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# 1 2 3 4 ve h 14 h 43 h 23 h 21 0 728 the net flow out

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1 2 3 4 +ve h 14 + h 43 - h 23 - h 21 = 0 (7.28) The net flow out of each junction must be equal to zero. 7 1 2 3 4 5 6 Q 17 + Q 47 - Q 76 - Q 75 - Q 72 - Q 73 = 0 (7.29)

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Fluid Mechanics Chapter 7 – Steady Flow in Pipes P.7-31 Method of Analysis Assume h i = k i *Q i 2 The sign convention is clockwise positive for the discharge and head loss. Initially a flow rate, Q is assumed. A correction for discharge, Q is then evaluated and the new flow rate is Q + Q. the head loss for each member can be approximated as h i ’ = k i *(Q i + Q) 2 = k i *[Q i 2 + 2Q i Q + ( Q) 2 ] k i *Q i 2 + 2 k i *Q i * Q Summation around the loop Σ h i ’ = Σ [k i *Q i 2 + 2k i *Q i * Q] = 0 (by assumption 1) i.e. Σ h i + 2 Q* Σ (h i / Q i ) = 0 hence Q = 1 2 Σ Σ h h Q i i i ( ) (7.30) Σ h i is the sum of head loss around a loop which can be +ve or -ve. Σ (h i / Q i ) is the sum of the ratio (head loss/flow) for each member of the loop. The ratio is a magnitude and therefore is +ve only. Q is the correction flow for a loop. Each pipe within the loop will have this correction. Any pipe belonging to 2 or more loops, the correction for that particular pipe will contribute from every loop.
Fluid Mechanics Chapter 7 – Steady Flow in Pipes P.7-32 Worked examples: 1. Determine the flow in each branch of the loop as shown below by using (i) equivalent pipe method, and (ii) Hardy-Cross method. 2m /s 3 2m /s 3 pipe 1, k pipe 2, 2k Q 1 Q 2 Answer (i) As a parallel system h f1 = h f2 k*Q 1 2 = 2k*Q 2 2 or Q 1 = 2 Q 2 As Q = 2m 3 /s, = Q 1 + Q 2 2 = 2 Q 2 + Q 2 (1+ 2 )Q 2 = 2 Q 2 = 0.8284 m 3 /s Q 1 = 2 Q 2 = 2 *0.8284 = 1.1716 m 3 /s

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Fluid Mechanics Chapter 7 – Steady Flow in Pipes P.7-33 (ii) Assume Q 1 = 1.5 m 3 /s Q 2 = 0.5 m 3 /s 2m /s 3 2m /s 3 1.5m 3 /s -0.5m 3 /s +ve 1 st iteration Pipe k i Q i ± h i abs(h i /Q i ) 1 1 1.5 2.25 1.5 2 2 -0.5 -0.5 1.0 Σ = 1.75 2.5 Q = 1 2 Σ Σ h h Q i i i ( ) = - 175 2 2 5 . * . m 3 /s = -0.35 m 3 /s 2 nd iteration Pipe k i Q i ± h i abs(h i /Q i ) 1 1 1.15 1.3225 1.15 2 2 -0.85 -1.445 1.70 Σ = -0.1225 2.85 Q = - 01225 2 2 85 . * . m 3 /s = 0.0215 m 3 /s
Fluid Mechanics Chapter 7 – Steady Flow in Pipes P.7-34 3 rd iteration Pipe k i Q i ± h i abs(h i /Q i ) 1 1 1.1715 1.3724 1.1715 2 2 -0.8285 -1.3728 1.655 Σ = -0.0004 2.8265 Q = - 0 0004 2 2 8265 . * . m 3 /s = 0.000071 m 3 /s Q 1 = 1.1715 m 3 /s Q 2 = 0.8285 m 3 /s (after 3 iterations)

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Fluid Mechanics Chapter 7 – Steady Flow in Pipes P.7-35 2. Find the flow in the pipeline using Hardy Cross Method. K for vertical members are 3 and horizontals are 5. A B C D E F 0.5m /s 3 0.1 0.2 0.2 m /s 3 m /s 3 m /s 3 300m 200mm 300m 200mm 300m 200mm 500m 200mm 500m 200mm 500m 200mm 500m 200mm Answer Assume the flow rates are A B C D E F 0.5m /s 3 0.1 0.2 0.2 m /s 3 m /s 3 m /s 3 0.2 0.05 0.2 0.15 0.05 0.3 0.15 m /s 3 m /s 3 m /s 3 m /s 3 m /s 3 m /s 3 m /s 3 1 st Iteration Pipe k i Q i ± h i Abs(h i /Q i ) AB 5 0.20 0.2000 1.00 BE 3 0.15 0.0675 0.45 DE 5 -0.20 -0.2000 1.00 AD 3 -0.30 -0.2700 0.90 -0.2025 3.35 BC 5 0.05 0.0125 0.25 CF 3 0.05 0.0075 0.15 EF 5 -0.15 -0.1125 0.75 BE 3 -0.15 -0.0675 0.45 -0.16 1.6 Q 1 = 02025 2 335 . * . = 0.03 m 3 /s Q 2 = 016 2 16 . * . = 0.05 m 3 /s
Fluid Mechanics Chapter 7 – Steady Flow in Pipes P.7-36 2nd Iteration Pipe k i Q i ± h i Abs(h i /Q i ) AB 5 0.230 0.2650 1.15 BE 3 0.130 0.0509 0.39 DE 5 -0.170 -0.1441 0.85 AD 3 -0.270 -0.2183 0.81 -0.04657 3.2 BC 5 0.100 0.05 0.5 CF 3 0.100 0.03 0.3 EF 5 -0.100 -0.05 0.5 BE 3 -0.130 -0.05087 0.390672 -0.02087 1.690672 Q 1 = 0 4657 2 32 .

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