NETWORKS 6.1 Multi-reservoirs with Branched Pipe System One of the important topics that the practical engineers dealt with in real life, is the design and\or analysis of pipeline systems that are composed of three or more reservoirs at various elevations. This system is referred as branching pipes since one or more pipes are separated into two or more pipes (or combine into to a single one) and are not coming together again at the downstream. The point of connection of more than two pipes at a single point is called Junction and that’s why sometimes these problems are referred as Junction Problems as well.
Transcript
NETWORKS6.1 Multi-reservoirs with Branched Pipe System
One of the important topics that the practicalengineers dealt with in real life, is the design and\oranalysis of pipeline systems that are composed of threeor more reservoirs at various elevations. This system isreferred as branching pipes since one or more pipes areseparated into two or more pipes (or combine into to asingle one) and are not coming together again at thedownstream.The point of connection of more than two pipes at asingle point is called Junction and that’s whysometimes these problems are referred as JunctionProblems as well.
In fact, at the Junction ‘J’, the mass balanceequation (Continuity) should be satisfied,
ΣQIN = ΣQOUT .
In general, if all the pipes are sufficiently long
• the minor losses and
• the velocity head just at the junction may beneglected.
Note that, the energy head at the junction ‘HJ’ hasa single common value.
The methodology for the solution depends on theavailable information related with:
• the reservoirs elevations,
• the pipes diameters,
• The pipes lengths and
• the pipes types (characteristic) etc.
If the flow directions are not known (which is the generalcase), an energy level for the junction point ‘HJ’ isassigned and hence the directions of the flows within thepipes are defined based on this assumption.
Then, by trial and error, until the equation of continuitysatisfies, the assumed junction level ‘HJ’ is reevaluated(changed).
Note that, the basic equation, which is the continuityequation within the system, is the one thatcontrols the flows direction.
Hence, the actual directions of the flow depends on:
the reservoir (or tank) pressures and/or elevations,
the diameters, lengths, and types of the pipes etc.
For the solution of this type of problems, generally a trial-and-error procedure is unavoidable.
The best procedure, as informed earlier, is the assumptionof the energy head level to the junction ‘HJ’ and thencomputing the flow rate for each pipeline and checkingthe continuity equation based on these values.
If the continuity equation is satisfied, then, thecomputed flow values are correct.
(i.e. the total flow entering equals to the total
flow leaving at the junction).
If the continuity equation is not satisfied, adifferent energy head level value at the junctionshould be reassumed.
Usually a satisfactory solution is obtained afterseveral trials.
Note 1:NEVER HJ > HIGHEST RESERVOIR WATER SUFACE LEVEL.
Under normal conditions (if no pump is installed).
Note 2:NEVER HJ < LOWEST RESERVOIR WATER SUFACE LEVEL.
Under normal conditions.
Note 3:After the first iteration assume:- HIGHER than the previously assumed value,
if at the Junction: flow entering > flow leaving.
- LOWER than the previously assumed value,
if at the Junction: flow entering < flow leaving.
Consider the possible simplest system that is consisted of1 Junction & 3 Reservoirs where the estimation of thedischarge in each pipe sections has to be determined, asshown in Fig. 6.1.
Figure 6.1: Three-reservoirs system.
J
hL AJ
hL JB
hL JC
pD
For the solution of these branch systems, the necessarynumbers of equations have to be obtained through energyand continuity concepts.
In these type of problems, basically one of these 3different parameters are unknown at the junction:1. the total energy head at the junction (HJ) is unknown.2. the flow direction of the middle reservoirs connecting
pipe (JB) is unknown.3. the flow amount of the middle reservoirs connecting
pipe (JB) is unknown.
To determine the total energy head, the flow amount and thedirection within that pipe (JB), the solution procedure iscarried out through the energy equation on the basis of theoutcome of the iterations.
First Iteration:a) Since it is difficult to incorporate the velocity head at the
junction (J), this type of problems can be solved, byassuming this velocity head to be a small quantity relativeto the other energy terms and may be ignored.
Hence, letting the total energy head (piezometric head, sincethe velocity head is neglected) at the junction (J) be equal tothe total energy head (usually the water surface elevation) ofthe reservoir surface B (middle reservoir), hence no flowwould occur within the pipe JB.
b) By writing the energy equation between A-J and J-C separately, one can obtain the values of discharges QAJ and QJC.
where the head losses are defined by using the Darcy-Weisbach equation.
AD
LD
DA hP
zz γ
CB
LCD
D hzP
z γ
zA=HJ+hLAJ=
HJ ≈
JJ
JJ
J
AJ
AJ
2 22 2
5 4
1
2 2 12.1 12.1DC
DC avDC avDC DCL DC DC DC DC
DC DC DC
L v v Lh f k f Q k Q
D g g D D
2 22 2
5 4
1
2 2 12.1 12.1AD
avAD avADAD ADL AD AD AD AD
AD AD AD
v vL Lh f k f Q k Q
D g g D D
AJ
JC
AJAJ
AJ AJ AJ
AJ
AJ
AJ
AJAJ
AJ
JC
JC
JC
JC JC
JC
JC
JC
JCJC
JC
c) Using continuity equation, compare the values of
discharges QAJ and QJC and hence, estimate the
direction of the flow along pipe J-B.
One of the below 2 equations is only correct.
AD DC DBQ Q Q
AD DC DBQ Q Q
or
AJ JC JB
AJ JC JB
Second iteration:a) Since the direction of flow along J-C is determined,suggest more reasonable (possible) total energy ‘HJ’value for that junction.
i- if QAJ > QJC → QJB is an outflow (from the junction J to reservoir B),
hence to satisfy that select: HA > HJ > HB.
(a value between zA and zB)
ii- if QAJ < QJC → QJB is an inflow (from reservoir B to the junction J),
hence to satisfy that select: HC < HJ < HB. (a value between zB and zC).
b) By writing the energy equation between A-J, B-J (or J-B) and J-C separately, obtain the values of discharges QAJ,QJD (or QJB) and QJC using the head losses defined byDarcy-Weisbach equation.
c) To check the correctness of this assumed junctionenergy level HJ, yse the properly establish correctcontinuity equation.
JBJCAJ QQQ ?
JCJBAJ QQQ?
or
d) If the continuity does not satisfy, then re-assumeanother new value for the junction head relative to theprevious assumption and the result of the continuity.
e) Repeat the above procedures from ‘a to c’ until thecontinuity at procedure ‘d’ satisfies.
Note: The head loss along a single pipe is:
hLi= = KTQi2
KT hLi = KT Qi2
ii
ii
ii
ii
Δ
Δ
Note that for any pipe headloss is:
minor5 4
L 1K f k
12.1 D 12.1 DT
if the minor loss effects are ignored, then the equation issimplified as:
5
LK f
12.1 DT
Pminor5 4 2
HL 1K f k
12.1 D 12.1 D QT
general headloss of any pipe having a pump installed over is:
ΔhL= KTQ2
Iteration Table for Branched pipes (correction for Junctions total head HJ)
Pipe
Name
Length
L
(m)
Dia.
D
(cm)
ks
(cm)
ks/D
Qi
(m3/s)
𝑅𝑒𝑖
*105
𝑓𝑖
KTi
(s2
m5)
∆hi
(m)
Qi
=∆hi
KTi
(m3/s)
𝑄𝑖
2∆ℎ𝑖
(m2/s)
𝛿h =−∑Qi
∑ Qi
2∆hi
(m)
∆hi+1= ∆hi + δh
(m)
HD(i+1)
(m)
HD(i+1)
(m)
Qi+1 =∆hi+1
KTi
(m3/s)
Iteration No: Hj=
Assume at the junction:Q (inflow) : + Q (outflow) : –
If along any pipeline portion contains a pump, than the effect of this pump should be included while calculating that portions Δh value for each iteration.
Example 6.1:For the multiple reservoir system shown, the water level elevations and
other flow parameters are detailed within the table. If the discharge of the pipe BJ,QBJ, is 100 lt/s and flows towards J and the topographic elevation of the junction atJ is zJ= 70 m; neglecting all the minor losses and assuming that the water depthswithin the reservoirs (A, B and C) remain constant; (take γ=9.8 kN/m3).
i. determine the total energy at the junction J?ii. determine the flow directions and compute the discharges of the pipes AJ and JC?
iii. determine the water elevation in the reservoir C?iv. compute the pressure head at J?v. Draw the energy grade line (EGL) of the system.
Pipe L(m) D(mm) f
AJ 1000 300 0.02
JB 1500 300 0.02
JC 2000 400 0.02
J
HJ=89.79 mQBJ=100 lt/sQAJ=87.5 lt/sQJC=187.5 lt/szC=78.44 m
Example 6.2:For the given multiple reservoir system, determine
the direction and the amount of the discharges passing ineach pipe. Water level elevations are at the reservoirs arezA=326.7 m, zB=240.2 m and zC=38.5 m. The length,diameter and type of the pipes are as tabulated. Ignore theminor losses (take γ = 9.8 kN/m3).
Example T:Water at 200 C flows between three reservoirs (A, B and C). Ignore minor lossesDetermine the discharge and the flow directions in each pipe for:a) No pump along pipe 3b) 1-pump along pipe 3 that pumps water from reservoir C,c) 2-pumps in series along pipe 3 that pumps water from reservoir C.
Example T: SolutionWater at 200 C flows between three reservoirs (A, B and C). Ignore minor lossesDetermine the discharge and the flow directions in each pipe for:a) No pump along pipe 3,b) 1-pump along pipe 3 that pumps water from reservoir C,c) 2-pumps in series along pipe 3 that pumps water from reservoir C.
Example T:Water at 200 C flows between three reservoirs (A, B and C). Ignore minor losses.Determine the discharge and the flow directions in each pipe for: a) 1 pump along pipe 3 that pumps water to reservoir C,b) 2-pumps in series along pipe 3 that pumps water to reservoir C,c) 3-pumps in series along pipe 3 that pumps water to reservoir C,d) 4-pumps in series along pipe 3 that pumps water to reservoir C.
Example T: SolutionWater at 200 C flows between three reservoirs (A, B and C). Ignore minor losses.Determine the discharge and the flow directions in each pipe for: a) 1 pump along pipe 3 that pumps water to reservoir C,b) 2-pumps in series along pipe 3 that pumps water to reservoir C,c) 3-pumps in series along pipe 3 that pumps water to reservoir C,d) 4-pumps in series along pipe 3 that pumps water to reservoir C.
a) Not possible since Hj > 38.0 m does not satisfy Q1 = Q2 + Q3
b) 2-pumps in series Hj ≈ 27.33 m where Q1 = Q2 + Q3
Q1 = 0.01498 m3/s, Q2 = 0.00856 m3/s, Q3 = 0.00642 m3/sc) 3-pumps in series Hj ≈ 20.34 m where Q1 = Q2 + Q3
Q1 = 0.01605 m3/s, Q2 = 0.00169 m3/s, Q3 = 0.01436 m3/s d) 4-pumps in series Hj ≈ 16.89 m where Q1 + Q2 = Q3 (continuity equation changed!)
Pipe Length ‘L’ (m) Diameter ‘ϕ’ (mm) Total Minor Loss ‘Σk’ [-]
1 263.2 500 7.82
2 789.3 320 2.6
3 2106.8 300 3.8
4 1551.8 350 3.6
Below given reservoir-pipeline-pumping system plan, water @ 13 °C is discharging from pipe 4 to the atmosphere at
a rate of Q4=63.50 lt/s. The system is functioning under the absolute atmospheric pressure Patm=101.17 kPa (abs).
a) Determine the discharges flowing within other pipes? (20 p)
b) The topographic water surface elevation of reservoir C? (5 p)
c) The required power ‘P’ for this pumping system. (8 p)
d) What should be the maximum topographic elevation at junction B ‘zB’ for no cavitation. (7 p)
Single pump ‘Hp – Q’ characteristic is:
Q (lt/s) 0 10 20 30 40 50 53
Hp (m) 10.75 10.37 9.23 7.33 4.67 1.25 0
η (%) 57.8 71.5 83.6 75.2 51.8 38.9 0
Equation of a Single pump: Hp = 10.75 - 3800Q2 (where Q in [m3/s] and Hp in [m])
Previous Exam Question:
If in a water distribution system all the pipes coming outfrom one reservoir and get increased by branches wherenone of them ever get reconnected or interconnectedagain so as to form loops are called BRANCHED (OPEN)NETWORK.This distribution network mainly used for municipalitiessince requires less number of pipes but having lots ofweaknesses, hence not recommended nowadays.Nowadays, the most popular system in municipalities isthe LOOP (CLOSED) NETWORK where all the pipes areinterconnected with each other by forming junctions andloops. This system having more advantages than branchednetworks.
6.2 Interconnected Pipeline Systems
6.2 Interconnected Pipeline Systems
This type of water distribution network analysis, providesthe basis for the design of new systems and the extensionof the existing systems.
In fact, an extension of pipes in parallel is a casefrequently encountered in municipal distribution systems,in which the pipes are interconnected, so that the flow tothe given outlet may come by several different paths.
Indeed, it is frequently impossible to tell by inspection,which direction the flow occurs. Nevertheless, the flow inany network, however is complicated and must satisfy thebasic relations: the continuity and the energy equations.
Design criteria are those specified as minimum flow ratesand pressure heads that must be attained at the outletpoints of the network.
It should be remembered that, the flow and pressuredistribution across a network are affected by thearrangement and the size of the pipes and thedistribution of the outflows.
Since a change of diameter in one pipe length will affectthe flow and pressure distribution everywhere, networkdesign is not an explicit process.
Hence, designing any section between any twojunctions can not be done by isolating thatportion! The whole network should beconsidered.
For the analysis, the determination of the pipe flowrates and pressure heads which satisfies thecontinuity and energy conservation equationsshould be used.
In literature the common solution techniques for this typeof connections are:I. The Head Balance Method (‘LOOP METHOD’)
[HARDY-CROSS METHOD]II. The Quantity Balance Method (‘NODAL METHOD’)III. The Newton – Raphson MethodIV. The Linear Theory Method.
Closed Network Sistems
SOLUTIONOF
INTERCONNECTED PIPELINE SYSTEMBY
HARDY-CROSS METHOD
It is the earliest (1936) systematic method used for network analysis
Solution procedure of Hardy-Cross method:1. Check the continuity of the whole system,
2. Name and/or label each and every node and pipe within the system,
3. Determine and give different names for each loop and also assign+ve flow direction for each loop. For the sign convection selectingthe clockwise flow as positive is a usual procedure.
1. Assign an arbitrary direction and a discharge value for one of thepipes in each loop and hence distribute the flow at each node byconsidering the continuity equation. It is a good practice NOT toassume a zero flow to any pipe. Note that: for problems of 2 loopsonly 2 (1 for each loop) arbitrary discharge values and directionswill be assinged.
+ Q
5. For each loop, obtain the energy equation by considering thesuggested directions and discharges. It is definite that, for thecommon pipe of two loops, if the error obtained from the previousloop is applied after correction to the other loop, then the number ofiteration needed to reach to the required solution will be reduced!
6. For the second trial, use the previously calculated discharge values ineach pipe of that Loop,
7. Until the required acceptable limit is reached for the whole system,the number iterations should be continued.
To determine the required correction, thebasic assumption comes from the headbalance criterion where the algebraic sumof the losses around any closed loop shouldbe zero; Δh=0.
KT (i)
But if the minor loss effects are ignored, then theequation is simplified as:
D 12.1
LfK
5
i
iiT(i)
2
minor5 4 2
L 1f k Q
12.1 D 12.1 D
PH
Q
The head loss along a single pipe is:
hL=2
minor5 4
L 1f k Q
12.1 D 12.1 D
= KT(i)Q(i)
2(i)
Δ
KT(i)
The head loss along a single pipe with a pump installed is:
hL(i)=Δ
For the estimation of flow error ΔQ:
2
(i)(i)(i)
2
(i)T(i)
2
(i)(i)T(i)L(i) QQQ2QKQQKh
Neglecting ΔQ(i)2, by assuming ΔQ(i) is a small value,
then:
(i)(i)
2
(i)T(i)L(i) QQ2QKh
So within a close loop since ΣhL= 0 from the continuity,
Neglecting the minor losses in the pipes, for the given data, using Hardy-Cross Method, determine the flow
in each pipe. Note that for simplicity, let f=0.0193 for pipes constant, do for ΔQ < 0.5 lt/s.
Pipe AB BC CD BD DE AE
Length (m) 600 580 215 480 810 335
Diameter (mm) 250 150 200 220 100 150
Example 6.4:Neglecting the minor losses in the pipes, for the given data, using Hardy-Cross Method, determine the flow
in each pipe. Note that for simplicity, let f=0.0193 for pipes constant, do for ΔQ < 0.5 lt/s.
Pipe AB BC CD BD DE AE
Length (m) 600 580 215 480 810 335
Diameter (mm) 250 150 200 220 100 150
Example 6.4:
115.696 lt/s
5.643 lt/s
54
.35
7 lt
/s
49.422 lt/s
34.076 lt/s
Example 6.5:
++
Iteration 1
Loop
I
pipe ks(cm) D(cm) ks/D L(m) Q(L/s) Re*105 f K KQ 𝑄 2K 𝑄 ∆Q(m
For the given flow network system, determine the discharge passing in each pipe.Take υ =1.004x10-6 m2/s. Solve until ΔQ < 0.50 lt/s.The pump characteristic in pipe 1 : Hp = 18.0 – 0.009Q2 (put Q as lt/s to this equation)
or : Hp = 18.0 – 9000Q2 (put Q as m3/s to this equation).
Example 6.7 T:
43.811
14.189
24.811
41.189
36.358
68.358
25.642
Answer 6.7 T:
ITERATION TABLE FOR MULTI-RESERVOIRS (BRANCHED-PIPES)
(CORRECTION BASED ON JUNCTION POINT TOTAL ENERGY HEAD ‘HJUNCTION’)
Iteration No:
Pipe
Name
Pipe
Length L
(m)
Diameter D
(cm)
k
s
(cm)
ks/D Q
i
(m3/s)
𝐑𝐞
*105 𝐟
KTi
(𝐬𝟐
𝐦𝟓)
∆𝐡𝐢
(m)
𝐐𝐢
= ∆𝐡𝐢
𝐊𝐓𝐢
(m3/s)
𝐐𝐢
𝟐∆𝐡𝐢
(m2/s)
𝛅𝐡
=−∑𝐐𝐢
∑ 𝐐𝐢
𝟐∆𝐡𝐢
(m)
∆𝐡𝐢+𝟏
= ∆𝐡𝐢 + 𝛅𝐡
(m)
𝐇𝐃(𝐢+𝟏)
(m)
𝐇𝐃(𝐢+𝟏)
(m)
𝐐𝐢+𝟏
= ∆𝐡𝐢+𝟏
𝐊𝐓𝐢
(m3/s)
Iteration No:
LOOP Diameter Material
# (φ) (ks)
Σ
KT KTQ│Q│ 2KT│Q│Re Length (L)Pipe Name Q
f
)
)(
(
Hardy-Cross Method Correction Table
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water quality behavior of pressurized pipe networks(both JUNCTION & LOOP problems).
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