drainage systems

2. When a pipe drainage system is being
designed, the following elements must be
determined:
-a- lay-out (alignment) of laterals and
collectors; this must be adapted to the
topographical features of the area and other
conditions.
b- spacing and depth of laterals; these are
primary factors in the control of the ground-
water table.
c- diameter and gradients of lateral and
collector pipes; these must ensure the
proper evacuation of the water taken up by
the laterals.
-

3. When the hydraulic design of a drain pipe
system is being considered one faces
such question as:
1- What area can be drained by a pipe line
of given diameter laid at a given slope,
assuming a certain specific discharge?
2- What pipe diameter is needed for a pipe
line of given length, laid at a given
slope, with given drain spacing and
specific discharge?

4. To provide answers to such questions, one must
consider the following items:
1- Basic flow equations (uniform flow) for
different types of drain pipes.
2- Flow equations that take into account the fact
that the flow in a drain pipe line increases in
the direction of flow as the drain takes up
water over its entire length (non-uniform
flow).
3- Gradient and slope of pipe line.
4- A safety factor to allow for some decrease in
capacity due to a certain degree of
sedimentation.
5- A drain composed of sections of increasing
diameter in the direction of flow.

5. 1- The case of uniform flow in circular
conduits running full.
2- The discharge and hydraulic gradient are
constants at all sections of the pipe.

6. 1- The flow rate Q gradually increases from Q=0 at the
upstream end to Q=qBL at the outflow where q is
the drainage coefficient, B is the width of area to
be drained pipe line.
2- The flow rate gradually increases in the direction of
flow.
3- The hydraulic gradient increases also.

7. 1- Uniform Flow:
A- Wesseling Equations (Smooth):
‫تستخدم‬
‫في‬
‫تصميم‬
‫مواسير‬
‫الصرف‬
‫الملساء‬
Clay, Cement, Smooth plastic pipes
Ql = qBL = f.s. ( 50 d2.714 S0.572 )
q: Drainage coefficient factor (m/sec)
B: Spacing between pipe drains.
L: Length of pipe drain. f.s.: Factor of Safety.
d: Pipe diameter (m).
S: Drain slope (dimensionless).

8. Factor of safety (f.s.)
f.s. = 60% for field drain d ≤15cm
f.s. = 75% for collector drain d >15cm
f.s. = 1for maximum length or maximum
drainage coefficient.
‫بسبب‬ ‫بالتدريج‬ ‫يقل‬ ‫المواسير‬ ‫قطر‬ ‫أن‬ ‫نتيجة‬ ‫اآلمان‬ ‫معامالت‬ ‫استخدام‬ ‫يتم‬
‫بداخله‬ ‫التربة‬ ‫حبيبات‬ ‫ترسيب‬
.

9. B- Manning Equations (Corrugated):
‫تستخدم‬
‫في‬
‫تصميم‬
‫مواسير‬
‫الصرف‬
‫البالستيك‬
‫المموجة‬
(
‫ذات‬
‫تع‬
‫اريج‬
)
Corrugated plastic PVC
Ql = qBL = f.s. ( 22 d2.667 S0.5 )
2- Non-Uniform Flow:
A- Wesseling Equations (Smooth):
Ql = qBL = f.s. ( 89 d2.714 S0.572 )
B- Manning Equations (Corrugated):
Ql = qBL = f.s. ( 38 d2.667 S0.5 )

10. ‫القطر‬ ‫فيكون‬ ‫المختلفة‬ ‫األقطار‬ ‫ذات‬ ‫المواسير‬ ‫من‬ ‫مجموعة‬ ‫نستخدم‬ ‫الطويلة‬ ‫المجمعات‬ ‫حالة‬ ‫في‬
‫المصب‬ ‫الى‬ ‫الماء‬ ‫سريان‬ ‫اتجاه‬ ‫في‬ ‫بالتدريج‬ ‫القطر‬ ‫يزداد‬ ‫ثم‬ ‫المجمع‬ ‫بداية‬ ‫األصغرفي‬
.
‫آخ‬ ‫معامل‬ ‫االعتبار‬ ‫في‬ ‫نأخذ‬ ‫فاننا‬ ‫المختلفة‬ ‫األقطار‬ ‫ذات‬ ‫المواسير‬ ‫مجموعة‬ ‫أطوال‬ ‫حساب‬ ‫عند‬
‫ر‬
‫ويسمى‬ ‫القانون‬ ‫في‬ ‫فعال‬ ‫للموجود‬ ‫باإلضافة‬
Reduction Factor (P)
P = 0.85 ‫اذا‬
‫كان‬
‫المجمع‬
‫يتكون‬
‫من‬
‫ماسورتين‬
P = 0.75 ‫اذا‬
‫كان‬
‫المجمع‬
‫يتكون‬
‫من‬
‫أكثر‬
‫من‬
‫ماسورتين‬

11. 1- The pipes used for lateral drains are
cement and corrugated PVC pipes with
internal diameters 100 and 72 mm
respectively. Find the maximum length
with each type for a drainage rate of
3mm/day and drain spacing 50m in the
following cases:
a- drain slope 0.1%
b- drain slope 0.2 %

12. 1- Cement Pipe
Ql = qBL = f.s. ( 89 d2.714 S0.572 )
(3*50*L)/(1000*24*3600) = 1*89 * (0.1)2.714
S0.572
L= 99040 S0.572
S (%)
1904.6
L (m)
0.1
2831.4
0.2

13. 2- PVC Pipe
Ql = qBL = f.s. ( 38 d2.667 S0.5 )
(3*50*L)/(1000*24*3600) = 1*38 * (0.072)2.667
S0.5
L= 1961.34 S0.5
L (m)
0.1
0.2
62
87
S (%)

14. Find the maximum drainage coefficient, which can be
drained from an area with corrugated plastic pipe of
a diameter 72mm at a spacing of 60m and length
200m if the slope of the drain is:
a- 10cm per 100m b- 0.2%
Solution
Ql = qBL = f.s. ( 38 d2.667 S0.5 )
a- q * 60 * 200 = 1* 38 * (0.072)2.667 * (10/10000)0.5
q = 9*10-8 m/sec * 1000 * 24 * 60 * 60 = 7.76 mm/day
b- q * 60 * 200 = 1 * 38 * (0.072)2.667 * (0.2/100)0.5
q = 1.3*10-7 m/sec * 1000 * 86400 = 0.97 mm/day

15. What is the maximum area which can be
drained with pipe of constant diameter
when the allowable slope should not
exceed 0.04% and the drainage
coefficient is 4mm/day? The pipes are of
100,150, and 200mm diameter and they
are:
a- corrugated plastic tubes.
b- smooth pipes.

16. Max. area f.s. =1.0
A- Corrugated Plastic tubes:
Ql = qBL = f.s. ( 38 d2.667 S0.5 )
(4/1000)*(BL / 24*60*60) = 38 d2.667 (0.04/100)0.5
d
BL (fed.)
0.1 0.15 0.2
0.75 BL
35340 104208
26505 78156
224447
168335.3

17. B- Smooth Pipes
Ql = qBL = f.s. ( 89 d2.714 S0.572 )
(4/1000)*(BL / 24*60*60) = 89 d2.714 (0.04/100)0.572
d
BL (fed.)
0.1 0.15 0.2
0.75 BL
42288 127096
31716 95322
277470
208102.5

18. Design a corrugated plastic collector
drain with a slope of 10cm per 100m
and increasing diameters 125, 160, and
200, if the drainage coefficient is
3mm/day. The length of laterals on
each side is 200m and the total length
of the collector is 650m. What is the
drop in elevation for a pipe of diameter
of 350mm to transport the flow of this
area to a lake at 300m from the outlet?

19. Ql = qBL = f.s. ( 38 d2.667 S0.5 )
(3 / 1000*24*60*60) * 400 * L = f.s. * 38 * d2.667 *
(0.001)0.5
L = f.s. (86519.916 d2.667 )
Diameter (m)
f.s.
Max. L (m)
0.75 L (m)
Approximate L (m)
Each length (m)
Total length (m)
150
150
151.98
0.125 0.16
0.75
250 250
202.64054
365 650
0.6
489.289
650
366.967 665.4
887.2
0.75
0.2

20. Transporting Case:
Ql = qBL = f.s. ( 22 d2.667 S0.5 )
(3 / 1000*24*60*60) * 400 * 650 = 0.75 * 22 *
(0.35)2.667 * (K/300)0.5
K = 0.0242804 m
The drop in elevation (k) = 24.28mm

21. Design a corrugated plastic collector
with increasing diameters 125, 150,
200 and 260mm are used and the
pipe slope is 0.08%. The drainage
rate is 4mm/day and the width of the
area served is 350m. What is the total
length of the collector in this case?

22. Ql = qBL = f.s. ( 38 d2.667 S0.5 )
(4 / 1000*24*60*60) * 350*L = 38 * d2.667 *(0.08/100) S0.5 * f.s.
Diameter (m) 0.125
f.s. 0.6
Max. L (m) 155.4
0.75 L (m) 116.55
Approximate L (m) 115
0.15
0.6
315.8
236
235
0.2 0.26
0.75
680.2
510
510
0.75
1369.3
1027
1025

23. A concrete collector with a diameter 20
cm, a length 600m laid at slope 0.04%
drains an area 300 m wide with
discharge rate 10mm/day. What will
be the over-pressure at the upstream
end of the collector if its capacity is
to be set at 75%.

24. Ql = qBL = f.s. ( 89 d2.714 S0.572 )
(10 / 1000*24*60*60) * 300 * 600 = 0.75 * 89 *
(0.2)2.714 S’0.572
S’ = 1.54*10-3
S’ = Z/L = Z/600
Z = 0.924
i = S = H/L
H = S*L = (0.04/100) * 600
H = 0.24m
Over Pressure = Z – H = 0.924 – 0.24 = 0.684m =
68.4cm

25. Calculate the maximum area to be saved
by a cement collector pipe in the tile
drainage system according to the
following data: drainage coefficient =
4mm/day, collector pipe diameter
=20cm, average slope = 4cm/100m.

26. Smooth pipe, q=4mm/day, d=20cm,
S= 4*10-4
Ql = qBL = f.s. ( 89 d2.714 S0.572 )
(4 / 1000*24*60*60) BL = 0.75 * 89 * (0.2)2.714
*(4*10-4)0.572
BL = 208103.2 m2 = 49.5 fed

27. A collector drain in a composite system has a
total length of 750m and slope 0.04%
serves an area with a width 400m. If the
drainage coefficient is 2.0mm/day and
pipes available are corrugated plastic
tubes with diameters equal to 150, 200,
and 250mm. Find the maximum length that
can be used of each size to make a
collector with increasing diameter.

28. S = 0.04%, B=400m, L=750m, q=2mm/day
Ql = qBL = f.s. ( 38 d2.667 S0.5 )
(2 / 1000*24*60*60)*400*L = 38*d2.667*(0.04/100)S0.5 * f.s.
1.22 * 10-5 L = f.s. * d2.667
Diameter (m)
f.s.
Max. L (m)
0.75 L (m)
Approximate L (m)
Each length (m)
Total length (m)
230
230
234.6
0.15 0.2
0.75
400 515
312.86
630 1145
0.6
842.2
750
631.66 1145.2
1526.9
0.75
0.25