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Unformatted text preview: Date: 27 Oct. 2010 CE340 Test 2
Last Name:
First Name:
COVER PAGE
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
4σ cos θ
dp
h=
Ev = −
= ρc2
γD
d V /V
I xc
ycp − y =
yA
∂p
al
−
+z =
∂l γ
g
p
a0 l
p
a0 l
p
ω 2 r2
p
ω 2 r2
+z+
=
+z+
,
+z−
=
+z−
γ
gA
γ
gB
γ
2g A
γ
2g B
2
∂ Vs
∂ Vs
V
∂u
∂u
∂u
∂u
a=
+ Vs
es + s er ,
a=
+u
+v
+w
∂t
∂s
r
∂t
∂x
∂y
∂z
dy
v
=
dx
u
for oriﬁces and Venturi meters,
for rectangular weirs,
for triangular weirs, 2gh†
1 − (d/D)4
2
3,
Q = Cwr Lw 2gH
Cwr = 0.611 + 0.075(H/P )
3
8
θ
5
Q = Cwt
tan
2gH
15
2
V2
hL = K
2g
Q = Cd A 1 Date: 27 Oct. 2010 CE340 Test 2
Last Name:
First Name: 1. (20 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.
(a) Three cases of water ﬂow (from left to right in all cases) in an open channel are shown in
proﬁle view, with gravity acting vertically downwards as shown. The bottom boundary
and the free surface can be taken as streamlines. In cases I and II, these are curved,
while in case III, the streamlines are straight and parallel everywhere. The depths in all
three cases are equal (namely, h), at the vertical section containing the point A shown. ❧
1 In all three cases, the pressure at the bottom boundary (at point A) is γ h, where γ
is the speciﬁc weight of water.
❧The pressure at A in case I is larger than the pressure at A in case II.
2
❧
3 Of all cases, the pressure at A in case III is least.
⑤
4 The pressure at A in case II is larger than the pressure at A in case III.
❧
5 The pressure at A in case I is larger than the pressure in case III.
❧
6 The piezometric head is constant throughout the vertical section containing point
A in all three cases.
❧1 and 6
7
❧
8 2 and 5
(b) For a given steady ﬂow over a sharpcrested weir,
❧
1 the magnitude of the ﬂow velocity at a point over the weir crest section increases in
the vertical direction.
❧
2 the pressure distribution at the weir crest section is assumed to be hydrostatic in
the derivation of the standard weir formulae.
❧
3 the nappe refers to the ﬂow upstream of the weir section.
❧
4 for accurate evaluation of the weir discharge using the standard weir formulae, it
is required that the water surface elevation at the weir crest section be measured
accurately.
❧the discharge coeﬃcient of the weir accounts only for the frictional eﬀects neglected
5
in the theoretical analysis.
⑤for accurate evaluation of the weir discharge using the standard weir formulae, the
6
nappe should be well aerated, i.e., surrounded on all sides by the atmosphere.
❧2 and 4
7
❧
8 3 and 6 2 CE340 Test 2 Date: 27 Oct. 2010 Last Name:
First Name:
(c) The analysis of practical ﬂows is generally based on a control volume, to which the
concepts of mass, momentum, or energy conservation are applied. For steady ﬂows,
❧
1 the ﬂows of mass, momentum, and energy into a control volume must be equal to
the corresponding ﬂows out of the control volume.
❧the control volume always contains the same physical material.
2
❧
3 for ideal ﬂuids, the ﬂow of momentum into a control volume may diﬀer from the ﬂow
of momentum out of the control volume only due to the action of pressure forces.
⑤the energy equation, applied in its usual form, may be interpreted in terms of the
4
diﬀerence in the ﬂows of a quantity, (p/γ ) + z + αV 2 /2g , out of and into a control
volume and relating this diﬀerence to the rate of frictional loss and energy input or
loss to shaft devices.
❧1 and 2
5
❧
6 2 and 3
❧
7 3 and 4
❧
8 1, 3, and 4
(d) Four pipe systems with real ﬂuids are shown in the ﬁgure, together with two piezometers
in each system. Systems I and II are constantdiameter systems, but in systems III and
IV, the crosssectional area at A is smaller than the crosssectional area at B. Streamlines
may be assumed straight and parallel where the piezometers are installed. In systems I
and III, the levels in the two piezometers at A and B are the same, while in systems II
and IV, the level in the piezometer at B is higher than the level in the piezometer at A.
Velocites may be assumed to be constant over all crosssections (i.e., α = 1). ❧
1 In system III, there cannot be any ﬂow.
❧
2 In system I, there cannot be any ﬂow.
❧
3 In system III, if there is ﬂow, it must be from B to A.
❧
4 In system II, the ﬂow must be from B to A.
❧
5 In system IV, the ﬂow may be from A to B or from B to A depending on the
magnitude of the diﬀerence in piezometric heads, and on the change in crosssectional
areas, between A and B.
❧In system III, the ﬂow may be from A to B or from B to A depending on the
6
magnitude of the diﬀerence in piezometric heads, and on the change in crosssectional
areas, between A and B.
❧2 and 3
7
❧
8 2 and 4
⑤
9 2, 4 and 5 3 Date: 27 Oct. 2010 CE340 Test 2
Last Name:
First Name:
2. (20 points)
A sharpedged oriﬁce of diameter 1.4 cm is
installed in a pipe of diameter 2 cm, inclined with respect
to the horizontal, as shown in the ﬁgure. A diﬀerential
manometer, with a manometer ﬂuid of speciﬁc gravity
of 0.75, is applied across the oriﬁce, and its deﬂection is
observed to be 0.2 m for a given water ﬂow. Determine
the water discharge in the pipe.
Solution: This is a straightforward application of the oriﬁce equation for a pipe, namely
2gh†
Q = Cd Ao
1 − (d/D)4 where Ao = π d2 /4 is the area of the oriﬁce (d is the oriﬁce diameter), and not the area of
the pipe, and h† = ∆[(p/γ ) + z ], the change in piezometric head across the oriﬁce. The usual
analysis of the manometer (see end of solution) leads to
h† = [1 − (γm /γ )]∆h = (1 − 0.75)(0.2 m) = 0.05 m where γm /γ = sm = 0.75, the speciﬁc gravity of the manometer ﬂuid, and ∆h = 0.2 m is
the manometer deﬂection. The discharge coeﬃcient, Cd , is correlated (see inset in correlation
ﬁgure) with the Reynolds number, Re = V D/ν , where V = Q/Apipe is the average velocity in
the pipe, D is the pipe diameter, and ν is the ﬂuid kinematic viscosity. Because Q is unknown,
we cannot evaluate directly Re, we guess an initial value of Cd based on the known value
of d/D =1.4 cm/2 cm = 0.7. From the given graphical correlation, Cd = 0.61 for large Re
(where the curve ﬂattens out, which is usually a good initial guess), so we now evaluate Q
with Cd = 0.61,
2(9.8 m/s2 )(0.05 m)
Q = 0.61(π (0.014 m2 /4)
= 9.2 × 10−5 m3 /s
1 − 0.54
We now check with our Cd correlation. We evaluate Re = (Q/Apipe )D/ν = 4Q/(νπ D) = 4(9.2 × 10−5 m3 /s/10−6 m2 /s = 5857 From our graphical correlation, this value of Re is unfortunately not within the range, but
we choose closest available value, namely Cd ≈= 0.645, so we make another iteration, with
Cd = 0.645 as the new guess. We ﬁnd Q = 9.8 × 10−5 m3 /s, Re = 6240, which is closer to
our desired value.
Manometer analysis: pA − p1 = −γ (zA − z1 ) p1 − p2 = −γm (z1 − z2 ) p2 − pB = −γ (z2 − zB ) =⇒ p
+z
γ A − p
+z
γ =h =
† B γm
1−
γ ∆h Note that if we performed the manometer analysis from B to A, i.e., assuming a downwards ﬂow,
we would have obtained a negative value, which would indicate that the assumed downwards ﬂow
direction was not correct. 4 Date: 27 Oct. 2010 CE340 Test 2
Last Name:
First Name: 3. (25 points) The pipe ﬁxture shown in the ﬁgure connects two inﬂow pipes, and combines them into a single
outﬂow pipe. All pipes of diﬀerent diameters are in a
horizontal plane. The ﬁxture is isolated mechanically by
ﬂexible joints such that the net forces on the pipe ﬁxture
due to the ﬂow are sustained entirely by the pipe support
shown. Pressure gages at the inlets (A and B) and the
outlet (C) of the ﬁxture record pressures of 725 lbf/ft2 ,
721 lbf/ft2 , and 720 lbf/ft2 . A discharge of 0.06 cfs ﬂows
in the inﬂow (2in diameter) pipe, while a discharge of
0.09 cfs ﬂows in the other (3in diameter) inﬂow pipe.
The ﬂow in the outlet (4in diameter) pipe is in a direction directly opposite to that in the 2in diameter inﬂow
pipe, while the other (3in diameter) inﬂow pipe lies at 50◦ angle to the other two pipes as
shown in the ﬁgure. Determine the magnitude and direction of the two components of the
net force acting on the pipe supports due to the ﬂow. Sketch and label carefully all control
volumes as well as any coordinate system used in your analysis.
Solution: We draw our control volume and coordinate system as shown. Note that pressure
forces are always normal and compressive. We assume a net external force on the ﬂow due to
the pipe (or pipe support) in the negative x and positive y  directions, and write the x and
y  component equations, with forces on the left hand side, and momentum ﬂows (analagous
to ma in Newton’s second law) on the right hand side, as :
−Fx + (pA AA )x + (pB AB )x + (pC AC )x = (ρQV )Cx − [(ρQV )Ax + ρQV )Bx ] Fy + (pA AA )y + (pB AB )y + (pC AC )y = (ρQV )Cy − [(ρQV )Ay + ρQV )By ]
We now simplify where possible (e.g, (pB AB )x = −pB AB sin θ,
(ρQV )Bx = ρQB VB (− sin θ), (pA AA )y = 0, (pC AC )y = 0,
(ρQV )Cy = 0, (ρQV )Ay = 0, (ρQV )By = ρQB VB (− cos θ),
choosing carefully the signs and the appropriate components
(θ = 50◦ ) according to our coordinate system,
Fx = pA AA − pB AB sin θ + pC AC − ρQC (−VC )+
[ρQA VA + ρQB VB (− sin θ)] Fy = −pB AB (− cos θ) − ρQB VB (− cos θ)
Because QC = QA + QB = 0.15 cfs, and in each pipe V = Q/A,
the terms on the right hand side of the equations can be evaluated, and hence
Fx = [725 lbf /ft2 × π (1/6 ft)2 /4] − [721 lbf /ft2 × π (1/4 ft)2 /4] sin(50◦ ) + [720 lbf /ft2 × π (1/3 ft)2 /4]
+ 1.94 slugs/ft3 × [(0.06 ft3 /s)2 /{π (1/6 ft)2 /4}] − 1.94 slugs/ft3 × [(0.09 ft3 /s)2 /{π (1/4 ft)2 /4}] sin(50◦ ) = 52.1 lbf Fy = (721 lbf /ft2 )[π (1/4 ft)2 /4] cos(50◦ ) + 1.94 slugs/ft3 × [(0.09 ft3 /s)2 /{π (1/4 ft)2 /4}] cos(50◦ )
= 23.0 lbf 5 Date: 27 Oct. 2010 CE340 Test 2
Last Name:
First Name: As the results for Fx and Fy are positive, the assumed directions are correct, namely that the
force components on the ﬂow due to the pipe (or pipe support) are in the negative x and
positive y directions. The force components on the pipe support are equal in magnitude but
opposite in direction to that shown in the control volume diagram (the force in the xdirection
is to the right, and that in the y direction is downwards). 6 Date: 27 Oct. 2010 CE340 Test 2
Last Name:
First Name:
4. (25 points)
A 5cm diameter steady round water jet discharges into the atmosphere, impinging on a stationary 15cm
diameter circular disk, and so is deﬂected to ﬂow radially (see
ﬁgure). The distance between the jet exit and the circular disk
is 20 cm, and the thickness of the ﬂow at the edge of the disk
is observed to be 0.5 cm. The total volume of water from the
jet exit to the edges of the disk is estimated to be 600 cm3 .
It may be assumed that the velocity magnitude at the jet exit
is constant over the jet exit section and similarly the velocity
magnitude at the disk edge is constant vertically, though the
velocity magnitude at the jet exit is not necessarily the same
as the velocity magnitude at the disk edge. The ﬂuid can also
be assumed to be ideal (i.e., frictionless). Sketch and label
carefully any control volume, as well as any coordinate system
or streamline used in your analysis.
(a) Determine the jet discharge.
(b) If the circular disk is maintained in place entirely due to its
weight (i.e., no other external force is applied), what must
be the weight of the circular disk?
Solution: a) This is a combination of various types of problems, and involves applying the Bernoulli equation and the
continuity (i.e., mass conservation) equation. We apply the
Bernoulli equation along the streamline between the points
A and B shown,
p
V2
p
V2
+z+
=
+z+
γ
2g A
γ
2g B At both points, the pressure is zero because the jet is in the atmosphere, and so rearranging,
we ﬁnd that
2
2
2
2
VA − V B
AB − A2
Q
1
1
Q2
A
= zB − zA = ∆ z
=⇒
−2=
= ∆z
2g
2g
A2
AB
2gA2
A2
A
A
B
If a diﬀerent streamline had been taken, e.g., say, along the edge of the jet from jet exit to
disk edge, from A to B , a slight diﬀerence in resulting exit velocity at the disk edge would
have arisen due to the slight diﬀerence in elevation (between B and B ), but we are assuming
that the velocity is constant over the control surface at the disk edge.
After some algebra, we can solve for Q in terms of the other variables as
2g ∆z
Q = AA
1 − (AA /AB )2
which is very similar to the oriﬁce equation (which should not be too surprising). It is clear
that AA = π d2 /4, where d is jet diameter, but AB = π Dh, where D is the disk diameter, and
7 Date: 27 Oct. 2010 CE340 Test 2
Last Name:
First Name: h is thickness of the radial jet at the disk edge (recall that the ﬂow is radially outwards at
the disk edge, and so the appropriate area is the area of the side of cylinder of diameter D
and height h). So eventually with AA /AB = (π d2 /4)/(π Dh) = d2 /4Dh, we can evaluate the
discharge as
π (0.05 m)2
2g (0.2 m)
Q=
= 7.04 × 10−3 m3 /s
2 /(4 × 0.15 m × 0.005 m)}2
4
1 − {(0.05 m)
b) To determine the weight, Wdisk , of the circular disk, we simply
need to consider the z momentum or force balance (because of
radial symmetry, no other force components is relevant). With
the control volume and the coordinate system shown, we write,
forces (with the proper signs) on the left hand side, and momentum ﬂows on the right hand side,
−Wdisk − Wﬂuid = [(ρQV )out ]z − [(ρQV )in ]z
Because the jet is in the atmosphere, pressures are all zero, and
so pressure forces do not arise, but the weight of the ﬂuid, Wﬂuid ,
within the control volume should be taken into consideration (especially since the volume of ﬂuid, Vﬂuid = 600 cm3 , in the control
volume is given). Because the z component of the velocity at the outlet is zero, i.e., (Vout )z =
0, the balance is very simple, namely
Wdisk = ρQVin − Wﬂuid = 103 kg/m3 × [(7.04 × 10−3 m3 /s)2 /{π (0.05 m)2 /4}] − (9.8 × 103 kN/m3 )(600 × 10−6 m3 )
= 19.5 N 8 ...
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