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Freshman Clinic II
Spring 20XX
Dr. Kauser Jahan
Hands on an Aquarium: Pump Design
(Teacher Version)
Objectives
1.
To learn about parameters which effect pump design
2.
To understand common terms such as
head, flow rate,
system
curves,
and pump
curve
s
.
Pumps
Basic Principles of Fluid Mechanics
Pumps are used to move fluids, such as water, oil, and air in a controlled flow at a desired rate. Pumps
are often needed to overcome gravity,
for example,
when moving water uphill, or to ove
rcome frictional
forces in fluid systems. Pumps work by adding energy to a fluid system in the form of pressure. Since all
fluids will travel from high pressure to low pressure, creating a continuous pressure difference will cause
fluids to flow at a con
stant rate. The amount of energy created by a pump is referred to as
head
. Head
can
be used to describe energy of a fluid system due to pressure, velocity, or potential energy. Friction
and other forms of energy loss are described as head loss.
Head is
measured in the units of distance,
such as feet or meters.
This is done to make it easier to equate all the forms of energy in a system, as
well as easily determine the amount of energy needed to move fluid vertically.
Pressure head
is the energy of a f
luid related to the force it exerts upon its container. Pressure head is
calculated by measuring the pressure of the fluid, and divid
ing it by
its
specific weight.
(eq. 1)
Where P is pressure, in either pounds per square foot or
Pascales
, and γ is the specific weight, in either
pounds per cubic foot or Newtons per cubic meter.
At standard conditions, γ is 62.4 lbs/cubic foot
or
9.81 kN/cubic meter
.
Velocity head
is the energy of a fluid related to the kinetic energy as it flows.
Velocity head can be
calculated by the following equation:
(eq. 2)
Where V is the linear velocity, in feet per second or meters per second, and g is the acceleration due to
gravity. The velocity of the fluid can be calculated using the
following equation:
(eq. 3)
Where V is the velocity, Q is the flow rate, in cubic feet or cubic meters per second, and A is the cross
sectional area, in square feet or square meters.
Elevation
Head
is the energy a fluid has
available
when it can flow from a higher elevation to a lower
one. This translates to potential energy
of the system, similar to how any object placed at a certain
elevation above a surface will turn its potential energy into kinetic energy once released
. Since t
his
value is already a distance which can be measured in feet or meters, it can be equated to pressure or
velocity head easily.
A pump’s efficiency can be determined by measuring the amount of energy put into the system, the
flow rate, and the head. Th
is is important for engineers to consider, since energy efficiency is an
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important topic and pumps in constant use can use large amounts of electricity. This energy use needs
to be factored into much it would cost for an establishment to run the pump on c
ertain intervals. The
following equation can be used to determine the efficiency of any system.
(eq. 4)
Where e is the efficiency in decimal, Q is the flow rate in cubic feet per second or cubic meters per
second, H
pum
p
is the head the pump produces, γ is the specific weight
of water, and bhp is the
horse
power the motor of the pump produces.
Types of Pumps
Centrifugal

The most commonly used pump for a broad range of applications is the centrifugal pump.
Centrifugal
pumps work using a turbine, which imparts centrifugal force on incoming fluid, pushing it to
the outer walls of the turbine casing. This fast moving, higher pressure fluid departs out a tangential
exit pipe. Centrifugal pumps are highly efficient and can
achieve a high flow rate.
Figure
1
: Centrifugal Pump
(
http://www.betterbricks.com/graphics/assets/images
)
Reciprocating

These pumps work similarly to the cylinder of an automobile engine. Using one way
valves, water is drawn
into a cylinder by a plunger on its down

stroke. On the up

stroke, another one
way valve allows fluid to exit. Reciprocating pumps deliver lower flow rates, but can
add
more head to a
system than a centrifugal pump. These pumps also create a pulsing fl
ow rate, since flow is only
delivered on one half of the motors rotation.
Figure
2
: Reciprocating Pump
(
http://www.lcresources.com/resources/getstart/pump.gif
)
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Peristaltic

Peristaltic pumps work by repeatedly squeezing flexi
ble hosing and pushing a volume of fluid
along the hose. This is accomplished by using two or more rollers attached along the edge of rotating
disk. As the disk rotates, the wheels squeeze the hose and push the volume of fluid trapped within the
rollers
in the direction of rotation. Since this utilizes no turbines or valves, these pumps can be used to
pump fluid in either direction of the hosing. Peristaltic pumps will produce a pulsing flow rate, since
flow is delivered for
every space between the roll
ers
.
Figure
3
: Peristaltic Pump
(
http://www.blue

white.com/images/per_pumphead.jpg
)
System and Pump Curves:
A system curve is used to describe the amount of head needed to have a desired flow rate for a given
system. These take into account head losses, which are frictional forces which must be overcome in
order to have flow. For the most part, system curves s
how
that
in o
rder to have a higher flow rate
one
must be able to contri
bute more energy to the system
.
Pump curves demonstrate how a given pump will perform at varying heads, flow rates, efficiencies, and
pump sizes. These are commonly used with centrifu
gal pumps, since they are most commonly used.
For a centrifugal pump, several factors will affect
or can be affected by how the pump is operated. For
example, use of larger impellers will not only increase the flow rate, but also the head. Operating at
certain flow rates will increase
the
efficiency of
a given
type of impeller. Below is an example
of a
system curve and pump curve.
Figure
4
: Pump Curve
Overlaid a System Curve
By plotting a certain pump curve over a system curve, one can find the o
pe
rating
flow rate for a system.
For the system above, the operating point is at 1100 cfs, with 30 feet of head.
0
10
20
30
40
50
60
0
500
1000
1500
2000
Head (ft)
Flowrate (cfs)
System Curve
Pump Curve
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Procedure
For this lab, yo
u will
use the apparatus seen in the below figure. It consists of a peristaltic pump
with
Easy Load II canister
(or
any other pump that is available that can connected to an electric flow control)
,
Master Flex
electronic flow controller, flexible tubing, ring stand, clamps, funnel, 2000mL beaker,
and
reservoir
.
All tests should be completed with the control knob indic
ating a power level of 10.
Figure
5
: Lab Apparatus
At least seven tests should be run at varying heights. Using a stopwatch, record the time it takes to fill
the beaker with 1000L for varying heights (between 0 and 30 inches). In order to do tests at
0 and 5
inches, you will need to hold the beaker near the edge of the table, since the beaker is too tall. Convert
the height from inches to feet, and the volume from liters to cubic feet (12in = 1 ft, 1 L = .0353 cubic
feet). Record the data on the fol
lowing table. Also, measure the diameter of the plastic tubing. This will
be used in calculating the velocity head. Pressure at the tube in the reservoir will change as the run
proceeds. After each run, empty the water that was filled into the beaker b
ack into the reservoir.
Returning the water to the reservoir will ensure that each run begins at the same water elevation in the
reservoir. Even though there is change of water elevation in the reservoir during a run, the change will
be so small it will
not need to be accounted for in the calculation of elevation head.
Height (ft)
Time (seconds)
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Post Lab Analysis
1)
Plot the data in excel of height in feet vs
.
flow rate in cubic feet per second. From the data,
predict the maximum height the pump could deliver water.
2)
From the analysis done in
step
1, what is the maximum pressure the pump could produce?
(
Hint: equate pressure head with
elevation
head
)
3)
Now cal
culate the velocity head
for each of
your flow rate
s
using equations 2 and 3
, and add
each value
to its corresponding height
. Make sure to check your units
. Compare t
hese
value
s
to
the maximum height
obtained in number 1.
Are they similar?
4)
Plot the dat
a obtained from Problem 1 on top of the following system curve
, which describes a
water filtration system for an aquarium
. The equation for the system curve is
displayed on the
below chart.
What is your operating point?
(i.e. at what point do they inter
sect
?
)
5)
It is known that the average aquarium requires the entire volume of water to be circulated
through the filter at least once every 5 hours. Using the operating flow rate, would the pump be
able to meet these requirements for a
4
cubic foot tank?
6)
Assuming the
pump motor produces
.003
horsepower
at operating speed
, determine the
efficiency of the pump at the operating flow rate and head.
Assuming it costs $2000 dollars a
year for an aquarium shop to run these pumps 24 hours a da
y, 365 days a year, how much
would it cost if the efficiency was increased by 20%?
y = 4E+06x
2
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
0
0.0002
0.0004
0.0006
0.0008
0.001
0.0012
Required Head (ft)
Flow Rate (cubic ft)
System Curve
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Solutions and Examples:
1)
Assuming the following data was recorded
by timing how long it took to fill one liter
:
Height
(in)
Height
(ft)
Time
(s)
Q (cfs)
0
0
42
0.00084
10
0.833333
68
0.000519
15
1.25
93
0.00038
20
1.666667
124
0.000285
25
2.083333
197
0.000179
30
2.5
340
0.000104
The flow rate, Q, was calculated by dividing the volume, .0353 cubic feet, by the number of seconds
measured
for each trial.
From this data a plot can be formed of the data.
2)
Using a linear trendline, find the equation from the data and determine the y intercept. This is
the maximum head the pump could produce. Set this equal to equation 1, setting gamma equal
to 62.4.
The
y intercept is 2.67.
P= 166.6 psf
3)
To calculate the velocity head, first determine the cross sectional area.
Now divide
the flow rate by the area, to find velocity.
And use the equation to find the velocity head
Doing this for all the flow rates, you get the following table
y =

3332.5x + 2.6702
0.5
0
0.5
1
1.5
2
2.5
3
0
0.0002
0.0004
0.0006
0.0008
0.001
Elevation (ft)
Flowrate (cfs)
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Q (cfs)
velocity
v

head (ft)
0.00084
1.095797
0.0186455
0.000519
0.676816
0.007113
0.00038
0.494876
0.0038028
0.000285
0.371157
0.0021391
0.000179
0.233622
0.0008475
0.000104
0.135363
0.0002845
By comparing the velocity head values to the elevation head, students should see as the
elevation head increases, the velocity head decreases. The reason that these values are not
equal, however, is due to the fact that frictional forces are a function of velocity head.
4)
By using the equation given in the question, a system curve can be plot
ted on the same chart as
the pump curve. The following figure shows what is obtained using data given in the example:
Using this chart, the intersection can be approximated near a flow rate of .00044 cfs.
5)
By dividing the volume of the tank by the flow rate found in part 4, the time it takes to circulate
the volume of the tank in seconds. Divide this by 3600sec/1hr to find the time in hours.
4 cubic feet / .00044 cfs = 9090 seconds
9090 seconds / 3600s
ec/1hr = 2.52 hours
Since this time is less than the 5 hour limit, it is acceptable.
6)
Using equation 4 and plugging in the flow rate and head for the operating point
as well as the
horsepower, an efficiency of .158 or 15.8%.
For part two of the question, simply multiply the $2000 by .158, and then divide it by .158+.20.
2000x.158/.358= 882.68 dollars.
0
1
2
3
4
5
6
7
0
0.0002
0.0004
0.0006
0.0008
0.001
0.0012
0.0014
Elevation (ft)
Flowrate (cfs)
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