CHEE 3363Spring 2014Handout 20

�Reading: Fox, 8.6--8.8

�1

1. �

�

3.

�

τturb → 0

Turbulence 1

Reynolds stress:u′ v′

wall layer �

u′ v′ x y

�3

�4

Defect law

viscous sublayer

Friction velocity

Recall ν = µ / ρ

Turbulence 3

Power-law equation

U

�5

Q − Ws − Wshear − Wother =∂

∂t

∫

CV

eρ dV +

∫

CS

(u + pv +v

2

2+ gz)ρv · dA

�1. �

�3. �4.

�6

�7

∫

A

v2

2ρv dA = α

∫

A

v2

2ρv dA = αm

v2

2

α s.t.:

3

g

�8

α1v2

1

2=

α2v2

2

2

Major losseshlm

�9

�

f

3Minor losses:

K1.

Le

K f

�11

�

�

f friction factor �

�

�

�

�

equivalent length of pipe Le

hl = f

L

D

V

2

2

f =64

Re

1√

f

= −2.0 log

(e/D

3.7+

2.51

Re√

f

)

hlm= K

V

2

2

hlm= f

Le

D

V

2

2

(p1

ρ

+ α1

V

2

1

2+ gz1

)−

(p2

ρ

+ α2

V

2

2

2+ gz2

)= hlT

=

∑hl +

∑hlm

�

1

Given Q 3

p p3; pipes are D

Determine ��

Assumptions �1. �

�3. �4.

�13

�14

Δp 1Given

QDetermine �

�

Assumptions �1. �

�3. �4.

�15

K = 0.56

Δp

�16

Δp 3

�17

1Given

Determine

Assumptions �1. �

�3. �4.

�18

�19

ν η ρ

ν = 1.08 × 10−5

ft2

s

f V

��

1. �

�

3. �

4. ��

�