ME 309 Fall 2008
Section 2 (Merkle)
Homework # 4
Due Fri. 5 September 2008
Problem 41.
Water enters a gardenhose nozzle with an average velocity of 1.8
m/s
.
The diameter at the nozzle entrance is 15
mm
. The nozzle is open such that the
effective flow area at the exit is an annulus with outer and inner diameters of 6
mm
and 5
mm
, respectively. What is the average water velocity at the exit annulus?
SOLUTION:
Known
:
V
in
, D
in
, D
out,eff
Find:
V
out
Sketch
:
Assumptions:
Use conservation of mass
Fluid is incompressible
Constant and uniform fluid properties
Steady State
Solution
:
Apply
conservation of mass principle:
dt
dM
m
m
m
CV
gen
out
in
=
+

Third and fourth terms are zero:
0
=

out
in
m
m
out
out
out
in
in
in
A
u
A
u
ρ
=
properties are constant—density is constant
( )
2
2
2
005
.
0
006
.
0
*
/
015
.
0
*
8
.
1
/

=
=
π
in
in
out
A
u
u
= 36.82
m/s
Problem
42
.
Consider
a
cylindrical tank with a diameter
D
of 1.0 m and a height
H
of 1.5
m as shown on the sketch. Water
(
ρ
= 1000 kg/m
3
) enters the tank
at the bottom through a tube with
diameter
d
of 5 mm at a constant
average
velocity
of
4
m/s.
Initially (time
t
= 0), the tank is
half
full
as
shown;
water
continues to enter until the tank
begins to overflow at time
t
full
.
Selecting
the
water
inside
the
tank at any instant as the control
volume
of
interest,
write
or
Sketch of Cylindrical Tank
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View Full Documentdevelop a formal explicit expression of mass conservation for this control volume. Your
result should be in the form of an ordinary differential equation. Solve your differential
equation for
t
full
and determine a numerical value using the data given.
SOLUTION:
Known
:
Cylindrical tank,
D, H,
ρ
,
d
,
u
in
Find
:
t
full
Sketch
:
Above
Assumptions
:
Constant inlet velocity
Constant density
No mass generation
Solution
:
Conservation of mass gives:
dt
dM
m
m
m
CV
gen
out
in
=
+

simplifications (no outflow; no generation) give:
dt
dM
m
CV
in
=
Incorporate variables from problem:
dt
dh
D
h
D
dt
d
u
d
in
4
4
4
2
2
2
π
=
=
Simplifying:
in
u
D
d
dt
dh
2
2
=
Integrate from h/2 to h:
( )
full
in
in
t
u
D
d
t
t
u
D
d
h
h
2
2
0
2
2
2
=

=

Solve for t
full
:
m
h
s
s
d
u
D
h
t
in
full
5
2
7500
005
.
0
*
4
1
2
5
.
1
2
2
2
2
2
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 Spring '08
 MERKLE
 Velocity, Patm

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