pipe3

Pipe#3Pipeflow Civil Engineering Fluid Mechanics

1

Covered in this lecture:

1. Minor losses2. Equivalent Lengths3. Noncircular pipes4. Multiple pipe systems - series

Minor Losses

S Losses due to entrances, expansions, bends,valves, fittings, etc.

S A’int necessarily minor, can be significant

S Caused by turbulence, flow separation

S Denoted by

hm or hlm = K V2

2gwhere K (or Ke, Kc, Kb, KE, etc.) is determined experimentally foreach fitting.

Sudden Expansion

he = KV2

12g = [1 - (D1

D2)2] V2

12g

2

in which

K = [1 - (D1D2

)2]2


2

Sudden Contraction

S The process of converting pressure to velocity(shown 1 to 0) is very efficient, the vena con-tracta is point 0

S The process of converting velocity back to pres-sure to is not as efficient (shown 0 to 2)

V1 V2

1

0 2

S Head loss is expressed in terms of V2

hc = KcV2

22g

S Values are available in Tables

S Entrance loss Ke from reservoir varies (0.78 re-entrant, 0.5 for square- edged, to 0.04 for wellrounded)

S Exit loss to a reservoir is always KE = 1.0


3

Equivalent Length

The head loss through a component is expressed in terms of alength of pipe Le needed for the same head loss, so

hm = f LeD

V2

2gNow if K = S? i’s in the section, and we know f and D, we can write

Le = KDf

Example: If all the minor loss add up to K = 20 in 5000 ft of lineof 12 inch diameter line, and if f = 0.020, the equivalent length ofthe minor losses

Le = KDf = 20 (12 in 1 ft/12 in )

0.02 = 1000 ft

So 5000 ft + 1000 ft = 6000 ft of pipe without the minor losseswould have the same amount of head loss as the pipe section withthe minor losses.

Noncircular Pipes

The D used in our equations is actually the hydraulic diameter,

Dh = 4AP

where A is the area of the flow, and P is the wetted perimeter of thepipe section. For a circular pipe

Dh = 4AP =

4p D2

42pD/2

= D


4

Example: What are the Reynolds number, relative roughness, andhead loss for 1000 ft for a 8 in by 1 ft rectangular cast iron pipe (e= 0.00085 ft carrying water (? = 1 x 10- 5 ft2/s) at 9 ft/s?

Dh = 4AP = 4(8/12 ft) × 1 ft

2 × (8/12 ft) + 2(1 ft)= 0.80 ft

Using Dh as D

Re = VD? = (9 ft/s)(0.80 ft)

1 × 10- 5 ft2/s= 720, 000

Also use Dh for D in the calculation of relative roughness

ÁD = 0.00085 ft

0.80 ft= 0.001063

From Moody, f = 0.022, so

hf = f LD

V2

2g

hf = 0.0221000 ft0.80 ft

(9 ft/s)2

2 × 32.2 ft/s2 = 34.6 ft


5

Example: If steel pipe with D=15 cm, Find i) the dischargethrough the pipe in the flowing figure for H = 10 m , and ii) deter-mine the head loss H for Q = 60 l/s.

Standard Elbows

Water at 20 deg C

Square- edgedentrance

1

2

H = ?

Q = ?12 m

30 m 60 m

Globe Valve

Part i) Writing the energy equation between points 1 and 2, using2 as the datum (z = 0) and including all losses

H + 0 + 0 =V2

22g + 0 + 0 + 1

2V2

22g + f

102 m0.15 m

V22

2g + 2 × 0.9V2

22g + 10

V22

2g + 1.0V2

22g

velocityhead at 1

pressurehead at 1

pressurehead at 2

elevationhead at 2 entrance

losselbowloss valve

lossfrictionloss

L = 30 + 12 + 60 m

exitloss

or

H =V2

22g (13.3 + 680 f )

Guess f = 0.022, so

10 =V2

22g (13.3 + 680 f )

V = 2.63 m/s


6

So for ? = 1.01 µm2/s, e/D = 0.0017, then Moody gives

Re = (2.63 m/s )(0.15m)1.01 µm2/s

= 391, 000

For which f = 0.023. Repeating the procedure gives V2 = 2.60 m/s,Re = 380,000 and f = 0.023. The discharge is

Q = V2A2 = (2.60 m/s)(p4)(0.15 m)2 = 0.0459 m3/s = 45.9 l/s

Part ii) Q is known, solution is straightforward

V2 = QA = 0.06 m3/s

(p /4)(0.15 m)2 = 3.40 m/s Re = 505, 000 f = 0.023

and

H =V2

22g (13.3 + 680 f )

H = (3.4 m/s)2

2 × 9.8 m/s2 (13.3 + 680 × 0.023 ) = 17.06 m

Using equivalent lengths

Using the technique of equivalent lengths, the value of f is approxi-mated, say f = 0.022. The sum of minor losses is 13.3, and the ki-netic energy at point 2 is considered a minor loss, so

Le = KDf = 13.3 × 0.15

0.022 = 90.7 m

So the total pipe length is 30 m + 12 m + 60 m +90.7 m = 192.7 m.

Solving for Part i) of the problem

10 m = f L + LeD

V22

2g = f192.7 m0.15 m

(V2 m/s)2

2g m/s2

If f = 0.022, V2 = 2.63 m/s, Re = 391,000, so from Moody new f= 0.023.


7

If f = 0.023, V2 = 2.58, so Q = 45.6 l/s.

S Normally it is not necessary to use the new f toimprove Le, unless your first guess if f is verypoor.


8

Multiple pipe systems

When fluid flows from one pipe to another of a different diameter,it is considered as connected in series.

ABH··

Ke

1

2

Not to scaleIn general two type of problems

S H is desired for a Q, or

S Q is desired for a given H

Writing the energy equation from A to B including all minor losses

H + 0 + 0 = 0 + 0 + 0 + KeV2

12g + f1

L1D1

V21

2g +velocity

head at B

pressurehead at A

pressurehead at B

elevationhead at B

entranceloss

friction

loss Pipe 1

+ KExpV2

12g + f2

L2D2

V22

2g + KExitV2

22g

friction

loss Pipe 2

Sudden Expansion

loss Pipe 1- 2

Expansion

Exit loss Pipe 2

velocityhead at A


9

Using the continuity equation V1D1 = V2D2 to eliminate V2 fromthe above equation, along with our diameter calculation for Kexp= [1 - (D1/D2) 2 ]2 and setting Kexit = 1, we get

H =V2

12g [Ke +

f1L1D1

+ [1 - (D1D2

)2]2 +f2L2D2

(D1D2

)4 + (D1D2

)4]For pipes of known lengths and sizes, we can reduce this to

H =V2

12g (C1 + C2 f1 + C3 f2)

Where C1, C2, C3 are known.

S With Q given, Re is computed and f read fromMoody

S With H given, V1, f1, f2 are unknown. Assumevalues, solve for V, compute estimate of Re,solve for f ’s, etc.

Example: Suppose for our Figure,Ke = 0.5 L1 = 1000 ft D1 = 2 ft e1 = 0.005 ft

L2 = 800 ft D2 = 3 ft e2 = 0.001 ft? = 0.00001 ft2/s,H = 20 ftWe get

H =V2

12g [0.5 + f1(1000

2 ) + (1 - (23)2)2 + f2(8003 )(23)4 + (23)4]

20 =V2

12g (1.01 + 500 f1 + 52.6 f2)


10

S Now for e1/D1 = 0.0025, and e2/D2 = 0.00033,From the Flat part (complete turbulent) ofMoody

. f1 = 0.025 f2 = 0.015

Solving for V1, V1 = 9.49 ft/s, and so V2 = 4.21 ft/s, for which

Re1 = 9.49 ft/s × 2 ft0.00001 ft

= 1, 898, 000

Re2 = 4.21 ft/s × 3 ft0.00001 ft

= 1, 263, 000

S With these values, from Moody

. f1 = 0.025 f2 = 0.016

So, V1 = 9.46 ft/s, so

Q = 9.46 ft/s × p4 (2 ft)2 = 29.8 cfs

Using Equivalent Pipes

S Equivalent lengths can be used since two pipesystems are said to be equivalent when the samehead loss produces the same discharge in bothsystems

Writing the head loss for the first pipe

hf 1 = f1L1D1

Q21

(D21p /4)22g

=f1L1D5

1

8Q21

p 2g


11

Writing the head loss for the second pipe

hf 2 =f2L2D5

2

8Q22

p 2g

S For equvilent pipe, flow is equal, an the head lossis the same then:

For the two to be equivalent pipeshf 1 = hf 2 Q1 = Q2

Setting hf1 = hf2, and simplifyingf1L1D5

1=

f2L2D5

2Solving for L2 gives

L2 = L1f1f2

(D2D1

)5

S This is the length of a second pipe to be equiva-lent to that of the first pipe

Example: Replace 300 m of 25 cm pipe with an equivalent lengthof 15 cm pipe

S Approximate values of f1 and f2, assuming com-plete turbulence

If for this example f1 = 0.020 and f2 = 0.018, then

L2 = 300 m 0.0200.018(15 cm

25 cm)5 = 25.9 m

S For the assumed conditions, 25.9 m of 15 cmpipe has the same head loss as 300 m of 25 cmpipe.


12

Example: Solve the previous reservoir example using equivalentpipes

S First express minor losses in terms of equivalentlengths, using f1 and f2 as fully turbulent

For pipe 1

K1 = 0.5 + [1 - (2/3)2]2 = 0.809entrance expansion

Le1 =K1D1

f1= 0.809 × 2

0.025 = 65 ft

For pipe 2

K2 = 1 Le2 =K2D2

f2= 1 × 3

0.015 = 200 ft

So we now have

1000 ft + 65 ft = 1065 ft of 2 ft diam. pipe800 ft + 200 ft = 1000 ft of 3 ft diam. pipeExpressing the 3 ft diam. pipe in terms of 2 ft diam. pipe

Le = 1000 ft × 0.0150.025 (2 ft

3 ft)5 = 79 ft

Le = L2 = L1f1f2

(D2D1

)5

By adding this to the 2 ft pipe, the problem is now equivalent to

1065 ft + 79 ft = 1144 ft of 2 ft diam. pipeSo if e = 0.05 ft, and H = 20 ft,With f = 0.025, V = 9.5 ft/s, Re = 1,900,000. Converges, so Q=9.5p = 29.9 ft3/s


13

H = f1144 ft

2 ftV2

2g

Documents

pipe3