Head loss in a pipe—using the Darcy-Weisbach Equation

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