# Hydraulic Fluids Components and Symbols

**Topics:**Valve, Valves, Fluid dynamics

**Pages:**38 (845 words)

**Published:**April 20, 2015

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

4...

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