Measurement and Control Basics 3rd Edition (complete book)
.pdfChapter 2 – Process Control Loops 
65 
EXERCISES
2.1Explain the operation of the singleloop feedback control loop shown in Figure 21.
2.2Define the term lag in the context of process control.
2.3A sensor measures temperature linearly with a transfer function of 22 mv/°C and a 0.5s time constant. Find the sensor output 1 s after the input changes rapidly from 20°C to 22°C.
2.4Show that the time constant (τ ) in an electrical resistance and capacitance circuit system has the units of time.
2.5Show that the time constant (τ ) in a thermal system has the units of time.
2.6The tank shown in Figure 211 has an operating head (h) of 5 ft and
a normal output flow of 5 ft3/s. The crosssectional area of the tank is 16 ft2. Find the system time constant (τ ).
2.7Explain the concept of dead time in a process.
2.8Determine the dead time for the process shown in Figure 215 if the temperature detector is located 5 m from the water tank and the velocity of the water in the discharge pipe is 5 m/s.
2.9The ZieglerNichols ultimate method was used to determine an ultimate sensitivity of 0.6 psi/ft and an ultimate period of 2 min for a level control loop. Determine the PID controller settings needed for a onequarter damping response.
2.10Using the ZieglerNichols openloop method, determine the PID controller settings if the following values are obtained from the
reaction curve for a temperature control loop: Lr = 1.0 min, LrRr = 15°C/psi, and KC = 26°C /psi.
BIBLIOGRAPHY
1.Chemical Engineering magazine staff. Practical Process Instrumentation and Control, New York: McGrawHill, 1980.
2.Considine, Douglas M. (ed.). Handbook of Applied Instrumentation, New York: McGrawHill, 1964.
3.The Foxboro Company. Introduction to Process Control. Foxboro Company, 1986.
4.Johnson, C. D. Process Control Instrumentation Technology, 2d ed., New York:
John Wiley & Sons, 1982.
5.Kirk, F. W., and N. R. Rimboi. Instrumentation, 3d ed., Homewood, IL: American Technical Publishers, 1975.
66Measurement and Control Basics
6.Miller, J. A., A. M. Lopez, C. L. Smith, and P. W. Murrill. “A Comparison of Controller Tuning Techniques,” Control Engineering, Dec. 1967, 7275.
7.Murrill, P. W. Fundamentals of Process Control Theory, 3d ed., Research Triangle Park, NC: ISA, 2000.
8.Ogata, K. Modern Control Engineering, Englewood Cliffs, NJ: PrenticeHall, 1970.
9.Spitzer, D. W. Regulatory and Advanced Regulatory Control: Application Techniques, Research Triangle Park, NC: ISA, 1994.
10.St. Clair, D. W. Controller Tuning & Control Loop Performance: A Primer, Newark, DE: StraightLine Control, 1989.
11.Wade, H. L. Regulatory and Advanced Regulatory Control: System Development, Research Triangle Park, NC: ISA, 1994.
12.Weyrick, R. C. Fundamentals of Automatic Control, New York: McGrawHill, 1975.
13.Ziegler, J. G., and N. B. Nichols. “Optimum Settings for Automatic Controllers,” ASME Transactions 64 (1942): 75968.
3
Electrical and Electronic Fundamentals
Introduction
Designing process control systems requires a thorough knowledge of the basic principles of electricity and electronics. This chapter will discuss the fundamentals of electricity and then examine the electrical and electronic circuits that are commonly encountered in the instrumentation and control field. We will also discuss the operation and purpose of electrical and electronic devices, such as power supplies, relays, solenoid valves, contactors, control switches, and indicating lights.
Fundamentals of Electricity
Electricity is a fundamental force in nature that can produce heat, motion, light, and many other physical effects. The force is an attraction or repulsion between electric charges called electrons and protons. An electron is a small atomic particle having a negative electric charge. A proton is a basic atomic particle with a positive charge. It is the arrangement of electrons and protons that determines the electrical characteristics of all substances. As an example, this page of paper contains electrons and protons, but you cannot detect any evidence of electricity because the number of electrons equals the number of protons. In this case, the opposite electrical charges cancel, making the paper electrically neutral.
If we want to use the electrical forces that are associated with positive and negative charges, work must be performed that separates the electrons and protons. For example, an electric battery can do such work because its chemical energy separates electrical charges to produce an excess of elec
67
68 Measurement and Control Basics
trons at its negative terminal and an excess of protons at its positive terminal.
The basic terms encountered in electricity are electric charge (Q), current (I), voltage (V), and resistance (R). For most common applications of electricity a charge of a very large number of electrons and protons is required. It is therefore convenient to define a practical unit called the coulomb (C) that is equal to the charge of 6.25 × 1018 electrons or protons. The symbol for electric charge is Q, which stands for quantity of charge, and a charge of 6.25 × 1018 electrons or protons is stated as Q = 1 C. Voltage is the potential difference or force produced by the differences in opposite electrical charges. Fundamentally, the volt is a measure of the work required to move an electric charge. When one joule of work is required to move one coulomb between two points, the potential difference is 1 volt (V).
When the potential difference between two different charges forces a third charge to move, the charge in motion is called electric current. If the charges move past a given point at the rate of one coulomb per second, the amount of current is defined as one ampere (A). In equation form, I = dQ/dt, where I is the instantaneous current in amperes and dQ is the differential amount charge in coulombs passing a given point during the time period (dt) in seconds. If the current flow is constant, it is simply given by the following equation:
I = 
Q 
(31) 

t 



where Q is the amount of current in coulombs flowing past a given point in t seconds.
Example 31 shows how to calculate current flow.
EXAMPLE 31
Problem: A steady flow of 12 coulombs of charge passes a given point in an electrical conductor every 3 seconds. What is the current flow?
Solution: Since constant current flow is defined as the amount of charge (Q) that passes a given point per period of time (t) in seconds, we have
= Q = 12 coulombs =
I 4 amps t 3 seconds
Chapter 3 – Electrical and Electronic Fundamentals 
69 
The fact that a conductor carrying electric current can become hot is evidence that the work done by the applied voltage in producing current must be meeting some form of opposition. This opposition, which limits current flow, is called resistance.
Conductivity, Resistivity, and Ohm's Law
An important physical property of some material is conductivity, that is, the ability to pass electric current. Suppose we have an electric wire (conductor) of length L and crosssectional area A, and we apply a voltage V between the ends of the wire. If V is in volts and L is in meters, we can define the voltage gradient (E), as follows:
E = 
V 
(32) 

L 



Now, if a current I in amperes flows through a wire of area A in meters squared (m2), we can define the current density J as follows:
J = 
I 
(33) 

A 



The conductivity C is defined as the current density per unit voltage gradient E or, in equation form, we have
C = 
J 
(34) 

E
Using the definitions for current density J and voltage gradient E we obtain conductivity in a different form:
C = 
I/A 

(35) 





V/L 


Resistivity (r) is defined as the inverse of conductivity, or 


r = 
1 

(36) 

C 




The fact that resistivity is a natural property of certain materials leads to the basic principle of electricity called Ohm's law.
70 Measurement and Control Basics
Consider a wire of length L and crosssectional area A. If the wire has resistivity, r, then its resistance, R, is
R = r 
L 
(37) 

A 



The unit of resistance is the ohm, which is denoted by the Greek letter omega, Ω . Since resistivity, r, is the reciprocal of conductivity, we obtain the following:
r = 
V/L 
(38) 

I/A 



When Equation 37 is substituted into Equation 36, we obtain the following equation:
R = 
V 
(39) 

I 



This relationship in its three forms — R = V/I or I = V/R or V = IR — is called Ohm's law. It assumes that the resistance of the material that is used to carry the current flow will have a linear relationship between the voltage applied and the current flow. That is, if the voltage across the resistance is doubled, the current through it also doubles. The resistance of materials such as carbon, aluminum, copper, silver, gold, and iron is linear and follows Ohm's law. Carbon is the material most commonly used to manufacture a device that has fixed resistance. This device is called a resistor.
Wire Resistance
To make it possible to compare the resistance and size of one conductor with another, the United States established a standard or unit size for conductors. The standard unit of measurement for the diameter of a circular wire or conductor is the mil (0.001 inch), and the standard unit of wire length is the foot. The standard unit of wire size in the United States is the milfoot. That is, a wire is said to have a unit size if it has a diameter of 1 mil and a length of 1 foot.
The circular mil is the standard unit of crosssectional area for wire used in U.S. wiresizing tables. Because the diameter of circular wire is normally only a small fraction of an inch, it is convenient to express these diameters in mils, to avoid the use of decimals. For example, the diameter of a 0.010inch conductor is expressed as 10 mils instead of 0.010 inch. The circular mil area is defined as the square of the mil diameter of a conductor, or area
Chapter 3 – Electrical and Electronic Fundamentals 
71 
(cmil) = d2 (mil2). Example 32 illustrates how to calculate conductor area in circular mils.
EXAMPLE 32
Problem: Calculate the area in circular mils of a conductor with a diameter of 0.002 inch.
Solution: First, we must convert the diameter in inches to a diameter in mils. Since we defined 0.001 in. = 1 mil, this implies that 0.002 in. = 2 mils, so the circular mil area is (2 mil)2 or 4 cmil.
A circularmil per foot is a unit conductor 1 foot in length that has a crosssectional area of 1 circular mil. The unit cmil per foot is useful for comparing the resistivity of various conductors. Table 31 gives the specific resistance (r) in cmilohms per foot of some common solid metals at 20°C.
Conductors that are used to carry electric current are normally manufactured out of copper because of its low resistance and relatively low cost. We can use the specific resistance (r) given in Table 31 to find the resistance of a conductor of length (L) in feet and crosssectional area (A) in cmil by using Equation 37, R=rL/A.
Table 31. Specific Resistivity.
Material 
r, resistivity, cmilΩ /ft 


Silver 
9.8 
Copper (drawn) 
10.37 
Gold 
14.70 
Aluminum 
17.02 
Tungsten 
33.20 
Brass 
42.10 
Steel 
95.80 


Example 33 illustrates how to find the resistance of copper wire with a fixed conductor area in circular mils.
The equation for the resistance of a conductor, R = rL/A, can be used in many applications. For example, if you know R, r, and A, you can determine the length by a simple mathematical transformation of the equation for the resistance of a conductor. A typical application is locating a problem ground point in a telephone line. To find ground faults, the telephone company uses specially designed test equipment. This equipment oper
72 Measurement and Control Basics
EXAMPLE 33
Problem: Find the resistance of 1,000 feet of copper (drawn) wire that has a crosssectional area of 10,370 cmil and a wire temperature of 20°C.
Solution: From Table 31, the specific resistance of copper wire is 10.37 cmilohms/ft. By inserting the known values into Equation 36, the resistance is determined as follows:
R = r 
L 



A 







1000 ft 


R = (10.37 cmil  Ω 
/ft) 




10, 400 cmil
R = 1Ω
ates on the principle that the resistance of a conductor varies in direct relation to distance, so the distance between a test point and a fault can be measured accurately with properly designed equipment.
Example 34 illustrates how to find the distance to a short to ground in a telephone line.
EXAMPLE 34
Problem: The resistance to ground on a faulty underground telephone line is 5Ω . Calculate the distance to the point where the wire is shorted to ground if the line is a copper conductor that has a crosssectional area of 1,020 cmil and the ambient temperature of the conductor is 20°C.
Solution: To calculate the distance to the point where the wire is shorted to ground, we transform Equation 36 as follows:
L = RA r
Since R = 5Ω , A = 1,020 cmils, and r = 10.37 cmilΩ /ft, we have
= (5Ω )(1020cmil) =
L 492 ft
10.37cmil − Ω / ft
Chapter 3 – Electrical and Electronic Fundamentals 
73 
Wire Gauge Sizes
Electrical conductors are manufactured in sizes that are numbered according to a system known as the American Wire Gauge (AWG). As Table 32 shows, the wire diameters become smaller as the gauge numbers increase, and the resistance per one thousand feet increases as the wire diameter decreases. The largest wire size listed in the table is “08,” and the smallest is “24.” This is the normal range of wire sizes typically encountered in process control applications. The complete AWG table goes from 0000 to 40; the larger and smaller sizes not listed in Table 32 are manufactured but are not commonly encountered in process control.
Table 32. American Wire Gauge for Copper Wire
AWG Number 
Diameter, mil 
Area cmil 
Ω 
/1000 ft at 
Ω /1000 ft at 

25°C 
65°C 












08 
128.0 
16500 

0.641 
0.739 
10 
102.0 
10400 

1.020 
1.180 
12 
81.0 
6530 

1.620 
1.870 
14 
64.0 
4110 

2.58 
2.97 
16 
51.0 
2580 

4.09 
4.73 
18 
40.0 
1620 

6.51 
7.51 
20 
32.0 
1020 

10.4 
11.9 
22 
25.3 
642 

16.5 
19.0 
24 
20.1 
404 

26.2 
30.2 






Example 35 shows how to calculate the resistance of a typical conductor.
EXAMPLE 35
Problem: Determine the resistance of 2,500 feet of 14 AWG copper wire. Assume the wire temperature is 25°C.
Solution: Using Table 32, we see that 14 AWG wire has a resistance of 2.58Ω per 1,000 ft at 25°C. So the resistance of 2,500 ft is calculated as follows:
R = (2.58Ω /1000 ft)(2500 ft) = 6.5Ω
74 Measurement and Control Basics
Direct and Alternating Current
Basically, two types of voltage signals are encountered in process control and measurement: direct current (dc) and alternating current (ac). In direct current, the flow of charges is in just one direction. A battery is one example of a dc power source. Figure 31a shows a graph of a dc voltage over time, and Figure 31b shows the symbol for a battery.
Voltage 
DC Voltage 
+V 
Time 
a) Graph of DC Voltage 
b) Battery Symbol 
Figure 31. DC voltage graph and symbol
An alternating voltage source periodically reverses its polarity. Therefore, the resulting current flow in a closed circuit will reverse direction periodically. Figure 32a shows a sinewave example of an ac voltage signal over time. Figure 32b shows the schematic symbol for an ac power source.
Voltage 
+ 
0 
 
a) Graph of AC Voltage
Time
b) Symbol for AC Voltage Source
Figure 32. AC voltage graph and symbol
The 60cycle ac power that is used in homes and industry in the United States is a common example of ac power. The term 60 cycle means that the voltage polarity and current direction go through 60 reversals or changes per second. The signal is said to have a frequency of 60 cycles per second. The unit for 1 cycle per second (cps) is called 1 hertz (Hz). Therefore, a 60 cyclepersecond signal has a frequency of 60 Hz.