5 / 1 X Re x 37 . 0 X Re x 5 7 / 1 X Re x 16 . 0 5 / 1 X f Re 058 . 0 c X 2 f Re 06 . 0 ln 455 . 0 c X f Re 664 . 0 c Key Boundary Layer Equations Normal transition from Laminar to Turbulent x x U Re 0 x Boundary layer thickness (m) at distance x down plate = ) x ( 5 x 10 x 5 Re Shear stress on plate at distance x down plate 2 U c 2 f 0 U 0 free stream vel. kinematic visco. Rough tip –induced turbulence
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Key Boundary Layer Equations Normal transition from Laminar to Turbulent x Boundary layer thickness (m) at distance x down plate = Shear stress on plate.
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5/1XRe
x37.0
XRe
x5 7/1
XRe
x16.0
5/1X
fRe
058.0c
X2f
Re06.0ln
455.0c
Xf
Re
664.0c
Key Boundary Layer Equations
Normal transition from Laminar to Turbulent
x
xURe 0
x
Boundary layer thickness (m) at distance x down plate = )x(
5x 10x5Re
Shear stress on plate at distance x down plate
2
Uc
2
f0 U0 free stream vel.
kinematic visco.
Rough tip –induced turbulence
Shear Resistance due to flow of a viscous fluid of density and free stream vel = Uo
Over a plate Length L Breath B
L
0x
20
fs 2
UBLCdxBF
Flow in Conduits --Pipes
+ -
LT
22
22
P
21
11 hh
g2
Vz
ph
g2
Vz
p
Head IN from pumpNote pump power
PP
hQP
Head OUT from TurbineNote power recovered
TT hQP
Q discharge
0< <1 efficiency
Heat Loss
Our concern is to calculate this term
The nature of Flow in Pipes
Development of flow in a pipe
We use energy Eq.—assume = 1
If we select the points [a] and [b] to be at the top of the tanks Eq. 1Simplifies to
(1)
HhL
We can not measure H BUT we can estimate the head loss hL
There are a number of items that contribute to the head loss hL
In current problem Three components for head loss
In Example problem
Minor Losses
Note formDimensionless No X
V2/2g
See Table 10.3 in Crowe, Elger and Robinson
41.0K,6.DD
87.0K,2.DD
1K,0DD
E21
E21
E21
In this case reduces to
Head loss in a pipe
Head loss in a pipe
=0 by continuity
Rearrange
(1)
(2)
Wetted perimeter
(1) And (2)
Introduce a Dimensionless friction factor
Then
In a full circular pipe
So to find head loss hL Need to find friction factor f