WATER BUDGET EQUATION

Water Budget Equation is one of the very important topics from Hydrology and from Civil Engineering point of perspective. Not only in semester examinations of Civil Engineering Students, but also Water Budget Equation is the key topic for many of the competitive examinations like GATE, UPSC, and any company’s entry exams. Here we have a clear cut explanation of Water Budget Theory, to make it practical a few examples are also included in this online study material. Please share this post with all your Civil Engineering friends. Now lets drive through the topic – Water Budget Equation, its theory, related problems and solutions.

WHAT IS CATCHMENT AREA IN WATER BUDGET EQUATION?

The area of land draining into a stream or a water course at a given location is known as the catchment area. It is also called drainage area or drainage basin. In the USA, it is known as watershed. A catchment area is separated form its neighbouring areas by a ridge called divide in the USA and watershed in UK. The areal extent of the catchment is obtained by tracing the ridge on a topographic map to delineate the catchment and measuring the area by a planimeter. It is obvious that for a river while mentioning the catchment area, the station to which it pertains must also be mentioned. It is normal to assume the groundwater divide to coincide with the surface divide. Thus, the catchment area affords a logical and convenient unit to study various aspects relating to the hydrology and water resources of a region.

Further it is probably the single most important drainage characteristic used in hydrological analysis and design. Since now we are well aware about Catchment Area, let us direclty jump into the Water Budget Equation and then we can proceed to solve some problems.

For a given problem area, say a catchment, in an interval of time Δt, the continuity equation for water in its various phases is written as

Mass Inflow – Mass Outflow = Change in mass storage

If the density of the inflow, outflow and storage volumes are the same,

V_i – V₀ = ΔS

where V_i = Inflow volume of water into the problem area during the time period,

V₀ = Outflow volume of water from the problem area during the time period, and

ΔS = Change in the storage of the water volume over and under the given area during the given period.

In applying this continuity equation to the paths of the hydrologic cycle involving change of state, the volumes considered are the equivalent volumes of water at a reference temperature. In hydrologic calculations, the volumes are often expressed as average depths over the catchment area. Thus for example, if the annual stream flow from a 10 sq.km catchment is 10⁷ cubic.m, it corresponds to a depth of (10⁷/10×10⁶) = 1 m = 100 cm

Rainfall, evaporation and often runoff volumes are expressed in units of depth over the catchment.

While realising that all the terms in a hydrological water budget may not be known to the same degree of accuracy, an expression for the water budget of a catchment for a time interval Δt is written as

P – R – G – E – T = ΔS

In this equation,

P = Precipitation
R = Surface runoff
G = Net groundwater flow out of the catchment
E = Evaporation
T = Transpiration
ΔS = Change in storage

The storage S consists of three components as

S = S_s+ S_sm + S_g

where S_s = surface water storage
S_sm = water in storage as soil moisture
S_g = water in storage as groundwater

Thus in the equation

ΔS = ΔS_s+ ΔS_sm + ΔS_g

All terms in the above equation have the dimensions of volume. Note that all these terms can be expressed as depth over the catchment area (eg. in cm) and in fact this is a very common unit.

In terms of rainfall-runoff relationship, the above equation can be represented as

R = P – L

where L = losses = water not available to runoff due to infiltration (causing addition to soil moisture and groundwater storage), evaporation, transpiration and surface storage.

EXAMPLE PROBLEMS AND SOLUTIONS

EXAMPLE 1 – A lake had a water surface elevation of 103.200 m above datum at the beginning of a certain month. In that month, the lake received an average inflow of 6.0 m³/s from surface runoff sources. In the same period, the outflow from the lake had an average value of 6.5 m/s. Further, in that month, the lake received a rainfall of 145 mm and the evaporation from the lake surface was estimated as 6.10 cm. Write the water-budget equation for the lake and calculate the water surface elevation of the lake at the end of the month. The average lake-surface area can be taken as 5000 ha. Assume that there is no contribution to or from the groundwater storage.

SOLUTION

In a time interval Δt, the water budget for the lake can be written as

Input volume – Output volume = Change in storage of the lake

(IΔt + PA) – (QΔt + EA) = ΔS

Where I = average rate of inflow of water into the lake
Q = average rate of outflow from the lake

ΔS = change in storage volume of the lake.

Here Δt = 1 month = 30 X 24 X 60 X 60 = 2.592 X 10⁶ s = 2.592 Ms

In one month,

Inflow Volume = IΔt = 6.0 X 2.592 = 15.552 M m³
Outflow Volume = QΔt = 6.5 X 2.592 = 16.848 Mm³

Input due to precipitation = PA = (145 X 5000 X 100 X 100)/ (1000 X 106) m³ = 7.25 Mm3
Output due to evaporation = EA = (6.10/100 x 5000 X 100 X 100)/10⁶ = 3.05 m³

Hence ΔS = 15.552 + 7.25 – 16.848 – 3.05 = 2.904 m³

Change in elevation Δz = ΔS/A = (2.904 X 10⁶)/(5000 X 100 X 100) = 0.058 m

New water surface elevation at the end of the month = 103.200 + 0.058 = 103.258 m above the datum

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EXAMPLE 2 – A small catchment of 150 ha area received a rainfall of 10.5 cm in 90 minutes due to a storm. At the outlet of the catchment, the stream draining the catchment was dry before the storm and experienced a runoff lasting for 10 hours with an average discharge of 1.5 m³/s. The stream was again dry after the runoff event.
(a) What is the amount of water which was not available to runoff due to combined effect of infiltration, evaporation and transpiration?
(b) What is the ratio of runoff to precipitation?

SOLUTION

The water budget equation for the catchment in a time Δt is

R = P – L

where L = losses = water not available to runoff due to infiltration (causing addition to soil moisture and groundwater storage), evaporation, transpiration and surface storage

In the present case, Δt = duration of the duration of the runoff = 10 hours

Rainfall occurred in the first 90 minutes and for the rest 8.5 hours, the precipitation was zero

(a) P = Inout due to precipitation in 10 hours
= 150 X 100 X 100 X (10.5/100) = 157,500 m³

R = Runoff volume = outflow volume at the catchment outlet in 10 hours
= 1.5 X 10 X 60 X 60 = 54,000 m³

(b) Runoff/rainfall = 54,000/157,500 = 0.343
(This ratio is known as runoff coefficient)

EXAMPLE 3 – A catchment area of 140 km2 received 120 cm of rainfall in a year. At the outlet of the catchment, the flow in the stream draining the catchment was found to have an average rate of (i) 1.5 m3/s for the first 3 months, (ii) 2.0 m3/s for 6 months and (iii) 3.5 m3/s for the remaining 3 months. (a) What is the runoff coefficient of the catchment? (ii) If the afforestation of the catchment reduces the runoff coefficient to 0.35, what is the increase in the abstraction from precipitation due to infiltration, evaporation and transpiration for the same annual rainfall of 120 cm?

SOLUTION

(i) Before Afforestation

Consider a period = Δt = 1 year
Input volume to the catchment through precipitation = V_i = 140 X 10⁶ X 120/100 = 168 Mm³

Runoff = Output volume = V_o = (1.5 X 3) + (2 X 6) + (3.5 X 3) = 27 Mm³

= 27 (365/12) X 24 X 60 X 60 = 70.956 X 106 m³ = 70.956 Mm³

Runoff coefficient = 70.956/168.0 = 0.4224

Abstraction volume = 168.0 – 70.956 = 97.044 Mm³

(ii) After Afforestation

Runoff = 0.35 X 168 = 58.8 Mm³
New abstraction volume = 168.0 – 58.8 = 109.2 Mm³
Increase in abstraction = 109.20 – 97.044 = 12.156 Mm³