Bernoulli and Torricelli Theorem Basics.
Bernoulli and Torricelli Theorem Basics.
Bernoulli and Torricelli Theorem Basics.
+ z
A
+ v
v
v
A
A
A
2
2
2
/
/
/
2
2
2
g
g
g
= p
p
p
F
F
F
/
/
/
+ z
F
+ v
F
2
/ 2g
z
A
= z
F
+ v
F
2
/ 2g
26
v
F
= [(z
A
- z
F
)*2g]
1/2
v
F
= [(3.0)*2*9.81]
1/2
v
F
= 7.67 m/s
Q
F
= A
F
* v
F
Q
F
= 3.77 x 10
-3
m
3
/s
The same Q will apply at points B, C, D, E!
27
Now lets find the pressures at the various points.
Set up the equation Points A & B
p
A
/ + z
A
+ v
A
2
/ 2g = p
B
/ + z
B
+ v
B
2
/ 2g
p
p
p
A
A
A
/
/
/
+ z
A
+ v
v
v
A
A
A
2
2
2
/
/
/
2
2
2
g
g
g
= p
B
/ + z
B
+ v
B
2
/ 2g
p
B
= [ (z
A
- z
B
) - v
B
2
/ 2g]
what is (z
A
- z
B
)????
what is v
B
?
v
B
= Q/A
B
28
v
B
= 3.77 x 10
-3
/0.001257
= 3.0 m/s
Therefore pressure at B
p
B
= [ (z
A
- z
B
) - v
B
2
/ 2g]
p
B
= 9.81* [ 0 (3)
2
/ 2*9.81]
p
B
= -4.50 kPa
*** Hmmmmm pressure at B is negative!!!!!
A and B are at the same level in the fluid why arent the
pressures the same??????????
Didnt we learn that in the previous lecture????????
SO WHAT IS DIFFERENT HERE?????
29
Set up the equation Points A & C
p
A
/ + z
A
+ v
A
2
/ 2g = p
C
/ + z
C
+ v
C
2
/ 2g
p
p
p
A
A
A
/
/
/
+ z
A
+ v
v
v
A
A
A
2
2
2
/
/
/
2
2
2
g
g
g
= p
C
/ + z
C
+ v
C
2
/ 2g
p
C
= [ (z
A
- z
C
) v
C
2
/ 2g]
what is (z
A
- z
C
)????
= -1.2 m
what is v
C
?????????
30
v
C
= Q/Ac
= ????
Pressure at C
p
C
= [ (z
A
- z
C
) v
C
2
/ 2g]
p
C
= 9.81* [ -1.2 -0.459]
p
C
= -16.27 kPa
What about the pressure at D?????????????
Any guesses????????????????
And
WHY????
31
pD =
Set up the equation Points A & E
p
A
/ + z
A
+ v
A
2
/ 2g = p
E
/ + z
E
+ v
E
2
/ 2g
p
p
p
A
A
A
/
/
/
+ z
A
+ v
v
v
A
A
A
2
2
2
/
/
/
2
2
2
g
g
g
= p
E
/ + z
E
+ v
E
2
/ 2g
p
E
= [ (z
A
- z
E
) v
E
2
/ 2g]
what is (z
A
- z
E
)????
= 3.0 m
what is v
E
?????????
32
v
E
= 3.0 m/s
pressure at E
p
E
= [ (z
A
- z
E
) v
E
2
/ 2g]
p
E
= 9.81* [ 3.0 3.0
2
/ 2*9.81]
p
E
= 24.93 kPa
33
Important take-home points from this problem
The velocity of the siphon and the flow rate out of the
siphon depends on the elevation difference between free
surface of fluid and level of siphon.
The pressure at B is negative (below atmospheric) even
though it is at the same level as A in the fluid!
The decreased pressure energy (negative) at B got
converted to what??????
The velocity of flow is the same at all points if the pipe size
does not change (B, C, D, E)
Pressure at C is the lowest, since its at the highest elevation
34
Pressure at D and B are the same since the elevation and
velocity heads for these two locations are the same (the
other two terms in the Bernoullis equation)
Pressure at point E is the highest in the system since it is at
the lowest position.
35
Torricellis Theorum (year - 1645):
If Bernoullis equation is applied to a tank with
an orifice.
p
1
/ + z
1
+ v
1
2
/ 2g = p
2
/ + z
2
+ v
2
2
/ 2g
p
p
p
1
1
1
/
/
/
+ z
1
+ v
v
v
1
1
1
2
2
2
/
/
/
2
2
2
g
g
g
= p
p
p
2
2
2
/
/
/
+ z
2
+ v
2
2
/ 2g
v
2
=
[2g (z
1
- z
2
)]
1/2
v
2
=
[2gh]
1/2
VELOCITY at ORIFICE dependent of fluid elevation!
36
Velocity will change as h changes!
Velocity and flow rate will decrease with time as the tank
drains! why because the value of h decreases!
Problem 6.13:
Compute the flow rate Q from the nozzle if:
h starts at 3.0 m and for increments of 0.5 m.
diameter of nozzle 50mm
area of nozzle = 0.001963 m
2
for h = 3.0 m
v = [2* 9.81* 3.0]
1/2
v = 7.67 m/s
37
therefore Q = Av = 0.001963 * 7.67 = 0.0151 m
3
/s
compute Q for decreasing 0.5 m increments and plot
Excel spreadsheet computations -
h v A Q
3 7.672 0.001963 0.015
2.5 7.004 0.001963 0.014
2 6.264 0.001963 0.012
1.5 5.425 0.001963 0.011
1 4.429 0.001963 0.009
0.5 3.132 0.001963 0.006
0 0.000 0.001963 0.000
38
0.000
1.000
2.000
3.000
4.000
5.000
6.000
7.000
8.000
9.000
0 1 2 3 4
h (m)
v
e
l
o
c
i
t
y
(
v
)
0.000
0.002
0.004
0.006
0.008
0.010
0.012
0.014
0.016
d
i
s
c
h
a
r
g
e
Q
v
Q
39
Another case of Torricellis theorem
Under the situation given above, the jet should rise to level h!!!
What do you have to do if you want the jet to go beyond h????
40
Example 6.14:
Compute the air pressure above the water column required to
cause the jet to rise 40 ft from the nozzle.
Water column h = 6.0 ft.
p1 =????????
Apply the bernoullis equation between water level and
nozzle
p
1
/ + z
1
+ v
1
2
/ 2g = p
2
/ + z
2
+ v
2
2
/ 2g
p
1
/ + z
1
= z
2
+ v
2
2
/ 2g
v
2
2
/ 2g = p
1
/ + (z
1
- z
2
)
v
2
2
/ 2g = p
1
/ + h
Now apply equation between points 2 and 3
p
2
/ + z
2
+ v
2
2
/ 2g = p
3
/ + z
3
+ v
3
2
/ 2g
z
2
+ v
2
2
/ 2g = z
3
or v
2
2
/ 2g = z
3
- z
2
41
p
1
/ + h = 40
p
1
/ = 40-h
p1 = 62.4 * (40-6)
= 2121 lb/ft2
Or 14.73 psig
42
Torricellis experiment time take for head to fall from h1
to h2
) (
2
) / ( 2
2 / 1
2
2 / 1
1 1 2
h h
g
A A
t t
j t
=
At - cross sectional area of tank
Aj cross sectional area of nozzle
43
Considering a 2 m diameter tank with a 50 mm nozzle
h v A Q time
m m/s m2 m3/s s
3 7.672 0.001963 0.015
2.5 7.004 0.001963 0.014 109
2 6.264 0.001963 0.012 121
1.5 5.425 0.001963 0.011 137
1 4.429 0.001963 0.009 162
0.5 3.132 0.001963 0.006 212
0 0.000 0.001963 0.000 511
0
100
200
300
400
500
600
0 1 2 3 4
h (m)
t
i
m
e
(
s
)
v
44
Venturimeter
Instrument that utilizes the Bernoullis principle and the
manometer principle to determine the pressure in pipes.
45
Determine velocity of flow at A, and the volume flow rate.
Apply the Bernoullis equation for points A and B
p
A
/ + z
A
+ v
A
2
/ 2g = p
B
/ + z
B
+ v
B
2
/ 2g
46
we know the pressure difference between A and B
because of the manometer
we know the elevation difference
but we dont know any velocities (and we want to find
velocity at A)
Rearrange the equation
(p
A
- p
B
) / + (z
A
- z
B
) = (v
B
2
- v
A
2
) / 2g
What is (z
A
- z
B
) ?????
So we know the elevation head/term.
Now what about the pressures determine from the
manometer
SG of the fluid = 1.25
g
of the fluid = 1.25*9.81 kN/m3 = 12.26 kN/m3
of water at 60C = 9.65 kN/m3.
47
(p
A
- p
B
) = (0.46 1.18) +
g
(1.18)
(p
A
- p
B
) / = (0.46 1.18) +
g
(1.18) /
= 0.78 m
OK what about the velocities we still have 2
unknowns????????????
(A
A
= 0.07068 m
2
; A
B
= 0.03141 m
2
)
v
B
2
= 5.06 v
A
2
48
NOW put all terms back into the original equation!
(p
A
- p
B
) / + (z
A
- z
B
) = (v
B
2
- v
A
2
) / 2g
Substituting the known, we get
0.78 m - 0.46 m = (4.06 v
A
2
) / 2g
Solve for v
A
!
v
A
= 1.24 m/s
Q = v
A
* A = 0.0877 m
3
/s
We are DONE!
49
ASSIGNMENT # 5:
1. 6.41M
2. 6.60M
3. 6.65M
4. 6.72M
5. 6.78M
6. 6.87E