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On noting that. We can write. In this case we can calculate the resistance r for a given pipe and use the form. to analyses flow in pipe networks. Head loss in a pipe—using the Darcy-Weisbach Equation. - PowerPoint PPT Presentation
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Head loss in a pipe—using the Darcy-Weisbach Equation
g2
V
D
Lfh
2
On noting that 16/DVAVQ 422222
We can writegD
fL8rwhere,rQh
252
If flow is fully turbulent f will not depend on Reynolds number and r can be considered aconstant—i.e. NOT a function of velocity V
In this case we can calculate the resistance r for a given pipe and use the form 2rQh
to analyses flow in pipe networks
Flow in a Parallel Pipe System
A B
Consider flow from A to B through the three pipes in the directions shown.If the flow upstream of A is Q m3/s how is it split between pipes 1, 2 and 3.
1
2
3
The diagram has Two NODES A and B(points where pipe join)
And Two LOOPS
-From A along pipe 1 to B and back along pipe 2 to A
-From A along pipe 2 to B and back along pipe 3 to A
Head loss will increase as we move in direction of flow and decrease as we move against flow
2rQh
So for loop 1 we have 0QrQr 222
211 (1)
There can be NO net change in head around a closed loop.
For loop 2 0QrQr 233
222 (2)
Continuity 321 QQQQ (3)
IF Q is known we can solve the three equations (Using SOLVER in EXCELL) to obtainValues for the Qi Download excel file “parbal.xls” in notes section
We can extend the ideas to general Pipe Networks exercise 1 in the pipe network lab
4 NODES A, B, C, D5 Pipes2 Loops
Loop 1:R1*QQ1*ABS (QQ1) +RR4*QQ4*ABS (QQ4) + RR5*QQ5*ABS (QQ5) = 0Loop 2: -RR2*QQ2*ABS(QQ2) - RR3*QQ3*ABS(QQ3) + RR4*QQ4*ABS(QQ4) = 0Balance Node A: QQA + QQ5 – QQ1 = 0Balance Node B: QQ1 – QQ2 – QQB – QQ4 = 0Balance Node C: QQ2 – QQ3 – QQC = 0Balance Node D: QQ3 + QQ4 + QQ5 – QQD = 0Overall Balance: QQA – QQB – QQC – QQD = 0
More equations than we need—But SOLVER can handle them
Note if you “Guess” a wrong direction fro flow the discharge value will benegative
For given Q inputs at A, B, C and D and r’s determinePipe discharges
Pipe systems with reservoirs and pumps
One Node And ??? Loops
El. 20 mEl. 25 m
El. 0
PL1
L2
L3
El. 20 mEl. 25 m
El. 0
PL1
L2
L3
Create “pseudo loops” with zero flows (Q = 0) between reservoirs surfaces
Then head-loss loop equations are
0hQrQr251
0QrQr2520
P211
233
222
233
Note: directions around loopsThere is a positive head loss when we movethrough a pump opposite to flow direction
Continuity0QQQ 321
El. 20 mEl. 25 m
El. 0
PL1
L2
L3
More on the pump
The head provided by the pump is a function of the discharge Q through it
Q
hp
The shutoff head the maximum head that can be provided—the pump can lift water tothis height BUT water can not flow (Q = 0)
The free-delivery This is the maximum flow through the pump. It can onlybe achieved if no pipe is attached tothe pump ( hP = 0).
The efficiency of the pump is also a function of Q
pQh
Power
It will be zero at “shutoff” and free-delivery and attaint a maximum
< 100% for deliveryfreeQQ0
It is important to choose a pumpThat is efficient for required Q