Chapter 5. Infiltration – Elementary Engineering Hydrology

5

Infiltration

Chapter Outline

5.1 DEFINITION

Infiltration may be defined as entry and movement of water through the land surface into the substrata below.

5.1.1 Infiltration and Percolation

Infiltration includes entry of water into the soil surface and its movement, while percolation is the movement of water under gravity. These two phenomena are confusing as they are closely related, but technically there is a difference. Percolation starts after infiltration.

5.1.2 Abstractions

When there is precipitation, it may or may not result in overland flow into a stream depending upon its intensity and duration. The part of precipitation that is not available as surface runoff is referred to as precipitation loss or abstraction. The abstractions include (1) interception, (2) depression storage, (3) evaporation and (4) infiltration.

Generally, interception and depression storages are together termed as ‘surface retention’ Knowledge of these abstractions, their rate, and so on, is necessary for the determination of the surface flow.

5.1.3 Dominant Abstraction

Out of these abstractions, evaporation during precipitation is very negligible. Interception and depression storage are comparatively small as compared to infiltration. Hence, infiltration is called as dominant abstraction.

5.1.4 Excess Rainfall

The part of precipitation that is available in the form of surface flow after meeting all the abstractions is known as excess rainfall. It is also known as effective rainfall.

5.2 PROCESS OF INFILTRATION

When rainwater falls on the ground, there is some resistance offered by the soil surface for the entry of rainwater and also to the flow of water through the soil. There are cracks, vertical as well as lateral in the soil, so also there are some voids between the soil particles, which are ordinarily occupied by air or water. Water flows through these cracks and gaps until it reaches the saturated zone below. Naturally, the passage of water experiences some resistance.

The rate at which water enters the ground surface and then flows downwards is known as infiltration rate. This rate is high in the beginning because it has to meet the requirements of the dry soil. However, it attains a steady constant lower value after a passage of time.

The unit for the rate of infiltration is mm/h.

5.3 FACTORS AFFECTING INFILTRATION

The factors affecting infiltration are discussed below:

5.3.1 Rainfall Characteristics

The duration as well as the intensity of rainfall influences infiltration. If the intensity of rainfall is more than the infiltration rate, then only surface runoff is noticed. On the contrary, if the intensity of rainfall is less than the infiltration rate, no surface flow is observed and all the rainfall is abstracted as infiltration.

The rate of infiltration is high in the beginning and goes on reducing and attains a steady state after some time. At earlier stage the intensity of rainfall may be less than the rate of infiltration. Under this condition, the ground will absorb all the rainfall and there will not be any surface flow.

However, when the infiltration rate reduces and becomes less than the intensity of precipitation, surface flow will be noticed, which might increase till the infiltration rate stabilizes. In addition to these considerations, rainfall has some additional effects as follows:

  • When rainwater strikes the bare soil, there is mechanical compaction of soil due to the impact, which may reduce the infiltration rate.
  • Fine soil particles are carried down due to rain water resulting in choking of the pore spaces in the soil, and consequently resulting in reducing the infiltration rate.

5.3.2 Ground Surface Condition

The land surface that receives rain may be bare, vegetated or covered with mulch or litter. The bare ground receiving rainfall may be subjected to the effects of impact, and so on. If impervious material is exposed at the surface, naturally infiltration is small or negligible.

The vegetated ground reduces the impact effect of the raindrops, avoids dislodging of particles by mechanical binding action of the roots and hence provides a high rate of infiltration by maintaining open soil structure and also slows down the rate of runoff.

Thus, due to vegetated ground the rate of infiltration increases.

Mulch cover has got a similar effect. Because of mulch and litter, the infiltration rate increases. This has been proved by the field experiments. Contour ploughing and terracing in the agricultural land delays the surface flow and increases the infiltration rate.

Small surface slope has little effect on the infiltration rate. If the slope of the ground is steeper than 16°, then the rate of infiltration reduces due to the velocity of water causing more overland flow.

5.3.3 Soil Characteristics

Soil characteristics have definite effect on infiltration.

A uniformly graded material will have more pores and hence the infiltration rate will be more than a well-graded material. In clayey soils, because of the removal of moisture due to evaporation, some shrinkage cracks may be observed. These are termed as Sun cracks. Because of these cracks, infiltration rate may increase in the initial stage. Subsequently, when the soil gets wet, the cracks get closed and may not affect the infiltration rate.

Because of the presence of water, some soils have a tendency to form aggregation that reduces the infiltration rate. Also, the organic material present helps to promote soil aggregation. In some cases, puddle is formed at the surface where the soil has a fair proportion of clay, which becomes relatively impervious when it dries.

In the case of urbanization, because of concrete buildings, asphalt pavement, and so on, the infiltration rate is considerably reduced.

5.3.4 Soil Moisture

Even if the soil contains some moisture, there is no effect, practically, on the rate of infiltration except that the rate is reduced at the initial stage, as shown in Fig. 5.1.

5.3.5 Human Activities

Cultivation of land disturbs the soil structure, closes the openings made by burrowing animals and insects as well as decaying roots and thus reduces the rate of infiltration.

5.3.6 Climatic Conditions

The flow of water through the soil is laminar. Change in temperature may cause change in the viscosity of water and consequently may cause change in the velocity of water, and thus may affect the rate of infiltration.

Fig. 5.1 Rate of infiltration w.r.t. time

5.3.7 Entrapped Air

If infiltration process covers a large area, there may not be an exit passage to the entrapped air in the soil. Also, because of the downward passage of water, the air entrapped may get compressed and may offer more resistance to the flow, which may result in reducing the infiltration rate.

5.3.8 Other Minor Factors

Other factors, such as depth of water on the ground, groundwater table, and so on, have practically no effect on the infiltration rate.

5.4 MEASUREMENT OF INFILTRATION

The rate of infiltration is initially high. It goes on reducing with time and after some time it becomes steady. A usual graph of the rate of infiltration is shown in Fig. 5.1. The rate of infiltration for a soil is measured in the field as well as in the laboratory. These are known as infiltrometers.

The most common types are the following:

  • Flooding-type infiltrometers
  • Sprinkling-type infiltrometers or rain simulators

5.4.1 Flooding-Type Infiltrometers

There are two types of flooding-type infiltrometers:

  1. Single-tube flooding infiltrometer
  2. Double-tube flooding infiltrometer

Fig. 5.2 Single-tube infiltrometer

5.4.1.1 Single-tube flooding infiltrometer

The single-tube flooding infiltrometer consists of a metal tube 250–300 mm in diameter with both ends open. This tube is driven in a vertical position into an open level ground surface up to a depth of 500 mm, leaving about 100 mm above the ground. The tube is so driven into the ground that the soil is disturbed to a minimum.

Water is then added to this tube to maintain a constant level, sufficiently deep to submerge the plant or the grass crowns. If the soil is bare, i.e., there is no vegetation or grass, the soil is protected by a perforated metal disc to avoid turbidity.

A pointer gauge is used to measure the water level accurately. As the infiltration starts, the water level may drop down but it is maintained at a constant level by adding a measured quantity of water at successive time intervals, till a constant rate of infiltration is achieved.

Knowing the quantity of water added, the time interval and the area of tube, the rate of infiltration can be worked out. The infiltration rate thus worked out is more than the actual value.

Figure 5.2. shows a single-tube infiltrometer. It is also known as single-ring infiltrometer.

5.4.1.2 Double-tube flooding infiltrometer

In the case of a single-tube infiltrometer, water may flow sideways. Such loss is controlled to some extent by using one more ring outside the test ring. Instead of one tube, two concentric circular tubes are used. These may be of size 300 and 600 mm in diameter. These are driven 150 mm in the ground in a vertical position leaving 100 mm above the ground level. The outside metal ring is used to avoid the side and border effects.

The procedure followed for single tube is followed in this double-tube case also. Water is added into both the tubes and the same level of water is maintained in both tubes by adding the measured quantity of water at successive time intervals.

The observations for the inner tube are used for working out the infiltration rate. Knowing the water added, the time interval and the area of cross section of the inner tube, the infiltration rate with respect to time can be calculated.

Figure 5.3. shows a double-ring infiltrometer. It is also known as double-ring infiltrometer.

Fig. 5.3 Double-tube infiltrometer

5.4.1.3 Drawbacks in tube infiltrometers

The main drawbacks in the tube infiltrometers are as follows:

  • The soil is disturbed to some extent when the tubes are driven into the soil.
  • There may be lateral flow of infiltrated water. In double-tube infiltrometers, efforts are made to reduce it, but some lateral flow still occurs.
  • Similarly, air entrapped in the soil may escape laterally. In a double-tube infiltrometer, this effect persists.
  • Effect of the raindrop impact on soil is not accounted.
  • Effect of the slope of ground is not accounted.
  • Experiments cannot be conducted on soil with boulders etc.
  • The infiltration is affected because of the ring size. Smaller the diameter of the ring, more will be the rate of infiltration.

Example 5.1

The quantity of water added to a double-ring infiltrometer of 1.00-m diameter at 30 min interval to keep the water level constant is as follows:

Find:

  1. Rate of infiltration for every 30 min and plot the graph
  2. Average rate of infiltration

Fig. 5.4 Estimation of infiltration rate

Solution:

The graph of time in hours versus infiltration rate is as shown in Fig. 5.4.

Total quantity of water added in 150 min till a steady state was achieved = 10.0 + 9.2 + 8.6 + 8.2 + 8.0 = 44.0 lit

Therefore, average rate of infiltration

5.4.2 Sprinkling-Type Infiltrometer

This experiment is conducted in the laboratory under controlled conditions.

For this purpose, a very small catchment area, about 5 m2, is selected. Water is sprinkled over the catchment area to represent rainfall at a uniform rate of about 50 mm/h. Sprinkling of water may be from a height of 5 m. The trial starts with the sprinkling of water. It is, therefore, called sprinkling- type infiltrometer. Sprinkling of water represents rain, and hence it is also called rain simulator.

Fig. 5.5 Selected catchment area

The surface runoff is measured very accurately at the catchment outlet and thus knowing all the abstractions such as ‘evaporation, interception, depression storage, surface detention’ and surface runoff, the infiltration rate during the time of experiment can be worked out.

Consider the following particulars of a specific experiment.

A small catchment was selected for the experiment as shown in Fig. 5.5.

The experiment started at 8:00 a.m. by sprinkling water over the entire catchment area at a uniform rate, which was stopped at 12:00 noon. There was no surface runoff noticed up to 8:30 a.m. The discharge was noticed at 8:30 a.m. and it went on increasing up to 10:00 a.m. and remained constant up to 12:00 noon. When the sprinkling of water, i.e. rainfall, was stopped, the discharge went on reducing and finally stopped at 2:00 p.m. as shown in Fig. 5.6.

Fig. 5.6 Observed discharge

The rainfall P may be accounted for as follows:

 

P = E + I + DS + F + SD + Q

 

where, P = Precipitation at a uniform rate over the entire catchment in mm/h

      E = Evaporation in mm/h

      I = Interception in mm/h

      DS = Depression storage in mm/h

      F = Infiltration in mm/h

      SD = Surface detention (water stored temporarily in channel)

      Q = Surface runoff in m3/s converted to mm/h over the catchment area

 

     Thus, F = P − (E + I + DS + SD + Q)

(All these quantities are measured in mm uniformly spread over the entire catchment area.) The parameters in the bracket were calculated separately, combined together and then subtracted from the precipitation and thus the infiltration rate was calculated. Infiltration started immediately at 8:00 a.m. The following assumptions were made:

  1. Evaporation being negligible during the experiment was neglected.
  2. As usual, interception and depression storage were combined together and treated as SR surface retention. This abstraction started immediately at 8:00 a.m.

The precipitation was started at 8:00 a.m. and stopped at 12:00 noon. The precipitation graph with respect to time and its mass curve is shown in Fig. 5.7.

A judgement was taken as to how much water was lost as surface retention, and its requirement would be first met with completely from 8:00 a.m. to 8.30 a.m. and there would be no addition in this abstraction after 8:30 a.m. Surface runoff started only after the requirement of the surface retention was met with. The surface retention and its mass curve is shown in Fig. 5.8.

Fig. 5.7 Mass curve of precipitation

Fig. 5.8 Surface retention

The discharge was observed from 8:30 a.m. onwards. It went on increasing up to 10:00 a.m. It was constant from 10:00 a.m. to 12:00 noon and then reduced from 12:00 noon to 2:00 p.m. The discharge as well as its mass curve is shown in Fig. 5.9.

The SD is in mm/s, i.e. water stored in the channel is a function of depth in a channel. More the discharge in the channel, more will be the depth in the channel and more will be the storage in the channel. Thus, indirectly SDwill be a function of discharge. Here the discharge was started at 8:30 a.m. Similarly, SD also started at 8:30 a.m. It went on increasing up to 10:00 a.m. It remained constant from 10:00 a.m. to 12:00 noon and then was reduced from 12:00 noon to 2:00 p.m.

Fig. 5.9 Mass curve of discharge

Fig. 5.10 Estimation of surface detention

Since initially and also after the experiment the channel carried no water, the water stored in the channel from 8:30 a.m. to 12:00 a.m. was drained completely from 12:00 noon to 2:00 p.m.

When the precipitation stopped at 12:00 a.m., there was still flow in the stream. This was due to draining the water stored in the channel, i.e. SD. Immediately after the precipitation stopped, the discharge was maximum. SD was also maximum. The total volume of water stored in the channel was the area under the recession curve. As the discharge reduced, so also did the storage in the channel. The storage in the channel will be the area under the recession curve. The recession curve is shown in Fig. 5.10.

The area under the recession curve is the total channel storage.

 

A = A1 + A2 + A3

 

When the discharge is Q1, the channel storage SD is A1.

When the discharge is Q2, the channel storage SD is A1+ A2.

When the discharge is Q3, the channel storage SD is A1+ A2+ A3, which is the area under the recession curve.

Thus, the relation between Q and SD can be established as shown in Fig. 5.11.

SD can be calculated from this graph and then correlated with time. SD thus calculated will be in terms of m3 and will have to be worked out in terms of mm over the entire catchment area for further calculations. The surface detention and its mass curve are shown in Fig. 5.12.

A combined mass curve of (SR+ Q + SD) and mass curve of P is plotted on the graph paper and the difference between these two mass curves is the infiltration rate w.r.t. time as shown in Fig. 5.13.

From this figure, the infiltration rate w.r.t. time can be plotted. The infiltration rate is more at the beginning and goes on reducing, and after some time it becomes steady as shown in Fig. 5.14.

Fig. 5.11 Surface detention w.r.t. discharge

Fig. 5.12 Mass curve of surface detention

5.4.2.1 Drawbacks in sprinkling-type infiltrometer

The main drawbacks in the sprinkling-type infiltrometer are as follows:

  • It is very difficult to have a uniform sprinkling of water of a uniform intensity over the entire catchment area for a sufficient duration.
  • The discharge has to be measured very accurately.
  • The assumption of SR requires good judgement.
  • The area under the experiment being small, lateral flow may be significant.

Fig. 5.13 Estimation of infiltration by mass curves

 

Example 5.2

 

A field experiment to assess the infiltration capacity of an area was conducted by following the Sprinkling type infiltrometer technique’. The discharge was measured accurately over a notch at 10 minutes interval.

Fig. 5.14 Horton’s equation of infiltration rate

The details are as follows.

1. Catchment area

4.5 m2

2. Experiment started at

8:00 a.m.

3. Experiment concluded at

12:50 a.m.

4. Uniform rainfall intensity used

4 cm/h

The discharge measured is as follows.

The discharge measured on the notch in c. c. /s, was converted for simplicity, into depth in cm /h uniformly distributed over the entire catchment area of 4.5 m2, for these calculations. 12.5 c.c./s is equivalent to 1.00 cm /h as follows

The discharge measured in c.c./s converted in cm / h is entered in col. 4 in the statement.

The calculations done to evaluate the infiltration rate are tabulated in the following statement.

* Read from the graph in Fig. 5.16.

** Assumed

 

The experiment was started at 8:00 a.m. by starting sprinkling water at the rate of 4.00 cm / h over the entire catchment at a uniform rate. There was no flow of water upto 8:10 a.m. since the requirement of SR [Interception and depression storage] was to be met with.

It was assumed plot SR is equivalent to 2.35 cm. The discharge started flowing at 8:20 a.m. but was very low because the requirement of SR was not fully met with. It was assumed that the requirement of SR of 2.35 cm. was completely met with upto 8:20 a.m.

The discharge as it started at 8:20 went on increasing upto 10:30 a.m. and reached 2.50 cm and remained constant from 10:30 a.m. upto 10:50 a.m.. This indicated that the infiltration rate has reached a steady state.

From 10:50 a.m., sprinkling of water was stopped.

Naturally the discharge started reducing from 10:50 a.m. and was completely stopped at 12:50 p.m. Fig. 5.15. shows the discharge observed.

The storage capacity SD of the channel w.r.t. discharge flowing in the channel is calculated from the falling graph of discharge from 10:50 a.m. onwards and is shown in Fig. 5.16.

Fig. 5.15 Discharge

Fig. 5.16 Discharge-Storage curve

Fig. 5.17 Infiltration rate

The infiltration rate of the soil will be as

F = P − [Q + SD + SR]

Since SD is a mass curve, all calculations for F are done by preparing mass curves of P and SD Q + SR. The mass curve of the infiltration thus worked is entered in col. 9 of the statement. The infiltration rate w.r.t. time is entered in col. 10.

The infiltration rate thus calculated is shown in Fig. 5.17.

The infiltration rate ‘4 cm/h’ is maximum at the beginning. It goes on reducing following a curved path with concavity upwards. It is constant ‘1.5 cm/h’ for some time indicating a steady state. Then it goes on reducing. This curve is also having concavity upwards and finally it reduces to zero.

5.5 EXPRESSION OF INFILTRATION

The rate of infiltration is expressed as follows:

  • An equation for the graph of the rate of infiltration
  • Infiltration index

5.5.1 Equation

Infiltration rate is higher at the beginning. It goes on reducing w.r.t. time, and finally it attains a steady rate as shown in Fig. 5.14.

Horton has suggested the following equation for the curve

 

f = fc + (fofcekt

 

where, f = Infiltration rate at time t

      fo = Initial infiltration rate

      fc = Final infiltration rate

      k = A constant

      t = Time in hours

when t = 0, e−kt = 1 then f = fo

when t = ∞, e−kt = 0 then f = fc

Value of k is normally between 2 to 5.

 

Example 5.3

 

The value of k in the Horton’s equation for infiltration is 2. The maximum and the minimum rates of infiltration are 2 cm/h and 0.5 cm/h. Plot the infiltration rate curve.

 

Solution:

 

The Horton’s equation will be:

f = 0.5 + (2.0 0.5) e−2t = 0.5 + 1.5−2t and the hourly rate of infiltration will be as under:

The infiltration rate curve will be as shown in Fig. 5.18.

Fig. 5.18 Estimation of infiltration rate

5.5.2 Infiltration Indices

For simplicity in calculations, the infiltration rate is expressed as a uniform average rate. This is expressed in the following ways:

  • ϕ index
  • W index
  • Wmin index
  • fAV

5.5.2.1 ϕ index

The most common way of expressing infiltration rate is ϕ index.

Now, P = E + SR + SD + Q + I

 or, I = P − (E + SR + SD) − Q

where, P = Precipitation

       E = Evaporation

      SR = Surface retention (interception + depression storage)

      SD = Surface detention (finally it flows as surface runoff)

      Q = Surface runoff

      I = Infiltration rate

Neglecting E, SR and SD, one can say that

 

ϕ index = PQ

 

Indirectly, ϕ index includes all abstractions, e.g. evaporation, surface retention, and so on. Figure 5.19 shows definition sketch of ϕ index.

Fig. 5.19 Definition sketch of ϕ index

5.5.2.2 W index

W index is the rate of infiltration when all other abstractions are accounted, viz,

 

W index = P − (E + SR+ SD + Q)

 

Therefore, W index < ϕ index

5.5.2.3 Wmin index

It is the minimum rate of infiltration when a uniform stage after stabilization is attained.

Wmin = ϕ index after it is stabilized

       = index, where k will be always less than one

5.5.2.4 fAV

This approach is a slight modification of ϕ index. In a storm, when there is no precipitation or when the precipitation is very low, infiltration is still going on because of the previous high precipitation. A provision is, thus, made for this infiltration when precipitation is very low from the previous precipitation when it is high. Consider a storm as shown in Fig. 5.20.

Fig. 5.20 Definition diagram of fAV

A provision for infiltration when there is no precipitation, i.e. between 4 to 5 h, is made from the storm precipitation between 3 to 4 h and hence is deducted from the excess rainfall from 3 to 4 h. fAV is calculated and then the excess rainfall is worked out. It is slightly less than ϕ index.

 

Example 5.4

 

The average precipitation during a storm over a catchment area of 10 km2 is as follow:

  1. 40 mm/h for 1 h
  2. 60 mm/h for 1 h
  3. 30 mm/h for 1 h

The resulting hydrograph was plotted on a graph paper with the following scale.

1 cm = 1 h on x-axis

1 cm = 10 m3/s on y-axis

If the area of the hydrograph was measured as 30 cm2, find the ϕ index of infiltration.

 

Solution:

 

Total runoff observed = 30 × 10 × 3600

                                 = 1.08 × 106 m3

Runoff from the storm assuming rate of infiltration to be ϕ

Therefore, (130 − 3ϕ) × 104 = 1.08 × 106

                                         3ϕ = 22

                                           ϕ = 7.33 mm/h

 

Example 5.5

 

The storm over a catchment of 50 km2 was having the following intensity:

  1. 40 mm/h for 1 h
  2. 70 mm/h for 2 h
  3. 30 mm/h for 1 h

The catchment area had infiltration rate as follows:

  1. 20% area ϕ =10 mm/h
  2. 60% area ϕ =15 mm/h
  3. Balance impervious

Find the runoff due to the storm.

 

Solution:

 

  1. The runoff from the area having ϕ = 10 mm/h will be
  2. The runoff from the area having ϕ = 15 mm/h will be
  3. The runoff from impermeable area will be

Therefore, total runoff = 1.7 × 106 + 4.5 × 106 + 2.1 × 106

                                  = 8.3 × 106 m3

Example 5.6

A catchment area having an average rate of infiltration of 15 mm/h experienced the storm of the following intensities.

  1. 50 mm/h for 2 h
  2. 30 mm/h for h

The resulting runoff was 10 × 106 m3.

Find the catchment area.

Solution:

Example 5.7

For a catchment area of 12 km2, a 7-h storm was as follows:

The discharge observed at the gauging site was as follows:

Assume evaporation loss to be 0.6 mm/h/m2and seepage loss equal to 50% of the evaporation loss. Find the ϕ index fAV index and W index.

Solution:

Total precipitation = (20 + 40 + 0 + 30 + 50 + 40 +5) = 185 mm

∑ Discharge observed =

(0+ 7+ 17 + 32 + 66 + 55 + 45 + 36 + 28 + 21 + 15 + 10 + 6 + 3 + 1 + 0) = 342 m3/s

ϕ index

Assume ϕ index = x mm/h

Assuming that the value of ϕ index will be more than 5 mm/s, the last rainfall slab of 5 mm/s is neglected.

fAV

Assume fAV = y mm/h

Even though there is no precipitation between 2 and 3 h, there will be infiltration. This is taken into consideration. So also the precipitation between 6 and 7 h is far less than the fAV (expected). Infiltration during this period is also taken into consideration.

      Therefore, fAV = y = 11.77 mm/h

W index

Assume W index = z mm/h

Evaporation loss + seepage loss = 1.5 × 5 × 0.6 = 4.5 mm

      

Therefore, W index = z =14.58 mm/h

Figure 5.21 shows the three indices thus worked.

Fig. 5.21 Estimation of indices

Example 5.8

The rate of precipitation in mm/h observed over a catchment of 30 km2 for successive 30 min is as follows:

        16, 20, 24, 36, 28, 12, 4

If the value of ϕ = 22 mm/h, find

(1) Total precipitation and (2) Runoff in ha. m

Solution:

Since the precipitation observed is for a duration of 1/2 hour, the total precipitation will be as follows:

Since the rate of infiltration is higher than the first, second, sixth and seventh slab, all the precipitation will be lost in the infiltration and the runoff will be only from third, fourth and fifth slab. It will be as follows:

REVIEW QUESTIONS
  1. Define infiltration. Explain the process of infiltration.
  2. Discuss the factors affecting infiltration.
  3. Explain with the help of neat sketches the flooding-type infiltrometers. What are their advantages and disadvantages?
  4. Explain sprinkling-type infiltrometer.
  5. Discuss the infiltration indices.
  6. Write short notes on the following:
    1. Dominant abstraction
    2. Disadvantages of tube infiltrometers
    3. Horton’s equation of infiltration rate
    4. Estimation of ϕ index
    5. Excess rainfall
    6. Limitations of sprinkling-type infiltrometer
  7. Differentiate between the following:
    1. Infiltration and percolation
    2. Single-tube infiltrometer and double-tube infiltrometer
    3. ϕ index and fAV index
    4. W index and Wmin index
NUMERICAL QUESTIONS
  1. The quantity of water added to a double-ring infiltrometer with 1.0-m inside diameter at 30-min interval to keep the water level constant is as under:

    Find,

    1. Rate of infiltration for every 30 min and plot the graph
    2. Average rate of infiltration

    Ans: (ii) 11.49 mm/h/m2

  2. The value of K in the Horton’s equation for infiltration is 2.2. And the maximum and minimum rates of infiltration are 2.1 cm/h and 0.5 cm/h. Plot the infiltration rate curve.
  3. A catchment area of 13 km2 experiences a storm of average precipitation as: (1) 45 mm/h for 1 h, (2) 50 mm/h for 1.5 h, (3) 25 mm/h for 1 h and (4) 8 mm/h for 1 h.

    The resulting hydrograph was plotted on a graph with the following scales:

        1 cm = 1 h on x-axis

        1 cm = 10 m3/s on y-axis

    The area under the hydrograph thus plotted was measured and was found to be 40 cm2. Find the ϕ index of infiltration.

    Ans: 10 mm/h

  4. The storm over a catchment of 60 km2 was having the intensity as: (1) 45 mm/h for 1 h, (2) 60 mm/h for 2 h and (3) 35 mm/h for 1 h.

    The catchment area had infiltration rates as under:

    1. ϕ for 25% area = 10 mm/h
    2. ϕ for 50% area = 15 mm/h
    3. Balance impervious

    Find the runoff due to this storm.

    Ans: 8.53 × 106 m3

  5. For a catchment area of 12 km2, a storm of 6-h duration was observed as under:

    The discharge observed at the gauging site was measured as follows:

    Assume the evaporation loss to be 1 mm/day/m2 and the seepage loss equal to 50% of the evaporation loss. Find, ϕ index, fAV and W index.

    Ans: ϕ index = 10.95 mm/h
    fAV index = 7.30 mm/h
    W index = 9.45 mm/h

MULTIPLE CHOICE QUESTIONS
  1. Because of mulch and litter, the infiltration rate
    1. Reduces
    2. Increases
    3. Has no effect
  2. When the slope of the ground is more than 16°, it has an effect in
    1. Increasing the rate of infiltration
    2. Reducing the rate of infiltration
    3. Has no effect on the rate of infiltration
  3. The rate of infiltration observed on a single-tube infiltrometer is
    1. Less than that observed on a double-ring infiltrometer.
    2. More than that observed on a double-ring infiltrometer.
    3. Equal to that observed on a double-ring infiltrometer.
  4. The curve of the infiltration rate is
    1. Concave upwards
    2. Concave downwards
    3. A straight line
  5. Wmin index will always be
    1. Less than ϕ index
    2. More than ϕ index
    3. Equal to ϕ index
    4. None of the above
  6. Which of the following will have the maximum rate of infiltration?
    1. Forest area
    2. Grazed land
    3. Rock outcrops
    4. Concrete Pavement
  7. Horton’s equation of infiltration rate is given by
    1. f = fc + (fofc) ekt
    2. f = fc + (fo − f) e-kt
    3. f = fc − (fofo) ekt
    4. f = fc + (fo + fc) ekt
ANSWERS TO MULTIPLE CHOICE QUESTIONS

1. b      2. b      3. b      4. a      5. a      6. a      7. a