Lecture 3 The energy equation Dr Tim Gough: t.gough@bradford.ac.uk
General information Lab groups now assigned Timetable up to week 6 published Is there anyone not yet on the list? Week 3 Week 4 Week 5 Week 6 Tuesday 14.00-17.00 01 October 2013 08 October 2013 15 October 2013 22 October 2013 Friday 14.00-17.00 04 October 2013 11 October 2013 18 October 2013 25 October 2013 Dr Tim Gough ENG 2038 M 2 J1 2 J2 Fluid Mechanics 2 Chesham Building C.01.14 18 18 18 Can anyone not see Blackboard yet? Please see me at interval and I ll sort this out.
Lecture 2 recap Streamlines etc Flow classifications Viscous and inviscid flows Discharge and mean velocity Flow continuity Flow continuity problems
Lecture 2 recap Flow Classifications Conditions in a body of fluid can vary from point to point and, at any given point, can vary from one moment of time to the next. Flow is described as uniform if the velocity at a given instant is the same in magnitude and direction at every point in the fluid. If, at a given instant, the velocity changes from point to point, the flow is described as non uniform. A steady flow is one in which the velocity, pressure and cross section of the stream may vary from point to point but do not vary with time. If, at any given point, conditions do change with time, the flow is described as unsteady.
Lecture 2 recap Inviscid and viscous flows Velocity profiles for inviscid flow Velocity profiles for viscous flow
Lecture 2 recap Discharge and mean velocity V r r R r V a) Laminar flow b) Turbulent flow In many problems we simply assume a constant velocity equal to the mean velocity to give: /
Lecture 2 recap Mass continuity For a streamtube (no fluid crosses boundary): 1 2 Area = A 1 Velocity = V 1 Density = 1 Area = A 2 Velocity = V 2 Density = 2 Or for an incompressible fluid where 1 = 2 this reduces to:
Lecture 2 recap Examples Example 1 Branched pipes Example 2 Porous walls
Lecture 2 recap Examples Example 3 Surge tank Example 4 Turbojet engine
The energy equation
Mechanical energy of a flowing fluid B B Element of fluid will possess Potential Energy (PE) due to its height, z, above the datum. It possesses Kinetic Energy (KE) due to its velocity, v. Datum level z A mg A For an element of weight, mg. Potential energy = mgz Cross sectional area A Potential energy per unit weight = z Kinetic energy = Kinetic energy per unit weight =
Mechanical energy of a flowing fluid B B A steadily flowing stream of fluid can also do work because of its pressure. Datum level z A mg A Cross sectional area A At any given cross section the pressure generates a force and, as this cross section moves forward work will be done. If the pressure at a cross section AB is p and the area of the crosssection is A then: Force exerted on AB = pa After a weight mg of fluid has flowed along the streamtube, section AB will have moved to A B : Volume passing AB = mg/g= m/
Mechanical energy of a flowing fluid Therefore: B B Distance AA = m/a Datum level z A mg A Cross sectional area A And work done = force x distance AA = pa x m/a Work done per unit weight = p/g The term p/g is known as the flow work or the pressure energy. This can be viewed as a potential energy in transit. Bernoulli
Mechanical energy of a flowing fluid Each of these terms has the unit of a length, or head and are often referred to as the pressure head p/g, the velocity head v 2 /2g, the potential head, z and the total head, H. Between any two points, on a streamline, we can write these as: That is: Total energy per unit weight at 1 = Total energy per unit weight at 2
Mechanical energy of a flowing fluid For the flow of a single fluid undergoing no density changes (i.e. no compressibility and no chemical reaction) we can simplify further to: Bernoulli The term metres. The term metres. is known as the pressure head and has dimensions of is known as the velocity head and also has dimensions of
Bernoulli's theorem So looking at points 1 and 2 for this water flow (known as a Venturi meter). Q If fluid height at point 1 is 100mm and at point 2 it is 20mm and the velocity at point 1 is 2.5 m/s. What is the velocity at point 2???
Bernoulli's theorem p 1 = gh 1 p 2 = gh 2 1 2 1 2 1 2
Bernoulli's theorem V 1 = 2.5m/s h 1 = 0.1m h 2 = 0.02m g = 9.81m/s 2 2 9.81 0.1 0.02 2.5. /
Benoulli s principle High pressure, low speed Accelerating flow Low pressure, high speed Decelerating flow
Bernoulli's theorem Fluids 1 lab revision Bernoulli can be used to take flowrate measurements in the field. In the Fluids 1 laboratory you performed measurements of flowrate using two techniques: a) Thin plate weir and b) Venturi meter. Both of these techniques use Bernoulli s principle to measure volumetric flowrate (or discharge). Both have no moving parts so are pretty much failsafe. Thin plate weir Venturi meter
Bernoulli's theorem Venturi meter So we use Bernoulli s principle which is simply another method of stating that energy is conserved. A Venturi meter is simply a pipe with a gradually converging section, with a narrow throat followed by a gradually diverging section. Venturi meter The energy per unit weight of a fluid is called the specific energy and has units of Joules / Newtons.!
Bernoulli's theorem Venturi meter Assuming conservation of energy and no density changes: Bernoulli Looking at a horizontal Venturi, z 1 = z 2 so:
Bernoulli's theorem Venturi meter Now we know that piezometers (the tubes!) measure static pressure difference through a head of fluid, h, at each location so: Where H is difference between heads at 1 and 2 And by mass continuity we can say:
Bernoulli's theorem Venturi meter 1 2 Substituting 2 into 1: 3 And then 3 into this: Gives:
Bernoulli's theorem Venturi meter Rearranging for v 2 : Since :
Bernoulli's theorem Venturi meter Now this assumes no energy losses, and we discovered that this was not truly accurate. We had to introduce the coefficient of discharge Cd (which you calculated).
Bernoulli's theorem Venturi meter So why did we have to introduce Cd? Where did the other energy go? Dissipated into heat and sound due to friction at the walls and the viscosity of the fluid. Thus we always get energy loss through flows, either due to friction or other head losses due to fittings, enlargements, contractions, bends, valves etc etc.
Energy equation In formulating this we assume that no energy has been supplied to, or taken away, from the fluid between points 1 and 2. However, energy could be supplied by introducing a pump, or energy could be lost by doing work against friction or in a machine such as a turbine. This we can expand Bernoulli s equation further: This is a form of the steady flow energy equation.
Energy equation Initial energy Energy supplied A Pump Final energy Energy loss Energy loss B So we accept that we lose energy to heat, sound etc as we flow through a system. We will be analysing this and quantifying these losses over the next few weeks. Clearly we can also put energy into the system (if necessary) through the use of, for example, a pump or a fan.
Energy equation example
Energy equation example A fire engine pump develops a head of 50 m, i.e. it increases the energy per unit weight of water passing through it by 50 N m N 1. The pump draws water from a sump at A (atmospheric pressure) through a 150 mm diameter pipe in which there is a loss of energy per unit weight due to friction of varying with the mean velocity u 1 in the pipe. The water is discharged through a 75 mm nozzle at C, 30 m above the pump, at the end of a 100 mm diameter delivery pipe in which there is a loss of energy per unit weight of. Calculate a) the velocity of the jet issuing from the nozzle at C and b) the pressure in the suction pipe at the inlet to the pump at B.
Energy equation example All energies are per unit weight If sump is large, v A = 0 and p A = 0 (atmospheric) p C = 0 (atmospheric), z 3 = 30 + 2 = 32 m Loss in inlet pipe Loss in discharge pipe Energy supplied by pump = 50 m
Energy equation example Loss in inlet pipe Energy supplied by pump = 50 m Loss in discharge pipe Rearranging: Eqn. 1 From continuity of flow equation:
Energy equation example Substituting into equation 1: Eqn. 1... /
Energy equation example b) the pressure in the suction pipe at the inlet to the pump at B. Apply Bernoulli between A and B: Where z 2 = 2 m, 0.25 8.314 2.079 / Thus:........ N/m 2 below atmospheric pressure