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# Answer for parallel pipes h f g 2 v d l f 2 1 1 1 1 g

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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 = 004 3000 06 2 981 1 2 .* . v = 0 04 2000 03 2 2 . v Therefore, v 1 = 0.767 m/s v 2 = 0.664 m/s Hence Q 1 = A 1 v 1 = π *. 4 0 767 2 = 0.217 m 3 /s Q 2 = A 2 v 2 = π 4 0664 2 = 0.047 m 3 /s Total discharge, Q = Q 1 + Q 2 = 0.217 + 0.047 m 3 /s = 0.264 m 3 /s

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Fluid Mechanics Chapter 7 – Steady Flow in Pipes P.7-26 7.5.2 Branched-pipe Problem h 1 h 3 h 2 J reservoir 1 reservoir 3 reservoir 2 pipe1, k 1 pipe2, k 2 pipe 3, k 3 Q 1 Q 2 Q 3 ± Assume h 1 > h 2 > h 3 and the 3 pipes intersect at junction J. ± As h 1 is the highest head, the flow in pipe 1 must be toward J. ± As h 3 is the lowest head, Q 3 is flowing from J to the reservoir 3. ± The flow Q 2 ’s direction is unknown because it depends on the head at junction J. ± If h J be the head at junction J. There are two possible cases (i) h 1 > h J > h 2 , or (ii) h 2 > h J > h 3 ± For case (i), h 1 > h J > h 2 , Q 2 is from J to reservoir 2. Q 1 - Q 2 - Q 3 = 0 h 1 – h J = k 1 *Q 1 2 h J - h 2 = k 2 *Q 2 2 ( 7 . 2 4 ) h J - h 3 = k 3 *Q 3 2 ± For case (ii), h 2 > h J > h 3 , Q 2 is from reservoir 2 to J. Q 1 + Q 2 - Q 3 = 0 h 1 – h J = k 1 *Q 1 2 h 2 – h J = k 2 *Q 2 2 ( 7 . 2 5 ) h J - h 3 = k 3 *Q 3 2
Fluid Mechanics Chapter 7 – Steady Flow in Pipes P.7-27 ± Both sets of equations have 4 unknowns Q 1 , Q 2 , Q 3 and h J . We have to determine which case controls the problem. ± It is determined by assuming h J = h 2 , i.e. no flow from J to reservoir 2. Therefore Q 1 = hh k 12 1 ( 7 . 2 6 ) Q 3 = k 23 3 ( 7 . 2 7 ) ± If Q 1 ’ > Q 3 ’, Q 2 is from J to reservoir 2 - case (i). ± If Q 1 ’ < Q 3 ’, Q 2 is from reservoir 2 to J - case (ii).

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Fluid Mechanics Chapter 7 – Steady Flow in Pipes P.7-28 Worked example: Three reservoirs are connected as the figure below. Determine the flow, Q 1 , Q 2 and Q 3 with k 1 = 3.058, k 2 = 8.860 and k 3 = 0.403 s 2 /m 5 and h f = k i *Q i 2 . J reservoir 1 reservoir 3 reservoir 2 Q 1 Q 2 Q 3 200m 140m 180m Answer Step 1 Pipe h i (m) k i (s 2 /m 5 ) 1 200 3.058 2 180 8.860 3 140 0.403 Step 2 - calculate Q 1 ’ and Q 3 Q 1 ’ = hh k 12 1 = 200 180 3058 . = 2.557 m 3 /s Q 3 ’ = k 23 3 = 180 140 0403 . = 9.963 m 3 /s Q 1 ’ < Q 3 case (ii) i.e. h 2 > h > h 3
Fluid Mechanics Chapter 7 – Steady Flow in Pipes P.7-29 Step 3 - set up the equations For case (ii), we have 200 - h = 3.058Q 1 2 or Q 1 = 200 3058 h . m 3 /s 180 - h = 8.860Q 2 2 or Q 2 = 180 8860 h . m 3 /s h - 140 = 0.403Q 3 2 or Q 3 = h 140 0 403 . m 3 /s Since Q 1 + Q 2 - Q 3 = 0 therefore 200 h . + 180 h . - h 140 0 403 . = 0 Step 4 - solve for h (180 < h < 140) by iterations h (m) Q 1 + Q 2 + Q 3 (m 3 /s) Error 180 2.557 + 0 – 9.963 -7.406 160 3.617 + 1.502 – 7.045 -1.926 150 4.044 + 1.840 - 4.981 +0.903 153 3.920 + 1.746 - 5.680 -0.014 152.9 3.925 + 1.749 - 5.658 +0.016 152.95 3.922 + 1.747 - 5.669 +0.000 Therefore, the head h at the junction is 152.95 m and Q 1 = 3.922 m 3 /s (towards J) Q 2 = 1.747 m 3 /s (towards J) Q 3 = 5.669 m 3 /s (towards reservoir C)

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Fluid Mechanics Chapter 7 – Steady Flow in Pipes P.7-30 7.5.3 Hardy-Cross Method (Reference only) loop 1 loop 2 loop 3 loop 4 loop 5 loop 6 water in water out ± The supply of water system to a city is a complicated network of pipelines. The commonly used technique is the Hardy-Cross method.
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Answer For parallel pipes h f g 2 v d L f 2 1 1 1 1 g 2 v d...

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