Chapter 5. Pipe System Learning Outcomes: Understand the concept of hydraulic grade line and energy grade line, Analyse flow in single pipe, pipe is series and pipe in parallel. Analyse loop pipe network using Hardy-Cross method.

5.1 Hydraulic Grade Line and Energy Grade Line Hydraulic grade line (HGL) or piezometric line shows the level of piezometric head, i.e. Energy grade line (EGL) shows the total energy head, i.e.

Entrance loss Friction loss Discharge loss

Friction loss Entrance loss Expansion loss Discharge loss

Friction loss Contraction loss Discharge loss Entrance loss

Example 5.1 In a fire fighting system, a pipeline with a pump leads to a nozzle as shown in Figure. Find the flow rate when the pump develops a head of 80 ft, given that we may express the friction head loss in the 6-in diameter pipe by hf6 = 5V62/2g, and the friction head loss in the 4-in diameter pipe by hf4 = 12V42/2g. Neglect minor losses. Sketch the energy line and hydraulic grade line, (b) Find the pressure head at the suction side of the pump, (c) Find the power delivered to the water by the pump, and (d) Compute the power of the jet.

C Energy equation between A and C: pA = 0; pC = 0; VA = 0; zA = 70 ft; zC = 80 ft; hp = 80 ft Express V4 and V6 in terms of VC using the continuity equation:

Energy equation between A and C: Substitute and where g = 32.2 ft/s2 (a) Sketch the energy line and hydraulic grade line.

145.72 ft 141.38 ft EGL HGL 93.70 ft 89.36 ft 70 ft 65.72 ft 64.86 ft

(b) Find the pressure head at the suction side of the pump Energy equation between A and B:

(c) Find the power delivered to the water by the pump (d) Compute the power of the jet

5.2 Branching Pipes Considering only friction losses, the elevation of P must lie between the surfaces of reservoirs A and C. If P is level with the surface of reservoir B, then hf2 = h2 = 0 and Q2 = 0. If P is above the surface of reservoir B, then water must flow in B and If P is below the surface of reservoir B, then the flow must be out of B and

5.3 Pipes in Series According to continuity and the energy equations, the following relations apply to the pipes in series:

Example 5.2 Pipes 1, 2 and 3 are 300 m of 300 mm diameter, 150 m of 200 mm diameter, and 250 m of 250 mm diameter, respectively, of new cast iron and are conveying 15C water. If z = 10 m, find the rate of flow from A to B. Assume fully turbulent flow and neglect minor losses. = 10 m L1 = 300 m D1 = 0.3 m L2 = 150 m D2 = 0.2 m L3 = 250 m D3 = 0.25 m

Pipe 1 2 3 L (m) 300 150 250 D (m) 0.3 0.2 0.25 e/D 0.00083 0.00125 0.00100 f 0.019 0.021 0.020 From continuity: Energy equation between A and B:

5.4 Pipes in Parallel For the parallel or looping pipes of the following figure, the continuity and energy equations provide the following relations: This is because pressures at A and B are common to all pipes.

Example 5.3 Three pipes A, B and C are interconnected as in figure below. The pipe characteristics are as follows: Pipe D (in) L (ft) f A 6 2000 0.020 B 4 1600 0.032 C 8 4000 0.024 Find the rate at which water will flow in each pipe. Find also the pressure at point P. All pipe lengths are much greater than 1000 diameters, so neglect minor losses.

1 2 Energy equation between 1 and 2: From continuity equation: For pipes in parallel:

1 2 Energy equation between 1 and 2: Check the continuity:

1 2 Pressure at point P Energy equation between 1 and P:

5.5 Pipe Networks Three simple methods to solve for pipe networks in loop configuration are: Hardy Cross (using either Darcy-Weisbach or Hazen-Williams equation), Linear theory, and Newton-Raphson. The most popular is the Hardy Cross method which involves a series of successive approximations and corrections to flows in individual pipes.

According to the Darcy-Weisbach equation, According to the Hazen-Williams equation, where, and n = 2 (Darcy-Weisbach in S.I. unit) and n = 1.85 (Hazen-Williams in S.I. unit) The sum of head losses around any closed loop is zero, i.e.

Roughness values for pipes Pipe material Equivalent roughness e (ft) Hazen-Williams coefficient C Brass, copper, aluminium Smooth 140 PVC, plastic 150 Cast iron (new) 8.0  104 130 Cast iron (old)  100 Galvanized iron 5.0  104 120 Asphalted iron 4.0  104 Wrought iron 1.5  104 Commercial and welded steel Riveted steel 60.0  104 110 Concrete 40.0  104 Wood stave 20.0  104

Equivalent resistance K for pipe Formula Units of Measurement K Hazen-Williams Q (cfs), L (ft), d (ft), hf (ft) Q (gpm), L (ft), d (in), hf (ft) Q (m3/s), L (m), d (m), hf (m) Darcy-Weisbach

Consider that Qa is an assumed pipe discharge that varies from pipe to pipe of a loop to satisfy the continuity of flow. If  is the correction made in the assumed flow of all pipes of a loop to satisfy then, Expansion by binomial theorem and retaining only the first two terms yield,

Procedure of Hardy-Cross method: Divide the network into a number of closed loops. Computations are made for one loop at a time. Compute K for each pipe. Assume a discharge Qa and its direction in each pipe of the loop. At each joint, the total flow in should equal to the flow out. Consider the clock-wise flow to be positive and the counter clockwise flow to be negative. Compute hf for each pipe and retain the sign of the flow direction. Compute hf/Q for each pipe without regard to the sign. Determine the correction . Apply the correction algebraically to the discharge of each member of the loop. For common members among two loops, both  corrections should be made, one for each loop. For the adjusted Q, repeat steps 4 to 7 until  becomes very small for all loops.

Example 5.4 Find the discharge in each pipe of the welded steel pipe network shown in Figure below. All pipes are 4 in. in diameter. The pressure head at A is 50 ft. Determine the pressure at the different nodes. 0.6 cfs B 100 ft D 100 ft A I II 1.8 cfs 50 ft 50 ft 100 ft 100 ft E C 0.4 cfs 0.8 cfs

+ + 0.6 cfs Assumed discharges B 0.8 cfs D 1.0 cfs A 0.2 cfs 0.2 cfs I II + 1.8 cfs 0.8 cfs 0.6 cfs E C 0.4 cfs 0.8 cfs

1st iteration (1) (2) (3) (4) (5) (6) (7) Loop Pipeline K Qa (cfs) hf = KQa1.85 1 AB 14.25 +1.0 +14.25 +0.9 BC 7.12 +0.2 +0.36 1.80 +0.185 CA 0.8 9.43 11.79 0.9 +5.18 27.84 2 BD +0.8 +9.43 +0.715 DE +0.115 EC 0.6 5.54 9.23 0.685 CB 0.2 0.36 0.185 +3.89 24.62 For loop 1, For loop 2,

2nd iteration (1) (2) (3) (4) (5) (6) (7) Loop Pipeline K Qa (cfs) hf = KQa1.85 1 AB 14.25 +0.9 +11.73 13.03 BC 7.12 +0.185 +0.31 1.68 CA 0.9 11.73 27.74 2 BD +0.715 +7.66 10.71 DE +0.115 +0.13 1.13 EC 0.685 7.08 10.34 CB 0.185 0.31 +0.40 23.86 For loop 1, For loop 2,

+ + Final discharges 0.6 cfs B 0.715 cfs D 0.9 cfs A I II 1.8 cfs

Project Question A water distribution network is shown as in figure. Elevation of A is 61 m, with pressure head = 45.72 m. Elevation of D is 30.5 m. Using f = 0.015, find: Flow rate through each pipe, and Pressure head at node D. 0.3 m3/s B 1000 m -  500 mm 1000 m -  250 mm 0.6 m3/s A 1500 m -  500 mm 0.5 m3/s D 1200 m -  750 mm 700 m -  500 mm C 0.2 m3/s

Quiz 2 cumec 0.8 cumec A C + B D 0.7 cumec 0.5 cumec