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

• 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/s

2)

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/m

3) 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 (Joules/m

3)

Where 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 (Joules/m3)

Where P = Pressure (Bar)

To convert metres head pressure to Pascal’s, you use the following

ρgz (Metres head) (Joules/m3)

Where 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 = PB + ρgHB + ½ ρvB

2

Where 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 ρvB

2

Where the pressure energy is metres head

ρgzA + ρgHA + ½ ρvA2 = ρgzB + ρgHB + ½ ρvB

2

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/sec

2

Height (h) m

Metres head (z) m

Energy (joule) kg.m2/s

2

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 (Bar) (Joules/m3)

Giving 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 m

3/sec = 0.033333m

3/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 m

2

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 ρvB

2

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 ρvB

2

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 m

2

Area at point 2 = AB 0.7854 x d2 = 0.7854 x 0.1 x 0.1 = 0.007854 m

2

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 39.0625vA

2

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 ρvB

2

Now we need to determine the velocity energy of the water using the continuity equation

Q = VA

Therefore by transposing V = Q/A

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 0.7854 d2

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

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 ρvB

2

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 ρvB

2

Then substitute in known values – velocities and density

1200000 + ½ x 1200 x 2.4872 = PB x 100,000 + ½ x 1200 x 39.81

2

Multiply out

1200000 + 3711 = PB x 100,000 + 950902

Transpose the formula giving

1200000 + 3711 – (950902) = PB x 100,000

252809 = 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 m

2

Area at point B = 0.7854 x d2 = 0.7854 x 0.05 x 0.05 = 0.001963 m

2

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 ρvB

2

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 m

2

Area at point B = AB 0.7854 x d2 = 0.7854 x 0.025 x 0.025 = 0.0004908 m

2

Now we need to determine the velocity at Pont A and Point B using VA = Q/AA and VB =

Q/AB

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 ρvB

2

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 ρvB

2

Now straight away you can cancel out the potential energy on both sides giving

ρgzA + ½ ρvA2 = ρgzB + ½ ρvB

2

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 ρvB

2

7848 + ½ ρvA2 = 1962 + ½ ρvB

2

By transposition taking off 1962 both sides gives

5886 + ½ ρvA2 = ½ ρvB

2

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 = ½ ρ(4vA)

2

Multiply out the brackets gives you

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 m

2

Q = VA = 0.89 x 0.002 = 0.00178 m/sec

Rate of flow = 1.78 litres second