SECTION 25 HYDRAULIC AND PNEUMATICSYSTEMS DESIGN Determining Response Time of Pilot-Operated Solenoid-Energized Spool Valves in Hydraulic Systems 25.1 Hydraulic-System Reservoir and Hea
Trang 1SECTION 25 HYDRAULIC AND PNEUMATIC
SYSTEMS DESIGN
Determining Response Time of
Pilot-Operated Solenoid-Energized Spool
Valves in Hydraulic Systems 25.1
Hydraulic-System Reservoir and Heat
Exchanger Selection and Sizing 25.12
Choosing Gaskets for Industrial
Hydraulic Piping Systems 25.19
Computing Friction Loss in Industrial
Hydraulic System Piping 25.26
Hydraulic-Cylinder Clearance for
Damping End-of-Stroke Forces 25.29
Hydraulic System Pump and Driver
Selection 25.32
Hydraulic Piston Acceleration,
Deceleration, Force, Flow, and Size
Determination 25.36
Hydropneumatic Accumulator Design
for High Force Levels 25.39
Membrane Vibration in Hydraulic
Pressure-Measuring Devices 25.41
Power Savings Achievable in Industrial
Hydraulic Systems 25.42
Pneumatic-Circuit Analysis Using
Various Equations and Coefficients
25.44
Air Flow Through Close-Clearance
Orifices in Pneumatic Systems 25.49
Labyrinth Shaft Seal Leakage
Determination 25.58
Pipe-Wall Thickness for Hydraulic Systems without Fluid Shock 25.67 Pipe-Wall Thickness for Hydraulic Systems with Fluid Shock 25.68 Allowable Stress in Hydraulic System Piping 25.68
Hydraulic Fluid Compressibility and System Bulk Modulus 25.69 Selection of Fluids for Oil Hydraulic and Control Systems 25.69
Effect of Trapped Air on Hydraulic System Bulk Modulus 25.71 Surge Pressure in Hydraulic Cylinders 25.72
Sizing a Hydraulic System Fluid Reservoir 25.72
Required Volume of Bladder-Type Accumulator 25.73
Determining Hydraulic Accumulator Demand Volume 25.74
Effective Force Developed by a Acting Hydraulic Cylinder 25.74 Hydraulic Cylinder Oil Consumption and Extension Time 25.75
Double-Hydraulic Cylinder Power Output 25.76 Selecting Hydraulic Motors and Pumps
by Using Manufacturer’s Size Tables 25.76
DETERMINING RESPONSE TIME OF
PILOT-OPERATED SOLENOID-ENERGIZED SPOOL
VALVES IN HYDRAULIC SYSTEMS
A pilot-operated solenoid-energized spool control valve in a hydraulic system hasthe dimensions, operating pressures, and performance given in Table 1 Pilot supplypressure is 100 lb / in2 (689 kPa); main supply pressure is 500 lb / in2(3445 kPa).Find the maximum velocity of this valve, its acceleration time, and the total re-sponse time Next, using the same dimensions and main operating pressure, find
Trang 225.2 DESIGN ENGINEERING
TABLE 1 Dimensions and Operating Conditions*
Pilot Spool Main spool Diameter, in d⫽ 0.25 D⫽ 2.5
Mass, lb-sec 2 /in m⫽ 0.0002 M⫽ 0.05
Density, lb-sec 2 /in 4 0.000,085 0.000,085
*SI values given in calculation procedure.
the same unknowns when the pilot pressure is made equal to the main operatingpressure i.e., 500 lb / in2(3445 kPa) As a further modification, a small actuatingpiston is placed at each end of the main spool, Fig 3, to increase the longitudinalvelocity for a given pilot-fluid flow rate Trial and error would normally be used
to calculate the most effective diameter for the actuating piston In this procedure
we will use a diameter d x⫽1.4 in (3.56 cm) for this small actuating piston If thedimensions and operating pressures are unchanged, analyze the valve for the sameunknowns as above
Calculation Procedure:
The forces acting on the main spool at maximum velocity are: Pilot backpressure,
p B ; viscous damping force, D V ; and radial jet force P , Fig 2 From the equation,rad
P as⫽2F ƒA r M⌬P
where the symbols are as given, Table 2 Then, P ax⫽2(0.04)(0.6)(1.2)(450)⫽26
lb (115.6 N) converting to pressure by dividing by the area of the main spool valveend, we have 26 / 4.9⫽ 5.3 lb / in2(36.5 kPa)
Provisionally, estimate that D Vis equivalent to 3.2 lb / in2(22 kPa) and P B⫽valve backpressure ⫽ 20 lb / in2 (138.8 kPa) The combined hydrodynamic resis-tance is then the sum of: Radial pressure, lb / in2(kPa)⫹Viscous drag, lb / in2(kPa)
pilot-⫹Pilot-valve backpressure, lb / in2(kPa) Or combined hydrodynamic resistance⫽
5.3⫹3.2⫹20 ⫽28.5 lb / in2(196.4 kPa)
Trang 3HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN 25.3 TABLE 2 Valve Symbology*
Pilot
Actuating piston
Main valve Spool
Dimensions
and Mass
Diameter, in Cross-sectional area, in 2
Mass, lb-sec 2 /in Stroke: Intermediate Full Engagement (length in contact), in Land length, in (total)
Spool-to-bore radial clearance, in
d
— m x s
—
— C
—
S
—
L C
Solenoid
Forces
Initial, lb Gradient, lb/in Final, lb
Ratio, A/B
A B
FSOLr
Forces
Back pressure, psi Viscous drag, lb (or psi) Radial jet, lb
Coefficient of friction Axial jet, lb Acceleration force, lb
Flow rate, in 3 /sec Oil velocity, in/sec (through port) Oil mass flow, lb-sec/in Oil-jet deflection angle, deg
Valve
response
Acceleration time, sec Shifting velocity, in/sec Shifting time, sec (after energization)
*SI values given in calculation procedure.
Trang 425.4 DESIGN ENGINEERING
The pilot-valve pressure differential, delta P ⫽ 100⫺ 28.5 ⫽71.5 lb / in2(492.6kPa) Hence, the valve flow rate is, using the equation below
2⌬p
q⫽ƒa o冪
where q⫽flow, in3/ s (mL / s); ƒ⫽flow factor, dimensionless, ranging from 0.55
to 0.70 depending on valve type; a o⫽cross-sectional area, in2(cm2); of the imum port opening—usually the drilled port hole; ⌬ p ⫽ p ⫺ p1 ⫽ differentialpressure, lb / in2 (kPa) measured across the pilot inlet and outlet ports; p ⫽ fluidmass density, lb-s2/ in4, normally 0.000085 for oil Substituting, q ⫽ 0.6(0.5)[(2)(71.5) / 0.000085)]0.5⫽40 in3/ s (656 mL / s), using a value of ƒ⫽0.6 forthis valve
is nearly impossible to compute the viscous resistance Substituting, D V ⫽
2.5(6)(8.2)(80) / (0.0003)(6.9⫻106)⫽3.05 lb / in2(21 kPa) Thus, the provisional
estimate of D V ⫽ 3.2 was close enough (within 4.9 percent) and recalculation isnot necessary
The forces acting upon the spool during acceleration are: p R , P ax , D V , and F, where
F⫽Ma Assuming a mean value for initial port opening A M⫽0.4 in2(2.58 cm2),then from
P as⫽2ƒA M⌬P cos␣
where ␣normally varies from 70 degrees at initial opening to 90 degrees at fullopening In calculations, use the axial jet pressure during initial opening, and theaxial component of radial pressure during the remainder of travel Substituting,
P ax⫽2(0.6)(0.4)(450)(0.26) ⫽56 lb (248.1 N) Then 56 / 4.9⫽11.4 lb / in2(78.5kPa),␣ ⫽75 deg; cos␣ ⫽0.26
Viscous drag will be the average: D V⫽3.2 / 2⫽1.6 lb / in2(11 kPa)
Backpres-sure is still p B⫽20 lb / in2(137.8 kPa) So the total is 11.4⫹ 1.6⫹20⫽ 33 lb /
in2(227.4 kPa)
Therefore, accelerating pressure ⫽ 100⫺ 33 ⫽ 67 lb / in2 (461.6 kPa) verting to force, we have 67 (4.9)⫽ 328 lb (1441.2 N) The acceleration time, t a
Con-s⫽MV / F⫽0.05 (8.2) / 328 ⫽0.0013 s
Trang 625.6 DESIGN ENGINEERING
FIGURE 2 Jet-force drag in pilot-operated spool valves.
The displacement of the main spool during the acceleration period is negligible,being less than 1 percent of the total stroke Time for the total stroke of 1.5 in (3.8cm) is 1.5 / 8.2 ⫽0.182 s, and the time interval from energization of the solenoid
to completion of the main valve stroke, T⫽ 0.190 s
Much larger flow will pass through the pilot valve because of the higher pressure
Maximum velocity period: P ax⫽5.3 lb / in2(36.5 kPa), the same as before; D V⫽
7.7 lb / in2(53.1 kPa)—a higher estimate, proportional to the anticipated velocity;
p B⫽ 20.0 lb / in2(137.8 kPa), the same as before The total is 33.0 lb / in2(227.5kPa)
The new ⌬ P ⫽ 467 lb / in2 (3217.6 kPa), and Q ⫽ 4.7 (467)0.5 ⫽ 102 in3/ s
ob-a mob-arked effect on the operob-ationob-al speed of the mob-ain vob-alve
Note that increasing the pilot pressure fivefold, from 100 lb / in2 to 500 lb / in2(689 kPa to 3445 kPa) only doubles the speed of response, from 0.19 s to 0.08 s.Increasing the port area can result in an nearly proportional gain in speed, and no
Trang 7HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN 25.7
additional pressure is necessary, saving on pumping costs Costs of producing a0.375-in (9.52-mm) pilot spool are not much greater than those for a 0.25-in (6.35-mm) spool The increase in capacity is 50 percent without the additional heat lossesentailed by an increase in pressure
For the valve, Fig 3, with the small actuating pistons, taking the summation of theviscous drag, 兺D V, and inserting the known optimum values in parentheses afterthe computed values, we have:
The total time, T⫽0.100⫹0.009⫽0.109 s Using a pilot pressure of 500 pst
(3445 kPa), V⫽20.8 in / s (52.8 cm / s) and d⫽1.06 in (2.69 cm) Then:
Trang 9HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN 25.9
TABLE 3 Effect of Adding Actuating Pistons*
Pilot
pressure p,
psi
Main valve diameter, in
Maximum valve velocity, without piston in/sec
Total shift time
T,
sec
Maximum valve velocity, with piston in sec
Piston diameter, in
Total shift time
common valves used in industrial hydraulic systems Speed of response of thesevalves is important during the design and operation of any hydraulic system Theprocedure given here analyzes the speed of response of a typical valve in terms ofthe fluid flow rate, characteristic force-vs-airgap curve of the solenoid; shape, size,clearance, and displacement of each spool; and the fluid viscosity
The method given in this procedure relates the above parameters for the valve
in Fig 1 and can be applied to any other pilot-operated spool valve And theprocedure includes a special technique for a large spool valve, Fig 3, actuated by
a small auxiliary piston
In the sequence of operation of solenoid-energized pilot-operated spool valve,here is what happens The solenoid is energized, the pilot spool moves quickly tothe full open position, Fig 5, and the main spool is shifted at a rate determined bythe amount of fluid that can move through the pilot ports against these five resistingforces: (1) pilot system backpressure, lb / in2(kPa); (2) viscous damping force, lb(N); (3) radial jet force, lb (N); (4) axial jet force, lb (N); (5) acceleration force, lb(N)
Trang 1025.10 DESIGN ENGINEERING
SI Values in./sec cm/sec in.2 cm2
FIGURE 4 Effect of varying piston diameter.
FIGURE 5 Before energization and after full stroke of a solenoid-energized spool valve.
Trang 11HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN 25.11
FIGURE 6 Time-displacement graph for solenoid-energized pilot-operated spool valve.
Figure 6 shows a time-displacement chart for the spool as it shifts to a newposition This chart is for a 0.25-in (0.635-cm) diameter closed-center pilot spoolcontrolling flow to a 2.5-in (6.35-cm) main spool
Pilot pressure chosen for this procedure is 100 lb / in2(689 kPa), which is lessthan the main supply pressure A separate supply for the pilot is required Thereare no hard and fast rules for establishing pilot pressures, but if possible keep thepressure in the range of a few hundred lb / in2(kPa) if this range will do the job,
to avoid possible distortion or leakage in the pilot system
In analyzing any pilot-operated valves, some simplifying assumptions must bemade; otherwise there is no practical mathematical solution For one, assume thatthe backpressure of the pilot system, set by the pilot exhaust valve, is constant.Ignore line resistance because the connecting lines are short Neglect viscous damp-ing except at the full-velocity portion of the stroke Then the five dynamic resis-tances can be handled with the simple equations presented in this procedure
The pilot-system backpressure is usually 5 to 7 percent of the pilot pressure, p.
Select the higher value if the operating pressures are over 200 lb / in2(1378 kPa),because it adds a margin of safety that compensates for spool rubbing friction Thefriction is from metal-to-metal contact at points where the oil film is partially de-stroyed
Above 400-lb / in2(2756 kPa) operating pressure, a separate pilot supply usually
is provided Pilot pressure in these instances ought to be at least 7 percent of themain operating pressure to ensure adequate force to move the main spool
This procedure is the work of Louis Dodge, Hydraulics Consultant, as reported
in Product Engineering magazine.
Trang 1225.12 DESIGN ENGINEERING
FIGURE 7 Well-designed hydraulic-system reservoir and circuit diagram for it.
HYDRAULIC-SYSTEM RESERVOIR AND HEAT
EXCHANGER SELECTION AND SIZING
(1) Determine if a ‘‘first-pass’’ reservoir choice, Fig 7, can dissipate enough heat
to keep oil temperature below 120⬚F (48.9⬚C) in a 70⬚-F (21.1⬚-C) ambient (50⬚-F[27.8⬚-C]) temperature difference, T D The source of heat is a 20-gal / min (1.26-L -
Trang 13HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN 25.13
FIGURE 8 Single-pass shell-and-tube heat exchanger for industrial
hy-draulic system.
/ s) constant-delivery pump operating in a 60-percent overall efficiency system Thehydraulic working unit, including the piping, is not helping to dissipate heat Res-ervoir tank size is small with a cooling surface of 21 ft2 (1.95 m2), based on areservoir volume of twice pump flow—or 2 ⫻ 20 ⫽ 40 gal (75.8 L) Systempressure⫽750 lb / in2(5167.5 kPa); overall heat transfer coefficient, k⫽5 Btu / ft2
hr⬚F (28.4 W / m2⬚C), a conservative value (2) The return flow of the hydraulicfluid in another industrial hydraulic system must be cooled continuously to 125⬚F(51.7⬚C) The hottest uncooled drain temperature of the fluid is 140⬚F (60.0⬚C).Flow of the hydraulic fluid through the system is 12 gal / min (0.76 L / s); the cool-ing-water inlet temperature⫽ 65⬚F (18.3⬚C); outlet temperature⫽85⬚F (29.4⬚C);
k ⫽ 90 Btu / hr-ft2 ⬚F (511.2 W / m2 ⬚C) Use a counterflow, single-pass heat changer, Fig 8, in this analysis (3) Lastly, calculate the temperature of a standard60-gal (227.4 L) reservoir after 5 hr of operation Pump discharge is 20 gal / min(1.26 L / s) at 750 lb / in2(5.17 MPa) Cooling surface is 28 ft2(0.792 sq m) An
ex-attached heat-dissipating working unit weighs W ⫽ 800 lb (362.2 kg) and its
ef-fective surface area A⫽5.5 ft2(0.156 m2) With an initial system oil temperature,
Toil⫽70⬚F (21.1⬚C), and an ambient temperature, Tair ⫽50⬚F (10.0⬚C), the initial
temperature-over-ambient, T p ⫽ 70 ⫺ 50 ⫽ 20⬚F(11.1⬚C) The estimated median
value of k ⫽4 Btu / ft2hr⬚F(22.7 W / sq m⬚C) The 60 gallons of oil weigh 444
lb (201.6 kg)
Calculation Procedure:
Find the total heat generated in the system using the equation
E L⫽1.48⫻Q⫻P(1⫺)where the symbols are as given below Substituting
2
E L⫽1.48(20 gal / min)(750 lb / in )(1⫺0.60)⫽8880 Btu / hr (2601.1 W)
Trang 1425.14 DESIGN ENGINEERING
FIGURE 9 Hydraulic oil and cooling water temperature plot for heat exchanger in Fig 8.
Use the equation
T ⫽E /兺kA
max D L and solve for the required area, A Or, A ⫽ 8880 / (50 ⫻ 5) ⫽ 35.5 ft2(3.3 m2).Since this reservoir has only 21 ft2(1.95 m2) of cooling surface, the tank area isnot large enough to dissipate the heat generated Hence, a larger reservoir coolingarea must be provided for this installation
Use the equation
Eexch⫽ ⌬Toil⫻Qoil⫻210
⫽ ⌬Twater⫻Qwater⫻500
to find the heat exchanger heat load Substituting, we have Eexch⫽ (140 ⫺ 125)(12)(210) ⫽ 37,800 Btu / hr (11.1 kW) The maximum temperature difference,
⌬Tmax⫽ 125⫺ 65⫽60⬚F (51.7⬚C) Minimum temperature difference⫽ ⌬Tmin⫽
140⫺85⫽55⬚F (30.6⬚C) Log-mean temperature difference, computed as shownelsewhere in this handbook (see index) is⌬Tmean ⫽57.5⬚(31.9⬚C) Figure 9 showsthe oil and water temperature changes in a generalized manner
Find the required heat-transfer area from
Eexch⫽kA⌬TmeanBtu / hr
solving for A Or A⫽ 37,800 / (90⫻57.5)⫽ 7.3 ft2(0.68 m2)
Trang 15HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN 25.15
To find the required cooling-water flow rate, use Qwater⫽Eexch/ (500⫻gal / min),where 500⫽a constant to convert gal / min to gph Substituting, Qwater⫽38,800 /(500⫻ 20)⫽3.78 gal / min (0.2385 L / s)
Refer to manufacturer’s catalogs for the size of a heat exchanger with the propersurface area Be certain to check the heat exchanger pressure rating if the systempressure exceeds 150 lb / in2(1033.5 kPa)
Use the same relation as in step 1, above, to find E L ⫽8880 Btu / hr (2601.1 W)
The heat balance for the oil and attached heat dissipating working unit are given
by cW ⫽ 0.4(444) ⫹ 0.1(800) ⫽ 257.6 Btu /⬚F (489.4 kJ /⬚C) For the tank and
working unit, the kA⫽4(28⫹5.5)⫽134 Btu / hr⬚F (254.6 kJ / hr⬚C)
Use the equation
The maximum operating temperature over ambient, maxT D ⫽ E L / kA ⫽ 8880 /
134⫽66.3⬚F (19⬚C) Then, the oil temperature⫽50⫹66.3⫽116.3⬚F (46.8⬚C).Based on these results, no heat exchanger is required
Note that the result of this calculation depends on the correct evaluation of k,
which depends on air circulation around the reservoir and attached heat-dissipatingunit The influence of the initial temperature difference is minor Practical experi-ence with system design is most important
con-ventional hydraulic systems reservoir is 120⬚F (48.9⬚C) The procedure presentedabove shows ways to prevent the oil temperature from exceeding that level Thereare certain exceptions to the rule given above Some conventional hydraulic systemsare designed to operate at 150⬚F (65.6⬚C) So-called super-systems with specialfluids and seals can operate at 500⬚F (260⬚C), and higher But for any level ofoperating temperature, the same heat-transfer principles apply
In designing a fluid system’s heat-transfer, after you’ve established the basicsystem and reservoir design, follow the simple heat-balance method given above
If the reservoir’s peak temperature calculated this way is less than 120⬚F (48.9⬚C)(or some other desired temperature), no further work is necessary That’s why aheat exchanger was not required in step 6, above If the calculated reservoir tem-perature is higher than desired, you have two alternatives: (1) improve heat dissi-pation by modifying the reservoir tank, components, or piping; (2) add a heatexchanger, using the rating method given in this procedure
Most of the heat in industrial hydraulic systems comes from in-the-system ponents Exceptions are systems in hot environments or adjacent to heat-producingequipment, but the same heat-balance principles apply
com-Heat is generated whenever hydraulic oil is throttled or otherwise restricted.Examples of heat-producing devices include pressure regulators, relief valves, un-
Trang 1625.16 DESIGN ENGINEERING
SI Values psi MPa
250 1.72
500 3.44
750 5.17
1000 6.89
FIGURE 10 Typical duty cycle for industrial hydraulic system
showing the mean effective pressure, system pressure, and pump
discharge pressure.
dersize piping, dirty or undersize filters, leakage points, and areas of turbulenceanywhere in the system
A good measure of internally generated heat loss is the difference between
pump input power, Btu / hr (W), and system useful work, Btu / hr (W) The energy loss, E L⫽E (1in ⫺) where ⫽system efficiency
Figure 10 shows a typical duty cycle in an industrial hydraulic system Theutilization pressure, measured at the inlet to the working device (fluid motor orcylinder) will always be lower than the source pump pressure, depending on theamount of throttling or other regulation required in the system
The difference in energies—pumped vs utilized—must be absorbed during sients and eventually dissipated by the system An approximate measure of overallsystem efficiency, , is the ratio of the mean effective pressure, Fig 10, to pumppressure, where the mean effective pressure is calculated from the area under thecurve, Fig 10, divided by the time base
tran-Figure 11 compares reservoir tank temperature for two different pumps: (1) afixed-delivery pump, with constant flow and pressure and a full-flow relief valvefor bypass flow during idling of the workload, and (2) a variable-delivery pump,with pressure and flow automatically varied to match load requirements
Note that in the constant-pressure system, Fig 11, the greatest rate of heatgeneration is during idling of the workload; all the flow is throttled back to thereservoir and does no useful work By comparison, the variable-delivery pump doesnot waste energy at idle because the flow is automatically reduced to nearly zero.Both Fig 10 and 11 indicate that savings in energy are possible if only theneeded oil is pumped Excess capacity is forced back to the reservoir through therelief valve, and the energy is converted to waste heat Auxiliary pumps are greatoffenders if they are operated when not needed Additional useful guides for res-ervoir selection and sizing are given below
Trang 17HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN 25.17
to a stop during emergency shutdown if a return line breaks; (4) Thermal capacitywill be available to absorb unexpected heat for short periods or to store heat duringidle periods in a cold environment; (5) Enough surface area (reservoir tank walls)
is available for natural cooling during normal operation
If the reservoir tank volume in gallons (L) is less than twice pump flow in gallonsper min (L / min)—that is, if the tank can be pumped dry in less than 2minutes—add a heat exchanger to the system circuit to avoid excessive temperaturefluctuations For any size reservoir tank, specify an oil-level indicator or sight glass,
in addition to whatever automatic level controls are provided
When designing a reservoir tank, include each device shown in Fig 7 to providereliable service for the system The suction-line filter should be1⁄2to3⁄4in (1.3 to1.9 cm) above the tank bottom Strainer oil flow capacity should be two to fourtimes the pump capacity A vacuum gage on the pump suction will show if thestrainer is clogged A permanent-magnet filter can be specified as a drain plug ormounted on the baffle plate in a region of concentrated return oil flow
The main return oil flow should discharge below the reservoir oil level aboutone inch (2.5 cm) above the tank bottom Backpressure in the return line will be
5 to 10 lb / in2 (34.5 to 68.9 kPa), or higher Atmospheric return lines, includingseal-leakage lines, are at zero pressure and should be discharged above the hydraulicoil level
If the atmospheric lines have high flow and a high air content, they should bedischarged above the oil level into a chute sloping gradually (5 to 10 degrees) into
Trang 1825.18 DESIGN ENGINEERING
the tank fluid The chute slows and fans out the flow, enabling the oil to free itself
of air This is important, because oil saturated with air and operating at high sures will run 25 percent hotter than air-free oil This is caused partly by the heat
pres-of compression pres-of the air and partly by its low thermal conductivity
Internal baffles between the return pipe and pump inlet will slow the fluid culation, help settle out dirt particles, give air the chance to escape, and allowdissipation of heat The top of the baffle should be submerged about 30 percentbelow the surface of the fluid
cir-Keep the hydraulic oil temperature between 120 and 150⬚F (54 and
66⬚C)—preferably at the lower value for oil viscosities from 100 to 300 SSU based
on 100⬚F (38⬚C) Temperatures up to 160⬚F (71⬚C) are permissible if the hydraulicfluid viscosity is from 300 to 750 SSU, based on 100⬚F (38⬚C) Higher operatingtemperatures require special design
Tank walls should be thin to permit good thermal conductivity Make themapproximately1⁄16in (0.16 cm) for tank capacities up to 25 gal (95 L);1⁄8in (0.32cm) for capacities up to 100 gal (379 L);1⁄4in (0.64 cm) for 100 gal (379 L) ormore Use slightly heavier plate for the bottom Give the top plate four times wallthickness to assure vibration-free operation and to hold alignment of pump andmotor shafts Specify a thermometer to be mounted on the tank top where theoperator can see it
Avoid designing industrial hydraulic system machines with integral tanks It isbetter to have a separate reservoir, accessible from all sides Small reservoir tankscan even be mounted on castors Tanks within the machine frame are troublesome
to maintain; be sure to work out maintenance details of such a tank before mitting yourself to the design
com-Equip the reservoir tank with cleanout doors and slope the bottom toward thedoors Provide a drain cock or discharge valve at the low point of the bottom and
at other low points if needed for complete drainage Put a manhole cover on thetank top for removing filters and strainers Design a connection for hooking to aportable filtration unit
If the reservoir tank is made of cast iron, don’t paint the interior surface Besure that all grit and core sand are removed before putting the tank into service.Surfaces must be sandblasted
This procedure is the work of Louis dodge, Hydraulics Consultant, as reported
in Product Engineering magazine SI values were added by the handbook editor.
Heat-transfer terminology and symbols
Heat loss and efficiency
⫽ System efficiency, Eused/ E , %in
Eused ⫽ Energy utilized in system, Btu / hr (W)
Ein ⫽ Pump input power, Btu / hr (W)
E L⫽ Heat loss generated in system, Btu/hr (W)
EA⫽ Heat absorbed by oil, tank and components, Btu/hr (W)
ED⫽ Heat dissipated to atmosphere or coolant, Btu/hr (W)
Eexch ⫽ Heat exchanger load, Btu / hr (W)
Fluid conditions and flow
t⫽ Operating time, hr
Q⫽ Flow, gal/min (L/s)
P⫽ Pump gage pressure, lb/in 2 (kPa)
T D ⫽ Temperature-over-ambient for oil, ⬚F: T D ⫽ T ⫺ T (values are mean) (C)oil air
⌬T ⫽ Heat exchanger only: ⌬Twater ⫽ Tout⫺ T ; ⌬T ⫽ T ⫺ T ; ⌬Tin oil in out mean ⫽ log-mean
⌬T, oil-to-water (C)
Trang 19HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN 25.19
TABLE 4 Pressure-Temperature Value for Gasket Materials*
Material Max P i ⫻ T Max Temp, ⬚F
Vegetable fiber 40,000 250
Rubberized cloth 125,000 400
Compressed asbestos** 250,000 850
Metal types ⬎250,000 depends on metal
*SI values given in procedure.
**Or acceptable substitute.
Equation constants
c⫽ Specific heat (mean value), Btu/lb-⬚F (kJ/kg ⬚C)
k ⫽ Overall heat-transfer coefficient, Btu/sq ft-hr-⬚F (W/m2 ⬚C)
W⫽ Combined weight of oil and system components, lb (kg)
A⫽ Surface area for dissipating heat, sq ft (sq m)
e⫽ Base for natural logs ⫽ 2.718
兺 ⫽ Summation sign 兺cW ⫽ effective cW for all of system components
Typical values for c and k
c, Btu / lb-⬚F: (kJ/kg ⬚C) Oil, 0.40; aluminum, 0.18; iron, 0.11; copper, 0.09
k, Btu / sq ft-hr-⬚F: (W mm/m 2 ⬚C)
2 to 5—Tank inside machine or with inhibited air circulation
5 to 10—Steel tank in normal air
10 to 13—Tank with good air circulation (guided air current)
25 to 60—Forced air cooling or oil-to-air heat exchanger
80 to 100—Oil-to-water heat exchanger (k values increase slightly with temperature)
CHOOSING GASKETS FOR INDUSTRIAL
HYDRAULIC PIPING SYSTEMS
Choose a suitable gasket to seal industrial hydraulic fluid at 1200 lb / in2(8.27 MPa)and 180⬚F (82.2⬚C) Flanges are 1.5-in (3.8-cm) raised-face, 600-lb (2668.8-N)weld-neck type made from Type 304 stainless steel There are four bolts, 0.75-in(1.9-cm) 10 NC, made from ASTM A193 grade B7 alloy steel Hydrotest pressure
is specified as 2.5 times operating pressure
Calculation Procedure:
Assuming that torque wrenches will be used to check this installation, as is almostuniversally done today, select the bolt-stress method to calculate the total bolt force.This method uses the equation,
F b⫽N S A b b b
where the symbols are as given below
Trang 2025.20 DESIGN ENGINEERING
TABLE 5 Stress Areas for Flange Bolts*
Nominal dia, in
Coarse Threads Threads
per in
Stress area,
sq in
Fine Threads Threads
per in
Stress area,
sq in 0.125 (No 5) 40 0.0079 44 0.0082
*SI values given in procedure.
From Table 5, the stress area for 3 / 3-10 NC bolt is 0.3340 sq in (2.15 sq cm).The bolt material specified can easily take a stress of 30,000 lb / in2 (206.7 MPa)without yielding This can be verified from
F b⫽16,000D / A b b which gives F b⫽(16,000)(0.75 / 0.3340)⫽36,000 lb (160.1 kN)
From the bolt-stress equation, F b⫽4(30,000)(0.3340)⫽40,080 lb (178.3 kN).The torque required to produce this stress level at installation is given by,
T⫽0.2D S A b b b
Or, T⫽0.2(0.75)(30,000)(0.3340) / 12⫽125 ft-lb (169.4 Nm) This torque will bespecified on the system assembly drawings so it is used during construction
The pressure-temperature relation for this installation is 1200⫻180⫽216,000 inUSCS units and 6798 in SI units This, from Table 4, suggests choosing a com-pressed-asbestos (or acceptable substitute) type gasket This would be compatiblewith industrial hydraulic fluid
The gasket area is 4.73 sq in (30.5 sq cm), calculated from an outside diameter
of 27⁄8in (7.07 cm), the same as the OD of the raised flange, per ASA-B16.5, and
an ID of 1.5 in (3.8 cm) The seating stress is computed from
Trang 21HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN 25.21
S g⫽F / A b g
Or, S g⫽40,080 / 4.73⫽8470 lb / in2(58.4 MPa)
Table 6 shows this stress will easily seat the selected compressed-asbestos (oracceptable substitute) gasket Tentatively choose asbestos (or acceptable substitute)with CR (neoprene) binder for oil resistance; thickness1⁄32in (0.079 cm)
The mean area acted upon by the pressure in the hydraulic line is defined by adiameter of (2.875 ⫹1.5) / 2 ⫽ 3.74 sq in (24.1 sq cm) Selecting a safety factor
of 1.5 from Table 8, the end force is calculated and balanced against the total boltforce by the equation,
F b⭌KP A t m
Or,
Thus, there is no end-force balance problem with bolts stressed to 30,000 lb / in2(206.7 MPa)
Table 6 shows that a concentric-serrated surface finish on the flange is best omy may dictate a conventional spiral-serrated surface, which Table 6 shows willwork in this case
Include in the specifications the material type, dimensions, and bolt-torque datacomputed in step 1, above For greater torque-wrench accuracy, specify uniform fit
on all bolts and lubrication before installation
Related Calculations. While the procedure given here is directed at industrialhydraulic systems, the steps and data are valid for choosing gaskets for any pipingsystem: steam, condensate, oil, fuel, etc Just be certain that the pressures andtemperatures are within the ranges in the tables and equations presented here.Three main design factors govern the selection of a gasket material—whethersheet packing, metal, or a combination of materials These factors are:
Fluid compatability at the pressure-temperature condition being designed for
must be checked first Refer to data available from gasket manufacturers—there ismuch of it available free to designers
The pressure-temperature combination determines whether the gasket material
is inherently strong enough to resist blow-out One rule-of-thumb criterion is the
product of operating pressure, P i , and operating temperature, T Table 4 lists values
of this product for several basic types of gasket material These figures are based
on experience, test data, and analysis of current technical literature
The total bolt force at installation must be sufficient to: (1) flow the gasket
surface into the flange surface to make an effective seal; (2) prevent the internalpressure from opening the flanges This demands careful matching of gasket ma-terial, seating area, bolt selection, and flange-surface finish The procedure pre-sented here gives a logical way to achieve the right balance among these factorsfor the majority of gasket joint applications
Where asbestos is the recommended gasket material in this procedure, the signer must review the environmental aspects of the design Acceptable substitutematerials may be required by local environmental regulations Hence, these regu-lations must be carefully studied before a final design choice is made
Trang 2625.26 DESIGN ENGINEERING
TABLE 8 Safety Factors for Gasketed Joints
This procedure is the work of J J Whalen, Staff Engineer, Johns-Manville
Corporation, as reported in Product Engineering magazine.
COMPUTING FRICTION LOSS IN INDUSTRIAL
HYDRAULIC SYSTEM PIPING
(1) Determine the friction loss for hydraulic-system fluid having a viscosity of 110centistokes at 120⬚F (48.9⬚C) when flowing through a 1-in ID (25.4 mm) pipe 50
ft (15.2 m) long at a velocity of 20 ft / s (6.1 m / s); specific gravity⫽0.88 (2) Findthe pressure loss for a ‘‘light’’ hydraulic oil having a viscosity of 32 centistokes at
100⬚F (37.8⬚C) when flowing through a 100-ft (30.5-m) long 2-in (50.8-mm) IDcommercial steel pipe at a velocity of 30 ft / s (9.1 m / s); specific gravity⫽0.88
Calculation Procedure:
Use the relation
where R⫽ Reynolds number, dimensionless; V ⫽ hydraulic fluid velocity in the
piping, ft / s (m / s); D ⫽ pipe diameter, in (mm); v ⫽ kinematic viscosity of
hy-draulic fluid, centistokes Substituting, R⫽(7740)(20)(1) / 110⫽ 1407.3
Since the Reynolds number for this piping is less than 2000, roughness of the pipedoes not enter into the calculation See Fig 12
Trang 2825.28 DESIGN ENGINEERING
Use Fig 12 and the Reynolds number Since R is less than 2000, ƒ ⫽ frictionfactor⫽ 64 / R⫽64 / 1407.3⫽0.045
Use the relation pƒ⫽ 0.0808 ƒ (L / D) V2(s), where pƒ⫽ pressure loss in piping,
lb / in2 (kPa); L⫽ pipe length, ft (m); s⫽ specific gravity of the hydraulic fluid;
other symbols as before Substituting, pƒ⫽0.0808 (0.045)(50)(400)(0.88)⫽63.99
lb / in2; say 64 lb / in2(440.9 kPa)
Use the same equation as in step 1, R⫽(7740)(30)(2) / 32⫽14,512.5
Use Fig 13, entering at the pipe diameter at the bottom and projecting vertically
upwards to the commercial steel pipe curve to read the relative roughness, e / D as
0.0009
7. Find the friction factor, ƒ
Enter Fig 12 at Reynolds number and relative roughness to read ƒ⫽ 0.03
Using the same equation as in step 4, we have pƒ ⫽ (0.0808)(0.03)(100)(900)(0.88) / 2⫽95.99 lb / in2(661.4 kPa)
hy-draulic oils in Fig 14 The Herschel relationship, which expresses the temperature function between two viscosities,, and,0, existing at temperatures
tempera-is part of Fig 14, gives exponents for the hydraulic oils shown in the chart, covering
a temperature range of 70 to 130⬚F (21 to 54⬚C) As a fair approximation, theviscosity of oils most commonly used in hydraulic work changes with the thirdpower of the temperature gradient within the normal operating range With a com-monly encountered temperature gradient of about 2:1 between cold start and max-imum operating temperature, viscosities vary as 8:1
A graph representing the value of the friction factor, ƒ, as a function of the
Reynolds number, R, is often called the Stanton chart, after its developer, who was
the first to employ this representation of the friction factor
A chart taking advantage of the functional relationships established by researchwas drawn up by Lewis F Moody, and is reproduced in Fig 12 in a form convenientfor the user of this handbook In Fig 12, the friction factor, ƒ, is shown as a function
of the Reynolds number, R, and the relative roughness, e / D, e being a linear
quan-tity in feet or meters representing the absolute roughness An auxiliary chart is
given in Fig 13, from which e / D can be taken for any size and type of pipe.
The procedure given here is valid for industrial hydraulic systems used in draulic presses; drilling, boring, and honing machines; planers; grinders; milling,transfer, and broaching machines; die-casting and plastic molding machines; hy-
Trang 29hy-HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN 25.29
HYDRAULIC-CYLINDER CLEARANCE FOR
DAMPING END-OF-STROKE FORCES
An undamped hydraulic cylinder, Fig 15, is fitted with an annular clearance in thecavity of the cylinder cap, Fig 16, as a flow restriction The cylinder is then pro-
Trang 31HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN 25.31
FIGURE 15 Undamped cylinder experiences shock forces at, and during,
re-versal of mass.
FIGURE 16 Annular clearance in cavity of cylinder serves as flow
restriction for dashpot Hydraulic fluid then escapes at outlet port.
vided with a piston having a velocity of 1.1 ft / s (0.34 m / s) having a mass, M, of
64 lb-s2/ ft (95.3 kg-s2/ m), a length, L⫽4 in (10.16 cm), a dashpot radius of R⫽
1 in (2.54 cm), a dashpot capacity of 12.5 cu in (204.8 cu cm), a coefficient ofdischarge, CD ⫽ 0.62, and a pressure differential, ⌬P, of 30 lb / in2 (206.7 kPa).What annular clearance is needed when handling hydraulic fluid with a specificgravity of 0.85?
Calculation Procedure:
Use the relation, F⫽MV2/ 2L, where F⫽mean dashpot resistance, lb (N); other
symbols as given above Substituting, F⫽(64)(1.1)2(12) / (2)(4)⫽116.2 lb (516.9N)
Use the relation, a ⫽ F / M, where a ⫽ piston acceleration or deceleration, ft / s2;
other symbols as given earlier Substituting, a ⫽F / M⫽ 116.2 / 64⫽ 1.816 ft / s2(0.55 m / s2)
Trang 3225.32 DESIGN ENGINEERING
Use the relation, T⫽2 V / a, where T⫽stopping time, sec; other symbols as before
Substituting, T⫽(2)(1.1) / 1.81⫽ 1.22 s
Use the relation, Q D⫽0.26 B / T, where Q D⫽cylinder liquid discharge, gal / min
(L / s); B⫽dashpot capacity, cu in (cu cm); other symbols as before Substituting,
Q D⫽0.26(12.5) / (1.22)⫽2.66 gal / min (0.168 L / s)
Use the relation, Z ⫽ Q (S) G 0.5/ [238(RC D)(⌬P) ] where Z0.5 ⫽ annular clearance
required, in (cm); Q G⫽ liquid discharge, gal / min (L / s); S ⫽ specific gravity of
the hydraulic fluid handled; R⫽radius of dashpot, in (cm); other symbols as before
Substituting, Z⫽(2.66)(0.92) / (238)(1)(0.62)(5.47)⫽ 0.00303 in (0.0077 cm)
limits axial piston velocity where and when desired by trapping oil ahead of thepiston, then releasing it through a restriction, slowly and with predetermined con-trol In this procedure we assumed that at the start of dashpot action inertia forcesalone are dissipated through the ejection of dashpot oil Hence, the kinetic energy
of the moving parts equals the work done during penetration of the dashpot
The assumed value of the coefficient of discharge, C D, may be checked againstthe Reynolds number of the calculated flow and adjusted if it deviates too muchfrom what experience shows as reasonable Use the previous calculation procedure
to check the Reynolds number For Reynolds numbers below 100, C D may varyfrom 0.1 to 0.7
Hydraulic damping and shockless reversal can be obtained with flow-restriction
or pressure-reducing devices These include servo-controlled variable pumps, control valves, cylinder modifiers (orifices, special pistons), and other power cutoutsand reducers
flow-This procedure is the work of Louis Dodge, Hydraulics Consultant, as reported
in Product Engineering magazine SI values were added by the handbook editor.
HYDRAULIC SYSTEM PUMP AND DRIVER
SELECTION
Choose the pump and the driver horsepower for a rubber-tired tractor bulldozerhaving four-wheel drive The hydraulic system must propel the vehicle, operate thedozer, and drive the winch Each main wheel will be driven by a hydraulic motor
at a maximum wheel speed of 59.2 r / min and a maximum torque of 30,000 lb䡠in(3389.5 N䡠m) The wheel speed at maximum torque will be 29.6 r / min; maximumtorque at low speed will be 74,500 lb䡠in (8417.4 N䡠m) The tractor speed must
be adjustable in two ways: for overall forward and reverse motion and for turning,where the outside wheels turn at a faster rate than do the inside ones Other oper-ating details are given in the appropriate design steps below
Calculation Procedure:
Usual output requirements include speed, torque, force, and power for each function
of the system, through the full capacity range
Trang 33HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN 25.33
First analyze the propel power requirements For any propel condition, hp ⫽
Tn / 63,000, where hp⫽horsepower required; T⫽torque, lb䡠in, at n r / min Thus
at maximum speed, hp⫽(30,000)(59.2) / 63,000⫽28.2 horsepower (21.0 kW) At
maximum torque, hp ⫽ 74,500 ⫻ 29.6 / 63,000 ⫽ 35.0 (26.1 kW); at maximum
speed and maximum torque, hp⫽(74,500)(59.2) / 63,000⫽70.0 (52.2 kW).The drive arrangement for a bulldozer generally uses hydraulic motors geareddown to wheel speed Choose a 3000-r / min step-variable type of motor for eachwheel of the vehicle Then each motor will operate at either of two displacements
At maximum vehicle loads, the higher displacement is used to provide maximumtorque at low speed; at light loads, where a higher speed is desired, the lowerdisplacement, producing reduced torque, is used
Determine from a manufacturer’s engineering data the motor specifications Foreach of these motors the specifications might be: maximum displacement, 2.1 in3/
r (34.4 cm3/ r); rated pressure, 6000 lb / in2 (41,370.0 kPa); rated speed, 3000 r /min; power output at rated speed and pressure, 90.5 horsepower (67.5 kW); torque
at rated pressure, 1900 lb䡠in (214.7 N䡠m)
The gear reduction ratio GR between each motor and wheel ⫽(output torquerequired, lb䡠in) / (input torque, lb䡠in,⫻gear reduction efficiency) Assuming a 92percent gear reduction efficiency, a typical value, we find GR⫽74,500 / (1900 ⫻
0.92)⫽ 42.6:1 Hence, the maximum motor speed ⫽wheel speed⫻ GR⫽ 59.2
⫻42.6⫹2520 r / min At full torque the motor speed is, by the same relation, 29.6
⫻42.6⫽ 1260 r / min
The required oil flow for the four motors is, at 1260 r / min, in3/ r ⫻4 motors
⫻(r / min) / (231 in3/ gal)⫽ 2.1⫻4⫻ 1260 / 231⫽ 45.8 gal / min (2.9 L / s) With
a 10 percent leakage allowance, the required flow⫽50 gal / min (3.2 L / s), closely,
or 50 / 4⫽12.5 gal / min (0.8 L / s) per motor
As computed above, the power output per motor is 35 horsepower (26.1 kW).Thus, the four motors will have a total output of 4(35)⫽ 140 horsepower (104.4kW)
The dozer uses a linear power output Two hydraulic cylinders each furnish a imum force of 10,000 lb (44,482.2 N) to the dozer at a maximum speed of 10 in /
max-s (25.4 cm / max-s) Amax-smax-suming that the maximum operating premax-smax-sure of the max-symax-stem imax-s
3500 lb / in2(24,132.5 kPa), we see that the piston are a required per cylinder is:force developed, lb / operating pressure, lb / in2 ⫽ 10,000 / 3500 ⫽ 2.86 in2 (18.5
cm2), or about a 2-in (5.1-cm) cylinder bore With a 2-in (5.1-cm) bore, the ating pressure could be reduced in the inverse ratio of the piston areas Or, 2.86 /(22/ 4)⫽p / 3500, where p⫽cylinder operating pressure, lb / in2 Hence, p⫽3180
oper-lb / in2, say 3200 lb / in2(22,064.0 kPa)
By using a 2-in (5.1-cm) bore cylinder, the required oil flow, gal, to each der⫽(cylinder volume, in3)(stroke length, in) / (231 in3/ gal)⫽(22/ 4)(10) / 231⫽
cylin-0.1355 gal / s, or cylin-0.1355 gal / s, or cylin-0.1355⫻ (60 s / min)⫽8.15 gal / min (0.5 L / s),
or 16.3 gal / min (1.0 L / s) for two cylinders The power input to the two cylinders
is hp⫽16.3(3200) / 1714⫽ 30 / 4 horsepower (22.7 kW)
The winch will be turned by one hydraulic motor This winch must exert a mum line pull of 20,000 lb (88,964.4 N) at a maximum linear speed of 280 ft / min(1.4 m / s) with a maximum drum torque of 200,000 lb䡠in (22,597.0 N䡠m) at adrum speed of 53.5 r / min
Trang 34maxi-25.34 DESIGN ENGINEERING
Compute the drum horsepower from hp⫽ Tn / 63,000, where the symbols are
the same as in step 1 Or, hp⫽ (200,000)(53.5) / 63,000⫽170 horsepower (126.8kW)
Choose a hydraulic motor having these specifications: displacement ⫽ 6 in3/ r(98.3 cm3/ r); rated pressure⫽6000 lb / in2(41,370.0 kPa); rated speed⫽2500 r /min; output torque at rated pressure⫽5500 lb䡠in (621.4 N䡠m); power output atrated speed and pressure⫽218 horsepower (162.6 kW) This power output rating
is somewhat greater than the computed rating, but it allows some overloading.The gear reduction ratio GR between the hydraulic motor and winch drum, based
on the maximum motor torque, is GR⫽(output torque required, lb䡠in) / (torque atrated pressure, lb䡠in,⫻reduction gear efficiency)⫽20,000 / (5500⫻0.92)⫽39.5:
1 Hence, by using this ratio, the maximum motor speed ⫽53.5 ⫻ 39.5⫽ 2110
r / min Oil flow rate to the motor⫽in3/ r⫻(r / min) / 231⫽6⫻2110 / 231⫽54.8gal / min (3.5 L / s), without leakage With 5 percent leakage, flow rate⫽1.05(54.8)
⫽57.2 gal / min (3.6 L / s)
List the required outputs and the type of motion required—rotary or linear Thus:propel⫽rotary; dozer⫽linear; winch⫽rotary
There are two simultaneous functions: (a) propel motors and dozer cylinders; ( b)
propel motors at slow speed and drive winch
For function a, maximum oil flow ⫽ 50 ⫹ 16.3 ⫽ 66.3 gal / min (4.2 L / s);maximum propel motor pressure ⫽ 6000 lb / in2 (41,370.0 kPa); maximum dozercylinder pressure ⫽ 3200 lb / in2 (22,064.0 kPa) Data for function a came from
previous steps in this calculation procedure
For function b, the maximum oil flow need not be computed because it will be less than for function a.
These are the dozer, propel, and winch functions
These are the propel and dozer functions
The propel and dozer functions have priority over the winch function
Table 9 lists the normal functions required in this machine and the type of valvethat would be chosen for each function Each valve incorporates additional func-tions: The step variable selector valve has a built-in check valve; the propel direc-tional valve and winch directional valve have built-in relief valves and motor over-load valves; the dozer directional valve has a built-in relief valve and a fourth
position called float In the float position, all ports are interconnected, allowing the
dozer blade to move up or down as the ground contour varies
These are: Horsepower for propel and dozer ⫽ (gal / min)(pressure, lb / in2) / 1714for the propel and dozer functions, or (50)(6000) / 1714 ⫹ (16.3)(3200) / 1714 ⫽
205.4 horsepower (153.2 kW) Winch horsepower, by the same relation, is
Trang 35HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN 25.35 TABLE 9 Hydraulic-System Valving and Piping
FIGURE 17 Schematic of the piping, valves, and motors for bulldozer.
(57.2)(6000) / 1714⫽200 horsepower (149.1 kW) Since the propel-dozer functions
do not operate at the same time as the winch, the prime mover power need be only205.4 horsepower (153.2 kW)
To provide independent simultaneous flow to each of the four propel motors, plusthe dozer cylinders, choose two split-flow piston-type pumps having independentoutlet ports Split the discharge of each pump into three independent flows Twopumps rated at 66.3 / 2 ⫽ 33.15 gal / min (2.1 L / s) each at 6000 lb / in2 (41,370.0kPa) will provide the needed oil Figure 17 shows a schematic of the piping, valvesand motors for this bulldozer, while Fig 18 shows the valving
When the vehicle is steered, additional flow is required by the outside wheels.Design the circuit so oil will flow from three pump pistons to each wheel motor.Four pistons of one split-flow pump are connected through check valves to all fourmotors With this arrangement, oil will flow to the motors with the least resistance
To make use of all or part of the oil from the propel-dozer circuits for the winchcircuit, the outlet series ports of the propel and dozer valves are connected into the
winch circuit, since the winch circuit is inoperative only when both the propel and
Trang 3625.36 DESIGN ENGINEERING
FIGURE 18 Valving for bulldozer.
the dozer are operating When only the propel function is in operation, the winch
is able to operate slowly but at full torque
As computed in step 3, the winch gear ratio is based on torque Now, because aknown gal / min (gallons per minute) is available for the winch motor from thepropel and dozer circuits when these are not in use, the gear ratio can be based onthe motor speed resulting from the available gal / min
Flow from the propel and dozer circuit⫽66.3 gal / min (4.2 L / s); winch motorspeed ⫽ 2450 r / min; required winch drum speed ⫽ 53.5 r / min Thus, GR ⫽
2450 / 53.5⫽45.8:1
With the proposed circuit, the winch gear reduction should be increasedfrom 39.5:1 to 45.8:1 The winch circuit pressure can be reduced to (39.5 / 45.8)(6000)⫽5180 lb / in2(35,716.1 kPa) The required size of the winch oil tubing can
be reduced to 0.219 in (5.6 mm)
Using a mechanical efficiency of 89 percent, we see that the prime mover for thepumps should be rated at 205.4 / 0.89 ⫽ 230 horsepower (171.5 kW) The primemover chosen for vehicles of this type is usually a gasoline or diesel engine Figure
19 shows the final tractor-dozer hydraulic circuits
equip-ment using a hydraulic system, such as presses, punches, and balers Other cations for which the method can be used include aircraft, marine, and on-highwayvehicles Use the method presented in an earlier section of this handbook todetermine the required size of the connecting tubing
appli-The procedure presented above is the work of Wes Master, reported in Product
Engineering.
HYDRAULIC PISTON ACCELERATION,
DECELERATION, FORCE, FLOW, AND SIZE
DETERMINATION
What net acceleration force is needed by a horizontal cylinder having a 10,000-lb(4500kg) load and 500-lb (2.2-kN) friction force, if 1500 lb / in2(gage) (10,341 kPa)
Trang 37HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN 25.37
FIGURE 19 Final tractor-dozer hydraulic circuits.
Trang 3825.38 DESIGN ENGINEERING
is available at the cylinder port, there is zero initial piston velocity, and a 100-ft /min (30.5-m / min) terminal velocity is reached after 3-in (76.2-mm) travel at con-stant acceleration with the rod extending? Determine the required piston diameterand maximum fluid flow needed
What pressure will stop a piston and load within 2 in (50.8 mm) at constantdeceleration if the cylinder is horizontal, the rod is extending, the load is 5000 lb(2250 kg), there is a 500-lb (2224-N) friction force, the driving pressure at the headend is 800 lb / in2 (gage) (5515.2 kPa), and the initial velocity is 80 ft / min (24.4
m / min)? The rod diameter is 1 in (25.4 mm), and the piston diameter is 1.5 in(38.1 mm)
Calculation Procedure:
Use the relation F A⫽Ma⫽M⌬V /⌬t, where F A⫽net accelerating force, lb (N);
M ⫽ mass, slugs or lb䡠s2/ ft (N䡠s2/ m); a ⫽ linear acceleration, ft / s2(m / s2), sumed constant;⌬V⫽velocity change during acceleration, ft / s (m / s);⌬t⫽ time
as-to reach terminal velocity, s Substituting for this cylinder, we find M ⫽10,000 /32.17⫽ 310.85 slugs
Next⌬S ⫽3 in / (12 in / ft)⫽ 0.25 ft (76.2 mm) Also⌬V⫽ (100 ft / min) / (60
s / min)⫽1.667 ft / s (0.51 m / s) Then F A⫽0.5(310.85)(1.667)2/ 0.25⫽1727.6 lb(7684.4 N)
Add the friction force and compute the piston area and diameter thus: 兺F ⫽ F A
⫹ F F, where 兺F⫽sum of forces acting on piston, i.e., pressure, friction, inertia,
load, lb; F F⫽friction force, lb Substituting gives兺F⫽1727.6⫹ 500⫽2227.6
lb (9908.4 N)
Find the piston area from A⫽ 兺F / P, where P ⫽fluid gage pressure available
at the cylinder port, lb / in2 Or, A ⫽ 2227.6 / 1500 ⫽ 1.485 in2 (9.58 cm2) The
piston diameter D, then, is D⫽(4A /)0.5⫽ 1.375 in (34.93 mm)
The maximum fluid flow Q required is Q⫽VA / 231, where Q⫽ maximum flow,
gal / min; V⫽terminal velocity of the piston, in / s; A⫽piston area, in2 Substituting,
the effective driving force is F ED⫽ 1413.6⫺500⫽913.6 lb (4063.7 N)
The mass, in slugs, is M⫽F A/ 32.17, from the equation in step 1 By substituting,
M⫽5000 / 32.17⫽ 155.4 slugs
Next, the linear piston travel during deceleration is ⌬S ⫽ 2 in / (12 in / ft) ⫽
0.1667 ft (50.8 mm) The velocity change is ⌬V⫽80 / 60 ⫽1.333 ft / s (0.41 m /s) during deceleration
Trang 39HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN 25.39
The decelerating force F A⫽ 0.5M(⌬V2) /⌬S for the special case when the
ve-locity is zero at the start of acceleration or the end of deceleration Thus F A ⫽
0.5(155.4)(1.333)2/ 0.1667⫽828.2 lb (3684 N)
The total decelerating force is兺F⫽ F A⫹F ED⫽827.2 ⫹913.6 ⫽ 1741.8 lb(7748 N)
The cushioning pressure is P c⫽ 兺F / A, where A differential area⫽piston area⫺
rod area, both expressed in in2 For this piston, A⫽(1.5)2/ 4⫺(1.0)2/ 4⫽0.982
in2(6.34 cm2) Then P⫽ 兺F / A⫽1741.8 / 0.982⫺1773.7 lb / in2(gage) (12,227.9kPa)
Related Calculations. Most errors in applying hydraulic cylinders to accelerate
or decelerate loads are traceable to poor design or installation In the design area,miscalculation of acceleration and / or deceleration is a common cause of problems
in the field The above procedure for determining acceleration and decelerationshould eliminate one source of design errors
Rod buckling can also result from poor design A basic design rule is to allow
a compressive stress in the rod of 10,000 to 20,000 lb / in2(68,940 to 137,880 kPa)
as long as the effective rod length-to-diameter ratio does not exceed about 6:1 atfull extension A firmly guided rod can help prevent buckling and allow at leastfour times as much extension
With rotating hydraulic actuators, the net accelerating, or decelerating torque in
lb䡠ft (N䡠m) is given by T A ⫽ J␣ ⫽ MK2rad / s2 ⫽ 0.1047 MK ⌬N /⌬T ⫽ WK2
⌬N / (307) ⌬t, where J ⫽ mass moment of inertia, slugs䡠ft2, or lb䡠s2䡠ft; ␣ ⫽
angular acceleration (or deceleration), rad / s2; K⫽radius of gyration, ft;⌬N⫽r /min change during acceleration or deceleration; other symbols as given earlier.For the special case where the r / min is zero at the start of acceleration or end
of deceleration, T A⫽0.0008725MK2(⌬N)2/⌬revs; in this case,⌬revs⫽total olutions⫽ average r / min⫻ ⌬t / 60⫽0.5⌬N⌬T / 60;⌬t⫽ 120(⌬revs /⌬t) For the
rev-linear piston and cylinder where the piston velocity at the start of acceleration iszero, or at the end of deceleration is zero, ⌬t ⫽ ⌬S / average velocity⫽ ⌬S / (0.5
High water base fluids (HWBF) are gaining popularity in industrial fluid powercylinder applications because of lower cost, greater safety, and biodegradability.Cylinders function well on HWBF if the cylinder specifications are properly pre-pared for the specific application Some builders of cylinders and pumps offerdesigns that will operate at pressures up to several thousand pounds per squareinch, gage Most builders, however, recommend a 1000-lb / in2(gage) (6894-kPa)limit for cylinders and pumps today
Robotics is another relatively recent major application for hydraulic cylinders.There is nothing quite like hydrostatics for delivering high torque or force incramped spaces
This procedure is the work of Frank Yeaple, Editor, Design Engineering, as
reported in that publication
HYDROPNEUMATIC ACCUMULATOR DESIGN
FOR HIGH FORCE LEVELS
Design a hydropneumatic spring to absorb the mechanical shock created by a
300-lb (136.4-kg) load traveling at a velocity of 20 ft / s (6.1 m / s) Space available tostop the load is limited to 4 in (10.2 cm)
Trang 4025.40 DESIGN ENGINEERING
FIGURE 20 Typical hydropneumatic accumulator (Machine Design.)
Calculation Procedure:
Figure 20 shows a typical hydropneumatic accumulator which functions as a spring.The spring is a closed system made up of a single-acting cylinder (or sometimes arotary actuator) and a gas-filled accumulator As the load drives the piston, fluid(usually oil) compresses the gas in the flexible rubber bladder Once the load isremoved, either partially or completely, the gas pressure drives the piston back forthe return cycle
The flow-control valve limits the speed of the compression and return strokes
In custom-designed springs, flow-control valves are often combinations of checkvalves and fixed or variable orifices Depending on the orientation of the checkvalve, the compression speed can be high with low return speed, or vice versa.Within the pressure limits of the components, speed and stroke length can be varied
by changing the accumulator precharge Higher precharge pressure gives shorterstrokes, slower compression speed, and faster return speed
The kinetic energy that must be absorbed by the spring is given by E k ⫽
12WV2/ 2g, where E k ⫽kinetic energy that must be absorbed, in䡠lb; W ⫽weight
of load, lb; V ⫽ load velocity, ft / s; g ⫽ acceleration due to gravity, 32.2 ft / s2
From the given data, E k⫽12(300)(20)2/ 2(32.2)⫽22,360 in䡠lb (2526.3 N䡠m)
To find the final pressure of the gas in the accumulator, first we must assume anaccumulator size and pressure rating Then we check the pressure developed andthe piston stroke If they are within the allowable limits for the application, theassumptions were correct If the limits are exceeded, we must make new assump-tions and check the values again until a suitable design is obtained
For this application, based on the machine layout, assume that a 2.5-in cm) cylinder with a 60-in3(983.2-cm3) accumulator is chosen and that both arerated at 2000 lb / in2(13,788 kPa) with a 1000-lb / in2(abs) (6894-kPa) precharge.Check that the final loaded pressure and volume are suitable for the load
(6.35-The final load pressure p2 lb / in2 (abs) (kPa) is found from p (n⫺1) / n ⫽
2
p (n⫺1) / n2 {[E k (n⫺1) / ( p1v1)]⫹1}, where p1⫽precharge pressure of the accumulator,
lb / in2 (abs) (kPa); n ⫽the polytropic gas constant⫽ 1.4 for nitrogen, a popularcharging gas; v1 ⫽ accumulator capacity, in3 (cm3) Substituting gives p(1.4⫺1) / 1.42
⫽1000(1.4⫺1) / 1.4{[22,360(1.4⫺1) / (1000⫻60)]⫹1}⫽1626 lb / in2(abs) (11,213.1kPa) Since this is within the 2000-lb / in2(abs) limit selected, the accumulator isacceptable from a pressure standpoint