.036$$ m over $$6$$ hours
- Rainfall intensity: $$i_1 = \frac{36}{6} = 6$$ mm/h
- Rain period: hours $$3$$ to $$9$$ (duration $$6$$ hours)
- Baseflow: From data, initial flow $$= 3$$ m$$^3$$/s
- Peak flow observed: $$Q_{p1} = 43$$ m$$^3$$/s at $$t=15$$ hours
### Step 2: Calculate direct runoff volume from current storm
First, separate baseflow from total flow.
Baseflow estimation: Initial baseflow $$= 3$$ m$$^3$$/s. Assume constant baseflow or use recession method.
From data:
- At $$t=48$$, flow $$= 3.9$$ m$$^3$$/s $$\approx 4$$ m$$^3$$/s
- At beginning: $$3$$ m$$^3$$/s
- Assume baseflow increases slightly during event
For simplicity, assume baseflow $$= 3$$ m$$^3$$/s constant (or take average $$= 3.5$$ m$$^3$$/s).
Using constant baseflow $$Q_b = 3$$ m$$^3$$/s:
| Time (h) | Total $$Q$$ (m$$^3$$/s) | Baseflow (m$$^3$$/s) | Direct Runoff (m$$^3$$/s) |
|---|---|---|---|
| 0-3 | 3 | 3 | 0 |
| 3-6 | 10 | 3 | 7 |
| 6-9 | 25 | 3 | 22 |
| 9-12 | 39 | 3 | 36 |
| 12-15 | 43 | 3 | 40 |
| 15-18 | 37 | 3 | 34 |
| 18-21 | 30.5 | 3 | 27.5 |
| 21-24 | 25 | 3 | 22 |
| 24-27 | 21 | 3 | 18 |
| 27-30 | 17 | 3 | 14 |
Direct runoff volume $$V_d$$:
$$V_d = \sum (Q_{\text{direct}} \times \Delta t) \times 3600$$ (seconds)
Using 3-hour intervals:
$$V_d = (0+7+22+36+40+34+27.5+22+18+14) \times (3 \times 3600)$$ m$$^3V_d = 220.5 \times 10800 = 2,381,400$$ m$$^3$$
### Step 3: Calculate runoff coefficient
Rainfall volume $$V_r$$:
$$V_r = A \times P_1 = 100 \times 10^6 \times 0.036 = 3,600,000$$ m$$^3$$
Runoff coefficient $$C$$:
$$C = \frac{V_d}{V_r} = \frac{2,381,400}{3,600,000} = 0.6615 \approx 0.66$$
### Step 4: Apply to new storm conditions
New storm: $$P_2 = 50$$ mm $$= 0.050$$ m over $$3$$ hours
Rainfall intensity: $$i_2 = \frac{50}{3} = 16.67$$ mm/h
#### Option 1: Using runoff coefficient
Direct runoff volume for new storm:
$$V_{d2} = C \times A \times P_2 = 0.66 \times 100 \times 10^6 \times 0.050V_{d2} = 0.66 \times 5,000,000 = 3,300,000$$ m$$^3$$
But this assumes same temporal distribution.
#### Option 2: Using unit hydrograph principles
From the given hydrograph:
- Peak direct runoff $$= 43 - 3 = 40$$ m$$^3$$/s
- Rainfall depth $$= 36$$ mm
For unit hydrograph (1 mm of effective rainfall):
Peak flow per mm $$= \frac{40}{36} = 1.111$$ m$$^3$$/s per mm
### Step 5: Account for storm duration difference
Current storm: $$6$$ hours duration, intensity $$6$$ mm/h
New storm: $$3$$ hours duration, intensity $$16.67$$ mm/h
The shorter, more intense storm will produce a higher peak flow.
Using the rational method concept: $$Q_p = C i A$$
For same $$C$$:
$$\frac{Q_{p2}}{Q_{p1}} = \frac{i_2}{i_1} = \frac{16.67}{6} = 2.78$$
So: $$Q_{p2\text{(direct)}} = 40 \times 2.78 = 111.2$$ m$$^3$$/s (direct runoff)
Total peak: $$Q_{p2} = 111.2 + 3 = 114.2$$ m$$^3$$/s
### Step 6: More refined estimation using hydrograph scaling
From given hydrograph:
- Time to peak: $$15 - 6 = 9$$ hours from rain start (rain: 3-9 h, peak at 15 h)
- Base length: About $$45$$ hours total
For shorter storm (3 hours), time to peak will be shorter.
Typically: $$T_p \propto \sqrt{L}$$ where $$L$$ is storm duration
If $$T_{p1} = 9$$ h for 6 h storm,
$$T_{p2} = 9 \times \sqrt{\frac{3}{6}} = 9 \times 0.707 = 6.36$$ h

Peak flow inversely proportional to time to peak:
$$Q_{p2} = Q_{p1} \times \frac{T_{p1}}{T_{p2}} \times \frac{P_2}{P_1}Q_{p2\text{(direct)}} = 40 \times \frac{9}{6.36} \times \frac{50}{36}Q_{p2\text{(direct)}} = 40 \times 1.415 \times 1.389 = 40 \times 1.966 = 78.64$$ m$$^3$$/s

Total peak: $$Q_{p2} = 78.64 + 3 = 81.64$$ m$$^3$$/s

### Step 7: Consider SCS method concepts

For same catchment, peak flow $$\propto$$ (rainfall depth)$$^n$$ where $$n \approx 0.5-0.7$$
Using $$n = 0.6$$:
$$Q_{p2} = 43 \times \left(\frac{50}{36}\right)^{0.6} = 43 \times (1.389)^{0.6}= 43 \times 1.214 = 52.2$$ m$$^3$$/s

### Step 8: Final prediction with qualifications

Given the various methods, a reasonable estimate:

$$\boxed{Q_{p\text{max}} \approx 80-110 \text{ m}^3/\text{s}}$$

Most likely: $$\boxed{Q_{p\text{max}} \approx 90 \text{ m}^3/\text{s}}$$

## Qualifications and Assumptions

1. **Runoff coefficient assumed constant:** This may not hold for different intensities. Higher intensity may yield higher $$C$$.

2. **Catchment conditions:** The drought ending suggests dry initial conditions. For the new storm, antecedent conditions are unknown but likely wetter, giving higher runoff.

3. **Storm temporal pattern:** Uniform rainfall assumed, but actual pattern affects peak flow.

4. **Baseflow assumption:** Constant baseflow of $$3$$ m$$^3$$/s used. Actual baseflow might be higher for second storm if catchment is wetter.

5. **Hydrograph shape:** Assumed similar shape but compressed in time for shorter storm.

6. **No flooding effects:** Channel capacity limitations not considered.

7. **Spatial distribution:** Uniform rainfall over catchment assumed.

8. **Climate factors:** Season, temperature, evaporation not considered.

## Recommended Approach

For most accurate prediction:
1. Develop a unit hydrograph from the given data
2. Adjust for storm duration using S-curve method
3. Apply to new rainfall hyetograph
4. Add estimated baseflow

Given the data limitations, the prediction range of $$80-110$$ m$$^3$$/s is reasonable, with $$90$$ m$$^3$$/s as a best estimate.

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