ENVE4403 Assignment 1
Question 1 [6 marks]
A vertical belt moves upwards and draws a thin film of fluid out of a tank. The fluid film
maintains a constant thickness (h) along the belt. Assume the belt is very wide, and moves
steadily at speed Wbelt.
(a) If the ambient air is stationary, find the minimum belt velocity that has all the fluid in
the film moving upwards. [3.5 marks]
(b) For the value of the belt velocity in (a), sketch the velocity profile in the film,
providing as much detail as possible. [1.5 marks]
(c) If the belt has a height H and width b (into the page), what is the power required to
drive the belt at this speed? [1 mark]
Wbelt
Film thickness = h
z
x
Question 2 [9 marks]
Consider the wind-stress-driven oceanic flow generated by a large-scale high pressure
system in the atmosphere. When the wind blows entirely in the x-direction (as in the figure
elow), there is a constant wind stress vector (??????????, 0).
(Plan view)
WATER
LAND
??????????
For z positive upwards (z = 0 at the water surface), it has been proposed that the resulting
steady-state fluid motion in the ocean is given by:
??(??) = �
√2??????????
?????? �
????/?? cos �
??
?? −
??
4� ??
(??) = �
√2??????????
?????? �
????/?? sin �
??
?? −
??
4�
where the length scale ?? = �2??/|??| is known as the ‘Ekman layer depth’.
(a) Firstly, by considering the relative magnitudes of all the terms, simplify the Navier-
Stokes equations in the horizontal directions down to two-term equations. [1 mark]
(b) Verify that the solutions for flow velocities in the ocean (above) satisfy these Navier-
Stokes equations and all relevant boundary conditions. [5 marks]
(c) Consider this phenomenon off the coast of San Francisco, California. A typical
quantification of the wind stress on the ocean is 0.001??????102 , where ??10 is the wind speed
(at a height of 10 m) and ???? is the density of air.
For a wind speed of 10 m/s, plot (not sketch!) the velocity vector in the ocean, and its
variation over depth. Use this plot to explain why this flow is called the ‘Ekman spiral’.
[2.5 marks]
(d) How would your plot in (c) change if the flow was tu
ulent? [0.5 marks]
Question 3 [5 marks]
In tu
ulent flow over a smooth wall, the flow is typically separated into two regions. Near
the wall, viscosity damps tu
ulent motions such that the flow is laminar (this is known as
the “viscous sublayer”). Further from the wall (in the “tu
ulent layer”), tu
ulence
dominates the vertical transport of momentum. The shear stress is roughly constant across
the two layers, taking a value equal to the stress at the wall (τw). The shear stress at the wall
is often expressed as a shear velocity, defined as ??∗ = �????/??.
(a) For horizontal flow over a long, wide wall with no applied pressure gradient, solve the
elevant equation of motion to show that the mean velocity profile in the viscous sublayer is
given by ??�(??) = ??∗2?? ??⁄ . [1.5 marks]
(b) The thickness of the viscous sublayer (δv) is observed to be roughly 10ν/??∗. Determine
the mean velocity at the top of the viscous sublayer. [0.5 marks]
(c) In the tu
ulent layer, the eddy velocity scales on ??∗ and the size of the largest eddy
scales on the distance from the wall (z). Consequently, the eddy viscosity has the form
????(??) = ????∗??, where ?? is an O(1) constant.
Using your answer to (b) as a boundary condition for the velocity in the tu
ulent layer (and
the fact that the shear stress is constant), determine the mean velocity profile in the tu
ulent
layer. Explain why this layer is typically refe
ed to as the “logarithmic layer”. [3 marks]
z = 0
z
FLUID MECHANICS 200
1
TOPIC 1. PHYSICAL LAWS GOVERNING FLUID MOTION
Threshold concepts
1. The Navier-Stokes equations for fluid flow are powerful but complex (9
terms each, lots of partial derivatives). Their solution is made considerably
easier in situations where we can ignore the acceleration terms (LHS).
2. Solutions for controlled/engineered settings can (sometimes) be
straightforward. But there are many complicating factors for
environmental systems, which necessitates this unit!
3. To get around this complexity, we can evaluate dimensionless
parameters (ratios of magnitudes of two terms in the equation) to
eliminate negligible terms.
PART A: THE GOVERNING EQUATIONS
Kinematics
In mechanics there are two frames of reference used for describing motion:
Lagrangian (moving reference frame) and Eulerian (fixed reference frame).
The former is conventional in particle mechanics, but Eulerian frames are of
necessity used in fluid mechanics, due to our general inability to identify and
track individual fluid particles.
This has the effect of there being 2 components to fluid acceleration:
(i)
(ii)
2
In an Eulerian frame, consider a rectangular coordinate system (x, y, z) where
(u, v, w) are the fluid velocities at a point. Fluid acceleration is given by
???? =
????
????
=
????
????
+ ??
????
????
+ ??
????
????
+ ??
????
????
???? =
????
????
=
????
????
+ ??
????
????
+ ??
????
????
+ ??
????
????
???? =
????
????
=
????
????
+ ??
????
????
+ ??
????
????
+ ??
????
????
where the operator
D
Dt is called the material derivative. The material derivative
is relevant to Eulerian reference frames.
Physical Law #1: Conservation of Mass
Consider the flow of a fluid with density ρ. If pressure has no significant
effects, we say the flow is incompressible; fluid density does not change as
fluid particles move around.
Consider flow into and out of an infinitesimal control volume of dimensions
(dx, dy, dz).
3
Conservation of mass for this control volume can be expressed as
For incompressible fluid:
Therefore mass conservation states that:
Note: This is valid for both steady and unsteady flow (only assumption is
incompressibility).
This equation says that if the velocity in a given direction changes in that
direction, then it will also change the velocities in the other directions.
0u v w
x y z
∂ ∂ ∂
+ + =
∂ ∂ ∂
4
Physical Law #2: Conservation of Momentum
Again consider flow in and out of an infinitesimal control volume of
dimensions (dx, dy, dz).
Consider the momentum balance in the x direction. The forces that can act on
the control volume include:
(i) normal stresses: forces due to pressure (p) and viscous forces acting normal
to the surfaces of the control volume
(ii) viscous shear stresses acting on faces of the control volume.
(iii) gravity (a body force)
Recall how we describe fluid stresses in a Newtonian fluid: the stress acting in
the x direction on a surface in the y plane ?????? is given by
What does fluid viscosity represent?
5
Hence conservation of momentum can be written as:
This yields the Navier-Stokes equations:
Again, the only assumption here is that of incompressible flow. The LHS of each
equation represents acceleration in that direction. The RHS represents the sum
of the forces per unit mass acting in that direction.
2 2 2
2 2 2
2 2 2
2 2 2
2 2 2
2 2 2
1:
1:
1:
x
y
z
u u u u p u u ux u v w g
t x y z x x y z
v v v v p v v vy u v w g
t x y z y x y z
w w w w p w w wz u v w g
t x y z z x y z
ν
ρ
ν
ρ
ν
ρ
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
+ + + = − + + + + ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
+ + + = − + + + + ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
+ + + = − + + + + ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
6
Physical Law #3: Conservation of Energy - Bernoulli Equation
Recall: In steady, incompressible flow, when we can ignore viscous effects and
losses due to tu
ulence, then:
??2
2??
+
??
????
+ ?? = constant along a streamline
where ?? = √??2 + ??2 + ??2 is the fluid speed.
We can derive the Bernoulli Equation from conservation of momentum.
Consider flow in the vertical direction, for which the equation of motion is
(1)
If we consider only steady, inviscid, vertical (for now just 1D) flow, then (1)
simplifies to
w ∂w
∂z
= − 1
ρ
∂p
∂z
− g (2)
i.e