- Fluids can be treated as continuous in time and space, and a fluid takes up the shape of its container. Absolute and gauge pressures are used in engineering. The perfect gas equation can be used to obtain gas properties.

1. key points
• A stationary fluid cannot withstand a shear stress but
deforms under the action of a shear.
• Fluids can be treated as continuous in time and space.
• A fluid is shaped by external forces (i.e. a fluid takes up the
shape of its container)
• Gauge pressure equals the absolute pressure minus the
atmospheric pressure, or
• Absolute pressure equals the gauge pressure plus the
atmospheric pressure.

2. key points
• Absolute pressure (relative to vacuum) and gauge pressure
(relative to atmospheric pressure) are both used in engineering.
• Temperatures must be in kelvin in the perfect gas equation.
• To convert from C to K, add 273.15.
• Pressure in a fluid is associated with molecular motion.
• When pressure is constant over an area, (F = P . A) .

3. key points
• The perfect gas equation of state can be used to
obtain gas properties.
• Liquids can usually be treated as incompressible
but gases cannot.

4. Learning summary
By the end of this section you will be able to:
• Calculate the density of an object from its mass and
volume;
• Know the difference between and calculate gauge and
absolute pressure;
• Use the perfect gas equation of state to calculate the
properties of gases.

5. key points
• In a fluid at rest, pressure acts equally in all directions.
• Where a fluid is in contact with a surface, the pressure
gives rise to a force acting perpendicular to the surface.
• In a fluid at rest, pressure is constant along a horizontal
plane.

6. key points
• In a fluid at rest, pressure increases with depth according
to the relationship p = g. h
• The pressure at the base of a column of fluid of depth “h” is
equal to the pressure at the top + g.h

7. key points
• Pressures can be measured by manometers.
• Surface tension can affect the readings of manometers.
• On a submerged horizontal surface the pressure is
constant and the centre of pressure is also the centre of
area (centroid).
• On a submerged vertical surface the pressure increases
with depth and the centre of pressure is below the
centroid.

8. key points
• On an inclined flat or curved surface the horizontal force
and its line of action is equal to the horizontal force on the
vertical projection of the inclined or curved surface.
• On an inclined flat or curved surface, the vertical force is
equal to the weight of the volume of water vertically above
the surface and its line of action passes through the centre
of gravity of that volume.

9. key points
• A body fully immersed in a fluid experiences a vertical
upwards force equal to the weight of the volume of
fluid displaced.
• A floating body displaces its own weight in liquid.

10. Learning summary
By the end of this section you will be able to:
• Determine the pressure at any depth below the surface of a
liquid at rest;
• Calculate the pressure difference indicated by a manometer;
• Calculate by integration the magnitude and line of action of the
force due to fluid static pressure on a submerged, flat, horizontal
or vertical surface;

11. Learning summary
• Evaluate the horizontal and vertical components of force on
a submerged, inclined, flat or simple curved surface and
determine the resultant force and line of action for some
simple shapes;
• Calculate buoyancy forces on submerged and floating
objects and determine the conditions for equilibrium.

12. key points
• A steady flow is one that does not change with time.
• A uniform flow is one where the properties do not vary
across a plane or within a volume.
• A one-dimensional flow is one where flow properties only
vary in one direction.
• An ideal (inviscid) fluid has no viscosity, and is
incompressible. No real fluid is ideal but the simplification is
valid in some situations.

13. key points
• For a steady flow, the law of conservation of mass (continuity)
means that the flow entering a volume must equal the flow
leaving a volume.
• The Euler equation is a differential equation for the flow of an
ideal fluid.
• The Bernoulli equation expresses the relationship between
pressure, elevation and velocity in an ideal fluid for steady flow
along a streamline.
• The Bernoulli equation can be expressed in three ways, in terms
of specific energy, pressure or head.

14. key points
• The Bernoulli equation can be used for real fluids when
losses due to friction are negligible and the fluid is
incompressible.
• Viscosity in a fluid creates frictional drag in a fluid flow.
• The higher the viscosity of a fluid, the greater is the
resistance to motion between fluid layers, and, for a given
applied shear stress, the lower is the rate of shear
deformations between layers

15. key points
• There are two fundamental types of fluid flow laminar and
turbulent. They can be characterized by the Reynolds
number of the flow.
• The steady flow energy equation can be applied to the flow of
real fluids and leads to the extended Bernoulli equation that
can be used to describe the losses resulting from viscous
friction in a flow.
• The Moody Chart can be used to estimate the frictional
losses in pipe and duct flows.
• The SFEE “Steady Flow Energy Equation” can be used to
determine the performance of a pump needed in a pipe
system.

16. Learning summary
By the end of this section you will be able to:
• Calculate the mass flow rate of a steady flow in a pipe or
duct;
• Understand the three forms of the Bernoulli equation;
• Be able to apply the Bernoulli equation to calculate flows in
pipes including the performance of Venturi, nozzle and orifice
plate flow meters and a Pitot-static probe;
• Calculate the drag forces created by viscosity in thin films
between moving surfaces;

17. Learning summary
• Calculate the Reynolds number of flows in pipes and ducts
and determine whether the flow is likely to be laminar or
turbulent;
• Estimate the pressure losses in flows in pipes due to friction;
• Calculate the pressures and flows in simple single pipe
systems accounting for losses due to friction in pipes and
other components of pipe systems;
• Calculate the performance of a pump needed in a simple
pipe flow system.

18. key points
• The forces exerted by fluids when they change velocity
and direction can be evaluated using the momentum
equation derived from Newton’s second law of motion.
• Control volumes are the easiest way to analyse
momentum flow problems.
• The linear momentum equation states that for a steady
flow: The force in a particular direction on a control
volume is equal to the rate of change in momentum of
the fluid flowing in that direction.

19. key points
• The force on the control volume is the sum of all the
forces acting, including external pressure, gravity and
structural forces from solid objects crossing the control
volume boundaries.

20. Learning summary
 Calculate the forces generated when a fluid flow
impinges on an object or is constrained to change
direction.
By the end of this section you will be able to: