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ICE401: PROCESS INSTRUMENTATION
AND CONTROL
Class 11
Mathematical Modeling of Pneumatic and
Hydraulic Systems
Dr. S. Meenatchisundaram
Email: meenasundar@gmail.com
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015
Pneumatic System:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015
• Pneumatic system uses compressible fluid as working
medium and it is usually air.
• In pneumatic systems, compressibility effects of gas cannot
be neglected and hence dynamic equations are obtained using
conservation of mass.
• In pneumatic systems, change in fluid inertia energy and the
fluid’s internal thermal energy are assumed negligible.
• In pneumatic system, the mass and volume flow rates are not
readily interchangeable.
• Pneumatic devices involve the flow of gas or air, through
connected pipe lines and pressure vessels.
Pneumatic System:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015
• Hence, the variables of pneumatic system are mass flow rate,
qm, and pressure P.
• The mass flow rate is a through variable and it is analogous
to current. The pressure variable is across variable and is
analogous to voltage.
• The two basic elements of a pneumatic system are the
resistance and capacitance.
• The gas flow resistance, R is defined as the rate of change in
gas pressure difference for a change in gas flow rate.
( )
( )
2
/
/ sec
Changein gas pressuredifference N m
R
Changein gas flowrate Kg
=
Pneumatic System:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015
• Pneumatic capacitance is defined for a pressure vessel and
depends on the type of expansion process involved.
• The capacitance of a pressure vessel may be defined as the
ratio of change in gas stored for a change in gas pressure.
( )
( )2
/
Changein gas stored Kg
C
Changein gas pressure N m
=
Pneumatic System:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015
Pros and Cons of Pneumatic systems:
• Advantages
– The air or gas used is non inflammable and so it offers
safety from fire hazards.
– The air or gas has negligible viscosity, compared to high
viscosity of hydraulic fluids.
– No return pipelines are required and since air can be let
out at the end of work cycle.
• Disadvantage
– The response is slower than that of hydraulic systems
because of the compressibility of the working fluid.
Pneumatic System:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015
Applications of Pneumatic systems:
• Guided Missiles
• Aircraft systems
• Automation of production machines
• Automatic controllers
• Many more……
Pneumatic System:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015
Pi = air pressure of the source at steady state (newton/m2)
P0 = air pressure in the vessel at steady state (newton/m2)
∆ Pi = small change in air pressure of the source from its steady state
∆ P0 = small change in air pressure of the vessel from its steady state
Pneumatic System:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015
• Rate of gas storage in vessel = rate of gas inflow
• Applying Laplace and rearranging the terms, we get
0 0id P P PP
C
dt R R
∆ ∆ − ∆∆
= =
0 ( ) 1
( ) ( 1)i
P s
P s RCs
∆
=
∆ +
Pneumatic System:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015
Consider the system shown below:
Pneumatic System:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015
Let
Steady-state value of input air pressure
Increase in the pressure of air-source
Steady-state value of pressure inside the bellows
P = Increase in pressure inside the bellow
Steady state value of air flow rate
qm = Increase in air flow rate
A = Area of each flat surface of the bellows
R = Resistance of the restriction
C = Capacitance of the bellows.
x = Displacement of the movable surface of the bellows
iP =
ip =
P =
mQ =
Pneumatic System:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015
• Let the pressure of air source be increased from its steady
state value by an amount Pi.
• This results in an increase in air flow by qm and increase in
the pressure inside the bellows by p.
• Due to increase in pressure, there will be a displacement of
the movable surface of the bellows, by an amount x.
• Here, the terms Pi, qm, p and x are all functions of time, t and
therefore can be expressed as Pi(t), qm(t) p(t) and x(t) .
• The force exerted on the movable surface of the bellows is
proportional to increase in pressure inside the bellows,
i.e, Fb ∝ p(t)
Pneumatic System:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015
• Force exerted on the movable surface of the bellow,
Fb =A p(t) (8.1)
• The force opposing the movement of the flat surface of
bellow walls is proportional to the displacement, i.e.,
F0 ∝ x(t) (8.2)
• Force opposing the motion,
F0 =K x(t) (8.3)
• Where, K is a constant representing the stiffness of the
bellows. At steady state the above two forces are balanced,
Fb = F0 → A p(t) =K x(t) (8.4)
Pneumatic System:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015
• The resistance R can be given as,
• Rearranging, (8.5)
• The Capacitance C can be given as,
• Rearranging,
(8.6)
( ) ( )Difference between Change in pressure
Change in air flow rate ( )
i
m
p t p t
R
q t
−
= =
( ) ( )
( ) i
m
p t p t
q t
R
−
=
( )Change in air flow rate
( )Rate of change of pressure
mq t
C
dp t
dt
= =
( )
( )m
dp t
q t C
dt
=
Pneumatic System:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015
• Equating eqn. 8.5 and 8.6,
(8.7)
• Rearranging, (8.8)
• From the eqn. 8.4,
(8.9)
• Differentiating eqn. 8.9 yields,
(8.10)
• Substituting eqn. 8.10 into 8.8,
( ) ( )( ) ip t p tdp t
C
dt R
−
=
( )
( ) ( )i
dp t
RC p t p t
dt
+ =
( ) ( )
K
p t x t
A
=
( ) ( )dp t K dx t
dt A dt
=
Pneumatic System:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015
(8.11)
• Taking Laplace
(8.12)
• Rearranging,
(8.13)
• Where,
( )
( ) ( )i
K dx t K
RC x t p t
A dt A
+ =
( ) ( ) ( )i
K K
RC sX s X s P s
A A
+ =
( )
( ) 1 1i
A A
X s K K
P s RCs sτ
= =
+ +
RCτ =
Hydraulic Systems:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015
• They are used in hydraulic feedback systems and in combined
electro-mechanical-hydraulic systems.
• In hydraulic devices, power is transmitted through the action of
fluid flow under pressure and the fluid is incompressible.
• The fluid used are petroleum based oils and non – inflammable
synthetic oils.
• Hydraulic devices used in control systems are generally
classified as hydraulic motors and hydraulic linear actuators.
• The output of a hydraulic motor is rotary motion.
• The hydraulic motor is physically smaller in size than an
electric motor for the same output.
Hydraulic Systems:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015
• Applications are power steering and brakes in automobiles,
steering mechanisms in large ships, control of machine tools
etc.
Advantages:
• Hydraulic fluid acts as a lubricant and coolant.
Comparatively small sized hydraulic actuators can develop
large forces or torques.
• Hydraulic actuators can be operated under continuous,
intermittent reversing and stalled conditions without damage
• Hydraulic actuators have a higher speed of response. They
offer fast starts, stops and speed reversals.
Hydraulic Systems:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015
• With availability of both linear and rotary actuators, the
design has become more flexible.
• Because of low leakages in hydraulic actuators, when loads
are applied the drop in speed will be small.
• Hydraulic components are more rugged than their
counterparts.
Disadvantages:
• Hydraulic power is not readily available compared to electric
power.
• Inherent problems of leaks and sealing them against foreign
particles.
Hydraulic Systems:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015
• Operating noise
• Costs more compared to electrical system
• Fire and explosion hazards exist
• Hydraulic lines are not flexible as electric cables
• Highly non linear response
Hydraulic Systems:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015
Hydraulic Systems:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015
Let
• qp = Rate at which the oil flows from the pump
• qm = Oil flow rate through the motor
• qi = Leakage flow rate
• qc = Compressibility flow rate
• x = Input stroke length
• θ = Output angular displacement of motor
• P = Pressure drop across motor
Hydraulic Systems:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015
• The rate at which the oil flow from the pump is proportional to
stroke angle, i.e., qp ∝ x.
• Oil flow rate from the pump, qp = Kp x
• Where Kp is a constant and is equal to the ratio of rate of oil
flow to unit stroke angle.
• The rate of oil flow through the motor is proportional to motor
speed, i.e.,
• Therefore, Oil flow rate through motor
• Where, Km = Motor displacement constant.
m
d
q
dt
θ
∝
m m
d
q K
dt
θ
=
Hydraulic Systems:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015
• All the oil from the pump does not flow through the motor in
the proper channels.
• Due to back pressure in the motor, a portion of the ideal flow
from the pump leaks back past the pistons of motor and
pump.
• The back pressure is the pressure that is built up by the
hydraulic flow to overcome the resistance to free movement
offered by load on motor shaft.
• It is usually assumed that the leakage flow is proportional to
motor pressure, i.e. qi ∝ P
Hydraulic Systems:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015
• Therefore, Leakage flow rate, qi= Ki P
• Where Ki = constant.
• The back pressure built up by the motor not only causes
leakage flow in the motor and pump but also causes the oil in
the lines to compress.
• Volume compressibility flow is essentially proportional to
pressure and therefore the rate of flow is proportional to the
rate of change of pressure, i.e.,
→ Try the rest.
c
dP
q
dt
∝
References:
• Modern Control Engineering, 5th Edition, by Katsuhiko Ogata.
• Advanced Control Systems Engineering, Ronald Burns
• Control Systems, Nagoor Kani.
• A course in Electrical, Electronic Measurements and
Instrumentation, A.K. Sawhney.
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015

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Class 11 mathematical modeling of pneumatic and hydraulic systems

  • 1. ICE401: PROCESS INSTRUMENTATION AND CONTROL Class 11 Mathematical Modeling of Pneumatic and Hydraulic Systems Dr. S. Meenatchisundaram Email: meenasundar@gmail.com Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015
  • 2. Pneumatic System: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015 • Pneumatic system uses compressible fluid as working medium and it is usually air. • In pneumatic systems, compressibility effects of gas cannot be neglected and hence dynamic equations are obtained using conservation of mass. • In pneumatic systems, change in fluid inertia energy and the fluid’s internal thermal energy are assumed negligible. • In pneumatic system, the mass and volume flow rates are not readily interchangeable. • Pneumatic devices involve the flow of gas or air, through connected pipe lines and pressure vessels.
  • 3. Pneumatic System: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015 • Hence, the variables of pneumatic system are mass flow rate, qm, and pressure P. • The mass flow rate is a through variable and it is analogous to current. The pressure variable is across variable and is analogous to voltage. • The two basic elements of a pneumatic system are the resistance and capacitance. • The gas flow resistance, R is defined as the rate of change in gas pressure difference for a change in gas flow rate. ( ) ( ) 2 / / sec Changein gas pressuredifference N m R Changein gas flowrate Kg =
  • 4. Pneumatic System: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015 • Pneumatic capacitance is defined for a pressure vessel and depends on the type of expansion process involved. • The capacitance of a pressure vessel may be defined as the ratio of change in gas stored for a change in gas pressure. ( ) ( )2 / Changein gas stored Kg C Changein gas pressure N m =
  • 5. Pneumatic System: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015 Pros and Cons of Pneumatic systems: • Advantages – The air or gas used is non inflammable and so it offers safety from fire hazards. – The air or gas has negligible viscosity, compared to high viscosity of hydraulic fluids. – No return pipelines are required and since air can be let out at the end of work cycle. • Disadvantage – The response is slower than that of hydraulic systems because of the compressibility of the working fluid.
  • 6. Pneumatic System: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015 Applications of Pneumatic systems: • Guided Missiles • Aircraft systems • Automation of production machines • Automatic controllers • Many more……
  • 7. Pneumatic System: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015 Pi = air pressure of the source at steady state (newton/m2) P0 = air pressure in the vessel at steady state (newton/m2) ∆ Pi = small change in air pressure of the source from its steady state ∆ P0 = small change in air pressure of the vessel from its steady state
  • 8. Pneumatic System: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015 • Rate of gas storage in vessel = rate of gas inflow • Applying Laplace and rearranging the terms, we get 0 0id P P PP C dt R R ∆ ∆ − ∆∆ = = 0 ( ) 1 ( ) ( 1)i P s P s RCs ∆ = ∆ +
  • 9. Pneumatic System: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015 Consider the system shown below:
  • 10. Pneumatic System: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015 Let Steady-state value of input air pressure Increase in the pressure of air-source Steady-state value of pressure inside the bellows P = Increase in pressure inside the bellow Steady state value of air flow rate qm = Increase in air flow rate A = Area of each flat surface of the bellows R = Resistance of the restriction C = Capacitance of the bellows. x = Displacement of the movable surface of the bellows iP = ip = P = mQ =
  • 11. Pneumatic System: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015 • Let the pressure of air source be increased from its steady state value by an amount Pi. • This results in an increase in air flow by qm and increase in the pressure inside the bellows by p. • Due to increase in pressure, there will be a displacement of the movable surface of the bellows, by an amount x. • Here, the terms Pi, qm, p and x are all functions of time, t and therefore can be expressed as Pi(t), qm(t) p(t) and x(t) . • The force exerted on the movable surface of the bellows is proportional to increase in pressure inside the bellows, i.e, Fb ∝ p(t)
  • 12. Pneumatic System: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015 • Force exerted on the movable surface of the bellow, Fb =A p(t) (8.1) • The force opposing the movement of the flat surface of bellow walls is proportional to the displacement, i.e., F0 ∝ x(t) (8.2) • Force opposing the motion, F0 =K x(t) (8.3) • Where, K is a constant representing the stiffness of the bellows. At steady state the above two forces are balanced, Fb = F0 → A p(t) =K x(t) (8.4)
  • 13. Pneumatic System: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015 • The resistance R can be given as, • Rearranging, (8.5) • The Capacitance C can be given as, • Rearranging, (8.6) ( ) ( )Difference between Change in pressure Change in air flow rate ( ) i m p t p t R q t − = = ( ) ( ) ( ) i m p t p t q t R − = ( )Change in air flow rate ( )Rate of change of pressure mq t C dp t dt = = ( ) ( )m dp t q t C dt =
  • 14. Pneumatic System: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015 • Equating eqn. 8.5 and 8.6, (8.7) • Rearranging, (8.8) • From the eqn. 8.4, (8.9) • Differentiating eqn. 8.9 yields, (8.10) • Substituting eqn. 8.10 into 8.8, ( ) ( )( ) ip t p tdp t C dt R − = ( ) ( ) ( )i dp t RC p t p t dt + = ( ) ( ) K p t x t A = ( ) ( )dp t K dx t dt A dt =
  • 15. Pneumatic System: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015 (8.11) • Taking Laplace (8.12) • Rearranging, (8.13) • Where, ( ) ( ) ( )i K dx t K RC x t p t A dt A + = ( ) ( ) ( )i K K RC sX s X s P s A A + = ( ) ( ) 1 1i A A X s K K P s RCs sτ = = + + RCτ =
  • 16. Hydraulic Systems: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015 • They are used in hydraulic feedback systems and in combined electro-mechanical-hydraulic systems. • In hydraulic devices, power is transmitted through the action of fluid flow under pressure and the fluid is incompressible. • The fluid used are petroleum based oils and non – inflammable synthetic oils. • Hydraulic devices used in control systems are generally classified as hydraulic motors and hydraulic linear actuators. • The output of a hydraulic motor is rotary motion. • The hydraulic motor is physically smaller in size than an electric motor for the same output.
  • 17. Hydraulic Systems: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015 • Applications are power steering and brakes in automobiles, steering mechanisms in large ships, control of machine tools etc. Advantages: • Hydraulic fluid acts as a lubricant and coolant. Comparatively small sized hydraulic actuators can develop large forces or torques. • Hydraulic actuators can be operated under continuous, intermittent reversing and stalled conditions without damage • Hydraulic actuators have a higher speed of response. They offer fast starts, stops and speed reversals.
  • 18. Hydraulic Systems: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015 • With availability of both linear and rotary actuators, the design has become more flexible. • Because of low leakages in hydraulic actuators, when loads are applied the drop in speed will be small. • Hydraulic components are more rugged than their counterparts. Disadvantages: • Hydraulic power is not readily available compared to electric power. • Inherent problems of leaks and sealing them against foreign particles.
  • 19. Hydraulic Systems: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015 • Operating noise • Costs more compared to electrical system • Fire and explosion hazards exist • Hydraulic lines are not flexible as electric cables • Highly non linear response
  • 20. Hydraulic Systems: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015
  • 21. Hydraulic Systems: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015 Let • qp = Rate at which the oil flows from the pump • qm = Oil flow rate through the motor • qi = Leakage flow rate • qc = Compressibility flow rate • x = Input stroke length • θ = Output angular displacement of motor • P = Pressure drop across motor
  • 22. Hydraulic Systems: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015 • The rate at which the oil flow from the pump is proportional to stroke angle, i.e., qp ∝ x. • Oil flow rate from the pump, qp = Kp x • Where Kp is a constant and is equal to the ratio of rate of oil flow to unit stroke angle. • The rate of oil flow through the motor is proportional to motor speed, i.e., • Therefore, Oil flow rate through motor • Where, Km = Motor displacement constant. m d q dt θ ∝ m m d q K dt θ =
  • 23. Hydraulic Systems: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015 • All the oil from the pump does not flow through the motor in the proper channels. • Due to back pressure in the motor, a portion of the ideal flow from the pump leaks back past the pistons of motor and pump. • The back pressure is the pressure that is built up by the hydraulic flow to overcome the resistance to free movement offered by load on motor shaft. • It is usually assumed that the leakage flow is proportional to motor pressure, i.e. qi ∝ P
  • 24. Hydraulic Systems: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015 • Therefore, Leakage flow rate, qi= Ki P • Where Ki = constant. • The back pressure built up by the motor not only causes leakage flow in the motor and pump but also causes the oil in the lines to compress. • Volume compressibility flow is essentially proportional to pressure and therefore the rate of flow is proportional to the rate of change of pressure, i.e., → Try the rest. c dP q dt ∝
  • 25. References: • Modern Control Engineering, 5th Edition, by Katsuhiko Ogata. • Advanced Control Systems Engineering, Ronald Burns • Control Systems, Nagoor Kani. • A course in Electrical, Electronic Measurements and Instrumentation, A.K. Sawhney. Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015