Measurement and Control Basics 3rd Edition (complete book)

EXERCISES

2.1Explain the operation of the single-loop feedback control loop shown in Figure 2-1.

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 0.5-s 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 2-11 has an operating head (h) of 5 ft and

a normal output flow of 5 ft3/s. The cross-sectional 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 2-15 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 Ziegler-Nichols 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 one-quarter damping response.

2.10Using the Ziegler-Nichols open-loop 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: McGraw-Hill, 1980.

2.Considine, Douglas M. (ed.). Handbook of Applied Instrumentation, New York: McGraw-Hill, 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, 72-75.

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: Prentice-Hall, 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: Straight-Line 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: McGraw-Hill, 1975.

13.Ziegler, J. G., and N. B. Nichols. “Optimum Settings for Automatic Controllers,” ASME Transactions 64 (1942): 759-68.

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:

where Q is the amount of current in coulombs flowing past a given point in t seconds.

Example 3-1 shows how to calculate current flow.

EXAMPLE 3-1

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

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 cross-sectional 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:

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:

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

E

Using the definitions for current density J and voltage gradient E we obtain conductivity in a different form:

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 cross-sectional area A. If the wire has resistivity, r, then its resistance, R, is

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:

When Equation 3-7 is substituted into Equation 3-6, we obtain the following equation:

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 mil-foot. 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 cross-sectional area for wire used in U.S. wire-sizing 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

(cmil) = d2 (mil2). Example 3-2 illustrates how to calculate conductor area in circular mils.

EXAMPLE 3-2

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 circular-mil 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 3-1 gives the specific resistance (r) in cmil-ohms 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 3-1 to find the resistance of a conductor of length (L) in feet and cross-sectional area (A) in cmil by using Equation 3-7, R=rL/A.

Table 3-1. Specific Resistivity.

Example 3-3 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 3-3

Problem: Find the resistance of 1,000 feet of copper (drawn) wire that has a cross-sectional area of 10,370 cmil and a wire temperature of 20°C.

Solution: From Table 3-1, the specific resistance of copper wire is 10.37 cmilohms/ft. By inserting the known values into Equation 3-6, the resistance is determined as follows:

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 3-4 illustrates how to find the distance to a short to ground in a telephone line.

EXAMPLE 3-4

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 cross-sectional 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 3-6 as follows:

L = RA r

Since R = 5Ω , A = 1,020 cmils, and r = 10.37 cmil-Ω /ft, we have

= (5Ω )(1020cmil) =

L 492 ft

10.37cmil − Ω / ft

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 3-2 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 3-2 are manufactured but are not commonly encountered in process control.

Table 3-2. American Wire Gauge for Copper Wire

Example 3-5 shows how to calculate the resistance of a typical conductor.

EXAMPLE 3-5

Problem: Determine the resistance of 2,500 feet of 14 AWG copper wire. Assume the wire temperature is 25°C.

Solution: Using Table 3-2, 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