Lecture 31

ACCESSORIES USED IN FLUID POWER SYSTEMS [CONTINUED]

1.6 Filters and Strainers

For proper operation and long service life of a hydraulic system, oil cleanliness is of prime importance.

Hydraulic components are very sensitive to contamination. The cause of majority of hydraulic system

failures can be traced back to contamination. Hence, filtration of oil leads to proper operation and long

service life of a hydraulic system.

Strainers and filters are designed to remove foreign particles from the hydraulic fluid. They can be

differentiated by the following definitions:

1. Filters: They are devices whose primary function is the retention, by some fine porous medium, of

insoluble contaminants from fluid. Filters are used to pick up smaller contaminant particles because they

are able to accumulate them better than a strainer. Generally, a filter consists of fabricated steel housing

with an inlet and an outlet. The filter elements are held in position by springs or other retaining devices.

Because the filter element is not capable of being cleaned, that is, when the filter becomes dirty, it is

discarded and replaced by a new one. Particle sizes removed by filters are measured in microns. The

smallest sized particle that can be removed is as small as 1 µm. A strainer is a device whose function is to

remove large particles from a fluid using a wire screen. The smallest sized particle that can be removed

by a strainer is as small as 0.15 mm or 150 µm.

2. Hydraulic strainers: A strainer is a coarse filter. Fluid flows more or less straight through it. A

strainer is constructed of a fine wire mesh screen or of screening consisting of a specially processed wire

of varying thickness wrapped around metal frames. It does not provide as fine a screening action as filters

do, but offers less resistance to flow and is used in pump suction lines where pressure drop must be kept

to a minimum. A strainer should be as large as possible or wherever this is not practical, two or more may

be used in parallel.

1.6.1 Causes of Contamination

The causes of contamination are as follows:

o Contaminants left in the system during assembly or subsequent maintenance work.

o Contaminants generated when running the system such as wear particles, sludge and varnish due

to fluid oxidation and rust and water due to condensation.

o Contaminants introduced into the system from outside. These include using the wrong fluid when

topping up and dirt particles introduced by contaminated tools or repaired components.

1.6.2 Types of Filters

Filters may be classified as follows:

1. According to the filtering methods:

Mechanical filters: This type normally contains a metal or cloth screen or a series of metal disks

separated by thin spacers. Mechanical filters are capable of removing only relatively coarse

particles from the fluid.

Absorption filters: These filters are porous and permeable materials such as paper, wood pulp,

diatomaceous earth, cloth, cellulose and asbestos. Paper filters are impregnated with a resin to

provide added strength. In this type of filters, the particles are actually absorbed as the fluid

permeates the material. Hence, these filters are used for extremely small particle filtration.

Adsorbent filters: Adsorption is a surface phenomenon and refers to the tendency of particles to

cling to the surface of the filters. Thus, the capacity of such a filter depends on the amount of

surface area available. Adsorbent materials used include activated clay and chemically treated

paper.

2. According to the size of pores in the material:

Surface filters: These are nothing but simple screens used to clean oil passing through their

pores. The screen thickness is very thin and dirty unwanted particles are collected at the top

surface of the screen when the oil passes, for example, strainer.

Depth filters: These contain a thick-walled filter medium through which the oil is made to flow

and the undesirable foreign particles are retained. Much finer particles are arrested and the

capacity is much higher than surface filters.

3. According to the location of filters:

Intake or inline filters (suction strainers): These are provided first before the pump to protect

the pump against contaminations in the oil as shown in Fig. 1.15. These filters are designed to

give a low pressure drop,otherwise the pump will not be able to draw the fluid from the tank. To

achieve low pressure drop across the filters, a coarse mesh is used. These filters cannot filter out

small particles.

Figure 1.15Suction filter.

Advantages of suction filters:

(a) A suction filter protects the pump from dirt in the reservoir. Because the suction filter is outside the

reservoir, an indicator telling when the filter element is dirty can be used.

(b) The filter element can be serviced without dismantling the suction line or reservoir (easy to maintain).

Disadvantages of suction filters:

(a) A suction filter may starve the pump if not sized properly.

Pressure line filters (high-pressure filters): These are placed immediately after the pump to

protect valves and actuators and can be a finer and smaller mesh (Fig. 1.16). They should be able

to withstand the full system pressure. Most filters are pressure line filters.

Figure 1.16Pressure filter.

Advantages of a pressure line filter:

(a) A pressure filter can filter very fine contaminants because the system pressure is available to push

the fluid through the element.

(b) A pressure filter can protect a specific component from the harm of deteriorating particles

generated from an upstream component.

Disadvantages of a pressure line filter:

(a) The housing of a pressure filter must be designed for high pressure because it operates at full

system pressure. This makes the filter expensive.

(b) If pressure differential and fluid velocity are high enough, dirt can be pushed through the element

or the element may tear or collapse.

Return line filters (low-pressure filters): These filters filter the oil returning from the pressure-

relief valve or from the system, that is, the actuator to the tank (Fig. 1.17). They are generally

placed just before the tank. They may have a relatively high pressure drop and hence can be a fine

mesh. These filters have to withstand low pressure only and also protect the tank and pump from

contamination.

Figure 1.17Return line filter.

Advantages of a return line filter:

(a)A return line filter catches the dirt in the system before it enters the reservoir.

(b)The filter housing does not operate under full system pressure and is therefore less expensive

than a pressure filter.

Disadvantages of a return line filter:

(a) There is no direct protection for circuit components.

(b) In return line full flow filters, flow surges from discharging cylinders, actuators and accumulators

must be considered when sizing.

1. Depending on the amount of oil filtered by a filter:

Full flow filters: In this type, complete oil is filtered. Full flow of oil must enter the filter element

at its inlet and must be expelled through the outlet after crossing the filter element fully(Fig.

1.18). This is an efficient filter. However,it incurs large pressure drops. This pressure drop

increases as the filter gets blocked by contamination.

Figure 1.18Full flow filter.

Proportional filters (bypass filters): In some hydraulic system applications, only a portion of oil

is passed through the filter instead of entire volume and the main flow is directly passed without

filtration through a restricted passage (Fig. 1.19).

Figure 1.19Proportional filter.

1.6.3 Beta Ratio of Filters

Filters are rated according to the smallest size of particles they can trap. Filter ratings are identified by

nominal and absolute values in micrometers. A filter with a nominal rating of 10 μm is supposed to trap

up to 95% of the entering particles greater than 10 μm in size. The absolute rating represents the size of

the largest pore or opening in the filter and thus indicates the largest size particle that could go through.

Hence, absolute rating of a 10 μm nominal size filter would be greater than 10 μm.

A better parameter for establishing how well a filter traps particles is called the beta ratio or beta rating.

The beta ratio is determined during laboratory testing of a filter receiving a steady-state flow containing a

fine dust of selected particle size. The test begins with a clean filter and ends when pressure drop across

the filter reaches a specified value indicating that the filter has reached the saturation point. This occurs

when contaminant capacity has been reached.

By mathematical definition, the beta ratio equals the number of upstream particles of size greater than

Nμm divided by the number of downstream particles having size greater than Nμm where N is the

selected particle size for the given filter. The ratio is represented by the following equation:

Beta ratio = No. of upstream particles of size >  μm

No. of downstream particles of size > μm 

N

N (1.4)

A beta ratio of 1 would mean that no particle above specified N are trapped by the filter. A beta ratio of

50 means that 50 particles are trapped for every one that gets through. Most filters have a beta ratio

greater than 75:

Beta efficiency = No. of upstream particles No. of downstre am particles

No. of upstream particles

Thus,

Beta efficiency = 1 − 1

Beta ratio (1.5)

1.7 Heat Exchangers

The heating up of hydraulic oil beyond tolerable limits in an otherwise well-designed hydraulic systemis

usually a phenomenon associated with high-pressure, high-flow systems cycling at high frequencies. The

input power in such systems is usually more than 40 kW.

Heat is generated in hydraulic systems because no component can operate at 100% efficiency. Significant

sources of heat include the pump, pressure-relief valves and flow control valves. Heat can cause the

hydraulic fluid temperature to exceed its normal operating range of 35–70C. Excessive temperature

hastens the oxidation of the hydraulic oil and causes it to become too thin. This promotes deterioration of

seals and packing and accelerates wear between closely fitting parts of hydraulic components of valves,

pumps and actuators.

The steady-state temperature of fluid of a hydraulic system depends on the heat-generation rate and the

heat-dissipation rate of the system. If the fluid operating temperature in a hydraulic system becomes

excessive, it means that the heat-generation rate is too large relative to the heat-dissipation rate. Assuming

that the system is reasonably efficient, the solution is to increase the heat-dissipation rate. This is

accomplished by the use of coolers, which are commonly called “heat exchangers.”

It is basically a problem associated with fluid power systems using constant delivery pumps in which, for

most part of the cycle, the fluid is dumped to the tank through the relief valves at high pressures resulting

in wasted power. The consequent heating of oil should not be a cause for concern if oil temperatures do

not exceed 30C above ambient even after 8 h of continuous operation.

Injection molding machines and huge hydraulic presses are typical examples for high-pressure, high-flow

systems. If the frequency of the operating cycle is also high, then heat builds up because the high

operating cycle does not permit heat dissipation. The consequent temperature raise shall be between 40

and 60C above ambient. In places where the ambient temperatures are above 30C, this heating up of oil

would be a cause for concern. Electrical valves begin to malfunction and leakage past the seals and spool

valves due to viscosity changes would affect the system performance. A poorly designed hydraulic

system is another cause for heating up of oil. The glow ratings of all valves and conductors must be

adequate to handle the pump flow. In a double-acting cylinder with a large piston area to annular area

ratio, valves must be selected to handle the discharge out of the rod end of the cylinder and not based on

the flow into the cap end. Heat exchangers are designed primarily to overcome this problem. If installed

at an appropriate location, which is the tank return line, heat exchangers help the system to dissipate heat

and maintain oil temperatures within normal limits.

In some applications, the fluid must be heated to produce a satisfactory value of viscosity. This is typical

when, for example, mobile hydraulic equipment is to operate below 0C. In these cases, the heat

exchangers are called “heaters.” However, for most hydraulic systems, the natural heat-generation rate is

sufficient to produce high enough temperatures after an initial warm-up period.

Basically, there are two types of heat exchangers: liquid-to-liquid and liquid-to-air type. Liquid-to-liquid

types are of shell and tube construction consisting of a bundle of small tubes held inside a shell. The

coolant flows through the small tubes, while the hydraulic oil passes around and between these tubes. The

tubes can be either of the straight-type or of U-type. Straight-type units have higher thermal efficiency

and are less expensive. U-types are best suited for high-temperature and high-pressure applications. They

can accommodate thermal expansion but the tube banks are difficult to clean. The heat exchangers can be

of single pass or double pass and can be of parallel-flow or counter-flow type. The double-pass counter-

flow type provides maximum heat transfer for a given size.

For hydraulic applications, the commonly used coolant is water. Standard liquid-to-liquid units can

handle pressures up to 15 bar and temperatures up to 150C although the actual temperature difference

between oil and water should not exceed 90C. The mechanism of action of liquid-to-liquid heat

exchangers is one of evaporation and condensation. The tubes are basically an enclosure containing a

fluid that can be vaporized and a material that provides capillary action. In what is basically an isothermal

process, heat at the input end causes the fluid to evaporate. Vapor travels through the tube to the

condenser or output end. Here the vapor condenses giving up its latent heat and returns to the fluid state

and travels back to the system through the evaporator end. Clean and soft water should be used to prevent

corrosion and scaling in the tubes. If the water is hard and has excessive salt, it can lead to clogging of the

insides of the narrow tube due to scale and dirt deposits. Liquid-to-air heat exchangers transfer heat from

the hydraulic fluid to the atmosphere, just like an automobile radiator. The air passes over-finned tubes

made of either copper or aluminum in which the hot hydraulic fluid circulates.

Heat exchangers are available as of the shelf components in various sizes and configurations. They are

available with temperature control valves, water strainers and bypass check valves. Bypass valves protect

heat exchangers by diverting the excess oil to the tank during surge and peak pressures.

1.7.1 Sizing of Heat Exchangers

The selection of a suitable heat exchanger involves determining the total heat equivalent to wasted energy

or the heat load. This can be assessed by analyzing the duty cycle of the hydraulic machine. The total time

period of a cycle must be broken up into its phases consisting of idling, approach, work, return, etc. and

the wasted energy calculated for each.

The sum of the energy less the energy dissipated through the reservoirs and flow lines by convection and

radiation constitutes the net energy that the heat exchanger must dissipate or the rate of heat transfer. The

maximum oil temperature at the inlet point to the heat exchanger can now be determined by using the

equation

p 1 0( )H mC T T (1.6)

Here H is the heat transfer rate in kJ/s;Cp is the specific heat at constant pressure that for hydraulic oil =

1.97 kJ/kgK;T1 is the oil temperature at the inlet to the heat exchanger, which typically lies between 55

and 65C;T0 is the oil temperature at the outlet, which is the desired oil temperature that the heat

exchanger must maintain. A suitable value may be assumed forT0which is consistent with the system

requirement. Also m is the mass flow rate determined by the equation

  (kg /s)m q (1.7)

where is the density of hydraulic oil in kg/m3 and q is the oil flow rate in m

3/s. If the difference between

T1 and T0 is large, then it justifies the need for a heat exchanger. If the difference is only marginal, the heat

exchanger may be dispensed with. By sizing the surface area of the reservoir, so that it can transfer

enough heat from oil to the atmosphere by conduction, convection or radiation, the net wasted energy that

the heat exchanger must dissipate can be reduced, thus reducing the size of heat exchanger.

After ascertaining the need for the heat exchanger, the size of heat exchanger in terms of quantity of water

required to dissipate the heat and maintain the desired temperature must be determined based on the

equation

w

1 w

(860)( )PQ

T T

(1.8)

wherePw is the wasted energy in kW,Q is the water flow rate through the heat exchanger in L/min, 1Tis the

temperature of the oil and wTis the water temperature. When the flow capacity as determined from the

equation is much larger than the capacity of available heat exchangers, then two smaller heat exchangers

of equal sizes can be used in parallel. Standard heat exchangers would have the oil-to-water ratio of either

1:1 or 2:1.This means that for every liter of oil circulated, one or half a liter of water must be circulated to

achieve the desired oil temperature. The size and therefore the cost of a heat exchanger depend largely on

the initial temperature difference between the inlet oil and water. If this difference decreases, the cost

decreases. The pressure drop that the hydraulic system can permit is another factor that has a direct

bearing on the size and cost of the heat exchanger. Ideal allowable pressure drops should be between 3.5

and 5 bar.

If it becomes necessary to install a heat exchanger in an existing system, then the waste energy may safely

be assumed equal to 20% of the connected power. The value of T1 may be obtained directly by measuring

the maximum temperature of the oil in the tank and heat exchanger size, based on Eq. (1.8) above. Shell

and tube heat exchangers can be mounted either vertically or horizontally.

Example 1.1

In a hydraulic system operating at 200 bar, the pump delivery is 25 LPM and input power to the pump

drive is 10 kW. The system cycle is such that the pump is unloaded at 60% of the operating time. The

overall efficiency of the systems when it is on load is 65%. If the ambient temperature is 15C and the

maximum permissible fluid temperature in the reservoir is 50C, calculate a suitable size for the fluid

reservoir assuming that

(a) Normal air circulation around the fluid reservoir doubles the cooling owing to the natural radiation.

(b) The fluid reservoir is of square section of side a with a length of 2a.

Solution: Heat dissipation from the vertical plate:

v v   3.6H h A T

where T = 50 – 15 = 35C and A = 6a2. Also 1/4

v

1/435

  1.42  3.45 ( )h aa

So

1/4 2

v 3.45 ( ) 6   35  3.6 (W)H a a = 2608 7/4( )a (W)

Heat dissipation from the horizontal top plate:

H H   3.6H h A T

where T = 50 – 15 = 35C and A = 6a2. Also

1/4

1/4

H

35  1.32  3.21 ( )h a

a

So

1/4 2

H 3.21 ( ) 2  35 3.6H a a = 809 7/4( )  Wa

Heat dissipation owing to the natural radiation:

v HH H 2608 7/4( )a 809 7/4( )a =3417 7/4( )a

Heat dissipation with normal air circulation may be taken as twice that owing to natural radiation and is

equal to 6834 7/4( )a W.

Heat input to the system during “on load” part of cycle is

10 × (1 − 0.65) = 3.5 kW

But the system is only on load for 40% of the time; thus the average heat energy input to the fluid is given

by

Average input = 0.4 × 3.5 = 1.4 kW = 1400 W

For thermal equilibrium, the heat energy entering the system must be equal to the heat energy dissipated

from the system; therefore,

6834 7/4( ) 1400a

Solving, we get a = 0.404 m = 40.4 cm.

Assume the tank material measurement to be 0.4 m wide, 0.4 m high and 0.8 m long; then the volume of

the fluid contained is given by

Q = 0.4 × 0.4 × 0.8= 0.128 m3 = 128 L

In practice, the tank has to be higher than 0.4 m as there must be clearance volume above the oil. Size the

reservoir from the rule of thumb formula, that is, the reservoir capacity is equal to three to four times the

pump delivery per minute.

Reservoir capacity = −75 to 100 L

This value is similar to the value calculated.

If a very large fluid reservoir is needed to dissipate the heat energy, it may be advantageous to use a fluid

cooler.

Example 1.2

Determine the beta ratio of a filter when, during test operation, 30000 particles greater than 20 m enter

the filter and 1050 of these particles pass through the filter. What is the beta efficiency?

Solution: We have

No. of upstream particles of size   μm 30000Beta ratio 28.6

 No. of downstream paricles of size  μm  1  050 

N

N

Beta efficiency = No. of upstream particles No. of downstream particles 30000 1050

96.5%No. of upstream particles 30000

Beta efficiency could also be calculated as

Beta efficiency = 1 − 1 1

1 96.5%Beta ratio 28.6

Example 1.3 A hydraulic machine has the following duty cycle: Idle at 15 bar for 2 s, clamp workpiece at 100 bar for 5

s, approach at 15 bar for 2 s, perform work at 300 bar for 3 s, declamp, return at 15 bar for 2 s.The pump

flow is 100 LPM, the total surface area of the oil reservoir is 2.5 m2 and the hydraulic pipeline is

25 mm × 2500 mm. Calculate the net wasted energy that needs to be dissipated and recommend a suitable

heat exchanger if necessary. Assume the following values: Ambient temperature = 20C, volumetric

efficiency of the pump = 0.85, density of oil = 860 kg/m3, 1 kWh = 3.6 × 10

6 J.

Solution: Power dissipated in kW is given by the equation

600

pQ × η

wherep is the pressure in bar, Q is the flow in LPM and η is the volumetric efficiency of the pump.

Power dissipated during the idle cycle while passing oil through valves and pipelines is

15 × 100

600 × 0.85 = 2.94 kW

Because this phase lasts for 30 s, the energy consumed is

2.94 × 30

36000.0245 kWh or 0.0245 (3.6×10

6)= 8.82 J

Power dissipated during clamping is 100 × 100

600 × 0.85 = 19.6 kW

Because this phase lasts for 5 s, the energy consumed is

19.6 × 15

3600 = 0.027 kWh

= 0.027 × (3.6 × 106)

= 9.72×104 J

Power dissipated during approach is 100 × 15

600× 0.85 = 2.94 kW

Because this phase lasts for 2 s, the energy consumed is

2.94 × 2

3600 = 0.001 kWh

=0.001 × (3.6 × 106)

= 3.6 × 103 J

Power dissipated during work phase is

100 × 300

600 × 0.85 = 58.8 kW

Because this phase lasts for 3 s, the energy consumed is

58.8 × 3

3600 = 0.049 kWh

= 0.049 × (3.6×106)

= 17.64 × 104 J

Power dissipated during return is

100 × 15

600 × 0.85 = 2.94 kW

Because this phase lasts for 2 s, the energy consumed is

2.94 × 2

3600 = 0.001 kWh

= 0.001 × (3.6 × 106)

= 3.6 × 103 J

Total power dissipated in 42 s = 2.1×105 J.In terms of average wasted power, this will be

2.1 × 5

6

10

3.6 10 = 0.058 kWh

= 0.058 ×3600

42

= 5 kW.

Available hydraulic power is

300 ×30

3600 × 0.85 = 58.8 kW

Wasted power is 8.5% of available power.

Now

(T1 − T0) = H

m × Cp (1.9)

We have

H = 2.1 × 510

1000 42 = 5 kJ/s

m = 860 × 0.001= 0.86 kg/s

Substituting the above values in Eq. (1.9), we get

(T1−T0) = 5

0.86 × 1.97 = 2.95

Because this difference is negligible, there is no need for a heat exchanger. Heat dissipated by the oil

reservoir and hydraulic pipelines can be calculated as follows: Heat dissipated by the oil reservoir

kW3600

KT A

where ΔT is the difference between the maximum oil temperature and the ambient temperature in C, A is

the surface area of the reservoir in m2 and K the average rate of convection from the reservoir is 6–10

kJ/cm2 h, depending upon the condition of the surroundings. For poor air circulation, assume a value

equal to 6. The surface area of hydraulic pipelines is given by pdl, where d is the outside diameter of the

pipe and l is the length in appropriate units.

Example 1.4

Oil at 49C and 69 bar is flowing through a pressure-relief valve at 38 LPM. What is the downstream oil

temperature?

Solution: First we calculate the power lost/wasted:

Power lost = p (bar) × Q (LPM)

= (69 × 105 N/m

2) (38 × 10

3 m

3/min) × 1/60 (s)

= 4370 W = 4.37 kW

Next we calculate the oil flow rate in units of kg/s and the temperature increase.

Oil flow rate (kg/s) = 895 × Oil flow rate (m3/s)

= 895 × 38 × 310

60

= 0.6 kg/s

Now

Temperature ( C ) =

Heat generation rate (kW) 

kJOil specific heat  °C  Oil flow rate (kg / s)

kg

4.374°C

1.8 0.6

Downstream oil temperature = 49 + 4 = 53 C

Example 1.5 The hydrostatic transmission receives 6 kW from the engine at the pump. The transmission output

delivers 3.7 kW to the rear wheel output drive. Assuming that the difference is lost to heat, compute the

total heat loss over the period of 4 h of operation.

Solution: , Energy loss is given by the difference between the energy supplied(to pump) and energy

taken out( at the actuator), mathematically, we get

Heat energy loss = Operation time (Pump input power – Actuator output power)

= 4 h (6 − 3.7) = 9.2 kWh

The temperature increase resulting from the heat generated in the fluid computed in C/kg of fluid is

attributed to discharging fluid across such a component as pressure-relief valves and from the mechanical

friction of moving parts. In most hydraulic systems, mechanical friction accounts for less than 20% of the

input power.

The formula for computing temperature increase due to heat generation by a fluid passing through

restriction is

Heat generated (kW)

Temperature increase ( C)  

Oil specific weight  C × OikJ

kl flow rate ( kg/s)

g

where the specific heat of oil is given as 1.8 kJ/kgC. . When the flow rate is given in 3m /s and the mass

density of the oil is taken as 895 3kg/m , the mass flow rate of oil is computed from

Mass flow rate (kg/s) = 895 3kg/m Oil flow rate (m3/s)

Example 1.6

The pressure drop across a sticking control valve is observed to be 68.9 bar. If the fluid has a specific

gravity of 0.895 and a flow rate of 0.19 LPS, estimate the rise in temperature of the fluid that can be

attributed to the control valve.

Solution: The heat generated by the pressure drop across the control valve is

Heat loss generated = 68.9 × 105 × 0.19 × 10

–3 = 1309 W/s

The mass flow rate of fluid through the valve is computed as

Mass flow rate  kg/s = 895 3kg/m × 0.19 × 103

m3/s = 0.17  kg/s

Solving for the temperature rise in C of the fluid we get

Heat generated (kW)Temperature increase ( C)  

Oil specific weight  C  Oil flow rate ( kg/s)

13094.28 C

0.17 1.8 1000

kJ

kg

A heat exchanger relieves the hydraulic fluid of excess heat to lower its operating temperature. In gross

terms, the amount of heat to be removed and transferred to the cooling medium equals the difference

between the input to the hydraulic pump and the output of all system actuators. This assumes, of course,

that the existing ambient temperature is appropriate for the system operation and the environmental

conditions do not add or subtract heat from the fluid. This is seldom the case. In extreme environments,

both cold and hot heat exchangers may be employed to counter act primarily environmental

circumstances rather than operating conditions to maintain the temperature of oil within operable limits.

For example, in a cold region, heat is frequently added to the fluid to decrease viscosity, whereas locating

pumps, transmission lines and actuators near furnaces require that heat be subtracted from the fluid to

increase viscosity and reduce the temperature. This is a different problem than lowering the temperature

of fluids that are heated by working cycle itself, for example, by being discharged across a relief valve.

Example 1.7 A hydraulic machine has the following duty cycle: Idle at 12 bar for 30 s, clamp workpiece at 98 bar for 6

s, approach at 16 bar for 2 s, perform work at 300 bar for 3 s, declamp, return at 15 bar for 2 s.The pump

flow is 100 LPM, the total surface area of the oil reservoir is 2.5 m2 and the hydraulic pipeline is 25 mm

in diameter and 2500 mm in length. Calculate the net wasted energy that needs to be dissipated and

recommend a suitable heat exchanger if necessary. The room temperature is 22C. The volumetric

efficiency of pump = 0.86. The density of the oil used = 860 kg/m3.

Solution: Power dissipated in kW

12 100

2.33 kW600 600 0.86

P Q

η

Energy consumed during 30 s is

2.33 30

0.0194 kWh3600

Power dissipated during clamping is

98 100

18.99 kW600 0.86

Because clamping takes place in 5 s, the energy consumed is

18.99 6

0.032 kWh3600

Power dissipated during approach is

100 16

3.1  kW600 0.86

Because approach takes place in 2 s, the energy consumed is

3.1 2

0.0017 kWh3600

Power dissipated during forward (work) is

100 300

58.4 kW600 0.86

Because work takes place in 3 s, the energy consumed is

58.4 3

0.0485 kWh3600

Power dissipated during return is

100 15

2.91 kW600 0.86

Because return takes place in 2 s, the energy consumed is

2.91 2

0.0016 kWh3600

Total power dissipated in 59 s = 0.1032 kWh. We can express this in kW as

0.058 36006.29 kW

59

Also

Mass flow rate = 860 0.001 = 0.86 kg/s

We have

Heat generation rate (kW) Temperature C

kJOil specific heat  C  Oil flow rate (kg / s)

kg

6.294.06  C

1.8 0.86

( )

Temperature rise is small; therefore, there is no need for a heat exchanger.

Example 1.8 A hydraulic pump operates at 140 bar and delivers oil at 0.001 m

3/s to a hydraulic actuator. Oil discharges

through a pressure relief valve during 60% of the cycle time. The pump has an overall efficiency of 82%

and 15% of power is lost due to frictional pressure losses in the hydraulic lines. What heat exchanger

rating is required to dissipate all the generated heat?

Solution: We have

Pump power loss   Pump  power input Pump power output

Power output

Pump power loss     Power outputOverall efficiency

0

1Pump power loss    1 pump power output

η

51 140 10 0.001

    1 3.073 kW82 1000

5140 10 0.001

 (0.60) 8.4 kW1000

Also

5140 10 0.001

Line average loss  {  0.60} 0.15 1.26  kW1000

Therefore

Total loss = 3.073 + 8.4 + 1.26 = 12.77 kW

Select heat exchanger rating of 12.77 kW.

Example 1.9 What would be an adequate size of a reservoir for a hydraulic system using 0.0005 m

3/s pump?

Solution: Size of the reservoir is three to four times the capacity of the pump, that is Size of reservoir = 4 capacity of pump (LPM)

= 4  0.0005 4 30 (LPM) 120 L tank is required

Example 1.10 A pump delivers oil to a hydraulic motor at 20 LPM at a pressure of 15 MPa. If the motor delivers 4 kW

and 80% of the power loss is due to internal leakage, which heats the oil, calculate the heat-generation

rate in kJ/min.

Solution: We have 0.02

Pump power     150000 5 kW60

Motor delivers 4 kW; therefore loss is 1 kW. Now  Power loss due to leakage  0.8 1 0.8 kW

 Power loss due to leakage  0.8 60 48 kJ / min

Example 1.11 A certain static O-ring has a diameter of 11 mm and is given an initial squeeze of 10%. The swell of the

seal is equal to 20% increase in squeeze while the compression set is 10%. Calculate the final squeeze

percentage of the seal.

Solution: We have

Final squeeze = Initial squeeze – Compression set + Swell

= 11 (0.10−0.10 +.20) = 2.2 mm

So

Percentage of final squeeze = 2.2

100 20%11

Objective-Type Questions

Fill in the Blanks

1. Seals are used in hydraulic systems to prevent excessive internal and external leakage and to keep out

________.

2. ________rings are seals designed to prevent foreign abrasive or corrosive materials from entering a

cylinder.

3.________ is the most widely used plastic for seals of hydraulic systems.

4. ________ almost a standard material for elastomer-type seals for use at elevated temperatures up to

240C.

5. ________is an instrument used to measure the indentation hardness of rubber and rubber-like materials.

6. A vented ________ allows the tank to breathe as the oil level changes due to system demand

requirements.

7. The pressurized reservoir usually operates at between 0.35 and ________ bar.

8. The reservoir capacity should be ________ times the pump delivery per minute. This may well be too

high a volume for mobile application.

9. A hydraulic strainer is a ________ filter.

10. ________ is a surface phenomenon and refers to the tendency of particles to cling to the surface of the

filters.

11. ________ filters are porous and permeable materials such as paper, wood pulp, diatomaceous earth,

cloth, cellulose and asbestos.

12. Mostly filters have a beta ratio ________ than 75.

13. For hydraulic applications, the commonly used coolant in a heat exchanger is ________.

State True or False

1. Internal leakage does not cause loss of fluid because the fluid returns to the reservoir. 2. A non-positive seal allows a small amount of internal leakage.

3. Dynamic seals are subject to less wear compared to static seals.

4. V- and U-ring seals are compression-type seals used in virtually all types of rotary motion applications.

5. T-ring seals are static seals.

6. Natural rubber is commonly used as a seal material.

7. Standard liquid-to-liquid units can handle pressures up to 15 bar and temperatures up to 150°C

although the actual temperature difference between oil and water should not exceed 90C.

Review Questions

1. Explain the two types of leakages in a hydraulic system. In what way do they affect the performance of

a fluid system?

2. What is a seal and what are its functions?

3. How are hydraulic seals classified? What is meant by positive sealing and non-positive sealing?

4. Distinguish between a static seal and a dynamic seal.

5. How are seals classified based on geometrical cross-section?

6. Explain the different types of sealing materials commonly used.

7. What are the primary and secondary functions of a reservoir system?

8. Explain the important elements of a reservoir system and explain the function of each.

9. What is a filter and how is it classified?

10. What are surface and depth filters?

11. What are the important locations of filters? Explain the advantages and disadvantages of each

location.

12. Why should the temperature of a working fluid be properly maintained?

13. What is the purpose of seals in a hydraulic system and how are they classified?

14. List out the most commonly used types of seal configuration.

15. What is the difference between a strainer and a filter?

16. Name various filter media.

17. List out the basic types of filtering methods used in a fluid system.

18. List various locations where filters are installed in fluid power systems.

19. What are the main criteria in the design of a hydraulic system?

1. What is the purpose of a baffle plate in the fluid power pack?

.

Answers

Fill in the Blanks

1. Contamination

2. Wiper

3. Tetrafluoroethylene

4. Viton

5. A durometer

6. Breather cap

7. 1.4 bar

8. Three–four times

9. Coarse

10. Adsorption

11. Absorption

12. Greater

13. Water

State Tue or False

1 True

2. True

3. False

4. False

5. False

6. False

7. True