ENG2038M – Fluid Mechanics 2
General information
• Lab groups now assigned
• Timetable up to week 6 published
• Is there anyone not yet on the list?
• Can anyone not see Blackboard yet?
• Please see me at interval and I’ll sort this out.
Week 3 Week 4 Week 5 Week 6
Tuesday 14.00 - 17.00 01 October 2013 08 October 2013 15 October 2013 22 October 2013Friday 14.00 - 17.00 04 October 2013 11 October 2013 18 October 2013 25 October 2013
Dr Tim GoughENG 2038 M 2 J1 2 J2
Fluid Mechanics 2Chesham Building C.01.14
18 18 18
ENG2038M – Fluid Mechanics 2
Lecture 2 recap
• Streamlines etc
• Flow classifications
• Viscous and inviscid flows
• Discharge and mean velocity
• Flow continuity
• Flow continuity problems
ENG2038M – Fluid Mechanics 2
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.
ENG2038M – Fluid Mechanics 2
Lecture 2 recap
Inviscid and viscous flows
Velocity profiles for viscous flowVelocity profiles for inviscid flow
ENG2038M – Fluid Mechanics 2
Lecture 2 recap
Discharge and mean velocity
/
In many problems we simply assume a constant velocity equal to the mean velocity to give:
b) Turbulent flow
Vrr
R
Vr
a) Laminar flow
ENG2038M – Fluid Mechanics 2
12
Area = A1Velocity = V1Density = 1
Area = A2Velocity = V2Density = 2
• For a streamtube (no fluid crosses boundary):
Lecture 2 recapMass continuity
• Or for an incompressible fluid where 1 = 2 this reduces to:
ENG2038M – Fluid Mechanics 2
Lecture 2 recapExamples
Example 1 ‐ Branched pipes
Example 2 – Porous walls
ENG2038M – Fluid Mechanics 2
Lecture 2 recapExamples
Example 3 – Surge tank
Example 4 – Turbojet engine
ENG2038M – Fluid Mechanics 2
Mechanical energy of a flowing fluid
• 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.
• For an element of weight, mg.
• Potential energy = mgz
• Potential energy per unit weight = z
• Kinetic energy =
• Kinetic energy per unit weight =
A
B
A’
B’
mg
Cross‐sectional area A
z
Datum level
ENG2038M – Fluid Mechanics 2
A
B
A’
B’
Datum level
mg
Cross‐sectional area A
z
• A steadily flowing stream of fluid can also do work because of its pressure.
• 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 cross‐section 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
ENG2038M – Fluid Mechanics 2
Mechanical energy of a flowing fluid
A
B
A’
B’
Datum level
mg
Cross‐sectional area A
z
• Therefore:
Distance AA’ = m/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
ENG2038M – Fluid Mechanics 2
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 v2/2g, 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
ENG2038M – Fluid Mechanics 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 is known as the pressure head and has dimensions of
metres.
• The term is known as the velocity head and also has dimensions of
metres.
ENG2038M – Fluid Mechanics 2
Bernoulli's theorem
• So looking at points 1 and 2 for this water flow (known as a Venturimeter).
• 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???
ENG2038M – Fluid Mechanics 2
Bernoulli's theorem
V1 = 2.5m/sh1 = 0.1mh2 = 0.02mg = 9.81m/s2
2 9.81 0.1 0.02 2.5
. /
ENG2038M – Fluid Mechanics 2
Benoulli’s principle
Accelerating flow Decelerating flow
Low pressure,high speed
High pressure,low speed
ENG2038M – Fluid Mechanics 2
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
ENG2038M – Fluid Mechanics 2
Bernoulli's theorem – Venturi meter
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.
• The energy per unit weight of a fluid is called the ‘specific energy’ and has units of Joules / Newtons.
!
ENG2038M – Fluid Mechanics 2
Bernoulli's theorem – Venturi meter
• Assuming conservation of energy and no density changes:
Bernoulli
• Looking at a horizontal Venturi, z1 = z2 so:
ENG2038M – Fluid Mechanics 2
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:
ENG2038M – Fluid Mechanics 2
Bernoulli's theorem – Venturi meter
1
2
3• Substituting 2 into 1:
• And then 3 into this:
• Gives:
ENG2038M – Fluid Mechanics 2
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).
ENG2038M – Fluid Mechanics 2
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.
ENG2038M – Fluid Mechanics 2
• 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:
Energy equation
• This is a form of the steady flow energy equation.
ENG2038M – Fluid Mechanics 2
Pump
A
B
Initial energy Energy supplied
Energy loss
Energy loss
Final energy
Energy equation
• 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.
ENG2038M – Fluid Mechanics 2
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 u1 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
ENG2038M – Fluid Mechanics 2
Energy equation example
All energies are per unit weight
If sump is large, vA = 0 and pA = 0 (atmospheric)
pC = 0 (atmospheric), z3 = 30 + 2 = 32 m
Loss in inlet pipe
Loss in discharge pipe
Energy supplied by pump = 50 m
ENG2038M – Fluid Mechanics 2
Energy equation example
Loss in inlet pipe
Loss in discharge pipe
Energy supplied by pump = 50 m
Rearranging:
From continuity of flow equation:
Eqn. 1