4B-Velocity Profiles (2021)
4B-Velocity Profiles (2021)
4B-Velocity Profiles (2021)
https://www.engineeringtoolbox.com/dynamic-viscosity-d_571.html
In Navier-Stoke equation, the convective acceleration in
the inertial term is non-linear.
𝑑𝑑 (𝑚𝑚𝑣𝑣𝑥𝑥 )
and apply Newtons Second Law, =𝐹𝐹𝑥𝑥 . In this case even though
𝑑𝑑𝑑𝑑
𝑑𝑑 (𝑚𝑚𝑣𝑣𝑥𝑥 )
the fluid is moving, but it’s not accelerating, so that =0.
𝑑𝑑𝑑𝑑
This force balance can be written around a fluid element of unit area
to determine the r/ship τ and ΔP as :
( P + ∆P )(πr ) − P (πr ) = τ (2πrL)
2 2
- - -(4.6)
which can give
∆P 2τ
= - - -(4.7)
l r
Motion of cylindrical fluid element inside a pipe/tube.
∆𝑃𝑃𝑃𝑃
Solving for 𝜏𝜏 yield: 𝜏𝜏 = - - -(4.8)
2𝐿𝐿
∆𝑃𝑃𝑃𝑃
and at the tube wall: 𝜏𝜏𝑤𝑤 = - - -(4.9)
2𝐿𝐿
4�
𝜏𝜏𝑤𝑤
The ∆𝑃𝑃 and 𝜏𝜏𝑤𝑤 are related by : ∆𝑃𝑃 = - - -(4.12)
𝐷𝐷
To determine a dough viscosity, dough at constant temperature is
forced through a long, round capillary/slit, & data on the flow rate
resulting from different applied P is taken. From the P measurements,
shear stress can be calculated at the wall using eqn. (4).
∆𝑃𝑃𝑃𝑃
and at the tube wall: 𝜏𝜏𝑤𝑤 = - - -(4.9)
2𝐿𝐿
Shear rate can be calculated from the flow:
4�
The ∆𝑃𝑃 and 𝜏𝜏𝑤𝑤 are related by : 𝜏𝜏𝑤𝑤
∆𝑃𝑃 =
𝐷𝐷
- - -(4.12)
v r
τW
∫O dv = − µR ∫ rdr
R
or τW
v= (R 2 − r 2 ) - - -(4.15)
2 Rµ
Recognising the maximum velocity will occur at r = 0,
v max = τ W R / 2 µ - - -(4.16)
v = m / ρs = 1 / s ∫ vds - - -(4.17)
@
v 2
= 2[1 − (r / R ) ] - - -(4.19)
v
Eqn. 4.19 is the eqn. for a parabola, which is the characteristics
velocity profile of a Newtonian fluid in a circular conduit.
Velocity profile
for a power law
fluid
The above sequence of steps can also be performed using the power
law model, Eqn. 4.5 to express the r/ship between 𝜏𝜏 and 𝛾𝛾̇ . When
this is done, the following results
v 3n + 1
n +1
- - -(4.20)
= [1 − (r / R) n
]
v n +1
indicating how v changes as a fn of r for a power law fluid.
n +1
v 3n + 1 - - -(4.20)
= [1 − (r / R) n
]
v n +1
Note:
The slit is very wide (infinite) compared to its thickness 2C, so that
the effects of the sides of the slit can be ignored.
@
τ = ∆Pc / L - - -(4.22)
Rewriting the balance for the τ at the wall gives
τ W = ∆PC / L - - -(4.23)
Substituting Newton’s law of viscosity (Eqn. 4.2) into Eqn. 4.23 &
integrating yields
v = τ W C / 2 µ[1 − (c / C ) ] 2
- - -(4.24)
Recognising that
v = τ W C / 3µ - - -(4.25)
So
v 3
= [1 − (c / C ) 2 ] - - -(4.26)
v 2
Replacing Newton’s law of viscosity with the power law model in the
above derivation gives
v 2n + 1 n +1
= 1 − (c / C )
n - - -(4.27)
v 2n