Tài liệu Handbook of Mechanical Engineering Calculations P25 pdf

25.1
SECTION 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 Double-
Acting Hydraulic Cylinder 25.74
Hydraulic Cylinder Oil Consumption and
Extension Time 25.75
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 has
the dimensions, operating pressures, and performance given in Table 1. Pilot supply
pressure is 100 lb/in
2
(689 kPa); main supply pressure is 500 lb/in
2
(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
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Source: HANDBOOK OF MECHANICAL ENGINEERING CALCULATIONS
25.2 DESIGN ENGINEERING
TABLE 1 Dimensions and Operating Conditions*
Pilot Spool Main spool
Diameter, in d ϭ 0.25 D ϭ 2.5
Mass, lb-sec
2
/in m ϭ 0.0002 M ϭ 0.05
Stroke, in s
ϭ 0.375 S ϭ 1.5
Land length, in — L
ϭ 6.0
Radial clearance, in — C
ϭ 0.0003
Coefficient of friction — F

R
ϭ 0.04
Solenoid force, lb (initial; final) F
ϭ 1; 8.5
SOL
—
Back pressure, lb / in
2
— p
B
ϭ 20
Supply pressure, lb/in
2
p ϭ 100 P ϭ 500
Differential pressure, lb/in
2
⌬p ϭ 70 (approx) ⌬P ϭ 450 (approx)
Port area, in
2
a
0
ϭ 0.05 A
M
ϭ 1.2
Flow coefficient ƒ
ϭ 0.6 ƒ ϭ 0.6
Viscosity, CP
␮
ϭ 80
␮

ϭ 80
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 operating
pressure i.e., 500 lb /in
2
(3445 kPa). As a further modification, a small actuating
piston is placed at each end of the main spool, Fig. 3, to increase the longitudinal
velocity 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
ϭ 1.4 in (3.56 cm) for this small actuating piston. If the
dimensions and operating pressures are unchanged, analyze the valve for the same
unknowns as above.
Calculation Procedure:
1. Compute the axial force on the main spool of this valve
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 ϭ 2F ƒA ⌬P
as r M

where the symbols are as given, Table 2. Then, P
ax
ϭ 2(0.04)(0.6)(1.2)(450) ϭ 26
lb (115.6 N). converting to pressure by dividing by the area of the main spool valve
end, we have 26/4.9
ϭ 5.3 lb / in
2
(36.5 kPa).
2. Compute the combined hydrodynamic resistance of the valve
Provisionally, estimate that D
V
is equivalent to 3.2 lb/in
2
(22 kPa) and P
B
ϭ pilot-
valve backpressure
ϭ 20 lb/in
2
(138.8 kPa). The combined hydrodynamic resis-
tance is then the sum of: Radial pressure, lb/in
2
(kPa) ϩ Viscous drag, lb/in
2
(kPa)
ϩ Pilot-valve backpressure, lb/in
2
(kPa). Or combined hydrodynamic resistance ϭ
5.3 ϩ 3.2 ϩ 20 ϭ 28.5 lb / in
2

(196.4 kPa).
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HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN
HYDRAULIC 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
d
x
a
p
—
—
S
a
l
p
—
C
D
A
s
M
—
S
—
L
C
Solenoid
Forces
Initial, lb
Gradient, lb/in
Final, lb

Ratio, A/B
A
B
F
SOL
r
—
—
—
—
—
—
—
—
Drag
Forces
Back pressure, psi
Viscous drag, lb (or psi)
Radial jet, lb
Coefficient of friction
Axial jet, lb
Acceleration force, lb
p
B
—
—
—
—
F ϭ ma
p

B
d
V
P
rad
F
R
P
ax
—
p
B
D
V
P
rad
F
R
P
zx
F ϭ Ma
Oil
pressure,
flow, and
port size
Pressure:
Supply, psi
Pilot downstream, psi
Differential, psi
Port area, in

2
(effective orifice)
Flow coefficient (0.55 to 0.70)
Viscosity, centipoise
Oil density, lb-sec
2
/in
4
Flow rate, in
3
/sec
Oil velocity, in/sec (through port)
Oil mass flow, lb-sec/in
Oil-jet deflection angle, deg
p
p
1
⌬p
a
0
ƒ
␮
␳
q
—
—
—
p
p
1

⌬p
—
ƒ
␮
␳
q
—
—
—
P
p
1
⌬P
A
M
ƒ
␮
␳
—
V
0
M
ƒ
a
Valve
response
Acceleration time, sec
Shifting velocity, in/sec
Shifting time, sec
(after energization)

t
a
v
p
t
T
a
v
T
T
a
V
T
*SI values given in calculation procedure.
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HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN
25.4 DESIGN ENGINEERING
3. Calculate the pilot-valve flow rate
The pilot-valve pressure differential, delta P
ϭ 100 Ϫ 28.5 ϭ 71.5 lb/in
2
(492.6
kPa). Hence, the valve flow rate is, using the equation below
2⌬p
q ϭ ƒa
o
Ί
␳

where q ϭ flow, in
3
/s (mL / s); ƒ ϭ flow factor, dimensionless, ranging from 0.55
to 0.70 depending on valve type; a
o
ϭ cross-sectional area, in
2
(cm
2
); of the min-
imum port opening—usually the drilled port hole;
⌬ p ϭ p Ϫ p
1
ϭ differential
pressure, lb/in
2
(kPa) measured across the pilot inlet and outlet ports; p ϭ fluid
mass density, lb-s
2
/in
4
, normally 0.000085 for oil. Substituting, q ϭ 0.6
(0.5)[(2)(71.5)/0.000085)]
ϭ 40 in
3
/s (656 mL /s), using a value of ƒ ϭ 0.6 for
0.5
this valve.
4. Determine the maximum velocity of the main spool and the viscous damping
force

The maximum velocity of the main spool, Fig. 1, V
ϭ (flow rate, in
3
/s/(area of
spool end, in
2
) ϭ 40 / 4.9 ϭ 8.2 in/s (20.8 cm/s). Knowing the velocity, we can
find the damping force, D
V
, from
D
␲
LV
␮
D ϭ
V
6
C ϫ 6.9 ϫ 10
where D
ϭ spool diameter, in (cm); L ϭ length of spool lands, in (cm); V ϭ main
spool velocity, in / s (cm/s); mu
ϭ absolute viscosity, centipoise; C ϭ spool-to-bore
radial clearance, in (cm). If the temperature varies more than 30 to 50 degrees, it
is nearly impossible to compute the viscous resistance. Substituting, D
V
ϭ
2.5
␲
(6)(8.2)(80)/(0.0003)(6.9 ϫ 10
6

) ϭ 3.05 lb/in
2
(21 kPa). Thus, the provisional
estimate of D
V
ϭ 3.2 was close enough (within 4.9 percent) and recalculation is
not necessary.
5. Find the accelerating pressure and acceleration time of the spool
The forces acting upon the spool during acceleration are: p
R
, P
ax
, D
V
, and F, where
F
ϭ Ma. Assuming a mean value for initial port opening A
M
ϭ 0.4 in
2
(2.58 cm
2
),
then from
P
ϭ 2ƒA ⌬P cos
␣
as M
where
␣

normally varies from 70 degrees at initial opening to 90 degrees at full
opening. In calculations, use the axial jet pressure during initial opening, and the
axial component of radial pressure during the remainder of travel. Substituting,
P
ax
ϭ 2(0.6)(0.4)(450)(0.26) ϭ 56 lb (248.1 N). Then 56 / 4.9 ϭ 11.4 lb / in
2
(78.5
kPa),
␣
ϭ 75 deg; cos
␣
ϭ 0.26.
Viscous drag will be the average: D
V
ϭ 3.2/2 ϭ 1.6 lb/in
2
(11 kPa). Backpres-
sure is still p
B
ϭ 20 lb / in
2
(137.8 kPa). So the total is 11.4 ϩ 1.6 ϩ 20 ϭ 33 lb /
in
2
(227.4 kPa).
Therefore, accelerating pressure
ϭ 100 Ϫ 33 ϭ 67 lb / in
2
(461.6 kPa). Con-

verting to force, we have 67 (4.9)
ϭ 328 lb (1441.2 N). The acceleration time, t
a
s ϭ MV /F ϭ 0.05 (8.2)/328 ϭ 0.0013 s.
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HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN
25.5
FIGURE 1 Typical solenoid-energized pilot-operated spool valve.
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HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN
25.6 DESIGN ENGINEERING
FIGURE 2 Jet-force drag in pilot-operated spool valves.
6. Determine the main spool displacement and the energization time interval
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 (3.8
cm) is 1.5/8.2
ϭ 0.182 s, and the time interval from energization of the solenoid
to completion of the main valve stroke, T
ϭ 0.190 s.
7. Analyze the valve with the higher pilot pressure
Much larger flow will pass through the pilot valve because of the higher pressure.
Maximum velocity period: P
ax
ϭ 5.3 lb/in
2
(36.5 kPa), the same as before; D

V
ϭ
7.7 lb/in
2
(53.1 kPa)— a higher estimate, proportional to the anticipated velocity;
p
B
ϭ 20.0 lb/in
2
(137.8 kPa), the same as before. The total is 33.0 lb / in
2
(227.5
kPa).
The new
⌬ P ϭ 467 lb/in
2
(3217.6 kPa), and Q ϭ 4.7 (467) ϭ 102 in
3
/s
0.5
(1671.5 mL/s); V ϭ 102/4.9 ϭ 20.8 in / s (52.1 cm/s); D
V
ϭ 1.82 (20.8) ϭ 37.8
lb (168.1 N)
ϭ 7.75 lb / in
2
(53.4 kPa), which proves out the assumption of 7.7 lb
/in
2
(53.1 kPa).

Accelerating time, t
a
ϭ (0.05)(20.8)/(2280) ϭ 0.0005 s. The 1.5-in (3.81-cm)
stroke takes 1.5/20.8
ϭ 0.072 s. Total time ϭ 0.081 s.
The flow rate of the pilot oil is more important than pressure intensity in ob-
taining a fast-acting valve. A slightly larger pilot valve and enlarged porting have
a marked effect on the operational speed of the main valve.
Note that increasing the pilot pressure fivefold, from 100 lb/in
2
to 500 lb / in
2
(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
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HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN
HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN 25.7
additional pressure is necessary, saving on pumping costs. Costs of producing a
0.375-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 losses
entailed by an increase in pressure.
8. Analyze the valve fitted with actuating pistons
For the valve, Fig. 3, with the small actuating pistons, taking the summation of the
viscous drag,
͚D
V
, and inserting the known optimum values in parentheses after
the computed values, we have:

8.2
ϫ 4.9 ϫ 4 ϫ 80
͚D ϭ
V
6
0.0003 ϫ 6l9 ϫ 10 ϫ 1.4
2.5
ϫ 6
ϫϩ2 ϫ 1.5
ͩͪ
1.4
12800 15
ϭϩ3
ͩͪ
2075 ϫ 1.4 1.4
ϭ 60.5 lb (269.1 N) [59.0 lb optimum; 262.4 N]
Introducing the value of
͚ D
V
in the equation,
3k 3(P
ϩ ͚D )
3 ax V
a ϭϭ
p
2k 2(p Ϫ p )
2 B
we have
3(26
ϩ 60.5)

22
a ϭϭ1.62 in (10.45 sq cm) [1.59 in ; 10.26 cm ]
p
2(100 Ϫ 20)
d
ϭ 1.44 in (3.66 cm) [1.425 in, 3.62 cm].
With optimum a
p
ϭ 1.59 cu in (26.06 mL), piston velocity using
2⌬p
q ϭ ƒa
o
Ί
␳
kk
23
v ϭ k Ϫ
1
23
Ί
aa
pp
22
kk kk
12 13
ϭϪ
23
Ί
aa
pp

80 (26 ϩ 59)
v ϭ 4.6 Ϫis
p
Ί
2.53 4.0
ϭ 15.0 in/s (38.1 cm / s)
The total time, T
ϭ 0.100 ϩ 0.009 ϭ 0.109 s. Using a pilot pressure of 500 pst
(3445 kPa), V
ϭ 20.8 in / s (52.8 cm / s) and d ϭ 1.06 in (2.69 cm). Then:
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HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN
25.8
FIGURE 3 Piston-operated solenoid-energized spool valve.
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HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN
HYDRAULIC 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
T,
sec
100 2.50 8.2 0.190 15.0 1.425 0.109
500 2.50 20.8 0.081 55.0 1.06 0.036
*SI values given in calculation procedure.
20.8 ϫ 4.9 ϫ 4.80
͚DV ϭ
6
0.0003 ϫ 6.9 ϫ 10 ϫ 1.06

15
ϫϩ3 ϭ 258 lb
ͩͪ
1.06
3(26
ϩ 258)
2
a ϭϭ0.886 in
p
2(500 Ϫ 20)
d
ϭ 1.06 in
z
480 284
v ϭ 4.6 Ϫ
p
Ί
0.784 0.61
ϭ 55 in/s
T
ϭ 0.036 s
Table 3 and Fig. 4 depict the effect of actuating-piston area upon the spool shifting
velocity and shifting time.
Related Calculations. Pilot-operated flow-control valves are probably the most
common valves used in industrial hydraulic systems. Speed of response of these
valves is important during the design and operation of any hydraulic system. The
procedure given here analyzes the speed of response of a typical valve in terms of
the 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 the
procedure 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 to
the full open position, Fig. 5, and the main spool is shifted at a rate determined by
the amount of fluid that can move through the pilot ports against these five resisting
forces: (1) pilot system backpressure, lb/in
2
(kPa); (2) viscous damping force, lb
(N); (3) radial jet force, lb (N); (4) axial jet force, lb (N); (5) acceleration force, lb
(N).
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HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN
25.10 DESIGN ENGINEERING
SI Values
in./sec cm/sec in.
2
cm
2
5 12.7 1 6.45
10 25.4 2 12.9
15 38.1 3 19.4
20 50.8 4 25.8
25 63.5 5 32.3
FIGURE 4 Effect of varying piston diameter.
FIGURE 5 Before energization and after full stroke of a solenoid-energized spool valve.
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HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN
HYDRAULIC 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 new
position. This chart is for a 0.25-in (0.635-cm) diameter closed-center pilot spool
controlling flow to a 2.5-in (6.35-cm) main spool.
Pilot pressure chosen for this procedure is 100 lb/in
2
(689 kPa), which is less
than the main supply pressure. A separate supply for the pilot is required. There
are no hard and fast rules for establishing pilot pressures, but if possible keep the
pressure in the range of a few hundred lb/in
2
(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 be
made; otherwise there is no practical mathematical solution. For one, assume that
the 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/in
2
(1378 kPa),
because it adds a margin of safety that compensates for spool rubbing friction. The
friction is from metal-to-metal contact at points where the oil film is partially de-
stroyed.

Above 400-lb/in
2
(2756 kPa) operating pressure, a separate pilot supply usually
is provided. Pilot pressure in these instances ought to be at least 7 percent of the
main 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.
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HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN
25.12 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
ЊF (48.9ЊC) in a 70Њ-F (21.1Њ-C) ambient (50Њ-F
[27.8
Њ-C]) temperature difference, T
D
. The source of heat is a 20-gal/min (1.26-L-
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HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN
HYDRAULIC 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. The

hydraulic working unit, including the piping, is not helping to dissipate heat. Res-
ervoir tank size is small with a cooling surface of 21 ft
2
(1.95 m
2
), based on a
reservoir volume of twice pump flow—or 2
ϫ 20 ϭ 40 gal (75.8 L). System
pressure
ϭ 750 lb / in
2
(5167.5 kPa); overall heat transfer coefficient, k ϭ 5 Btu/ft
2
hr ЊF (28.4 W/m
2
ЊC), a conservative value. (2) The return flow of the hydraulic
fluid in another industrial hydraulic system must be cooled continuously to 125
ЊF
(51.7
ЊC). The hottest uncooled drain temperature of the fluid is 140ЊF (60.0ЊC).
Flow of the hydraulic fluid through the system is 12 gal/min (0.76 L/s); the cool-
ing-water inlet temperature
ϭ 65ЊF (18.3ЊC); outlet temperature ϭ 85ЊF (29.4ЊC);
k
ϭ 90 Btu/hr-ft
2
ЊF (511.2 W/m
2
ЊC). Use a counterflow, single-pass heat ex-
changer, Fig. 8, in this analysis. (3) Lastly, calculate the temperature of a standard

60-gal (227.4 L) reservoir after 5 hr of operation. Pump discharge is 20 gal/min
(1.26 L/s) at 750 lb/in
2
(5.17 MPa). Cooling surface is 28 ft
2
(0.792 sq m). An
attached heat-dissipating working unit weighs W
ϭ 800 lb (362.2 kg) and its ef-
fective surface area A
ϭ 5.5 ft
2
(0.156 m
2
). With an initial system oil temperature,
T
ϭ 70ЊF (21.1ЊC), and an ambient temperature, T ϭ 50ЊF (10.0ЊC), the initial
oil air
temperature-over-ambient, T
p
ϭ 70 Ϫ 50 ϭ 20ЊF(11.1ЊC). The estimated median
value of k
ϭ 4 Btu/ft
2
hr ЊF(22.7 W/sq m ЊC). The 60 gallons of oil weigh 444
lb (201.6 kg).
Calculation Procedure:
1. Find the total heat generated in the system
Find the total heat generated in the system using the equation
E
ϭ 1.48 ϫ Q ϫ P(1 Ϫ

␮
)
L
where the symbols are as given below. Substituting
2
E ϭ 1.48(20 gal/min)(750 lb/in )(1 Ϫ 0.60) ϭ 8880 Btu/hr (2601.1 W).
L
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HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN
25.14 DESIGN ENGINEERING
FIGURE 9 Hydraulic oil and cooling water temperature plot
for heat exchanger in Fig. 8.
2. Compute the required reservoir cooling area
Use the equation
T
ϭ E /͚kA
max DL
and solve for the required area, A.Or,A ϭ 8880/(50 ϫ 5) ϭ 35.5 ft
2
(3.3 m
2
).
Since this reservoir has only 21 ft
2
(1.95 m
2
) of cooling surface, the tank area is
not large enough to dissipate the heat generated. Hence, a larger reservoir cooling

area must be provided for this installation.
3. Determine the heat-transfer area and cooling-water flow rate required
Use the equation
E
ϭ ⌬T ϫ Q ϫ 210
exch oil oil
ϭ ⌬T ϫ Q ϫ 500
water water
to find the heat exchanger heat load. Substituting, we have E ϭ (140 Ϫ 125)
exch
(12)(210) ϭ 37,800 Btu/hr (11.1 kW). The maximum temperature difference,
⌬T ϭ 125 Ϫ 65 ϭ 60ЊF (51.7ЊC). Minimum temperature difference ϭ ⌬T ϭ
max min
140 Ϫ 85 ϭ 55ЊF (30.6ЊC). Log-mean temperature difference, computed as shown
elsewhere in this handbook (see index) is
⌬T ϭ 57.5Њ (31.9ЊC). Figure 9 shows
mean
the oil and water temperature changes in a generalized manner.
Find the required heat-transfer area from
E
ϭ kA⌬T Btu/hr
exch mean
solving for A.OrA ϭ 37,800/(90 ϫ 57.5) ϭ 7.3 ft
2
(0.68 m
2
).
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HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN
HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN 25.15
To find the required cooling-water flow rate, use Q ϭ E /(500 ϫ gal/min),
water exch
where 500 ϭ a constant to convert gal/min to gph. Substituting, Q ϭ 38,800/
water
(500 ϫ 20) ϭ 3.78 gal/min (0.2385 L /s).
Refer to manufacturer’s catalogs for the size of a heat exchanger with the proper
surface area. Be certain to check the heat exchanger pressure rating if the system
pressure exceeds 150 lb / in
2
(1033.5 kPa).
4. Find the heat exchanger heat load
Use the same relation as in step 1, above, to find E
L
ϭ 8880 Btu / hr (2601.1 W).
5. Compute the heating variables for the fluid and reservoir
The heat balance for the oil and attached heat dissipating working unit are given
by cW
ϭ 0.4(444) ϩ 0.1(800) ϭ 257.6 Btu/ЊF (489.4 kJ/ЊC). For the tank and
working unit, the kA
ϭ 4(28 ϩ 5.5) ϭ 134 Btu / hrЊF (254.6 kJ / hrЊC).
6. Determine the temperature above ambient for the hydraulic fluid
Use the equation
E
tt
L
Ϫ͚
kA/
͚

cW
Ϫ͚
kA/
͚
cW
T ϭ (1 Ϫ e ) ϩ ( T )e
D init D
͚kA
8880
Ϫ
134/257
ϫ
5
Ϫ
134/257
ϫ
5
T ϭ (1 Ϫ e ) ϩ 20 e
D
134
Substituting, we have T
D
ϭ 62.8ЊF (17.1ЊC).
The maximum operating temperature over ambient, T
D
ϭ E
L
/kA ϭ 8880 /
max
134 ϭ 66.3ЊF (19ЊC). Then, the oil temperature ϭ 50 ϩ 66.3 ϭ 116.3ЊF (46.8ЊC).

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-dissipating
unit. The influence of the initial temperature difference is minor. Practical experi-
ence with system design is most important.
Related Calculations. The highest recommended temperature for oil in a con-
ventional hydraulic systems reservoir is 120
ЊF (48.9ЊC). The procedure presented
above shows ways to prevent the oil temperature from exceeding that level. There
are certain exceptions to the rule given above. Some conventional hydraulic systems
are designed to operate at 150
ЊF (65.6ЊC). So-called super-systems with special
fluids and seals can operate at 500
ЊF (260ЊC), and higher. But for any level of
operating temperature, the same heat-transfer principles apply.
In designing a fluid system’s heat-transfer, after you’ve established the basic
system and reservoir design, follow the simple heat-balance method given above.
If the reservoir’s peak temperature calculated this way is less than 120
ЊF (48.9ЊC)
(or some other desired temperature), no further work is necessary. That’s why a
heat 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 heat
exchanger, using the rating method given in this procedure.
Most of the heat in industrial hydraulic systems comes from in-the-system com-
ponents. Exceptions are systems in hot environments or adjacent to heat-producing
equipment, but the same heat-balance principles apply.
Heat is generated whenever hydraulic oil is throttled or otherwise restricted.
Examples of heat-producing devices include pressure regulators, relief valves, un-
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HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN
25.16 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 turbulence
anywhere 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
ϭ E (1 Ϫ
␮
) where
␮
ϭ system efficiency.
in
Figure 10 shows a typical duty cycle in an industrial hydraulic system. The
utilization pressure, measured at the inlet to the working device (fluid motor or
cylinder) will always be lower than the source pump pressure, depending on the
amount of throttling or other regulation required in the system.
The difference in energies—pumped vs utilized—must be absorbed during tran-

sients and eventually dissipated by the system. An approximate measure of overall
system efficiency,
␮
, is the ratio of the mean effective pressure, Fig. 10, to pump
pressure, where the mean effective pressure is calculated from the area under the
curve, Fig. 10, divided by the time base.
Figure 11 compares reservoir tank temperature for two different pumps: (1) a
fixed-delivery pump, with constant flow and pressure and a full-flow relief valve
for 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 heat
generation is during idling of the workload; all the flow is throttled back to the
reservoir and does no useful work. By comparison, the variable-delivery pump does
not 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 the
needed oil is pumped. Excess capacity is forced back to the reservoir through the
relief valve, and the energy is converted to waste heat. Auxiliary pumps are great
offenders if they are operated when not needed. Additional useful guides for res-
ervoir selection and sizing are given below.
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HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN
HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN 25.17
SI Values
FC
50 10
100 37.8
150 65.6
FIGURE 11 Proper pump choice can conserve energy and re-

duce system temperature.
Plan for a reservoir capacity that can feed the pump system for two or three
minutes, neglecting any return flow. With a tank that big, several related require-
ments usually are met automatically: (1) There will be enough reserve fluid to fill
the hydraulic system at startup without exposing the filter and strainer; (2) A fairly
stable oil level will be maintained despite normal fluctuations in flow; (3) Enough
hydraulic fluid will be available to sustain the system while the rotating parts coast
to a stop during emergency shutdown if a return line breaks; (4) Thermal capacity
will be available to absorb unexpected heat for short periods or to store heat during
idle 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 gallons
per min (L/min)—that is, if the tank can be pumped dry in less than 2
minutes—add a heat exchanger to the system circuit to avoid excessive temperature
fluctuations. 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 provide
reliable service for the system. The suction-line filter should be
1
⁄
2
to
3
⁄
4
in (1.3 to
1.9 cm) above the tank bottom. Strainer oil flow capacity should be two to four
times the pump capacity. A vacuum gage on the pump suction will show if the
strainer is clogged. A permanent-magnet filter can be specified as a drain plug or
mounted on the baffle plate in a region of concentrated return oil flow.

The main return oil flow should discharge below the reservoir oil level about
one inch (2.5 cm) above the tank bottom. Backpressure in the return line will be
5to10lb/in
2
(34.5 to 68.9 kPa), or higher. Atmospheric return lines, including
seal-leakage lines, are at zero pressure and should be discharged above the hydraulic
oil level.
If the atmospheric lines have high flow and a high air content, they should be
discharged above the oil level into a chute sloping gradually (5 to 10 degrees) into
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HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN
25.18 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 pres-
sures will run 25 percent hotter than air-free oil. This is caused partly by the heat
of compression of the air and partly by its low thermal conductivity.
Internal baffles between the return pipe and pump inlet will slow the fluid cir-
culation, help settle out dirt particles, give air the chance to escape, and allow
dissipation of heat. The top of the baffle should be submerged about 30 percent
below the surface of the fluid.
Keep the hydraulic oil temperature between 120 and 150
ЊF (54 and
66
ЊC)—preferably at the lower value for oil viscosities from 100 to 300 SSU based
on 100
ЊF (38ЊC). Temperatures up to 160ЊF (71ЊC) are permissible if the hydraulic
fluid viscosity is from 300 to 750 SSU, based on 100
ЊF (38ЊC). Higher operating

temperatures require special design.
Tank walls should be thin to permit good thermal conductivity. Make them
approximately
1
⁄
16
in (0.16 cm) for tank capacities up to 25 gal (95 L);
1
⁄
8
in (0.32
cm) for capacities up to 100 gal (379 L);
1
⁄
4
in (0.64 cm) for 100 gal (379 L) or
more. Use slightly heavier plate for the bottom. Give the top plate four times wall
thickness to assure vibration-free operation and to hold alignment of pump and
motor shafts. Specify a thermometer to be mounted on the tank top where the
operator can see it.
Avoid designing industrial hydraulic system machines with integral tanks. It is
better to have a separate reservoir, accessible from all sides. Small reservoir tanks
can 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 com-
mitting yourself to the design.
Equip the reservoir tank with cleanout doors and slope the bottom toward the
doors. 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 the
tank top for removing filters and strainers. Design a connection for hooking to a
portable filtration unit.

If the reservoir tank is made of cast iron, don’t paint the interior surface. Be
sure 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
␮
ϭ System efficiency, E / E ,%
used in
E ϭ
used
Energy utilized in system, Btu / hr (W)
E
ϭ
in
Pump input power, Btu/hr (W)
E
L
ϭ Heat loss generated in system, Btu / hr (W)
E
A
ϭ Heat absorbed by oil, tank and components, Btu / hr (W)
E
D
ϭ Heat dissipated to atmosphere or coolant, Btu/hr (W)
E
ϭ
exch
Heat exchanger load, Btu / hr (W)

Fluid conditions and flow
t
ϭ Operating time, hr
Q
ϭ Flow, gal / min (L / s)
P
ϭ Pump gage pressure, lb / in
2
(kPa)
T
D
ϭ Temperature-over-ambient for oil, ЊF: T
D
ϭ T Ϫ T (values are mean) (C)
oil air
⌬T ϭ Heat exchanger only: ⌬T ϭ T Ϫ T ; ⌬T ϭ T Ϫ T ; ⌬T ϭ log-mean
water out in oil in out mean
⌬T, oil-to-water (C)
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HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN
HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN 25.19
TABLE 4 Pressure-Temperature Value for Gasket Materials*
Material Max P
i
ϫ T Max Temp, ЊF
Rubber 15,000 300
Vegetable fiber 40,000 250
Rubberized cloth 125,000 400

Compressed asbestos** 250,000 850
Metal types
Ͼ250,000 depends on metal
*SI values given in procedure.
**Or acceptable substitute.
Equation constants
c
ϭ Specific heat (mean value), Btu/lb-ЊF (kJ/kg ЊC)
k
ϭ Overall heat-transfer coefficient, Btu / sq ft-hr-ЊF(W /m
2
ЊC)
W
ϭ Combined weight of oil and system components, lb (kg)
A
ϭ Surface area for dissipating heat, sq ft (sq m)
e
ϭ Base for natural logs ϭ 2.718
͚ ϭ Summation sign. ͚cW ϭ effective cW for all of system components
Typical values for c and k
c, Btu/lb-
ЊF: (kJ/kg ЊC) Oil, 0.40; aluminum, 0.18; iron, 0.11; copper, 0.09
k, Btu/sq ft-hr-
ЊF: (W mm/m
2
ЊC)
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 / in
2
(8.27 MPa)
and 180
ЊF (82.2ЊC). 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:
1. Determine the total bolt force and torque for these flanges
Assuming that torque wrenches will be used to check this installation, as is almost
universally done today, select the bolt-stress method to calculate the total bolt force.
This method uses the equation,
F
ϭ NS A
bbbb
where the symbols are as given below.
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HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN
25.20 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
0.138 (No. 6) 32 0.0090 40 0.0101
0.164 (No. 8) 32 0.0139 36 0.0146
0.190 (No. 10) 24 0.0174 32 0.0199
0.216 (No. 12) 24 0.0240 28 0.0257
1
⁄
4
20 0.0317 28 0.0362
5
⁄
16
18 0.0522 24 0.0579
3
⁄
8
16 0.0773 24 0.0876
7
⁄
16
14 0.1060 20 0.1185
1
⁄

2
13 0.1416 20 0.1597
1
⁄
2
12 0.1374
9
⁄
16
12 0.1816 18 0.2026
5
⁄
8
11 0.2256 18 0.2555
3
⁄
4
10 0.3340 16 0.3724
7
⁄
8
9 0.4612 14 0.5088
1 8 0.6051 12 0.6624
1
1
⁄
8
7 0.7627 12 0.8549
1
1

⁄
4
7 0.9684 12 0.0721
1
3
⁄
8
6 1.1538 12 1.3137
1
1
⁄
2
6 1.4041 12 1.5799
*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/in
2
(206.7 MPa)
without yielding. This can be verified from
F
ϭ 16,000D / A
bbb
which gives F
b
ϭ (16,000)(0.75/0.3340) ϭ 36,000 lb (160.1 kN).
From the bolt-stress equation, F
b
ϭ 4(30,000)(0.3340) ϭ 40,080 lb (178.3 kN).
The torque required to produce this stress level at installation is given by,
T

ϭ 0.2DS A
bb b
Or, T ϭ 0.2(0.75)(30,000)(0.3340)/12 ϭ 125 ft-lb (169.4 Nm). This torque will be
specified on the system assembly drawings so it is used during construction.
2. Choose a suitable gasket material
The pressure-temperature relation for this installation is 1200
ϫ 180 ϭ 216,000 in
USCS units and 6798 in SI units. This, from Table 4, suggests choosing a com-
pressed-asbestos (or acceptable substitute) type gasket. This would be compatible
with industrial hydraulic fluid.
The gasket area is 4.73 sq in. (30.5 sq cm), calculated from an outside diameter
of 2
7
⁄
8
in (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
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HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN
HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN 25.21
S ϭ F / A
gbg
Or, S
g
ϭ 40,080/4.73 ϭ 8470 lb/in
2
(58.4 MPa).
Table 6 shows this stress will easily seat the selected compressed-asbestos (or

acceptable substitute) gasket. Tentatively choose asbestos (or acceptable substitute)
with CR (neoprene) binder for oil resistance; thickness
1
⁄
32
in (0.079 cm).
3. Determine the hydrostatic end force for the chosen gasket
The mean area acted upon by the pressure in the hydraulic line is defined by a
diameter of (2.875
ϩ 1.5)/2 ϭ 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 bolt
force by the equation,
F
м KP A
btm
Or,
Thus, there is no end-force balance problem with bolts stressed to 30,000 lb / in
2
(206.7 MPa).
4. Select a suitable surface finish for the flange
Table 6 shows that a concentric-serrated surface finish on the flange is best. Econ-
omy may dictate a conventional spiral-serrated surface, which Table 6 shows will
work in this case.
5. Prepare the final specification for the gasket
Include in the specifications the material type, dimensions, and bolt-torque data
computed 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 industrial
hydraulic systems, the steps and data are valid for choosing gaskets for any piping
system: steam, condensate, oil, fuel, etc. Just be certain that the pressures and

temperatures are within the ranges in the tables and equations presented here.
Three main design factors govern the selection of a gasket material—whether
sheet 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 is
much 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 internal
pressure 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 factors
for the majority of gasket joint applications.
Where asbestos is the recommended gasket material in this procedure, the de-
signer must review the environmental aspects of the design. Acceptable substitute
materials may be required by local environmental regulations. Hence, these regu-
lations must be carefully studied before a final design choice is made.
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HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN
25.22
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HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN
25.23
TABLE 6 Minimum Seating Stresses for Typical Gasket Materials*
Material
Type (see
Table IV)
Thickness,
in
Minimum seating
stress, lb/in
2
Flange-surface finish
1st choice
2nd
choice
N
O
N
M
E
T
A
L
Rubber sheet
SBR (75 Durometer)
CR (60 Durometer)
Compressed asbestos**
SBR binder
CR binder

Rubberized cloth
Vegetable-fiber sheet
Fluorocarbon (TFE)
Virgin
Glass-filled
Asbestos cloth (impreg.)**
Flat
Flat
Flat
Flat
Folded
Flat
Flat
Flat
Flat
1
⁄
32
and up
1
⁄
32
and up
1
⁄
64
1
⁄
32
1

⁄
16
1
⁄
8
1
⁄
64
1
⁄
32
1
⁄
16
1
⁄
8
2-ply
3-ply
4-ply
all
1
⁄
64
1
⁄
32
1
⁄
16

1
⁄
8
1
⁄
64
1
⁄
32
1
⁄
16
1
⁄
8
3
⁄
32
200
175
3,000
2,000
1,600
1,200
3,750
2,500
2,000
1,500
2,500
2,100

1,800
750
14,000
6,500
3,700
1,600
14,000
11,000
6,000
3,000
1,600
Concentric-
serrated
80 rms
(
1
⁄
64
in only)
Concentric-
serrated
(all other
thicknesses)
All
other
types
All
other
types
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HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN
25.24
M
E
T
A
L
Flat metals
Aluminum
Copper
Carbon steel
Monel
Stainless
Stamped metals
Lead
Aluminum
Copper
Carbon steel
Monel
Stainless
Machined metals
Aluminum
Copper
Carbon steel
Monel
Stainless
Flat
Corrugated

jacket-
metal
filler
Profile
(
1
⁄
8
-in pitch)
1
⁄
32
and
1
⁄
16
9
⁄
64
only
All
thicknesses
20,000
45,000
68,750
81,250
93,750
500
1,000
2,500

3,500
4,500
6,000
25,000
35,000
55,000
65,000
75,000
Concentric-
serrated
80 rms
or
less
80
rms
or
less
150
to
200
rms
M
E
T
A
L
A
S
B
E

S
T
O
S
Metal (asbestos-filler)
Aluminum
Copper
Carbon steel
Stainless
Monel
Lead
Aluminum
Copper
Carbon steel
Nickel
Monel
Stainless
Same as stamped metals
Stainless
Corrugated
and
corded
Plain
metal
jacket
Corrugated
jacket
Spiral-wound
1
⁄

8
only
1
⁄
8
only
0.125
and
0.175
2,000
2,500
3,000
3,500
4,000
500
2,500
4,000
6,000
6,000
7,500
10,000
3,000
to
30,000
150 to
200
rms
80 rms
or
less

Concen-
tric-
serrated
150
to
200
rms
*SI values given in procedure.
**Or acceptable substitute.
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HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN
25.25
TABLE 7 Typical Gasket Cross-Sections
Type (see Table 4) Cross-section Description
Flat
Folded
Plain metal jacket
Corrugated jacket:
metal-filled
asbestos-filled*
Corrugated and
corded*
Profile
Spiral-wound
Simplest gasket form; available in wide variety of materials for
different conditions of fluid media, temperature and pres-
sure; most easily manufactured for nonsymmetrical shapes.
Fabric-based gaskets especially designed for extremely rough

and wavy flanges with low bolting loads.
Asbestos filler enclosed partially or totally by metal jacket;
combines easy compression with resistance to high pressure.
Temperature resistance depends on jacket material chosen.
Corrugated metal or asbestos core enclosed by a corrugated-
metal jacket and top washer. Better than flat gaskets for
rough flanges—corrugations give greater resilience.
Deeply corrugated metal with twisted asbestos cord cemented
into the corrugations on both faces; aluminized for non-
sticking. Well-suited to rough or warped flanges.
Heavy solid metal with concentric V-shape contact ridges that
provide multiple seating surfaces. Each ridge is 0.010 in
wide. For N ridges, effective arera is then:
␲
ϫ mean con-
tact dia
ϫ 0.010 ϫ N. Seating stress increases with decrease
in pitch.
Interlocked plies of preformed metal strip are spiral-wound
with an interleaving cushion of asbestos or fluorocarbon
plastic strip. Ratio of metal to filler can be closely con-
trolled to vary seating stress over wide range. Well suited to
fluctuating-temperature applications.
*Or acceptable substitute.
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HYDRAULIC AND PNEUMATIC SYSTEMS DESIGN