hydraulic subtopic

1. Pressure, Force
& Energy

2. Mass and force
• Force – Arising from gravitational
attraction between the mass of an
object and the earth.
• This force is A.K.A. weight
F = W = mg [kgms-2 or N]
• m: mass [kg]

3. • Pressure in fluids: The force acting per unit area,
P = F/A [Pa or Nm-2]
• 100 kPa = 1 atm = 1 bar
• Increase force, increase pressure.
• Decrease area, increase pressure.
• Example: Force F is applied to and enclosed fluid
via piston of area A. Pressure P is produced.
Force and pressure

4. • Pressure arising in fluid from weight of fluid: Head
pressure.
• Dependent of height (h) and density (ρ),
P = ρgh
Pressure and weight

5. • What happen to the pressure in the system?
A – P1>P2
B – P1<P2
C – P1=P2
Pressure transfer

6. Pressure measurements
Differential
pressure
Gauge
pressure
Absolute
pressure

7. • Almost all pressure transducers measure the
pressure difference between two input ports.
• Pressure transmitter indicates P1-P2 (= ΔP)
Differential pressure

8. • Almost universally used in hydraulic and pneumatic
systems.
• Low pressure input port is open to atmosphere.
Pressure transmitter indicates pressure above
atmospheric pressure.
Gauge pressure

9. • Pressure transmitter measuring pressure with respect
to vacuum.
• Important when compression of gases are
considered.
Absolute pressure

10. Gauge pressure and
absolute pressure

11. Example
• A lifting is to lift a load of 15kN and is to have
a system pressure of 75 bar. How large does
the piston surface need to be?
Solution:
P = F/A
A = F/P
= 15000N/(75x105 Pa)
= 0.002 m2

12. • Work (W) is done/energy transferred when an object is
moved at a certain distance (s) against a force (F),
W = F × s [J or Nm]
• Power : Rate of work,
Power = W/t (time) [Js-1 or Watt]
• 1 kW = 1.34 Hp
• Given Flow rate (Q) = Volume [m3]/t [s],
Derive Power = P × Q
• Prove that Power = P × Q = W/t
Work, energy & power
Pipe area A

13. • The concepts of hydraulic energy, power, and
power transformation are simply explained in the
following: Consider a forklift that lifts a load vertically
for a distance y during a time period Δt.

14. • To fulfill this function, the forklift acts on the load by
a vertical force F. If the friction is negligible, then in
the steady state, this force equals the total weight
of the displaced parts (F=mg). The work done by
the forklift is
W=Fy
The energy delivered to the lifted body per unit of
time is the delivered power N, where
N = Fy/Δt = Fv
N=Mechanical power delivered to the load, W
v=Lifting speed, m/s

15. • The load is lifted by a hydraulic
cylinder. This cylinder acts on
the lifted body by a force F and
drives it with a speed v.
• The pressurized oil flows to the
hydraulic cylinder at a flow rate
Q (volumetric flow rate, m3/s)
and its pressure is p. Neglecting
the friction in the cylinder, the
pressure force which drives the
piston in the extension direction
is given by F = pAp.

16. Flowrate
• During the time period, Δt, the piston travels
vertically a distance y. The volume of oil that
entered the cylinder during this period is V=Apy.
• Then, the oil flow rate that entered the cylinder is
• Assuming an ideal cylinder, then the hydraulic
power inlet to the cylinder is

17. • Torque (T) is a rotary force, a product of force (F)
and the effective radius (r),
T = F × r
Torque
r

18. END OF LECTURE