Solutions to Problem Set 4
1
Solutions to Problem Set 4
2
Solutions to Problem Set 4
3
Solutions to Problem Set 4
4
Solutions to Problem Set 4
5
Solutions to Problem Set 4
6
Solutions to Problem Set 4
7
Solutions to Problem Set 4
8
So
whi29346_ch03_138-227.qxd
10/29/09
15:52
Page 169
3.5 Frictionless Flow: The Bernoulli Equation
V(t)
169
Solution
V(t)
The appropriate control volume in Fig. E3.12 encloses the rocket, cuts through the exit jet,
and accelerates upward at rocket speed V(t)
whi29346_ch03_138-227.qxd
10/28/09
16:50
Page 151 Debd 208:MHDQ176:whi29346:0073529346:whi29346_pagefiles:
3.3 Conservation of Mass
151
This is the integral mass conservation law for a deformable control volume. For a
xed control volume, we have
d
t
CV
(
whi29346_ch03_138-227.qxd
172
10/28/09
16:58
Page 172 Debd 208:MHDQ176:whi29346:0073529346:whi29346_pagefiles:
Chapter 3 Integral Relations for a Control Volume
Ambient
air
Valid
Model
Valid,
new
constant
Valid
Valid
Invalid
Invalid
(a)
(b)
Valid, new
con
Exam 2
Friday, 10/16/2015
12:05 to 12:55 pm.
CEE 304GB v Fluid Mechanics ~ Fall 2015
Georgia Institute of Technology
Instructor: Dr. Hermann M. Fritz
NAME:
This is a closed book exam. 1 additional sheet of US-Letter paper with personal notes/equations
o
whi29346_ch03_138-227.qxd
10/28/09
16:54
Page 163 Debd 208:MHDQ176:whi29346:0073529346:whi29346_pagefiles:
3.4 The Linear Momentum Equation
163
CV
gage
pressure
F
gh 1
gh 2
=0
E3.10b
Assume steady incompressible ow with no variation across the width b.
whi29346_ch03_138-227.qxd
10/28/09
16:52
Page 157 Debd 208:MHDQ176:whi29346:0073529346:whi29346_pagefiles:
3.4 The Linear Momentum Equation
157
the external pressure force on a surface is normal to the surface and inward. Since
the unit vector n is dened
CEE 3040 Solution to Sample Final Problems
Name:
Consider a rectangular oil tank (2.6m x 9.5m) with a water layer at the bottom. How long
will it take for the water to drain through a 0.02m diameter drain hole?
1.9m
0.7m
Oil SG = 0.87
Water
0.02m dia
Page
whi29346_ch03_138-227.qxd
190
10/28/09
17:07
Page 190 Debd 208:MHDQ176:whi29346:0073529346:whi29346_pagefiles:
Chapter 3 Integral Relations for a Control Volume
The friction head is larger than the elevation change z, and the pump must drive the ow
agains
whi29346_ch03_138-227.qxd
142
10/28/09
16:46
Page 142 Debd 208:MHDQ176:whi29346:0073529346:whi29346_pagefiles:
Chapter 3 Integral Relations for a Control Volume
Wherever necessary to complete the analysis we also introduce a state relation such
as the per
whi29346_ch03_138-227.qxd
178
10/28/09
17:00
Page 178 Debd 208:MHDQ176:whi29346:0073529346:whi29346_pagefiles:
Chapter 3 Integral Relations for a Control Volume
The ow from 1 to 2 is a constriction exactly similar in effect to the venturi in Example 3.15,
whi29346_ch03_138-227.qxd
184
10/28/09
17:03
Page 184 Debd 208:MHDQ176:whi29346:0073529346:whi29346_pagefiles:
Chapter 3 Integral Relations for a Control Volume
Equation (1) thus becomes
Tok Q(R2 RV0)k
To
Vo
R
QR2
Ans.
The result may surprise you: Even if
whi29346_ch04_228-291.qxd
10/28/09
17:19
Page 235 Debd 208:MHDQ176:whi29346:0073529346:whi29346_pagefiles:
4.2 The Differential Equation of Mass Conservation
Incompressible Flow
235
A special case that affords great simplication is incompressible ow, wher
whi29346_ch03_138-227.qxd
202
10/29/09
15:52
Page 202
Chapter 3 Integral Relations for a Control Volume
P3.41 In Fig. P3.41 the vane turns the water jet completely
around. Find an expression for the maximum jet velocity
V0 if the maximum possible support
whi29346_ch04_228-291.qxd
250
10/28/09
17:29
Page 250 Debd 208:MHDQ176:whi29346:0073529346:whi29346_pagefiles:
Chapter 4 Differential Relations for Fluid Flow
Z
Liquidgas interface z = (x, y, t):
pliq = pgas (R1 + R1)
x
y
d
wliq = wgas =
dt
Equality of q
whi29346_ch06_346-455.qxd
11/4/09
10:39
Page 379 Debd 208:MHDQ176:whi29346:0073529346:whi29346_pagefiles:
6.8 Flow in Noncircular Ducts
379
piping head loss. If minor losses are neglected, the (horizontal) pipe length follows
from Darcys formula (6.10):
h
i ,
é! ﬂzig ~trm -f
[ 0553040 Quiz I Namezmxo o 86 re&
Closed Book and Notes February 6, 2008
Avg 7'
1) The weight shown falls at a constant speed of 50 mm/s under the
action of gravity. The thin gap between the weight and the cylinder
wall is ﬁlled with
7;?
CEE3040 Quiz l | Name: \c/ 0 EC ecJ
Closed Book and Notes March 7, 2008
k 1‘ olele ﬁle?
1) The triangular concrete gate is 3 ft wide and 4 ft tall. it is hinged at the top /
and rests on a wall at the bottom. The speciﬁc weight of the concrete is 150
CEE 3040A: FLUID MECHANICS
EXAM 2
Fall 1999
(1 1/8/1999)
Closed Book
50 minutes
Name:
SN#: 3.; g; ﬁﬁ‘b’ﬁf
30/50 m
(30 points)
The velocity ﬁeld of a two-dimensional ﬂow is given by: V = 6yi + 4j.
a) Determine the equation for the streamline that passes th
3.3l Water ﬂows from the faucet on the ﬁrst ﬂoor of the
building shown in Fig. P33! with a maximum velocity of 20
ft/s. For steady inviscid ﬂow, determine the maximum water
velocity from the basement faucet and from the faucet 0n the
second ﬂoor (assume e
2. 2.7 Bourdon gages (see Video V14-and Fig. 2.13) are
commonly used to measure pressure. When such a gage is
attached to the closed water tank of Fig. P217 the gage reads
5 psi. What is the absolute air pressure in the tank? Assume
standard atmospheric p
i 5.37%]
5.8 7 Water enters a rotating lawn sprinkler
through its base at the steady rate of 16 gal/min r a 8 in Nozzle exit
as shown in Fig. P5.87. The exit cross section area ’ area = 013%?
of each of the two nozzies is (3.04 in.2 and the flow /r—~\ /
l
5.2 An - incompressible ﬂuid flows
horizontally; in the x-y plane with a velocity
given by
u = 30 mm“ m/s, v z 0
where y and h are in meters and h is a
constant Detexmine the average velocity for
the portion of the ﬂow between y T— 0 and y i
h.
From [7, 5