1) The velocity distribution for the flow of an incompressible fluid is given by u = 3-x, v = 4+2y, w = 2-z. Show that this satisfies the requirements requirements of the continuity equation.
For three-dimensional flow of an incompressible fluid, the continuity equation simplifies to ,
SOLUTION:
2) Consider steady flow of water through an axisymmetric garden hose nozzle. Along the centerline of the nozzle, the water speed increases from uentrance to uexit as sketched. Measurements reveal that the centerline water speed increases parabolically through the nozzle. Write an equation for centerline speed u( x ), based on the parameters given here, x ), from x = 0 to x = L.
Solution We are to write an equation for centerline speed through a nozzle, given that the flow speed increases parabolically. Assumptions 1 The flow is steady. 2 The flow is axisymmetric. 3 The water is
incompressible. Analysis
A general equation for a parabola in the x direction direction is
General parabolic equation:
u = a + b ( x − c)2
(1)
We have two boundary conditions, namely at x = 0, u = uentrance and at x = L, u = uexit. By inspection, Eq. 1 is satisfied by setting c = 0, a = uentrance and b = (uexit - uentrance)/L2. Thus, Eq. 1 becomes
Parabolic speed :
3) Consider the following steady, two-dimensional velocity field as below. Is there a stagnation point in this flow field? If so, where is it?
For a given velocity field we are to find out if there is a stagnation point. If so, Solution we are to calculate its location. Assumptions.1 The flow is steady. 2 The flow is two-dimensional in the x-y plane. Analysis
The velocity field is (1)
At a stagnation point, both u and v must equal zero. At any point ( x ,y ) in the flow field, the velocity components u and v are obtained from above equation,
If the flow were three-dimensional, we would have to set w = 0 as well to determine the location of the stagnation point. In some flow fields there is more than one stagnation point. 4) Consider steady, incompressible, two-dimensional flow through a converging duct (as shown in figure). A simple approximate velocity field for this flow is as given below. In velocity field U 0 is the horizontal speed at x = 0. Note that this equation ignores viscous effects along the walls but is a reasonable approximation throughout the majority of the flow field. Calculate the material acceleration for fluid particles passing through this duct. Give your answer in two ways:(1) as acceleration components a x and ay and (2) as acceleration vector ⃗
Solution We are to calculate the material acceleration for a given velocity field.
Assumptions 1 The flow is steady. 2 The flow is incompressible. 3 The flow is two-
dimensional in the x-y plane. Analysis The velocity field is
For positive x and b, fluid particles accelerate in the positive x direction. Even though this flow is steady, there is still a non-zero acceleration field.
5) A steady, incompressible, two-dimensional velocity field is given by the following components in the xy plane:
Calculate the acceleration field (find expressions for acceleration components a x and ay ) and calculate the acceleration at the point ( x , y ) = (1, 2).
6) Converging duct flow (As shown in figure) is modeled by the steady, two-dimensional velocity field as given below. Generate an analytical expression for the flow streamlines.
Solution
For a given velocity field we are to generate an equation for the streamlines.
Assumptions 1 The flow is steady. 2 The flow is two-dimensional in the x-y plane.
7) A general equation for a steady, two-dimensional velocity field that is linear in both spatial directions ( x and y ) is where U and V and the coefficients are constants. Their dimensions are assumed to be appropriately defined. Calculate the x - and y -components of the acceleration field.
8) A cylindrical tank of water rotates in solid-body rotation, counterclockwise about its vertical axis at angular speed n = 360 rpm. Calculate the vorticity of fluid particles in the tank.
9) Consider the following steady, three-dimensional velocity field: Calculate the vorticity vector as a function of space ( x , y , z).
10) Consider the general form of the Reynolds transport theorem (RTT) given by
where ⃗r is the velocity of the fluid relative to the control surface. Let Bsys be the mass m of a system of fluid particles. We know that for a system, dm/dt = 0 since no mass can enter or leave the system by definition. Use the given equation to derive the equation of conservation of mass for a control volume.
11) Consider the general form of the Reynolds transport theorem (RTT). Let Bsys be the linear momentum m⃗of a system of fluid particles. We know thatfor a system, Newton’s second law
is
Use the RTT Equation and this equation to derive the equation of conservation of linear momentum for a control volume.
12) Consider steady flow of air through the diffuser portion of a wind tunnel. Along the centerline of the diffuser, the air speed decreases from uentrance to uexit as sketched. Measurements reveal that the centerline air speed decreases parabolically through the diffuser. Write an equation for centerline speed u( x ), based on the parameters given here, from x = 0 to x = L.
Solution
We are to write an equation for centerli ne speed through a diffuser, given that the
flow speed decreases parabolically.
Assumptions 1 The flow is steady. 2 The flow is axisymmetric. Analysis A general equation for a parabola in x is General parabolic equation:
(1)
We have two boundary conditions, namely at x = 0, u = uentrance and at x = L, u = uexit. By inspection, Eq. 1 is satisfied by setting c = 0, a = uentrance and b = (uexit - uentrance)/ L2. Thus, Eq. 1 becomes
13) For the velocity field of above problem 12, calculate the fluid acceleration along the diffuser centerline as a function of x and the given parameters. For L = 2.0 m, uentrance = 30.0 m/s,and uexit = 5.0 m/s, calculate the acceleration at x = 0 and x = 1.0 m. Solution
We are to generate an expression for the fluid acceleration for a given velocity,
and then calculate its value at two x locations.
Assumptions 1 The flow is steady. 2 The flow is axisymmetric. Analysis In the previous problem, we found that along the centerline,
Discussion: a x is negative implying that fluid particles are decelerated along the centerline of
the diffuser, even though the flow is steady. Because of the parabolic nature of the velocity field, the acceleration is zero at the entrance of the diffuser, but its magnitude increases rapidly downstream.
14) Given the eulerian velocity vector field as below. find the total acceleration of a particle. Solution:
15) Determine the family of
V
( x 2
2
y ) i
functions that will yield the velocity field
2 xy j .
Solution
Velocity field: V
( x
2
2
y ) i
Assumptions Steady state condition Incompressible fluid flow 2 - D problem
2 xy j
u
x
2
2
y & v
2 xy
Stream function (incompressible fluid flow version) definition: u
&
v
y
x
With the definition of stream function and the given velocity components: u
2
x y
y
v
x
2
… (1)
2 xy … (2)
2
2
(1) ( x y ) y
2 2 ( x y ) y
1
x 2 y y 3 f ( x) , 3
where f(x) is any function of x including constants. Using this obtained, take
Comparing (2) with (3),
Finally,
x
2
1
y
3
y
3
that will be
x
df ( x ) dx
x
2 xy 0
df ( x ) dx
… (3)
0 , that represents f(x) = constant .
const .
The stream function, , exists; therefore, the velocity field satisfies the continuity equation, u x
v y
0 , and also it can be said that it is a valid velocity field.
16) An incompressible velocity field is given by u=a(x2 - y2) , v = unknown , w = b where a and b are constants. What must the form of the velocity component v be? Solution Applying continuity equation
This is the only possible form for v that satisfies the incompressible continuity equation. The function of integration f is entirely arbitrary since it vanishes when v is differentiated with respect to y . 17)
