Sd282 Asg1

FLUID MECHANICS I Solution 1 Question 1: Example 4.1 p227 The solution is given in p227 of the textbook.

Question2 For a fluid flow with a velocity field: ˆ ~ (x, y, z, t) = 2xˆi − 2yˆj + 0k[m/s] V estimate the fluid velocity and acceleration at the point (2, 1, 0)[m]. Calculation of the velocity: u(x, y, z, t) = 2x[m/s] u(2, 1, 0, t) = 4[m/s] or

v(x, y, z, t) = −2y[m/s] v(2, 1, 0, t) = −2[m/s]

w(x, y, z, t) = 0m/s w(2, 1, 0, t) = 0[m/s]

ˆ ~ (2, 1, 0, t) = 4ˆi − 2ˆj + 0k[m/s] V Calculation of the acceleration: ax

≡

ay

≡

az

≡

Du ∂u ∂u ∂u ∂u = +u +v +w = 4x[m/s2 ] Dt ∂t ∂x ∂y ∂z ∂v ∂v ∂v ∂v Dv = +u +v +w = 4y[m/s2 ] Dt ∂t ∂x ∂y ∂z Dw ∂w ∂w ∂w ∂w = +u +v +w = 0[m/s2 ] Dt ∂t ∂x ∂y ∂z

Therefore ax (2, 1, 0, t) =

8[m/s2 ]

ay (2, 1, 0, t) = 4[m/s2 ] az (2, 1, 0, t) = 0[m/s2 ] or

2 ˆ ~a(2, 1, 0, t) = 8ˆi + 4ˆj + 0k[m/s ]

Note that the plots were produced using a math package (Matlab) for illustration. However, use of such a package will not be required for the exams. 1

Velocity Field Plot 5

4

y [m]

3

2

1

0

−1 −6

−4

−2

0 x [m]

2

4

6

4

6

Figure 1: Velocity field plot Acceleration Field Plot 6

5

4

y [m]

3

2

1

0

−1 −6

−4

−2

0 x [m]

2

Figure 2: Acceleration field plot

2

Question 3: P4.5 ~ (x, y, z, t) = Uo x ˆi − Uo y ˆj + 0 kˆ V L L with Uo and L constants. a) Acceleration components ³ x ´ µU ¶ ³ ∂u ∂u ∂u ∂u y´ o ax = +u +v +w = (0) + U0 + −Uo (0) + 0 ∂t ∂x ∂y ∂z L L L µ ¶ ³ x´ ³ U0 ∂v ∂v ∂v ∂v y´ − ay = +u +v +w = (0) + U0 (0) + −Uo +0 ∂t ∂x ∂y ∂z L L L ∂w ∂w ∂w ∂w az = +u +v +w = 0 m.s−2 ∂t ∂x ∂y ∂z The resultant acceleration is ~a =

U02 ˆ U02 ˆ U02 x i + 2 y j = 2 ~r L2 L L (1)

b) |a| = 25 =

U02 p 2 1 + 12 1.52

U0 = 6.3 m/s.

Question 4: P4.6 ~ = V0 (1 + 2 x ) ˆi + 0 ˆj + 0kˆ V L a) The acceleration components are ∂u ∂u ∂u ∂u 2V 2 ax = +u +v +w = 0 ∂t ∂x ∂y ∂z L

µ ¶ 2x 1+ L

ay = 0.0 m.s−2 az = 0.0 m.s−2 At x=L, the acceleration is ~a =

6V02 ˆ i L 3

Question 5: P4.7 µ ¶ 3 ~ = u ˆi = U0 1 + R V x3 a) The acceleration components are µ ¶µ ¶ R3 R3 ax = U0 1 + 3 −3U0 4 x x ay = 0.0 m.s−2 az = 0.0 m.s−2 The maximum occurs where dax =0 dx i.e. at x = −(7R3 /4)1/3 ≈ −1.205 R.

Question 6: Example 1.13 p41 The solution is given in p41 of the textbook.

Question 7: P1.81 ~ = Kxt ˆi − Kytˆj + 0 kˆ V The flow is unsteady and two-dimensional. To find the streamline equation use dx dy = u v dx dy = Kxt −Kyt The terms K and t both vanish and leave us with the same results as in Example 1.13 p41, that is, Z x Z y dx −dy = y x x y0 µ0 ¶ µ ¶ x y0 ⇒ ln = ln x0 y ⇒ xy = x0 y0 = Constant Note that you integrate from x0 ,y0 to x,y which corresponds to the streamline passing through a point (x0 , y0 ). The streamlines have exactly the same shape as in Fig. 1.13 p 41. However the flow is accelerating with increasing time. 4

Question 8: P1.82 ~ = V0 cosθ ˆi + V0 sinθ ˆj + 0 kˆ V where V0 and θ are constants. Streamline equation passing through point (x0 , y0 ) dx dy dx dy = = = u v V0 cosθ V0 sinθ dy ⇒ = tanθ dx Integrate such as Z

Z

y

x

dy = y0

tanθdx x0

⇒ y = (tanθ)x + y0 − (tanθ)x0 = (tanθ) x + Constant The streamlines are straight parallel lines which make an angle θ with the x axis. The velocity field represents a uniform stream moving upward at angle θ.

Question 9 The fluid flow in the region above the two planes : y = x and y = −x can be approximated by the velocity field: ~ (x, y, z, t) = −0.004 V

x ˆ y ˆ ˆ i − 0.004 2 j + 0k[m/s] x2 + y 2 x + y2

1. Make a sketch of the region in which the fluid flows for 0[m] < y < 1.0[m], See Figure 3. 1.2 1

y [m]

0.8 0.6 0.4 0.2 0 −0.2 −1

−0.5

0 x [m]

0.5

Figure 3: Region of interest

5

1

2. Plot the streamlines in the flow region ( 0[m] < y < 1.0[m]), Along any streamline that goes through the point (x0 , y0 ) in the flow region: vdx − udy x y dx + 0.004 2 dy −0.004 2 x + y2 x + y2 dx dy − x y

=

0 or

=

0 or

=

0

Now integrate from the point (x0 , y0 ) to another point (x, y) on the streamline: Z x Z y dy y x dx − =0⇒ = x y y x 0 0 y0 x0 This is the equation of a straight line which goes through the origin and the point (x0 , y0 ). Figure 4 shows a set of streamlines obtained by varying the reference point (x0 , y0 ). 1.2

1

0.8

y [m]

0.6

0.4

0.2

0

−0.2 −1

−0.8

−0.6

−0.4

−0.2

0 x [m]

0.2

0.4

0.6

0.8

1

Figure 4: Streamline plot 3. What is the fluid acceleration at the points (0.0[m], 0.5[m]) and (0.1[m], 0.5[m])? ax

= =

ay

= =

∂u ∂u ∂u ∂u +u +v +w ∂t ∂x ∂y ∂z x 2 2 −(0.004) 2 [m/s ] (x2 + y 2 ) ∂v ∂v ∂v ∂v +u +v +w ∂t ∂x ∂y ∂z y 2 2 −(0.004) 2 [m/s ] (x2 + y 2 ) 6

These expressions were evaluated with Maple. Therefore ~a(0.0[m], 0.5[m])

=

~a(0.1[m], 0.5[m])

=

2 ˆ 0ˆi − 1.28 · 10−4 ˆj + 0k[m/s ] −5ˆ −4 ˆ 2 ˆ −2.37 · 10 i − 1.18 · 10 j + 0k[m/s ]

Notice that the fluid parcels accelerate towards the origin. 4. Is the velocity field realistic in the region of the origin, (0.0[m], 0.0[m])? At the origin the fluid velocity and acceleration approach ∞ so the velocity field is NOT realistic in the region of the origin.

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