cve 372 hydromechanics - 2. flow in closed …users.metu.edu.tr/bertug/cve372/cve 372 hydromechanics...

CVE 372 Hydromechanics 1/47

Dr. Bertuğ AkıntuğDepartment of Civil EngineeringMiddle East Technical University

Northern Cyprus Campus

CVE 372CVE 372HYDROMECHANICSHYDROMECHANICS

FLOW IN CLOSED CONDUITS FLOW IN CLOSED CONDUITS IVIV


2. FLOW IN CLOSED CONDUITS2. FLOW IN CLOSED CONDUITS

Overview

Pipeline SystemsPipes in SeriesPipes in ParallelBranching PipesNetwork Solutions: Hardy-Cross MethodHydraulics and Operation of Pumped Discharge Lines



Pipeline Systems

2.5.1 Pipes in Series

Friction factors will be different for each pipe since Re and ε/D will be different.More trial and error will be required in the solutions.

321 QQQ ==

31 2 LLLL hhhHBA

++=−



Pipeline Systems

2.5.1 Pipes in Series - Equivalent Pipes:CBA QQQ ==

A B C

555

2

2

52

2

52

2

5

888

C

CC

B

BB

A

AA

C

CC

B

BB

A

AA

LLL

Dlf

Dlf

Dlf

gQ

Dlf

gQ

Dlf

gQ

Dlf

hhhCBA

=+

=+

=+

πππ

fC: equivalent friction factorlC: equivalent pipe length, and

DC: equivalent pipe diameter

If fA=fB=fC, then 555C

C

B

B

A

A

Dl

Dl

Dl

=+

51

5i

in

ieq

eq

Dl

Dl

∑=

=

For n pipes:



Overview




Pipeline Systems

2.5.2 Pipes in Parallel

321 QQQQ ++=

321 LLL hhh ==



Pipeline Systems

2.5.2 Pipes in Parallel - Equivalent Pipes

2/1522/152

8 and

8

=

=

BB

BLB

AA

ALA lf

DghQ

lfDgh

Q BAππ

BA QQQ +=

BA LLL hhh ==−21

QA

QB

QC

(2)(1)

2/152/152/15

+

=

BB

B

AA

A

CC

C

lfD

lfD

lfD



Pipe Systems

2.5.2 Pipes in Parallel - Equivalent Pipes

2/152/152/15

+

=

B

B

A

A

C

C

lD

lD

lD

BA QQQ +=

BA LLL hhh ==−21

QA

QB

QC

(2)(1)

If fA=fB=fC, then For n pipes:

∑=

=

n

i i

i

eq

eq

lD

lD

1

2/152/15



Overview




Pipe Systems

2.5.3 Branching Pipes

321 QQQ +=

2122

22

LLBB

BAA

A hhgVpz

gVpz ++++=++

γγ

3122

22

LLBB

BAA

A hhgVpz

gVpz ++++=++

γγ

A fluid particle traveling throughpipe (1) and (2).

A fluid particle traveling throughpipe (1) and (3).

32 LL hh =



Pipeline Systems

ExampleGiven pipe system with pipes AC and CB in series and AD and DB in series. Assume f=0.03 in all pipes and the flowrate Q=50 lt/s, choose an equivalent pipe and determine the head loss between A and B.

A

B

C

D

L=100 mD=0.1 m

L=200 mD=0.2 m

L=100 mD=0.15 m

L=200 mD=0.15 m

Q

Q



Pipeline Systems

ExampleTwo reservoirs are connected by a pipe of 0.2 m diameter with ε=0.03 mm and 5000 m length. The head difference between reservoirs is 40 m.a) Taking friction, entry and exit losses into account, determine the steady discharge between them.

b) If the discharge is to be increased to 50 lt/s without increase in gross head, determine the length of the secondary parallel pipe of the same characteristics. Ignore minor losses.

Kentrance=0.4H=40 m L=5000 m

D=0.2 mε=0.00003 mν=1 x 10-6 m2/s

Kexit=1.0

ExampleTwo reservoirs are connected by a pipe of 0.2 m diameter with ε=0.03 mm and 5000 m length. The head difference between reservoirs is 40 m.a) Taking friction, entry and exit losses into account, determine the steady discharge between them.

b) If the discharge is to be increased to 50 lt/s without increase in gross head, determine the length of the secondary parallel pipe of the same characteristics. Ignore minor losses.



Pipeline Systems

2.5.3 Branching PipeshL1

Hj

for Q1=Q2+Q3HA < Hj < HB

hL2

hL3

E.G.L



Pipeline Systems

2.5.3 Branching PipeshL1

Hj

for Q1 + Q2 = Q3HB < Hj < HC

hL2

hL3

E.G.L



Pipeline Systems

Example Find the water discharges for the junction problem shown in the figure.

1

2

3

327 m

281 m

19 m

0.030.641030.030.4291420.030.538531ε (cm)D (m)L (m)Pipe

ν = 1 x 10-6 m2/s



Overview




Pipeline Systems

2.5.4 Network Solutions: Hardy-Cross MethodThe ultimate in multiple pipe systems is a network of pipes.Network like these often occur in city water distribution systems.They have multiple inlets and outlets.The direction of flow may vary in time.



Pipeline Systems

2.5.4 Network Solutions: Hardy-Cross Method

Conservation of MassΣ Flow INJUNCTION = Σ Flow OUTJUNCTION

Energy ConservationAlgebraic sum of head losses around each and every loop must be zero“-Q” implies “-hL”.

04321=+−+=∑ LLLLL hhhhh

1 2

A B C

D E F

LOOP (clockwise)

A, B, …, F Junctions (nodes)

AB, BC, … Link (branch)

+

1

2

3

4



Pipeline Systems

2.5.4 Network Solutions: Hardy-Cross MethodHead Loss: (Darcy-Weisbach Equation)

link,link,link, mfL

mfL

hhhhhh

+=

+=

∑∑ +=

==

==

link 42link 52link

242

2

252

2

88

82

82

DgK

DgflK

QDgK

gVKh

QDgfl

gV

Dlfh

m

mmm

f

ππ

π

π

linklinklinklink, QQKhL =



Pipeline Systems

2.5.4 Network Solutions: Hardy-Cross MethodExample: Given is the network shown in figure below.Find the discharges in each pipe.

20 m3/s 50 m3/s

100 m3/s 30 m3/s

A B

D C

K=1

K=2

K=5

K=3K=6 K’s in s2/m5



Pipeline Systems

2.5.4 Network Solutions: Hardy-Cross MethodExample-Solution:Step-1: Initial Guess. The continuity equation must be satisfied at each and every node.

20 m3/s 50 m3/s

100 m3/s 30 m3/s

A B

D C

K=1

K=2

K=5

K=3K=6 K’s in s2/m5

70 m3/s

15 m3/s

35 m3/s

30 m3/s

35 m3/s

Loop1

Loop2



Pipeline Systems

2.5.4 Network Solutions: Hardy-Cross MethodExample-Solution:Step-2: Determination of ∆Q for each loop.

The conservation of energy condition must be satisfied for each and every loop.

First IterationLoop1:

Σ = 1350Σ = 285752x5x30=3005x-30x30=-45005-30CD2x3x35=2103x35x35=3675335AC2x6x70=8406x70x70=29400670AD2 K |Q|K Q |Q|KQ (m3/s)Pipe

∆Q= - 28575 / 1350 = -21.17

∆Q= - KQ|Q| / 2K|Q|



Pipeline Systems


The conservation of energy condition must be satisfied for each and every loop.

First IterationLoop2:

Σ = 253Σ = -27992x3x13.83=833x-13.83x13.83=-5743-13.83AC2x2x35=1402x-35x35=-24502-35BC2x1x15=3001X15X15=225115AB2 K |Q|K Q |Q|KQ (m3/s)Pipe

∆Q= - (-2799 / 253) = 11.06QAC=35-21.17=13.83



Pipeline Systems


20 m3/s 50 m3/s

100 m3/s 30 m3/s

A B

D C

K=1

K=2

K=5K=3

K=6

K’s in s2/m5

70-21.17=48.83

15+11.06=26.06

35-11.06=23.94

30+21.17=51.17

35-21.17-11.06=2.77

Loop1

Loop2



Pipeline Systems

2.5.4 Network Solutions: Hardy-Cross MethodExample-Solution:Step-2: Determination of ∆Q for each loop. Second IterationLoop1:

Σ = 1114Σ = 12412x5x51.17=5115x-51.17x51.17=-130905-51.17CD2x3x2.77=173x2.77x2.77=2332.77AC2x6x48.83=5866x48.83x48.83=14308648.83AD2 K |Q|K Q |Q|KQ (m3/s)Pipe

∆Q= -1241 / 1114 = -1.114



Pipeline Systems

2.5.4 Network Solutions: Hardy-Cross MethodExample-Solution:Step-2: Determination of ∆Q for each loop. Second IterationLoop2:

Σ = 158Σ = -4752x3x1.656=103x-1.656x1.656=-83-1.656AC2x2x23.94=962x-23.94x23.94=-11462-23.94BC2x1x26.06=521x26.06x26.06=679126.06AB2 K |Q|K Q |Q|KQ (m3/s)Pipe

∆Q= - 475 / 158 = 3.006QAC=2.77-1.114=-1.656



Pipeline Systems


20 m3/s 50 m3/s

100 m3/s 30 m3/s

A B

D C

K=1

K=2

K=5K=3

K=6

K’s in s2/m5

48.83-1.114=47.716

26.06+3.006=29.066

23.94-3.006=20.934

51.17+1.114=52.284

1.656-3.006=-1.45

Loop1

Loop2



Pipeline Systems


20 m3/s 50 m3/s

100 m3/s 30 m3/s

A B

D C

K=1

K=2

K=5K=3

K=6

K’s in s2/m5

48.83-1.114=47.716

26.06+3.006=29.066

23.94-3.006=20.934

51.17+1.114=52.284

1.45

Loop1

Loop2

Since ∆Q1 and ∆Q2 are not small, at least one more iteration is needed.



Overview




Pipeline Systems

2.5.5 Hydraulics and Operation of Pumped Discharged LinesA pump is a mechanical device which adds energy to the fluid.Pp: The power required to derive the motor is known as input power.Pf: The power gained by fluid.

wherePf: Power transmitted to fluid in Watts (N.m/s). Commonly used pump

power unit is hoursepower (hp). 1 hp = 745.7 Wattsγ: Specific weight of the fluid (N/m3)Q: Discharge (m3/s)Hp: Head supplied by pump (m)

pf HQP γ=



Pipeline Systems

2.5.5 Hydraulics and Operation of Pumped Discharged LinesIf there were no losses Pp=PfHowever there are losses:

Mechanical losses in bearing and sealsLosses due to the leakage of fluidOther frictional and minor losses.The efficiency (ηp)

P

p

P

fp P

HQPP γ

η ==p

pP

HQP

ηγ

=



Pipeline Systems

2.5.5 Hydraulics and Operation of Pumped Discharged Lines

21 HhHH Lp =−+ ∑

p

pP

HQP

ηγ

=

E.G.L

Hp

(1)

(2)



Pipeline Systems


gVPzhH

gVPz

HhHH

Lp

Lp

22

222

2

211

1

21

++=−+++

=−+

∑

∑

γγ

p

pP

HQP

ηγ

= HP

(2)

(1)



Pipeline Systems


Pumps in Parallel:

∑∑=P

pp P

QHγη



Pipeline Systems


Pumps in Series:

∑∑=

P

pp P

HQγη



Pipeline Systems


Example:The following pumped discharge pipelines are given;a) Determine the discharge when only one pump is operating (as shown

in Figure 1) and compute the power consumption.b) Determine the discharge when both pumps are operating in series (as

shown in Figure 2) and compute the power consumption.c) Determine the discharge when both pumps are operating in parallel

(as shown in Figure 3) and compute the power consumption.



Pipeline Systems

2.5.5 Hydraulics and Operation of Pumped Discharged LinesExample:



Pipeline Systems

2.5.5 Hydraulics and Operation of Pumped Discharged LinesExample: Solution



Pipeline Systems




Pipeline Systems


Example: Water is pumped from reservoir A to reservoirs C and D. The system and pump characteristics are given below. Determine:a) The discharge in each pipe,b) The head added to the system by pump,c) The head loss in each pipe for the computed discharges, d) Complete the HGL, e) Calculate the power of the pump for η=0.80.



Pipeline Systems


2. FLOW IN CLOSED CONDUITS2. FLOW IN CLOSED CONDUITS1

2

3

4

5

cve 372 hydromechanics - 2. flow in closed …users.metu.edu.tr/bertug/cve372/cve 372 hydromechanics...

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