midterm exam 31 march 2009 15:00 anfi gweb.iku.edu.tr/~asenturk/pipe iii.pdfsolution procedure...
TRANSCRIPT
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.