Week 1 2_pressure_force_and_energy3

Pressure, Force & Energy

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]

• 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

• Pressure arising in fluid from weight of fluid: Head

pressure.

• Dependent of height (h) and density (ρ),

P = ρgh

Pressure and weight

• What happen to the pressure in the system?

A – P1>P2

B – P1<P2

C – P1=P2

Pressure transfer

Pressure measurements

Differential pressure

Gauge pressure

Absolute pressure

• Almost all pressure transducers measure the

pressure difference between two input ports.

• Pressure transmitter indicates P1-P2 (= ΔP)

Differential pressure

• 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

• Pressure transmitter measuring pressure with respect

to vacuum.

• Important when compression of gases are

considered.

Absolute pressure

Gauge pressure and absolute pressure

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

• 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

• 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.

• 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

• 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.

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

• Torque (T) is a rotary force, a product of force (F)

and the effective radius (r),

T = F × r

Torque

r

END OF LECTURE