MCEN3002 2017 Tutorial 2 Solution PDF

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Practice Tutorial Question and Detailed Solution For Applied Fluid Mechanics. A unit in Mechanical Engineering...

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Curtin University Department of Mechanical Engineering

MCEN3002 Applied Fluid Mechanics Semester 2 – 2017 Tutorial 2 – Solution

Problem 1

Solution ~u = (3y + 2)iˆ+ (x − 8) jˆ + 5zkˆ

⇒

u = 3y + 2 v = x−8

w = 5z The flow speed U is the magnitude of the velocity vector: p U = |~u| = u2 + v2 + w2

At the point (x, y, z) = (0, 0, 0), we obtain: q U = (2)2 + (−8)2 + (02 ) ≈ 8.25 m.s−1

At the line x = z = 0, we obtain: U=

q

p

(3y + 2)2 + (−8)2 + (02 )

9y2 + 12y + 4 + 64 p = 9y2 + 12y + 68 m.s−1 =

Problem 2

Solution

with y in m

−V0 y V0 x u= p ,v=p 2 2 x +y x2 + y2

The flow speed U is the magnitude of the velocity vector: p U = |~u| = u2 + v2

Where in the flowfield is the speed equal to V0 ? p U = V0 ⇒ u2 + v2 = V0 v !2 u u −V y 0 + ⇒ t p x2 + y2 s V02 (y2 + x2 ) ⇒ = V0 x2 + y2 q ⇒ V 02 = V0 ⇒

V0 = V0

1

V0 x p

x2 + y2

!2

= V0

Thus, U = V0 throughout the entire flow field The slope at any point along the streamline is given by: V0 x dy v = dx u

p

dy x2 + y2 = dx p−V0 y x2 + y2 x dy =− dx y y dy = −x dx

⇒ ⇒ ⇒

x2 y2 + constant = − + constant 2 2 y2 + x2 = constant

⇒ ⇒

Thus, the fluid flow with circular streamlines and the speed is constant Problem 3

Solution u = x2 y

v = −xy2

The slope at any point along the streamline is given by: dy v = dx u

⇒ ⇒ ⇒ ⇒ ⇒

Problem 4

−xy2 dy = 2 x y dx y dy =− dx x 1 1 dy = − dx y x ln y + constant = − ln x + constant ln x + ln y = constant

⇒

ln xy = constant

⇒

xy = constant

Solution v = x2 − y2

u = 2xy If the two-dimensional flow is irrotational, then: ωz =

  1 ∂v ∂u =0 − 2 ∂x ∂y

For the given velocity components: ωz =

1 2



∂v ∂u − ∂x ∂y



1 = (2x − 2x) = 0 2

Thus, the flow is irrotational If the two-dimensional flow satisfies the conservation of mass, then: ∂u ∂v + =0 ∂x ∂y

2

For the given velocity components: ∂u ∂v = 2y + (−2y) = 0 + ∂x ∂y Thus, the flow satisfies the conservation of mass Problem 5

Solution ψ = 3x2 y − y3

(a) Sketch the streamlines passing through the origin: Lines of constant ψ are streamlines. The streamline passing through the origin (x = y = 0) has a value ψ = 0. Therefore, the equation for this streamline is: 3x2 y − y3 = 0

⇒

√ y = ± 3x

y=0

and

y √ 3

ψ =0

ψ =0 x

1

√ − 3

ψ =0

(b) Determine the rate of flow across the straight path AB: The volume rate of flow q crossing the line AB can be calculated from the streamlines passing through the points A and B: q = ψB − ψA At B: x = 0 and y = 1 m, so that ψB = 3(0)2 (1) − (1)3 = −1 m2 .s−1 At A: x = 1 and y = 0 m, so that ψA = 3(1)2 (0) − (0)3 = 0 m2 .s−1

Thus q = −1 m2 .s−1 The negative sign indicates that the flow is from right to left as we look from A to B y 1

B q

A 1

3

x...