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Exercises I of General Laws

Module by: Liqun Wang. E-mail the author

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Summary: The exercises I of general laws.

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Exercise 1

For the steady, incompressible flow shown, find the relation between the inlet and outlet velocities, V 1 V 1 and V 2 V 2 . Apply conservation of mass in integral form.

Figure 1:
Figure 1 (fig13.png)

Solution

In the diagram above, the control volume is denoted by a dotted line. Note that we chose the control volume such that no fluid passes through the sides of the volume because they coincide with the walls of the channel. When integration is performed, then, only the inlet and outlet of the channel need be considered.

The steady conservation of mass in integral form is

∫c.v.∂∂tρdV+∫c.s.a.ρ V → · n → dA=0 V c.v. t ρ A c.s.a. ρ V → · n → 0 (1)
We know that ∂∂tρ≡0 t ρ 0 because the fluid is incompressible, so
∫c.v.∂∂tρdV=0 V c.v. t ρ 0 (2)
As a result,
∫c.s.a.ρ V → · n → dA=0 A c.s.a. ρ V → · n → 0 (3)
In order to evaluate all the terms in this integral, we will walk around the control surface (which is represented by the dotted line in the above diagram). The density ρ drops out of the integral because it is constant. Again, remember that the sides of the control surface parallel to the walls of the channel contribute nothing because V → ⊥ n ^ V → ⊥ n ^ along these sides, which makes their dot product equal to zero.
∫c.s.a. V → · n → dA= ∫ c.s.a. ⁢ V → · n → ⁢ dA in + ∫ c.s.a. ⁢ V → · n → ⁢ dA out A c.s.a. V → · n → ∫ c.s.a. ⁢ V → · n → ⁢ dA in ∫ c.s.a. ⁢ V → · n → ⁢ dA out (4)
so
∫c.s.a. V → · n → dA= V → 1 · n → 1 A 1 + V → 2 · n → 2 A 2 =- V 1 A 1 + V 2 A 2 =0 A c.s.a. V → · n → V → 1 · n → 1 A 1 V → 2 · n → 2 A 2 V 1 A 1 V 2 A 2 0 (5)
or
V 1 = A 2 A 1 V 2 V 1 A 2 A 1 V 2 (6)

Exercise 2

The velocity field of an incompressible steady flow is given by

u= V ox L u V ox L (7)
v=0 v 0 (8)
w=- V oz L w V oz L (9)
Apply the integral form of the conservation of energy and find the flow volume across each cross-sectional area. The unit depth is b.

Solution

Figure 2:
Figure 2 (fig14.png)

V → =u i ^ +w k ^ V → u i ^ w k ^ (10)
V → · n → 1 = u ⁢ i ^ + w ⁢ k ^ · k ^ z=L = - V oz L z=L =- V 0 V → · n → 1 u ⁢ i ^ + w ⁢ k ^ · k ^ z=L - V oz L z=L V 0 (11)
Q 1 =∫(1) V → · n → 1 d A 1 =∫0L- V 0 bdx=- V 0 bL Q 1 A 1 (1) V → · n → 1 x 0 L V 0 b V 0 b L (12)
(Here the minus sign indicates that the flow is entering the volume in question.)
Q 2 =∫(2) V → · n → 2 d A 2 =∫x=0L u i ^ +w k ^ ·12i−k 2dx Q 2 A 2 (2) V → · n → 2 x x=0 L u i ^ w k ^ · 1 2 i k 2 (13)

Figure 3:
Figure 3 (fig15.png)

As seen in the above figure,

dL=2dx dL 2 2 dx (14)
so, with z = x (equation of surface (2)).
Q 2 =∫0L V 0 Lx+zdx Q 2 x 0 L V 0 L x z (15)
and because
∫ 0 L 2x= x 2 0 L =L2 ∫ 0 L 2 x x 2 0 L L 2 (16)
therefore,
Q 2 = V 0 bL Q 2 V 0 b L (17)
Q 3 =∫(3) V → · n → 3 d A 3 =u i ^ +w k ^ ·i=u= V 0 ⁢ X L x=0 =0 Q 3 A 3 (3) V → · n → 3 u i ^ w k ^ · i u V 0 ⁢ X L x=0 0 (18)

Let's check this answer. Since the fluid is incompressible, no fluid may accumulate inside of our fixed control volume. Therefore, the conservation of mass yields Q 1 + Q 2 + Q 3 =0 Q 1 Q 2 Q 3 0 , which is satisfied by the answers given above.

Exercise 3

Water, which may be assumed incompressible, is poured at a constant rate Q 1 Q 1 into a container. Water also leaks out through a hole at the bottom of the container at a speed proportional to the height of water in the container: V 2 =kh V 2 k h .

Figure 4: figure 16
Figure 4 (fig16.png)

  • (a) Find h(t).
  • (b) What is the steady state elevation, h ∞ h ∞ , that will be reached as t→∞ t ∞ ?
  • (c) Obtain a solution for h(t) for when the tank is empty before the flow begins.

Solution

Solution of (a): Conservation of mass ρ = const.

ddtV= Q 1 − Q 2 t V Q 1 Q 2 (19)
Q 1 − V 2 A 2 = A 1 ddth Q 1 V 2 A 2 A 1 t h (20)
V 2 =kh V 2 k h (21)
Q 1 = A 1 ddth+k A 2 h Q 1 A 1 t h k A 2 h (22)

Figure 5:
Figure 5 (fig17.png)

Solution of (b): h ∞ h ∞ , that will be reached as t→∞ t ∞ ?

At SS:

ddth=0 t h 0 (23)
h= h ∞ h h ∞ (24)
Q 1 =k A 2 h Q 1 k A 2 h (25)
h ∞ = Q 1 k A 2 h ∞ Q 1 k A 2 (26)

Solution of (c): As found in part (a), the governing differential equation of this system is

A 1 ddth+k A 2 h= Q 1 A 1 t h k A 2 h Q 1 (27)
The initial condition is when t = 0,
h=0 h 0 (28)
The general solution to this differential equation is easily found, after which the constant of integration is found by applying the initial condition.
h= c 1 ⅇ-k A 2 A 1 t+ Q 1 k A 2 h c 1 k A 2 A 1 t Q 1 k A 2 (29)
because h = 0 at t = 0,
0= c 1 + Q 1 k A 2 0 c 1 Q 1 k A 2 (30)
so,
c 1 =- Q 1 k A 2 c 1 Q 1 k A 2 (31)
therefore,
h= Q 1 k A 2 1−ⅇ-k A 2 A 1 t h Q 1 k A 2 1 k A 2 A 1 t (32)

Note that, as t→∞ t ∞ , the solution converges to the same steady-state height h ∞ h ∞ as was found in part (b). This helps confirm this solution.

Exercise 4

Consider a flow field of an incompressible fluid (for which divV= ∇ → · V → =0 V ∇ → · V → 0 ). Using a cylindrical coordinate system, we are given

V r =Aωsθr2 V r A ω s θ r 2 (33)
and
V z ≡0 V z 0 (34)
PROBLEM: Determine V θ rθ V θ r θ such that continuity is satisfied.

Solution

SOLUTION: We apply continuity in cylindrical coordinates for an incompressible fluid. (Refer back to Table if you need help remembering the ∇ → ∇ → operator in cylindrical coordinates! )

divV= ∇ → · V → =1r∂∂rr V r +1r∂∂θ V θ =0 V ∇ → · V → 1 r r r V r 1 r θ V θ 0 (35)
Separating the unknown V θ V θ ,
∂∂θ V θ =-∂∂rr V r =-∂∂rAcosθr=Acosθr2 θ V θ r r V r r A θ r A θ r 2 (36)
The value of V θ V θ is then found by integration. Remember that as a result of the integration over θθ, some function of r only is required.
V θ =Asinθr2+fr V θ A θ r 2 f r (37)

Exercise 5

PROBLEM: A flow field in polar coordinates has V z =0 V z 0 , V θ =0 V θ 0 , and V r ≠0 V r 0 . Apply continuity to find the permissible velocity distribution for an incompressible fluid.

Solution

SOLUTION: Continuity for an incompressible fluid in polar coordinates is

∇ → · V → =0 ∇ → · V → 0 (38)
or
1r∂∂rr V r +1r∂∂θ V θ +∂ V z ∂z=0 1 r r r V r 1 r θ V θ ∂ V z ∂z 0 (39)
Since V θ V θ and V z V z are both equal to zero, we are left with
1r∂∂rr V r =0 1 r r r V r 0 (40)
Integrating gives
r V r =constant r V r constant (41)
or
V r =constantr V r constant r (42)

Exercise 6

For an incompressible flow, the velocity field is V → =2xi−2yj V → 2 x i 2 y j . Determine and plot the stream function.

Solution

u=∂∂yψ=2x u y ψ 2 x (43)
v=-∂∂xψ=-2y v x ψ 2 y (44)
Integrate to find ψψ:
ψ=∫∂∂yψdy+fx=2xy+fx ψ y y ψ f x 2 x y f x (45)
v=-∂∂xψ=-2y−∂∂xf=-2y v x ψ 2 y x f 2 y (46)
Therefore,
∂∂xf=0 x f 0 (47)
or
fx=constant f x constant (48)
Strictly speaking, the value of the constant does not matter; even in finding flow rates through areas, the important quantity is the difference in the value of the stream function across the area. Hence, in this case, we may set the constant equal to zero. Results:

Figure 6:
Figure 6 (fig22.png)

Exercise 7

A uniform, steady incompressible flow field has a velocity component u of 1 m/s and a velocity component v of 3 m/s. Determine the expression for the stream function φφ.

Solution

SOLUTION: The stream function is defined in terms of two partial differential equations, one with respect to x and the other with respect to y. Integrate both over the appropriate direction; this will yield two expressions for φφ and the constants of integration may be easily found.

u=1m/s u 1 m/s (49)
v=3m/s v 3 m/s (50)
u=∂∂yψ u y ψ (51)
ψ=∫udy+fx ψ y u f x (52)
ψ=∫1dy+fx ψ y 1 f x (53)
ψ=y+fx ψ y f x (54)
v=-∂∂xψ v x ψ (55)
ψ=-∫vdx+fy ψ x v f y (56)
ψ=-3x+fy ψ 3 x f y (57)
Since ψ=y+fx=-3x+fy ψ y f x 3 x f y ,
ψ=y−3x ψ y 3 x (58)

Exercise 8

The velocity components of a two-dimensional incompressible flow are u=2x+y u 2 x y and v=-x−2y v x 2 y . Is the continuity satisfied? Find the stream function.

Solution

SOLUTION: We are given u=2x+y u 2 x y and v=-x−2y v x 2 y

∂∂xu+∂∂yv=0 x u y v 0 (59)
2−2=0 2 2 0 (60)
Clearly, continuity is satisfied.
u=2x+y=∂∂yψ u 2 x y y ψ (61)
so can infer into
ψ=2xy+y22+fx ψ 2 x y y 2 2 f x (62)
∂∂xψ=2y+∂∂xfx x ψ 2 y x f x (63)
∂∂xψ=x+2y x ψ x 2 y (64)
∴∂∂xfx=x ∴ x f x x (65)
Integrate, remembering to add a constant of integration:
fx=x22+C f x x 2 2 C (66)
Substitution yields
ψ=2xy+y22+x22+C ψ 2 x y y 2 2 x 2 2 C (67)

Exercise 9

The velocity components of a steady, two-dimensional incompressible flow field are u = 2y, v = 4x. Determine the corresponding stream function and show on a sketch several streamlines. Indicate the direction of flow along the streamlines.

Solution

The solution is as follows:

u=2y=∂∂yψ u 2 y y ψ (68)
So,
ψ=2y22+ f 1 x ψ 2 y 2 2 f 1 x (69)
v=4x=-∂∂xψ v 4 x x ψ (70)
So,
ψ=-2x2+ f 2 x ψ 2 x 2 f 2 x (71)
In order to satisfy both expressions for the stream function, we have
ψ=-2x2+y2+c ψ 2 x 2 y 2 c (72)
with c = 0, which is an arbitrary constant,
ψ=-2x2+y2 ψ 2 x 2 y 2 (73)
When ψ=0 ψ 0 , streamline,
0=-2x2+y2 0 2 x 2 y 2 (74)
So,
y=±2x y ± 2 x (75)
When ψ≠0 ψ 0 , streamline
1=-x2ψ2+y2ψ 1 x 2 ψ 2 y 2 ψ (76)
hyperbola. Thus, the streamlines are a family of hyperbolas with the ψ=0 ψ 0 streamline as asymtotes.

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