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Dr. Sanghamitra Kundu
Assistant Professor
Civil Engineering Department
BITS Pilani
� Types of flow
� Darcy-Weisbach equation
� Hydraulic Grade Line and Energy Grade Line
� Flow through long pipes
� Pipes in series or compound pipe
� Equivalent pipe
� Pipes in parallel
� Branched pipes
� Transmission of power through pipes
� Flow through nozzle at the end of a pipe
� Water hammer in pipes
� Pipe networks
Laminar: highly ordered fluid motion
with smooth streamlines.
Turbulent: highly disordered fluid motion characterized by velocity fluctuations and eddies.
Transitional: a flow that contains both
laminar and turbulent regions
Reynolds number, is the key parameter in determining whether flow is laminar or turbulent.
� Given:� L1, L2, L3
� D1, D2, D3
� f1, f2, f3
� Assumptions� Flow steady
� Minor losses small; hence neglected
� Reservoirs are large enough; so their water surface levels are constant
� Basic equations used to solve� Continuity equation
� Bernoulli’s equation
� Darcy- Weisbach Equation
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� Type 1� Given:
� D, f, L for all
� Q1, ZA, ZB
� To determine:� Q2, Q3, ZC
� Type 2� Given:
� D, f, L for all
� ZA, ZC, Q2
� To determine:� ZB, Q1, Q3
� Type 3� Given:
� D, f, L for all
� ZA, ZB, ZC
� To determine:� Q1, Q2, Q3
� Water flows from a reservoir A through a pipe of diameter D1 = 120 mm andlength L1 = 120 m to a junction at D, from which a pipe of diameter D2 = 75 mmand length L2 = 60 m leads to reservoir B in which the water level is 16m belowthat in reservoir A. A third pipe, of diameter D3 = 60 mm and length L3 = 40 m,leads from D to reservoir C, in which the water level is 24 m below that inreservoir A. Taking f = 0.01 for all the pipes and neglecting all losses other thanthose due to friction, determine the volume rates of flow in each pipe.
� L1 = 1500 m, D1 = 90 cm; L2 = 450 m; D2 = 60 cm; L3 = 1200 m; D3 = 45 cm; ZA =90 m; ZB = 75 m; Q1 = 1.42 m3/s; ZC = ?? [f = 0.02 for all pipes]
� A pipe having a length of 6000 m and dia. 70 cm connects tworeservoirs A and B, the difference between their water levels is 30m. Halfway along the pipe there is a branch through which watercan be supplied to a third reservoir C. Taking f = 0.024,determine the rate of flow of reservoir when (a) no water isdischarged to reservoir C; (b) the quantity of water discharged toreservoir C is 0.15 m3/s. Neglect minor losses.
� The water levels in the two reservoirs A and B are respectively 66m and 61.5 m above datum. A pipe joins each to a common pointD where the pressure is 103 KN/m2 gauge and height is 45 mabove datum. Another pipe connects D to another tank C. Whatwill be the height of water level in assuming the same value offriction factor for all pipes?� LAD = 2400 m; DAD = 0.30 m� LBD = 2700 m; DBD = 0.45 m� LCD = 3000 m; DCD = 0.60 m
� Software: Pipe flow expert (Demo) for 3-reservoir problem� http://www.youtube.com/watch?v=RFH0E1wC-qI
� Pipes carrying water under pressure from one point to another may be used to transmit hydraulic power.
� Used for functioning of several hydraulic machines
� P = f (Q, H)
� The power (or energy per sec) available at the outlet of the pipe is� P = wt. of water/sec × head available
� P = wQ(H-hf)
� Max. power transmitted when H = 3hf
� A nozzle is a gradually converging short tube which isfitted at the outlet end of a pipe for the purpose ofconverting the total energy of flowing water intovelocity energy
� E.g. fire extinguishers, impulse turbines
� Conditions for:� Maximum power available from a nozzle
� H = 3hf
� Dia. Of the Nozzle for transmitting maximum power
2A fL
a D=
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� A turbine operating under a total fall of 36 m issupplied water through 90 m long, 0.2 m dia. pipe. Ifthe flow rate is such that 9.5 m of head is lost infriction in the pipe, what will be the power delivered tothe turbine? Take f = 0.022.
� An interconnected system of pipes is called a pipe network
� Water distribution systems for municipalities
� Multiple sources and multiple sinks connected with aninterconnected network of pipes.
� The main problem in a pipe flow network is to determinethe distribution of flow through the various pipes of thenetwork such that all conditions of flow are satisfied andall the circuits are then balanced
� Computer solutions!
� KYpipes
� WaterCAD
� CyberNET
� EPANET
� Pipe network divided into loops with nodes� The conditions to be satisfied in any network of pipes:
� At each node, continuity may be applied:
� Around any loop, the sum of head losses must be zero:
� The relationship between head loss and discharge must be maintained for each pipe � Darcy-Weisbach equation
� [n value ranges from 1.72 to 2.00]
∑=
=
n
iiq
1
0
∑=
=
m
ifih
1
0
n
fh rQ=
� Solution techniques� Hardy Cross loop-balancing (optimizes correction)
� Use a numeric solver (Solver in Excel) to find a change in flow that will give zero head loss around the loop
� Use Network Analysis software (EPANET)
� In this method:� A most suitable flow is assumed in such a way as to
satisfy continuity at each node� With the assumed values of Q, compute the head losses
for each pipe� Consider different loops and compute the net head loss
around each circuit� Head loss clockwise round a loop are positive andanticlockwise are negative
� If assumed distribution of flow is correct, then� If , then the assumed flows are corrected by
introducing a correction factor ∆Q, till the loop isbalanced.
∑ = 0fih
0fi
h ≠∑
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� In the above eq., the denominator is a sum of absolute terms, so has no sign.
� If head losses in the CW direction are more than the CCW, ∆Q will be negative.� Subtracted from flow in CW direction� Added to flow in CCW direction
� If head losses in the CW direction are less than the CCW, ∆Q will be positive.� Subtracted from flow in CCW direction� Added to flow in CW direction
� For pipes common to two loops, corrections from both the loops are to be applied
� With the applied corrections to the flows in all pipes, a 2nd trial is made for all loops and the process repeated until the corrections become negligible
0
1
0
n
n
rQQ
rnQ −∆ = −
∑∑
� For the network shown below, the head loss is given byhf = rQ2. The values of r for each pipe, and thedischarge into or out of various nodes are shown in thesketch. The discharges are in an arbitrary unit. Obtainthe distribution of discharge in the network.
B C
A Dr = 7
r = 8
r = 4
r = 6r = 6
20 15
45 40
1st trial
LOOP ABC
Line r Q rQ2 2rQ
AB 6 15 -1350 180
CB 4 30 3600 240
CA 8 5 -200 80
∑ = 2050 500
Correction for loop ABC
∆Q = -4.1≈ - 4
LOOP ADC
Line r Q rQ2 2rQ
CD 6 5 -150 60
AD 7 10 700 140
CA 8 5 200 80
∑ = 750 280
Correction for loop ABC
∆Q = -2.67857≈ - 3
20
45 40
A D10
5
30
515
15
B C
20
45 40
A D7
9-3 = 6
26
819
15
B C
2nd trial
LOOP ABC
Line r Q rQ2 2rQ
AB 6 19 -2166 228
CB 4 26 2704 208
CA 8 6 -288 96
∑ = 250 532
Correction for loop ABC
∆Q = -0.46992≈ - 0.5
LOOP ADC
Line r Q rQ2 2rQ
CD 6 8 -384 96
AD 7 7 343 98
CA 8 6 288 96
∑ = 247 290
Correction for loop ABC
∆Q = -0.85172≈ - 0.9
20
45 40
A D7
9-3 = 6
26
819
15
B C
20
45 40
A D6.1
6.5-0.9 = 5.6
25.5
8.919.5
15
B C
3rd trial
LOOP ABC
Line r Q rQ2 2rQ
AB 6 19.5 -2281.5 234
CB 4 25.5 2601 204
CA 8 5.6 -250.88 89.6
∑ = 68.62 527.6
Correction for loop ABC
∆Q = -0.13006≈ - 0.13
LOOP ADC
Line r Q rQ2 2rQ
CD 6 8.9 -475.26 106.8
AD 7 6.1 260.47 85.4
CA 8 5.6 250.88 89.6
∑ = 36.09 281.8
Correction for loop ABC
∆Q = -0.12807≈ - 0.13
20
45 40
A D6.1
6.5-0.9 = 5.6
25.5
8.919.5
15
B C
20
45 40
A D5.97
5.73-0.13 = 5.6
25.37
9.0319.63
15
B C
