3rd Homework Solution Manual
Problem 1 Part A: (4 points for derivation, 0.5 points for the answer and 0.5 points for the units)
The free body diagram for this problem is presented above. In order for the door to swing open the
moment created by the water pressure has to offset the weight moment. Figure above shows the
positive moment (M), x, y, orientations water height (H) and pressure vectors (P) but you can define
these as you want as long as you stay consistent with your definition.
Because the door in this example is rectangular the weight is applied at the centroid and the moment
can be calculated as:
𝑀𝑑 = 𝑊𝑡 ∗
𝐿
2
The moment generated by the water pressure is calculated bellow
𝐿
𝑀𝑤 = 𝐹 ∗ 𝑥 = 𝑃 ∗ 𝐴 ∗ 𝑥 = ∫ 𝑥 ∗ 𝑝 𝑑𝐴 = ∫ 𝑝 𝑤 𝑥 𝑑𝑥 = 𝑝𝑤 ∫ 𝑥 𝑑𝑥
0
Setting the weight moment equal to water pressure moment would yield the following
𝑝=
𝑊𝑡
𝐿𝑤
Since the pressure in a static incompressible fluid equals
𝑃 = 𝜌𝑔𝐻
the height of the water needed to open the door can be computed as:
𝐻=
𝑊𝑡
=
𝐿 𝑤 𝜌 𝑔 (3𝑓𝑡)(3𝑓𝑡)
400𝑙𝑏𝑚
= 0.71𝑓𝑡
𝑠𝑙𝑢𝑔𝑠
𝑙𝑏𝑚
(1.94
)
(32.174
)
𝑠𝑙𝑢𝑔
𝑓𝑡 3
Part B: (4 points for derivation, 0.5 points for the answer and 0.5 points for the units)
This is similar to part A but now but because the hinge is replaced by a rail, the door can only move
up/down and the door will open when the force created by the water will offset the door’s weight.
𝐿
𝑊𝑡 = 𝑃 ∗ 𝐴 = ∫ 𝑝 𝑤 𝑑𝑥
0
Again, since the fluid is incompressible we can rewrite equation above as
𝐻=
𝑊𝑡
400𝑙𝑏𝑚
=
= 0.71𝑓𝑡
𝐿 𝑤 𝜌 𝑔 (3𝑓𝑡)(3𝑓𝑡) (1.94 𝑠𝑙𝑢𝑔𝑠) (32.174 𝑙𝑏𝑚 )
𝑠𝑙𝑢𝑔
𝑓𝑡 3
Note: answers for parts A and B are identical only because the door was rectangular and horizontal.
Problem 2 Part A (2 for a, d, e, f and 1 point for b, c)
u  F (r ) x means that the velocity is a product of the position vector and a scalar function F. To
multiply a vector by a scalar, multiply each component of a vector by that scalar.
The position vector is:
𝑥⃗ = 𝑥 𝑖 + 𝑦 𝑗 + 𝑧𝑘
And F * x becomes
𝐹 ∗ 𝑥 = 𝐹𝑥 𝑖 + 𝐹𝑦 𝑗 + 𝐹𝑧 𝑘
Since:
𝑟 = √𝑥 2 + 𝑦 2 + 𝑧 2
i,j and k components of velocity become:
𝑢 = 𝐹 (√𝑥 2 + 𝑦 2 + 𝑧 2 ) 𝑥
𝑣 = 𝐹 (√𝑥 2 + 𝑦 2 + 𝑧 2 ) 𝑦
𝑤 = 𝐹 (√𝑥 2 + 𝑦 2 + 𝑧 2 ) 𝑧
Part B
This vector field below was plotted using Maple. As can be seen, the fluid is moving away from the origin
at an increasing rate.
Part C
This vector field below was plotted using Maple. As can be seen, the fluid is moving toward the origin at
a decreasing rate.
Part D
The divergence of U is defined as:
∇∗𝑉 =
𝑑𝑢 𝑑𝑣 𝑑𝑤
+
+
𝑑𝑥 𝑑𝑦 𝑑𝑧
Since F is constant
∇∗𝑉 =
𝑑(𝐹𝑥) 𝑑(𝐹𝑦) 𝑑(𝐹𝑧)
+
+
= 𝐹 + 𝐹 + 𝐹 = 3𝐹
𝑑𝑥
𝑑𝑦
𝑑𝑧
Positive values value of velocity divergence means that the fluid is expanding. See the 2nd homework
help session notes for more information about the divergence.
Part E
Since F is defined as a function of r, you should use the spherical coordinates to solve this part and
define X as the following:
𝑥⃗ = 𝑟 𝑟⃗
After you multiply x by F you should get
𝑢
⃗⃗ = 𝐴 𝑟 −𝑛 ∗ 𝑟
The divergence of u in spherical coordinates is
Since we only have the radial component of u, the divergence of u will become
∇∗𝑈 =
(𝐴 𝑟 3−𝑛 )
1
𝑟 2 (𝐴 𝑟1−𝑛 )
1
1
(
𝑑
=
(
𝑑
(
))
(
)) = 2 ( (3 − 𝑛)𝐴𝑟 2−𝑛 ) = (3 − 𝑛)𝐴𝑟 −𝑛
2
2
𝑟
𝑑𝑟
𝑟
𝑑𝑟
𝑟
The fluid will be incompressible if n = 3 or A = 0
The fluid will be expanding if A > 0 and n <3 or A < 0 and n > 3
The fluid will be contracting if A > 0 and n >3 or A < 0 and n <3
Part F
The volumetric flow rate is
𝑄 = ∮ 𝑢 𝑑𝐴
And the Gausses’ theorem states
∮ 𝑢 𝑑𝐴 = ∭ ∇𝑢 𝑑𝑉
Since the volume of a sphere is
4
𝑉 = 𝜋 𝑟3
3
And a derivative of that is
𝑑𝑉 = 4𝜋𝑟 2 𝑑𝑟
We can calculate the flow rate as
𝑅
𝑅
𝑅
𝑄 = ∫ ∇𝑢 4𝜋𝑟 2 𝑑𝑟 = ∫ ∇𝑢 4𝜋𝑟 2 𝑑𝑟 = 4𝜋𝐴 ∫ (3 − 𝑛)𝑟 2−𝑛 𝑑𝑟 = 4𝜋𝐴𝑅 3−𝑛
0
0
0
Problem 3(Part A -> 3 pts for derivations, 1 pts for answer, Part B -> 5pts for derivation, 1 pt 6 points for
answer)
Schematic
3
1
2
Part A
To solve this problem simply apply the definition of Volume Flow Rate
10𝑐𝑚3
𝑄
12.73𝑐𝑚
𝑠
𝑄 =𝑉∗𝐴 → =𝑉 =
=
2
𝐴
0.25𝜋 𝑐𝑚
𝑠
Part B
Apply the conservation of volume (since density is constant) to the control volume shown in a schematic
above. This control volume has 3 surfaces. The first is a hemisphere, with a radius of 10cm. The second
is the circular hole and the third is a circle with a hole in the middle. Remember that if the fluid is leaked
through one of the surfaces of this control volume then some other fluid has to replace it through
another control volume. Since fluid cannot flow through the third control surface, there is a wall there;
the flow rate through the 1st control surface has to match the flow rate through the 2nd control surface.
𝑄1 = 𝑄2 = 𝑉2 ∗ 𝐴2
Area 2 is the surface area of the hemisphere (2πr2) and V2 is the mean velocity flowing perpendicular to
the hemisphere. Therefore
10𝑐𝑚3
𝑄1
𝑐𝑚
𝑠
𝑉2 =
=
= 0.016
2
𝐴2
2𝜋(10𝑐𝑚)
𝑠
Problem 4(2 points for the plot, 4 points for correct integral setup, 1 point for flow rate answer, 2 points
for part c derivation, 1 point for correct answer on part c)
Schematic
Part A
Velocity Profile
12
Velocity (cm/s)
10
8
6
4
2
0
-2
-1
0
1
Distance from the center (cm)
Part B
𝑟 2
𝑄 = ∫( 10 ( 1 − ( ) ) (2πr dr) = 20𝜋 𝑐𝑚3 /𝑠
𝑅
Part C
𝑉=
𝑄 20𝜋 𝑐𝑚3 /𝑠
=
= 5 𝑐𝑚/𝑠
𝐴 𝜋(2𝑐𝑚)^2
Note: Centerline velocity is twice the average velocity in a round pipes.
2
Problem 5(2 points for the plot, 4 points for correct integral setup, 1 point for flow rate answer, 2 points
for part c derivation, 1 point for correct answer on part c)
Part A
Velocity Profile
Velocity (cm/s)
12
10
8
6
4
2
0
0
0.5
1
1.5
Distance from the lower surface (cm)
Part B
2
𝑄 = ∫ 10𝑦 (2 − 𝑦)𝑑𝑦 =
0
40
𝑐𝑚2
= 13.33
3
𝑠
Part C
𝑉=
𝑄 13.3 20
𝑐𝑚
=
=
= 6.67
𝐴
2
3
𝑠
Note: Q in this problem is volume flow rate per width of the channel.
2