Water budget of catchments in the
Tafilalt model
Theo Olsthoorn, 140103
Intro
The catchments or sub-catchments are characterized by their initial loss after
precipitation, and their surface runoff and their groundwater runoff. Another factor
is storage in the catchments. Catchments may be mutually connected in a sense that
upstream ones can discharge both their surface runoff and their groundwater
outflow towards the receiving downstream neighbor. The computation of the
available water in any catchment is done recursively, that is starting from any
catchments the water budgets are computed by visiting upstream catchments first,
to obtain their surface and groundwater runoff to the catchment in question. If done
recursively every catchment is visited. This will work for any tree-type system of
connected catchments in a sense that any catchment can receive water from an
arbitrary number of upstream catchments, but can only discharge to one
downstream neighbor.
The procedure getRech
Traversing the hierarchical tree of catchments is done for an arbitrary starting
catchment along connected branches and twigs of the tree from an arbitrary starting
point until a leaf is reached.
When an arbitrary catchment in the tree is thus visited, this catchment questions all
its upstream neighbors in turn for their the outflowing data, i.e. outflowing runoff
and outflow groundwater.
This data is available if the upstream catchment was visited before. If it wasn’t, then
we deal with the upstream catchment first. By questioning the outflowing data of its
upstream neighbors in turn.
While we do this in turn, we end up with the end leaves of the tree. That is, in a
catchment that has no upstream neighbors. For such a catchment we can compute
the entire water budget including its runoff and groundwater outflow. We store
these budget values inside the catchment’s UserData. We also store the fact that this
catchment has been visited.
Then we fall back to the downstream catchment from which we came. We can now
obtain the outflow of our upstream neighbor and continue questioning our next
upstream neighbor in turn.
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This way all catchments will get their water budget data during the traversing of the
catchment tree by the routine getRech.
Because each catchment then holds all its water budget data, we can readily verify it
and make sure no water is lost in the procedure.
Water budget requirement
The beauty of recursive programming is that it can traverse connected trees of
arbitrary size with a very small routine, as is the case with getRech. The difficulty is
that it is sometimes tricky, at is calls itself numerous times. Therefore severe
checking is necessary. The computation of the water budget of each leaf of the tree
and the tree as a whole is a powerful method, which will be applied here. The
routine must ensure that the water budget of each catchment is maintained at all
times, that is for each leaf in the tree the budget must agree, which implies that the
budget of all upstream catchments also must agree. We may write a routine that
checks the budget of each catchments including all its upstream neighbors, that is
each subtree as a whole starting at an arbitrary node. This is the ultimate
verification.
Water budget computation for an arbitrary catchment
The water budget of a catchment, given index i, reads: “in “is “out” plus “change of
storage in the catchment” during time period k:
Ii,k = Oi,k + DSi,k
Where
Ni
Ni
j=1
j=1
I i,k = Pi,k + å RIn,k + å GIn,k
(1.1)
with Rin,k is the runin (contrary to runoff) coming from one of the connected
upstream catchments and Gin,k is the inflowing groundwater from the same
upstream catchment. The sum is carried out over all the N i catchments that are
upstream of catchment i.
The outflow or loss terms are
Oi,k = Li,k + Ri,k + Gi,k
(1.2)
There is no summation because we assume that each catchment has at most one
downstream catchment. Li,k is the initial loss of catchment i in time interval k. This
in, in fact, immediate evapotranspiration or interception. The terms R and G are as
explained before. As there is only little vegetation in the catchments, we ignore
further evapotranspiration. We may add it later, though.
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The initial loss will be set to the minimum of I i,k and an initial loss value given for
each catchment as a property of the catchment.
Li,k = min ( I i,k ,li )
(1.3)
The runoff equals a runoff factor, which is another catchment property times the
amount of inflowing water minis initial loss
Ri,k = ri ( I i,k - Li,k )
(1.4)
The groundwater runoff will be related to the storage, that is, the amount of
groundwater present in the catchment, at the beginning of the time interval.
Gi,k = gi Si,k-1
(1.5)
The change of storage is what makes the water budget match:
DSi,k = Si,k - Si,k-1 = Ii,k - Oi,k
(1.6)
Si,k = Si,k-1 + Ii,k - Oi,k
(1.7)
Hence,
This makes up the water balance for the catchments.
The storage and the groundwater runoff
There is uncertainty how to handle the storage and the groundwater outflow and
what numeric values to apply. Although this uncertainty cannot be completely
resolved without extended local field research, which is beyond the possibilities of
this project, we can establish reasonable values though.
By assuming all flow terms to be constant during the considered time interval,
which must therefore not be too long, we can take a continuum approach by writing
for an arbitrary catchment (no index used here)
dS
= ( P + Rin + Gin - Rout - L ) - Gout
dt
= N - Gout
(1.8)
Where N = P + Rin + Gin - Rout - L is the groundwater recharge during time interval
k.
Or
dS
= N - gS
dt
(1.9)
Because we assume N to be constant during the considered time interval, this
ordinary differential equation may be integrated:
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d ( N - gS )
= N - gS
dt
(1.10)
ln ( N - gS ) = t + C
(1.11)
which yields
The integration constant can be resolved by stating that we have S = S0 when t = t 0
where the zero indicates the beginning of time interval k or time k-1. Hence
ln ( N - gS ) = ln ( N - gS0 ) ( t - t0 )
(1.12)
N - gS = ( N - gS0 ) e- g(t-t0 )
(1.13)
or
which yields
S = S0 e- g(t-t0 ) +
(
N
1- e- g(t-t0 )
g
)
(1.14)
We observe that g is a time constant, which is another catchment property.
Therefore we prefer writing the previous equation as follows:
Si,t = Sk-1e
-
t-t k
Ti
t-t
- kö
æ
+ Ti N i,k ç 1- e Ti ÷
è
ø
(1.15)
in which we added the index of the catchment and the index of the time interval.
The first term determines the drainage of the groundwater from the catchment and
the second its filling. Note that without initial storage, and with perpetual constant
recharge N then after infinite time, the amount of water in the catchment has
reached a maximum equal to
Smax,i = Ti N
(1.16)
which indicates immediately a useful value for the characteristic time Ti . It relates
the maximum storage to the average inflow rate. We might for instance estimate T
to be equal to 10 years, indicating that the maximum storage in the aquifer of the
catchment is 10 times the yearly influx. We can also look at T in terms of the
halftime of the drainage of the basin in case no recharge is available. This halftime
equals
0.5Si,k = Si,k e
-
Ti ,0.5
Ti
® Ti,0.5 = Ti ln 2 » 0.69Ti
(1.17)
or
Ti » 1.4Ti,0.5
(1.18)
Therefore, if we would be able to estimate the half time, for instance from a
discharge decay curve at the exit of the catchment, we have the sought time constant
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as well. We might be tempted to estimate the halftime at 5 years. In that case
Ti » 7 yr , which we may accept as a crude first approximation as long as no other
data is available. The average groundwater discharge from t = t k-1 to t = t k due to
decay of the storage, i.e. without any recharge equals
Gi,tk-1®tk =
Si,k-1 - Si,k
Dt k
(1.19)
where Dt k = t k - t k-1 . Hence
Dt
- kö
Si,k-1 æ
Gi,k =
1- e Ti ÷
Dt k çè
ø
-
(1.20)
Dt k
1- e Ti
With Ti / Dt k = 7 and Dt k = 1yr we get with g =
= 0.13 / yr = 3.65 ´ 10 -4 / day
Dt k
Gi,k » 3.65 ´10 -4 Si,k-1
(1.21)
in m/d.
Hence, in case the storage in the reservoir is the equivalent of 1 m water, the
average yearly groundwater loss would be 130 mm. It is clear that this value is too
large for the circumstances in the Tafilalt where the total yearly precipitation is in
the order of 100 mm/year and we expect a recharge and, therefore, a maximum
discharge of around 5 mm/year. The total storage would then be about 0.0376 m of
water, of an average water layer thickness at 25 percent porosity of around 0.15 m
(15 cm), which is a very small amount indeed. Given that the actual alluvial aquifers
are concentrated in the valley bottoms and are much narrower than the catchment,
perhaps even 10 to 20 times narrower, we would expect the real depth of the
groundwater in the center of the valleys to be much greater than 15 cm, i.e. 1.5 to
several meters. We might even extrapolate this reasoning towards estimating the
transmissivity at the narrow outlet of the catchments where we know their width
and their topographical gradient and total flow. This, however, is not necessary for
further analysis of the model.
The water budget can be computed from the data stored in the catchment’s
UserData as
Bi,k = Pi,k + Riin,i,k + Gin,i,k - Li,k - Rout ,i,k - Gout ,i,k - ( Si,k - Si,k-1 ) = 0
(1.22)
which must be zero for all catchments and all time intervals.
Properties required for the catchments
The properties that are required for the computation of the water budget of the
catchments are
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1.
2.
3.
4.
L: [m] initial loss [m]
r: [-] runoff coefficient
T: [d] characteristic time of storage decay of catchment
S0: [m] the initial amount of water in the catchment
The precipitation is given per monthly stress period. Then the initial loss of
subtracted from the monthly recharge value. The initial loss should be in the order
of 10 to 22 mm. Then the monthly value is converted to average daily precipitation.
The runoff coefficient is unknown, but we estimate it between 25 and 75% of the
recharge plus incoming runoff form upstream catchments. It can be varied until the
model produces a reasonable outcome. Note that what is left over, i.e. the value of J
in equation (1.8) is recharge, Gout s the groundwater discharge.
We have to make sure that the time constant is in the right dimensions because of
the way the recharge is computed from monthly values. This implies that the time
constant has to be in days. We estimated 7 years for this time constant, which is
around 2500 days or .
-0-0-0-
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