3.

Principles of the Ernst equation

3.1 The original drain spacing equation of Ernst
The general principle underlying Ernst's basic equation (1962) is that the
flow of groundwater to parallel drains, and consequently the corresponding avail-
able total hydraulic head (h), can be divided into three components: a vertical
(v), a horizontal (h), and a radial (r) component or

where q is the flow rate and R is the resistance.

Working out various resistance terms, we can write the Ernst equation as

h = q -DV+ q - +L2 q-In- aD2

(3)
KV
8KD nK2
where

h, q, K2, D2, L = notation Hooghoudt's equation (Section 2.1)

Dv = thickness of the layer over which vertical flow is considered; in most
cases this component is small and may be ignored (m)
K = hydraulic conductivity for vertical flow (m/day)
V
KD = the sum of the product of the permeability (K) and thickness (D) of the
various layers for the horizontal flow component according to the
hydraulic situation:
one pervious layer below drain depth: KD = K D (Fig.2a)
1 1 + K2D2
two pervious layers below drain depth:KD = K I D l + K2D2 + K3D3 (Fig.2b)
a = geometry factor for radial flow depending on the hydraulic situation:
KD = K D
1 1
+ K2D2, a = I
KD = K I D l + K2D2 + K3D3, the a-value depends on the K2/K3 and D2/D3
ratios (see the auxiliary graph Ia)
u = wetted section of the drain (m); for pipe drains u = nr.

Eq.(3) shows that the radial flow is taken into account for the total flow (q),
whereas the Hooghoudt equation considers radial flow only for the layer below
drain level (q2).

It should be noted that Eq.(3) has been developed for a drainage situation
where K << K2 ( a clay layer on a sandy substratum) and can therefore only be
I
used where the flow above drain level is relatively small. However, if one uses

14
this equation for the case that K >> K2 (a sandy layer on a clay layer, a not
I
uncommon situation), the result is a considerable underestimate of drain spacing
compared with the result obtained when the Hooghoudt equation is used.
According to Ernst (1962) and Van Beers (1965), no acceptable formula has
been found for the special case that KI >> K2, and it was formerly recommended
that the Hooghoudt equation be used for this drainage situation. Since that
time, as we shall describe below, a generalized equation has been developed,which
covers all K /K ratios.
1 2

K3 C0.l K2
K3D3 neglible

Fig.2a. S o i l below t h e d r a i n : Fig.2b. S o i l below the d r a i n :

o n l y one p e r v i o u s l a y e r (D2) two p e r v i o u s l a y e r s (D2, D3)
(KD = KIDl + K2D2) (KD = KIDl + K2D2 + K3D3)

.... i m a g i n a r y boundary
- - - - r e a l boundary

Fig.2.Geometry of the Ernst equation if the vertical resistance may be ignored.

3.2 The generalized or the Hooghoudt-Ernst equation

This new equation is based on a combination of the approach of Hooghoudt
(radial flow only for the flow below the drains, q2) and the equation of Ernst
for the radial flow component.

Neglecting the resistance to vertical flow and rewriting Eq.(3), we obtain

8KDh

If we consider only horizontal flow above the drains (q ) and both horizontal
1
and radial flow below the drains (q2), we may write

15
 q=-+ (4)
L2 8 aD2
L * + - D
TI
LI^-

Introducing an equivalent drain spacing (L ),i.e. a drain spacing that would

be found if horizontal flow only is considered, we get

2 - 8KDh
Lo - - (5)
4

Substituting Eq.(5) into E q . ( 4 ) yields

8KDh 8KiDlh 8KzD2h

After a somewhat complicated re-arrangement (see Appendix A) the generalized

equation reads

where

L = drain spacing based on both horizontal and radial flow (m)

Lo = drain spacing based on horizontal flow only (m)

aD 2
c = Dz In - , a radial resistance factor (m)
KiDi
B = the flow above the drain as a fraction of the total horizontal
= KD
flow

GRAPH I
To avoid a complicated trial-and-error method as required by the Hooghoudt
approach, Graph I has been prepared. For different c/Lo and B-values, it gives
the corresponding L/L -value which, multiplied by LO' gives the required drain
spacing (L-value). However, as will appear furtheron, this graph is usually
only needed for the following specific situations:

16
 K, >> K2 or B > 0.1;

there are two pervious layers below drain level, K3D3 > K2D2, and
no Scientific Pocket Calculator (SPC) is available;

one wants to compare the generalized equation with other drainage

equations or one wants to prepare a specific graph ( s e e Section 4.1
and Appendix C 1 ) .

NOTE: Comparing t h e Hooghoudt equation u i t h t h e E r n s t equation and using

u = 4r i n s t e a d o f u = m, a g r e a t e r s i m i t a r i t y is o b t a i n e d .

3.3 The modified Hooghoudt-Ernst equation

In most actual situations the B-factor (K D /KD ratio) will be small and there-
1 1
fore the last term in Eq.(7) has little influence on the computed drain spacing.
Neglecting B, and rewriting Eq. (7) (multiplying by L3/L and substituting
8KDh/q for L 2 ) , we obtain the equation for the B = O line of Graph I:

DP
8L Dz In -
L2 + - - 8KDh/q = O
TI

If we compare Eq.(8) with the original Ernst equation (3a), it can be seen
KD
that the factor - of Eq. (3a) has changed into
KP KP
, which equals DP. However,
for a drainage situation with two pervious layers below drain level, - KzD2
KZ
becomes K2D2 + K3D3 and the equation f o r this situation reads
KP

8L KZDP + K3D3 aDs

L2 + - In - - 8 KD h/q = O
TI KP

A s regards the use of Eq.(9) it can be said that, in practice, Eq.(9) is very
useful if an SPC is available; if not, Graph I has to be used.

Eq.(8), on the other hand, will seldom be used. The simple reason for this
is that for most drainage situations with a barrier (only K D ) , a simplified
2 2
equation can be used.

17
3.4 The simplified Hooghoudt-Ernst equation
After Graph I had been prepared on linear paper, it was found that for c/L -
values < 0.3 and B-values < 0.1, the following relations hold

LILo = 1 - CfL or L = L - c (see Graph I) (10)

The question then arose whether these conditions were normal or whether they
were rather exceptional.In practice,the simplified equation proved to be almost al-
ways applicable.In addition,it was found from the calculations needed for the prepa-
ration of Graph II(r = 0.10 m, all K- and D-values) that, except for some un-
common situations (K = 0.25 m/day, D > 5 m), the equation L = L - c is a
reliable one, where C
D
= D2 In .
Note: Many years ago W.T.Moody, an engineer w i t h t h e U.S.5ureau of Reclamation
D
proposed a similar correction iD l n G)to be subtracted from t h e calculated
spacing (Maasland 1 9 5 6 , D m 1 3 6 0 ) . The only d i f f e r e n c e between t h e correction
proposed by Moody and t h a t i n this b u l l e t i n is t h a t now t h e conditions under
which t h e correction may be applied are p r e c i s e l y defined.

18
4. Application of the generalized equation and
corresponding graphs
There are many different drainage situations, five of which will be handled
in this section.

SITUATION 1 : Homogeneous soil; D < 4L; pipe drainage

K I = K2

SITUATION 2 : Slight differences between soil permeability above and below

KI 2 Ka drain level; differences in depth to barrier ( D 5 iL); pipe
and ditch drainage

SITUATION 3 : A highly pervious layer above drain level and a poorly

K I > > K2 pervious layer below drain level

SITUATION 4 : A heavy clay layer of varying thickness overlying a sandy sub-

K << K stratum; the vertical resistance has to be taken in account
1 2

SITUATION 5: Soil below drain level consists of two pervious layers

< (K2D2, K D ) ; the occurrence of an aquifer (K3 >> K 2 ) at
K3 > K 3 3
2
various depths below drain level.

4.1 Drainage situations

SITUATION 1: Homogeneous soil; O < &L; p i p e drains

K = K The use of t h e simplified equation and Graph 11

This is the most simple drainage situation; the required preparatory calcu-
lations are limited and a graph is usually not needed. For comparison with
other drain spacing equations, we shall use the example given by Wesseling ( 1 9 7 3 ) .

Note that - for reasons of convenience - in this and the other examples the
units in which the various values are expressed have been omitted with the
exception of the L-value. However, for values of h, E , and 5 , read metres; for
q and -
- K , r e a d m/day and for KD, read m2/day.

19
 PREPARATORY
DATA G I V E N COMPIJTATION D R A I N S P A C I N G ( L )
CALCULATIONS

L2 = KD 8h/q

h = 0.600 h/q = 300 Lo = 100.9 L = L - c = 8 7 m

q = 0.002 8h/q = 2400 c = 13.8

~~~~

K =0.8
1
n 1=0.30 K I D l = 0.24 c < 0.3 L Eq.Hooghoudt: L = 87 m

K =0.8 D =5.0 K2D2 = 4.0 B < 0.10 E q .Ernst: L = 8 4 m

2 2
Eq.Kirkham: L = 85 m
r = 0.10 KD = 4.24 (Eq.10 may be E q .Dagan: L = 8 8 m
u = nr B = 0.06 used) (Wesseling 1973)

NOTE: writing h = 0.600 instead of 0.6, is not meant to suggest accuracy,

but is only f o r convenience in determining the h/q value.
D2
If no SPC is available, the c-value (D2 In -1 can be obtained from Graph 111.

The simplified formula is very convenient if we want to know the influence

that different U-values will have on the drain spacing. For example

r = 0.05 m, then c = 17.3 and L = 84 m

u = 1.50 m, then c = 6.0 and L = 95 m

The influence of different K or D-values is also easy to find. However, if

the U-value is fixed, it is better to use Graph I1 or a similar graph, which
gives a very quick answer to many questions.

GRAPH I 1

This graph is extremely useful for the following purposes and where the fol-
lowing conditions prevail:

Purposes

A great number of drain spacing computations have to be made, for instance,

for averaging the L-values in a project area instead of performing one
calculation of the drain spacing L with the average K- or KD-value;

20
 - One wants to find out quickly the influence of a possible error in the K-
value or the influence of the depth of a barrier (D-value);

One wants to demonstrate to non-drainage specialists the need for borings

deeper than 2 . 1 0 m below ground level because a boring to a depth of
2 . 1 0 m results in a D -value of about I m, being 2 . 1 0 m minus
2
drain depth.

Conditions

Homogeneous soil below the drains (only K D ) ;

2 2
The wetted perimeter of the drains has a fixed value, say pipe drains with
r = 0.10 m or ditches that have a certain U-value;

An error o f 3 to 5% in the computed drain spacing is allowable.

Considering these purposes and conditions, it will be clear that the avail-
ability of Graph I1 or similar graph is highly desirable, except when:

exact theoretical computations are required

K I >> K2

there are two permeable layers below drain level instead of one (K D K D ).
2 2 ' 3 3

For these three conditions Graph I1 cannot be used and one must resort to
Graph I.

Other q-, h- or KI/K2 values than those given on t h e graph

Graph I1 has been prepared for the following conditions:

h = 0.6 m, q = 0.006 m/day or h/q = 100, K l = K 2 m/day, and r = 0.10 m

For the specific purposes and conditions for which this graph has been pre-
pared (approximate L-values suffice) an adjustment is only required for different
h/q values, through a change in the K 2 -values to be used. For instance, in the
example given for Situation 1 we have a h/q value of 300. Therefore

and we read for D = 5 m, L = 87 m.

21
 However, if many computations have to be done, it is preferable to prepare
a graph or graphs for the prevailing specific situation. For instance, for the
conditions prevailing in The Netherlands, three graphs for pipe drains would be
desirable: q = 7 "/day and h = 0.3, 0.5, and 0.7 m.

For irrigation projects a q-value of 2 "/day is usually applied.

If one wants to know the magnitude of an introduced error (h # 0.6, K I # K2),

the extra correction factor (f) for K; can be approximated with the formula

where

D l = 0.30 and D i = 0.5 h' .

For example:

h' = 0.90 D ; = 0.45

K 1 /K2 = 0.5, D = 5 f = 5 + ( 0 . 5 x 0.45) = o.986 or

2 5 + 0.30
about 1 % difference in the L-value

When the assumption of a homogeneous aquifer contains a rather large error (e.g.
K = 2 K ) , and moreover a larger hydraulic head being available (k'=0.9 m), we get
1 2
f=l.lI or 5% difference in the L-value.

Preparation
The preparation of a speficic graph is very simple, and takes only a few
hours. There are two methods of preparation.

Appendix C 1 gives an example of how it is done if the d-tables of Hooghoudt

are available. This is the easiest way.

Appendix C 2 shows a preparation based on the generalized equation in combina-

tion with Graph I. This method gives the same result, but requires more calcu-
lations.

Finally note that Graph I1 demonstrates clearly that if, in a drainage project,
augerholes of only 2 m depth are made (D
2
2 1 m), considerable errors in the
required drain spacing can result.

22
For example:

Given:
h/q = 300 ( i r r i g a t i o n p r o j e c t ) , K 2 = 0 . 8 , t h e n K; = 2.4;
d r a i n depth = 1.5 m

Depth t o b a r r i e r Flow d e p t h Spacing h/q = 100, K2 = 0.5

(D2) (L)
2.50 m I m 50 m L = 2 2 m
3.50 m 2 m 63 m 28 m
6.50 m 5m 87 m 34 m
11.50 m IO m 105 m 37 m
m m 130 m 37 m

T h i s example may show t h a t :

Graph I1 i s v e r y s u i t e d t o c a r r y o u t a s e n s i t i v i t y a n a l y s i s on t h e
i n f l u e n c e of t h e d e p t h of a b a r r i e r , e t c .

- The need f o r d r i l l i n g d e e p e r t h a n 2 m. Note t h a t h e r e o n l y t h e v a l u e D

2
h a s been c o n s i d e r e d . However, t h e r e can a l s o b e a c o n s i d e r a b l e change
i n t h e K -value.
2
The r e l a t i v e i n f l u e n c e of t h e D -value changes w i t h t h e s p a c i n g o b t a i n e d .
2

SITUATION 2: Slight differences between soil permeability above and below

KI K2 drain level (KI 3 K2);
Differences in depth to a barrier iD 5 1/4L);
pipe and ditch drainage

F o r t h e s i t u a t i o n D < { L and d r a i n a g e by d i t c h e s , t h e computation of t h e

d r a i n s p a c i n g , as w e l l a s t h e computation s h e e t i s t h e same a s have been g i v e n
i n S i t u a t i o n I . Only i f K I > K2 and D2 i s s m a l l s h o u l d s p e c i a l a t t e n t i o n be p a i d
t o t h e q u e s t i o n whether B < 0.10.

For t h e s i t u a t i o n D > a L , t h e f o l l o w i n g e q u a t i o n ( E r n s t 1 9 6 2 ) can b e u s e d :

23
The use of this equation will be demonstrated below and will be followed by a
sensitivity analysis for the u and D2 factor.

GIVEN COMPUTATION
L
h = 0.800 h/q = 400 L In -= 71 x 0.8 X 400 = 1005

q = 0.002 Graph 111, for u = 1.50 + L = 205 m

K I = 0.40 If a SPC is available, the L-value can also
K2 = 0.80 u = 1.50 be obtained by a simple trial- and error-method
error method

Equation (11) does not take the horizontal resistance into account because it
is negligible compared with the radial resistance; nor is the flow above the
drain considered. Only if the computed L-value is small, say about 40 m or less,
is a small error introduced.

If Graph I only is available, or one wants to check the computed L-value by

using the generalized equation, the following procedure can be followed:

Estimate the drain spacing and assume a value for D2 between i L and $L
(beyond this limit, the computed spacing would be too small); compare the two
L-values (control method) o r check whether the assumed D -value > 4L and < 0.5L
2
(computation method).

For the above example we get:

Given: D 2 = 80
Assume K2 = 0.8 KD = 64 Lo = 452 c/Lo = 0.70
Assume u = 1.5 8 h/q=3200 c = 318 Graph I: L / L o = 0.45 + L = 204 m

Using a SPC and Eq.(8) -f L = 202 m

Note: It may be useful - by way of exercise - t o t r y other D

2 ualues and to
compare the resulting L-ualues.

24
Sensitivity analysis for U-value and depth to a barrier (D-value)

D, = m
D 2 = 5 m
L
KD = 0.16 + 4.0 = 4.16 8 h/q = 3200 Lo=115 m
u = l m+L=192 m u = l m + c = 8 m + L = 107m
u = 1.5 m + L = 205 m u=1.5mc=6 m + L = 109m
u = 2 m+L=215 m u = 2 m c=4.5m+L= 110m
u = IO m + L = 300 m
u = 0.30 m + L = 162 m u = 0.30 m + L = 101 m; u = 0.20 + L = 101 m
u = 0.40 -+ L = 103 m

u-va Zue

This sensitivity analysis shows that the influence of difference in the

U-value increases as L increases. However, differences of 50% or more are gene-
rally of little importance. Therefore the U-values of pipe drains can be appro-
ximated by taking r = 0.10 m (u = 0.30) and the U-values of ditches approxima-
ted by taking the width of the ditch and two times the water depth (usually
2 x 0.30 m).

The slope of the ditch need not be taken into account because of the reasons
mentioned. Moreover, neither the water level in the ditch nor the drain width are
constant factors.

D-vazue

The large error made by assuming D2 = 5 m instead of D2 = m,as in the above

example (L = 109 m instead of 205 m), is easily made if, in an irrigation project

(q = 0.002, h/q and L-value very large), no hydro-geological investigations are

carried out.

SITUATION 3 : A highZy pervious Zayer above drain Zevel and a poorly pervious
Kl>>K2 Zayer beZow drain ZeveZ
The use o f Graph I

This particular situation is frequently found. It may be of interest to use

the data given below to compute drain spacings with other drainage equations and
then to compare the results with those obtained with the equation of Hooghoudt
or the generalized Hooghoudt-Ernst equation.

25
 PREPARATORY
DATA G I V E N COMPUTATION
CALCULATIONS

h = 1 .O00 h/q = 200 Lo = 53.7 m

q = 0.005 8h/q = 3200 c = 12.6 m

K = 1.6 D =0.50 KIDl = 0.8 c/L =0.235 L / L o = 0.88 (Graph I)

I 1
K = 0.2 D =5.0 K2D2 = 1.0 B = 0.44 L = 0.88 x 53.7=47.3 m
2 2

r = 0.10 KD = 1.8

u=O .40
* B = 0.44

Note: The computation s h e e t used here i s t h e same a s t h a t used f o r S i t u a t i o n I,

except t h a t no c/L -0alue i s needed i n S i t u a t i o n 1 .
O

*For t h e o r e t i c a l comparisons w i t h t h e r e s u l t s of t h e equation of Iiooghoudt,

it i s p r e f e r a b l e t o use u = 4 r instead o f u = T r

Comparison o f the E r n s t e q u a t i o n s with t h e e q u a t i o n o f Hooghoudt

Original equation of Ernst (Eq.3a) L = 32 m

Modified equation of Ernst (Eq.8) L = 39.9 m
Generalized equation of Ernst (Eq.7) L = 47.2 m

Equation Hooghoudt (Eq.2) L = 47.2 m

Graph 1I:K; = 0.2 X 2 X 1.7 = 0.68, D2 = 5 m + L + 41 m

T.t should be born in mind that the original equation of Ernst never has been
recommended for the considered situation with a major part of the flow through
the upper part of the soil above drain level (K <<K ) . Therefore it is not sur-
2 1
prising that the unjustified use of this formula will result in a pronounced
underestimating of the drain spacing; the modified equation is somewhat better,
while the generalized equation gives the same results as the equation of
Hooghoudt. In addition,it is demonstrated that the use of Graph I1 for the KI>>K2
situation also results in an underestimate of the drain spacing.

26
SITUATION 4: A heavy clay layer of varying t h i c k n e s s overZying a sandy
K <<K substratwn; the v e r t i c a l resistance has t o be taken i n t o account
1 2

This is another drainage situation that occurs frequently. Because the thick-
ness of the clay layer can vary, three different drainage situations can result
(see Fig.3). In this example, it is assumed that the maximum drain depth is
-1.40 m, in view o f outlet conditions, and that the land is used for arable
farming (h = drain depth - 0.50 m = 0.90 m).
U
V
The computation of the vertical component (hv = q , see Section 3.1) is
V
somewhat complicated because the D -values varies with the location of the drain
V
with respect to the more permeable layer. However, Fig.3 and the corresponding
calculations may illustrate sufficiently clearly how to handle the specific drai-
nage situations. It may be noted that as far as the author is aware the solution
given by Ernst for this drainage situation is the only existing one.

q=O.OlO
m /dav

D v = h + y Dv = h D = h - D '

h" = k"; Dv
h = q
v %
h
h" = t Dv

h ' = h - h h ' = h - h h ' = h - h

KD = K D
2 2 t K3D3 KD = K2D2 KD = K i D i + K2D2

c=:1nu aD2 D2 c = D l n -D2

c = D ln-
2 u 2 u

Ex.4a Ex.4b Ex.4c

D r a i n l e v e l above Drain level coincides D r a i n l e v e l below

t h e boundary w i t h t h e boundary t h e boundary o f t h e
two s o i l l a y e r s

Fig.3. Geometry of t h e Ernst equation i f v e r t i c a l r e s i s t a n c e has t o be taken

i n t o account (K <( KZI.
1
27
 Procedure

9 Determine D according to the specific situation;

V

Calculate h the loss of hydraulic head due to the vertical resistance,

V'
D
by using h = - ;

Calculate h' from h' = h - h where h' is the remaining available hydraulic
V
head for the horizontal and radial flow

Calculate the h'/q and KD-value. Note that the horizontal flow in the upper
layer with low permeability may be ignored;

Compute L; f o r Example 1 , Graph Ia is required in addition to Graph I or an

SPC and Eq.(9); for Examples 2 and 3, use L = L - c.

The following examples are intended to illustrate the procedure and the layout
of a computation sheet. (The data used have been taken from Fig.3).

Example 4a

Dv = h+y = 0.90 + 0.30 = 1.20 h/q = 66

hv = q/K1 x Dv = 0.2 X 1.2 = 0.24 h'=h-h =0.90-0.24=0.66 8 h/q 528
V

Kv = 0.05 D =0.80 K2D2 = 0.04 L0=50.6 m c / L =1.45 LILo = 0.25

2
K3 = 2.0 D =2.40 K D - 4.80 -+ c =?3.3 m B = 0 L = 12.6
3 3 3 -

K IK =20 D3/D2=3 KD = 484

3 2
U
a = 4.0 ( s e e Ex.5)

AZternativas

Pipe drains (u=0.30) + L = 5 m

- Ditch bottom i n the more permeable layer (ditch depth at 2.20) +

u = 0.90 and h = 1.70 + Lo = 80 m; c = 2m -fL = 78 m

Pipe drains at -2.20 m + Lo = 80 m; c = 5m + L = 75 m

Note: The Zast t#o a l t e r n a t i v e s mean t h a t t h e drainage water w i Z l haue t o be

discharged by pumping.

28
Example 46

Dv = h = 0 . 9 0 h/q = 72
hv = 0.2 X 0.90 = 0.18 h ' = 0.90 - 0.18 = 0.72 8 h/q = 576

K2 = 2.0 D2 = 3.20 KD = 6.40 Lo = 60.7 m L = 58.3 m

c = 2.4 m

Example 4c

Dv = h - D i = 0.90 - 0.40 = 0.50 h' = 0.80 8 h/q = 640

h = 0.2 x 0.50 = 0.10
V

K' = 2.0 D' = 0.40* K'D'=0.80 Lo = 67.9 m L = 65.5 m

1 1 1 1
K = 2.0 D2 = 3.20 K D =6.40 c = 2.40 m
2 2 2

KD = 7.20

* The available cross-section f o r horizontaZ f l o w = thickness of t h e more

pemeabZe Zayer above drain depth.

Remarks

I f we compare t h e computed d r a i n s p a c i n g f o r Example 4b (L=58 m) w i t h t h a t

of Example 4 c (L=65 m ) , we c a n c o n c l u d e t h a t f o r a g i v e n d r a i n d e p t h t h e e x a c t
t h i c k n e s s of t h e heavy c l a y l a y e r i s of minor importance a s l o n g as t h e bottom
1111
of t h e d r a i n i s l o c a t e d i n t h e more permeable l a y e r .

I f , however, t h e c l a y l a y e r c o n t i n u e s below d r a i n d e p t h , a s i n Example 4 a

(L=13 m), d r a i n s p a c i n g s would h a v e t o be v e r y narrow i n d e e d and t h e a r e a w i l l
s c a r c e l y b e d r a i n a b l e ( f o r p i p e d r a i n s , L=5 m , f o r a d i t c h , L=13 m ) .

The o n l y way o u t h e r e i s t o u s e deep d i t c h e s (L = 78 m) o r p i p e d r a i n s

(L = 75 m) t h a t r e a c h i n t o t h e permeable l a y e r , and t o d i s c h a r g e t h e d r a i n a g e
w a t e r by pumping.

29
SITUATION 5: SoiZ below drain depth c o n s i s t s o f two pervious layers
(K2D2, KgDgj.

Graph I a .
The occurrence o f an aquifer ( K 3 >> K2) a t v a r i o u s depths
below d r a i n level

Hydrologically speaking, this drainage situation is very complicated. Up to

now the problem could only be solved by using an additional graph (Ia) (Ernst
1962) or by the construction of various graphs for various drainage situations
(ToksÖz and Kirkham, 1971 ) .
The graph of Ernst that can be used for all situations (various K3/K2 and
D /D ratios) gives the results he obtained by applying the r e h x a t i o n method.
3 2
A somewhat modified form of this graph has been published by Van Beers (1965).

The reZiabiZity and importance o f Graph I a

Considering the method by which this graph has been constructed, the question
arises as to how reliable it is. The correctness of an equation can easily be
checked, but not the product of the relaxation method.

Fortunately, the results obtained with this graph could be compared with the
results obtained with 36 special graphs, each one constructed for a specific
drainage situation (ToksÖz and Kirkham, 1971). It appeared that both methods gave
the same results (Appendix B). Thus the conclusion can be drawn that the gene-
ralized graph of Ernst is both a reliable and an important contribution to the
theory and practice of drainage investigations. It is particularly useful in drai-
nage situations where there is an aquifer (highly pervious layer) at some depth
( I to 10 m or more) below drain level, a situation often found in irrigation
projects.

When there are two pervious soil layers below drain level, the two most common
drainage situations will be: K3<< K2, and K3>> K
2'

S i t u a t i o n Kg << K 2

The availability of Graph Ia enables us to investigate whether we are correct

in assuming that, if K g < 0.1 K 2 , we can regard the second layer below drain depth
(K D ) as being impervious. If we consider the L-values for this situation, as
3 3

30
given in Appendix B , we can conclude that although the layer K3D3 for K 3 < 0 . 1 K2
has some influence on the computed drain spacing, it is generally so small that
it can be neglected. However, if one is not sure whether the second layer below
drain level can be regarded as impervious, the means (equation and graph) are
now available to check it.

Situation K 3 >> K 2

This situation is of more importance than the previous one because it occurs
more frequently than is generally realized and has much more influence on the
required drain spacing.The examples will therefore be confined to this situation.

Examples (see Fig.4)

Given: The soil or an irrigation area consists of a loess deposit (K = 0.50
m/day) of varying thickness. In certain parts of the area an aquifer occurs
(sand and gravel, K = 10 mlday, thickness 5 m).

In the first set of examples (A) the loess deposit is underlain by an imper-
vious layer at a certain depth, varying from 3 to 40 m. In the second set
of examples (B), instead of an impervious layer, an aquifer is found at a
depth of 3 m and 8 m, whereas Examples C give alternative solutions in rela-
tion to drain depth and the use o f pipe drains instead of ditches.

It is intended to drain the area by means of ditches (drain level = 1.80 m,

wetted perimeter (u = 2 m). The maximum allowable height of the water table
is 1 m below surface (h = 0.80 m). The design discharge is 0.002 m/day
(h/q = 400).

A. Influence of t h e l o c a t i o n of an impervious l a y e r . Homogeneous s o i l

For this simple drainage situation, only the result of the drain spacing
computations will be given.

Drain spacing
Example Depth b a r r i e r
Ditch (u=2 m ) Pipe drain (u=0.30 m )

Al 3 m 50 m 49 m
A 2 8 m 96 m 84 m
A 3 40 m 146 m 107 m

31
 These results show that the depth of a barrier has a great influence on the
drain spacing and that the influence of the wetted perimeter of the drains ( u )
can vary from very small to considerable, depending on the depth of the barrier.

DRAIN SPACINGS A N D D R A I N A G E SURVEY N E E D S IN IRRIGATION

PROJECTS
Influence location of
A an impervious layer I B an aquifer
1 2 3 1 la 2

- - -- - - - - -
-.

q-O 002 m/day

loess K=050m/day

impervious layer O O
(fine textured
a IIu vial depos its )

, I
,
I
I
i
I ,
I
I I
I
I
1I
I
I I ,
ditch (uZ2m) L z ' 5 0 m b6m \46m 305 m I
625m I
I
200 m
pipedrain L ~ 4 9 m 84m 107m 180m 615 m
(u I 0.30)

1 1
'
KI= hydr. cond. above drain level lntluence KD aquifer
K ~ Z ., .. below ,. ,, (first layer) 1000 m2/day 1000 m*/day
K j i . . .. .. ,, .. (second layer) L~625m Lz245m I
u : wetted perimeter 1 much influence l i t t i e influence I

Fig.4. Drain spacings and drainage survey needs in irrigation projects.

B. Influence o f t h e location of an a q u i f e r
The various situations that will be handled here are:

Example Depth a q u i f e r Drain level Ditches Pipe drains Spacing

B I - 3 m - 1.80 m + 305 m
B la - 3 m - 1.80 m + 180 m
B 2 - 3 m - 3 m + 625 m
B Za - 3 m - 3 m + 615 m
B 3 - 8 m - 1.80 m + 200 m

32
Example B 1
Aquifer at -3 m (1.20 m below drain level)

h = 0.800 h/q = 400

q = 0.002 8h/q = 3200

K = 0.50 D2 = 1.20 K2D2=0.60 Lo = 402 m L = L -c=313m

2
K = IO D3 = IO K3d3 = 50 c = 89m SPC: L = 305 m
3

K /K = 20 D /D =4 KD = 50.6 c < 0.3 Lo

3 2 3 2
a = 4.0 (c/L0=O.22)

Note: The fZow aboue t h e drains can be neglected i n t h i s drainage s i t u a t i o n

KD aD2
K >> K2). Therefore, KD = K D + K3D3, whereas c = - In __
3 2 2 K2

3
If KD-values are high (here KD = 50 m /day), the flow in the K2D2 layer can also
be neglected and KD = K3D3 , the more so because the KD-value of the aquifer is
a very approximate value.
The L-value can be determined in two ways: either by using Graph I or by using
Eq.(9c) in combination with an SPC. It is recommended that both methods be used
to allow a check on any calculation errors. Small differences may occur in the
results of the two methods, but this is of no practical importance.

Example B l a
Drainage by pipe drains (u=0.30 m), instead of ditches

Lo = 402 m (see Ex.B I ) c/Lo = 0.69

c = 277 m LILo = 0.45 + L = 180 m

Note t h a t i n t h i s s i t u a t i o n t h e use of pipe drains instead of d i t c h e s has a

great i n f l u e n c e on the r e s u l t i n g drain spacing.

Example B 2‘
Ditch bottom in the aquifer (u=2 m); drain level -3 m.

h = 2.000 8 h/q = 8.000 ~ ~ = 6 3 2 mL = L -c=627m

q = 0.002
KD = 50 c = 5 m

33
Example B Za

Pipe drain in the aquifer (u=0.30 m), drain level -3 m.

Lo = 6 3 2 m (see E x . B 2) L = L -c=618m

c = 14m

Note t h a t i n t h i s s i t u a t i o n t h e use of pipe drains instead of d i t c h e s has very

Z i t t l e influence on the drain spacing, because here t h e radial r e s i s t a n c e i s
very small.

Example B 3

The aquifer at - 8 m (6.20 m below drain depth) ; KD = 50 m2/day;

ditch (u = 2 m); drain level -1.80 m.

h = 0.800 h/q = 400

q = 0.002 8q/h =3200

K2 = 0.50 D2=6.20 K2D2 = 3.10 Lo = 412 m c/L =0.61

K3 = 10.0 D3 = 5.0 K3D3 = 50.0 c = 253 m L/Lo=0.49 + L=202 m

K K =20 D3/D2=0.8 KD = 53.1

3 2
a = 3.5

C. I n f l u e n c e of t h e KD-value o f an a q u i f e r

Example C 1

In Example B 1 (aquifer at -3 m, KD-value = 50 m2/day + L = 305 m, the

KD-value has been estimated from borings to be at least 50 m2/day. Now the
question arises whether it is worthwhile to carry out pumping tests to obtain
a better estimate.

If, in a certain drainage situation, one wants to analyse the influence of

the magnitude of the KD-value on the spacing, it is convenient to calculate
I
firstly, - In
KP
*
, which in this case equals 1.75.

34
Assume KD = 100

8 h/q = 3 , 2 0 0 Lo = 566 c / L o = 0.31

KD = 100 c = 100 x 1 . 7 5 = 175 L = L - c = 300 m

KD = 500 + L = 570 m

KD = 1000 + L = 625 m

These computations show that in this case it will indeed be worthwhile to

carry out pumping tests.

Example C 2
In Example B 3 , with the depth of the aquifer at -8 m and KD = 5 0 , L = 200 m.
Making the same computations as for Example C 1, we get:

8 h/q = 3200 Lo = 566 m c/Lo = 0 . 8 9

KD = 100 c = 504 m L / L ~= 0 . 3 8 L = 215 m

~~

KD = 1000

8 h/q = 3200 Lo = 1789 m c/Lo = 2.82 SPC and Eq.9

TTLt
KD = 1000 c = 5040 m L = -= 250 m L = 245 m
8c

These results show that here an estimate of the KD-values will suffice and
therefore - in contrast to Example B 1 - no pumping tests are required.

Importance of geohydroZogicaZ investigations

If we compare Situation B 3 with that of A 2 (Fig.4), we get:

A 2 : impervious layer at -8 m + L = 100 m

B 2 : instead of an impervious layer, an aquifer at -8 m + L 200 m

35
 This comparison of drain spacings shows clearly that a drain spacing can be
considerably influenced by layers beyond the reach of a soil auger.

From Fig.4 it will be clear that if in the given situation drainage investi-
gations are only conducted to a depth of 2 m and a barrier at 3 m is assumed, the
recommending drain spacing will be 50 m.

If, however, geo-hydrological investigations are conducted, they will reveal

that parts of the area can be drained with spacings of 300 m (drain level -1.8 m)
or 600 m if the drain level is -3 m.

36
4.2 Summary of graphs and equations
G r a p h s
G.1 : c/Lo-, B- and L / L - v a l u e s + L-values (drain spacing)

G.Ia : K /K
3 2
- and D / D - v a l u e s
3 2
+ a-value ( a u x i l i a r y graph f o r
radial resistance)

G.11 : Homogeneous s o i l and p i p e d r a i n s -f L-value (for a l l D

2
- and
K -values)
2
D L
G.III: D In 2 or L In - ( a u x i l i a r y g r a p h i f a SPC i s n o t a v a i l a b l e )
2 u

E q u a t i o n s
Only one p e r v i o u s l a y e r below d r a i n d e p t h

D2 < $ L
U S E

Eq.(3a) L2 + E
TlK-
L In r_2
U
- 8 KD h / q = O o r i g i n a l equation o u t of u s e
L

Eq. ( 7 ) (t7 O
+ (so)
(kr- - B (
$)=O
O
g e n e r a l i z e d eq. G. 1

2 8 D
Eq.(8) L + - L D In - 8 KD h / q = O modified e q u a t i o n SPC
T l 2 u

Eq.(lO) L = L - c s i m p l i f i e d eq. f o r c <0.3

B 10.1
where Lo = 8 KD h / q
D
c = D I n 1
2 u

L
Eq.(ll) L I n - = n K2 h/q G.111 or
SPC
~~

Two p e r v i o u s l a y e r s below d r a i n d e p t h

Eq.(9) L
2 + -
8 L -
T
KD I n
K2
a
u
- 8 KD h / q = O G . 1 o r SPC

where KD = K D + K2D2 + K3D3 or for K

3
>> K 2 : G.1a

KD = K D
2 2 + K3D3

a = f (K~/K*, D ~ / D ~ )

37
4.3 Programmes Scientific Pocket Calculator (SPC)

Note: These p r o g r m e s should be adjusted i f necessary, t o suite t h e s p e c i f i c

D 2 8 KD aD2
E q . (8): L 2 s LD2 In 2
71 U
- 8KD h/q=O Eq. ( 9 ) : L +- L
v
-
Kz
In -- 8KD h/q = O

KD = K D KD = K2D2 + K D
1 1 + K2D2 3 3

4bZ-D
TI
D
In 2 ib -Tu _
KD
K7
aD2
I n __
U

Programme examples

KD ENT 8 h / q (x) KD ENT STO 8 h / q (X)

D2 ENT u (+) ( I n ) o r T ( + ) u ( + ) ( l n ) a ENT D2 (X) u (+) ( I n )

D2 (XI 4 (XI n (+) RCL (X) K2 (+) 4 (X) TI (+)

STO ENT (X) (+) (&) RCL (-) STO ENT (x) (+) (&) RCL (-)

or

D2 = 5 KD = 4 . 2 4 a = 4.6 KD = 17.3
L=87.07
u = a4 8 h/q = 2.400 D2 = 1.6 K = 1.2 L = 73.22
2
u = nr 8 h / q = 800
r = 0.10

38