A steady, three-dimensional velocity field is given by V → = ( u , v , w ) = ( 2.49 + 1.36 x − 0.867 y ) i → + ( 1.95 x − 1.36 y ) j → + ( − 0.458 x y ) k → Calculate me vorticity vector as a function of space variables (x, y, z).
A steady, three-dimensional velocity field is given by V → = ( u , v , w ) = ( 2.49 + 1.36 x − 0.867 y ) i → + ( 1.95 x − 1.36 y ) j → + ( − 0.458 x y ) k → Calculate me vorticity vector as a function of space variables (x, y, z).
A steady, three-dimensional velocity field is given by
V
→
=
(
u
,
v
,
w
)
=
(
2.49
+
1.36
x
−
0.867
y
)
i
→
+
(
1.95
x
−
1.36
y
)
j
→
+
(
−
0.458
x
y
)
k
→
Calculate me vorticity vector as a function of space variables (x, y, z).
Quantities that have magnitude and direction but not position. Some examples of vectors are velocity, displacement, acceleration, and force. They are sometimes called Euclidean or spatial vectors.
The velocity field for a fluid flow is given by following expression:
=(0.2x² + 2y+2.5)î +(0.5x+2y² – 6) ĵ+(0.15x² + 3y° + z)k
The strain tensor at (2,1,–1) will be:
0.8
1.25 0.30
a) | -1.25
-4
0.30
-1
(0.8 1.25 0.70
b) | 1.25
2
0.30
-2
1
0.8 1.25 0.30)
c) | 1.25
4
-2
0.30
-2
1
0.8 1.25 0.30
d) | 1.25
8.
-2
0.8
2
1
Consider a three-dimensional, steady velocity field given by
V = (u, v, w) = (3.2 + 1.4x)i + (2.4 – 2.1y)j + (w)k.
If the w-velocity is only a function of z, and the magnitude of w-velocity at z = 0 is 5, find the
velocity field of w if the flow is known to be incompressible.
A fluid has a velocity field defined by u = x + 2y and v = 4 -y. In the domain where x and y vary from -10 to 10, where is there a stagnation
point? Units for u and v are in meters/second, and x and y are in meters.
Ox = 2 m. y = 1 m
x = 2 m, y = 0
No stagnation point exists
x = -8 m, y = 4 m
Ox = 1 m, y = -1 m
QUESTION 6
A one-dimensional flow through a nozzle has a velocity field of u = 3x + 2. What is the acceleration of a fluid particle through the nozzle?
Assume u, x and the acceleration are all in consistent units.
O 3 du/dt
9x + 6
1.5 x2 + 2x
O O
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