1Fluid Mechanics – Lecture 9

Fluid mechanics(wb1225)

Lecture 9:flow losses

2Fluid Mechanics – Lecture 9

Laminar vs turbulent pipe flow

Laminar flow

Turbulent flow

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3Fluid Mechanics – Lecture 9

Friction factor for turbulent flow

u(r ) = u0 1 −r

R

1 n

V

u0

=2n 2

(2n + 1)(n + 1)=

49

60= 0.817 for n = 7

Re = 4x103 2x104 1x105 1x106 2x106 3x106

n = 6.0 6.6 7.0 8.8 10 10

f =8τ w

ρV 2 ⇒f

4=

ρu∗2

12 ρV 2 = 2

u∗

V

2

Fanning friction factor

u(y)

u∗

= Cyu∗

ν

1/ 7

⇒u0

u∗

= CRu∗

ν

1/ 7

⇒V

u∗

= ′CRu∗

ν

1/ 7

V

u∗

7

= ′′CRu∗

ν⇒

V 8

u∗8

= ′′CRV

ν⇒ 4 f = 2

u∗

V

2

= C ⋅ Re−1 4

4Fluid Mechanics – Lecture 9

Moody diagram

f =

64 Re

0.316 Re−1 4

1.8 log Re 6.9( ) −2

Re < 2300

4000 < Re < 10 5

Re > 10 5

Hagen-Poiseuille

Blasius

Colebrook

implicit:1

f= 2.0 log Re f( )− 0.8 Prandtl

5Fluid Mechanics – Lecture 9

Effect of roughness

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6Fluid Mechanics – Lecture 9

Hydraulic diameter

p1p2

L

A τ w

momentum balance:

14 π D 2

area{ ⋅ ∆p = τ w ⋅ π D

perim.{ ⋅ L ⇒

∆p

L=

4τ w

Dh

hydraulic diameter: Dh = 4surface area

wetted perimeter

flow between 2 parallel plates at distance 2h:

f =24

Re h

⇒ f =96

Re Dh

Dh = 42hw

4h + 2w= 4h for w → ∞

cf. pipe flow: f =64

Re

7Fluid Mechanics – Lecture 9

Other losses in pipe systems• entrance / exit

• sudden expansion / contraction

• bends / elbows / tees

• valves

• gradual expansion / contraction

K =∆P

12 ρV 2

⇒ ∆Ptot = 12 ρV 2 f

L

D+ K i

i∑

8Fluid Mechanics – Lecture 9

Sudden expansion

∆p = K ⋅ 12 ρV 2 ⋅ S

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9Fluid Mechanics – Lecture 9

Sudden expansionconservation of mass: S1V1 = S2V2 ⇒ V2 =

S1

S2

V1

momentum balance:

p1S1 + p1(S2 − S1 ) + &mV1 = p2S2 + &mV2 with &m = ρS1V1 = ρS2V2

p1S2 = p2S2 + ρS2V2 (V2 − V1 )

p1 = p2 + ρ(V22 − V2V1 )

Energy equation:

p1 = p2 + 12 ρ(V2

2 − V12 ) + K 1

2 ρV12

0 = 0 + 12 ρ(2V2

2 − 2V2V1 ) − 12 ρ(V2

2 − V12 ) − K 1

2 ρV12

K 12 ρV1

2 = 12 ρ(2V2

2 − 2V2V1 − V22 + V1

2 )

K 12 ρV1

2 = 12 ρ(V2

2 − 2V2V1 + V12 )

K 12 ρV1

2 = 12 ρV1

2 V22

V12 − 2

V2

V1

+ 1

⇒ K =

V2

V1

− 1

2

=S1

S2

− 1

2

=d 2

D 2 − 1

2

10Fluid Mechanics – Lecture 9

Commercial valve

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11Fluid Mechanics – Lecture 9

Other valves

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12Fluid Mechanics – Lecture 9

Bends

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13Fluid Mechanics – Lecture 9

Loss coefficient table

14Fluid Mechanics – Lecture 9

Example 6-16Water is pumped between two reservoirs at 5.4 liter/sec through 120 m of a 5-cmdiameter pipe with several minor losses. The roughness ratio is 0.001. Computethe required power for the pump.

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15Fluid Mechanics – Lecture 9

Diffuser

[1]

16Fluid Mechanics – Lecture 9

Diffuser performance

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17Fluid Mechanics – Lecture 9

Summary• Chapter :

• Examples:

• Problems:

18Fluid Mechanics – Lecture 9

Source

1. Frank M. White, Fluid Mechanics, McGraw-Hill Series in Mechanical Engineering