MIDTERM EXAM

31 MARCH 2009

15:00

ANFI G

BRANCHING PIPES

Junction Problems

Junction Problems

Q1+Q2+Q3=0

hj=zj+pj/γ

hl,i= hf,i + hm,i=zi-hj

(flows towards the junction are considered to be positive!)

hl,i= hf,i + hm,i=zi-hj

j j

Identify the problem• Given

�Reservoir elevations, z1,z2, z3

�Pipe characteristics: L1,L2,L3,D1,D2,D3,ε1, ε2 ,ε3

• Determine

�The flow rate Q1,Q2, Q3

�and direction on each pipe

�Junction head: hj (piozometer head)

Solution Procedure�Assume an appropriate direction for the flow in each pipe

�Assume a junction head

If friction factor is unknownMake best assumption for the initial value of the friction factor (fully turbulent rough flow)

�Write energy equation from each reservoir to the junctionCalculate the flow rate for each pipe by using energy equation

�Put into continuity equation the determined discharge values

�Check that the continuity equation is satisfied or not

�If not change the junction head

Example

• Assume Q1 and Q2 “+” and Q3 “-“

• Neglect all minor losses

Energy conservation b/w A&J: hA= hJ+h l,AJ= hJ+hf,AJ Energy conservation b/w B&J: hB= hJ+h l,BJ= hJ+hf,BJ Energy conservation b/w J&C: hJ= hC+h l,JC= hC+hf,JC

Assume hJ and check if Q1 + Q2 + Q3 = 0.

If not, iterate by assuming new hJ until ∑ Qi = 0 checks.

Assume

hJ = 100m (elevation+50 m of pressure head)

hf,AJ = 51

2

21

gD

fLQ8

π=9.92 f1 Q1

2

hence b/w A and J 160 - 100 = 60 = 9.92 f1 Q12

b/w B and J 140 - 100 = 40 = 0.22 f2 Q22

b/w J and C 100 - 0 = 100 = 9.92 f3 Q32

* Assume hydraulically rough flow, f=f(ε/D)

h f,BJ = 0.22 f2 Q22

h f,CJ = 9.92 f3 Q32

ε/D fi Q i

0.001 0.02 17.39

0.0005 0.018 100.50 0.001 0.02 22.57

17.39 + 100.50 >> 22.57 Therefore increase hJ

hJ = 130m

12.30 + 50.25 > 25.29Therefore increase hJ

fi Qi

0.02 12.30

0.018 50.25

0.02 25.59

hJ = 139m

fi Qi

0.02 10.29

0.018 15.89

0.02 26.47

10.29+ 15.89 = 26.18 ≅ 26.47

hJ = 100m Concrete pipe ε = 0.0001 Q1+Q2 =Q3

NETWORK

OF PIPES

NETWORK OF PIPES

LOOP “+”clockwise

A,...,I:JUNCTIONS/ NODES

EF:LINK/BRANCH

1 2

34

A

B C

E

GH

I

C

F

Conservation of Mass: Flow into each junction

must be equal to the flow out of the junction

Energy Conservation: Algebraic sum of head

losses around each and every loop must be zero.

Head Loss Equations

linklinklinkl

2

link

2

m

2

f2

2

42

2

5

mf

QQKh

QKQKQKg

Q8

Dg

Q8

D

Lfh

hhh

linklink

=

=+=π

+π

=

+=

l

l

Km

Conservation of Energy around any loop

∑

∑−=∆

=

=∆+⇒=

+∆+=

∆+==

∑∑

∑∑

∑∑∑

=

−

=

=

−

=

==

=

=

N

i

o

N

i

oo

QK2

QKQ

N

1i

1no

N

1i

noloop

N

1i

1n

o

N

1i

n

o

n

oi

N

1İ

n

ii

N

1iloopl

nkelementsnumberofliN

1i

Q

2nIf

oQQnKKQ0h

QQnKKQ

)QQ(KQKh

KK

n=2 for Darcy-Weisbach

Solution Procedure

1) Assume flow distribution

Condition: Flow distribution should satisfy conservation of mass at each node/junction

Notation: Qo

2) Calculate ∆∆∆∆Q for each loop

3) Check ∆∆∆∆Q

small magnitude

YES: stop

NO: Step 4

4) Correct Q values

Qo,new=Qo,old+∆Qloop

Example

Given is the network shown in figure below. Find the discharges in each

and every link.

Initial Guess

K Q |Q| 2 K |Q|

AD 6*70*70=29400 2*6*70=840

AC 3*35*35 =3675 2*3*35=210

DC *-5*30*30 =-4500 2*5*30=300

Σ=28575Σ=28575Σ=28575Σ=28575 Σ=1350Σ=1350Σ=1350Σ=1350

17.211350

28575Q1 −=−=∆ AC 35-21.17

LOOP 1

Continuity Flow into each junction

must be equal flow out of the junction

K Q |Q| 2 K |Q|

AB 1*15*15 =225 2*1*15=300

BC 2*-35*35=-2450 2*2*35=140

AC 3*-13.83*13.83 =-574 2*3*13.83=83

Σ=Σ=Σ=Σ=−−−−2799279927992799 Σ=253Σ=253Σ=253Σ=253

LOOP 2

Initial Guess

17.211350

28575Q1 −=−=∆

AC 35-21.17

06.11253

27992Q ==∆

LOOP 1

LOOP 2

Flow Distribution After First Iteration

Second iteration

LOOP 1

K Q |Q| 2 K |Q|

AD 6*48.83*48.83 =14308 2*6*48.83 =586

AC 3*2.77*2.77 =23 2*3*2.77 =17

CD 5*-51.17*51.17 =-13090 2*5*51.17 =511

Σ=1241Σ=1241Σ=1241Σ=1241 Σ=1114Σ=1114Σ=1114Σ=1114

11411114

12411 .Q −−−−====−−−−====∆∆∆∆

LOOP 2

K Q |Q| 2 K |Q|

AB 1*26.06*26.06 =679 2*1*26.06=52

BC 2*-23.94*23.94=-1146 2*2*23.94=96

AC 3*-1.656*1.656 =-8 2*3*1.656=10

Σ=Σ=Σ=Σ=−−−−475475475475 Σ=158Σ=158Σ=158Σ=158

AC 2.77-1.114 0063158

4752 .Q ========∆∆∆∆

LOOP 1

11411114

12411 .Q −−−−====−−−−====∆∆∆∆

LOOP 2

AC 2.77-1.114=1.656

0063158

4752 .Q ========∆∆∆∆

Attention: One more iteration is needed

Final Distribution of Second Iteration

HYDRAULICS AND OPERATION OF

PUMPED DISCHARGE LINES

• Pump: adds energy to a

fluid, resulting in an increase in pressure across the pump.

• Turbine: extracts energy from the fluid, resulting in

a decrease in pressure across the turbine.

Turbine

Types of Pumps

(a) Radial flow pump(b) Axial-flow pump

Centrifugal Pumps

The power gained ⇒ Pf = γγγγ Q Hp

Hp Head supply by pump (m)Q Discharge (m3/s)Pf in Watts (Nm/s)

(in practice hp –horsepower- is used)

1hp =745.7 Watts

7.745

QH P p

f

γ=

A PUMP IS A MECHANICAL DEVICE

WHICH ADDS ENERGY TO THE FLUID

TRANSFER

into fluid

ENERGY

ENERGY

(Supply)MOTOR

S

H

A

F

T

BLADES

ROTOR

Pf = γ Q Hp

•Axial

•Radial

•Mixed

There is a casing which provides passageway to the

fluid.

The power

required to

derive the motor is bhp=ωT

bhp: brakehorsepower

ω:shaft angular velocity

T: shaft torque

Static Suction Lift

Suction lift- Static head when

the pump is located above

the suction tank

Static Suction Head

Static head when the pump is located below the suction tank

Pump Head• Net Head

• Water horsepower

• Brake horsepower

• Pump efficiency

Efficiency

• If there were no losses : bhp = Pf

Yet, there are losses:

• mechanical losses in bearings and seals

• losses due to leakage of fluid

• other frictional and minor losses

p

p

p

ff

P

H Q

P

P

bhp

P γ===η

Pump Power and efficiency

η

γ=

p

p

H Q P

where η is the efficiency.

So η must be maximized over a wide range of Q.

Pump Performance

Q

Hp

η

Pp

Qr

shut off head

rated capacity

(normal/design lowrate)

ηηηη

HpPp

Q (lt/s) 0 10 20 30 40 50 60

H (m) 70 67 62.5 57.5 51 43 32

ηηηη (%) 45 63 75 82 79 71

SELECTION OF A PUMP

Valve

Pump

SELECTION OF A PUMP

• Pump Hp vs Q must be known !

• System hydraulics must be known !

Consider a system with two reservoirs; water is supplied to the reservoir with higher elevation by means of a pumped line !

Hs

Total StaticHead

2

1

Static

Suction

HeadP

∆z

±

0

valve

H1= H2 – Hp + hf2

p

2

222

2

111 KQ H-

g2

V

Pz

g2

V

Pz ++

γ+=+

γ+

Hp= ∆z + KQ2

Hp= Hs + KQ2

Pumps in Parallel • Head across each pipe (pump) is identical

•The total discharge through the pumping system is equal to the sum of individual discharges through each pump.

Pumps in Parallel

∑∑γ

=P

D

pP

QHη

Head across each pipe (pump) is identicalThe total discharge through the pumping system is equal tothe sum of individual discharges through each pump.

A BH

Pump B

SystemDemand

Q

Pump A

Pump in Series

Construction of Pump Curve

A BH

Pump B

System Demand

Pumps A and B combined

Qi

Hi,A

Hi,B

Pump A

Hi(A+B)=Hi,A+Hi,B

For each discharge the head of the pump A and B are added each other.

Then A new curve is constructed

( )

∑∑

=P

DP

P P

QHγη

Discharge through each pump is identical!!!

A BH

Pump B

System Demand

Pumps A and B combined

QD=QA=Q

B

Pump A

Pump in Series

Pipeline CharacteristicsL=2000 m; D=0.2 m; εεεε=0.0002m; νννν=1* 10-6m2/s

Determine the discharge and compute the power consumption.

a) when only one pump is operating

b) when both pumps are operating in series

c) when both pumps are operating in parallel

EXAMPLE :

60 m

100 m

P

Fig. 1

100 m

60 m

P P

Fig. 2

100 m

60 m

P

P

Fig. 3

Q (lt/s) 0 10 20 30 40 50 60

H (m) 70 67 62.5 57.5 51 43 32

ηηηη (%) 45 63 75 82 79 71

Pump Charecteristics

Single Pump

L (m) 2000

D (m) 0.2

εεεε (m) 0.0002

υυυυ (m2/s) 0.000001

Q Re f hl system

0 0 0 0.00 40.00

0.01 63662 0.0234 1.21 41.21

0.02 127324 0.0219 4.52 44.52

0.03 190986 0.0212 9.88 49.88

0.04 254648 0.0209 17.28 57.28

0.05 318310 0.0207 26.72 66.72

0.06 381972 0.0205 38.19 78.19

Case 1: Single Pump

0102030405060708090

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07Q(m

3/s)

H(m

)

system Single Pump eff.

Single Pump

H Q

70 0

67 0.01

62.5 0.02

57.5 0.03

51 0.04

43 0.05

32 0.06

νπ=

ν D

Q4VD = Re g2

V

D

Lfh

2

f =

Case 1: Single Pump

0102030405060708090

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07Q(m

3/s)

H(m

)

system Single Pump eff.

SOLUTIONQ=35 lt/s

ηηηη=80%(for Q=35 llll/s)

H=54m

P=23.2kW

∑∑γ

=P

D

pP

QHη

H(m) Q(m3/s)

140 0

134 0.01

125 0.02

115 0.03

102 0.04

86 0.05

64 0.06

Case 2: Pump in Series

0102030405060708090

100110120130140150

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07Q(m

3/s)

H(m

)

system Single Pump eff. p. in s.

SOLUTIONQ=56 llll/s

ηηηη=75%(for Q=56 llll/s)

∑H=72m∑P=52.8kW

Single Pump

H Q

70 0

67 0.01

62.5 0.02

57.5 0.03

51 0.04

43 0.05

32 0.06

Pump in Parallel H Q

70 0

67 0.02

62.5 0.04

57.5 0.06

51 0.08

43 0.1

32 0.12

Case 3: Pump in Parallel

0102030405060708090

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14Q(m

3/s)

H(m

)

system Single Pump eff. p.i.p

SOLUTIONQ=46 llll/s

ηηηη=68%(for Q=23llll/s)

∑H=61m∑P=40.5kW

Single Pump

H Q

70 0

67 0.01

62.5 0.02

57.5 0.03

51 0.04

43 0.05

32 0.06

summary

single series paralel

Q (lt/s) 35 56 46 (2*23)

η (%) 80 75 68

H (m) 54 72 61

P (kW) 23.2 52.7 40.5 (2*20.24)

P/Q (kW.s/lt) 0.66 0.94 0.88

GRAVITY PIPELINES

The pipelines through which flow is maintained by the action of gravity are known as gravity pipelines.

hvalve

control valve

0.5 m/s < V < 2 m/s

(p/γγγγ) > 5m

A control valve might be used to control the discharge and pressure

Energy Equation between points (A) & (B) is:

HA=HB+hL, or in open form:

hg2

VPhz

g2

VPhz

L

2

BB

BB

2

AA

resA++

γ++=+

γ++

Qin

hmax

hres zA

spillway

±0

A

B;valve

Because of limits set forth on velocity and pressure head there are lower and upper bounds for the discharge through a gravity pipeline. Consider the following schematic representation of a gravity pipeline system shown below

PA=PB=0, VA=0, and zB=0.

the maximum value of V=2 m/s.

neglected becan �ˆ m 2.081.9x2

2¡Ü

g2

V 22

B =

2f52

2252f

KQ=h �ËgDπ

Lf8=K where

,KQ=QgDπ

Lf8=h

hg2

VPhz

g2

VPhz

L

2

BB

BB

2

AA

resA++

γ++=+

γ++

Qin

hma

xhres zA

spillway

±0

A

B;valve

( )[ ]

)neglected! are heads velocity m/s 2V since that (note

K/zhQ �ËKQhzh2/1

Aresout

2

outfAres

<

+===+

Therefore, Energy Equation becomes:

A relationship between Q and the reservoir level can be obtained as:

K

zhQ Ares

+=

Qin

hma

xhres zA

spillway

±0

A

B;valve

( )

K

z+h=Q

constants areK and z since Q=Q h=h

Amaxmax

maxmaxA

For a full reservoir hres= hmax Q=Qmax

maximum discharge that may occur = in the pipeline system is called

The system capacity

For empty reservoir hres= 0 Q = Qmin= Q0

Q0 is the minimum flow rate, which will pressurize the pipe, in other words, it is the minimum flow rate for a full pipe flow.

K

z=Q Therefore A

0

Qin

hmax

hres zA

spill

±0

A

Valve

Q=Qmax

Qin< Q< Qmax

Q=Q0

A

B

• If Qmax< Qin h=hmax; Qspill=Qin-Qmax

• If Qmin< Qin <Qmax 0< hres<hmax; Qspill=0

• If Qin< Qmin to prevent free flow the valve is partially closed

The free surface flow may be prevented by use of a valve at the pipe exit.

Valves control the flow rate by providing a mean to adjust the over all system loss coefficient to the desired value.

Valve and free surface

• Consider the same system with a valve at the

end of the pipe:

A

B

Datum zA

hA

Energy Equation between points (A) & (B):

HA=HB+hf+hV , or in open form:

hhg2

VPhz

g2

VPhz

Vf

2

BB

BB

2

AA

resA+++

γ++=+

γ++

PA=PB=0, VA=VB= 0, and zB=0.

Also hf=KQ2 & the local loss can also be written as hV= KVQ2

Therefore:

zA+hA=(K+KV)Q2, solving for Q:

V

resA

KK

hzQ

+

+=

A

BDatum z

A

hA

Valve Conditions

• when valve is closed KV →∝∝∝∝ Therefore Q=0.

• Openining of the valve reduces the value of KV, producing a desired flow rate.

V

resA

KK

hzQ

+

+=

• The head loss in valves is mainly a result of dissipation of kinetic energy of a high speed portion of flow.

• when Qin<Qmin , the control valve at the pipe exit will pressurize the flow in the pipe by dissipating the excess kinetic energy, hV, and in turn increasing the water level in the reservoir.

• The magnitude of the valve loss depends on how much one needs to presurize the flow, that is the water depth in the reservoir.

• If the water level is to be kept constant to obtain a certain discharge, a control valve can be chosen accordingly.

Example Consider the reservoir-pipe system given below, with following values.

Reservoir depth hmax=5m, zA=5m, L=2000m, D=0.8m, f=0.02. Determine:

a) The system capacity, Qmax.b) Minimum flowrate, Q0.c) Spill flowrate, if Qin =1.2m3/s.

d) Valve loss, hv, if Qin =0.5m3/s.

hA

ZA

hmax

A

Q

B

Q=Q0

Q0<Q<Qmax

Q=Qmax

valve

Datum

SolutionIf minor losses except hv (valve loss) are neglected,

2/1

Aresout

K

zhQ

+= 09.10

Dg

fL8K

52=

π=

s/m 0.109.10

55Q 3

2/1

max ≅

+=

s/m 70.009.10

5Q 3

2/1

min ≅

=

m 48.2h

)5.0.(09.105h

KQzhKQhzh

V

2V

2inAV

2inVAres

≥

−≥

−≥⇒+=+

If inflow is 1.2 m3 / s ======� Qspill = 0.20 m3 / s

Selection of Pipe Diameter: Cost

• In design of gravity pipelines, an optimum diameter is selected that minimizes the capital cost (The estimated total cost) and operation cost.

Cost=Capital+Operation cost

Selection of Pipe Diameter: Velocity

• Depending on the type of pipe material a lower and an upper limits are set for the velocity as:

0.5 m/s < V < 2 m/s;

Selection of Pipe Diameter: Pressure

• To prevent air entrainment minimum pressure

head, (p/γγγγ)min permitted along the pipeline is 5m.

)m(80/P53 ≤γ≤−

)m(80/P3020 ≤γ≤−

• The Range of Pressure

Transmission line

City Network

Computation of Pipe Diameter For

Gravity Pipelines

A

B

C

D

E

F

S3

S1

S2

pipeline

A

B

C

D

E

F

S3

S1S2

pipeline

• Determine the control points (C,E,F) and their topographic elevations zc, zE, zF

• Add the minimum required pressure head to

these elevations (zc and ze).

• Determine the minimum energy grade line slope,

between the reservoir and control points

A

B

C

D

E

F

S3

S1S2

pipeline

Minimum Energy Slope

AC

mincA

1 L

)/pz(HS

γ+−=

321SSS <<

Pipe Diameter for A-C

• Compute the pipe diameter for line A-C using

Darcy-Weisbach equation and select the nearest

commercially available dameter

)Sg

fQ8(D

1

2

2

computedπ

=

Velocity Check

• Compute the Velocity

4/D

QV

2π=

• If Vmin< V < Vmax then the selected diameter is used in the project

• If V < Vmin a booster pump must be installed at the reservoir site to increse the velocity to Vmin

Booster Pump Head

• If booster pump is installed, the head supplied by

booster pump will be computed as

g2

V

D

Lf

pzHH

2

minmin

cPA+

γ+=+

Maximum Velocity Check

• If V > Vmax then the pipe diameter is increased to reduce the velocity

QV4

Dmax

2

=×π

Pressure Check

• If p/γγγγ < pmax then the selected pipe diameter is used

• If p/γγγγ > pmax A pressure reduction pump must be installled

Determination of diameter(C-D-E)

• Compute the piozemeter head at C

• Calculate the diameter for C-D-E-F segment

• Note that E is the control point

• Compute the diameter of C-D-E

• Compute the diameter of the segment E-F

Control valve

• Install a control valve at point F and determine the necassary headloss at the valve to maintin the pressurized flow at segment E-F.