Unravelling the mystery surrounding Bernoulli s Theorem

Transcription

1 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.m /s ) 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/m 3 ). 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 metre 3 (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 m 3 of fluid can be considered as Where ρgh (Joules/m 3 ) p = density (kg/m 3 ) g = acceleration due to gravity (9.81 m/sec ) H = height (m)

2 Kinetic energy The kinetic energy is due to the fluid being in motion. The Kinetic energy can be considered as Where p = density (kg/m 3 ) V = Velocity (m/sec) ½ ρv (Joules/m 3 ) 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 Where P = Pressure (Bar) P X 100,000 (Joules/m 3 ) To convert metres head pressure to Pascal s, you use the following ρgz (Metres head) (Joules/m 3 ) Where ρ = Density (kg/m 3 ) g = acceleration due to gravity 9.81 m/sec z = height (m) Bernoulli Pascal s Where the pressure energy is Pascal s P A + ρgh A + ½ ρv A = P B + ρgh B + ½ ρv B Where P A is pressure energy at point A (joules) ρgh A 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 ρv A is the kinetic energy at point A (joules)

3 V = velocity m/sec Where the pressure energy is Bar P A x 100,000+ ρgh A + ½ ρv A = P B x100, 000+ ρgh B + ½ ρv B Where the pressure energy is metres head ρgz A + ρgh A + ½ ρv A = ρgz B + ρgh B + ½ ρv B 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 (m 3 /sec) can be considered as constant. Q = VA Where Q = Rate of flow (m 3 /sec) V = Velocity (m/sec) A = Area (m ) If the flow is constant then Q = V A A A = V B A B This is shown here in this diagram showing a pipe

4 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.

5 SI unit Length (L) m Area (A) m Velocity (V) Acceleration due to gravity (g) Height (h) Metres head (z) m/sec 9.81 m/sec m m Energy (joule) kg.m /s Pressure (Pascal) n/m Volumetric flow (Q) m 3 /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.

6 Question 1 A pump is pumping m 3 /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 () At the top of the throw of the jet B) Explain why (1) and () are not equal (acceleration due to gravity is 9.81m/sec) What is the first thing you should do? Make a sketch and fill all details as shown here

7 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 ρgz A + ρgh A + ½ ρv A Next we need to determine the potential energy per kilogram at the pump. The potential energy per m 3 of fluid can be considered as Therefore the potential energy per m 3 is 1000 X 9.81 x 5 = 49,050 joules ρgz (Joules/m 3 ) However the question asks per kilogram so you divide by the density of water which is 1000kgm 3

8 This gives you joules/kg Now let us determine the pressure energy at the pump/kg using the following P X 100,000 (Bar) (Joules/m 3 ) Giving 8 x 100,000 = 800,000 joules/m 3 This needs to be converted per kg 1 m 3 = 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 ρv So we need to calculate the velocity using Q = VA Therefore, by transposition V = Q/A Convert m 3 /min = /60 m 3 /sec = m 3 /s Now we need to determine the area at the outlet using x d = x 0.1 x 0.1 = m Using V = Q/A V = 0.33/ = 4.4m/sec Next we need to determine the kinetic energy using Kinetic energy = ½ (4.4) = 8.98 joules/kg Total energy = = joules That is the energy per kg at the pump

9 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 m 3 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 1000kgm 3 This gives you joules/kg Therefore the energies are not the same. The energy at the pump is The energy at the jet is joules/kg 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 Water is flowing horizontally through a 50mm diameter pipe and into a constriction of 100mm diameter. The pressure difference is measured as 3.5mm of mercury. Using Bernoulli s theorem, calculate the rate of flow. (Density of mercury = 13,600 kg/m 3 ) (Density of water = 1000 kg/m 3 ) ( 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 ρgz A + ρgh A + ½ ρv A = ρgz B + ρgh B + ½ ρv B Remember the first thing to do is to produce a sketch with all the relevant information on.

10 Pressure difference = 3.5mm mercury Density of mercury = 13,600 kg/m 3 ) 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; ρgz A + ½ ρv A = ρgz B + ½ ρv B 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 unknowns then you need to establish a relationship between the two points. Q = VA and V A A A = V B A B Now let s determine the area at the two points using x d Area at point 1 = A A x d = x 0.5 x 0.5 = m Area at point = A B x d = x 0.1 x 0.1 = m Now if you divide it out / you find that A A is 6.5 A B

11 This is the relationship we needed to determine. The area at point A A is 6.5 times bigger than the area at point A B. Now we know that V A A A = V B A B So now although we do not know the velocity in the system we do know that the velocity at point A B is 6.5 times higher than the velocity at point A A So V B = 6.5V A Now all you do is substitute this principle in the formula ρgz A + ½ ρv A = ρgz B + ½ ρ(6.5v A ) Now multiply out the brackets gives you ρgz A + ½ ρv A = ρgz B + ½ ρ v A 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 ρgz A - ρgz B = ½ ρ v A Now we need to put in the pressure difference. In this question the pressure difference is ρgz A - ρgz B The question tells you that the pressure difference is 7.5mm of mercury.(density of mercury = 13,600 kg/m 3 ) Replace this with the pressure energy gives you 13,600 x 9.81 x = ½ ρ v A This gives you = ½ ρ v A Transpose both sides knowing density of water is 1000 kg/m 3 (divide both sides by 1000) = ½ v A

12 To remove the ½ multiply both sides by gives = v A Divide both sides by gives = v A Finally use the square root of each side, gives = V A m/sec Now we have the velocity at point A we can determine the rate of flow Therefore using Q = V A A A Substitute in known values Q = X = m 3 /sec This equates to roughly 1 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 0mm diameter. 750 lts/min of concentrate (Density 100 kg/m) is flowing through the generator and the pressure inlet is 1 Bar. What is the pressure at the point where the diameter is 0mm? Remember the first thing to do is to produce a sketch with all the relevant information on.

13 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 P A x 100,000+ ρgh A + ½ ρv A = P B x100,000+ ρgh B + ½ ρv B 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 = M 3 /sec Q = 750 litres/min Q = 750 / 1000 to get m 3 and then divide by 60 to convert minutes into seconds Q = 750 / 1000 / 60 Q = m 3 /sec Now we need to determine the velocity energy using Q = VA transposing gives V = Q/A At point A we get V A = Q/A A Now we need to determine the area in m To determine the area at point A we use the formula d Therefore A A = x 0.08 x 0.08 = m Therefore using V A = Q/A A V A = 0.015/ V A =.487 m/second To determine the area at point B we use the formula

14 d Therefore x 0.0 x 0.0 = m Therefore using V B = Q/A B V B = 0.015/ V B = m/second Substitute all known values in Bernoulli formula P A x100,000 + ρgh A + ½ ρv A = P B x 100,000 + ρgh B + ½ ρv B We can cancel out the potential energy at both points because the datum you chose is the centre of the pipe. 1 x100,000 + ½ ρv A = P B x 100,000 + ½ ρv B Then substitute in known values velocities and density ½ x 100 x.487 = P B x 100,000 + ½ x 100 x Multiply out = P B x 100, Transpose the formula giving (95090) = P B x 100, = P B x 100,000 Therefore transposing again (divide both sides by 100,000) P B =.58 bar or 5818 Pascal s Question 4 Water is flowing in a vertical tapering pipe 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?

15 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 m 3 /second So we convert giving 1300 x 60 / 1000 Q = m 3 /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 = x d = x 0.1 x 0.1 = m Area at point B = x d = x 0.05 x 0.05 = m Then using Then using V A = Q/A A V A = 0.017/ V A =.76 m/second V B = Q/A B V B = 0.017/ V B = m/second

16 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/m 3 ) P A + ρgh A + ½ ρv A = P B + ρgh B + ½ ρv B Then substitute in all known values P A x 9.81 x + ½ x 1000 x (.76) = P B ½ 1000 x (11.07) Multiply out P A = P B + 61,7.45 Transpose by taking away from both sides gives P A = P B Therefore by transposition P A - P B = 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 5mm 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

17 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 = m 3 /sec = 500 lts/min = 500/1000/60 Q = m 3 /sec Next we need to calculate the velocity at Point A and Point B Area at point A = A A x d = x x = m Area at point B = A B x d = x 0.05 x 0.05 = m Now we need to determine the velocity at Pont A and Point B using V A = Q/A A and V B = Q/A B V A = / V A = 5.4 m/second V B = / V B = m/second Now substitute all known values in Bernoulli s theorem. The question quotes the pressure in bar so we use the following formula. P A x 100,000+ ρgh A + ½ ρv A = P B x100,000+ ρgh B + ½ ρv B

18 Substitute in known values 4.8 x 100, ½ x 1000 x (5.4) = P B x100, x 9.81 x 15 + ½ x 1000 x (16.97) Multiply out gives 480, = P B x100, Multiply out gives 493,78.8 = P B x100, , Therefore P B x100,000 = 0, Divide both sides by 100,000 P B =.0 Bar Question 6 If the manometer readings are 800mm and 00mm, what is the flow? (Density of water = 1000kg/m 3 )

19 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 ρgz A + ρgh A + ½ ρv A = ρgz B + ρgh B + ½ ρv B Now straight away you can cancel out the potential energy on both sides giving ρgz A + ½ ρv A = ρgz B + ½ ρv B Next you can insert the pressure energy in the formula 1000 x 9.81 x ½ ρv A ½ ρv A = 1000 x 9.81 x 0.+ ½ ρv B = ½ ρv B By transposition taking off 196 both sides gives ½ ρv A = ½ ρv B Next we need to determine the velocity of the water. Using Q = VA Now because we have 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 V A A A = V B A B Now the area at point A is four times the area at point B. Therefore based on V A A A = V B A B. The velocity at point B must be four times the velocity at point A In other words V B = 4V A Then all you do is substitute in 4V A Instead of V B This gives you ½ ρv A = ½ ρ(4v A ) Multiply out the brackets gives you

20 ½ ρv A = ½ ρ16v A Then transpose the formula gives you 5886 = ½ ρ15v A Then substitute in the density (1000 kg/m 3 ) 5886 = ½ x 1000 x 15 x v A Multiply out 5886 = 7500 x v A Then divide both sides by 7500 gives you 5886 = 7500 x v A = v A Now square root both sides giving you = v A 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 = 000mm = 0.00 m Q = VA = 0.89 x 0.00 = m/sec Rate of flow = 1.78 litres second