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457.562 Special Issue on River Mechanics (Sediment Transport)

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Seoul National University

457.562 Special Issue on River Mechanics (Sediment Transport)

.02 Review of Fluid Mechanics

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Seoul National University

§ Shear stress between adjacent layers in a simple parallel flow is det ermined by the viscosity and the velocity gradient

§ If laminar flow is disturbed by an obstacle or roughness, then disturb ances must be damped by the molecular viscosity (stable)

§ However, in turbulent flow, inertia overcomes the viscous force and disturbance grows and becomes random motion.

§ Therefore, in turbulent motion rapid mixing of fluid particles during fl ow cause significant diffusion.

§ What is the actual meaning of viscosity? Is there any other name of viscosity?

1. Flow velocity distribution : law of the wall

τ = µ dv dy

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Seoul National University

§ Consider a steady, turbulent, uniform, open-channel flow having – H : a mean depth

U : a mean flow depth

B : a mean width (much greater than depth) – S : a mean slope

ks : effective height

– In a wide channel (B/H >>1)

1. Flow velocity distribution : law of the wall

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Seoul National University

§ The roughness height (or effective height) will be proportional to a m ean size or diameter D.

§ The bed shear stress

§ The above equation is the one-dimensional momentum conservatio n equation

§ Using the boundary shear stress, the shear velocity can be defined as

§ The sear velocity (the boundary shear stress) provides a direct measure of the flow intensity able to entrain and transport sediment particles.

1. Flow velocity distribution : law of the wall

u* = τb / ρ

τb = ρgHS (ρ = water density, g=gravity)

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Seoul National University

§ The above figure shows typical variation of total shear stress in a tur bulent pipe flow.

§ Based on the linear relationship of shear profile (in the previous clas s), and the Prandtl relationship,

1. Flow velocity distribution : law of the wall

τ = τb 1z H

⎛⎝⎜ ⎞

⎠⎟ = ρl2 du dz

⎛⎝⎜ ⎞

⎠⎟

2

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Seoul National University

§ Near wall, l=kz, and mixing length in a pipe flow

§ Then the previous equation can be written

1. Flow velocity distribution : law of the wall

l = kz 1− z H

⎛⎝⎜ ⎞

⎠⎟

1/2

τb

ρκ 2z2 =

du dz

⎛⎝⎜ ⎞

⎠⎟

2

du

dz = τb / ρ κz =

u* κz u

u* = 1 κ ln

z z0

⎝⎜

⎠⎟

u = time -averaged flow velocity at a distance z above the bed

z0 = bed roughness length (distance above the bed where u(z0) = 0) κ = is von Karman's constant ~0.41.

Only in a relatively thin layer (z/H=0.2) (1)

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§ Through intensive experiment works,

§ When the effective roughness height becomes by (1) and (2)

§ Very near the smooth wall (viscous sublayer), velocity is linear with depth

§ Then two equations above need to match

1. Flow velocity distribution : law of the wall

u

u* = 1 κ ln

u*z

ν +5.5

u

u* = u*δν ν =

1 κ ln

u*δν

ν + 5.5 (at z = δν) u*δν

ν =11.6 δν =11.6 ν u u

u* = u*z

ν (until z =δν) zb/ 9u*

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Hydraulic smooth condition of flow

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Seoul National University

1. Flow velocity distribution : law of the wall

§ Velocity distribution near a smooth wall

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Seoul National University

1. Flow velocity distribution : law of the wall

§ But, most boundaries in river flow are rough. Therefore we need to c onsider roughness height.

§ If ksν>1, then viscous sublayer will not exist.

§ The corresponding logarithmic velocity profile is given by

§ It follows that

§ Then what happens in the intermediate condition?

u

u* = 1 κ ln

z ks

⎝⎜

⎠⎟ +8.5 = 1

κ ln 30 z ks

⎝⎜

⎠⎟

z0 = ks / 30 Hydraulically rough flow

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Seoul National University

1. Flow velocity distribution : law of the wall

§ Hydraulically rough flow

§ Hydraulically smooth flow

§ Where “ the Roghness Reynolds number”

§ Yalin (1992) proposed a way to represent both ranges and intermedi ate ranges as

ks

δν 1, Re* 11.6 ks

δν 1, Re* 11.6

Re

*

= u

*

k

s

ν , (R

*C

= 11.6)

u

u

*

= 1 κ ln

z k

s

⎝⎜

⎠⎟ + B

s

B

s

= 8.5 + ⎡⎣ 2.5 ln Re ( )* − 3 ⎤⎦ e−0.121 ln Re[ *]2.42

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Seoul National University

1. Flow velocity distribution : law of the wall

§ Alternative way of writing,

u

u

*

= 1

κ ln A

s

z k

s

⎝⎜

⎠⎟

A

s

= e

κBs
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Seoul National University

2. Velocity-Defect and Log-Wakes laws

§ Equation (1) requires some knowledge of the bed roughness charact eristics.

§ An alternative formulation can be obtained if the flow depth H is intro duced as the relevant length scale.

§ Assuming that the maximum flow velocity umax takes places at the w ater surface z=H, then equation (1) can be manipulated to obtain the velocity-defect law (outer low of the wall).

§ But, there were so many arguments and defects or log law working only in z/H<0.2. Nezu and Nakagawa added a wake function.

umaxu

u* = − 1 κ ln

z H

u

u* = 1

κ ln u*z

ν

⎝⎜

⎠⎟ + 5.5+w z H

⎝⎜

⎠⎟

(3)

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Seoul National University

2. Velocity-Defect and Log-Wakes laws

§ W(z/H) is the wake function first proposed by Coles (1956) for turbul ent boundary-layer flows,

§ W0 is the Coles wake parameter, expressing the strength of the wak e function.

§ Equation (3) can also be written in log-wake form

§ Coles wake parameter varies depending on the condition and incras es with increasing sediment concentration raging from 0.191 to 0.86 1.

w z H

⎝⎜

⎠⎟ = 2W0

κ sin2 π

2 z H

⎝⎜

⎠⎟

umax u

u* = 1

κ ln z H

⎝⎜

⎠⎟ + 2W0

κ cos2 πz 2H

⎝⎜

⎠⎟

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Seoul National University

2. Relations for channel resistance

§ Most river flows are indeed hydraulically rough.

§ The previous log law is used to obtain an appropriate expression for depth-averaged velocity U.

§ This relation is known as Keulegan’s resistance law for rough flow.

U = 1

H u dz

0 H

U

u* = 1 H

1 κ ln

z ks

⎝⎜

⎠⎟ +8.5

dz

ks H

U u* = 1

κ ln H ks

⎝⎜

⎠⎟ +6 = 1

κ ln 11 H ks

⎝⎜

⎠⎟

(Avoiding z=0 : singular point)

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Seoul National University

2. Relations for channel resistance

§ Manning-Strickler form.

§ Log - > power : was first presented by Keulegan.

u

u* = 1

κ ln 30 z ks

⎝⎜

⎠⎟ ≅ 9.34 z ks

⎝⎜

⎠⎟

1/6

U

u* = 1

κ ln 11H ks

⎝⎜

⎠⎟ ≅ 8.1 H ks

⎝⎜

⎠⎟

1/6

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Seoul National University

2. Relations for channel resistance

§ Bed shear stress can be found as

§ In two-dimensional case τb = ρu*2 = ρCfU2

Cf = 1

κ ln 11H ks

⎝⎜

⎠⎟

⎣⎢ ⎤

⎦⎥

−2

or 8.1 H ks

⎝⎜

⎠⎟

1/6

⎣⎢

⎦⎥

−2

τbsbn

( )

= ρCf ⎡⎣U2 +V2⎤⎦1/2

(

U,V

)

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