8/18/2012

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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