Unravelling the mystery surrounding Bernoulli’s Theorem
This presentation has been prepared to assist those sitting the science paper of the
Membership Examination. I am going to dispel the mystery surrounding Bernoulli’s
Theorem in an easily understood manner and then to show how to apply this to previous
examination questions by giving model answers.
Daniel Bernoulli was a Dutch born Swiss Scientist, who discovered basic principles of fluids.
The Bernoulli principle is that a fluid (liquid or gas) in motion can have three types of energy
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•
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Potential energy
Kinetic energy
Pressure energy
These can be interchanged but unless energy is taken out (e.g. turbulence or friction) or
energy is put in (e.g. pump) then the total energy must be constant.
The frictional loss is neglected in calculations due to being small compared to the total
energy, however, you must consider frictional loss in certain circumstances e.g. sprinkler
calculations
To use Bernoulli’s theorem in calculations it is important to have all three forms of energy in
the same units. The Systems International (SI) unit for energy is the Joule (kg.m2/s2)
however; when using Bernoulli the energy is expressed per unit mass or per unit volume.
Therefore, there are different forms of the Bernoulli Equation depending on whether we are
working with either joules per kilogram (j/kg) or joules per meter cubed (j/m3). In order to
simplify the matter I am only going to use the Bernoulli Equation that expresses the energy
in the form of joules per metre3 (j/m3) which I believe is easier to apply to IFE examination
questions.
Potential energy
This is the energy due to the potential above the datum line from which all the energies are
measured. The potential energy per m3 of fluid can be considered as
ρgH (Joules/m3)
Where
p = density (kg/m3)
g = acceleration due to gravity (9.81 m/sec2)
H = height (m)
Kinetic energy
The kinetic energy is due to the fluid being in motion. The Kinetic energy can be considered
as
½ ρv2
Where
(Joules/m3)
p = density (kg/m3)
V = Velocity (m/sec)
Pressure energy
The pressure energy is due to being under pressure. The SI unit of pressure is Pascal but in
the Fire Service, the Bar and metres head are still used. Therefore, you must remember to
use the correct formula
To convert from Bar to Pascal’s you use the following
P X 100,000
Where
(Joules/m3)
P = Pressure (Bar)
To convert metres head pressure to Pascal’s, you use the following
ρgz
Where
(Metres head) (Joules/m3)
ρ = Density (kg/m3)
g = acceleration due to gravity 9.81 m/sec2
z = height (m)
Bernoulli – Pascal’s
Where the pressure energy is Pascal’s
PA + ρgHA + ½ ρvA2
Where
= PB + ρgHB + ½ ρvB2
PA is pressure energy at point A (joules)
ρgHA is the potential energy at point A ( joules)
ρ = Density of fluid (kg/m3)
g = Acceleration due to gravity 9.81 m/sec
H = height of column of water
½ ρvA2 is the kinetic energy at point A (joules)
V = velocity m/sec
Where the pressure energy is Bar
PA x 100,000+ ρgHA + ½ ρvA2 = PB x100, 000+ ρgHB + ½ ρvB2
Where the pressure energy is metres head
ρgzA + ρgHA + ½ ρvA2
= ρgzB + ρgHB + ½ ρvB2
Continuity equation
When considering Bernoulli it is also very important to understand the continuity equation.
This is due to the fact that in a closed system the rate of flow Q (m3 /sec) can be considered
as constant.
Q = VA
Where
Q = Rate of flow (m3/sec)
V = Velocity (m/sec)
A = Area (m2)
If the flow is constant then
Q = VAAA = VBAB
This is shown here in this diagram showing a pipe
As the water flows down the pipe and it tapers out what you will find is that the waters
velocity will reduce. In other words as the area increases the velocity falls. This is a very
important relationship when attempting Bernoulli calculations as will be shown later.
Before we attempt questions involving the use of these equations, I would like to give you a
few tips to ensure mistakes are not made.
Tip one
Produce a sketch and enter all the details given in the question first. This will make
understanding the problem much clearer.
Tip two
Always convert ALL units to SI units before attempting to answer the question. Many
candidates make mistakes because they don’t convert the units and simply place the
number in the formula.
Here is a list of the most common units.
SI unit
Length (L)
m
Area (A)
m2
Velocity (V)
m/sec
Acceleration due to gravity
(g)
9.81 m/sec2
Height (h)
m
Metres head (z)
m
Energy (joule)
kg.m2/s2
Pressure (Pascal)
n/m2
Volumetric flow (Q)
m3/sec
Tip three
You have to place a datum line which is where you are measuring the energies from. Now if
this is a horizontal pipe you always put the datum in the centre of the pipe because in this
way you have zero potential at both points. This is because the potential energy above and
below the datum cancels out. Now if the situation is not in a horizontal pipe for example
like this example.
What you do is always place your datum line at the lowest point in the system. In this way
only one of the points will have potential energy and it makes it easier to answer the
question.
Question 1
A pump is pumping 2m3/min of water the surface of which is 5m below the pump inlet. At
the outlet the pump has a diameter of 100mm and at this point the pressure is 8 Bar. From
the nozzle (which is at the same level as the pump outlet) the jet rises 35m.
A) Calculate the energy/kg of the water
(1) At the outlet of the pump
(2) At the top of the throw of the jet
B) Explain why (1) and (2) are not equal
(acceleration due to gravity is 9.81m/sec2)
What is the first thing you should do?
Make a sketch and fill all details as shown here
This question deals with energy. Now
Potential energy is due to its position
Kinetic energy is due to movement
Pressure energy is in the fluid
Using Bernoulli the total energy is
Pressure energy + Potential energy + Kinetic energy
ρgzA + ρgHA + ½ ρvA2
Next we need to determine the potential energy per kilogram at the pump.
The potential energy per m3 of fluid can be considered as
ρgz (Joules/m3)
Therefore the potential energy per m3 is
1000 X 9.81 x 5 = 49,050 joules
However the question asks per kilogram so you divide by the density of water which is
1000kgm3
This gives you 49.050 joules/kg
Now let us determine the pressure energy at the pump/kg using the following
P X 100,000
Giving
(Bar)
(Joules/m3)
8 x 100,000 = 800,000 joules/m3
This needs to be converted per kg 1 m3 = 1000kg
So finally the energy per kg is 800 joules
The next part of the question is asking for energy at the outlet of the pump.
Kinetic Energy = ½ ρV2
So we need to calculate the velocity using
Q = VA
Therefore, by transposition
V = Q/A
Convert 2m3/min = 2/60 m3/sec = 0.033333m3/s
Now we need to determine the area at the outlet using
0.7854 x d2 = 0.7854 x 0.1 x 0.1 = 0.007854 m2
Using
V = Q/A
V = 0.33/0.007854
= 4.24m/sec
Next we need to determine the kinetic energy using
Kinetic energy = ½ (4.24) 2 = 8.98 joules/kg
Total energy = 50 + 800 + 8.98 = 858.98 joules
That is the energy per kg at the pump
Next we need to determine the energy at the nozzle
Now ALL the energy is converted to potential energy. The water has a potential height of
35m.
Therefore the potential energy per m3 is
1000 X 9.81 x 35 = 343,350 joules
However the question asks per kilogram so you divide by the density of water which is
1000kgm3
This gives you 343.350 joules/kg
Therefore the energies are not the same.
The energy at the pump is
858.98 joules/kg
The energy at the jet is
343.5 Joules/kg
They are different because of all the energy losses such as frictional loss through the hose,
frictional losses in the nozzle. You also have losses due to turbulence in the nozzle and
friction and turbulence as the water travels through the air.
Question 2
Water is flowing horizontally through a 250mm diameter pipe and into a constriction of
100mm diameter. The pressure difference is measured as 23.5mm of mercury. Using
Bernoulli’s theorem, calculate the rate of flow.
(Density of mercury = 13,600 kg/m3)
(Density of water = 1000 kg/m3)
( g = 9.81)
Next you need to write down Bernoulli’s equation. The pressure energy is quoted in metres
head and so the formula to apply is
ρgzA + ρgHA + ½ ρvA2 = ρgzB + ρgHB + ½ ρvB2
Remember the first thing to do is to produce a sketch with all the relevant information on.
Pressure difference = 23.5mm mercury
Density of mercury = 13,600 kg/m3)
Now place the datum in the centre of the pipe as you always do with a horizontal pipe and
the potential energy is zero giving;
ρgzA + ½ ρvA2 = ρgzB + ½ ρvB2
Now we need to determine the velocity energy of the water using the continuity equation
Q = VA
Now in this situation we can only determine one parameter , which is the area. Can you see
that you don’t know the velocity and you don’t know the quantity of water flowing.
In this case what you have to do is identify a relationship between the velocity at the two
different points. This type of question comes up a lot so remember if you have 2 unknowns
then you need to establish a relationship between the two points.
Q = VA
and
VAAA = VBAB
Now let’s determine the area at the two points using
0.7854 x d2
Area at point 1 = AA 0.7854 x d2 = 0.7854 x 0.25 x 0.25 = 0.0490875 m2
Area at point 2 = AB 0.7854 x d2 = 0.7854 x 0.1 x 0.1 = 0.007854 m2
Now if you divide it out 0.0490875/0.007854
you find that
AA is 6.25 AB
This is the relationship we needed to determine. The area at point AA is 6.25 times bigger
than the area at point AB. Now we know that
VAAA = VBAB
So now although we do not know the velocity in the system we do know that the velocity at
point AB is 6.25 times higher than the velocity at point AA
So
VB = 6.25VA
Now all you do is substitute this principle in the formula
ρgzA + ½ ρvA2 = ρgzB + ½ ρ(6.25vA)2
Now multiply out the brackets gives you
ρgzA + ½ ρvA2 = ρgzB + ½ ρ 39.0625vA2
Now you need to transpose the formula out. Don’t forget to transpose a formula all you do
is whatever you do to one end you also do to the other
This gives you
ρgzA - ρgzB = ½ ρ 38.0625vA2
Now we need to put in the pressure difference.
In this question the pressure difference is
ρgzA - ρgzB
The question tells you that the pressure difference is 27.5mm of mercury.(Density of
mercury = 13,600 kg/m3) Replace this with the pressure energy gives you
13,600 x 9.81 x 0.0275 = ½ ρ 38.0625vA2
This gives you
3668.94 = ½ ρ 38.0625vA2
Transpose both sides knowing density of water is 1000 kg/m3 (divide both sides by 1000)
3.66894 = ½ 38.0625vA2
To remove the ½ multiply both sides by 2 gives
7.33788 = 38.0625vA2
Divide both sides by 38.0625 gives
0.1929688 = vA2
Finally use the square root of each side, gives
VA = 0.43928 m/sec
Now we have the velocity at point A we can determine the rate of flow
Therefore using Q = VAAA
Substitute in known values
Q = 0.43928 X 0.0490875 = 0.021563 m3/sec
This equates to roughly 21 litres second
Question 3
A foam generator consists of a horizontal tube of circular cross section which tapers from an
input of 80mm internal diameter to 20mm diameter. 750 lts/min of concentrate (Density
1200 kg/m) is flowing through the generator and the pressure inlet is 12 Bar.
What is the pressure at the point where the diameter is 20mm?
Remember the first thing to do is to produce a sketch with all the relevant information on.
Next you need to write down Bernoulli’s equation in the correct format. The question is
quoted in Bars so the correct formula to apply is
PA x 100,000+ ρgHA + ½ ρvA2 = PB x100,000+ ρgHB + ½ ρvB2
Now we need to determine the velocity energy of the water using the continuity equation
Q = VA
V = Q/A
Therefore by transposing
Now let’s determine the velocity energy at point A
Now don’t forget to convert to SI units
SI unit of Q = M3/sec
Q = 750 litres/min
Q = 750 / 1000 to get m3 and then divide by 60 to convert minutes into seconds
Q = 750 / 1000 / 60
Q = 0.0125 m3/sec
Now we need to determine the velocity energy using
Q = VA transposing gives
V = Q/A
At point A we get
VA = Q/AA
Now we need to determine the area in m2
To determine the area at point A we use the formula
Therefore AA = 0.7854 x 0.08 x 0.08 = 0.005027m2
Therefore using
VA = Q/AA
VA = 0.0125/0.005027
VA = 2.487 m/second
To determine the area at point B we use the formula
0.7854 d2
0.7854 d2
Therefore 0.7854 x 0.02 x 0.02 = 0.000314m2
Therefore using
VB = Q/AB
VB = 0.0125/0.000314
VB = 39.81 m/second
Substitute all known values in Bernoulli formula
PAx100,000 + ρghA + ½ ρvA2 = PB x 100,000 + ρghB + ½ ρvB2
We can cancel out the potential energy at both points because the datum you chose is the
centre of the pipe.
12 x100,000 + ½ ρvA2 = PB x 100,000 + ½ ρvB2
Then substitute in known values – velocities and density
1200000 + ½ x 1200 x 2.4872 = PB x 100,000 + ½ x 1200 x 39.812
Multiply out
1200000 + 3711
= PB x 100,000 + 950902
Transpose the formula giving
1200000 + 3711 – (950902)
252809
= PB x 100,000
= PB x 100,000
Therefore transposing again (divide both sides by 100,000)
PB = 2.528 bar or 252818 Pascal’s
Question 4
Water is flowing in a vertical tapering pipe 2 metres in length. The top of the pipe is 100mm
diameter and the bottom is 50mm diameter. The quantity of water flowing is 1300
litres/minute.
Calculate the pressure difference between the top and the bottom of the pipe?
Remember the first thing to do is to produce a sketch with all the relevant information on.
Then convert all units into SI units.
Q = 1300 litres minute SI is m3/second
So we convert giving 1300 x 60 / 1000
Q = 0.0217 m3/sec
Now we can determine the velocity of the water at both point A and point B.
Firstly we need to determine the area at points A and B using
Area at point A = 0.7854 x d2 = 0.7854 x 0.1 x 0.1 = 0.007854 m2
Area at point B = 0.7854 x d2 = 0.7854 x 0.05 x 0.05 = 0.001963 m2
Then using
VA = Q/AA
VA = 0.0217/0.007854
VA = 2.76 m/second
Then using
VB = Q/AB
VB = 0.0217/0.001963
VB = 11.07 m/second
Now we can insert all known values in the formula which in this case because it is just asking
for pressure difference we will choose the formula with Pascal’s. (Note- Density of water =
1000kg/m3)
PA + ρgHA + ½ ρvA2
= PB + ρgHB + ½ ρvB2
Then substitute in all known values
PA + 1000 x 9.81 x 2 + ½ x 1000 x (2.76)2
= PB + 0 + ½ 1000 x (11.07)2
Multiply out
PA + 23428.8 = PB + 61,272.45
Transpose by taking away 23428.8 from both sides gives
PA = PB + 37843.65
Therefore by transposition
PA - PB = 37843.65 Pascal’s
This means that the pressure difference is 37,843 Pascal’s
Question 5
A pump supplies 4kw of energy to the water flowing through a 45mm hose. The water flows
15m vertically and through a 25mm branch at a rate of 500 litres/minute. Use Bernoulli’s
theorem and find the pressure at the branch.
Make a sketch and fill all details as shown here
The first thing to do to solve this problem is to determine the pressure at the outlet. In this
case you also need to know another formula which is
WP = 100 LP / 60
Where
WP = Water power (Watts)
L = Flow (litres min)
P = Pressure (Bar)
Therefore by transposition
P = 60 X WP / 100 X L
P = 60 X 4000 / 100 X 500
P = 4.8 Bar
Now you have the pressure at the outlet. We need to determine the flow in the correct
units
Q = m3/sec = 500 lts/min = 500/1000/60
Q = 0.00833 m3/sec
Next we need to calculate the velocity at Point A and Point B
Area at point A = AA 0.7854 x d2 = 0.7854 x 0.045 x 0.045 = 0.00159 m2
Area at point B = AB 0.7854 x d2 = 0.7854 x 0.025 x 0.025 = 0.0004908 m2
Now we need to determine the velocity at Pont A and Point B using VA = Q/AA
Q/AB
and VB
=
VA = 0.00833/0.00159
VA = 5.24 m/second
VB = 0.00833/0.0004908
VB = 16.97 m/second
Now substitute all known values in Bernoulli’s theorem. The question quotes the pressure
in bar so we use the following formula.
PA x 100,000+ ρgHA + ½ ρvA2 = PB x100,000+ ρgHB + ½ ρvB2
Substitute in known values
4.8 x 100,000+ 0+ ½ x 1000 x (5.24)2 = PB x100,000+ 1000 x 9.81 x 15 + ½ x 1000 x (16.97)2
Multiply out gives
480,000 + 13728.8 = PB x100,000 + 147150 + 143990.45
Multiply out gives
493,728.8 = PB x100,000 + 291,140.45
Therefore
PB x100,000 = 202,588.35
Divide both sides by 100,000
PB = 2.02 Bar
Question 6
If the manometer readings are 800mm and 200mm, what is the flow ?
(Density of water = 1000kg/m3)
Firstly you need to select the format of Bernoulli you are working in. In this case the
pressure energy is expressed in metres head and therefore the formula to apply is
ρgzA + ρgHA + ½ ρvA2
= ρgzB + ρgHB + ½ ρvB2
Now straight away you can cancel out the potential energy on both sides giving
ρgzA + ½ ρvA2
= ρgzB + ½ ρvB2
Next you can insert the pressure energy in the formula
1000 x 9.81 x 0.8 + ½ ρvA2
= 1000 x 9.81 x 0.2+ ½ ρvB2
7848 + ½ ρvA2
= 1962 + ½ ρvB2
By transposition taking off 1962 both sides gives
5886 + ½ ρvA2
= ½ ρvB2
Next we need to determine the velocity of the water.
Using
Q = VA
Now because we have 2 unknowns in the formula we need to create a relationship between
the two points in the problem. Then we can substitute the value in.
I will show you how this works now
We know that
Q = VA
and
VAAA = VBAB
Now the area at point A is four times the area at point B. Therefore based on VAAA = VBAB .
The velocity at point B must be four times the velocity at point A
In other words
VB = 4VA
Then all you do is substitute in 4VA Instead of VB
This gives you
5886 + ½ ρvA2
Multiply out the brackets gives you
= ½ ρ(4vA) 2
5886 + ½ ρvA2
= ½ ρ16vA 2
Then transpose the formula gives you
5886 = ½ ρ15vA 2
Then substitute in the density (1000 kg/m3)
5886 = ½ x 1000 x 15 x vA 2
Multiply out
5886 = 7500 x vA 2
Then divide both sides by 7500 gives you
5886 = 7500 x vA 2
0.7848 = vA 2
Now square root both sides giving you
0.886 = vA
Therefore the velocity of the water at point A is 0.89 metres/second
Now we can determine the rate of flow using
Q = VA
A = 2000mm2 = 0.002 m2
Q = VA = 0.89 x 0.002 = 0.00178 m/sec
Rate of flow = 1.78 litres second