# 751 resistance coefficients for pipelines in series

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7.5.1 Resistance Coefficients for Pipelines in Series and Parallel In general, the equation of head loss can be expressed as h f = k*Q 2 (7.19) Pipes are in series if they are connected end to end so the fluid flow in a continuous line is a constant. Q Q Q Q Q h 1 h 2 h 3 h n By continuity of flow, Q is same for each pipe. The total loss of the system is given as h f = h 1 + h 2 + h 3 +…+ h n = k 1 *Q 2 + k 2 *Q 2 + k 3 *Q 2 + … + k n *Q 2 = (k 1 + k 2 + k 3 + … + k n )*Q 2 (7.20) The effective resistance coefficient is k eff = k 1 + k 2 + k 3 + … + k n (7.21) i.e. the total head loss is the summation of the individual pipe.

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Fluid Mechanics Chapter 7 – Steady Flow in Pipes P.7-22 For pipes connected in parallel , the fluid can flow from one to the other by a number of alternative routines. Q 1 Q 2 Q 3 Q n Q Q h f The head loss for individual pipe is the same as the total head loss. The total flow rate is the summation of the individual pipe. Since h f = k i *Q i 2 or Q i = h k f i Total Q = Q 1 + Q 2 + Q 3 + … + Q n (7.22) = h k f 1 + h k f 2 + h k f 3 +…+ h k f n = ( 1 1 k + 1 2 k + 1 3 k +…+ 1 k n ) h f = h k f eff Hence 1 k eff = 1 1 k + 1 2 k + 1 3 k +…+ 1 k n (7.23)
Fluid Mechanics Chapter 7 – Steady Flow in Pipes P.7-23 Worked examples: 1. Two reservoirs are connected by a pipeline which is 150 mm in diameter for the first 6 m and 225 mm in diameter for the remaining 15 m. The water surface in the upper reservoir is 6 m above that in the lower. By neglecting any minor losses, calculate the rate of flow in m3/s. Friction coefficient f is 0.04 for both pipes. Answer The velocities v 1 and v 2 are related by the continuity equation. i.e. A 1 v 1 = A 2 v 2 v 1 = v 2 A A 2 1 = v 2 d d 2 1 2 = 225 150 2 v 2 = 2.25 v 2 Friction in the 150 mm pipe h f1 = g 2 v d L f 2 1 1 1 1 = 0 04 6 015 2 1 2 . * . v g = 1.6 v g 1 2 2 = 1.6*2.25 2 * v g 2 2 2 = 8.1 v g 2 2 2 Similarly, friction in the 225 mm pipe h f2 = g 2 v d L f 2 2 2 2 2 = 0 04 15 0 225 2 2 2 . * . v g = 2.67 v g 2 2 2 Hence, total head loss = h f1 + h f2 = 10.77 v g 2 2 2 Applying Bernoulli’s equation between the two top water surfaces, p 1 = p 2 = 0 (P atm ) v 1 = v 2 = 0 (water surfaces) z 1 = 6 m; z 2 = 0 2 2 2 2 1 2 1 1 z + g 2 v + p = z + g 2 v + p γ γ +h L

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Fluid Mechanics Chapter 7 – Steady Flow in Pipes P.7-24 or z 1 – z 2 = h L 6 = 10.77 v g 2 2 2 or v 2 = 6 2 9 81 10 77 * * . . = 3.31 m/s Hence Q = A 2 v 2 = π * . * . 0 225 4 3 31 2 m 3 /s = 0.132 m 3 /s
Fluid Mechanics Chapter 7 – Steady Flow in Pipes P.7-25 2. Two reservoirs have a difference of level of 6 m and are connected by two pipes laid in parallel. The first pipe is 600 mm diameter of 3000 m long and the second one is 300 mm diameter of 2000 m long. By neglecting all the minor losses, calculate the total discharge if f = 0.04 for both pipes. Answer For parallel pipes, h f = g 2 v d L f 2 1 1 1 1 = g 2 v d L f 2 2 2 2 2 Apply Bernoulli’s equation to the points on the free surfaces and from the result of the previous worked example, level difference = head loss H = g 2 v d L f 2 1 1 1 1 = g 2 v d L f 2 2 2 2 2 6 = 0 04 3000 0 6 2 9 81 1 2 . * . * .

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