PROCESS INSTRUMENTATION I
MODULE CODE: EIPIN1B
STUDY PROGRAM: UNIT 2
VUT
Vaal University of Technology
2/10
EIPINI Chapter 4: Level Measurement Page 4-1
4. LEVEL MEASUREMENT
This chapter aims to introduce students to some of the level measurement technologies
used to measure the level of liquids and granulars in a container.
4.1 INTRODUCTION
On the 28th of March, 1979, thousands of people fled from Three Mile Island when
the cooling system of a nuclear reactor failed. This dangerous situation developed
because the level controls turned off the coolant flow to the reactor when they detected
the presence of cooling water near the top of the tank. Unfortunately, the water
reached the top of the reactor vessel not because there was too much water in the tank,
but because there was so little that it boiled and swelled to the top. From this example,
we can see that level measurement is more complex than simply the determination of
the presence or absence of a fluid at a particular elevation.
Most level measurement techniques can be categorized into one of two groups –
direct and indirect (inferred) methods. Direct methods involve measuring the height of
fluid directly as for example with a dipstick, overflow pipe, float or sight glass. With
indirect methods, another variable is measured that correlate to the liquid level.
Measuring the weight of a substance in a container or the pressure exerted on the
bottom of a tank or transmitting an ultrasonic beam to the level surface and measuring
the time of flight of the transmitted and received signal, is indirectly related to the
level height.
Level measurement may also be continuous or “point-level”. A continuous method
refers to a technique whereby the device measures level on a constant basis, displaying
or transmitting the actual level of the liquid as it changes. Point-level devices measure
liquid at specific points within the tank. As the liquid level rises and falls, it passes
through definite points during it's transit. Continuous methods may however be
programmed to output alarms at specific points also, in addition to transmitting a
continuous level measurement.
Level measuring devices are also described as either a contact or non-contact type of
instrument. A contact type device, such as a float, makes physical contact with the
liquid in the container, in order to determine the level. A non-contact device, such as
ultrasonic or radar, does not require contact with the material in the container to
measure the level.
4.2 THE SIGHT GLASS
The sight glass consists of a strengthened glass tube, attached to the container as
shown in Figure 4-1, through which the fluid level in the container can easily be
observed by the operator. Monitoring the level from a distance, is facilitated when a
magnetic float inside the tube, is allowed to rise and fall with the liquid level, causing
metallic flags (or louvers) to flip and expose a different colour, indicating the level.
These devices are available with continuous monitoring equipment, allowing both a
local visible indication as well as an external signal for remote monitoring.
EIPINI Chapter 4: Level Measurement Page 4-2
Valve
Sight glass
Magnetic
float
Metallic
flaps
Valve
Container with sight glass
Sight glass with magnetic indicator
Figure 4-1
4.3 FLOAT TYPE LEVEL INDICATORS
4.3.1
Chain Float
This type of float is linked to a
rotating drum, by means of a
chain, as shown in Figure 4-2.
The chain engages a sprocket,
which turns the drum, and with
it the level indicator. A tape,
that wraps around the drum, is
also used, instead of a chain. A
weight is attached to the other
end of the chain or tape, to
keep the chain pulled straight
while the float moves up or
down with the changing level.
4.3.2
Ball Float
The float is attached to a rod
and rotary shaft, operating
through a packing and bearing
in the container wall, to the
level indicator (Figure 4-3).
Practical considerations, limit
the shaft rotation to ± 30º from
the horizontal, and therefore the
range of the instrument as well.
Level indicator
Drum
Seal
Chain
Weight
Float
Figure 4-2
Level
indicator
Rotary shaft
Packing and
bearing
Float
Figure 4-3
EIPINI Chapter 4: Level Measurement Page 4-3
4.3.3
Magnetic Float (Magnetic Coupled Float and Follower) Level Meter
For this type of indicator, a dip
pipe, made of non-magnetic
material, is permanently installed in
the container and sealed off, as
shown in Figure 4-4. A doughnut
shaped float with high strength
magnet, is fitted around the dip
pipe, to rise and fall vertically with
the liquid level. A similar magnet
that magnetically bonds to the
outer magnet, is suspended within
the dip pipe and attached to a
rod, chain and pulley or similar
arrangement, to the level indicator.
This system is advantageous where
leakages cannot be tolerated – such
as with toxic, explosive or
flammable liquids, as no packing
or stuffing box is required and
only the dip tube is required to
withstand vessel pressure and
temperature conditions.
4.3.4
Level indicator
Indicator rod
Non-magnetic
dip tube
Doughnut float
with
outer magnet
Follower
with
inner magnet
Figure 4-4
Magnetic Float Switch
The
magnetic
float
switch
(Figure 4-5) is a point level device.
When the level reaches a certain
point, the float magnet activates
the magnetic reed switch. The
electric contacts are safely isolated
from the inside (wet side) of
the container by non-magnetic
material that allows magnetic
interaction between the float
magnet and magnetic switch. The
contacts may be used to switch a
pump on or off, to sound an alarm
or for other control purposes.
Float magnet
Non-magnetic
housing
Magnetic
reed switch
Swivel pin
Float
Figure 4-5
EIPINI Chapter 4: Level Measurement Page 4-4
4.3.5
Flexure Tube Displacer (Torque tube) Level Meter
The displacer type liquid level measuring instrument is not a float as such, for the
displacer is heavier than the process fluid and the displacer moves very little during
changes in tank level (a definite advantage over other float types). According to
Archimedes’s law (illustrated in the figure below), the apparent weight of the
displacer when immersed in a liquid, is its nominal weight in air minus the weight
of the displaced liquid. The weight of the displacer will thus vary linearly from its
weight in air (when the tank is empty) to its apparent weight when fully immersed
in the liquid (when the tank is full). The weight of the displacer acting on the
torque arm, will cause an angular displacement of the free end of the flexible torque
tube and this movement will be transmitted to the outside world, by the torque rod.
Torque tube
Archimedes’s Principle
Torque tube
flange
Level indicator
Torque rod
Torque arm
Chain
Displacer
Figure 4-6
Note: This instrument may also
be fitted in a separate
measuring chamber, fitted
to the main vessel.
Main vessel
Measuring
chamber with
displacer, torque
tube and measuring
equipment
Example 4-1
During calibration of a displacer type level meter, it is found that the torque registered
by the meter when the tank is empty, is 10 N-m. If the torque arm is 0.1 m, calculate
the torque that will be generated when the displacer, with volume 0.002 m3, is fully
immersed in the process fluid with density 1000 kg/m3.
Remembering that torque is given by T = Fr, the weight of the displacer in air is
10/0.1 = 100 N. Using Archimedes’ law, the apparent weight of the displacer when it
is completely immersed in the liquid, is 100 – 0.002×1000×9.81 = 80.38 N. The torque
generated then, will be 80.38×0.1 = 8.038 N-m.
EIPINI Chapter 4: Level Measurement Page 4-5
4.4 DIFFERENTIAL PRESSURE MEASUREMENT OF LEVEL
Perhaps the most frequently used technique of measuring level, is the method of
measuring differential pressure. The primary benefit of this method is that the
equipment measuring the differential pressure, can be externally installed or
retrofitted to an existing vessel. It can also be isolated safely from the process
using block valves for maintenance and testing.
There are, however, some disadvantages to this method. One vessel penetrations
near the bottom of the vessel is needed, where leak paths could be the cause of
many problems. Measurement errors also occur due to changes in liquid density.
Density variations are caused by temperature changes or change of product. These
variations must always be compensated for, to maintain accurate measurements.
4.4.1 Open Containers
To determine the fluid level H (Figure 4-7) in an open container, it is only necessary to
measure the difference between the pressure at the bottom of the container and
atmospheric pressure. In Figure 4-7 (a), the pressure difference is measured with a
differential pressure transmitter while in Figure 4-7 (b), a u tube manometer is used.
The pressure instruments may be installed with their zero lines level with the bottom
of the container or at a distance z, below the bottom of the container.
Patm
Patm
H
H
Patm
z
HP
Patm
LP
Zero line
z
HP
LP
DP
transmitter
U tube
manometer
Figure 4-7 (a)
Figure 4-7 (b)
h
EIPINI Chapter 4: Level Measurement Page 4-6
Example 4-2
A mercury (density 13600
kg/m3) u-tube manometer is
used to measure the level of a
liquid (density 1000 kg/m3) in
a 5 meter high container, with
the zero level of the
manometer, exactly in line
with the bottom of the tank.
Calculate the manometer
reading h, if the tank level is 3
meter.
Patm
Patm
5m
3m
h
X
Zero line
Y
U-tube
Equating pressures on the XY line:
P X = PY
∴Patm + 1000×(3 + ½h)×9.81 = Patm + 13600×h×9.81
∴3 + ½h = 13.6h
∴13.1h = 3
∴h = 0.229 m
= 229 mm
Example 4-3
A mercury (density 13600
kg/m3) well type manometer
with A2/A1 = 0.01 is used to
measure the level of a liquid 5 m
(density 1000 kg/m3) in a 5
3m
meter high container, with the
zero level of the manometer,
exactly in line with the
bottom of the tank. Calculate
the manometer reading h if
the tank level is 3 meter.
P2 = PATM
PATM
P1
X
d
h
Zero line
Y
Well type manometer
(Students must please take
note of the orientation of
the manometer. The well is
always connected to the
higher pressure.)
We can approach this problem in three ways:
a) Compare the pressures on the XY line, the
same way as with the u tube manometer.
b) Calculate P1 and P2 exactly on the well type
manometer liquid meniscuses and utilise
Equation 2-5 (Chapter 2).
c) Ignore the level change (d) in the well (because it is so small), assume P1 acts on
the zero line and P2 on the manometer liquid in the tube, and use Equation 2-5 to
arrive at an approximate answer for h.
Methods a) and b) will produce the exact value of h while method c) will give us an
approximate value of h, although very nearly correct.
EIPINI Chapter 4: Level Measurement Page 4-7
We will now use method a), b) and c) to solve this problem, but we will prefer the
approximate but easier method c) for the remaining well type manometer problems
and also for closed tanks involving well type manometers.
a) Comparing pressures on the XY line:
PATM + 1000×3×9.81 + 1000×d×9.81 = PATM + 13600×h×9.81 + 13600×d×9.81
∴1000×3×9.81 + 1000×d×9.81 = 13600×h×9.81 + 13600×d×9.81
∴29430 + 9810d = 133416h + 133416d
∴29430 = 133416h + 123606d………………………………………………..(a)
But from Equation (2), Chapter 2, Page 2-7: d = (A2/A1)h
∴d = 0.01h……………………………………………………………………..(b)
(b) in (a):
29430 = 133416h + 123606×(0.01h)
∴29430 = 133416h + 1236.06h
∴134652.06h = 29430
∴h = 0.21856 m
b) P1 = PATM + 1000×3×9.81 + 1000×d×9.81
= PATM + 29430 + 9810×d = PATM + 29430 + 9810×(0.01h)
∴P1 = PATM + 29430 + 98.1h …………………………………….…………… (a)
And P2 = PATM ………………………………………………………………… (b)
From (a) and (b):
P1 – P2 = 29430 + 98.1h ………………………………………………………. (c)
But from Equation 2-5 (Chapter 2):
P1 – P2 = ρhg(1 + A2/A1) …………………………………...…………………. (d)
(c) in (d):
29430 + 98.1h = 13600×h×9.81(1 + 0.01)
∴29430 + 98.1h = 133416×h×1.01
∴29430 + 98.1h = 134750.16h
∴134652.06h = 29430
∴h = 0.21856 m
c) P1 = PATM + 1000×3×9.81 (neglecting d)
∴P1 = PATM + 29430 ………………………………………….…………….… (a)
And P2 = PATM …………………………………………………….……...…… (b)
From (a) and (b):
P1 – P2 = 29430 ………………………………………………….……………. (c)
And from Equation 2-5 (Chapter 2):
P1 – P2 = ρhg(1 + A2/A1) …………………………………...…………………. (d)
(c) in (d):
29430 = 13600×h×9.81(1 + 0.01)
∴29430 = 134750.16h
∴h = 0.2184 m
EIPINI Chapter 4: Level Measurement Page 4-8
Example 4-4
A DP transmitter must be calibrated to measure the level of a liquid in an open tank.
The density of the liquid is 1000 kg/m3. The DP transmitter will be mounted one meter
below the bottom of the tank. The tank is full when the height of the liquid in the tank
is 5 meter and it is empty when there is only liquid in the high pressure line connected
to the DP transmitter. Determine the necessary calibration specifications for an output
signal of 4 to 20 mA.
Empty:
P1 = Patm + ρhg
= Patm + 1000×1×9.81
= Patm + 9810
∴P1 – P2 = 9810 Pa
Full:
P1 = Patm + ρhg
= Patm + 1000×6×9.81
= Patm + 58860
∴P1 – P2 = 58860 Pa
Empty
Full
Patm
Patm
5m
1m
4 mA
P1
Patm
P2
1m
DP cell
20 mA
P1
Patm
P2
DP cell
Calibration specification: Output = 4 mA when input = 9810 Pa (empty condition)
Output = 20 mA when input = 58860 Pa (full condition)
Example 4-5
Patm
If, in example 4-4, the DP cell is replaced by a Empty
u-tube manometer using mercury (density 13600
kg/m3), with its zero level 1 meter below the
bottom of the container, calculate the full and
empty readings, hempty and hfull.
1m
ZL
Empty:
X
Equating pressures on the XY line:
Patm+1000×(1+½h)×9.81 = Patm+13600×h×9.81
∴1 + ½h = 13.6h
Patm
Full
∴hempty = 0.07634 m = 76.34 mm.
Full:
Equating pressures on the XY line:
Patm+1000×(6+½h)×9.81 = Patm+13600×h×9.81
∴6 + ½h = 13.6h
∴hfull = 0.458 m. = 458 mm.
Patm
hempty
Y
U tube
Patm
5m
ZL
1m
hfull
X
Y
U tube
EIPINI Chapter 4: Level Measurement Page 4-9
Example 4-6
A DP transmitter, correctly calibrated to measure the level of
a liquid in a 5 meter high open container, delivers an output
pressure of 50 kPa. Calculate the liquid level in the tank.
100
50
0
H
5 Level
Patm
Patm
5m
H
1m
Comparing pressure on the XY line:
Patm+1000×(H+1+0.15)×9.81=Patm+13600×0.3×9.81
∴H + 1.15 = 13.6×0.3
∴H = 4.08 – 1.15
= 2.93 meter
Example 4-8
A mercury (density 13600 kg/m3) well type
manometer with A2/A1 = 0.01, is used to measure
the level of a liquid (density 1000 kg/m3) in a
5 meter high container, with the zero level of the
manometer, 1 meter below the bottom of the tank.
Calculate the level in the tank if the manometer
reading is 200 mm (0.2 m).
Po=16×Level+20
20
Liquid level H = [(50-20)/(100-20)]×5 = 1.875 m.
Or alternatively: 50 = 16×H + 20 ⇒ H = 1.875 m
Example 4-7
A mercury (density 13600 kg/m3) u tube manometer
is used to measure the level of a liquid (density
1000 kg/m3) in a 5 meter high container, with the
zero level of the manometer, 1 meter below the
bottom of the tank. Calculate the level in the tank if
the manometer reading is 300 mm (0.3 m).
Po
0.3 m
ZL
X
Y
U tube
Patm
H
5m
P2=Patm
1m
The input pressure, P1, on the well side of the
0.2 m
P1
manometer, with respect to the zero line, is:
ZL
P1 = Patm + 1000×(H + 1)×9.81
and
P2 = Patm
Well type
∴P1 – P2 = Patm + 1000×(H + 1)×9.81 – Patm
d
manometer
= 1000×(H+1)×9.81 = 9810(H+1)
Furthermore, from Equation 2-5 (Chapter 2):
P1 – P2 = ρhg(1 + A2/A1) = 13600×0.2×9.81×(1 + 0.01) = 2747×9.81 = 26948
Therefore 9810(H+1) = 26948 ⇒ H+1 = 2.747
∴H = 1.747 m.
Note: If we did wish to include d, then d = (A2/A1)h = 0.01×0.2 = 0.002 m
P1=Patm+1000×(H+1+d)×9.81=Patm+1000×(H+1+0.002)×9.81=9810(H+1.002)
And P1 – P2 = 9810(H+1.002) = 26948 ⇒ H + 1.002 = 2.747
∴H = 1.745 m
EIPINI Chapter 4: Level Measurement Page 4-10
4.4.2 Closed Containers
For a closed container (Figure 4-8), the outer leg of the differential pressure
measuring equipment, can not be open to atmosphere, as the tank may be
pressurised and the tank pressure Pt, could differ from atmospheric pressure. The
outer leg is therefore brought back into the top of the vessel, to ensure that the
pressure inside the tank and the outer leg pressure, are both equal to Pt. Liquid
vapour will typically condense in the outer leg and if an effort is made to keep this
leg devoid of process fluid, it will result in what is called a dry leg system. If, on
the other hand, the outer leg is allowed to be completely filled with the same
process fluid as in the tank, a wet leg system will result. We will focus on wet leg
closed tank systems.
Differential pressure level measurement in a closed vessel (wet leg), is
fundamentally different from level measurement in an open vessel. The high
pressure input will always be connected to the outer leg while the low pressure
input is obtained from the bottom of the tank. Another difference is that the
differential pressure is zero when the tank is full, while the maximum differential
pressure is obtained when the tank is empty. Students must ensure that they
understand this critical difference between differential pressure measurement in an
open vessel and a closed vessel, very clearly.
Figure 4-8(a) illustrates a DP transmitter measuring the pressure difference between
the outer leg and the tank. It may be connected in line with the bottom of the tank or
below. In Figure 4-8 (b), a u tube manometer is used to measure the pressure
difference. In the case where a manometers is used, it is clear from Figure 4-8 (b), that
the zero line must be lower than the bottom of the tank.
Pt
Pt
Pt
Pt
L
H
L
H
z
LP
HP
z
LP
DP transmitter
ZL
h
U tube
manometer
Figure 4-8 (a)
Figure 4-8 (b)
HP
EIPINI Chapter 4: Level Measurement Page 4-11
Example 4-9
A DP transmitter is used to measure the level of a liquid, with density 1000 kg/m3 in a
closed tank, that can store liquid to a maximum level of 5 meter. The DP transmitter is
installed 1 meter below the bottom of the tank. Calculate the input differential pressure
to the DP transmitter when the tank is empty and when it is full.
Pt
Pt
Pt
Pt
5m
5m
P2
1m
P1
DP transmitter
Empty:
P1 = Pt + 1000×(5 + 1)×9.81
P2 = Pt + 1000×1×9.81
P1 – P2 = 49050 Pa = 49.05 kPa.
P2
1m
P1
DP transmitter
Full:
P1 = Pt + 1000×(5 + 1)×9.81
P2 = Pt + 1000×(5 + 1)×9.81
P1 – P2 = 0 Pa.
Example 4-10
A u tube mercury (δ = 13.6) manometer is used to measure the level of a liquid (δ=1),
in a closed tank, that can store liquid to a maximum level of 5 meter. The zero level of
the manometer is 1 meter below the bottom of the tank. Calculate the manometer
readings when the tank is empty and full.
Pt
Pt
P
t
Pt
5m
5m
1m
1m
hempty
u-tube
X
ZL
u-tube
Y
hfull
X
ZL
Y
Empty:
Full:
Equating pressures on the XY line:
Pt+1000×(1 - ½h)×9.81 + 13600×h×9.81
= Pt + 1000×(5 + 1 + ½h)×9.81
∴(1 - ½h) + 13.6h = 6 + ½h
∴13.6h – ½h - ½h = 6 – 1 ⇒ 12.6h = 5
∴hempty = 0.3968 m = 396.8 mm.
Equating pressures on the XY line:
Pt+1000×(5+1-½h)×9.81+13600×h×9.81
= Pt + 1000×(5 + 1 + ½h)×9.81
∴(6 - ½h) + 13.6h = 6 + ½h
∴13.6h – ½h - ½h = 0 ⇒ 12.6h = 0
∴hfull = 0 (as we knew all along)
EIPINI Chapter 4: Level Measurement Page 4-12
Example 4-11
The level of a liquid (density 1000
kg/m3) in a closed tank, 5 meter
high, is measured with a mercury
(density 13600 kg/m3) well type
manometer (A2/A1=0.01). The
manometer is installed with its
zero line 1 meter below the bottom
of the tank. If the reading on the
manometer is 0.2 m, calculate the
level H of the fluid in the tank.
Pt
Pt
5m
H
P2
1m
0.2 m
Please note that
the well (high
pressure input)
is connected to
the outside tube
P1
Zero line
d=.002
P1 = Pt + 1000×(5+1)×9.81 = Pt + 58860
X
Y
P2 = Pt + 1000×[H+(1-0.2)]×9.81 = Pt + 9810×(H+0.8)
∴P1 – P2 = 58860 – 9810(H+0.8) = 58860 – 9810H – 7848 = 51012 – 9810H…(1)
And P1 – P2 = ρhg(1 + A2/A1) = 13600×0.2×9.81×(1 + 0.01) = 26950 ………… (2)
From (1) and (2): 51012 – 9810H = 26950 ⇒ 9810H = 51012 – 26950 = 24062
∴H = 2.453 m.
(The exact value of H may be obtained if we equate pressures on the XY line:
Pt+1000×[H+(1–0.2)]×9.81+13600×(0.2+0.002)×9.81=Pt+1000×(5+1+0.002)×9.81
∴[H+0.8] + 13.6×0.202 = 6.002 ⇒ H + 0.8 + 2.747 = 6.002 ⇒ H + 3.547 = 6.002
∴H = 2.455 m.)
Example 4-12
The level of a liquid, with density ρℓ, stored in a
closed tank, L meters high, is measured with a u
tube manometer, using a manometer liquid, with
density ρm (ρm > ρℓ). The manometer is installed
with its zero line a distance z meter below the
bottom of the tank. When the fluid level in the
container is H meter, the reading obtained from
the manometer is h meter. Derive an equation
that can be used to calculate the liquid level H, in
terms of the manometer reading h and in terms of
the constants ρℓ, ρm, L, and z.
Pt
Pt
L
H
ρℓ
z
h
X
ZL
Y
ρm
Equating pressures at points X and Y:
Pt + ρℓ×(H + z - ½h)×g + ρm×h×g = Pt + ρℓ×(L + z + ½h)×g
∴ρℓ×(H + z - ½h)×g + ρm×h×g = ρℓ×(L + z + ½h)×g
∴ρℓ×(H + z - ½h) + ρm×h = ρℓ×(L + z + ½h)
∴ρℓH + ρℓz - ½ρℓh + ρmh = ρℓL + ρℓz + ½ρℓh
∴ρℓH = ρℓL + ρℓz + ½ρℓh - ρℓz + ½ρℓh - ρmh = ρℓL + ρℓh - ρmh
⎛ ρ ⎞
ρ
⎜
⎟
∴H = L + h - m h = L + ⎜1 − m ⎟h (interestingly, z doesn’t play a role)
ρ
ρ ⎟
⎜
l ⎠
l
⎝
EIPINI Chapter 4: Level Measurement Page 4-13
4.5 BUBBLE METERS
A bubbler level meter (air purge system or gas flushing system), provides a simple
and inexpensive but less accurate (±1-2%) level measurement system for virtually
any liquid, and typically for corrosive or slurry-type applications. A constant flow
of compressed air or an inert gas (usually nitrogen), is introduced through a dip
tube. The pressure required to force bubbles through the tube at a constant rate
(normally 60 bubbles per minute), is proportional to the hydrostatic pressure at the
bottom end of the dip tube. A bubbler sight glass, facilitate with adjusting the
bubble rate. The level is determined from the difference between the bubbler
pressure (back pressure) and the pressure at the surface of the liquid, by means of a
DP transmitter or manometer. Disadvantages of the bubbler system are 1) that the
liquid’s density, influences measurement accuracy and 2) the need for compressed
air or gas. The end of the dip tube may become plugged or clogged, and the dip
tube is periodically purged (release of large amounts of air to clear the tube), to
combat this situation. Figure 4-9 (a) and (b), depicts the measurement arrangement
for an open tank and a closed tank respectively, using a u-tube manometer.
Pressure Filter
regulator
Air/gas
supply
Bubbler
sight glass
Dip tube
Figure 4-9 (a)
Pressure Filter
regulator
Air/gas
supply
Bubbler
sight glass
Dip tube
Figure 4-9 (b)
EIPINI Chapter 5: Temperature Measurement Page 5-1
5. TEMPERATURE MEASUREMENT
The purpose of this chapter is to introduce students to fundamental concepts related to
temperature and to discuss important industrial methods used to measure temperature.
5.1 INTRODUCTION
Galileo Galilei (picture to the left), was born in the year
1564 in Pisa, Italy. This brilliant scientist and astronomer is
most famous for his early development of the telescope,
being the first to see the moons of Jupiter and other celestial
objects. Through his work in astronomy, Galileo supported
the theory originated by Copernicus, that the earth moved
around the Sun. Eventually, and under immense pressure
from the Church, he publicly retracted his support for the
Copernicus theory. Even so, Galileo was placed under house
arrest and spent the last 10 years of his life in almost
complete seclusion, having dared to offend the Church.
Galileo was indeed an incredible man. He was the first to realise how the swinging
of a pendulum, could be used to measure time and he is also credited to have
developed the first device to indicate changes in temperature.
This instrument did not measure temperature as such, and is
therefore called a thermoscope. (Vincenzo Viviani states in
his Vita di Galileo, that the thermoscope was conceived by
Galileo in 1597.) Galileo’s thermoscope was based on the
fascinating idea that air will expand when heated and contract
when cooled. The instrument consisted of a glass bottle about
the size of an egg, with a long glass neck. Air in the bottle
was heated with the hands and the open end side of the tube
immersed partially in a vessel containing water or wine. As
the air in the tube cools, it contracts and the liquid is drawn
Cold
Hot
into the tube. Once an equilibrium point is established, a rise
in temperature increases the volume of the air, forcing fluid back down; a fall in
temperature reduces the volume of the air, drawing more fluid into the tube.
Early in the 1700’s, Gabriel Daniel Fahrenheit developed a
more reliable temperature measuring device. A thin glass
tube holds a volume of mercury in a bulb at the bottom.
When heated, the mercury expands upward in the tube. The
tube could be sealed and allowed for accurate, reproducible
temperature readings. Because this device could be marked
with a numerical scale, it was called a thermometer.
Cold
Hot
EIPINI Chapter 5: Temperature Measurement Page 5-2
As the inventor of a reliable thermometer, Fahrenheit was afforded the luxury of
developing a temperature scale to his liking. He defined a temperature scale on which
0 was the coldest temperature he could reproduce in the lab (the temperature of an icesalt mixture) and 12 as his body temperature (as 12 inches in a foot). Fahrenheit soon
discovered that 12 divisions between the two set points were not fine enough for good
measurement so he doubled the divisions first to 24, then to 48 and finally to 96 (the
average person’s body temperature is actually 98.6ºF). On the Fahrenheit temperature
scale water freezes at 32º and boils at 212º. Indeed the unit, degrees, may arise from
the fact that there are 180 increments between freezing and boiling of water.
The Swedish astronomer, Anders Celsius, arbitrarily selected 100 to be the freezing
point of water and 0 to be the boiling point of water. Later the end points were
reversed and the centigrade scale was born (centigrade since there were 100
increments between the freezing point and boiling point). In 1948 the name was
officially changed from the centigrade scale to the Celsius scale.
The Fahrenheit and Celsius scales are termed relative scales. An object can be colder
than zero on the Fahrenheit scale or the Celsius scale and negative temperatures on
both scales are common. Since temperature is related to the motion of atoms or
molecules, zero should represent the absolute zero of temperature. Such a scale would
be an absolute scale. Two absolute temperature scales do exist. The Kelvin scale
(named after William Thomson, later Lord Kelvin) uses the same size increments as
the Celsius scale, but has its zero at absolute zero temperature. The Rankine scale
(named after William John Macquorn Rankine) is the absolute scale whose increments
are the same size as those in the Fahrenheit scale.
5.2 UNITS OF TEMPERATURE
Definition of Temperature
Temperature is defined as the degree of heat of a body. The SI unit for
temperature is Kelvin (K).
To set a temperature scale, two fixed temperature points must be defined. The distance
between the two fixed points, is called the fundamental interval. The two fixed points
used, are the following:
Bottom fixed point: The temperature of ice (prepared from distilled water)
mixed with distilled water, at standard atmospheric pressure of 760
mm. mercury.
Top fixed point: The temperature of distilled water that boils at standard
atmospheric pressure of 760 mm. mercury.
The Celsius scale
The bottom fixed point is 0 degrees Celsius (0 °C).
The top fixed point is 100 degrees Celsius (100 °C)
The fundamental interval of the Celsius scale is thus divided into 100 even parts
to express 1 °C.
EIPINI Chapter 5: Temperature Measurement Page 5-3
The Fahrenheit scale
The bottom fixed point is 32 degrees Fahrenheit (32 °F).
The top fixed point is 212 degrees Fahrenheit (212 °F).
The fundamental interval of the Fahrenheit scale is thus divided into 180 even
parts to express 1 °F.
The Kelvin scale
Zero Kelvin (-273.15 °C) corresponds to the absolute zero of temperature. The
melting point of ice is 273.15 Kelvin (K) and the boiling point of water is 373.15 K.
A temperature change of 1 °C, corresponds to a temperature change of 1 K.
The Rankine scale
Zero Rankin (-459.67 °F) corresponds to the absolute zero of temperature. The
melting point of ice is 491.67 Rankine (R) and the boiling point of water is 671.67 R.
A temperature change of 1 °F, corresponds to a temperature change of 1 R.
212 °F
672 R
Water boils
100 °C
373 K
68 °F
528 R
Room temperature
20 °C
293 K
32 °F
492 R
Water freezes
0 °C
273 K
-460 °F
0R
Absolute zero
-273 °C
0K
Fahrenheit
Rankine
Celsius
Kelvin
Relationship between Fahrenheit and Celsius
To determine the relationship between Fahrenheit and Celsius, we could draw a graph
of F (temperature in °F) versus C (temperature in °C), for C in the range 0 ≤ C ≤ 100.
Using the general equation for a
straight line:
y = m x + c,
180
C + 32
F=
100
∴F =
9
C + 32
5
5
(F – 32)
9
212
Equation 5-1
We can now also express Celsius in
terms of Fahrenheit.
From Equation 5-1:
9
C = F – 32
5
∴C =
F (°Fahrenheit)
Equation 5-2
180
32
100
0
100
C
(°Celsius)
EIPINI Chapter 5: Temperature Measurement Page 5-4
Example 5-1
Convert the absolute zero of temperature (–273.15 °C), to degrees Fahrenheit.
F = (9/5)C + 32 = (9/5)×(-273.15) + 32 = -459.67 °F ≈ -460 °F
Example 5-2
Convert 86 °F into degrees Celsius, Rankine and Kelvin.
C = (5/9)(F – 32) = (5/9)×(86 – 32) = 30 °C
R = F + 460 = 86 + 460 = 546 R
K = C + 273 = 30 + 273 = 303 K
Example 5-3
Determine the temperature in degrees Fahrenheit that would translate into the same
value in degrees Celsius.
In Equation 5-2 (or Equation 5-1), set C = F:
∴F = (5/9)(F – 32) ⇒ 9F = 5F – 160 ⇒ 4F = -160 ⇒ F = -40 °F
Example 5-4
Convert 0 °F to degrees Celsius.
C = (5/9)(F – 32) = (5/9)(0 – 32) = -17.78 °C.
5.3 THE INTERNATIONAL TEMPERATURE SCALE
This scale is based on a number of fixed temperature points and serves as calibration
reference. All temperatures are measured at standard atmospheric pressure of 760 mm.
mercury.
•
•
•
•
•
•
The oxygen point:
The ice point:
The boiling point:
The sulphur point:
The silver point:
The gold point:
The boiling point of liquid oxygen: –182.97 °C.
The melting point of pure ice: 0 °C.
The boiling point of pure water: 100 °C.
The boiling point of pure sulphur: 444.6 °C.
The melting point of silver: 961.78 °C.
The melting point of gold: 1064.18 °C.
5.4 LIQUID IN GLASS THERMOMETERS
The liquid in glass thermometer, is the most commonly used device to measure
temperature and it is inexpensive to make and easy to use. The liquid in glass
thermometer has a glass bulb attached to a sealed glass tube (also called the stem
or capillary tube). A very thin opening, called a bore, exists from the bulb and
extends down the centre of the tube. The bulb is typically filled with either
mercury or red-coloured alcohol and is free to expand and rise up into the tube
when the temperature increases, and to contract and move down the tube when
the temperature decreases.
EIPINI Chapter 5: Temperature Measurement Page 5-5
The background of the glass tube is covered with white enamel and the front of
the glass tube forms a magnifying glass that enlarges the liquid column and
facilitates with reading the temperature.
In Figure 5-1 (a), an all glass thermometer is depicted, with its scale etched
into the stem. Liquid in glass thermometers are fragile and for industrial use, the
thermometer is mounted in a protective housing and the scale is engraved on a
separate plate that is part of the protective case. An industrial thermometer is
shown in Figure 5-1 (b).
Protective case
Scale
(etched)
Scale
(on plate)
Bore
Bore
Lens front capillary
Tube (stem)
Lens front capillary
tube (stem)
Liquid column
Liquid column
Socket
Bulb
Bulb chamber
Bulb
Figure 5-1 (a)
Figure 5-1 (b)
Heat conducting
medium
Liquids used in glass thermometers
Liquid
Mercury
Alcohol
Pentane
Toluene
Creosote
Temperature range (Celsius)
-35
to +510
-80
to +70
-200 to +30
-80
to +100
-5
to +200
EIPINI Chapter 5: Temperature Measurement Page 5-6
5.5 MERCURY IN STEEL THERMOMETERS
Pointer
The mercury in steel thermometer
and scale
system, allows for rugged construction
and is used extensively in industrial
Bourdon
applications. The thermometer consists
tube
of a steel bulb, a steel capillary tube
and Bourdon tube, as shown in Figure
5-2. An advantage of the mercury in
steel thermometer is that measurements
can be taken a distance away from the
Steel tube
application, as the steel tube can be
(capillary)
made fairly long and flexible . The
whole system is completely filled with
mercury under pressure and sealed off.
When the temperature around the bulb
Mercury
increases, the mercury inside the bulb
will expand. The effect of the mercury Figure 5-2
Steel bulb
trying to increase its volume within a
confined space, will be an increase in pressure, transmitted via the capillary tube,
to the coiled Bourdon tube. The increase in mercury volume and pressure inside
the coiled Bourdon tube, will result in the Bourdon tube starting to uncoil,
proportional to the temperature. A pointer, linked to the free end of the Bourdon
tube, will subsequently move over the scale to indicate the temperature.
5.6 GAS FILLED THERMOMETERS
The thermometer consists of a steel
bulb, a steel tube and Bourdon tube, as
shown in Figure 5-3. The whole system
is filled with a gas at a high pressure
and sealed off. The gas commonly used
is nitrogen which is an inert gas, with a
high cubical expansion coefficient and
which is readily available. It may be
assumed that the volume of the gas
remains nearly constant. For an ideal
gas, the gas law applies:
PV = nRT.
It is clear from this equation that if V is
constant, the pressure is proportional to
the temperature T. Temperature is thus
converted to pressure, which is
Figure 5-3
detected by the Bourdon tube.
Pointer
and scale
Bourdon
tube
Steel tube
(capillary)
Gas
(nitrogen)
Steel bulb
EIPINI Chapter 5: Temperature Measurement Page 5-7
5.7 VAPOUR PRESSURE THERMOMETERS
The thermometer consists of a steel
bulb, a steel tube and Bourdon tube, as
shown in Figure 5-4. The bulb is
partly filled with a volatile liquid,
evacuated and sealed off. With rising
temperature, the average velocity of
the molecules in the liquid will
increase. As a result, more liquid
molecules will acquire enough energy
to escape from the surface of the liquid
and saturate the evacuated area. The
result will be an increase in vapour
pressure in the capillary tube and
Bourdon tube. The Bourdon tube,
sensitive to the changes in pressure,
will record the temperature via the
pointer that is linked to the moving
Figure 5-4
end of the Bourdon tube.
Pointer and
scale
Bourdon
tube
Steel tube
(capillary)
Vapour
Steel bulb
Volatile
liquid
5.8 BI-METAL THERMOMETERS
Bimetal strips consists of two different
metals welded together to form a cantilever
as shown in Figure 5-5 (a). When heated,
both metals expand, but the bottom strip
(assuming that the lower strip has the higher
expansion coefficient), expands more than
the top metal. The result is that the bottom
of the strip becomes longer than the top, and
the cantilever curls upwards as shown in
Figure 5-5 (b).
Pointer
and scale
Socket
Bearing
Shaft
Figure 5-5 (a)
Figure 5-5 (b)
A bi-metal thermometer uses a bi-metal strip, shaped in
a helix or spiral form, as shown in Figure 5-6. The one
end is fixed and the other end is free to rotate as the helix
curls in or out with changing temperature. A shaft and
pointer is linked to the rotating helix, to indicate the
temperature. The stem is filled with silicone fluid, to
provide damping and thermal conductivity between the
stem and bi-metal strip.
Guide
Stem
Helical
bi-metal
element
Figure 5-6
EIPINI Chapter 5: Temperature Measurement Page 5-8
5.9 RESISTANCE TEMPERATURE DETECTORS (RTD’s)
5.9.1 Temperature coefficient of resistance (TCR)
Resistance thermometers are based on the principle that the resistance of a metal
increases with temperature. The temperature coefficient of resistance (TCR) for
resistance thermometers (denoted by αo), is normally defined as the average resistance
change per °C over the range 0 °C to 100 °C, divided by the resistance of the
thermometer, Ro, at 0 °C.
Note: Starting with a straight line
R
− R0
αo = 100
R 0 ×100
Equation 5-3
where,
R0 = resistance of wire at 0 °C (ohm), and
R100 = resistance of wire at 100 °C (ohm),
As a first approximation, the relationship
between resistance and temperature, may
then be expressed as (see Figure 5-7):
Rt = Ro(1 + αot),
where:
Equation 5-4
Rt (Ω)
in standard form, y = mx + c:
Rt = mt + c where c = Ro
and m = (R100 - Ro)/100 = Roαo.
∴Rt = Roαot + Ro = Ro(1 + αot)
R100
Ro
Rt = Ro(1+αot)
t (°C)
0
100
Figure 5-7
Rt = resistance of wire at
temperature t (ohm),
Ro = resistance of wire at 0 °C (ohm), and
αo = temperature coefficient of resistance (TCR) at 0 °C (per °C)
Example 5-5
A platinum thermometer measures 100 Ω at 0 °C and 139.1 Ω at 100 °C.
a) Calculate the TCR for platinum.
b) Use Equation 5-4 and calculate the resistance of the thermometer at 50 °C.
c) Use Equation 5-4 and calculate the temperature when the resistance is 110 Ω.
a) From Equation 5-3: αo =
R100 − R 0
139.1 - 100
=
= 0.00391 /°C.
100 × 100
R 0 × 100
b) From Equation 5-4: R50 = Ro(1 + αot) = 100(1 + 0.00391×50) = 119.55 Ω
c) From Equation 5-4: Rt = Ro(1 + αot) ⇒ 110 = 100(1 + 0.00391t)
∴1 + 0.00391t = 1.1 ⇒ 0.00391t = 0.1 ⇒ t = 25.58 °C.
5.9.2 Resistance curves
In general, the relationship between resistance and temperature, is not linear and
Equation 5-4 does not describe the resistance / temperature behaviour of metals,
adequately. In Figure 5-8, the resistance curves for nickel, copper and platinum (with
Ro = 100 Ω) are shown and it is immediately clear that higher order equations must be
used to describe the resistance / temperature relationship. The resistance / temperature
characteristic for a platinum thermometer, for example, can be described by the
equation:
EIPINI Chapter 5: Temperature Measurement Page 5-9
⎧ R [1 + At + Bt 2 + Ct 3 ( t − 100)]
o
- 200 < t < 0 °C
R t = ⎪⎨
⎪⎩ R o [1 + At + Bt 2 ]
where:
0 ≤ t < 850 °C
,
Rt is the resistance at temperature t
Ro is the ice point resistance and
A, B and C are coefficients describing the particular thermometer.
R (ohm)
800
Nickel
600
400
Copper
200
Platinum
-200 0
t (°C)
200 400 600 800
Figure 5-8
An excerpt from the resistance table for a Minco
platinum thermometer (code PB), is given in Table 5-1.
Example 5-6
In the temperature range 0 °C ≤ t < 850 °C, the
resistance / temperature relation of a platinum
thermometer with ice point resistance of 100 Ω, is
described by the second order expression:
Partial temperature/
resistance table for
platinum (Ro=100Ω)
Temperature Resistance
(°C)
(Ω )
-200
17.26
-100
59.64
-80
67.83
-60
75.96
-40
84.03
-20
92.04
0
100.00
20
107.92
40
115.78
60
123.60
80
131.38
100
139.11
200
177.04
300
213.81
400
249.41
500
283.84
600
317.09
700
349.18
Rt = 100×[1 + 0.0039692×t – (5.8495×10-7)×t2].
Table 5-1
a) Calculate the resistance of the thermometer at 50 °C.
b) Calculate the temperature when a resistance of 110 Ω is measured.
a) R50 = 100×(1 + 0.0039692×50 – 5.8495×10-7×502)
= 100×(1 + 0.19846 - 0.001462375) = 119.7 Ω.
b) Rt = 100[1 + (3.9692×10-3)×t – (5.8495×10-7)t2].
∴110 = 100[1 + (3.9692×10-3)×t – (5.8495×10-7)t2]
∴110 = 100 + 0.39692t – 5.8495×10-5t2
∴(5.8495×10-5)t2 + (– 0.39692)t + 10 = 0
∴t =
=
0.39692 ±
( 0.39692 )
2
− 4 × ( 5.8495 × 10
2 × ( 5.8495 × 10
0.39692 ± 0.39396
116.99 × 10
−6
−5
−5
) × 10
)
= 25.301 °C (ignoring 6760 °C.)
[ax2+bx+c=0]
[x= − b ±
b 2 − 4ac
2a
]
EIPINI Chapter 5: Temperature Measurement Page 5-10
5.9.3 Thermometer construction
Depending
upon
their
intended
application, resistance thermometers are
available in different shapes and forms.
A typical structure is shown in Figure
5-9. The resistance winding is located in
the lower end of the protecting tube or
stem, which is sealed. The upper part of
the stem terminates in the terminal
housing (head) with the resistor winding
leads. The resistance winding must be in
good thermal contact with the stem, for
fast heat transfer from the medium to the
winding while electrical isolation must
be ensured. The fundamental design
problem with resistance thermometers is
to achieve high electrical insulation and
minimum thermal insulation.
Terminal
cap
Connector
conduit
Socket
Stem
Leads
Resistor bulb
(resistance winding
or thin film)
Figure 5-9
5.9.4 Measuring temperature with resistance thermometers
The simplest way to measure the
resistance of a RTD, is to inject a
constant current into the thermometer
R2
and to measure the voltage that develops
M
across the thermometer. A Wheatstone
E
bridge circuit, shown in Figure 5-10, is
B
however generally used to detect the
changes in resistance of a resistance
thermometer. The values of the fixed
R3
resistors, R1,R2 and R3, are very
C
accurately known, while RT represents
the resistance thermometer with leads a
and b. The bridge is said to be in null
Figure 5-10
balance, when the voltage across points
R1
A
b
a
RT
R
A and B is zero. This occurs when RT = R3× 1 , causing VAC = VBC, resulting in
R2
the reading on M, to become zero. The zero condition would correspond to the zero
point or set point of the resistance thermometer output. As the temperature
increases, the resistance RT, of the resistance thermometer will increase, causing
the bridge to become unbalanced, and meter M to show a reading. The meter M
may be calibrated in temperature units or VAB may be converted into a standard 4 to
20 mA or 1 to 5 V signal. The current flowing through the thermometer must be
kept as low as possible (< 1 mA) to minimise errors caused by I2R losses and
associated temperature rise in the thermometer itself.
EIPINI Chapter 5: Temperature Measurement Page 5-11
Example 5-7
The Wheatstone bridge in Figure 5-10, is supplied from a 10 V battery, and used to
measure temperature with a platinum resistance RT. The value of each of the fixed
resistors R1, R2 and R3, is 100 Ω (in a practical circuit, R1 and R2 would be chosen
much higher to improve bridge linearity and minimize errors caused by I2R losses in
the thermometer). The meter M measures the thermometer output voltage V, which
is given by: V = VAB = VAC - VBC. At 0 °C, the resistance RT, of the platinum RTD,
is 100 Ω, while at 100 °C, the resistance RT increases to 139.1 Ω. Calculate the
output voltage V, of the thermometer, when the temperature is 0 °C and when the
temperature is 100 °C. Assume that the measuring device draws negligible current
from the circuit.
I2
I1
0 °C:
R2
R1
I1 = 10/(100 + 100) = 50.00 mA.
10V
V 100Ω
100Ω
∴VAC = RT×I1 = 100×(50×10-3)
A
B
= 5 volt
R3
I2 = 10/(100 + 100) = 50.00 mA.
VAB RT
VBC
VAC
-3
∴VBC = R3×I2 = 100×(50×10 )
100Ω
100Ω
= 5 volt
C
C
∴V = VAB = VAC – VBC (using Kirchoff’s
0 °C
= 5 – 5 = 0 volt
voltage law: VAB-VAC+VBC=0)
We of course expected this result, as the bridge is indeed balanced at 0 °C.
100 °C:
I1=10/(100+139.1)=41.82 mA.
∴VAC = RT×I1 = 139.1×(41.82×10-3)
= 5.817 volt
I2 = 10/(100 + 100) = 50.00 mA.
∴VBC = R3×I2 = 100×(50×10-3)
= 5 volt
∴V = VAB = VAC – VBC
= 5.817 – 5.000 = 0.817 volt
I2
R2
100Ω
10V
R1
100Ω
V
A
B
R3
VBC
100Ω
C
I1
RT
VAC
139.1Ω
100 °C
C
Example 5-8
The Wheatstone bridge in Figure 5-10, is supplied from a 10 V battery, and used to
measure temperature with a platinum resistance RT. The value of each of the fixed
resistors R1, R2 and R3, is 100 Ω. The meter M measures the thermometer output
voltage V, which is given by: V = VAB = VAC - VBC.
a) Derive an expression for the resistance RT of the platinum element, in terms of
the thermometer output voltage V.
b) Derive an expression for the temperature t in terms of RT, if in the temperature
range 0 °C ≤ t < 850 °C, the relationship between the resistance RT of the
thermometer and the temperature t, is given by:
RT = 100[1 + (3.9692×10-3)×t – (5.8495×10-7)t2].
c) Calculate the measured temperature when the bridge output voltage is 0, 0.2, 0.4,
0.6, 0.8, and 1.0 volt.
EIPINI Chapter 5: Temperature Measurement Page 5-12
10
ampere
I2
I1
RT + 100
R
R
2
1
10 × RT
∴VAC = I1×RT =
volt
RT + 100
10V
100Ω V 100Ω
10
A
B
and I2 =
= 50 milliamp.
100 + 100
R3
VBC
VAC
∴VBC = I2×R3 = (50×10-3)×100
100Ω
RT
= 5 volt
∴V = VAB = VAC - VBC
C
C
10 × RT
10 × RT
∴V =
-5⇒
= V + 5 ⇒ 10×RT = (RT + 100)×(V + 5)
RT + 100
RT + 100
∴10×RT = V×RT + 100×V + 5×RT + 100×5
∴10×RT – 5×RT - V×RT = 100×V + 100×5
∴(10 – 5 – V)×RT = 100×(V + 5) ⇒ (5 – V)×RT = 100×(V + 5)
100(V + 5)
∴RT =
…………………………….………………. Equation (a)
5-V
a) I1 =
b) RT = 100[1 + (3.9692×10-3)×t – (5.8495×10-7)t2]
∴RT = 100 + 0.39692×t – (5.8495×10-5)×t2
∴(5.8495×10-5)×t2 - 0.39692×t + RT – 100 = 0
0.39692
1
5.8495 ×10 - 5 2
∴
×t ×t +
×(RT-100) = 0
5.8495 ×10 - 5
5.8495 ×10- 5
5.8495 ×10- 5
∴ t2 - 6786t + [17095×(RT-100)] = 0
[x2+bx+c=0]
6786 − (6786) 2 − 4 × 17095 × (RT − 100)
∴t =
2
[x= − b ±
b 2 − 4c
2
]
6786 − 46.05 × 106 − 68380(RT − 100)
…….………….. Equation (b)
2
(Note: From Example 5-6 we already know that we can ignore the second
solution associated with the positive square root.)
∴t =
c)
V = 0:
V = 0.2:
V = 0.4:
V = 0.6:
V = 0.8:
From Equation (a): RT = 100(0 + 5)/(5 - 0) = 100 Ω
From Equation (b): t = {6786 - √[46.05×106 - 68380(100 - 100)]}/2
= [6786 - 6786]/2 = 0 °C
From Equation (a): RT = 100(0.2 + 5)/(5 - 0.2) = 108.33 Ω
From Equation (b): t = {6786 - √[46.05×106-68380(108.33-100)]}/2
= [6786 - √45.48×106]/2 = 21.05 °C
RT = 100(0.4 + 5)/(5 - 0.4) = 117.39 Ω
t = {6786 - √[46.05×106 - 68380(117.39 - 100)]}/2 = 44.09 °C
RT = 100(0.6 + 5)/(5 – 0.6) = 127.27 Ω
t = {6786 - √[46.05×106 - 68380(127.27 - 100)]}/2 = 69.40 °C
RT = 100(0.8 + 5)/(5 - 0.8) = 138.1 Ω
t = {6786 - √[46.05×106 - 68380(138.1 - 100)]}/2 = 97.37 °C
EIPINI Chapter 5: Temperature Measurement Page 5-13
V = 1.0:
RT = 100(1 + 5)/(5 - 1) = 150 Ω
t = {6786 - √[46.05×106 - 68380(150 - 100)]}/2 = 128.4 °C
The results from Example
t (°C)
5-8, highlight again that the
140
relationship
between
the
thermometer output and the
120
measured temperature, is not
100
linear (a graph of our
calculated temperatures versus
80
thermometer output, is shown
to the right). The non-linear
60
behaviour arises from the non40
linear relationship between RT
and V in Equation (a) and
20
V (volt)
from
the
non-linear
relationship
between
the
0
0.2
0.4
0.6
0.8
1.0
temperature t and RT in
Equation (b).
In Example 5-8, temperature and resistance were related by a simple quadratic
equation. When converting resistance to temperature, thermometer manufacturers may
in certain cases, be required to find the root of a third or fourth order polynomial. In
the case of nickel, individual manufacturers have developed different step-wise
approximations to calculate temperature from resistance. Fortunately today, many
instruments use microprocessors and manufacturers are using resistance / temperature
lookup tables stored in read only memory, with linear interpolation for measurements
falling between table entries.
5.9.5
Ambient temperature compensation
If it is necessary to perform the temperature measurement, some distance away
from where the resistance thermometer bulb is installed, ambient temperature can
have a detrimental effect on the integrity of the measurement. The reason is that the
leads that connect the thermometer to the instrument, will have resistance of their
own, and what is more, this resistance will change with changing ambient
temperature. The error introduced by lead resistance, can of course be minimised,
if leads with very low resistance is used.
An ingenious approach to the lead resistance problem, is the three wire method,
illustrated in Figure 5-11. The three wire method attempts to cancel the effect of
the lead resistance, by introducing the same resistance in both branches (wires c
and a) of the Wheatstone bridge. The three wire method is used extensively for
industrial RTD’s. Three leads a, b and c, connect to the resistance thermometer and
the resistance of wires a and c must be matched. This method assumes that the
meter is a high impedance device, and that essentially no current is flowing in the b
wire, which only acts as a voltage sense lead.
EIPINI Chapter 5: Temperature Measurement Page 5-14
R1
R2
c
Rlead
b
Rlead
a
Rlead
V
E
A
B
RT
R3
C
Figure 5-11
The four wire method, depicted in Figure 5-12, alleviates many problems
associated with the Wheatstone bridge. Instead of using a Wheatstone bridge
configuration, a current source is employed to supply a constant current I, to the
thermometer, through wires a and d. A high impedance voltmeter M, measures the
voltage developed across the thermometer, via wires b and c. The measured
voltage is directly proportional to the resistance of the thermometer, so only the
conversion from resistance to temperature is necessary. Wires b and c only act as
voltage sense leads, and with M a high impedance meter, virtually no current flows
in wires b and c, and therefore no voltage drop in these leads and thus no lead
resistance error in the measurement. The disadvantage of the four wire method is
that we need one more wire than with the three wire method, but it is a small price
to pay if we are at all concerned with the accuracy of the temperature
measurement. The four wire method is however gaining in popularity in industry.
I
a
Rlead
b
Rlead
c
Rlead
d
Rlead
M
RT
Figure 5-12
Example 5-9
A Wheatstone bridge, with fixed resistors R1, R2 and R3, of 100 Ω and supplied from a
10 V battery, is used to measure a temperature of 100 °C with a platinum resistance
RT, that has a resistance of 139.1 Ω at 100 °C . The thermometer is connected to the
bridge with leadwires that have a resistance Rlead of 10 Ω at 20 °C and the thermometer
is correctly calibrated for an ambient temperature of 20 °C. Assume that the lead
resistance changes to 11 Ω when the ambient temperature rises to 30 °C.
EIPINI Chapter 5: Temperature Measurement Page 5-15
Calculate the temperature error introduced at an ambient temperature of 30 °C, when
using a) the two wire method and b) the three wire method
a) Measuring with two leads
Calibrated bridge voltage at
20 °C ambient temperature
I2 100Ω
I1 100Ω
I2 100Ω
V
10V
Bridge output voltage at
30 °C ambient temperature
10Ω
b
100Ω
a
Process
temberature
100°C
I1 100Ω
V
10V
11Ω
b
139.1Ω
RT
10Ω
100Ω
a
Process
temberature
100°C
139.1Ω
RT
11Ω
I1 = 10/(100+10+139.1+10) = 0.0386 A
I1 = 10/(100+11+139.1+11) = 0.0383 A
I2 = 10/(100+100) = 0.05 A
I2 = 10/(100+100) = 0.05 A
V = (11+139.1+11)×0.0383 - 100×0.05
V = (10+139.1+10)×0.0386 - 100×0.05
= 6.141 – 5 = 1.141 V (calibrated 100°C)
= 6.17 – 5 = 1.17 V (105°C)
Therefore, if the bridge was correctly calibrated for an ambient temperature of 20 °C,
an error of 0.029 V will occur when the ambient temperature rises to 30 °C. So instead
of reading 100 °C, it would measure approximately 105 °C, an error of 5 °C or 5%.
b) Measuring with three leads
Calibrated bridge voltage at
20 °C ambient temperature
I2
I1 100Ω
100Ω
10Ω
Bridge output voltage at
30 °C ambient temperature
I2
Process
temberature
100°C
I1 100Ω
100Ω
11Ω
c
100Ω
c
10Ω
V
10V
Process
temberature
100°C
b
139.1Ω
a
RT
10Ω
V
10V
100Ω
11Ω
b
139.1Ω
a
RT
11Ω
I1 = 10/(100+10+139.1+10) = 0.0386 A
I1 = 10/(100+11+139.1+11) = 0.0383 A
I2 = 10/(100+100) = 0.05 A
I2 = 10/(100+100) = 0.05 A
V = (139.1 + 11)×0.0383 - 100×0.05
V = (139.1 + 10)×0.0386 - 100×0.05
= 5.755 – 5 = 0.755 V (calibrated 100°C)
= 5.749 – 5 = 0.749 V (99.13°C)
Therefore, measuring with the three wire method, an error of only 0.006 V will
occur. If the thermometer was correctly calibrated to read 100 °C at an ambient
temperature of 20 °C, then if the ambient temperature rises to 30 °C, the reading of
0.749 V would correspond to a value of 138.76 Ω for RT with no rise in ambient
temperature, or in terms of temperature, 99.13 °C, an error of 0.87 °C or 0.87%. This
is clearly a remarkable improvement over the two wire method.
EIPINI Chapter 5: Temperature Measurement Page 5-16
5.10 THERMOCOUPLES
5.10.1 The Seebeck effect
If two dissimilar metals are joined together to form a closed loop, and if one
junction is kept at a different temperature from the other, an electromotive force is
generated and electric current will flow in the closed loop.
Metal A
Metal B
Metal B
Seebeck emf
This very important discovery in the field of thermometry, was made by T.J. Seebeck
in the year 1821. Experiments by Seebeck and others, have shown that the generated
emf (called the Seebeck voltage in his honor), is relative, in a predictable manner, to
the difference in temperature between the two junctions. So, if the temperature of one
junction is kept at a known value, the temperature of the other junction can be
determined by the amount of voltage produced. This discovery resulted in the
temperature sensor that we know today as the thermocouple.
The Seebeck voltage is made up of two components: the Peltier voltage generated at
the junctions, plus the Thomson voltage generated in the wires by the temperature
gradient. The Peltier voltage is proportional to the temperature of each junction while
the Thomson voltage is proportional to the square of the temperature difference
between the two junctions. It is the Thomson voltage that accounts for most of the
observed voltage and non-linearity in thermocouple response.
5.10.2 Thermocouple laws
The following empirically derived thermocouple laws, are useful to understand,
diagnose and utilise thermocouples.
a) Law of homogeneous circuits
If two thermocouple junctions are at T1 and T2, then the thermal emf generated is
independent and unaffected by any temperature distribution along the wires
T3
T1
T2
T4
Figure 5-13
In Figure 5-13, a thermocouple is shown with junction temperatures at T1 and T2.
Along the thermocouple wires, the temperature is T3 and T4. The thermocouple emf is,
however, still a function of only the temperature gradient T2 – T1.
EIPINI Chapter 5: Temperature Measurement Page 5-17
b) Law of intermediate metals
The law of intermediate metals states that a third metal may be inserted into a
thermocouple system without affecting the emf generated, if, and only if, the
junctions with the third metal are kept at the same temperature.
Metal A
T1
T2
Metal B
Metal C
Metal B
T3
Figure 5-14
When thermocouples are used, it is usually necessary to introduce additional metals
into the circuit This happens when an instrument is used to measure the emf, and when
the junction is soldered or welded. It would seem that the introduction of other metals
would modify the emf developed by the thermocouple and destroy its calibration.
However, the law of intermediate metals states that the introduction of a third metal
into the circuit will have no effect upon the emf generated so long as the junctions of
the third metal are at the same temperature, as shown in Figure 5-14.
c) Law of intermediate temperatures
The law of intermediate temperatures states that the sum of the emf developed by a
thermocouple with its junctions at temperatures T1 and T2, and with its junctions at
temperatures T2 and T3, will be the same as the emf developed if the thermocouple
junctions are at temperatures T1 and T3.
T1
T2
emf (T2-T1)
T2
T3
emf (T3-T2)
T1
T3
emf (T3-T1)
Figure 5-15
This law, illustrated in Figure 5-15, is useful in practice because it helps in giving a
suitable correction in case a reference junction temperature other than 0 °C is
employed. For example, if a thermocouple is calibrated for a reference junction
temperature of 0 °C and used with a junction temperature of 20 °C, then the correction
required for the observation would be the emf produced by the thermocouple between
0 °C and 20 °C.
5.10.3 Thermocouple types
Any two dissimilar metals can in theory be made into a thermocouple. However,
certain metals have been selected over time that make ideal thermocouples for various
applications. These metals have been chosen for their emf output and their ability to
operate under various conditions. There are several types of these “standard”
thermocouples in use today. Some are listed Table 5-2.
EIPINI Chapter 5: Temperature Measurement Page 5-18
Type Positive
element
K Chromel
Isolation Negative
colour
element
Yellow Alumel
Isolation Outer
Temperature
colour isolation range (°C)
Red
Yellow
-200 to 1200
E
Chromel
Purple
Constantan
Red
Purple
-200 to 800
J
Iron
White
Constantan
Red
Black
-200 to 750
T
Copper
Blue
90% Platinum
Black
10% Rhodium
Constantan
Red
Blue
-200 to 350
Platinum
Red
Green
0 to 1500
S
Table 5-2
Types K, E, J and T, are the 'general purpose' thermocouples. Type K is the most
popular thermocouple in use today while type T is used for lower temperature
applications. The conventional type J thermocouple, even with its unfavourable iron
lead, is still popular, mainly because of its widespread use in older instruments. Type
E is the most sensitive of the standard thermocouples (68μV/°C). Noble metal type S,
has low sensitivity(10μV/°C), is expensive and used in the higher temperature range.
Thermocouple alloys referred to in Table 5-2, are chromel (chrome and nickel), alumel
(aluminium and nickel) and constantan (copper and nickel).
Different colour codes have been adopted for thermocouple wire and product
identification and the colour codes listed in Table 5-2, are used in the United States.
5.10.4 Thermocouple construction
Thermocouple thermometers are available in different shapes and in different
coverings. Three basic types of construction can be recognised:
Wire construction
The most basic construction of the thermocouple
is the two dissimilar metals joined (welded)
together to form the measuring junction. In this
form, the exposed junction offers good response
times, but may suffer from environmental
damage. A bare thermocouple element is shown
in Figure 5-16.
Sheathed construction
To improve mechanical strength, mineral
insulated thermocouples were developed. The
thermocouple wires are embedded in compressed
mineral oxide powder and enclosed in a metal
sheath, usually stainless steel or inconel (nickelchromium-iron alloy). A completely insulated
thermocouple is shown in Figure 5-17 (a).
Bare thermocouple junction
Figure 5-16
Magnesium
oxide
Thermocouple
wires
Insulated junction
Inconel
sheath
Figure 5-17 (a)
EIPINI Chapter 5: Temperature Measurement Page 5-19
To improve response time, the junction may be grounded as shown in Figure
5-17 (b), or the thermocouple tip could be exposed as shown in Figure 5-17 (c).
Grounded junction
Exposed junction
Figure 5-17 (b)
Figure 5-17 (c)
Protecting tube construction
Cover
Protecting tubes (or thermowells)
are used to shield thermocouple
sensing
elements
against
mechanical damage, thermal shock
and corrosive or contaminating
atmospheres. Various types of
constructions are available. A
typical
construction
method,
similar to that of RTD’s, is shown
in Figure 5-18. Cast iron protection
tubes are used for example, in
molten aluminum, magnesium and
zinc applications while ceramic
tubes are used in industries such as
iron and steel, glass, cement and
lime processing.
Conduit
entrance
Socket
Metal or ceramic
protecting tube
Sheathed
thermocouple
Figure 5-18
5.10.5 Measuring temperature with thermocouples
The voltage generated by a thermocouple is a function of the temperature difference
between the measurement and reference junctions. Traditionally the reference junction
was held at 0 °C by an ice bath, as shown in Figure 5-19. The thermocouple emf is
measured with a high impedance voltmeter.
Hot junction
v
Cold junction
Ice bath
Figure 5-19
EIPINI Chapter 5: Temperature Measurement Page 5-20
The relationship between thermocouple voltage and temperature is unfortunately not
linear, and it is necessary to use thermocouple temperature conversion tables to find
temperature from the measured voltage. An extract from the voltage / temperature
table for a type K thermocouple (0 °C reference), is given in Table 5-3.
Temperature (°C) versus emf (μV) for type K thermocouple with 0 °C reference.
deg C
0
10
20
30
40
50
60
70
80
90
100
0
0
397
798
1203
1612
2023
2436
2851
3267
3682
4096
1
39
437
838
1244
1653
2064
2478
2893
3308
3723
4138
2
79
477
879
1285
1694
2106
2519
2934
3350
3765
4179
3
119
517
919
1326
1735
2147
2561
2976
3391
3806
4220
4
158
557
960
1366
1776
2188
2602
3017
3433
3848
4262
5
198
597
1000
1407
1817
2230
2644
3059
3474
3889
4303
6
238
637
1041
1448
1858
2271
2685
3100
3516
3931
4344
7
277
677
1081
1489
1899
2312
2727
3142
3557
3972
4385
8
317
718
1122
1530
1941
2354
2768
3184
3599
4013
4427
9
357
758
1163
1571
1982
2395
2810
3225
3640
4055
4468
Table 5-3
We could store these look-up table values in a computer and use the table to convert
between emf and temperature. A more viable approach used by manufacturers
however, is to approximate the table values using a power series polynomial (see
Example 5-14) and allow the instrument’s microprocessor or process computer, to
calculate temperature from emf or the emf from temperature (inverse polynomial).
The ice bath cold junction is not considered practical anymore. Instead the terminals
connecting the thermocouple to the measuring device, are now assumed to play the
role of the reference junction or 'cold junction', as it is still called today. The reference
junction temperature may now be kept, for example, at room temperature, where the
junction temperature is measured with an auxiliary temperature sensor, such as a RTD.
According to the law of intermediate temperatures, the thermocouple voltage that
corresponds to the cold junction temperature, may be added to the measured
thermocouple voltage. The true temperature of the hot junction, with respect to 0 °C,
can then be determined from this augmented voltage.
Example 5-10
Calculate the average sensitivity (μV/°C) of a type K thermocouple in the temperature
range 0 °C to 100 °C.
From Table 5-3 the change in emf developed by a type K thermocouple from 0 °C to
100 °C, is 4096 μV. The average sensitivity is therefore 4096/100 = 40.96 μV/°C.
EIPINI Chapter 5: Temperature Measurement Page 5-21
Example 5-11
The cold junction of a type K thermocouple is kept at 0 °C. Use Table 5-3 to
determine the temperature if the measured voltage is a) 798 μV and b) 2602 μV.
a) 20 °C
b) 64 °C
Example 5-12
The relationship between emf and temperature for a certain (imaginary) thermocouple,
is described by the relation: v = t2, where v is the generated thermocouple emf in
microvolt (μV), and t the temperature difference in °C, between the hot junction and
0 °C. If the thermocouple emf reading is 3000 microvolt and the temperature of the
cold junction is 25 °C, calculate the temperature of the hot junction.
Emf corresponding to (25 – 0) °C = 252 = 625 μV
Total emf (T - 0) = 3000 + 625 = 3625 μV
According to the law of intermediate temperatures:
Hot junction temperature T = √v = √3625 = 60.21 °C
(T is NOT = √3000 + 25 = 54.77 + 25 = 79.77 °C)
0°C
25°C
25°C
625μV
T
3000μV
0°C
T
3625μV
Example 5-13
An unknown temperature is measured with a type K thermocouple. A thermocouple
voltage of 2602 μV is measured. If the cold junction temperature is 20 °C, calculate
the process temperature, measured by the hot junction side of the thermocouple.
From Table 5-3, the cold junction (20°C) emf is
0°C
20°C
20°C
T
798 μV. According to the law of intermediate
798μV
2602μV
temperatures, the correction voltage of 798 μV
should be added to the measured voltage of
0°C
T
2602 μV, to obtain 3400 μV. The corrected voltage
3400μV
represents the thermocouple emf that would be
obtained, if the reference junction was kept at 0 °C.
Again from Table 5-3, the temperature that 84°C
corresponds to 3400 μV, is somewhere between
83 °C and 84 °C. To find the correct temperature,
1°C
we must use linear extrapolation between these
?
two values. The difference between 3433 μV
X
(84 °C) and 3391 μV (83 °C) is 42 μV, while
3400 μV is 9 μV more than 3391 μV (83 °C). 83°C
9μV
Therefore the temperature we are looking for is
42μV
83 °C plus (9/42) °C, which is 83.2143 °C.
The way NOT to calculate the hot junction
3391μV 3400μV
3433μV
temperature, is to look up the measured voltage
(2602 μV) as 64 °C and then to add the cold
1
9
X
×1
=
∴X
=
junction temperature of 20 °C, to obtain 84 °C.
42
9 42
This is NOT CORRECT.
EIPINI Chapter 5: Temperature Measurement Page 5-22
Example 5-14
The following equations are provided for a type K thermocouple, to calculate
temperature (t in °C) from emf (v in μV) in the temperature range 0 °C to 500 °C:
t = (2.508355×10-2)v + (7.860106×10-8)v2 - (2.503131×10-10)v3
+ (8.315270×10-14)v4 – (1.228034×10-17)v5 + (9.804036×10-22)v6
- (4.413030×10-26)v7 + (1.057734×10-30)v8 – (1.052755×10-35)v9 ,
Equation (a)
and to calculate emf (v in μV) from temperature (t in °C) in the temperature range
0 °C to 1372 °C:
v = -17.600413686 + 38.921204975t + (1.8558770032×10-2)t2
- (9.9457592874×10-5)t3 + (3.1840945719×10-7)t4
- (5.6072844889×10-10)t5 + (5.6075059059×10-13)t6
- (3.2020720003×10-16)t7 + (9.7151147152×10-20)t8
– (1.2104721275×10-23)t9 + 118.5976 e − 1.183432 ×10
−4
(t − 126.9686) 2
Equation (b)
a) Use Equation (a) to calculate the temperature when the emf is 2602 μV.
b) Use Equation (b) to calculate the emf when the temperature is 20 °C.
a) t = (2.508355×10-2)×2602 + (7.860106×10-8)×26022 - (2.503131×10-10)×26023
+ (8.315270×10-14)×26024 – (1.228034×10-17)×26025 + (9.804036×10-22)×26026
- (4.413030×10-26)×26027 + (1.057734×10-30)×26028 – (1.052755×10-35)×26029
= 65.26740 + 0.5321609 – 4.409664 + 3.811584 – 1.464694 + 0.3042627
- 0.03563592 + 0.002222464 – 0.05755631
= 63.95 °C
b) v = -17.600413686 + 38.921204975×20 + (1.8558770032×10-2)×202
- (9.9457592874×10-5)×203 + (3.1840945719×10-7)×204
- (5.6072844889×10-10)×205 + (5.6075059059×10-13)×206
- (3.2020720003×10-16)×207 + (9.7151147152×10-20)×208
−4
2
– (1.2104721275×10-23)×209 + 118.5976 e − 1.183432 ×10 (20 − 126.9686)
= -17.600414 + 778.4241 + 7.423508 – 0.7956607 + 0.05094551
- 0.001794331 + 35.88804×10-6 – 409.8652×10-9 + 2.487069×10-9
- 6.197617×10-12 + 30.61899
= 798.1 microvolt
(Compare these answers to those of Example 5-11)
5.10.6 Compensating leads
Thermocouple thermometers are normally installed some distance away from the
voltmeter or computer that measures the emf generated by the thermocouple. For
this purpose, cheaper and lower grade thermocouple wires, called extension wire or
compensating leads, are used to connect the thermocouple to the measuring device
at the reference junction. Compensating leads have the same thermoelectric
EIPINI Chapter 5: Temperature Measurement Page 5-23
properties as the thermocouple and do not introduce a significant error into the
temperature measurement. Compensating leads must be matched to the
thermocouple and for each type of thermocouple, corresponding extension leads
are available.
5.11 THERMISTORS
A thermistor is similar to a resistance thermometer, but a semiconductor material is
used instead of a metal. A distinct advantage of thermistors over resistance
thermometers, is that their temperature coefficient of resistance is approximately
ten times higher than that of a resistance thermometer, causing thermistors to be
much more sensitive temperature detectors. The resistance change with
temperature is however very nonlinear, and unlike RTD’s, most thermistors have a
negative temperature coefficient of resistance, that is, the resistance of a thermistor
decreases with increasing temperature.
In Figure 5-20, the resistance curve of a typical thermistor is shown. The following
equation, called the Steinhart and Hart equation, is used to describe the resistance /
temperature relation for thermistors:
1
= A + Bln(R) + ClnR 3 ,
T
where T is absolute temperature in Kelvin, R is the thermistor resistance at
temperature T and A, B and C are coefficients that describe a given thermistor.
Resistance R
Figure 5-20
Temperature T
The resistivity of thermistors is also much larger than that of RTD’s, and
thermistors can therefore be made very small which means they will respond
quickly to temperature changes.
Thermistors cannot be used to measure high temperatures, compared to RTDs. In
fact, the maximum temperature of operation is sometimes only 100 or 200 oC. The
high resistivity as well as high temperature coefficient of resistance of the
thermistor, makes it feasible to use the simpler two wire technique when measuring
thermistor resistance.
EIPINI Chapter 5: Temperature Measurement Page 5-24
For example, a common thermistor value is 5000 ohms at 25°C. With a typical
temperature coefficient of resistance of 0.04/°C, a measurement lead resistance of
10 Ω, produces only 0.05 °C error. This error is a factor of 500 times less than the
equivalent RTD error. Thermistors have not gained nearly the popularity of RTDs or
even thermocouples in industry due to their limited span as well as other
disadvantages. Since thermistors are semiconductor devices, they are quite susceptible
to permanent decalibration when exposed to high temperatures. In addition,
thermistors are quite fragile and great care must be taken to mount them so that they
are not exposed to shock or vibration.
5.12 RTD, Thermocouple or Thermistor ?
Resistance temperature detectors (RTDs)
An RTD sensing element consists of a wire coil or deposited film of pure metal. The
element’s resistance increases with temperature in a known and repeatable manner.
RTDs exhibit excellent accuracy over a wide temperature range and represent the
fastest growing segment among industrial temperature sensors. Their advantages
include:
• Temperature range: Models cover temperatures from -260 to 850°C.
• Repeatability and stability: Industrial RTDs typically drift less than 0.1°C/year.
• Sensitivity: The voltage drop across an RTD provides a much larger output than a
thermocouple.
• Linearity: Platinum and copper RTDs produce a more linear response than
thermocouples or thermistors. RTD non-linearities can be corrected through
proper design of resistive bridge networks.
• Low system cost: RTDs use ordinary copper extension leads and require no cold
junction compensation.
Thermocouples
A thermocouple consists of two wires of dissimilar metals welded together into a
junction. At the other end of the signal wires, usually as part of the input instrument, is
another junction called the reference junction. Heating the sensing junction generates a
thermoelectric potential (emf) proportional to the temperature difference between the
two junctions. This millivolt-level emf, when compensated for the known temperature
of the reference junction, indicates the temperature at the sensing tip. Published
millivolt tables assume the reference junction is at 0°C. Thermocouples are simple and
familiar. Designing them into systems, however, is complicated by the need for special
extension wires and reference junction compensation. Thermocouple advantages
include:
• Extremely high temperature capability: Thermocouples with precious metal
junctions may be rated as high as 1800°C.
• Ruggedness: The inherent simplicity of thermocouples makes them resistant to
shock and vibration.
• Small size/fast response: A fine-wire thermocouple junction takes up little space
and has low mass, making it suitable for point sensing and fast response.
EIPINI Chapter 5: Temperature Measurement Page 5-25
Thermistors
A thermistor is a resistive device composed of metal oxides formed into a bead and
encapsulated in epoxy or glass. A typical thermistor shows a large negative
temperature coefficient. Resistance drops dramatically and non-linearly with
temperature. Sensitivity is many times that of RTDs but useful temperature range is
limited. There are wide variations of performance and price between thermistors from
different sources. Typical benefits are:
• Low sensor cost: Basic thermistors are quite inexpensive. However, models with
tighter interchangeability or extended temperature ranges often cost more than
RTDs.
• High sensitivity: A thermistor may change resistance by tens of ohms per degree
temperature change, versus a fraction of an ohm for RTDs.
• Point sensing: A thermistor bead can be made the size of a pin head for small area
sensing.
Sensor
type
Temperature
range
Sensor
cost
System
cost
Stability
Sensitivity
Linearity
RTD
-260 to
850°C
Moderate
Moderate
Best
Moderate
Best
Thermocouple
-270 to
1800°C
Low
High
Low
Low
Moderate
Thermistor
-80 to
150°C
(typical)
Low
Moderate
Moderate
Best
Poor
Specify for:
General purpose
sensing
Highest accuracy
Highest
temperatures
Best sensitivity
Narrow ranges
(e.g. medical)
Point sensing
EIPINI Chapter 6: Process Control Page 6-1
6. PROCESS CONTROL
The purpose of this chapter is to introduce students to industrial process control
methods used in industry to ensure that manufactured products meet predetermined
quality requirements. This is accomplished by continuously monitoring the production
process and automatically correcting or minimizing any deviations from the required
specifications, that may be detected during the manufacturing process.
6.1 INTRODUCTION
The history of automatic control goes back many centuries. Water level control may
already be identified in water clocks used in the middle ages while mechanical clocks,
making their appearance in the 1200’s, used an escapement mechanism that may be
described in terms of feedback control. With the advent of the steam engine, it was
evident that some sort of speed control was needed. In what is generally considered a
major event in the history of feedback control, James Watt, completed
the design of the centrifugal flyball governor for regulating the
speed of the rotary steam engine, in 1788. This device
employed two pivoted rotating flyballs which were flung
outward by centrifugal force. As the speed of rotation increased,
the flyweights swung further out, operating a steam flow throttling
valve which slowed the engine down. Thus, a constant speed was
achieved automatically. From this point onwards, many new developments followed
to improve control systems as well as mathematical research to understand physical
processes. A major invention, however, was the introduction of proportional, integral
and derivative (PID) control, which was formulated in 1922 by Nicholas Minorsky
(1885-1970). Observing the way in which a helmsman steered a ship and compensate
for the disturbances from the ocean, motivated this threefold control strategy.
Although advanced tools, such as model predictive controllers, are available today,
the Proportional, Integral, Derivative (PID) control strategy, is still the most widely
used in modern industry, controlling more than 95% of closed loop industrial
processes.
6.1.1 A SIMPLE WATER LEVEL CONTROL
In Figure 6-1, an operator has the simple task of
keeping the water level in a tank at a prescribed
value, for instance half full. If the water level
should drop below the desired level due to
increased water outflow, the operator must
increase the water inflow by opening the inflow
valve more until the desired level is restored.
Similarly, if the water level should increase due
to a decrease in outflow, the operator must
decrease the inflow by closing the inflow valve
more, until the required level is restored.
Inflow
valve
Inflow
Outflow
valve
Water inlet
Outflow
Figure 6-1
EIPINI Chapter 6: Process Control Page 6-2
If the water level reaches the desired value (also called the set point), the operator
must try to keep the inflow equal to the outflow. This manual control example
illustrates essentially how we would expect an automatic control system to behave.
In Figure 6-2, an
Inflow
arrangement
for
automatic control
Inflow
Level
Level
Outflow
of the water level
valve
controller detector
valve
in the container,
is
shown.
A
level
measuring
Figure 6-2
instrument senses
Water level
Water inlet
Outflow
the water level in
control
the container, and
feeds this value back to the controller. The basic function of the controller is to
calculate the difference between the desired water level and the measured water level,
and to determine a sensible control action, that will minimize this difference. Once the
controller has calculated an appropriate control action, this value is transmitted as a
pneumatic signal to the inflow valve, which is adjusted accordingly. A well-designed
controller should respond quickly to changes in the water level and always steer the
level in the water container back to the required level, in a disciplined way.
6.1.2 CONTROL TERMINOLOGY
The water container system in Figure Water inflow
6-2 could be represented as an
Water level
Process
input/output system or process, as (Input)
(plant)
(Output)
shown in Figure 6-3. The quantity that
we want to regulate is the water level Water outflow
in the container, which is called the
Figure 6-3
controlled variable. The controlled
variable is also called the output variable, the process variable or the measured
variable. The rate of water inflow into the container (a system input) with which we
may change or manipulate the water level in the container, is called the manipulated
variable. The manipulated variable is also called the controller output or control
variable. A change in the rate at which the water flows out of the container (a system
input), will cause the water level in the container to change or to be disturbed, and is
called a disturbance variable. (Note: there could be other disturbance variables that
may influence the controlled variable, for instance a leak in the container.)
Three values are associated with the controlled variable. Firstly, the measured value,
which is the current water level in the container, determined by the level sensor.
Secondly, the desired value (normally locally programmed into the controller), which
is the water level that we need or require, for example 50% full. The desired value is
also called the set point or reference value. Thirdly the error value which is the
difference between the desired value and the measured value of the water level.
EIPINI Chapter 6: Process Control Page 6-3
1. CONTROLLED VARIABLE: Process output variable that is maintained
between specified limits.
2. MEASURED VALUE: Actual value of the controlled variable, as determined
by the instrumentation.
3. DESIRED VALUE: Required value of the controlled variable (set point).
4. ERROR VALUE: The difference between the desired value and the measured
value.
5. MANIPULATED VARIABLE: Process input variable that is adjusted, to steer
the controlled variable towards the desired value.
6. DISTURBANCE VARIABLE: Process input variable that can cause the
controlled variable to deviate from the desired value.
Example 6-1
A control system for a heat exchanger is shown in Figure 6-4. Cold water enters the
container at point 1 and is heated by steam, entering the steam line 3, at point 2. The
heated water leaves the system at point 7. The temperature of the hot water is
measured by a thermometer at point 6 and presented to the temperature controller,
block 5. The controller operates a steam valve 4, such that when the hot water is below
the required temperature, the steam flow rate through the steam line is increased,
allowing more heat transfer to the cold water, and when the hot water is above the
required temperature, the steam flow is reduced. Identify: a) the controlled variable, b)
the measured value, c) the manipulated variable and d) some disturbance variables.
2
Steam
supply
4
Steam
control
valve
Cold water
1
TIC
3
6
Steam outlet
5
Figure 6-4
Heat
exchanger
7
Hot water
a) Controlled variable: The temperature of the hot water delivered at 7.
b) Measured value: Temperature indicated by the temperature detector 6.
c) Manipulated variable: The steam flow rate through the steam line 3, adjusted by
valve 4.
d) Disturbance variables:
i) Change in hot water demand (this is an energy change at the output 7 and is
known as an output or demand load disturbance).
ii) Change in steam supply pressure (this is an energy change at the input 2, and
is known as an input or supply load disturbance).
iii) More heat loss through the heat exchanger walls, because of colder ambient
conditions.
EIPINI Chapter 6: Process Control Page 6-4
6.1.3 OPEN LOOP AND CLOSED LOOP SYSTEMS
Automatic control systems may operate under open or closed loop control. The input
of an open loop system is not influenced by its output. This is in contrast with a closed
loop system that monitors its output and then manipulates the input, to ensure that the
output meets set conditions. An example of an open loop system is a bullet fired from
a rifle and an example of a closed loop systems is a cruiser missile, fired from a ship,
continuously monitoring and adjusting its trajectory.
Open loop control strategies may perform well in situations where the behaviour of
the system is very well defined and modelled. For example a room temperature
controller set to a certain on/off heater schedule, will produce the desired room
temperature. However, it is difficult to suggest an open loop control strategy for the
water level control in Figure 6-2, as the factors that influence the water level are
difficult to predict, and a closed loop control strategy, as indicated in Figure 6-2, is
clearly the most sensible solution.
Open loop system: The input to the system is not determined by the output.
Closed loop system: The input to the system is determined by the output.
6.1.4 FEEDBACK AND FEEDFORWARD CONTROL
Feedback closed loop control systems measure changes in the controlled variable
directly and feed them back as input variables to the controller. Feedforward control
systems on the other hand, do not measure changes in the controlled variable but
rather other variables, termed intermediate variables, which are indicative of the
controlled variable.
The interrelationship between the elements of a general closed loop feedback control
system, is illustrated with the block diagram in Figure 6-5. The water level control
system in Figure 6-2 is an example of a closed loop feedback control system. The
water level provided by the level sensor, is fed back to the error detector which
calculates the error. The controller uses the error value to compute a suitable
command for the inflow valve, thereby regulating the water level close to the desired
value.
Disturbance
variables
Comparator
Desired
value
Measured
value
Error
value
Control
unit
Manipulated
variable
Process
Controlled
variable
Sensor
Figure 6-5 Block diagram of a feedback control system
Output
EIPINI Chapter 6: Process Control Page 6-5
Feedforward control systems, on the other hand, will measure disturbance variables
that directly influence the controlled variable. It is conceivable, for example, to
measure the outflow in the water level control system in Figure 6-2, and to base the
inflow control on the value of this disturbance variable. Because the source of the
level disturbance is monitored and acted upon, a change in water level is anticipated
and corrected before it even occurs.
Whereas feedback systems must wait for a deviation to occur before corrective
action is taken, the principle advantage of feedforward control systems is that they act
before deviations occur. It is clear that feedforward control alone would not be
adequate for the water level control system, as the water level would eventually drift
away from set point because of measuring and modelling errors. In general,
feedforward control is used to complement feedback control and to enhance system
performance.
Feedback control: Measure the controlled variable to determine the control strategy.
Feedforward control: Measure disturbance variables to determine the control
strategy.
6.1.5 DIRECT ACTING AND REVERSE ACTING CONTROL
A control valve that opens more when the pressure input is increased (air to open) is
called a reverse acting valve. For the water level control system depicted in Figure 6-2,
we have assumed that a reverse acting control valve is used. This means that when the
water level drops below set point, the controller must increase its output to the control
valve to increase the water inflow. This type of control action is called reverse acting,
because when the water level decreases, the controller output must increase. If a direct
acting (air to close) control valve was used, the control action would be direct acting,
as the controller output would decrease when the water level decreases.
A heating system would typically be reverse acting, as a decrease in temperature
would demand an increased controller output to the heating element. A cooling
system, on the other hand, would typically be direct acting because an increase in
temperature would require an increased output to the cooling element.
Direct acting control: A control arrangement in which the controller output
increases if the measured value rises above the set point.
Reverse acting control: A control arrangement in which the controller output
increases if the measured value drops below the set point.
6.1.6 PROCESS TIME LAGS
Most processes exhibit a time delay between its input and output. For example, if the
heating element of a temperature control system is switched on, the temperature will
not immediately rise to the maximum value. The inlet flow valve in the level control
system in Figure 6-2 is another example. When it receives a signal to open 100%, the
flow rate will not immediately increase to 100% due to delays inherent in the
operation of the valve.
EIPINI Chapter 6: Process Control Page 6-6
Figure 6-6 shows the typical response to a step input of a process such as a control
valve receiving a command to open from 50 % to 60 %, resulting in the increase in
inflow from 50 % to 60 %.
Input
60 %
50 %
Time
Output
60 %
Dead time and first order lag
50 %
Time
td
tf
Figure 6-6
The response curve in Figure 6-6 illustrates two kinds of time delays that may be
identified. Firstly a dead time td, when the system does not respond at all, and
secondly a first order delay tf, when the system output changes to its new final value
in an exponential fashion.
The dead time is associated with a delay in the transportation of material or
information. For example, it may take 20 minutes to send a signal to the Rover on
Mars. During this time, the vehicle can not respond to this control command, because
the control information is still being transported to the vehicle. The dead time is also
known as a distance-velocity lag, and the reason for this is illuminated in
Example 6-2. Another term used for the dead time, is transportation lag.
The first order lag is associated with a delay in the transfer of energy. For example,
when you step on the accelerator of a car, the car will take time to convert the
chemical energy in the fuel, into mechanical kinetic energy, and the speed increase
will be exponential in nature. The first order lag is also known as a transfer lag,
exponential lag or resistance-capacitance lag.
Dead time: Delay due to the time it takes information or material to be transported
from one point to another.
First order lag: Delay due to the time it takes energy to be transferred from one
point or form to another.
Example 6-2
In the system depicted Figure 6-7, a conveyor belt is used to transport material that
was deposited by a feeder, to the delivery station. A request for more material is given
to the feeder. Calculate the time it will take the sensor at the delivery site to detect a
reaction to the increased demand for material. The velocity of the conveyor belt is v
meter per second and the distance between the feeder and delivery point is d meter.
EIPINI Chapter 6: Process Control Page 6-7
Feeder
v
Delivery
station
Figure 6-7
Sensor
d
Any increase in material delivered, must travel a distance d with a velocity v, before it
reaches the delivery point, where the sensor will detect the increase. The time it takes
the material to be transported to the sensor is t = distance/velocity = d/v seconds. The
dead time lag in this case is called a distance-velocity lag and it is a problem not easily
handled by a PID controller.
6.2 CONTROL SCHEMAS
We will now discuss some of the common
Maximum inflow:
techniques (on-off and PID control modes)
0.01 m3/sec.
used to control closed loop feedback
QIN
Maximum
systems. For this purpose we will use the
level:
water level control system in Figure 6-2 as
5m
5 meter
basis. Central to our system is the water
H
container shown in Figure 6-8. We will use
a container with height 5 m and base area
of 1 m2. The level of the water in the
2
1
m
QOUT
container is H, with a maximum value of
5 m. The rate of water outflow is QOUT,
Maximum outflow:
3
with a maximum value of 0.01 m /sec. The Figure 6-8
0.01 m3/sec.
rate of water inflow is QIN. In order to have
any control over the water level in the tank whatsoever, the maximum inflow should
be more or equal to the maximum outflow. We will assume that they are equal, that is,
when the demand is a maximum of 0.01 m3/sec. and the inflow is a maximum of
0.01 m3/sec, the water level will remain constant.
To simplify matters, we will use percentage values for all variables. The variable M
will be used to represent the water level, expressed as a percentage value. The water
level will be measured by an electronic DP transmitter, delivering 4 mA when the tank
is empty (H = 0 m or M = 0 %) and 20 mA when the tank is full (H = 5 m or
M = 100 %). We will also assume that the level detector output M will immediately
(with no time delay) reflect the actual water level H.
The variable C will be used to represent the controller output, expressed as a
percentage value. The controller will transmit 20 kPa to the control valve to close the
valve completely (C = 0 %) and 100 kPa to open the valve fully (C = 100 %). We will
also assume that a linear relationship (with no time delay) exists between the
controller output and the water inflow. This simplification means that we can use the
same variable C for the controller output and the inflow (manipulated variable).
EIPINI Chapter 6: Process Control Page 6-8
Therefore, if the controller output C is 0 %, the water inflow QIN will immediately
cease (QIN = 0 m3/sec.) and when C is 100 %, the water inflow will immediately rise
to its maximum value of QIN = 0.01 m3/sec.
The outflow demand or load QOUT will be represented by the percentage variable L.
L will be 0 % when the outflow QOUT = 0 m3/sec and L will equal 100 % when the
outflow is the maximum value of QOUT = 0.01 m3/sec.
The complete water level control system is shown in Figure 6-9.
100%
Inflow C (0% – 100%)
Inflow
valve
C
Water supply
Level M
controller
Level
detector
Figure 6-9
Water level
control system
Water inflow C
Controller
Output C
100%
Water level M
(0% – 100%)
Outflow L (0 – 100%)
Before we can use our water tank, we must find out how it works. This means that
we need to understand how the inflow and outflow influence the water level in the
container. Clearly when C = L, the water level M will stay constant. When C > L, the
water level will rise and when C < L, the water level will drop. It is therefore easy to
ΔM
, is proportional to C – L. In
argue that the rate at which the water level changes,
Δt
the example given at the end of this chapter in Appendix 6-1, the exact relationship
between M, C and L, was found to be:
dM
= 0.002×(C – L) percent per second.
dt
Finally, the error value will play a fundamental part in the operation of on-off and
PID control. The percentage error value E is defined as:
E = S – M ………….…………..………..…………………… Equation 6-1
The set point is denoted by S and is expressed as a percentage of the maximum
measured value. For example, a set point value S = 50 %, will imply a required water
level of 2.5 m or half filled tank for the water level control system.
6.2.1 ON-OFF OR TWO-POSITION CONTROL
With on-off control mode (also called bang-bang control), the controller output C, can
take on only two values, on or off. Referring to the level control in Figure 6-9, either
EIPINI Chapter 6: Process Control Page 6-9
the inflow is switched on fully (100%) when the water level drops below a certain
level, say 40%, or it is closed completely (0%), when the water level rises above a
certain level, say 60%. With the aid of Equation 6-1, we could express the control law
for the water level on-off control as:
C=
0%
unchanged
100%
if E ≤ -10%
if -10% < E < 10%
if E ≥ 10%
(or M ≥ 60%)
(or 60% > M > 40%)
(or M ≤ 40%)
A possible response of the water level to on-off control as well as the controller
action, is shown in Figure 6-10. On average, the water level remains 50%.
Water level (M)
60%
40%
Time
Water inflow (C)
100%
0%
Time
Figure 6-10 Typical on/off control performance
On-off control will also work well for a room temperature control – the heater is
fully on, until the room temperature increases to a prescribed value, after which the
heater is switched off, and so on. Not all systems are suited to on-off control. Imagine
a speed cruise control in a car, set to 120 km/h, using on-off control. 118 km/h: full
acceleration, 122 km/h: no throttle. Surely not very comfortable driving.
A disadvantage of on-off control is the wear on the final control element, such as the
control valve in the present example that continually moves from one extreme position
to the other.
Example 6-3
A water container is 5 meter high and has a base area of 1 m2. The maximum water
inflow rate, as well as the maximum water outflow rate, is 0.01 m3/sec. On-off control
is used to regulate the water level in the tank, between 40% and 60% of its maximum
level (5 meter). Assume that the rate of water outflow, is regulated and constant at
60% of the maximum rate of outflow. Calculate:
a) The time period that the inflow valve is closed
b) The time period that the inflow valve is open.
Solution:
a) Water level when the inflow valve must close: 60% of 5 m = 0.6×5 = 3 meter.
Water level when the inflow valve must open: 40% of 5 m = 0.4×5 = 2 meter.
Constant demand from tank: 60% of 0.01 = 0.6×0.01 = 0.006 meter3/second.
EIPINI Chapter 6: Process Control Page 6-10
Max inflow (100%) = 0.01 m3/s
The inflow valve closes when the
water level reaches the 3 m mark, and
Max level=60%=3 m
stays closed until the water level
drops to the 2 m mark. Total volume 5m
Min level=40%=2 m
of water that must leave the container
before the 2 m mark is reached:
Controlled and
V = Ah = 1×1 = 1 m3.
2
constant outflow
1m
Rate of outflow Q, is 0.006 m3/s.
Outflow=60% of max (0.01) = 0.006 m3/s
But volume = flow rate×time
∴V = Qt [(m3/sec)×(sec)]
M
∴tclosed = V/Q = 1/0.006
= 166.7 sec = 2.778 min
60%
b) The inflow valve opens when the
water level reaches the 2 m mark and 40%
stays open until the water level rises
to the 3 m mark. Total volume of
t
water that must enter the container, is
C
1 m3. The resultant rate at which
water flows into the tank is the rate of 100%
inflow, minus the outflow rate.
0%
Q = 0.01–0.006 = 0.004 m3/sec.
t
2.778
4.167
∴topen = V/Q = 1/0.004
min
min
= 250 sec = 4.167 min.
On-off control: A control strategy in which the controller output switches the final
control element fully on or off to keep the controlled variable near set point.
6.2.2 PROPORTIONAL CONTROL
The proportional control strategy provides more smooth control than on-off control.
The control effort is proportional to the magnitude of the error, which makes sense if
one thinks about a person driving a car and trying to obey a 120 km/h speed limit. The
more the speed falls below 120 km/h during an uphill climb, the more he will step on
the accelerator, and when the speed tends to go above 120 km/h as he goes downhill,
he will start to release the accelerator. When reaching 120 km/h on a straight road, he
will keep the accelerator in just the right position to maintain his target speed of
120 km/h.
This is very important to understand, the driver does not release the accelerator
completely when the target is reached. For proportional control, this control effort that
is maintained when the error is zero, is called the bias. We will denote the bias by the
percentage variable R. When designing a proportional controller, a fixed bias R must
be chosen and this choice depends on the typical load on the system. For our water
level control example, we will assume that the typical outflow demand is 50%, and a
suitable choice for the bias, is R = 50%.
EIPINI Chapter 6: Process Control Page 6-11
A simple and logical proportional control law for the water level system can now be
formulated. The inflow equals 100% when the tank is empty, the inflow equals 50%
(bias) when the tank is half full (equal to the desired value) and the inflow equals 0%
when the tank is full (overflowing). This control strategy is depicted in Figure 6-11.
Water inflow C (%)
100 %
50 %
-50 %
Tank full,
M = 100%
Figure 6-11
Basic
proportional
control law
R (Bias)
0%
Tank half full,
M = 50%
Error (%)
(E=50-M)
50 %
Tank empty,
M = 0%
This is obviously a reverse acting control strategy, because the controller output
increases when the measured value decreases. A negative slope of the C-E line would
indicate a direct acting control strategy. It is now a simple matter to write down an
expression for this proportional control law, as the graph of C versus E, is a straight
line of the form: y = mx + c. From Figure 6-11: C = E + 50. The slope of this line is
clearly 1, because the controller output C, changes by 100% when the error E, changes
by 100%. The slope of the line is called the proportional gain KP, of the controller.
For the controller characteristic in Figure 6-11, the gain is therefore KP = 1.
If the controller gain, KP, is changed, the slope of the straight line, describing the
relationship between controller output C, and error value E, will also change. Graphs
of C versus E, for three different controller gains, KP = 2, KP = 1 and KP = 0.5, are
shown in Figure 6-12. (Note: Although a gain of 0.5, prescribes a theoretical error of
100% before the output reaches 100 %, such an error could of course not occur in our
water level control system.)
C (%)
100%
C (%)
C (%)
100%
75%
100%
R=50%
E(%)
-25% 0% 25%
-50%
KP = 2
R=50%
R=50%
25%
E(%)
0%
KP = 1
50%
-100%
-50%
0%
50%
E(%)
100%
KP = 0.5
Figure 6-12 Proportional control with different proportional gains
An alternative representation of the relationship between C and E, for KP = 2, KP = 1
and KP = 0.5, is shown in Figure 6-13. The movement of the left hand side of a
EIPINI Chapter 6: Process Control Page 6-12
Figure 6-13 Proportional band
E
100%
C
0%
E
25%
0
E
50%
50%
C
0%
0
50%
C
0%
0
50%
-25%
100%
KP = 2
-50%
100%
KP = 1
100%
-100%
KP = 0.5
pivoted beam, represents changes in the error E, while the right hand side, sweeps out
corresponding values of the controller output C. For KP = 2, an error range of 50%
(from -25% to 25%), sweeps out the complete output range of 100%. For KP = 1, an
error range of 100% results in an output change of 100%. For KP = 0.5, the error value
must change 200% to cover the whole output range. The error range that results in the
total change in output range, is called the proportional band, of the controller, and is
denoted by the percentage variable PB.
For a high gain (narrowband) system (KP =2 in Figure 6-13 for example), a small
error would cause a large reaction from the controller in his effort to correct the error.
A low gain (wideband) system (KP = 0.5 in Figure 6-13 for example), would act more
gently, because large errors, will cause a milder controller reaction. The relationship
between proportional band PB (in percent) and proportional gain KP, is given by:
PB =
100
percent. …………...……………………….. Equation 6-2
K
P
We can now obtain a general expression for the output C, of a proportional
controller, in terms of the proportional band PB, the set point S, the measured value M
and the bias R. In general, for any gain KP and bias R, the relationship between the
controller output C, and the process error E, is a straight line of the form, y = mx + c,
with the slope of the line equal to the controller gain KP and the y intersect, equal to
the controller bias R.
C (%)
∴C = KPE + R ………………..………...….… (1)
From Equation 6-1: E = S – M …...............….. (2)
R (%)
Slope=KP
100
And from Equation 6-2: KP =
………..…. (3)
PB
E (%)
(2) and (3) in (1):
C = KPE + R =
100
(S – M) + R ……………………..…..…. Equation 6-3
PB
In a typical application, S = 50 and R = 50, therefore C =
100
(50 – M) + 50.
PB
EIPINI Chapter 6: Process Control Page 6-13
Example 6-4
A process is controlled by a proportional controller. The controller is programmed for
a positive gain (reverse acting controller) and proportional band of 80 %, a set point of
50 % and a bias of 50 %. The system error is calculated from the difference between
set point and the measured value i.e. E = 50 – M.
a) Calculate the proportional gain of the controller.
b) If the measured value of the process is indicated as 36 %, calculate the output of
the controller at that instant.
c) If the measured value of the process is indicated as 65 %, calculate the output of
the controller at that instant.
d) For the given settings, give the proportional control law (C as a function of E).
e) Draw a graph of the controller control law (C versus E).
a) From Equation 6-2:
KP = 100/PB = 100/80 = 1.25.
b) From Equation 6-3:
100
C=
×(S – M) + R
PB
100
=
×(50 – 36) + 50 = 67.5%.
80
c) From Equation 6-3:
100
C=
×(S – M) + R
PB
100
=
×(50 – 65) + 50 = 31.25%.
80
d) C = KPE + Bias
∴ C = 1.25E + 50
e)
C
100%
ΔC
ΔE
100
=
80
= 1.25
Slope =
50%
E
-40%
0%
40%
Example 6-5 (students must study this important example very carefully)
The water level in a container is controlled by a proportional controller with KP = 1
and bias R = 50%. At a given moment, the outflow demand L, is 50% and the inflow
C, is 50%. The water level in the tank is also stable at its set point of 50% (the error is
zero, therefore the inflow is 50% which is equal to the outflow). The outflow demand
suddenly increases to L = 60%. Determine the new stable water level in the tank.
Solution: The moment
C=50%
C=60%
the outflow increases, the
water level will start to
S=50%
S=50% Offset
drop, and it will continue
(E=10%)
M=50%
to drop until the inflow C,
M=40%
also increases to 60%.
The inflow is driven by
the error value according
L=50%
L=60%
to C = KPE + R = E + 50.
Therefore the controller output will reach 60% when E = 10%. This means, from
E = 50–M, that the water level M, must drop to 40%. Although trying to keep the
water level close to 50%, this example clearly illustrates that a proportional controller
is not able to force the measured value back to set point, when the demand is
EIPINI Chapter 6: Process Control Page 6-14
different from the bias value. Decreasing the proportional band (increasing KP), will
however keep the level closer to set point. For example, if KP = 2, the controller will
need an error of 5 % to increase its output to 60 %. The water level will thus drop to
only 45%.
The difference between the set point and the measured value that may occur in
proportional control, is called the offset, and it is a serious disadvantage of
proportional only control. It is also important to note that we could have restored the
level to 50 %, by changing the bias value of the controller to 60 %, if the outflow
remained constant at 60 %.
Example 6-6
The water level in a container is controlled by a proportional controller and control
valve. The controller gain is 1 and the bias is 50 %. Assume that the water inflow
delivered by the valve is equal (no time delay) to the control signal C, and that the
level detector instantly reflects the level M, of the water in the container. The
relationship between the water level M, and the inflow C and outflow L, is given by:
dM
= 0.002×(C – L).
dt
a) Sketch the installation.
b) Draw a block diagram of the system.
c) The system is initially in the state: C = 50%, L = 50%, M = 50% and S = 50%.
When time equals zero, the outflow demand L is suddenly increased to 60%.
Determine a time expression for the water level M. (This will demonstrate the
system’s response to a disturbance and is also called regulatory control action).
a)
S=50
Inflow
valve
C
Inflow = C
M Level M
Level
controller
detector
KP=1 & R=50
Water supply
Outflow demand = L
b)
S=50
E=50-M Control unit
C=E+50
C
Control
valve
C
L
M
Water container
dM
=0.002(C–L)
dt
Controller
Sensor
c) The output of the proportional controller is given by:
C = KPE + R = KP×(S – M) + R = 1×(50 – M) + 50 = 100 – M.
and
dM
= 0.002(C – L) = 0.002[(100-M) – 60] = 0.08 – 0.002M
dt
M
EIPINI Chapter 6: Process Control Page 6-15
∴
dM
+0.002M = 0.08 and with M(0) = 50 gives the solution M = 40 + 10e-t/500
dt
A graph of the response of the water level M to the disturbance, is shown below.
Demand L
60 %
50 %
Time (sec.)
Water level M
50 %
M = 40 + 10e-t/500
43.7%
Offset
40 %
Time (sec.)
t=0
t=500 sec.
The behaviour of the water level may be compared with our findings in Example 6-5.
Proportional control: A control strategy in which the controller output is
proportional to the magnitude of the error. {C = KPE + R = (100/PB)×(S-M) + R}
Proportional gain: Ratio of controller output change to error value change ΔC/ΔE.
Proportional band: The error range that causes 100 % change in controller output.
Offset: The steady state difference between the set point and the measured value.
6.2.3 PROPORTIONAL AND INTEGRAL CONTROL
Examples 6-5 and 6-6 illustrated very clearly the
L
basic weakness of proportional only control. When
60%
a disturbance causes an error, proportional control
50%
uses that same error to combat the disturbance. By
its very nature, proportional control cannot
M
eliminate the offset caused by a disturbance. In the
50%
past, control engineers discovered that the offset
40%
could be eliminated by slowly changing the bias
value up or down until the measured value is equal
E
to the desired value. This adjustment was called
10%
resetting the controller. In Example 6-5, for
0%
instance, shifting the bias to 60 % would reset the
level back to 50 %. This manual reset is in essence
C
the same as automatic reset or integral reset. The
60%
offset problem is highlighted again in Figure 6-14.
50%
At instant A, the demand increases from 50% to
60%. With proportional only control, the level M
drops to 40 %, the error E rises to 10 % and the
inflow C, driven by the error value, rises to 60 %.
Time
Time
Time
A
B
Time
Figure 6-14
EIPINI Chapter 6: Process Control Page 6-16
The graphs from point B onwards, suggest a possible procedure with which we
could eliminate the offset (perhaps someone gave us a hosepipe with running water
and the task to restore the water level to 50 %). Firstly, we must keep the total inflow
more than 60 % so that the level starts returning to 50 %. While we are doing this, we
must remember that the error is getting smaller and we will get less and less assistance
from the proportional controller. Secondly, after the level is restored to 50 %, the
inflow from the proportional controller will only be 50 % and we need to supply the
extra 10 % to equal the demand of 60 % so that the level remains at 50 %.
Integral action gives an output proportional to the time integral of the error. This
action can accomplish the two tasks mentioned above, namely, restoring the measured
value to set point and providing the extra bias to match the demand. Integral control is
rarely used alone but proportional control together with integral control, or PI control,
is widely used in industry because it provides the important practical advantage of
eliminating the offset. The controller output C, of a PI controller, can be expressed as:
∫
C = KPE + KI Edt + R, ……………….……………………… Equation 6-4
where KP is the proportional gain, KI is defined as the integral gain and R is the bias.
(Note: For PI control mode the bias term is optional and may be disabled.)
Comparison of Integral Control with Proportional Control
In Figure 6-15 the load change behaviour of the water level control system with
proportional plus integral control, is compared with the behaviour of the same system
with only proportional control (refer to Example 6-6). For this illustration, S = 50 %,
KP = 1 and R = 50 % for both controllers. The integral gain KI of the PI controller,
was chosen as KI = 0.0005, which is small and produces a very sluggish but easilly
understandable response. The integral component is denoted by I = 0.0005 ∫ Edt .
The fundamental difference between P control and PI control is that with P control,
the error persists while with PI control, the error vanishes. With P control it is indeed
the error that sustains the final increase in inflow of 10 % to satisfy the increased
demand of 60 %. With PI control it is the area under the error curve, accumulated by
the integral action I = KI ∫ Edt , that sustains the final increase in inflow of 10 % to
satisfy the increased demand of 60 %. In a sense, the final scenario resembles a P only
controller that shifted its bias automatically from 50 % to 60 %. In contrast with P
control, the magnitude of the error has no effect on the inflow in the end, because the
error is erased by the integral action. It is however clear from Figure 6-15, that in the
early stage after the disturbance, the error and ascociated proportional action is
playing a critical part in assisting the integral action to provide the extra inflow needed
to replace the water lost during the initial drop in water level and to restore the water
level back to 50 %. For the integral action to carry this burden alone, would typically
result in the water level oscillating around the set point and may very well be one of
the reasons why integral action is rarely used without proportional action.
EIPINI Chapter 6: Process Control Page 6-17
Proportional Control
Demand L
60%
50%
Proportional plus Integral Control
Time
Water level M
50%
40%
Time
Error E=50-M
10%
Demand L
60%
50%
Time
Water level M
50%
40%
Time
Error E=50-M
10%
0%
Time
0%
Time
∫
I=0.0005 Edt
10%
0%
Bias R
50%
Time
Controller output
C=E+R
60%
50%
Time
Time
Bias R
50%
Time
Controller output
C=E+I+R
60%
50%
Time
Figure 6-15 P and PI control comparison
Repeat time and reset time
Many commercial PI controllers do not directly allow for the adjustment of the
integral gain KI, but rather the adjustment of the repeat time or reset time. Imagine an
error signal that suddenly changes from 0 to 10 % and remains constant after that, as
shown in Figure 6-16.
Error E
Area = 10×TR
10%
0%
Time
PI controller output
∫
C=KPE+KI Edt
∫
∫
Integral effect = KI Edt = KI 10dt = 10KITR
Proportional effect = KP×E = KP×10
50%
Time
TR
(Repeat or reset time)
Figure 6-16
Reset time
If this step in error signal is presented to the input of a PI controller, that initially
EIPINI Chapter 6: Process Control Page 6-18
had an output of 50 %, there would be a sudden increase in controller output because
the proportional term KPE tracks the error instantaneously. Furthermore, as the
integral term, KI ∫ Edt accumulates the area under the error curve, it wil gradually add
to the output signal, in a linear fashion. There will come a time when the contribution
of the integral term, equals the proportional contribution. The time it takes for this to
happen is called the repeat time, as it indicates the time it takes the integral action to
‘repeat’ the action of the proportional action.
It is immediately clear from Figure 6-16 that the integral and proportional effects are
equal when 10KITR = 10KP or KITR = KP. The repeat/reset time is therefore given by:
TR =
K
K
P or K = P ………..……….………..………….. Equation 6-5
I
K
T
I
R
Controller tuning
In a practical situation, different values for KP and KI would be chosen for the water
level control system so that the water level would be corrected much faster after a
disturbance. Choosing suitable values for KP and KI is called tuning of the controller.
Guidelines that help with controller tuning do exist and even self tuning systems are
available. One such rule of thumb suggests the following settings for a liquid level
control: PB < 100 % and reset time of 10 minutes. If we should choose a PB of 80 %
(KP = 1.25), then a reset time of 10 minutes (600 sec.) would translate into an integral
gain of KI = 0.002083 (from Equation 6-5: KI = KP/TR = 1.25/600). Using these values
for KP and KI, the plots of the water level M and controller output C, shown in
Figure 6-17, were obtained with Matlab. The water level M drops to 45% and returns
to setpoint within 25 minutes.
M(%)
51
C(%)
70
50
65
49
60
48
47
55
46
45
0
500
1000 1500 2000 2500 3000 3500 4000
a) Water level
t (sec)
50
0
500
1000 1500 2000 2500 3000 3500 4000
t (sec)
b) Controller output
Figure 6-17
System response to demand change from 50% to 60% with PI control (KP=1.25, KI =0.002083)
Integral control: A control strategy in which the controller output is proportional
to the integral of the error.
∫
⎛
{For PI control: C = KPE + KI Edt + R = KP ⎜ E +
⎜
⎝
⎞
1
Edt ⎟ + R}
⎟
TR
⎠
∫
Reset or repeat time: Time taken for the integral control action to equal the
proportional control action under the influence of a constant error. {TR = KP/KI}
EIPINI Chapter 6: Process Control Page 6-19
6.2.4 PROPORTIONAL AND DERIVATIVE CONTROL
Derivative or differential control action (also called rate control or pre-act control)
gives an output which is proportional to the derivative of the error. Because derivative
action depends on the slope of the error curve, it has no effect if the error is constant.
Derivative control action therefore only adds to the controller output when the error
changes, that is when the set point or the measured value changes. Derivative control
can thus not be used alone because the controller would not combat the error if it is not
changing, even when the error is large. Derivative control action is usually found in
combination with proportional control to form a PD controller. The output of a PD
controller can be expressed as:
dE
C = KPE + KD
+ R, …………….…………….…………… Equation 6-6
dt
where KD is defined as the derivative gain. The basic effect of derivative control is to
oppose any change in the error value (if the water level M, in our example, would
drop, dM/dt would be negative and dE/dt = d/dt(S-M) would be positive and the
inflow C would immediately increase). As such, derivative action will always have a
stabilizing effect on the system by combating aggressive and unstable behaviour.
Derivative control reacts to changes in the error and starts with corrective action much
earlier than the proportional (and integral) action. Derivative action was in fact named
‘pre-act’ because of its ability to ‘anticipate’ the future movement of the measured
variable and the direction in which the process is heading. A PD controller, however,
does not have the capability to eliminate the offset and will also add to the wear on the
final control element such as a control valve.
Rate time
Commercial PD controllers normally do not directly allow for the adjustment of the
derivative gain KD, but rather the adjustment of the rate time. Imagine a linear error
signal that gradually changes from 0 to 10 % in TD seconds, as shown in Figure 6-18.
Error E
Figure 6-18
10%
Rate time
10
0%
Time
TD
PD controller output
Proportional effect = KP×E = KP×10
dE
C=KPE+KD
10
dE
dt
Derivative effect = KD×
= KD×
dt
TD
50%
Time
TD
If this ramp error signal is presented to the input of a PD controller, that initially had
an output of 50 %, there would be a sudden increase in controller output because the
derivative term, KD
dE
, immediately takes on a value proportional to the slope of the
dt
error curve. Furthermore, as the proportional term, KPE, follows the error value, it will
increasingly add to the output signal, in a linear fashion. The time, TD, that it takes the
EIPINI Chapter 6: Process Control Page 6-20
contribution of the proportional term to equal the derivative contribution is called the
rate time, and it gives an indication of how much faster the derivative action is
compared to the proportional action.
It is immediately clear from Figure 6-18 that 10×KITR = 10×KP = 10×KD/TD or
KPTD = KD. The rate time is therefore given by:
K
TD = D or KD = KPTD ……..……….…..………………….. Equation 6-7
KP
Derivative control: A control strategy in which the controller output is
proportional to the derivative of the error.
dE ⎞
dE
⎛
{For PD control: C = KPE + KD
+ R = KP ⎜ E + TD
⎟ +R}
dt
dt ⎠
⎝
Rate time: Time taken for proportional action to equal derivative action under the
influence of a linearly changing error. {TD = KD/KP}
6.2.5 PROPORTIONAL, INTEGRAL AND DERIVATIVE CONTROL
A proportional, integral and derivative controller (PID controller), gives an output that
is determined by proportional action, integral action and derivative action.
Mathematically the output of a PID controller can be expressed as:
∫
C = KPE + KI Edt + KD
dE
+ R. …………….……………… Equation 6-8
dt
Controller tuning
Adding integral and derivative action to a controller increases the complexity of the
system. Choosing suitable P, I and D gains for a PID controller (tuning the controller),
becomes complicated. Even for the very simple water level control system in Figure
6-9, analysis of the complete PID system is difficult. We will however conclude with
one example of PID control of the water level control system. The controller gains
chosen were KP = 5, KI = 0.08 (reset time = 1.04 min.) and KD = 500 (rate time = 1.7
min.). The plots in Figure 6-19, were obtained with Matlab. The response curves in
Figure 6-19, show the water level (although oscillating), not even dropping below
49% and returning back to setpoint in less than seven minutes.
M(%) 50.6
C(%)
65
50.4
50.2
50
60
49.8
49.6
49.4
49.2
0
500
1000 1500 2000 2500 3000 3500 4000
a) Water level
Figure 6-19
t (sec)
55
0
500
1000 1500 2000 2500 3000 3500 4000
b) Controller output
System response to demand change from 50 % to 60 %
with PID control (KP = 5, KI = 0.08 and KD = 500)
t (sec)
EIPINI Chapter 6: Process Control Page 6-21
6.3 PID CONTROLLERS
6.3.1 Pneumatic PID controller
One may justifiably ask
whether pneumatic controllers
still have a place in a digital
age. The answer is yes and
some of the reasons that can be
put forward are familiarity,
simplicity, a large installed
base, no fear of electricity and
above all, they are inherently
explosion-proof.
Manual
A typical front panel of a
pushbutton
pneumatic controller is shown
(decrease
output)
in Figure 6-20. Depending on
the manufacturer, controllers
Manual
will have slightly different
button with
configurations of pushbuttons,
indicating
meters and switches.
light
100
90
80
70
60
50
40
30
20
10
0
MV
Set point
thumbwheel
SP
OUTPUT
0
50
CLOSE
MANUAL
100
OPEN
AUTOMATIC
Manual
pushbutton
(increase
output)
Automatic
button with
indicating
light
Figure 6-20
Pneumatic implementation of the PID function is achieved in a variety of ways using
diaphragms and valves in different mechanical designs. One possible arrangement
revolves around the force balance principle utilizing a flapper-nozzle and bellows
assembly. The pressure variable that represents the measured value is normally
connected to the force bar (beam) by means of a bourdon tube or a bellows element.
The set point value, the bias value and the proportional band settings are adjusted in
several ways by different manufacturers using various mechanical or pneumatic
techniques.
i) Proportional control
The controller arrangement for proportional control is shown in Figure 6-21. For the
purpose of this illustration, we will assume that all variables are applied to the force
balance system by means of bellows elements and that the pivot is positioned at a
distance ‘a’ units from the S/M bellows and at a distance ‘b’ units from the C/R
bellows. The position of the proportional bellows in Figure 6-21 will result in reverse
acting control because a decrease in measured value will cause the flapper to move
towards the nozzle and balance will be restored at a higher controller output pressure.
Given that the flapper and nozzle will always maintain equilibrium, the controller
operation can very easily be described by equating moments around the pivot point,
assuming that all the bellows elements have a cross-sectional area of one unit.
a×S + b×R = a×M + b×C ⇒ bC = a(S – M) + bR ⇒ C =
a
(S – M) + R
b
Using E = S-M and KP = (a/b), it follows that C = KPE + R.
The device therefore provides a proportional control output with the proportional
gain determined by the position of the pivot point.
EIPINI Chapter 6: Process Control Page 6-22
Pilot
relay
Set point S
(20-100 kPa)
50% = 60 kPa
Restriction
Proportional
bellows
S
Controller
output C
Air
supply
C
Beam
a
Pivot
M
b
Reset
bellows
R
Measured
value M
(20-100 kPa)
Flapper and
nozzle
Figure 6-21
Pneumatic
proportional
controller
Bias value R
(20-100 kPa)
ii) Proportional plus integral control
With proportional control, the fixed bias value R that is applied to the reset bellows, is
called ‘manual reset’ because one way to reset the controlled variable back to set point
in a proportional only control system, is to manually change the bias value. In order to
change the bias value automatically with the aim to reset the controlled variable
automatically to set point, the reset bellows must now obtain its pressure from the
controller output pressure C. To achieve automatic reset the output pressure must be
applied via an adjustable needle valve restriction, as shown in Figure 6-22.
C
Pilot
relay
Set point S
(20-100 kPa)
50% = 60 kPa
Restriction
Proportional
bellows
S
Controller
output C
Air
supply
C
Beam
a
Pivot
M
Measured
value M
(20-100 kPa)
b
R
Reset
bellows
Reset time
adjustment
(r)
C
Flapper and
nozzle
Automatic reset R
Needle
valve
R
Figure 6-22
Pneumatic
proportional
plus
integral
controller
EIPINI Chapter 6: Process Control Page 6-23
The restriction will prevent the pressure R in the reset bellows to immediately
follow the output pressure C and causes a delay in the pressure build up (or decay)
in the reset bellows which will depend on the needle valve setting. This delay in
reset pressure will cause integral action to be added to the controller and it can be
∫ (S - M)dt . Using E = S–M,
a
ra
= and K = , this equation then becomes C = K E + K Edt .
∫
b
b
shown that the controller output is C =
KP
a
ra
(S – M) +
b
b
I
P
I
The controller thus provides a proportional plus integral output. The dynamic
integral term effectively replaces the fixed bias with an ‘automatic bias’. Commercial
controllers do however allow a fixed bias to be enabled for PI control, if so required.
iii) Proportional plus derivative control
Derivative action may be added to the proportional controller in Figure 6-21, by
inserting a variable restriction in the pressure line that feeds the proportional bellows,
as shown in Figure 6-23. The proportional bellows now receives a pressure Pd that is a
delayed version of the output pressure C.
Pd
C
Needle valve
Rate time
adjustment
(d)
Set point S
(20-100 kPa)
50% = 60 kPa
Pilot
relay
Restriction
Proportional
bellows
S
Controller
output C
Air
supply
PD
Beam
a
M
Measured
value M
(20-100 kPa)
Pivot
b
Reset
bellows
R
Flapper and
nozzle
Bias value R
(20-100 kPa)
Figure 6-23
Pneumatic
proportional
plus
derivative
controller
Again it can be shown that for the configuration in Figure 6-23, the controller output
a d(S - M)
a
a
a
is given by C = (S – M) +
+ R. Using E = S - M, KP = and KD =
,
b
bd
dt
b
bd
dE
this equation then becomes C = KPE + KD
+ R.
dt
It is now clear that inserting a delay in the proportional pressure line, causes
derivative action to be added to the proportional action.
Adding delay actions simultaneously in both the reset and the feedback line, will
result in a complete PID controller and such a structure is shown in Appendix 6-2.
EIPINI Chapter 6: Process Control Page 6-24
6.3.2 Electronic PID controller
Figure 6-24 shows an electronic implementation of the basic PID function. The first
operational amplifier generates the error signal from the reference S and the measured
value M. The error signal is fed to the P, I and D blocks to create the three PID
functions which are added together to form the final PID control function C.
Proportional
RPF action
E
RPI
VP
Op-amp
Gnd
100k
R
VP = - PF E
R
PI
= -KPE
Error
M
100k
100k
CI
Op-amp
S
Integral
action
E
E
VI
RI
Op-amp
E=S-M
100k
Adder
100k
100k
C
Op-amp
100k
Gnd
Gnd
1
Edt
R C
I I
= -KI Edt
∫
VI= -
∫
Gnd
C = - VP – VI – VD
∫
=KPE + KI Edt + KD
RD
Derivative
action
P
I
dE
dt
D
E
CD
Op-amp
VD
dE
dt
dE
= -KD
dt
VD= -RDCD
Gnd
100k
Figure 6-24
Electronic PID
controller
EIPINI Chapter 6: Process Control Page 6-25
6.3.3 Digital PID controller
Digital or software implementation of the PID control algorithm uses a digital
computer to calculate the control action as shown in Figure 6-25. Computers are
discrete machines and not capable of calculating the control action on a continuous
basis as the analog structure in Figure 6-24. It will instead try to approximate Equation
6-8, using numerical methods. Therefore the measured value M must be sampled with
sufficiently small sampling period Ts and digitised with an analog to digital converter
to deliver the sampled values M0, M1, M2, M3, M4, etc. to the computer at time instants
0, 1, 2, 3, 4. At each time instant, the proportional, integral and derivative component
must be calculated and added together to form the sequence of corresponding
controller outputs C0, C1, C2, C3, C4, …..
Analog to digital
converter
M
Digital to analog
converter
Computer
TS
M0
C
TS
C3
C2
M1 M
2 M
3 M
4
C4
C1
C0
Figure 6-25
For instance, referring to Figure 6-26, at time instant 4, just after M4 was sampled,
the error E4 may be obtained from 50–M4, given that the set point is 50%. From E4 the
proportional action can be calculated as P4 = KP×E4. If we assume that the controller
started its operation at time equals zero, then a very simple method to calculate the
integral component I4 is to approximate the area A, under the error curve with
4×T
S Edt ≈ T ×E + T ×E + T ×E + T ×E . The integral
s
0
s
1
s
2
s
3
0
rectangles to obtain A= ∫
action I4 may then be expressed as I4 = KI×(Ts×E0 + Ts×E1 + Ts×E2 + Ts×E3) or
I4=KI×Ts×(E0+ E1+E2+E3). The derivative component D4, depends on the slope of the
E −E
error curve at time instant 4, and may be approximated with D4 = KD 4 3
T
s
M
E
S=50%
E
E1
M1
E3
M2 M3
1
Ts
E1
E2
M0
0
E3
E2
E0
Ts
Ts
E1×Ts
4 Time 0
Ts
P
E3×Ts
E4
E3
E0×Ts
C4 = KPE4 +
Figure 6-26
E0
M4
3
2
E4
E2×Ts
1
Ts
2
Ts
4 Time 0
3
Ts
Ts
KI×TS ×(E0 + E1 + E2 + E3)
I
1
Ts
+
2
Ts
4 Time
3
Ts
Ts
E −E
3
KD× 4
Ts
D
EIPINI Chapter 6: Process Control Page 6-26
6.4 CONTROL VALVES
Control valves are used to regulate the flow rate of a medium and
serve as the correcting element in many control systems.
A pneumatic control valve consists of a diaphragm actuator and a
valve body with plug and seat through which the fluid is
regulated. The principle of operation of a control valve is very
simple. A diaphragm is bolted to a dished metal head to form a
pressure tight compartment. A pneumatic signal is applied to this
compartment to move the diaphragm which is opposed by a
spring. Attached to the diaphragm is the valve stem and plug so
that any movement of the diaphragm, results in a corresponding
movement of the plug, thereby controlling the flow. A typical
reverse acting control valve assisted by a valve positioner (section
6.4.4) to enhance valve operation, is illustrated in Figure 6-28.
6.4.1 Actuators
Actuators may be categorized
as ‘direct acting’ or ‘reverse
acting’ and some configurations
are indicated in Figure 6-27. In
a reverse acting actuator an
increase in the pneumatic
pressure
applied
to
the
diaphragm lifts the valve stem
(in a normally seated valve this
will open the valve and is
called ‘air to open’). In a directacting actuator, an increase in
the pneumatic pressure applied
to the diaphragm extends the
valve stem (for a normally
seated valve this will close the
valve and is called ‘air to
close’). The choice of valve
action is dictated by safety
considerations. In one case it
may be desirable to have the
valve fail fully open when the
pneumatic supply fails. In
another application it may be
considered better if the valve
fails fully shut.
Reverse acting
Direct acting
Figure 6-27
EIPINI Chapter 6: Process Control Page 6-27
Spring tension adjust
(Valve closed at 20 kPa)
Spring
Diaphragm
plate
Actuator
(Motor)
Diaphragm
Air
supply
Stem connector
(stroke adjust
20 – 100 kPa)
Open
Output
½
Travel
indicator
Close
Stem
Valve
positioner
Instrument
signal
(20-100 kPa)
Connector
arm
Yoke
Gland and
packing
Bonnet (yoke
hold down) nut
Bonnet
Gasket
Plug
Valve
body
Seat
Figure 6-28
Reverse acting (air to open) pneumatic control valve
EIPINI Chapter 6: Process Control Page 6-28
6.4.2 Valve types
The essential function of a valve is to start, throttle and stop a fluid flow. Throttling,
as shown in Figure 6-29, means regulating or controlling the rate of fluid flow. Globe
valves, gate valves, needle valves, pinch valves and diaphragm
valves use a linear stem movement to operate. A gate valve is
illustrated in Figure 6-30 (a) as an example of linear stem
movement. Ball valves, plug valves and butterfly valves function
with a 90° rotational movement and the operation of a ball valve is
indicated in Figure 6-30 (b). The valve components inside the
valve body are collectively referred to as the trimming.
Closed
Throttling
Open
Figure 6-29
Ball valve
Gate valve
Figure 6-30 (a)
Figure 6-30 (b)
Linear valve movement
Rotational valve movement
EIPINI Chapter 6: Process Control Page 6-29
1. Globe valves
Globe valves are the most widely used in industry for flow control in both on/off and
throttling service. They typically have rounded bodies from
which their name is derived. They are linear motion valves
with a tapered plug or disk attached to the stem that closes onto
a seating surface to act as flow control element, as shown in
Figure 6-31. The liquid must make two 90° turns when passing
through the valve and because of that, the pressure drop in the
globe valve is significant, even when fully open. The high
Figure 6-31
pressure drop is the main disadvantage of the globe valve.
Globe valve
2. Gate valves
Gate valves (also known as knife valves or slide valves) are the most common valve
used for on/off service but could be used for throttling. They are linear motion valves
and a flat disk or wedge slides into the flow stream
Parallel
to act as flow control element, as shown in Figure
disk
6-32. The direction of fluid flow is not changed by
the valve. When fully open, the wedge completely
clears the flow path creating minimum pressure drop.
Gate valves are advantageous in applications
Wedge
involving slurries, as their ‘gates’ can cut right
through the slurry. They are also used in applications
Figure 6-32
that involve viscous liquids such as heavy oils, light
Knife
Gate valve
grease, varnish, molasses, honey and cream.
3. Needle valves
A needle valve, shown in Figure 6-33, is used to make relatively
fine adjustments to the fluid flow and can be used for on/off and
throttling service. They are linear motion valves with a tapered
‘needle like’ cone shaped plug that acts as the flow control
element. The fluid enters from below into the seat which forms
part of the flow path. The needle plug permits the flow opening
in this channel to be increased or decreased very gradually.
Needle valves are widely used for steam, air, gas, water or other
non-viscous liquids. Disadvantages of the needle valve are a
large pressure loss and possible clogging of the flow orifice.
Figure 6-33
Needle valve
4. Pinch valves
The relatively inexpensive pinch valve, illustrated
In Figure 6-34, is the simplest of all the valve
designs and may be used for on/off and throttling
service. Pinch valves are linear motion valves
that use a flexible tube or sleeve to effect valve
closure. Pinch valves are ideally suited for the
handling of slurries, liquids with large amounts
of suspended solids and corrosive chemicals.
Figure 6-34
Pinch valve
EIPINI Chapter 6: Process Control Page 6-30
5. Diaphragm valves
A diaphragm valve, illustrated in Figure 6-35, is related to pinch
valves and one of the oldest types of valve known. Leather
diaphragm valves were used by Greeks and Romans to control
the temperature of their hot baths. Diaphragm valves use linear
stem movement for on/off and throttling and are excellent for
controlling fluid flow containing suspended solids. A stud moves
a flexible and resilient diaphragm into the flow thereby acting as
flow control element. The diaphragm valve is used primarily for
handling viscous and corrosive fluids as well as slurries.
Figure 6-35
Diaphragm valve
6. Plug valves
Plug valves, illustrated
in Figure 6-36, are also
called plug cock or stop
cock valves and they
also date back to ancient
Close
Throttle
Open
Figure 6-36
times, when used by Romans in plumbing systems. Today they
Plug valve
remain one of the most widely used valves for both on/off and
throttling services. A plug valve is a rotary moving valve that uses a cylindrical or
tapered plug as the flow control element. The opening through the plug may be
rectangular or round. Flow is regulated from closed to open during a 90° turn. If the
opening is the same size or larger than the pipe’s inside diameter, it is referred to as a
full port otherwise as standard round port. Full port plug valves offer little resistance
to flow when fully open, resulting in small pressure loss. A disadvantage off plug
valves is its poor throttling characteristics.
7. Ball valves
Ball valves, illustrated in Figure 6-37, are related to plug valves
and are used in situations where tight shut-off is required. Ball
valves are rotational motion valves and are used for on/off and
throttling service. The flow control element is a sphere with a
round opening rotating in a spherical seat. Flow is regulated
from closed to open during a 90° turn. Full port ball valves
create a minimum pressure loss when fully open. A disadvantage
of the ball valve is its poor throttling characteristics.
8. Butterfly valves
A butterfly valve, illustrated in Figure 6-38, is a rotary
movement valve used for on/off flow control and especially
in throttling applications. The flow control element in
butterfly valves is a circular disk with its pivot axis at right
angles to the direction of flow. A 90° turn of the axis moves
the valve from closed to open. Butterfly valves are well
suited to handle large flows of liquids or gasses as well as
slurries and liquids with large amounts of suspended solids.
Figure 6-37
Ball valve
Figure 6-38
Butterfly valve
EIPINI Chapter 6: Process Control Page 6-31
6.4.3 Valve characteristics
In essence a valve simply functions as a restriction in a flow line,
p1
with flow areas A1 and A2 and pressure differential p1 - p2, as shown
p A2
q A1 2
in Figure 6-39. As such, the flow equation (Equation 3-8(e),
Chapter 3) will in principle describe the flow rate of a fluid through Figure 6-39
a valve. The form of this equation used by valve suppliers is:
q=C
Δp
Equation 6-9
G
where C is called the valve flow coefficient, q the flow rate of the liquid through the
valve, Δp = p1 - p2 the pressure difference across the valve and G the specific gravity
(relative density) of the fluid. For historical reasons, the flow rate q, in Equation 6-9, is
measured in gallons per minute (gpm) and the pressure difference Δp, in pounds per
square inch (psi). When we compare Equations 3-8(e) and 6-9, it becomes clear that C
is fundamentally determined by the effective flow areas A1 and A2 of the valve. The
value of C will therefore change from zero (when the valve is fully closed) to a
maximum value (when the valve is fully opened).
Assume that a valve is opened a certain amount x (with x denoting the fractional
valve opening or valve travel between 0 and 1). The flow rate q, of the liquid (we will
assume water with G = 1 in Equation 6-9), flowing through the valve, is now allowed
to increase until the pressure Δp across the valve, reaches a maximum of say 4 psi. We
may conceivably expect the results depicted in Figure 6-40, shown for x = 0 (valve
fully closed), x = 0.25 (valve one quarter open), x = 0.5 (valve half open), x = 0.75
(valve three quarters open) and x = 1 (valve fully open).
25% open
x=¼
0% open
x=0
q
q
50% open
x=½
q
75% open
x=¾
100% open
x=1
q
q
Δp
Δp
Δp
Δp
Δp
q=0× Δp
(C=0)
q=10× Δp
(C=10)
q=20× Δp
(C=20)
q=30× Δp
(C=30)
q=40× Δp
(C=40)
q (gpm)
q (gpm)
q (gpm)
q (gpm)
q (gpm)
80
60
40
20
0
4
Δp
psi
0
4
Δp
psi
0
4
Figure 6-40
Δp
psi
0
4
Δp
psi
0
4
Δp
psi
EIPINI Chapter 6: Process Control Page 6-32
The special value of C that corresponds to the valve fully open, is called the
characteristic flow coefficient of the valve and denoted by CV. The characteristic flow
coefficient CV, is an extremely important parameter of a specific valve, and is used
extensively when choosing the correct valve for a certain application. For the valve
characteristics in Figure 6-40, for example, CV = 40.
The way q varies when the pressure is kept constant (normally the rated pressure
drop across the valve for maximum flow) and the way C changes as the valve opening
x changes, may be available for a particular valve, provided by the manufacturer in
tabulated format. For example, the graph of q (with Δp = 4 psi) as a function of x and
the graph of C as a function of x, are shown in Figure 6-41 (a) and (b) respectively, for
the flow characteristics of our hypothetical valve, depicted in Figure 6-40.
Figure 6-41(a)
q (gpm)
80
Figure 6-41(b)
C
40
60
30
40
20
Valve
opening
x
20
0
¼
½
¾
1
(0%) (25%) (50%) (75%) (100%)
Valve
opening
x
10
0
¼
½
¾
1
(0%) (25%) (50%) (75%) (100%)
The different values of C obtained as the valve travels through its full range (or
stroke as it is also called) from closed to open, is generally expressed in the following
format, by valve manufacturers:
C = CV×f(x)
Equation 6-10
where f(x) is called the inherent valve characteristic of a particular valve. For
example, from the graph of C, in Figure 6-41(b), we could express C as:
C = 40×x,
and we conclude therefore from Equation 6-10, that CV = 40 and f(x) = x.
The function f(x) varies from 0 (valve closed with x equal to 0) to 1 (valve fully
open with x equal to 1). The inherent valve characteristic f(x), is intimately linked to
the flow rate q through a valve, as we can see if we replace C with CVf(x) (as per
Equation 6-10), to rewrite the valve Equation 6-9 in the form:
q = CVf(x)
ΔP
G
Equation 6-11
It is indeed clear from Equation 6-11 that if we keep Δp constant (for a given liquid, G
is constant as well), q and f(x) will have exactly the same shape.
EIPINI Chapter 6: Process Control Page 6-33
A valve for which f(x) = x, is called a linear valve because the flow rate will change
linearly with valve opening x. Depending however on the valve design, f(x) may also
reflect quick opening or equal percentage (slow opening) characteristics, as shown in
Figure 6-42.
q
100%
1
Graphs also
represent
flow rate q
(% of max.
flow) if Δp
is constant.
0%
Quick
opening
f(x)
Quick
opening
Figure 6-42
Inherent valve
characteristics
Linear
0
Equal
percentage
0
(0% open)
(Ratio of
valve travel
x to maximum
valve travel)
Linear
Equal
percentage
1
(100% open)
The flow behaviour of a valve is dictated by the manner in which the flow areas A1
and A2 changes with valve position and therefore by the style and design of the valve
trimming and in particular the design of the valve seat and closure member (plug). The
quick opening flow characteristic provides for maximum change in flow rate at low
valve travels with a nearly linear relationship. Additional increases in valve travel
gives sharply reduced changes in flow rate. The linear flow characteristic curve allows
the flow rate to be directly proportional to the valve travel (Δq/Δx equals a constant) or
in terms of the inherent valve characteristic, f(x) = x. An equal percentage valve starts
initially with a slow increase in flow rate with valve position which dramatically
increases as the valve opens more.
The term equal percentage for a slow opening characteristic curve may at first be
confused with the description of a linear characteristic curve. However, for an equal
percentage valve, Δq/Δx at any stage, is proportional to the flow rate q at that moment.
This is in contrast with a linear characteristic for which Δq/Δx is constant. That Δq/Δx
is proportional to q, may be rephrased as Δq/q is proportional to Δx. This means the
percentage change Δq with respect to the current flow rate q (that is (Δq/q)×100), is
equal at every valve travel position x for the same change in valve travel Δx, hence the
term ‘equal percentage’. The inherent valve characteristic for an equal percentage
valve is exponential in nature and is normally given by valve manufacturers in the
x-1
form f(x) = R , where R is a constant for the valve. The exponential behaviour of an
equal percentage valve is explored further in Example 6-7.
Example 6-7
An equal percentage valve delivers 8 gpm of water when the valve is 50% open
(x=0.5). When the valve is 60% open (x=0.6), the flow rate increases to 16 gpm.
EIPINI Chapter 6: Process Control Page 6-34
Estimate the flow rate through the valve when it is 70% open (x=0.7). Assume that the
pressure drop across the valve remains constant.
Answer: When the valve opens from
q (gpm.)
50% to 60%, the flow rate changes
from 8 gpm to 16 gpm. This means
=%
that Δq is 8 gpm. when Δx is 0.1 and
q = 8. Therefore Δq/q = 8/8 = 1 and
the percentage change is 100%
?
(Δq/q×100) at x = 0,5. For an equal
Δq=?
percentage valve, the percentage 16
Δq=8
x
change in flow rate when the valve
8
0
opens from 50% to 60%, (Δx = 0.1)
0
.5 .6 .7
1
must be equal to the percentage
change in flow rate when the valve
Δx=.1 Δx=.1
opens from 60% to 70% (the same
Δx of 0.1). Therefore Δq/q at x = 0.6 must also be 1 (or 100%) for Δx = 0,1. But q is
16 gpm. when the valve is 60% open and we conclude that Δq should be 16 gpm.
when the valve opening changes from 60% to 70%. The flow rate through the valve is
therefore approximately 32 gpm. at 70% valve opening.
Typical application of quick opening, linear and equal percentage valves
i) Quick opening valve:
a) Frequent on-off service.
b) Used for systems where ‘instant’ large flow is needed (safety or cooling
water systems).
ii) Linear valve:
a) Liquid level and flow control loops.
b) Used in systems where the pressure drop across the valve is expected to
remain fairly constant.
iii) Equal percentage valve (most commonly used valve):
a) Temperature and pressure control loops.
b) Used in systems where large changes in pressure drop across the valve are
expected.
Installed flow characteristic
When valves are installed with pumps, piping and fittings, and other process
equipment, the pressure drop across the valve will vary as the valve travel changes.
When the actual flow in a system is plotted against valve opening, the curve is called
the installed flow characteristic and it will differ from the inherent valve characteristic
which assumed constant pressure drop across the valve. When in service, a linear
valve will in general resemble a quick opening valve while an equal percentage valve
will in general resemble a linear valve.
EIPINI Chapter 6: Process Control Page 6-35
Valve sizing
The characteristic flow coefficient CV of a valve and the inherent valve characteristic
f(x), completely describe the flow-pressure profile of the valve and play an important
role during the valve selection or valve sizing process. Although the characteristic
flow coefficient (CV) is the most fundamental and important parameter influencing the
user when selecting a valve, it is not the only consideration. Conditions such as piping
particulars, fluid type, laminar or turbulent flow, may be included as additional factors
in Equation 6-9. Nevertheless, to select an appropriate valve for a certain
implementation, the user must specify the maximum flow rate required and the
pressure drop expected with the valve fully open. With this information, the necessary
characteristic valve flow coefficient CV, can be calculated from Equation 6-9 by
setting C = CV, q = QRated (rated maximum flow rate with valve fully opened) and
Δp = ΔPFull (pressure across fully opened valve at maximum rated flow rate):
CV = Q
Rated
ΔP
Full ………………...…………………. Equation 6-12
G
Example 6-8
A system is pumping water from one tank to another through a piping system with
total pressure drop of 150 psi. The maximum design flow rate is 150 gpm. Calculate
the characteristic flow coefficient required for a control valve to be used in this system.
Solution: The usual rule of thumb is that a valve should be designed to use 10-15% of
the total pressure drop or 10 psi, whichever is greater, when fully open. For this
system 10% of the total pressure drop is 15 psi, which is what we must use. From
Equation 6-12 (remember that specific gravity G is the same as relative density or in
other words, G = δ):
CV = QRated / ΔPFull /G = 150/√(15/1) = 38.72 ≈ 39
(Note: CV is normally given as a dimensionless number but it really is a dimensional
quantity with units [gallon-inch/minute-poundforce½] from Equation 6-12.)
Example 6-9
The user in Example 6-8 has a choice between a valve with CV = 35 which would be
too small and the next one with CV = 45 which could be too large. So he decides to
check what the valve travel positions will be for the larger valve if he wants to control
the water flow between a maximum flow rate of 150 gpm and a minimum flow rate of
40 gpm. He expects a pressure drop of 15 psi across the valve at full flow and a rise in
pressure drop to 25 psi, when the flow rate drops to 40 gpm. The valve he intends to
purchase is a linear valve with inherent valve characteristic, f(x) = x.
Solution: From Equation 6-11 (q = CVf(x) ΔP/G ),
with qmax = 150 gpm. and Δpqmax = 15 psi: 150 = 45×x × 15/1
∴x = 150/(45×√15) = 0.8607 (86%)
with qmin = 40 gpm. and Δpqmin = 25 psi: 40 = 45×x × 25/1
∴x = 40/(45×√25) = 0.1778 (18%)
This seems to be good enough, as the aim should be to keep valve movement between
20% and 80% of maximum travel.
EIPINI Chapter 6: Process Control Page 6-36
6.4.4 Valve positioners
A 20-100 kPa controller signal travelling perhaps a long distance to a control valve to
exercise the required control action, may possibly not be powerful enough to quickly
move a large valve to its new position. It may therefore be advantageous to allow the
controller signal to control the air supply locally at the valve rather than to operate the
valve by itself. The device that may be added to enhance the controller signal to a
valve, is called a valve positioner. A valve positioner is essentially a pneumatic
controller that senses the valve stem position, compares it to the incoming controller
signal and adjusts the pressure to the actuator until the stem position corresponds to
the controller signal. Positioners can be used to overcome stem friction, high fluid
pressure, viscous or dirty fluids, or to improve slow system dynamic response.
Valve actuator
Elastic
Forcebalance
beam
Pivot
Cam
Valve stem
Figure 6-43 Valve positioner
Instrument
bellows
Controller signal
(Instrument signal)
20-100 kPa
Flapper and
nozzle
Actuator
output
Pilot
relay
Air supply
Restriction
Figure 6-43 shows a typical pneumatic valve positioner, assisting a reverse acting
valve. The operation revolves around a force balance flapper and nozzle
arrangement. The valve stem position is communicated to the force balance beam
by means of a lever and cam. If the controller signal increases, with the intention to
open the valve more, the instrument bellows will push the flapper (a flexible plate)
towards the nozzle. This will increase the valve actuator pressure and the valve
stem will start to move upwards. This will turn the cam which will subsequently
push the flapper away from the nozzle, until balance is reached so that the valve
remains in the new more opened position. If the controller signal decreases, the
flapper will move away from the nozzle and the actuator pressure will be reduced.
The valve will begin to close and the valve stem will begin to move downwards,
relieving cam pressure on the beam that kept the flapper away from the nozzle. The
flapper will tend to move towards the nozzle thereby restoring balance with the
valve more closed than before.
EIPINI Chapter 6: Process Control Page 6-37
APPENDIX 6-1
A water container is 5 meter high and has a base area of 1 meter2. The rate, measured
in cubic meter per second, at which water flows into the container at any instant, is
QIN. The rate, measured in cubic meter per second, at which water flows out of the
container at any instant, is QOUT. The level, measured in meter, of the water in the
container at any instant, is H. The maximum value of QIN and QOUT, is 0.01 cubic
meter per second, and the maximum value of H is 5 meter.
a) Express H in terms of a percentage M, of the maximum level.
b) Express QIN in terms of a percentage C, of the maximum inflow.
c) Express QOUT in terms of a percentage L, of the maximum outflow.
d) Calculate the volume V, of the water in the container, in terms of M.
e) Calculate the resultant rate Q, at which water accumulates into the tank or drains
out of the container. Express your answer in terms of C and L.
ΔV
, at which the volume changes, in terms of C and L.
Δt
ΔM
, at which the water level changes, in terms of C and L.
g) Calculate the rate
Δt
C (inflow in %)
0.01m3/s
Solution:
Qin (inflow in m3/s)
Max
M
f) Calculate the rate
×5 = 0.05×M meter …...……… (1)
100
5m max
C
3
M
(%)
×0.01 = 0.0001×C m /s ….….. (2)
Qin =
100
H (m)
L
3
×0.01 = 0.0001×L m /s ….…. (3)
Qout =
L (outflow in %)
0.01m3/s
100
3
Qout (outflow in m /s)
max
Volume V, of water in tank:
V = Height×Area = H×1
Q=0.0001×(C-L)
C
From (1): V = 0.05×M×1
∴V = 0.05×M cubic meter ……..…..….. (4)
Resultant rate Q = Qin - Qout
M V=0.05M
From (2) and (3):
∴Q = 0.0001×C–0.0001×L
L
Area = 1 m2
∴Q = 0.0001(C – L) m3/s ….....……....... (5)
The rate at which the water volume in the container increases or decreases, is the
same as the rate at which water is accumulated in or drained from the tank.
a) H =
b)
c)
d)
e)
f)
∴
ΔV
ΔV
= Q. Therefore from (5):
= 0.0001(C - L) cubic meter per second.
Δt
Δt
∴ΔV = 0.0001(C - L)×Δt meter3 ….….... (6)
g) From (4): V = 0.05×M
∴ΔV = 0.05×ΔM meter3 ……….…......... (7)
Substituting ΔV from (7) into (6): 0.05×ΔM = 0.0001(C – L)×Δt
ΔM 0.0001
ΔM
dM
for
∴
=
×(C – L), or, using the proper notation
,
Δt
0.05
dt
Δt
dM
= 0.002×(C – L) percent per second.
dt
EIPINI Chapter 6: Process Control Page 6-38
APPENDIX 6-2
Pd
Needle valve
Rate time
adjustment
(d)
Pd
Set point S
(20-100 kPa)
50% = 60 kPa
Proportional
bellows
S
C
Pilot
relay
Restriction
Controller
output C
Air
supply
Pd
Beam
a
Pivot
M
Measured
value M
(20-100 kPa)
b
Reset
bellows
Reset time
adjustment
(r)
Pd
R
Flapper and
nozzle
Automatic reset R
Needle
valve
Pneumatic PID controller
R
a(r + d)
ar
a d(S - M)
(S-M) +
(S - M)dt (S-M) +
bd
b
bd
dt
a(r + d)
ar
a
Using E = S-M, KP =
, KI =
and KD =
:
bd
b
bd
∫
C=
∫
C = KPE + KI Edt + KD
dE
dt