This document provides an overview of hydrological budgets. It defines a hydrological budget as a measure of the amount of water entering and leaving a system. The continuity equation, I-O=ΔS (inflow - outflow = change in storage), is introduced. A water budget for a watershed is explained as P-R-G-E-T=ΔS, where P is precipitation, R is surface runoff, G is groundwater flow, E is evaporation, and T is transpiration. Residence time is defined as the average time for a water molecule to pass through a subsystem, calculated by dividing the storage volume by the flow rate through that subsystem. Examples of computing water budgets and residence times
4. 3
What is a hydrological budget
• A hydrological budget is a measure of the amount of water
entering and the amount of water leaving a system.
• It is a basic tool that can be used to evaluate the occurrence
and movement of water through the natural environment.
• Historically, there may be a difference in a particular year
between inflow and outflow of water from an area.
• Years with more inflow than outflow replenish the water in lakes, dams
and groundwater. Years with less inflow than outflow deplete these
resources
• However, if over a sufficiently long period (taken as 30 years for normal
precipitation), Inflow balance outflow then system is sustainable.
• Lately there are imbalances due to human interference in hydrological
cycle leading to climate change.
6. 5
Continuity Equation
• Simple continuity equation is
• I – O = ΔS (Inflow – outflow = Change in storage)
• However, we first need to define area under consideration and time
duration being considered
• We may take complete global water cycle, a regional water cycle or
maybe a local catchment area-based water cycle
• Duration is typically taken as annual, or monsoon months or a
particular rainfall event (lasting a few days)
9. 8
Water-budget for a catchment
• For a watershed water budget for a time interval (Δt, typically annual)
would be
• P-R-G-E-T = ΔS
• Where P is net precipitation, R is surface runoff from outlet point, G is net
groundwater flow out of the catchment, E is evaporation, T is transpiration
and ΔS is change in storage. All terms could be expressed as volumes as well
as depth over the catchment area.
• Not all terms in the equation may be known to the same degree of accuracy.
• Storage consists of following components:
• S = Ss + Ssm + Sg, where Ss is Surface water storage, Ssm is soil moisture storage,
and Sg is groundwater storage.
10. 9
Simplifications often used
• Over a large river basin, groundwater boundaries often follow
surface-water divide. Therefore, G could be neglected, particularly
when considered over long time period.
• Over a long period, seasonal excesses and deficits in storage often
cancels each other and therefore change in storage could be assumed
zero
• For such cases
• P – R = ET
• Precipitations and runoffs could be observed with good accuracies.
This gives a chance to verify ET models.
11. 10
Residence Time
• The residence time, Tr, is the average duration for a water molecule to pass
through the subsystem of the hydrologic cycle.
• It would be calculated by dividing volume of water in storage in that
subsystem by the flow rate through that subsystem
• Average residence time could be also calculated for global, regional or local
hydrological cycles
• Example: What is average residence time of global atmospheric moisture
• Volume of atmospheric moisture is 12,900 km3. The flow rate of moisture from
atmosphere is in form of precipitation, which is 458,000 (Over ocean) + 119,000
(Over land) so total 577,000 km3/yr.
• So, average residence time for moisture in atm. is = 0.022 yrs = 8.2 days.
13. 12
Example of Water Budget Computations
• Sambhar Lake
• Real data: The lake receives water from six rivers: Mantha, Rupangarh, Khari,
Khandela, Medtha and Samod. Lake has 5700 square km catchment area. It occupies
an area of 190 to 230 square kilometers based on the season. The lake is elliptically
shaped with a length of approximately 35.5 km and a breadth varying between 3 km
and 11 km. Its elevation is approximately 360 m above mean sea level.
• Fictious data for example: Let us assume its elevation was 360.2 m and surface area
was 200 km2 at the beginning of a July month.
• In that month there was 183 mm rainfall over the lake. In that month lake received
on average approximately 50 m3/s inflow from all the rivers draining to the lake.
Evaporation from the lake was approximately 6.7 cm during that month. Water was
pumped out at the average rate of approximately 27 m3/s by various industries and
users around the lake. We can neglect the flow to groundwater.
• Write water budget for the lake and calculate its water surface elevation at the end
of that month.
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Solution
• Input volume = I, Output volume = O. Where, I and O are net inflow and outflow volume
in L3 units (say m3). It can be obtained by multiplying inflow rate with Δt, which is
duration. In present exercise 1 month = 1x31x24x60x60 sec = 2678400 sec.
• So, I - O = ΔS (Change in storage in lake)
• Inflow are, Inflows from various rivers as well as precipitation directly into the lake.
• I1 = inflow from rivers = 50 m3/s x 2678400 = 1.34 x 108 m3.
• I2 = direct rainfall over lake = 0.183x200x1000000 m3= 3.66 x 107 m3.
• So, total I = (1.34 + 0.366) x 108 = 1.71 x 108 m3.
• O1 = outflow = 27 m3/s * 2678400 = 7.23 x 107 m3.
• O2 = Evaporation loss = 0.067x200x1000000 = 1.34 x 107 m3.
• So, total O = 8.57 x 107 m3. Therefore ΔS = 8.48 x 107 m3.
• Change in surface elevation of lake Δh = ΔS/Area of lake = 0.424 m (Increase). So surface
elevation after one month = 360.2 + 0.424 = 360.624 m above msl.
15. 14
Example - 2
• On river Mej (tributary of Banas River), a rainfall event of 3 days was
recorded. Mej catchment area up to the gauging station, where
discharge was measured everyday, has been calculated to be 5271
km2.
• There are two rain gauge stations in the catchment area of Mej River,
namely Hindoli and Keshawrai Patan.
• From Thiessen polygon analysis over the catchment area, influence
area of Hindoli rain gauge is calculated to be 3369 km2 and that of
Keshawraj Patan to be 1902 km2.
• It was observed that during that storm, 9 cm rainfall was recorded in
Hindoli RG and 11.2 in Keshwrai Patan RG.
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Example – 2 Contd.
• Discharge were recorded for 12 days at the outlet point of basin and
were as follows
• Compute the amount
of combined water
which infiltrated and
evapotranspired from
the catchment area.
• What is the ratio of
runoff to
precipitation?
S.
No.
Hours since
beginning of flood
Discharge
Recoded (m3/s)
1. 0 0
2. 24 70.7
3. 48 152.9
4. 72 249.7
5. 96 364.5
6. 120 444.8
7. 144 558.6
8. 168 660.1
9. 192 576.6
10. 216 462.9
11. 240 320.6
12. 264 151.6
13. 288 0
18. 17
Solution
• Total precipitation falling on the catchment area
• P on RG1 X Area of Influence of RG1 + P on RG2 X Area of Influence of RG2
• Volume of precipitation = 3369x106x0.09 + 1902x106x0.112 m3= 5.16x108 m3.
• Total runoff from the catchment area
• From the table, average runoff would be 308.7 m3/s for 12 days.
• Runoff volume = 308.7x12x24x60x60 = 3.20x108 m3.
• Losses L, due to infiltration and evapotranspiration = (5.16-3.20)x108 m3
• L = 1.96x108 m3.
• Runoff coefficient = Runoff/Precipitation = 0.62
19. 18
Example – 3 (Home exercise)
• 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 next 6 months and (iii) 3.5 m3/s for the
remaining 3 months.
• (a) What is the runoff coefficient of the catchment
• (b) 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 and evapotranspiration for the annual rainfall of 140
cm.
