FLUID FLOW /PRESSURE IN
STATIONARY FLUIDS
Energy
Forms of energy
•
Any fluid in motion possesses 4 kinds of energy
Form of Energy
1.
Potential or head
2.
Kinetic or velocity
3.
Static or Pressure
4.
Heat
Example
- water at top of
tank
- water doing work
on water wheel
- Pressure exerted
inside line
- Heat generated in
line due to friction
Examples of conversion of energy
Pumps
- mechanical to velocity;
mechanical to pressure
Steam siphon
- Pressure to velocity
Water hammer - Velocity to pressure
Filling tank
- Pressure to velocity to potential
Total Energy
• The total energy that a fluid in motion has at any time is equal to the
sum of the form listed above. Because energy can neither be
created or destroyed, it is therefore possible to convert one form of
energy to another, such as velocity into pressure.
Fluid Flow
Bernoulli’s Principle
• when a fluid is in motion, its pressure increases whenever its
velocity decreases.
• Likewise its pressure decreases whenever its velocity increases
Fluid
• A fluid is a substance which undergoes continuous deformation
when subjected to a shear stress.
Factors determining Fluid Flow
Viscosity
•
Property of fluid to resists any force that tends to
produce flow. This resistance to flow [or movement] is
known as viscosity
•
It affects the performance of pumps and the pressure
drop when pumping through a given pipe line.
Factors determining Fluid Flow
Pipe size
•
For a given through put, the pressure drop increases
rapidly with decreasing pipe diameter
Roughness
•
The degree of roughness of the internal wall of the pipe
affects fluid flow. Old rusty or pitted pipe would offer more
resistance to flow (higher pressure drop)
Pipe restrictions
•
The number of fittings, valves, etc affects friction loss.
Type of Flow
Streamline flow – fluid flows in smooth layers.
Flow is in straight unbroken lines
Re < 2000
Turbulent flow – there is an irregular random
motion of fluid particles
Re > 4000
Transition – between above.
Reynolds Number
Nature of flow in pipe depends on pipe diameter, the density and
viscosity of the flowing fluid, and the velocity of flow.
Re = dvp
u
Where; Re - Reynolds number
d - internal diameter of pipe
v - kinematic viscosity
p - weight density of fluid
u - dynamic viscosity
The term kinematics refers to the quantitative description of fluid
motion or deformation.
Kinematic viscosity v = u/p
u = viscosity
p = density
Pumping
• In plant operations, we quite often want the flow to be from a low to
a high pressure area. This requires use of pumps or compressors.
• In the operation of pumps, pressure drop is of utmost importance. To
operate a centrifugal pump, for example, the suction line can be and
often is more important than discharge line. Although the discharge
line may be designed so that the pump is capable of putting the
required amount of liquid through it,
Pumping
• If the suction line is not large enough or is improperly
installed, pumping difficulties will be experience.
• If pressure drop is such that liquid starts to vaporize when it
reaches the pump, or if suction conditions are not correct,
then pump will not fill up properly and will become ‘gas
bound’.
• The pumping rate and discharge pressure will fall off, and pump
will begin to cavitate and heat up.
Flow of Gases
• Gases are compressible [ liquids are non-compressible]
• In moving a gas through a line, the higher the pressure on the
system, the more pounds of gas that can be transferred for the
same volume rate of flow.
The effects on plant operations due to compressibility of a
gas are:
•
•
•
•
Increased pressure means reduced equipment size
Gas such as air, under pressure can be used to operate pneumatic
tools
Hydrocarbon gases can be compressed, cooled and liquefied
Refrigeration
Fluid Performing Work
A column of water will exert a pressure of 0.433 psi
for each foot of weight.
• Suppose we build a dam and raised the height of
the water to 200 feet. That would mean that the
pressure at the bottom of the dam would be 0.433 x
200 or 86 psi. This says that if turbines were installed
at the bottom of the dam we would have a driving
force of 86 psi to turn the turbines which can be used
to generate electricity.
Siphon
• The siphon is a bent tube with branches of unequal length open at
both ends and it is used to move a liquid from a higher point to a
lower point, over an intermediate point higher than either.
C
1 ft
2 ft
A
B
Siphon
• Since the upward force at A is greater than B, the flow will be
from A to B. Suppose the tube is filled with water and placed in
the vessels as shown in the figure above. The water in leg CA
exerts a head or downward pressure of 1 foot of water or 0.433
psi. The water in branch CB exerts a head of 3 x 0.433 or 1.299
psi. Therefore, the net downward pressure causing flow from A to
elevation B is 0.866 psi.
Fluid Rates Measurement
• When pressure is converted to velocity, pressure is reduced. The
difference in pressure between 2 points in the line of flow is the
principle on which common flow meters operate.
• Orifice plate
• Pitot tube
• Venturi tube
Pressure in Stationary Fluids
Force - means a push or pull
Weight - means heaviness
Pressure - The push or pull or weight or weight per unit area
of the surface acted upon.
Pressure = __ Force_____
Area Acted upon
=
__ Weight____
Area Acted Upon
Water weight 62.4 pounds per cubic ft, 1 foot high water would therefore
exert a pressure of:
62.4lbs
ft3
x 1 ft =
144
0.433 lbs
ins2
Pressure Produced by Liquids
• Since all fluids have weight, they create a pressure against the walls
that hold them.
• The pressure exerted by a liquid at any given point or location in a
vessel depends upon the height of liquid above it.
• This pressure is independent of the shape of the vessel
• The pressure on the btm of all vessels below will be the same
Pressure Produced by Liquid (con’t)
• Different liquids weigh different amount for the same volume and
therefore would create different pressures. 1 cubic foot of water
weighs 62.4 pounds. If this cubic foot of water were put in a box 1 foot
high, 1 foot wide, and 1 foot long, the total force on the bottom of the
box would be 62.4 pounds. This total force would be distributed over
an area one foot square or 144 square inches. The pressure on a
given square inch would be 62.4 pounds divided by 144 square inches
[0.433 psi].
• Pressure Produced by Liquid (con’t)
• If a box were filled with water that was 2 feet high and held 2 cubic feet
the pressure would equal 62.4 x 2 /144 or 0.866 psi.
• Each foot of water height will develop 0.433 psi. For example a 100 foot
tower, if filled with water, would create 43.3 psi at the bottom of the
tower.
Density
• Weight of one cubic foot of material is called the density of that
material
Pressures Produced By Gases
• The deep layer of air which blankets the earth exerts a pressure
much like the water pressure at the bottom of the ocean. This
pressure is known as atmospheric pressure and is about 14.7 PSI at
sea level. At higher elevation, the atmospheric pressure falls.
• The weight of air causes an atmospheric pressure of 14.7 psia. Less
than 14.7 psia is a vacuum.
Gauge Pressure
• Most of pressure gauges are calibrated so that “0” on the gauge is
atmospheric pressure.
• Reading therefore is called PSIG
Standard Pressure
• The volume of a given sample of a gas has almost no meaning
unless the pressure is specified.
• For uniformity, 760 mm of mercury, or 1 atm, has been adopted as
the standard pressure.
• All specific properties of gases such as density – are always at
standard pressure.
Vacuum
• When a pressure is less than atmospheric pressure, it is called a
vacuum. A vacuum is a lack of air or fluid.
Laws Governing Gas Under Pressure
Boyle Law
• At constant temperature, the volume of a fixed weight of a given gas
is inversely proportional to the pressure under which it is measured.
P x V = K (at constant temp)
P1V1 = P2V2 at constant temperature
If the pressure is doubled, volume is halved
If the pressure is halves, volume is doubled
Charles’ Law
• At constant pressure, the volume of a fixed weight of gas is directly
proportional to the absolute temperature
V = kT (at constant pressure)
V1 = T1 (at constant pressure)
V2 T2
Dalton’s Law of Partial Pressure
• Each gas in a gaseous mixture exerts a partial pressure equal to the
pressure which it would exert if it were the only gas present in the
same volume, and the total pressure of the mixture is the sum of the
partial pressures of all the component gases.
P total = P1 + P2 +P3 …….
Raoult’s Law
Ideal solution - exhibits no change in the properties of its constituents
beyond that of dilution.
- For our present purpose an ideal solution is most
usefully defined in terms of generalization known as
Raoult’s law.
Raoult’s Law
- At any given temperature, the vapor pressure of any component of a
solution (i.e, its partial pressure in the mixed vapor in equilibrium
with the solution) is equal to the product of the mole fraction of that
component in the solution and its vapor pressure in thepure liquid
state at the same temperature.
Raoult’s Law
Raoult’s Law (con’t)
Thus in a solution composed of 2 volatile liquid A and B, the vapor
pressure of which in the pure state are PAo and PBo respectively, we
may write:
PA = XAPAo
PB = XBPBo
Total vapor pressure P of an ideal solution:
P = XAPAo + XBPBo
Problem solving
1.
A sample of gas occupies a volume of 86.8 ml at a pressure of
730 mm of mercury and a temperature of 27oC. What will be its
volume at standard pressure and 27oC?
Solution:
Standard pressure = 760 mm hg
V1 = 86.8 ml
P1 = 730 mm
V2 = ? Ml
P2 = 760 mm
Pressure is increased from 730 to 760 mm, the volume is therefore
decreased, and the original volume must be multiplied by a fraction
less than 1.
V2 = 86.8 x 730/760 = 83.4 ml
Problem Solving
2.
Calculate the volume that 22.4 liters of a gas measured at the
standard pressure would occupy at a pressure of 732 mm hg.
V1 = 22.4 liters
V2 = ? Liters
P1 = 760 mm
P2 = 732 mm
Since the pressure is decreased to 730/760 of its original value,
the volume is increased by the factor 760/732
V2 = 22.4 liters x 760/732 = 23.3 liters
Problem Solving
3.
A gas measures 83.4 ml at 1 atm pressure and a temperature of
27 oC. Calculate its volume at standard condition.
V1 = 83.4 ml
V2 = ? Ml
T1 = 27 + 273 = 300 oK
T2 = 273 oK
Since the temperature is decreased, the volume will also be
decreased. The original volume must therefore br multiplied by a
fraction less than 1, namely 273/300:
V2 = 83.4 ml x 273/300 = 75.9 ml
Problem Solving
4.
A sample of gas occupies a volume of 86.8 ml at a pressure of 730
mm and a temperature of 27 oC. What will be its volume at
standard condition?
V1 = 86.8 ml
V2 = ? Ml
T1 = 300 oK
T2 = 273 oK
P1 = 730
P2 = 760
Pressure is increased from 730 to 760 mm, volume should
decrease, we therefore multiply the original volume by 730/760.
Temperature decrease from 300 to 273 oK, likewise volume
decreases, hence we multiply by the temperature fraction 273/300,
therefore:
V2 = 86.8 ml x 730/760 x 273/300 = 75.9 ml.
Problem Solving
5. A certain gas measures 546 ml at a pressure of 1 atm and a
temperature of -80 oC. Calculate the volume it would occupy at a
pressure of 1.5 atm and a temperature of 30 oC.
First we change the centigrade temperature to absolute:
- 80 oC = 80 + 273 = 193 oK
30 oC = 30 + 273 = 303 oK
V1 = 546 ml
V2 = ? ml
P1 = 1 atm
P2 = 1.5 atm
T1 = 193 oK
T2 = 303 oK
V2 = 546 ml x 1/1.5 x 303/193 = 571 ml
Joule-Thomson Effect
• Gases are cooled upon sudden expansion from high to low pressure,
even if they do no external work.
• As gas expands the molecules pull apart from each other, and work
must be done to overcome the cohesive forces that tend to hold the
molecules together. Since this work is done at the expense of the
kinetic energy of the molecules of the gas, the temperature is lowered
on expansion.
Standard Volume
• Volume measured at standard condition ie temp = 60 oF
• Pressure = 1 atm (14.7psi)
• At standard condition one (1) mole of gas occupies 22.4 liters