Design of Pipe Drainage

Design of Pipe Drainage 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.

Design of Pipe Drainage 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?

Design of Pipe Drainage 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.

Uniform Flow (Transporting Flow) 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.

Non-Uniform Flow (Dewatering Flow) 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.

A- Wesseling Equations (Smooth): Equations for Design 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 (m). L: Length of pipe drain (m). f.s.: Factor of Safety. d: Pipe diameter (m). S: Drain slope (dimensionless).

Equations for Design 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. يتم استخدام معاملات الآمان نتيجة أن قطر المواسير يقل بالتدريج بسبب ترسيب حبيبات التربة بداخله.

Equations for Design Ql = qBL = f.s. ( 22 d2.667 S0.5 ) B- Manning Equations (Corrugated): تستخدم في تصميم مواسير الصرف البلاستيك المموجة (ذات تعاريج) 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 ) Ql = qBL = f.s. ( 38 d2.667 S0.5 )

with Increasing Diameter Drain Pipe lines with Increasing Diameter في حالة المجمعات الطويلة نستخدم مجموعة من المواسير ذات الأقطار المختلفة فيكون القطر الأصغرفي بداية المجمع ثم يزداد القطر بالتدريج في اتجاه سريان الماء الى المصب. عند حساب أطوال مجموعة المواسير ذات الأقطار المختلفة فاننا نأخذ في الاعتبار معامل آخر بالإضافة للموجود فعلا في القانون ويسمى Reduction Factor (P) P = 0.85 اذا كان المجمع يتكون من ماسورتين P = 0.75 اذا كان المجمع يتكون من أكثر من ماسورتين

Problem (1) 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 %

Solution (1) S (%) L (m) 0.1 1904.6 0.2 2831.4 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 (%) L (m) 0.1 1904.6 0.2 2831.4

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 S (%) L (m) 0.1 62 0.2 87

Problem (2) 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

Problem (3) 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.

Solution (3) Ql = qBL = f.s. ( 38 d2.667 S0.5 ) 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 0.1 0.15 0.2 BL (fed.) 35340 104208 224447

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

Problem (4) 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?

Solution(4) Ql = qBL = f.s. ( 38 d2.667 S0.5 ) Diameter (m) 0.125 0.16 0.2 f.s. 0.6 0.75 0.75 Max. L (m) 202.64054 489.289 887.2 (0.75) L (m) 151.98 366.967 665.4 Approximate L (m) 150 365 650 Each length (m) 150 250 250 Total length (m) 650

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

Problem (5) 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?

Solution(5) 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 0.15 0.2 0.26 f.s. 0.6 0.6 0.75 0.75 Max. L (m) 155.4 315.8 680.2 1369.3 0.75 L (m) 116.55 236 510 1027 Approximate L (m) 115 235 510 1025

Problem (6) 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%.

Solution(6) Ql = qBL = f.s. ( 89 d2.714 S0.572 ) 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

Problem (7) Calculate the 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.

Smooth pipe, q=4mm/day, d=20cm, S= 4*10-4 Solution(7) 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

Problem (8) 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.

Solution(8) 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) 0.15 0.2 0.25 f.s. 0.6 0.75 0.75 Max. L (m) 312.86 842.2 1526.9 0.75 L (m) 234.6 631.66 1145.2 Approximate L (m) 230 630 1145 Each length (m) 230 400 515 Total length (m) 750