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Unformatted text preview: Date: 28 Sept. 2010 CE340 Test 1 Last Name: SOLUTION KEY
Write your name on each sheet of paper that you hand in.
Read all questions very carefully. If the problem statement is not clear, you should
ask the proctors present to clarify if possible the question.
Pay particular attention to consistent units.
Please attempt to do all work in an organized manner, so that the papers can be graded generously.
Draw and label all diagrams neatly if they are to be considered in the grading. Circle all main
results including the ﬁnal result, which you wish to be considered in the grading. If you write on
the back of a sheet, or on additional paper, please note on your ﬁrst sheet that you have done work
on additional sheets and where the additional work is to be found.
ALL RELEVANT WORK SHOULD BE SHOWN. MARKS MAY BE DEDUCTED
IF THE GRADER IS NOT ABLE TO UNDERSTAND HOW AN ANSWER WAS
OBTAINED.
Unless otherwise speciﬁed, the ﬂuid is water, and room temperature and standard atmospheric
pressure apply.
There are four (4) problems in total. It is highly recommended that all problems be
attempted, so budget your time accordingly.
Formulae that you may or may not ﬁnd useful
h= 4σ cos θ
γD Ev = − dp
= ρc2
d V /V I xc
yA
∂p
al
−
+z
=
∂l γ
g
p
a0 l
p
a0 l
+z+
=
+z+
γ
gA
γ
gB
ycp − y = p
ω 2 r2
+z−
γ
2g = A a=
=
dy
dx = p
ω 2 r2
+z−
γ
2g B ∂ Vs
∂ Vs
V2
+ Vs
es + s er
∂t
∂s
r
∂u
∂u
∂u
∂u
+u
+v
+w
∂t
∂x
∂y
∂z
v
u 1 Date: 28 Sept. 2010 CE340 Test 1 Last Name: SOLUTION KEY
1. (24 points)
For each of the following multiple choice questions, choose one and only one
response, which is the most correct answer. In addition to completing the bubble form, it
is recommended that you also indicate clearly on the test paper the answer that you have
selected. Correct answers are indicated by the blacked out circles.
(a) Consider the streakline started at point A at time t = 0 and ending at point B at t = 5T .
The velocity ﬁeld is everywhere the same magnitude at any instant of time, but changes
in direction at t = 2T , and t = 4T . ❧
1 The pathline corresponding to a particle released at t = 0 at
is line I.
❧The pathline corresponding to a particle released at t = 0 at
2
is line II.
❧
3 The pathline corresponding to a particle released at t = 0 at
is line III.
⑤The pathline corresponding to a particle released at t = 0 at
is line IV.
❧The pathline corresponding to a particle released at t = 0 at
5
is line V.
❧A streamline through point A at t = 0 is shown as line V.
6
❧
7 A streamline through point A at t = 5T is shown as line V.
❧
8 4 and 6. point A until t = 5T
point A until t = 5T
point A until t = 5T
point A until t = 5T
point A until t = 5T (b) Consider three points, A, B, and C, in closed container ﬁlled with a ﬂuid of
constant density. Further, assume that the container can only move in the
xz plane. Points A and B are at the same elevation (i.e., same z ), while
point B is vertically above point C (so points B and C are at the same x). ❧
1 If the piezometric heads at A, B, and C are the same, then the
container can only be moving at a constant (including zero) velocity.
❧If the piezometric head at point B is larger than the piezometric head
2
at A, then the container is accelerating in the positive x direction.
❧
3 If the piezometric head at point B is larger than the piezometric head at point C,
then the container is accelerating in the positive z direction.
❧If the pressure at point C is the same as the pressure at point B, then the container
4
is accelerating in the negative z direction.
❧1 and 2
5
❧
6 1 and 3
⑤
7 1 and 4
❧
8 2 and 4 2 Date: 28 Sept. 2010 CE340 Test 1 Last Name: SOLUTION KEY
(c) Consider a steady ﬂow with a streamline pattern shown in the
ﬁgure, generated by the velocity ﬁeld deﬁned by u = 1+ y , v = 1,
w = 0 (where u, v , and w are the velocity components in the x,
y and z directions). ❧
1 The convective acceleration is everywhere zero.
❧
2 The ﬂow is steady and uniform.
⑤
3 The component of velocity normal to all points on all streamlines is zero.
❧
4 The component of acceleration normal to all points on all streamlines is zero.
❧
5 The total acceleration is everywhere constant and is directed in the positive y direction.
❧1 and 3
6
❧
7 3 and 4
❧
8 3 and 5 (d) Consider the steady ﬂow of an ideal ﬂuid around
a circular (or cylindrical) object of radius R as
shown. In particular, consider the streamline
from A to B to C to D to E, where the portion
AB is straight and directly upstream of the
object, the portion DE is straight and directly downstream of the object, and the portion
BCD is on the surface of object. Given that all points are at the same elevation and at
the points A and E, the magnitude of the velocity is U0 , while along BCD, the velocity
component in the tangential (θ) direction is 2U0 sin θ (the radial component is zero),
where θ is measured as shown, with points A and B at θ = 0◦ , point C at θ = 90◦ , and
points D and E at θ = 180◦ . Note that points B and D are stagnation points.
⑤
1 The piezometric head at point A is larger than the piezometric head at point C.
❧
2 The acceleration along the streamline from points B to D is everywhere the same.
❧
3 The dynamic pressure at point A is larger than the dynamic pressure at point C.
⑤
4 Of all the points (A, B, C, D, and E), the lowest static pressure is found at point C.
❧
5 The component of acceleration normal to the streamline at A is equal to the component of acceleration normal to the streamline at C.
❧1 and 2
6
❧
7 2 and 3
❧
8 2 and 5 Note: Due to a typographical error, there is in fact two correct choices shown above; in
such a case, choosing only one or the other would be given full marks. 3 Date: 28 Sept. 2010 CE340 Test 1 Last Name: SOLUTION KEY
2. (20 points)
A steady laminar ﬂow of SAE 30W oil at
◦ C occurs between two horizontal plates. The upper
40
plate moves left at a velocity, ut (< 0, since it is to the left),
while the lower plate is stationary. The velocity proﬁle is
speciﬁed as
u(y ) = u0 y
−
H y
H 2 + ut y
H where u0 = 25 cm/s, ut = −2 cm/s, and H = 3 cm.
(a) Determine the magnitude of the ﬂuid shear stress at the lower plate.
(b) Determine the velocity at the location where the ﬂuid shear stress is zero.
Solution:
a) We need τ (y = 0) = µ(du/dy )y=0 . From the provided ﬁgure, we ﬁnd that, for SAE 30W
oil at 40◦ , µ ≈ 0.1 Ns/m2 . Taking the derivative of the velocity proﬁle,
du
u0 2u0 y ut
=
−
+
dy
H
H2
H
which we evaluate at y = 0 to give (du/dy )y=0 = (u0 + ut )/H . As a result,
u0 + ut
τ (y = 0) = µ
H = (0.1Ns/m2 ) (0.25 m/s − 0.02 m/s)
= 0.77 Pa
0.03 m b) The ﬂuid shear stress is zero where du/dy = 0. From the derivative that we have already
taken above, we ﬁnd that this condition is satisﬁed where
y
u0 + ut
0.25 m/s − 0.02 m/s
=
=
= 0.46
H
2u0
2 × 0.25 m/s
This is substituted in the expression for the velocity proﬁle:
u(y = 0.46H ) = 0.25 m/s (0.46) − (0.46)2 − 0.02 m/s(0.46) = 0.053 m/s 4 Date: 28 Sept. 2010 CE340 Test 1 Last Name: SOLUTION KEY
3. (20 points)
For the manometer system shown in
ﬁgure, determine the diﬀerence in piezometric head
between points A and B, i.e., PA − PB , where P denotes the piezometric head. The manometer ﬂuid has
a speciﬁc gravity of 0.85.
Solution:
We need the diﬀerence in piezometric head between
points A and B, which is
PA − PB = p
+z
γ A − p
+z
γ B We write the hydrostatic equations as
pA − p1 = −γ (zA − z1 ) p1 − p2 = −γm (z1 − z2 ) p2 − pB = −γ (z2 − zB ) where the levels 1 and 2 are the manometer levels as shown, γ and γm are the speciﬁc weights
of water and the manometer ﬂuid respectively. As usual, we add these with the result that
pA − pB = −γ (zA − z1 ) − γm (z1 − z2 ) − γ (z2 − zB )
Dividing through by γ and transferring zA and zB from the right to the left hand side, we
obtain
p
+z
γ p
−
+z
γ
A B γm
γm
= z1 −z2 − (z1 −z2 ) = (z1 −z2 ) 1 −
γ
γ 5 = −1.5 m(1−0.85) = −0.225 m Date: 28 Sept. 2010 CE340 Test 1 Last Name: SOLUTION KEY
4. (36 points)
A closed pressurized vessel has a spout
that is closed by an 10in diameter circular gate that is
hinged along one side as shown in the ﬁgure. The vessel has a hemispherical dome of radius 1.8 ft. The base
of the dome is 1.5 ft above the hinge of the gate. The
dome experiences a resultant upwards force of magnitude
500 lbf due to hydrostatic pressure.
(a) Determine the magnitude of the force on the circular
gate due to hydrostatic pressure.
(b) Determine the line of action of the force on the circular
gate due to hydrostatic pressure. Sketch clearly the
location of the line of action relative to the center of
the gate.
(c) Determine the minimum torque necessary at the hinge in order to maintain the gate
closed (neglect the weight of the gate and any friction on the hinge).
Soln.: a) The dome experiences a net vertical upwards force, Fv = 500 lbf, which implies that
the level at which the pressure is zero can be determined as
Fv = γ V = γ [π R2 h∗ − (2/3)π R3 ] = 500 lbf where V is the volume, represented by the crosshatched region in the ﬁgure, above the curved
surface (the dome) to the zero pressure level , and h∗ is the distance from the base of the
dome to the zero pressure level (again, see ﬁgure). With R = 1.8 ft, we determine
(Fv /γ ) + (2/3)π R3
500 lbf/(62.4 lbf/ft3 ) + (2/3)π (1.8 ft)3
h=
=
= 1.99 ft
π R2
π (1.8 ft)2
∗ With the zero pressure level determined, we apply our formulae for forces on plane surfaces.
The magnitude of the force is found as
F = γ hc A = 62.4 lbf/ft3 (h∗ + 1.5 ft + (3/5)(5/12) ft)π (5/12)2 ft2 = 127.2 lbf
where hc = (h∗ + 1.5 ft + (3/5)(5/12) ft) = 3.74 ft is the vertical distance (see ﬁgure) from the
zero pressure level to the centroid of the gate.
b) The line action is found from
ycp − y = 4
2
π rgate /4
3rgate /4
Ixc
3(5/12)2 ft2 /4
=
=
=
= 0.007 ft,
2
yA
5h c
5 × 3.74 ft
[hc /(3/5)]π rgate so the line of action or center of pressure lies a distance of 0.007 ft directly below the center
of gate (as shown). y is the distance from the origin, O, at the zero pressure level to the
centroid of the surface along the coordinate y aligned with the surface.
c) To solve the statics problem of determining the necessary torque, T , to keep the gate closed
against the hydrostatic pressure, we set the sum of moments about the hinge to zero:
T − F Lh = 0 =⇒ T = F Lh = 127.2 lbf × [(5/12) + 0.007] ft = 53.9 ft lbf where Lh = rgate + (ycp − y ) = 0.424 ft is the moment arm of the resultant hydrostatic force.
6 ...
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 Spring '11
 Troy

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