1. Engineering Hydrology (ECIV 4323)
CHAPTER FIVE Runoff
Overland flow
interflow
Saturated overland flow
Base flow
Instructor:
Dr. Said Ghabayen

2. 5.1 Introduction To Runoff

3. Runoff Parts Runoff normally applies to flow over a surface
Overland flow: is the surface runoff from tracts of land (or sheet flow) before it reaches a defined channel .
Stream flow is used to describe the drainage after it reaches a defined channel
Drainage basins, watersheds, sub-basins, catchments: Defined by the total area contributing to a specific point. This point varies with the objectives of the project.

4. Stream Flow Components:
Direct precipitation on the channel (typically incorporated into total basin area)
Overland flow: when soil moisture storage and depression storage are filled “excess” rainfall generates overland flow.
Interflow: all rainfall that infiltrates does not reach saturated zone (ground water). Under certain conditions infiltrated moisture can travel through shallow soil horizons. Usually only significant for highly permeable soils.
Baseflow: contribution to stream flow from groundwater
Overland flow
interflow
Saturated overland flow
Base flow

5. 5.2 Hydrograph
Streamflow in terms of discharge vs time is represented by the hydrograph.
Hydrograph components thus become,
surface runoff
interflow
baseflow
direct precipitation
Direct Runoff: all moisture that reaches the stream channel without first entering zone of saturation (overland flow + interflow)
Streamflow = direct runoff + baseflow

6. 5.3 Runoff characteristics of streams
Perennial Stream (Fig 5.2)
A stream which always carries some flow. It is always fed from groundwater base flow. Even during dry seasons the water table will be above the bed of the stream.
An intermittent stream (Fig 5.3)
Stream is fed from groundwater base only in the wet season. During the wet season the water table is above the stream bed. During dry seasons the water table drops to a level lower than that of the stream bed; and therefore the stream dries up.
An ephemeral steam (Fig 5.4)
No contribution from groundwater (base flow) to the stream (most rivers in arid zones)

7. Water Year
In annual runoff studies it is advantageous to consider a water year beginning from the time when the precipitation exceeds the average evaporation losses.

8. Stream is always fed from groundwater base flow

9. Stream is fed from Groundwater base only in the wet season

10. No contribution from groundwater (base flow) to the stream (most rivers in arid zones)

11. 5.4: Runoff Volume SCS-CN METHOD OF ESTIMATIONING RUNOFF VOLUME

12. BASIC THEORY P= Ia + F + Q (1)
The SCS-CN method is based on the water balance equation of the rainfall in a known interval of time Δt , which can be expressed as
P= Ia + F + Q (1)
Where:
P = total precipitation
Ia = initial abstraction
F = cumulative infiltration excluding Ia.
Q = direct surface runoff (all in units of volume occurring in time Δ t).

13. BASIC THEORY

14. BASIC THEORY

15. Curve Number CN
Parameter S: the potential max retention depends upon the soil –vegetation-landuse complex of the catchment and also upon the antecedent soil moisture condition in the catchment just prior to the commencement of the rainfall event.
For convenience in practical application the soil Conservation Services (SCS) of USA has expressed S(in mm) in terms of a dimensionless parameter CN(curve Number) as

16. Curve Number CN The constant 254 is used to express S in mm.
The curve number is now related to S as
And has a range of 100 ≥ CN ≥ 0.
A CN value of 100 represents a condition of Zero potential retention (i.e impervious catchment) and CN =0 represents an infinitely abstracting catchment with S=∞. This curve number CN depends upon:
Soil Type
Land use/ cover.
Antecedent moisture condition.

17. Determination of CN
Soils are classified into four classes A, B,Cand D based upon the infiltration and other characteristics:
1. Group A: (low Runoff Potential).
2. Group B: (Moderately Low Runoff Potential).
3. Group C: (Moderately high Runoff Potential).
4.Group D: (High Runoff Potential).

18. Antecedent Moisture Condition (AMC)
Antecedent Moisture Condition (AMC) refers to the moisture content present in the soil at the beginning of the rain-fall- runoff event under consideration. It is well Known that initial abstraction and infiltration are governed by AMC.
For Purposes of practical application three levels of AMC are recognized by SCS as follows:
AMC-I: Soils are dry but not to wilting point. Satisfactory cultivation has taken place.
AMC-II: Average conditions.
AMC-III: Sufficient rainfall has occurred within the immediate past 5 days. Saturated soil conditions prevail.

19. Limits of three AMC classes
Limits of Classes based on total rainfall in the previous 5 days, and also it depends upon the seasons: two seasons; growing season, dormant season

20. Antecedent Moisture Condition (AMC)

21. The conversion of CNII to other two AMC condition can be made through the use of the following correlation equations. For AMC-I: (5)

24. For AMC-III:
(6)
The equations (5), (6) are applicable in the
,range of 55 to 95 which covers most of practical range.

25. Value of λ: On the basis of extensive measurements in small size catchments SCS(1985) adopted Value λ=0.2 as a standard value. With eq. becomes: Q: daily runoff. P: daily rainfall. S: retention parameter, all in units of mm.

26. Example:5.5

33. 5.5 Flow-Duration curve Step of construction the curve
Is a plot of discharge against the percentage of time the flow was equaled or exceeded (Discharge-Frequency curve)
Step of construction the curve
Stream flow discharge is arranged in decreasing order.
Use class interval if we have very large number of records (example 5.4)
Use plotting position to compute the probability of equal or exceeded.
m= order number of discharge (class value)
N= number of data
Pp= percentage probability
Plot the discharge Q against Pp (Fig 5.6)
Or semi-log or log-log paper (Fig 5.8)

34. 5.5 Flow-Duration curve
The flow duration curve represents the cumulative frequency distribution
The percentage probability of time that flow magnitude in average year is equal or greater
Q50 = 100 m3/s
Example 5.4

35. Example 5.4
The daily flows of a river for three consecutive years are shown in the table below. For convenience the discharges are shown in class intervals and the number of days the flow belonged to the class is shown.
Calculate the 50 and 75% dependable flows for the river

38. Flow-Duration Curve for different river types

40. 5.6 Flow mass curve
The flow-mass curve is a plot of cumulative discharge against time plotted in sequential order
The flow-mass curve integral curve (summation curve) of the hydrograph
See figure 5.9

41. Flow mass curve (Fig. 5.9)
Check the excel example in class web page for more details on this figure

42. … Flow mass curve
Unit is volume million m3 or m3/s day, cm over a catchment area
Slope of the mass curve =dv/dt =Q = flow rate
Average flow rate in the time between tm and tn
Slope of line AB is the average over whole period records

43. Calculation of storage volume
Assume there is a reservoir on a stream where FM curve is shown in Fig 5.9 and it is full at the beginning of a dry period, i.e. when the inflow rate is less than the withdrawal (demand) rate, the storage (S) of the reservoir
VD =demand volume (withdrawal)
VS = supply volume (Inflow)
The storage “S” is the maximum cumulative deficiency in any dry season. It is obtained as the maximum difference in the ordinate between mass curves of supply & demand.
If there are two dry periods, the minimum storage volume required is the largest of S1 and S2 (Fig. 5.9)

44. … Calculation of storage volume
i.e. The minimum storage volume required by a reservoir is the largest value of S
Refer to Figure 5.9
CD is drawn tangential to the first part of curve
Qd slope of CD is a constant rate of withdrawal from the reservoir
The lowest capacity is reached at E where EF is tangential at E
S1: the water volume needed as storage to meet the maximum demand (reservoir is full)
S2: is for C’D’
Then the minimum reservoir storage required is the largest storage S2>S1

45. Example 5.5
The following table gives the mean monthly flows in a river during Calculate the minimum storage required to maintain a demand rate of 40 m3/s.
Dec
Nov
Oct
Sept
Aug
July
June
May
Apr
Mar
Feb
Jan
Month 70 80 90
105 50 22 15 25 35 45 60
Mean Flow (m3/s)

46. … example 5.5
The actual number of days in the month are used in calculating the monthly flow volume. Volumes are calculated in units of cumec. day (= 8.64×104 m3)

47. … Example 5.5

48. Refer to Example 5.5
Compute the monthly flow volume by using the actual days in each month
Compute the accumulated volume in (cumec – day) or in m3.
Plot the flow mass curve (accumulated flow volume and time)
Draw a tangential line with slope of 40 m3/s
Slope = y/x y=x(slope)
y= 60.8 days x 40 =2432 cumec-day
To draw the demand line, the number of days in a month can be taken as 30.4 days.

49. Calculation of storage volume
Refer to Example 5.5
Flow mass curve (Figure 5.10)
Draw parallel line to AB in the valley point A’B’
Vertical value S1 is the storage required to maintain the demand
From figure m3/s.day
Solve Example 5.6

50. Example 5.6
Work out the example 5.5 through arithmetic calculation without the use of the mass curve
Dec
Nov
Oct
Sept
Aug
July
June
May
Apr
Mar
Feb
Jan
Month 70 80 90
105 50 22 15 25 35 45 60
Mean Flow (m3/s)

51. … Example 5.6

52. Calculation of maintainable demand
The storage volume is given. Your job is now to determine the maximum demand that can be maintained by this given storage volume.
Fig. 5.11:
Tangents are drawn from the ridges and valleys of the mass curves at various slopes (since we still don’t know the demand!).
The demand line that requires just the given storage is the proper demand that can be sustained by the reservoir in that dry period.
Similar demand lines are drawn at other valleys.
The smallest of the various demand rates thus is the maximum solid demand that can be sustained by the given storage
Check points 1 & 2 of page 162

53. Variable demand
In practice the demand rate varies with time to meet various end uses (such as irrigation, power and water supply needs, etc.)
A mass curve of demand is prepared and superposed on the flow-mass curve.

54. Mass curve of Flow and Mass curve of Variable Demand

55. In the analysis of problems related to reservoirs it is necessary to account for evaporation, leakage, and other losses from the reservoir.
These losses can either be included in demand rates or deducted from inflow rates. If you used the second method, the mass curve may have negative slopes at some points.
Solve Example 5.8