# Hydraulic Fluids Components and Symbols

Topics: Valve, Valves, Fluid dynamics Pages: 38 (845 words) Published: April 20, 2015
Hydraulic Systems

Examples of hydraulic systems

Basic hydraulic system

Basic fluid power symbols

Basic fluid power ANSI/ISO symbols

Basic fluid power symbols

Hydraulic Pumps

Pump characteristic
Operating pressure
 Speeds
 Displacement volume (V) – volume of liquid per
revolution
 Volumetric flow rate: Q = n × V
where n : number of rotation (rpm)
V : displacement volume (per rev)

Pump Efficiency (Volumetric)
To determine performance of pump
 Divided into two:

Volumetric efficiency (ηvol) : Indicates amount of
leakage that takes place in the pump. Effected by
pressure.

ηvol = QA/QT
= (actual flow rate)/(theoretical flow rate)

Pump Efficiency (Mechanical)

Mechanical efficiency (ηm): Indicates amount of
energy losses due to reasons other than leakage.
ηm = pQT/TAN
= (pump output power, no leakage)/(actual power
delivered to pump)
where p : pump discharge pressure [Pa]
QT: pump theoretical flow rate [m3/s]
TA : theoretical torque delivered to pump [Nm]
N : pump speed [rpm]

Pump Efficiency (Mechanical)

Or
ηm = TT/TA

= (theoretical torque to operate pump)/(actual torque
delivered to pump)
where
TT [Nm] = (V [m3] × P [Pa])/2π
TA = (actual power delivered to pump [W])/(2πN/60 [rpm])

Total/overall efficiency

Total efficiency: ηtot = ηvol × ηm

where

ηtot : total efficiency
ηvol : volumetric efficiency
ηm : mechanical/motor efficiency

Example (1)
A leakage of oil from a pump is 6% at 230 bar. Calculate
the total efficiency if the flow rate at 0 bar is 10 dm3min-1 and the motor efficiency is 75%.
 Solution:
Q (P = 0 bar) = 10 dm3min-1
Q (P = 230 bar) = 10 × 0.94 = 9.4 dm3min-1
ηmotor = 0.75, ηvol = 9.4/10 = 0.94
Therefore
ηtot = ηmotor × ηvol = 0.705 (= 70.5 %)

Example (2)
A pump has a displacement volume of 100 cm3. It
delivers 0.0015 m3/s at 1000 rpm and 70 bars. The
prime mover input torque is 120 Nm.
a) What is the overall efficiency of the pump?
b) What is the theoretical torque required to operate
the pump?

Solution

a) From QT = V × n,
Given V = 100 cm3/rev
= 0.0001 m3/rev
QT = V × n
= 0.0001 m3/rev × (1000/60 revs-1)
= 0.00167 m3/s

Solution

Solve volumetric efficiency
ηvol = QA/QT
= 0.0015/0.00167 = 0.898 = 89.8%
Solve mechanical efficiency
ηm = PQT/TAN

= (70 × 105)(0.00167)/(120)(1000 × (2π/60))
= 0.93 = 93%
Therefore, ηtot = 0.93 × 0.898 = 0.835 = 83.5%

Solution

b) ηm = TT/TA
TT = ηm × TA = 0.93 × 120 = 112 Nm

Example (3)
The pump in Example 2 is driven by an electric motor
having an overall efficiency of 85%. The hydraulic
system operates 12 hours per day for 259 days per
year. The cost of electricity is RM0.11 per kWh.
Determine:
a) The yearly cost of electric to operate the hydraulics
system.
b) Amount of yearly cost of electricity that is due to the
inefficiencies of the electric motor and pump.

Hydraulic Valves

Hydraulic valves
Device for controlling the energy flow direction.
 4 basic valve types:

◦ Directional control valve
◦ Non-return valves
◦ Pressure valves
◦ Flow control valves

Basic valve symbols

Basic valve symbols

Go to:
http://www.airlinehyd.com/knowledgecenter/symbols.asp

1. Directional control valve
Change, open or close flow path in hydraulic system.
 Types:
1. 2 ports/2 way
2. 3 ports/2 way
3. 4 ports/2 way
4. 4 ports/3 way

2 ports/2 way valve
1 working port and 1 pressure port
 Control delivery by closing or opening the passage

3 ports/2 way valve
1 working port, 1 pressure port and 1 tank
connection
 Control delivery
o Normal position – P is closed and A to T is open
o Actuated position – T is closed, flow from P to A

4 ports/2 way valve
2 working port (A, B), 1 pressure port and 1 tank
connection
 Control delivery
o Normal position – flow from P to B and A to T
o Actuated position – flow from P to A and B to T

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