Process Dynamics pracs iitr

LAB MANUAL

Process Dynamics and Control Lab (Undergraduate Level)

Chemical Engineering Department, IIT Roorkee

CHN 303: CHEMICAL ENGINEERING LABORATORY PROCESS DYNAMICS AND CONTROL (UNDERGRADUATE LEVEL)

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iii

TABLE OF CONTENTS

Experiment No Experiment topics Page

No

1 Dynamics of stirred tank with proportional ON/OFF temperature controller 1

2 Dynamics of evacuated tank 7

3 Dynamics of thermometer 9

4 Control valve characteristics 17

(A) Study of control valve flow coefficient (Cv) 19

(B) Study of inherent characteristics of linear control valve 23

(C) Study of installed characteristics of control valve 27

(D) Study of Hysteresis of control valve 29

(E) Study of Rangeability 33

5 Dynamics of pressurized tank 35

6 Interacting and Non-Interacting System 39

(A) Step response of single capacity system 41

(B) Step response of first order systems arranged in non-interacting mode 47

(C) Impulse response of first order systems arranged in non-interacting mode 53

(D) Step response of first order systems arranged in interacting mode 57

(E) Impulse response of first order systems arranged in interacting mode 61

7 Dead weight pressure gauge calibrator 65


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1

EXPERIMENT NO. 1

DYNAMICS OF STIRRED TANK TEMPERATURE CONTROLLER

OBJECTIVE

1. To study the dynamics of a stirred tank system fitted with electric heating assembly.

2. To study the response of an ON/OFF temperature controller thermostat fitted to the tank.

3. To estimate temperature band within which it controls the temperature of the tank.

THEORY

When a step input in terms of heat (by switching the electrical heater ON) is applied to the stirrer tank

vessel its temperature gradually increases and finally attains a constant value depending upon the

magnitude of step input and flow rate of water through the stirred tank. By recording the temperature

history of the vessel, the time constant (τ ) of the system can be computed.

f, To

f, T

V

Q

T

ControllerThermocouple

Figure 1.1: Sketch of the stirred tank system for mass balance

Let,

(J/s)input heat of Rate Q)(kg/m fluid theofDensity K)(in re temperatuReference T

K)(in re temperatuFinal TK)(in re temperatuInitial T

/s)m(in fluid theof rate flow Volumetric (J/kg.K) tank theentering fluid theofcapacity heat Specific

in

3

o

3

==

=

===

=

∗

ρ

fC

Making an energy balance on the tank as shown in the Figure 1.1 above we get,

( ))( )( )( ∗∗∗ −=+−−− TTCVdtdQTTCfTTCf ino ρρρ ….(1)


2

dtdTCVQTCfTCf ino )( )( ρρρ =+− ….(2)

dtdT

CfQ

TT ino τ

ρ=+−

)( ….(3)

f

V=τ ….(4)

At steady state, 0=dtdT

0

)( =+−fC

QTT ins

sso ρ

….(5)

From equation (3) and equation (5) we get

)(

)()( sin

sinss

oo TTdtd

CfQQ

TTTT −=−

+−−− τρ

….(6)

Let’s we define the deviation variables,

os

oo TT θ=− ….(7)

θ=− sTT

….(8)

HQQ sinin =− ….(9)

( )dtd

fCH

oθτ

ρθθ =+− ….(10)

Suppose we assume there is no variation in temperature of input stream then 0=oθ then

dtd

fCH θτθρ

=− ….(11)

Now taking Laplace operator both sides we get,

)()()( sssfCsH θτθ

ρ=− ….(12)

)1()/1(

)()(

+=

sfC

sHs

τρθ

….(13)

For a step input to heat sAsH =)( ….(14)

)1( )/1()(

+=

ssfCAs

τρθ ….(15)


3

+

=)/1(

1)(ττρ

θssfC

As ….(16)

+

−=)/1(

11)(τρ

θssfC

As ....(17)

Taking inverse Laplace we get,

[ ]τρ

θ /1)( tefCAt −−= ….(18)

Hence from eq.(8) and eq.(18) )1( /τ

ρts e

fCATT −−+= ….(19)

When the thermostat is used to control the temperature of the stirrer tank vessel heated by electric heater

it controls the temperature within the band. By studying the temperature-time history of the stirrer tank

vessel when controller pressed into action this temperature band can be estimated.

APPARATUS

1. Overhead constant head water tank.

2. Tank fitted with variable speed stirrer.

3. Immersion type electrical heater (3 kW).

4. Electronic type ON/OFF temperature controller.

PROCEDURE

1. Open valve (VC) so as to fill the overhead tank. Once the water level in the overhead tank attains

a constant level, water from the overflow line starts flowing in the bottom reservoir from where it

is pumped back to the overhead tank.

2. Open the valve (VA) to allow the flow of water from overhead tank to stirred tank. Slowly the

water level in the tank rises and through valve (VB) it drains out.

3. Switch ON the stirrer and switch ON the electric heater. Fix the value of set point temperature in

the controller at 90 °C. This will ensure the heater to remain on as due to the low capacity of the

heater the final temperature will never reach 90 °C with time.

4. Monitor the rise in temperature in the stirrer tank. Continue monitoring the temperature and time

till the temperature in the tank stabilizes.

5. Switch OFF the heater and note the fall in the temperature with time till it attains a constant

value.


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VB

VA

VC

Cooling Water

Ove

rflo

w li

ne

Liquid Level Indicator

Over Head Tank

StirrerThermometer On/Off Type Temp. Controller

Electric Heater

To Drain

Main phase

Bulb

Thermostat (Thermal Switch)

Switch (Electric)

Bottom reservoirRecirculating pump

To

over

hea

d

Figure 1.2: Schematic diagram of the apparatus (Note: the dashed lines show suggested changes that

don’t currently exist)

6. Fix the set point temperature between (40-50) °C in the controller and start recording the

temperature of the stirred tank with time. After a few minutes the electric bulb attached to the

controller will stop glowing indicating that electric heater has been switched OFF by the

thermostat. Keep on noting the temperature-time history of the tank. After a few minutes the bulb

will again glow indicating that the heater has been switched ON. Keep on noting the temperature-

time history till the electric bulb stops glowing for the second time.


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OBSERVATION AND CALCULATIONS

A) When the electric heater was switched ON (without controller operating)

Time, in s

Temperature, in °C

Draw a plot between temperature and time for the case and from this plot compute the time

constant of the stirred tank.

B) When the electric heater was switched OFF (without controller operating)

Time, in s

Temperature, in °C

Draw a plot between temperature and time for the case and from this plot compute the time

constant of the stirred tank.

C) When the controller was pressed into service to control tank temperature

Time, in s

Temperature, in °C

Draw a plot between temperature and time for the case and compute the temperature control band

from the plot, i.e., the temperature drop in tank observed between the ON and OFF states of the

controller.

DISCUSSIONS

1. Explain how a controller works and what type of controller is it?

2. Explain the statement ______ “For high liquid flow rates from the tank the controller may fail to

control the temperature of the tank especially when the set point temperature is high”.

REFERENCES

Steven E. LeBlanc, Donald R. Coughanowr, “Process systems analysis and control”, 3rd Ed.,

McGraw Hill, NY, 2009, pg.168-172.


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7

EXPERIMENT NO. 2

DYNAMICS OF EVACUATED TANK

OBJECTIVE

1. To study the dynamics of evacuated tank

THEORY

The following figure shows a sketch of the major parts of the instruments.

Vacuum pressure gauge

Vent cock

Vacuum Tank

Mercury reservoir

Manometer

Vacuum pump

Figure 2.1: Schematic of the evacuated tank assembly

APPARATUS

1. Mercury Reservoir and U-tube 2. Vacuum Pressure Gauge (0 to -760 mmHg) 3. Vacuum tank with vent cock 4. Vacuum pump

PROCEDURE

1. Start the vacuum pump and observe the rise of mercury in the tube. The vent cock on the vacuum

tank should remain closed while vacuum is being created. Note the levels at some proper time

intervals till the maximum vacuum has been achieved (Condition-1).

2. After steady state is attained, note the level of the mercury in the U-tube manometer attached to

the reservoir and also the vacuum pressure indicated by the pressure gauge.

3. Crack open the vent cock approximately half-way suddenly, so that the vacuum starts falling very

slowly. Note the fall in vacuum in the vessel (pressure gauge) and inlet line (U-tube manometer),

as a function of time, till a new steady state is reached i.e. the pressure ceases to vary (Condition-

2).


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4. Give the next step input by suddenly closing the vent cock and observe the change in pressure

with time P(t) (Condition-3).

5. Plot the pressure as a function of time (i.e P(t) vs time(t)) and discuss the results with

mathematical analysis.

6. Report the time constants at three different conditions and compare them.

7. Repeat Condition-3 for another two valve opening and perform similar analysis.

OBSERVATIONS AND CALCULATIONS

Condition – 1: Initial evacuation of tank calibration of pressure gauge using mercury manometer.

S. No. Pressure, in mm Hg Time, in s Level of mercury, in mm Hg

Condition – 2: After crack opening the vent cock half way

S. No. Pressure, in mm Hg Time, in s

Condition – 3: Completely closing vent cock valve

S. No. Pressure, in mm Hg Time, in s

Repeat the tables for condition – 2 and condition – 3 with varied openings of vent cock.

DISCUSSIONS

1. How does the height of mercury in the reservoir/pot affect the manometer readings? What precaution should you take to avoid error associated with it?

REFERENCES


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EXPERIMENT NO. 3

DYNAMICS OF THERMOMETER

OBJECTIVE

1. To study the dynamics of a thermometer with and without thermowell.

THEORY

The following figure shows a schematic of the experimental set-up. The set up consists of a blower with

heater. The blower can be operated with/without the heater ON. The blower blows air into the channel

from one end and has a thermometer that is inserted and held in place using a rubber plug at the other end.

Optionally a thermowell can be inserted on to the thermometer to study higher order dynamics.

For Heater For Blower

Channel

Thermometer

Air blower

Thermowell

Figure 3.1: Schematic of the experimental set-up for studying dynamics of thermometer.

A simple mercury thermometer is a first order system under the following assumption.

1. All the resistance to heat transfer resides in the film surrounding the bulb.

2. All the thermal capacity is in the mercury.

3. The glass wall containing the mercury does not expand or contract during the transient response.

The transfer function for this system is as follows:

1 1

)()(

+=

ssXsY

τ

where, τ is the time constant given by (mC/hA), m is the mass of mercury, C is the specific heat capacity,

h is heat transfer coefficient and A is the area of heat transfer. The introduction of thermowell leads to a

second order system due to additional resistance of the oil. The transfer function is of the form:


10

)12(1

)()(

22 ++=

sssXsY

ζττ

The system is over damped with the values of τ and 𝜁𝜁calculated using methods such as slope and

intercept method, method of moments, method of Harriot, etc.

The time response of the first order system to a step input is given by

≥−

<=

=

− 0 when t)1A(

0 hen t w 0)(

bygiven is )(

τt

etY

sAsX

The step response of a second order system is as follows:

)(

)( 2121

12

−

−

+=−− ττ ττ

ττ

tteeAAtY

( ) ( )1 and 1- where 22

21 −+=−= ζζττζζττ

APPARATUS

1. Thermowell

2. Glass bulb thermometer

3. Hot air blower

PROCEDURE

1. Make sure the thermometer bulb is clear of oil and dirt.

2. Note the steady state temperature of the thermometer before switching on the hot air blower.

3. Switch on the hot air blower and note the change in temperature with respect to change in time till

the steady state value is attained.

4. Switch off the hot air blower and record the fall in the temperature with respect to time as in

previous step till the steady state is attained.

5. Repeat the above steps with the introduction of the thermowell.

6. Calculate the time constants for the first order as well as the second order system respectively

using graph.

CALCULATIONS

1. Thermometer without thermowell:


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During heating:

The initial steady state value of temperature is Ti = ________ °C.

Time, in s

Temperature, in °C

The final steady state value of temperature T is Tf = ________ °C.

Therefore, step change is

A = (Tf - Ti ) = _________ °C

And step response of thermometer without thermowell is

)e-(1A )Y( t- τ=t ,

where, Y is deviation variable. The value of Y(t) at different time (t)

==

==

==

=

τ

τ

τ

3 when C_________ )e-A(12 when C_________ )e-A(1

when C_________ )e-A(1)Y(

o3-

o2-

o-1

ttt

t

are noted from the graph of Y(t) vs t for experimental data, we can calculate different value of time (t).

s ________ 3s _______ 2

s ______

===

τττ

Therefore, s ________ 6

32=

++

=ττττ

During Cooling:

Similarly, the cooling curve is obtained with a forcing function of step input of negative magnitude.



Time, in s

Temperature, in °C

Therefore, step change is


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A = (Ti - Tf ) = _________ °C

)e-(1A - )(

)1( )(

t-τ

τ

=

+

−=

tY

ssAsY

where, Y(t) is deviation temperature.

==

==

==

=

τ

τ

τ

3 when C_________ )e-A(1-2 when C_________ )e-A(1-

when C_________ )e-A(1-)Y(

o3-

o2-

o-1

ttt

t

Now from cooling curve: we can calculate different value of time (t),

s ______ 3s _______ 2s ________

===

τττ

Therefore, s _________ 6

32=

++

=ττττ

2. Thermometer with thermowell



Time, in s

Temperature, in °C

Therefore step change is

A = (Tf - Ti ) = _________ °C

And step response of thermometer with thermowell is

)(

)( 2121

12

−

−

+=−− ττ ττ

ττ

tteeAAtY

( ) ( )1 and 1- where 22

21 −+=−= ζζττζζττ

The value of Y(t) at different time (t) can be obtained from the plot of experimental data. The following

method can be used for determining the time constants:

i) Slope and Intercept Method


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This method works if the time constants are very different from each other i.e. |τ1| >> |τ2|, the

larger the difference the better the method works.

a) Since, |τ1| >> |τ2|, we can ignore the first term in the second bracket and rewrite the

equation as

121 )(

1)(1 2121

τττ ττττ

ttteee

AtY −−−

≈

−

−

=−

For large values of time, the fractional incomplete response is close to a first order

system. Therefore, we fit an exponentially decreasing curve to data at large times and

extend to the y-axis and find the intercept (say P). The time corresponding to (0.368 P)

gives the larger of time constants. b) Next to find the smaller time constant, subtract the exponential fit from the overall curve

that the experimental data fits. From the resulting curve, find the intercept with y-axis

(say R). The time corresponding to (0.368R) gives the smaller time constant.

The procedure is illustrated in the plot shown in Figure 3.2.

Figure 3.2: Illustration of slope and intercept method

ii) Semi-Log Method

Follow the procedure given on page 297 in Process Systems Analysis and Control by D. R.

Coughanowr, 2nd Edition to identify the process transfer function.

iii) Method of Harriott

a) When the fractional response (Y(t)/A) is plotted with t/(τ1+τ2) for various time constant

ratios (τ2/τ1 = 0, 0.1, 0.2, …), at Y(t)/A = 0.73, all the curves intersect and the

-0.20

0.00

0.20

0.40

0.60

0.80

1.00

1.20

1.40

0 100 200 300 400 500 600 700 800 900 1000

Slope Intercept Method

Fractional Incomplete Response First order system Delta

Original Experimental Data Exponential Fit

Difference of the two curves


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corresponding value of t/(τ1+τ2) = 1.3. Therefore, from our experimental data, find the

time (t0.73) corresponding to (Y(t)/A = 0.73), then (τ1+τ2) = (t0.73)/1.3.

b) Another plot of values of (Y(t)/A) at t/(τ1+τ2) = 0.5 from the above plots Vs

corresponding (τ2/τ1) was prepared. Now calculate t = 0.5 x (τ1+τ2) using (τ1+τ2)

determined above. Find value of (Y(t)/A) for this time from data. Then find (τ2/τ1) from

the second plot corresponding to this (Y(t)/A).

c) From the sum of time constants and product of time constants determine the time

constants τ1 and τ2.

iv) Method of Oldenbourg and Sartorius

This method is based on the point of maximum slope and inflection in the fractional response

curve.

a) Plot the fractional response curve (Y(t)/A) Vs time. Draw a tangent at the point of

inflection and extend it to Y(t)/A = 0 and Y(t)/A = 1.0 (horizontal lines) to get

intersection points A and B respectively. The time duration from point A to point B is τA

and time duration from point of inflection to point B is τB. These time durations are

related to time constants as follows:

𝜏𝜏𝐴𝐴 = 𝜏𝜏1 �𝜏𝜏1

𝜏𝜏2�

𝜏𝜏2𝜏𝜏1−𝜏𝜏2

𝜏𝜏𝐵𝐵 = 𝜏𝜏1 + 𝜏𝜏2

b) Let 𝜏𝜏1𝜏𝜏2

= 𝑥𝑥, we can get the relationship

𝑥𝑥(1 − 𝑥𝑥) 𝑙𝑙𝑙𝑙 �

1𝑥𝑥� = 𝑙𝑙𝑙𝑙(1 + 𝑥𝑥) − 𝑙𝑙𝑙𝑙 �

𝜏𝜏𝐵𝐵𝜏𝜏𝐴𝐴�

c) Plot L.H.S and R.H.S and find the intersect points (which are also solutions satisfying the

above equation). The intersection points are 𝜏𝜏1 𝜏𝜏𝐴𝐴⁄ and 𝜏𝜏2 𝜏𝜏𝐴𝐴⁄ . Using the intersection

points and 𝜏𝜏𝐵𝐵 = 𝜏𝜏1 + 𝜏𝜏2, determine the time constants.

v) Method of Moments

a) Plot the fractional incomplete response vs time and find the area under the curve. The

value obtained is the zeroth moment (𝜇𝜇𝑜𝑜) = 𝜏𝜏1 + 𝜏𝜏2.

b) The mean value of the curve is the first moment (𝜇𝜇1) = 𝜏𝜏12 + 𝜏𝜏1𝜏𝜏2 + 𝜏𝜏2

2.

c) From the values of moments determine the time constants.

DISCUSSIONS

1. How does the thermowell affect the temperature readings? What will be effect of empty

thermowell and thermowell filled with oil? What about different material of thermowell?


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REFERENCES

Donald R. Coughanowr, “Process systems analysis and control”, 2nd Ed., McGraw Hill, NY,

1991, pg.297-299.




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17

EXPERIMENT NO. 4

CONTROL VALVE CHARACTERISTICS (THREE VALVES)

OBJECTIVE

1. To study control valve characteristics

INTRODUCTION

VALVE ACTION AND ACTUATOR MECHANISM

If the control valve is used to control fluid flow, some mechanism must physically open or close

the valve. Different types of actuators are used to control the stem travel of the valve, like

electrical actuator, pneumatic actuator, hydraulic actuator etc. In present set-up, pneumatic

actuators are used for control valves. Spring opposed diaphragm actuator positions the valve plug

in response to the controller signals. Mostly the controller signals are in the range of 3-15 psig.

There are two types of actuators described below.

1. Direct acting actuator (Air to close):

Direct acting actuators basically consists of a pressure tight housing sealed by a flexible

fabric reinforced elastomer diaphragm. A diaphragm plate is held against the diaphragm

by a heavy compression spring. Signal air pressure is applied to upper diaphragm case

that exerts force on the diaphragm and the actuator assembly. By selecting proper spring

rates or stiffness, load carrying capacity and initial compression, desired stem

displacement can be obtained for any given input signal.

2. Reverse acting actuator (Air to open):

In case of reverse acting actuators the stem gets retracted with increase in pressure.

The Figure 4.1 shows an “air to open” type of control valve to control the flow rate. The air to

move the diaphragm is regulated through the pressure regulator. The air is supplied from an air

compressor. The flow of water takes place from a constant head tank that is equipped with an

overflow line. For the valve plug type shown in the figure, increasing the air pressure from the

regulator, the valve stem is moved down thus opening the control valve.


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Air pressure regulator

Air compressor

Pump

Flow inlet valve

Figure 4.1: A general scheme showing an air actuated control valve.

TYPES OF CONTROL VALVE

Valve is essentially a variable orifice. Control valve is a valve with a pneumatic, hydraulic

electric (excluding solenoids) or other externally powered actuator that automatically, fully or

partially opens or closes the valve to a position dictated by signals transmitted from controlling

instruments. Control valves are used primarily to throttle energy in a fluid system and not for

shutoff purpose. Depending upon the valve plug design the control valves can be divided in three

categories as under:

1. Equal percentage type: Flow changes by a constant percentage of its instantaneous value for each unit of valve lift.

aybeQ =

Q = Flow at constant pressure drop y = Valve opening e = base of natural logarithm a, b = Constants

Constants a & b can be evaluated to

Qo= Flow at constant pressure drop at zero stroke


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give more convenient form xyy

R

oeQQ

= max

log

R = Flow range of valve, maximum to minimum at constant pressure drop

ymax= Maximum rated valve opening.

2. Linear type: Flow is directly proportional to valve lift.

maxmax yy

FF

= Q = Flow at constant pressure drop. y = Valve opening k = constant

kyQ =

3. Quick opening type (On/Off type): Flow increases rapidly with initial travel reaching near its maximum at a low lift. It is generally not defined mathematically.

DESCRIPTION

The present set-up consists of three control valves with pneumatic actuators as shown in the

Figure 4.2.

Figure 4.2: Experimental set-up with three control valves

One control valve (CV1) is with equal percentage characteristics (air to close type), second is

quick opening valve (CV2) with On/Off characteristics (air to open type) and third is linear

control valve (CV3) with linear characteristics (air to open type). Apparatus is self-contained


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water re-circulating unit. Compressed air supply is to be provided from user end. Water from

sump tank is sucked and delivered to a constant level overhead tank by means of a centrifugal

pump. Water flows back from overhead tank to sump tank. Each control valve is provided with a

ball valve at the inlet. Flow rate of water passing through the control valve is measured with the

help of rotameter (FI). A common water manometer is provided for the measurement of water

pressure head at the inlet of control valve. As outlet of control valve is open to the atmosphere,

the pressure at inlet of control valve is considered as pressure drop across the valve. Change in

the stem travel/ position of control valve results in the change in flow through the control valve.

This adjustment is done by regulating the air pressure inside the diaphragm of control valve (3 to

15 psig). For this purpose a pressure regulator (PR) with pressure gauge (PI) is provided in

pneumatic line. Pressure gauge facilitates to get direct reading of pressure inside the diaphragm

of control valve. Individual ball valves are provided to actuate the pneumatic line for each

control valve. For detecting the stem travel/position, scale is provided on each control valve.

UTILITIES REQUIRED

1. Electricity supply: Single phase, 220V AC, 50Hz, 5-15 A. Socket with earth connection.

2. Compressed air supply: 1 CMH at 2 bar

3. Water supply

4. Drain

PRECAUTION & MAINTAINANCE INSTRUCTION

1. Never run the apparatus if power supply is less than 180 volts & greater than 230 volts.

2. If the apparatus will not be in use for more than one month, drain the apparatus

completely and fill pump with cutting oil.

3. To prevent clogging of moving parts, run pump at least once in a fortnight.

4. Always use clean water.

5. Always keep apparatus free from dust.

TROUBLESHOOTING

1. If pump gets choked, open the back cover of pump and rotate the shaft manually.


21

REFERENCE


McGraw Hill, NY, 2009, pg. 441.


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23

EXPERIMENT NO. 4(A)

STUDY OF CONTROL VALVE FLOW CO-EFFICIENT (CV)

OBJECTIVE

1. To determine flow coefficient CV of the control valve.

THEORY

A control valve is used to control the flow rate in a fluid delivery system to control the process.

There is a close relation between pressure and the flow rate in fluid stream passing through the

pipe so that if the pressure is changed, the flow rate will also be changed. A control valve

changes the flow rate by changing the pressure in the flow system because it introduces the

constriction in the delivery system. So we can say that the flow rate through the constriction can

be given as:

P

KQ∆

= …….(1)

The most important factor associated with control valve is correction factor K of the above

equation. This correction factor allows selection of proper size of valve to accommodate the flow

that the system must support. This correction factor is called the valve coefficient and is used in

valve sizing.

CONTROL VALVE FLOW COEFFICIENT

The valve coefficient is measured as the number of USGPM (US Gallon per minute) that flows

through a fully open valve with a pressure drop of 1 psig. (1 US gallon = 3.785 Liters)

In SI units the formula for calculating CV is:

GP

KQ vV

∆= …….(2)

where, Q is flow rate, m3/hr and ΔPv is pressure drop across valve, kgf/cm2

Relation between KV & CV is:

VV CK 856.0= …….(3)


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P

GQCV ∆= 6.11 …….(4)

where, Q is discharge rate of fluid in m3/h, ΔP= Pressure drop across the wide-open valve, kPa,

G = Specific gravity relative to water. G=1 for water

Note: To convert ΔP in cm of H2O to kPa divide by 10.33

PROCEDURE

1. Start the setup for control valve of equal percentage characteristics.

2. Open pneumatic line for the control valve

3. Open the control valve fully. As the control valve is “air to close” so pressure in

diaphragm should be 0 psig.

4. Adjust the rotameter for 500 LPH flow by regulating the valve provided at the inlet line

of the control valve and wait for 5 minutes to steady the flow.

5. Record the rotameter and manometer readings.

OBSERVATION & CALCULATION

OBSERVATION TABLE:

S.NO. Valve Type ΔP, cm H2O Q, LPH Cv (on valve) Cv (calculated)

1. Equal Percentage Valve 2

2. Linear Valve 2

3. Quick Opening Valve 5

CALCULATION

Make the calculation for Cv as per equation (4) and tabulate in observation table above.

When conducting the experiment for the linear control valve keeping in mind that it is “air to

open” which means pressure in diaphragm should be more than 15 psig. Adjust this pressure in

diaphragm by opening the pressure regulator. Do not increase this pressure to more than 20 psig.

NOMENCLATURE

ΔP = Pressure drop in kPa.


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ΔPv= Pressure drop across valve, kgf/cm2

CV = Flow coefficient of control valve.

G = Specific gravity of fluid (for water G = 1).

Q = Flow rate, LPH.

REFERENCES


McGraw Hill, NY, 2009, pg. 423-440.


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27

EXPERIMENT NO. 4(B)

STUDY OF INHERENT CHARACTERISTIC OF CONTROL VALVES

OBJECTIVE

1. To study the inherent characteristics of control valves THEORY

The amount of flow passing through a valve at any time depends upon the opening between the

plug and the seat. Hence there is a relationship between stem position, plug position, and the rate

of flow, which is described in terms of the flow characteristics of a valve. Inherent and installed

are two types of flow characteristics of a control valve.

INHERENT CHARACTERISTIC

The inherent characteristic of control valve is the relation between the flow and the valve travel

at constant pressure drop across the valve. Following are the inherent characteristics for different

types of valves.

0 10 1009080706050403020

010

100

9080

7060

5040

3020

Valve lift % of full lift

Flow

% o

f max

imum

Quick opening

Lnear

Equal

Figure 4(B).1: Comparison of inherent characteristics of three types of control valves

PROCEDURE

1. Start the setup for linear control valve.


28

2. Open pneumatic line for the control valve and start increasing the pressure to open the

valve (as the control valve is “air to open”). Stop as soon as the valve is fully open (flow

rate and stem position will not change beyond this point).

3. Adjust the rotameter for 490 LPH flow by regulating the flow inlet valve provided at the

inlet line of the control valve and wait for 5 minutes to steady the flow. Make sure the air

compressor is running to maintain pressure in the diaphragm.

4. Record the manometer reading in cm of water and the rotameter reading.

5. Now slowly decrease the air pressure by regulating pressure in small steps as given in

observation table, so that the stem travels towards closing position.

6. The pressure drop across the valve will increase. Throttle the flow inlet valve at the inlet

of the control valve to maintain pressure drop constant.

7. Again note down the reading of rotameter and stem position.

8. Repeat the procedure till the valve is fully closed (pressure down to 0 psig).

9. Plot the graph of % of maximum flow vs. % of full lift to show inherent characteristic of

the control valve

10. Perform the same procedure for other two valves too.


OBSERVATION TABLE:

S.No. 1 2 3 4 5 6 7 8

Pressure Regulator

Reading, in psig

Stem lift in mm

Q, LPH

The constant pressure drop across the control valve ΔP, in cm H2O = ___________

CALCULATION

Perform the experiment for the other two valves. As the equal percentage control valve is “air to

close” pressure in diaphragm should be 0 psig for fully open condition. Repeat the experimental

procedure same as above, but pressure in the diaphragm will be increased gradually.


29

REFERENCES




30


31

EXPERIMENT NO. 4(C)

STUDY OF INSTALLED CHARACTERISTIC OF CONTROL VALVE

OBJECTIVE

1. To study installed characteristics of control valve THEORY

The amount of flow passing through a valve at any time depends upon the opening between the

plug and the seat. Hence there is a relationship between stem position, plug position, and the rate

of flow, which is described in terms of flow characteristics of a valve. Inherent and installed are

two types of flow characteristics of a control valve.

INSTALLED CHARACTERISTIC

The installed characteristic of control valves described is subjected to distortion due to variations

in pressure drop with flow. Line resistance distorts linear characteristics towards that quick

opening valve and equal percentage to that of linear control valve.

PROCEDURE

1. Start the setup for equal % control valve.

2. Open pneumatic line for the control valve and start increase the air flow to close the valve

(as the control valve is “air to close”). At fully open condition the pressure in diaphragm

should be 0 psig.

3. Adjust the rotameter for 500 LPH flow by regulating the flow inlet valve provided at the

inlet line of the control valve and wait for 5 minutes to steady the flow.

4. Record the manometer reading in cm of water and the rotameter reading.

5. Now slowly increase the air pressure by regulator in small steps as given in observation

table, so that the stem travels towards closing position.

6. Wait for 5 minutes at each step to steady the flow and note down the reading of

rotameter, manometer and stem travel.

7. Repeat the procedure till the valve is fully closed (pressure up to 15 psig).

8. Plot the graph of % of maximum flow vs. % of full lift to show installed characteristic of

the control valve.

9. Perform the same procedure for other two valves too.


32


OBSERVATION TABLE:

S.No. 1 2 3 4 5 6 7 8

Pressure Regulator

Reading, in psig

Stem lift in mm

Q, LPH

ΔP, in cm H2O

DISCUSSION:

Installed characteristics of the linear valve slightly approaches to the characteristics of quick

opening valve and that of equal percentage valve approaches to the linear characteristic because

of the pipe friction and other resistance to the flow.

REFERENCES




33

EXPERIMENT NO. 4(D)

STUDY OF HYSTERESIS OF CONTROL VALVE

OBJECTIVE

1. To study the hysteresis of the control valve.

THEORY

Hysteresis is the difference in reading while opening and closing the valve. In case of control

valves for same actuator signal different stem level (hence valve coefficients) are obtained

depending upon the direction of change. The maximum error in stem travel (or valve coefficient)

expressed in percent for same actuator pressure while opening and closing the valve is indicated

as hysteresis.

EXPERIMENTAL PROCEDURE

1. Start the setup for equal percentage control valve.

2. Open pneumatic line for the control valve

3. Open the control valve fully. As the control valve is “air to close” so pressure in

diaphragm should be 0 psig.

4. Adjust the rotameter for 500 LPH flow by regulating the valve provided at the inlet line

of the control valve and wait for 5 minutes to steady the flow.

5. Record the manometer reading in mm of water.

6. Record the rotameter reading.

7. Now slowly increase the air pressure using the regulator up to 3 psig.

8. Wait for 5 minutes to steady the flow and note down the reading of rotameter, manometer

and pressure in psig.

9. Repeat the procedure and take the reading at each at +3 psig till the valve is fully closed

(pressure up to 15 psig).

10. Now increase the pressure up to 20 psig and start decreasing the pressure gradually down

to 0 psig.

11. Wait for 5 minutes to steady the flow.

12. Record the manometer reading in mm of water.

13. Record the rotameter reading.


34

14. Repeat the procedure and take the reading at each at -3 psig till the valve is fully opened

(Pressure down to 0 psig).

15. Calculate the valve flow coefficient for actuator pressure for every reading.

16. Plot the graph of actuator pressure vs. flow coefficient. The ratio of maximum difference

between flow coefficients at same actuator pressure to that of maximum flow coefficient

is termed as hysteresis.

17. Repeat the experiment for the other valves.

OBSERVATION AND CALCULATION

OBSERVATION TABLE:

Pressure (psig) Increase pressure Decrease pressure

ΔP, mm H2O Q, LPH ΔP, mm H2O Q, LPH

3

6

9

12

15

CALCULATION

P

GQCV ∆= 6.11 ….(1)

100C Maximum

pressure increasingat C -pressure decreasingat %

V

V ×= VCHysteresis ….(2)

Repeat the experiment for the linear control valve. As the control valve is “air to open” so

pressure in diaphragm should be 15 psig.

NOMENCLATURE

ΔP = Pressure drop in kPa.


35


G = Specific gravity of fluid.

Q = Flow rate, LPH.

REFERENCES




36


37

EXPERIMENT NO. 4(E)

STUDY OF RANGEABILITY

OBJECTIVE

1. To study the rangeability of equal percentage valve.

THEORY

Equal percentage valve has characteristics such that flow changes by a constant percentage of its

instantaneous value for a given percentage change in stem position. Generally this type of valve

does not shut off the flow completely in its limit of stem travel. The rangeability (R) is defined as

the ratio of maximum to minimum controllable flow.

min

max

FF

R = …….(1)

where, Fmax is the flow when the valve stem is at nearly extreme open position foe maximum

controllable flow and Fmin is the flow when valve stem is at nearly extreme closed position for

minimum controllable flow. Fmax and Fmin represent flow rates measured at constant pressure

drop across control valve. Hence, rangeability R also can be defined as ratio of Cv,max to Cv,min.

For equal percentage valve flow has exponential characteristics of rangeability:

𝐹𝐹 = 𝑅𝑅𝑚𝑚−1 …….(2)

R is the rangeability of the valve and m is its fractional stem position.

PROCEDURE

1. Start the setup for equal percentage control valve.

2. Adjust the rotameter valve and set 500 LPH flow.

3. Set actuator air pressure to 3 psig.

4. Note down the flow rate and pressure at inlet of control valve.

5. Set actuator air pressure to 15 psig.

6. Note down the flow rate and pressure at inlet of control valve.


38


OBSERVATION TABLE:

Pressure (psig) ΔP, mm H2O Q, LPH Cv Remarks

Nearly 3

Nearly 15

CALCULATION

........min

max ==v

v

CC

R …….(3)

........min

max ==FF

R …….(4)

Repeat the experiment by keeping constant pressure drop across the control valve and note the

flow rates.

NOMENCLATURE

ΔP = Pressure drop in bar.


G = Specific gravity of fluid.

Q = Flow rate, LPH.

REFERENCES




39

EXPERIMENT NO - 5

DYNAMICS OF PRESSURIZED TANK

OBJECTIVE

1. To study the dynamics of a pressurized tank.

THEORY

The following figure shows the schematic of the experimental set-up. The objective is to study the

changing pressure dynamics of the tank under varied flow conditions and determine its time constant.

Air compressor

V1

V6

V5V4

V3V2

PG1

PG2

Relief Valve

Tank

Figure 5.1: Schematic diagram for the study of dynamics of pressurized tank system

The compressed air to the tank is provided through the valves V1, V2 and V3 and pressure in the inlet line

can be measured by the pressure gauge (PG1). The main instrument is the tank which is provided with a

pressure gauge (PG2), a relief valve (V4), outlet valves (V5 and V6). For pressurizing the tank, first the

pressure is built in the inlet line. The valve (V3) is closed and using valves (V1 and V2), bring the pressure

in the inlet line to say 20 psig. Then keeping both outlet valves (V5 and V6) closed, slightly open valve V3

to let air into the tank, thus pressurizing it. Caution: while letting air into the tank, the inlet line pressure

might fall. In order to maintain the constant pressure inlet condition, you may adjust valves V1 and V2.

Readings are taken until the pressure in the tank reaches the line pressure of say 20 psig.

The process can be continued with valve V5 slightly opened and until a new steady state is reached.

Experiment with different openings of V5 to get different steady states and determine the time constant for

the tank.


40

The transfer function for the system can be derived from the mass balance of air for the general flow

system where we will assume certain flow through valve V5. Consider the following simplified system for

purpose of mass balance of air

Figure 5.2: Simplified diagram for mass balance

Let the volume of the tank be V and gauge pressure inside the tank be P2. The inlet line gauge pressure be

P1. The inlet valve V3 has a valve coefficient of Kv3 and outlet valve V5 has a valve coefficient of Kv5.

Then the flow rate of air through these valves are related to pressure drops across them as follows

𝑓𝑓𝑖𝑖𝑙𝑙 =𝐾𝐾𝑣𝑣3

√𝐺𝐺�𝑃𝑃1 − 𝑃𝑃2

𝑓𝑓𝑜𝑜𝑜𝑜𝑜𝑜 =𝐾𝐾𝑣𝑣5

√𝐺𝐺�𝑃𝑃2

Assuming density of air is constant at 𝜌𝜌 = (𝑃𝑃2 + 𝑃𝑃𝑎𝑎)𝑉𝑉 𝑅𝑅𝑅𝑅⁄ . The mass balance then gives,

𝐾𝐾𝑣𝑣3

√𝐺𝐺�𝑃𝑃1 − 𝑃𝑃2 −

𝐾𝐾𝑣𝑣5

√𝐺𝐺�𝑃𝑃2 =

1(𝑃𝑃2 + 𝑃𝑃𝑎𝑎)

𝑑𝑑𝑃𝑃2

𝑑𝑑𝑜𝑜

Multiply by (P2 +Pa) and linearize this equation to get a first order system.

𝐾𝐾𝑣𝑣3

√𝐺𝐺��𝑃𝑃1 − 𝑃𝑃2,𝑠𝑠��𝑃𝑃2,𝑠𝑠 + 𝑃𝑃𝑎𝑎� + ��𝑃𝑃1 − 𝑃𝑃2,𝑠𝑠� −

�𝑃𝑃2,𝑠𝑠 + 𝑃𝑃𝑎𝑎�2��𝑃𝑃1 − 𝑃𝑃2,𝑠𝑠�

� �𝑃𝑃2 − 𝑃𝑃2,𝑠𝑠��

−𝐾𝐾𝑣𝑣5

√𝐺𝐺��𝑃𝑃2,𝑠𝑠�𝑃𝑃2,𝑠𝑠 + 𝑃𝑃𝑎𝑎� + �

�𝑃𝑃2,𝑠𝑠 + 𝑃𝑃𝑎𝑎�2�𝑃𝑃2,𝑠𝑠

+ �𝑃𝑃2,𝑠𝑠� �𝑃𝑃2 − 𝑃𝑃2,𝑠𝑠�� =𝑑𝑑�𝑃𝑃2 − 𝑃𝑃2,𝑠𝑠�

𝑑𝑑𝑜𝑜

As the experimenter will hold P1 constant and steady state values are constant we can simplify the

equation as follows with deviation variables for pressure in the tank as 𝑃𝑃2′

𝐾𝐾𝑣𝑣3

√𝐺𝐺[𝐴𝐴1 + 𝐵𝐵1𝑃𝑃2

′ ] −𝐾𝐾𝑣𝑣5

√𝐺𝐺[𝐴𝐴2 + 𝐵𝐵2𝑃𝑃2

′ ] =𝑑𝑑𝑃𝑃2

′

𝑑𝑑𝑜𝑜

P2, V

P1

P2

V3 V5

fout fin

Pa


41

Taking Laplace of the equation and simplify we get

𝑃𝑃2(𝑠𝑠) =𝐴𝐴

𝜏𝜏𝑠𝑠 + 1

where,

𝐴𝐴 =𝐾𝐾𝑣𝑣3𝐴𝐴1 − 𝐾𝐾𝑣𝑣5𝐴𝐴2

𝐾𝐾𝑣𝑣5𝐵𝐵2 − 𝐾𝐾𝑣𝑣3𝐵𝐵1 and 𝜏𝜏 =

√𝐺𝐺𝐾𝐾𝑣𝑣5𝐵𝐵2 − 𝐾𝐾𝑣𝑣3𝐵𝐵1

Taking inverse Laplace of the first order system, we get

𝑃𝑃2(𝑜𝑜) = 𝑃𝑃2,𝑠𝑠 +𝐴𝐴𝜏𝜏𝑒𝑒−𝑜𝑜 𝜏𝜏�

Using various experiments as detailed above we can experimentally find the time constant 𝜏𝜏 and compare

with the simplified first order approximation derived.

APPARATUS

1. Air Compressor

2. Tank to hold pressurized air

3. System of valves to regulate air

PROCEDURE

1. Initially the tank and entire line is supposed to be at atmospheric conditions.

2. Close inlet valve to tank (V3), outlet valves from the tank (V6, V5). Adjust the valves V1, V2 and

V3 in such a way that a pressure of 20 psig is maintained in pressure gauge (PG1).

3. With valve V5 closed, note pressure reading (PG2) vs. time till pressure in vessel is equal to line

pressure.

4. Provide a step change in outflow of air by crack opening V5 slightly. Note the value of vessel

pressure as a function of time as it falls and reaches a new steady state.

5. Again close valve V5 and note vessel pressure reading till it reaches steady state.

6. Repeat the step change in outflow (different magnitudes) two more times for quarter open and

half open valve V5.

OBSERVATIONS AND CALCULATIONS

Given data from the valve manufacturer:

Valve coefficient for valve (V3), 𝐾𝐾𝑣𝑣3 = __________________

Valve coefficient for valve (V5), 𝐾𝐾𝑣𝑣5 = __________________


42

Condition – 1: the valve V5 is completely closed

Take readings of tank pressure P2(t) with time (t) and obtain the steady state value (P2,s).

Time, t in s

Pressure, P2, in Pa

Condition – 2: the valve V5 is cracked open by a quarter turn.

Take readings of tank pressure P2(t) with time (t) and obtain new steady state value (P2,sq).

Time, t in s

Pressure, P2, in Pa

The change in outlet valve flow rate caused a pressure magnitude change of (P2,s - P2,sq).

Condition – 3: the valve V5 is closed and again tank pressure is brought back up to P2,s. After attaining

steady state value, the valve is cracked open one more time by half turn.

Take readings of tank pressure P2(t) with time (t) and obtain new steady state value (P2,sh).

Time, t in s

Pressure, P2, in Pa

The change in outlet valve flow rate caused a pressure magnitude change of (P2,s - P2,sh).

Plot vessel pressure (P2) vs. time for all the cases of valve openings you experiment with.

Compute time constant of the system using different methods for rising and decreasing pressures

separately for all the cases.

DISCUSSIONS

REFERENCES


43

EXPERIMENT NO. 6 INTERACTING AND NON-INTERACTING SYSTEMS

INTRODUCTION

The setup is design to study dynamic response of single and multi capacity process when connected in interacting and non-interacting mode. It is combined to study

1. Single capacity process 2. Non-interacting process and 3. Interacting process.

The observed step response of the tank level in different mode can be compared with mathematically predicted response.

Tank 1

Tank 2Tank 3

Rotameter

Pump

R3R2

R1

Supply tank

Figure 6.1: Overall schematic of the interacting and non-interacting tank system

APPARATUS

1. Supply tank 2. Pump for water circulation. 3. Rotameter for flow measurement. 4. Valves for controlling fluid flow. 5. Transparent tanks with graduated scales, which can be connected in interacting and non-

interacting mode.

The components are assembled on frame to frame tabletop mounting.


44


45

EXPERIMENT NO. 6(A)

STEP RESPONSE OF SINGLE CAPACITY SYSTEM OBJECTIVE

1. To study the response of a single capacity system to step input

THEORY

Step function: Mathematically, the step function of magnitude A can be expressed as

function. stepunit a is u(t) where,)()( tAutX =

It can be graphically represented as in Figure 6(A).1:

Figure 6(A).1: Step input function

To study the transient response for step function, consider the system consisting of a tank of

uniform cross sectional area A1 and flow resistance R1 such as for a valve. Qo, volumetric flow

rate through the resistance, is related to head h1 by a linear relationship.

q

A1

qoh1R1

Figure 6(A).2: Schematic of single capacity system

t

A

X(t)

0

where,

≥<

=0 when tA0 when t0

)(tX

and sAsX =)(


46

1

1

Rhqo = ……………(1)

Writing a transient mass balance around the tank:

(Mass flow in) – (Mass flow out) = Rate of accumulation of mass in the tank

dtdhtqtq o

11A )()( =−

……………(2)

Combining equation (1) and (2) to eliminate qo(t) gives the following linear differential equation:

11

1

dtdhA

Rhq =− …………..…(3)

Initially the process is operating at steady state, which means that 01 =dtdh

01 =−R

hq s

s ……………(4)

where, the subscript‘s’ indicates the steady state value of the variable.

Subtracting equation (4) from equation (3), we get

dt

hhdA

Rhh

qq sss

)()()( 11

11

11 −+

−=− …..…………(5)

Defining deviation variable: 111 and HhhQqq ss =−=− equation (5) can be written as:

dt

dHAR

HQ 11

1 += ……………(6)

Taking a transform of equation (6) gives,

)()()( 111 ssHAR

sHsQ += ..…………(7)

Equation (7) can be rearranged into standard form of first order system as:

. where)1()(

)(111

11 RAsR

sQsH

=+

= ττ

…………..(8)

For a step change of magnitude A, we can write,


47

sAsQtAutQ == )(or )()( …………(9)

Now from equation (8) we can write

+

=)1(

)(1

11 s

RsAsH

τ ………….(10)

By taking Laplace transformation of equation (10) we get,

−=

− τt

eARtH 1)( 11 ………….(11)

According to above equation (11) we can find the nature of curve as shown in Figure 6(A).3

below.

t in s

H1(

t) in

m Single tank

Figure 6(A).3: Transient response of single tank system

APPARATUS

See the list of apparatus in the introduction to experiment 6. In addition we need:

1. Stop watch

PROCEDURE

1. Start the setup by inserting the flexible pipe provided at the rotameter outlet in to the

cover of the top tank 1. Keep the outlet valve R1 of the tank 1 fully open and R2 of the

tank 2 slightly closed.

2. Switch on the pump. Adjust rotameter flow rates in steps of 10 LPH from 40 to 70 LPH

and note steady state levels for tank 1 against each flow rate.


48

3. From the data obtained select a suitable band for experimentation, say 50-60 LPH in

which we will be getting more readings of tank level.

4. Adjust the flow rate at lower value of the band selected, say 50 LPH and allow the level

of the tank 1 to reach steady state and record the flow and level at steady state.

5. Apply the step change by increasing the rotameter flow by 10 LPH.

6. Immediately start recording the level of the tank 1 at the interval of 15 sec, until the level

reaches at steady state.

7. Carry out the calculations as mentioned in calculation part and compare the predicted and

observed values of the tank level.

8. Repeat the experiment by throttling outlet valve (R1) to change resistance.

OBSERVATIONS Diameter of tank (in mm) = ID 92 mm

Initial flow rate (LPH) = ….

Initial steady state tank level (in mm) = ….

Final flow rate (LPH) = ….

Final steady state tank level ( in mm) = ….

Fill up columns H1(t) observed and H1(t) predicted after calculations: S.No. Time (s) Level (in mm) H1(t) observed (in mm) H1(t) predicted (in mm)

1 0 …. …. ….

2 15 …. …. ….

3 30 …. …. ….

4 45 …. …. ….

CALCULATIONS

Let

−=

/smin rate flow Initialinput stepafter Flowchange. step of Magnitude

3A

=

outlet.)at resistancelinear non gconsiderin(When

s/min resistance veOutlet val

1

2

1

dQdHR


49

= 21

2111

1 s/min veoutlet val of resistance is R and min tank theof area theis A whereAsin constant time

Rτ

where, dH1 = change in level (Final steady state level – Initial steady state level)

dQ = change in flow (Final flow rate after step change – Initial flow rate).

observedtH )(1 = (Level at time t – level at time t = 0)

−=

−11)( 11τ

t

predicted eARtH

= Level predicted at time t in meter.

Plot the graph of H1(t) vs. time(t) for observed and predicted levels.

DISCUSSIONS

Observed response fairly tallies with theoretical calculated response. Deviations observed may

be due to following factors:

1. Non-linearity of valve resistance.

2. Step change is not instantaneous.

3. Visual errors in recording observations.

4. Accuracy of Rotameter.

REFERENCES




50


51

EXPERIMENT NO. 6(B)

STEP RESPONSE OF FIRST ORDER SYSTEMS ARRANGED IN NON-INTERACTING MODE

OBJECTIVE

1. To study the step response of two first order systems arranged in non-interacting mode

THEORY

In non-interacting systems we assume the tanks have uniform cross sectional area and the flow resistance

is linear. To find out the transfer function of the system that relates h2 to q, writing a mass balance around

the tank, we proceed as follows.

Figure 6(B).1: Schematic of two first orders systems in non-interacting mode

We can write mass balance at tank 1

dtdhAqq 1

11 =− …..(1)

A mass balance at tank 2 is given as

dt

dhAqq 2221 =− …..(2)

The flow head relationships for the two linear resistances in non-interacting system are given by the expressions.

A2

q2h2R2

Non interacting system

q(t)

A1

q1h1R1


52

1

11 R

hq = .…(3)

2

22 R

hq = …..(4)

From equation (1) & (3) we get

1

1)()(

1

1

+=

ssQsQ

τ …..(5)

where

=−=−=

111

111

QRAqq

qqQ

s

s

τ …..(6)

From equation (2) & (4) we get

1)(

)(

2

2

1

2

+=

sR

sQsH

τ …..(7)

where

=−=

222

222

RAhhH s

τ …..(8)

Overall transfer function can be calculated as follows

)1( )1()(

)(

12

22

++=

ssR

sQsH

ττ …..(9)

For a step change of magnitude A )()( tAutQ = so …..(10)

sAsQ =)( …..(11)

)1( )1(

)(12

22 ++

=sss

ARsHττ

…..(12)

H2 at time t is given by

−

−−=

−−

21

1221

2122

111)( ττ

ττττττ

tt

eeARtH …..(13)

According to above equation we can find the nature of curve as shown below.


53

t in s

H2(

t) in

m

Non interacting tanks

Figure 6(B).2: Transient response of non-interacting system

PROCEDURE


cover of the top tank 1. Ensure that the valve (R3) between tank 2 and tank 3 is fully

closed.

2. Switch on the pump and adjust the flow rate to 40 LPH. Allow the level of both the tanks

(tank 1 and tank 2) to reach a steady state and record the initial flow and steady state

levels of both tanks.

3. Apply the step change with increasing the rotameter flow by 10 LPH.

4. Record the level of tank 2 at an interval of 15 s until the level reaches steady state.

5. Record final flow and steady state level of tank 1.




OBSERVATIONS Diameter of tanks (in mm) = ID 92 mm


Initial steady state level of tank 1 (in mm) = ….



Final steady state level of tank 1 ( in mm) = ….


54


Fill up columns H2(t) observed and H2(t) predicted after calculations: SL No. Time (sec) Level of tank 2

(in mm) H2(t) observed ( in mm) H2(t) predicted (in mm)

1 0 …. …. ….

2 30 …. …. ….

3 60 …. …. ….

4 90 …. …. ….

CALCULATIONS

Let

−=


3A

=

.outlet)at resistancelinear non gconsiderin(When

1 tank of s/min resistance veOutlet val

1

2

1

dQdHR

=

)outletat resistancelinear non gconsiderin(When


2

2

2

dQdHR

==

1 tank of )s/m(in veoutlet val

of resistance is R and min 1 tank theof area theis )(d4

whereA

.1 tank of s)(in constant time

2

122

11111πτ AR

==



whereA

.2 tank of )s(in constant time

2

222

22222πτ AR

Where, dH1 = change in level of tank 1

= (Final steady state level – Initial steady state level)

dH2 = change in level of tank 2



observedtH )(2 = (Level at time t – level at time t = 0) x 10-3


55

−

−−=

−−

21

1221

212Pr2

111)( ττ

ττττττ

tt

edicted eeARtH

= Level in tank 2 predicted at time t in meter.

Plot the graph of

1. H2(t) Predicted vs. time(t) and

2. H2(t) observed vs. time(t).

DISCUSSIONS Observed response fairly tallies with theoretical calculated response. Deviations observed may






REFERENCES




56


57

EXPERIMENT NO. 6(C)

IMPULSE RESPONSE OF FIRST ORDER SYSTEMS ARRANGED IN NON-INTERACTING MODE

OBJECTIVE

1. To study the impulse response of two first order systems arranged in non-interacting mode

THEORY

Mathematically, the impulse function of magnitude A is defined as

)()( tAtX δ= …..(1)

where, )(tδ is the unit impulse function. Graphically, it can be described as:

Figure 6(C).1: Impulse input function

Overall transfer function of the system as described in the previous experiment

)1( )1()(

)(

12

22

++=

ssR

sQsH

ττ

For a impulse change of magnitude V (Volume added to the system)

)()( tVtQ δ= …..(2)

VsQ =)( …..(3)

Hence we find )1( )1(

)(12

22 ++

=ss

VRsHττ

…..(4)

t

A/b

X(t)

0 b

Where,

>

<<

<

=

bt when 0

bt0 when

0 when t0

)(bAtX

and AtAL

tAtXLimb

=

=→

)]([

)()(0

δ

δ


58

For impulse change H2 at time t is given by

−−

=

−−

2122

21

)(ττ

ττtt

eeVRtH …..(5)

Considering linear resistance at outlet valve of the tank 2, the value of R2 can be calculated as:

Q

HR s,2

2 = …..(6)

where H2,s is the steady state level of tank 2 and Q is the steady flow through the system from the pump.

Before and after the impulse the tank levels and flow rates will return to this steady state value and hence

these are used for calculating resistance of the valve.

Put the values in equation (3) to find out H(t) predicted and plot the graph of

1. H2(t) predicted vs. time (t) and

2. H2(t) observed vs. time (t).

PROCEDURE


cover of the top tank 1. Ensure that the valve (R3) between tank 2 and tank 3 is fully

closed.


(tank 1 and tank 2) to reach at steady state and record the initial flow and steady state


3. Apply impulse input by adding 0.5 liters of water in tank 1.

4. Record the level of tank 2 at an interval of 15 s, until the level reaches steady state (same

as before impulse is added).









59

Volume added (liters) = ….

Fill up columns H2(t) observed and H2(t) predicted after calculations: SL No. Time (sec) Level of tank 2


1 0 …. …. ….

2 30 …. …. ….

3 60 …. …. ….

4 90 …. …. ….

:

CALCULATIONS

Let

=



,1

2

1

QHR s

=


2 tank ofs/min resistance veOutlet val

,2

2

2

QHR s

==



whereA


2

122

11111πτ AR

==



whereA


2

222

22222πτ AR

Where, H1,s , H2,s = steady state level of tanks 1 and 2

Q = steady state flow through the system.

observedtH )(2 = (Level at time t – Level at time t = 0) x 10-3

−−

=

−−

212Pr2

21

)(ττ

ττtt

edictedeeVRtH


60

V= Volume of liquid added as an impulse input (in m3).

Plot the graph of



DISCUSSIONS






4. Accuracy of rotameter.

REFERENCES


McGraw Hill, NY, 2009, 123-130.


61

EXPERIMENT NO. 6(D)

STEP RESPONSE OF FIRST ORDER SYSTEMS ARRANGED IN INTERACTING MODE

OBJECTIVE

1. To study the step response of two first order systems arranged in interacting mode

THEORY

Assuming the tanks of uniform cross sectional area and valves with linear flow resistance the transfer

function of interacting system can be written as:

A2

q2h2R2

q3

R3

q(t)

A3

h3

Figure 6(D).1: Schematic of two first order systems in interacting mode

We can find the relation between H2(s) and Q(s) as follows:

1 )( )(

)(

23232

23

22

++++=

sRAsR

sQsH

ττττ …..(1)

Let,

23

23

23

11ττττRA

b ++= …..(2)

−

+

−=

23

2 122 ττ

α bb …..(3)

−

−

−=

23

2 122 ττ

β bb …..(4)

For a step change of magnitude A

−−

−=]/1/1[

] )/1[(] )/1[(1)( 22 βαβα βα tt eeARtH …..(5)


62

In terms of transient response the interacting system is more sluggish than the non-interacting system.

According to above equation we can find the nature of curve as shown below.

t in s

H2(

t) in

m

Non interacting tanks

Interacting tanks

Figure 6(D).2: Transient response of interacting system

PROCEDURE 1. Start the setup by inserting the flexible pipe provided at the rotameter outlet in to the

cover of the top tank 3. Keep the outlet valve (R2) of tank 2 slightly closed. Ensure that

the valve (R3) between tank 2 and tank 3 is also slightly closed.


(tank 2 and tank 3) to reach at steady state and record the initial flow and steady state


3. Apply the step change with increasing the rotameter flow by 10 LPH.

4. Record the level of tank 2 at the interval of 15 s, until the level reaches steady state.

5. Record final steady state flow and level of tank 3.








63





Fill up columns H2(t) observed and H2(t) predicted after calculations: S.No. Time (s) Level of tank 2


1 0 …. …. ….

2 30 …. …. ….

3 60 …. …. ….

4 90 …. …. ….

CALCULATIONS

Let

−=


3A

=



2

2

2

dQdHR

=



3

2

3

dQdHR

==



whereA


2

222

22222πτ AR

==



whereA


2

322

33333πτ AR

Where, dH2 = change in level of tank 2


dH3 = change in level of tank 3


64



observedtH )(2 = (Level at time t – Level at time t = 0) x 10-3

−−

−=]/1/1[

] )/1[(] )/1[(1)( 2Pr2 βαβα βα tt

edictedeeARtH

= Level in tank 2 predicted at time t in meters.

Calculate the value of b, βα and from equations given in the theory part.

Plot the graph of



DISCUSSIONS







REFERENCES




65

EXPERIMENT NO. 6(E)

IMPULSE RESPONSE OF FIRST ORDER SYSTEMS ARRANGED IN INTERACTING MODE

OBJECTIVE

1. To study the impulse response of two first order systems arranged in interacting mode

THEORY

Mathematically, the impulse function of magnitude A is defined as

)()( tAtX δ= …..(1)

where, )(tδ is the unit impulse function. Graphically, it can be described as:

Figure 6(E).1: Impulse input function

Overall transfer function of the system is

1 )( )(

)(

23232

23

22

++++=

sRAsR

sQsH

ττττ …..(2)

For a impulse change of magnitude V (Volume added to the system)

)()( tVtQ δ= …..(3)

VsQ =)( …..(4)

1 )(

)(2323

223

22 ++++

=sRAs

VRsHττττ

…..(5)

For impulse change H2 at time t is given by

][)(

)(23

22

tt eeVRtH βα

βαττ−

−= …..(6)

t

A/b

X(t)

0 b

Where,

>

<<

<

=

bt when 0

bt0 when

0 when t0

)(bAtX

and AtA

tAtXLimb

=

=→

)](L[

)()(0

δ

δ


66

where,

23

23

23

11ττττRA

b ++= …..(7)

−

+

−=

23

2 122 ττ

α bb …..(8)

−

−

−=

23

2 122 ττ

β bb …..(9)

PROCEDURE 1. Start the setup by inserting the flexible pipe provided at the rotameter outlet in to the

cover of the top tank 3. Keep the outlet valve (R2) of tank 2 slightly closed. Ensure that

the valve (R3) between tank 2 and tank 3 is also slightly closed.


(tank 2 and tank 3) to reach steady state and record the initial flow and steady state levels

of both tanks.

3. Apply impulse input by adding 0.5 liters of water in tank 3.

4. Record the level of tank 2 at an interval of 15 s, until the level reaches steady state (same

as before impulse is added).








Volume added (lit.) = ….

Fill up columns H2(t) observed and H2(t) predicted after calculations: S. No. Time (sec) Level of tank 2



67

1 0 …. …. ….

2 30 …. …. ….

3 60 …. …. ….

4 90 …. …. ….

:

CALCULATIONS

Let

=



,2

2

2

QHR s

=



,3

2

3

QHR s

==



whereA


2

222

22222πτ AR

==



whereA


2

322

33333πτ AR

where, H3,s , H2,s = steady state levels of tank 3 and 2

Q = steady flow through the system.

observedtH )(2 = (Level at time t – level at time t = 0) x 10-3

][

)()(

23

2Pr2

ttedicted eeVRtH βα

βαττ−

−=

V= Volume of liquid added as an impulse input (in m3).

Plot the graph of




68

DISCUSSIONS







REFERENCES




69

EXPERIMENT NO. 7

DEAD WEIGHT TESTER FOR PRESSURE GAUGE CALIBRATION

OBJECTIVE

1. Calibration of pressure gauge by Dead weight tester

THEORY

C

A

KI

G

H

F

D

E

A Weights table D Reservoir Cap G Screw press J

B E Oil Reservoir H Levelling Screws K Gauge Post

C F Valve I Table

APPARATUS

1. Weights

2. Pressure gauge for calibration


70

3. Testing apparatus

PROCEDURE

1. Set level by placing a spirit level on the weight table (A) and adjusting the leveling

screws (H).

2. Wind capstan handle of screw press (G) fully in, i.e., clockwise.

3. Remove reservoir cap (D) and pour oil (only castor oil technical grade free from acids

and water traces, not Castrol until nearly full and open valve (F).

4. Screw capstan (G) in and out until bubbles stop appearing in the reservoir.

5. Screw capstan (G) fully out.

6. Close the valve (F).

7. Select correct adopter and fit gauge/instrument (under test) at gauge post (K) using a

proper copper washer or bonded seal. Gauge should never be mounted until the priming

operation (step – 4) is done.

8. Ensure that the gauge /instruments to be tested are internally clean, if water, dirt or

chemicals are sucked into the system from contaminated instruments, the ram system will

be seriously damaged.

9. Load the weight table (A) with the weights. The equivalent of the weight table must also

be taken into account.

10. Apply the pressure with the screw press (G) until the weights begin to rise. Spin weight

table by hand and adjust the capstan until weight table skirt floats within the indicator red

band. The red band is easily visible on the piston cylinder from a seated position.

11. It is essential that the weights should be spinning freely whilst taking readings.

12. Never take off weights without removing all pressure in the system by unscrewing the

press (G).

13. After the test is finished stop the weights spinning, wind back the screw press (G) fully

and open the valve (F). The tester is now ready for another test.

14. In high pressure testers, a slight leakage through the cylinder vent hole on the side of the

ram cylinder is normal.


71

OBSERVATION AND CALCULATIONS

Record the weights and pressure readings

S.No. Weights

(decreasing)

Pressure Gauge

Reading

Weights

(increasing)

Pressure Gauge

Reading

1.

2.

:

The weight of the weights table = ___________ kgf/cm2.

DISCUSSIONS

REFERENCES


72

Documents

Process Dynamics pracs iitr