Measurement and Control Basics 3rd Edition (complete book)
.pdfChapter 2 – Process Control Loops |
35 |
response is the maximum in the beginning and continuously decreases from that time on.
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100% |
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63% |
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Time Constants (τ ) |
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Figure 2-6. Response of a first-order system to step
The time constant also provides an insight into the physical response of a system. Figure 2-7 shows two different response curves for first-order systems, but each has a different time constant. It can be seen from the two different curves that as the time constant increased, the response of the system to a step change becomes slower. The first system has a time constant of one second, and it takes five seconds to reach 99 percent of fullscale output. The second system has a time constant of two seconds and takes ten seconds to 99 percent of full scale. So, the second system responds more slowly to a step change at the input.
Comparison of Basic Physical Systems
To understand the effect of system time constants on process behavior, we will now look at three common physical systems: electrical, liquid, and thermal. All three systems can be said to have resistance and capacitance. It will become clear that the resistance multiplied by the capacitance of a system produces the process time constant.
Electrical Systems
The characteristics of an electric circuit that has pure resistance and pure capacitance in series are analogous to the resistance and capacitance of
36 Measurement and Control Basics
Output (%) τ = 1s τ = 2s
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Time (t) in seconds |
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Figure 2-7. Comparison of system time constants
most liquid and thermal systems. Resistance in a purely resistive circuit is defined by Ohm's law, which states that the potential or voltage required to force an electric current through a resistor is equal to the current times the resistance. In equation form, Ohm's law is
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V = IR |
(2-8) |
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= electrical potential in volts |
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= current in amps or coulombs per second |
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R |
= resistance in ohms |
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The electrical resistance is |
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R = V/I = Potential/Flow |
(2-9) |
The relationship for capacitance states that the charge on a capacitor is equal to the capacitance times the voltage (potential) across the capacitor, or
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q = CV |
(2-10) |
where |
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q |
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charge in coulombs |
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C |
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capacitance in farads |
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V |
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potential in volts |
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Chapter 2 – Process Control Loops |
37 |
Thus, the capacitance is given by
C = |
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Charge quantity |
(2-11) |
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A series electrical RC network is shown in Figure 2-8.
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Switch |
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Resistor |
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S1 |
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+ vr |
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VS |
Battery |
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Figure 2-8. RC series circuit
A step input to the circuit is provided by moving switch 1 from position 1 to position 2. This applies the battery voltage to the series circuit and current flows to charge the capacitor (C) to the applied battery voltage Vs. The voltage across the capacitor, vc(t), is the output of the system and the voltage vs(t) is the input to the system. The RC series circuit is the system or transfer function. When switch 1 is placed into position 2, current flows in the circuit and the capacitor charges up to the applied battery voltage Vs as shown in Figure 2-9. To understand the concept of time constant, we will next derive the system equation for the RC circuit.
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Vs = Battery Voltage |
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Figure 2-9. Voltage across capacitor in RC circuit
38 Measurement and Control Basics
After switch 1 is placed into position 2, the sum of the voltages in the RC section of the circuit equals the applied voltage Vs as follows
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vs(t) = vr(t) + vc(t) |
(2-12) |
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vs |
= the instantaneous voltage measured from the common |
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terminal on switch 1 to the negative terminal of the battery |
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vr(t) |
= the instantaneous voltage drop across the resistor |
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vc(t) |
= the instantaneous voltage across the capacitor. |
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According to Ohm's law, the voltage drop across the resistor is given by
vr(t) = i(t)R |
(2-13) |
Therefore, system Equation 2-12 becomes
vs(t) = i(t)R + vc(t) |
(2-14) |
The equation for charge on a capacitor is given by Equation 2-10, where
q = CV. To find the instantaneous rate of change of the charge on the capacitor as it charges, the derivative of Equation 2-10 must be taken as follows:
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dvc |
(2-15) |
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i ( t ) = |
d q |
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(2-16) |
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Substituting the current in Equation 2-15 into the circuit or system Equation 2-14, gives the following:
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=RC |
dvc |
+ vc |
(2-17) |
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Therefore, |
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(2-18) |
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dt |
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where τ = RC is the time constant for the system.
We can use unit analysis to show that resistance multiplied by capacitance results in seconds. The units of resistance (R) are [volts/coulombs/seconds] and the units of capacitance (C) are [coulombs/volts]. If we multiply the units of R by the units of C, we obtain seconds.
Chapter 2 – Process Control Loops |
39 |
The electric circuit RC time constant is the time required to charge the capacitor to 63.2 percent of its maximum value in the series circuit after the switch S1 is moved from position 1 to position 2. We can perform a Laplace transform on system Equation 2-18 to obtain
Vs = τ sVc + Vc |
(2-19) |
or
Vc |
= |
1 |
(2-20) |
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Note that the transformed system equation has the same form as the “firstorder lag” systems we discussed earlier.
To investigate the concept of time constants in other physical systems, we will next discuss forms of resistance and capacitance in liquid and thermal systems.
Liquid Systems
Figure 2-10 shows liquid flow in a pipe, with a restricting device (a valve) providing a hydraulic resistance (Rh) to the flow. Note that the walls of the pipe will also provide a small amount of resistance to flow, depending on how rough they are.
Flow in, qi |
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Hydraulic Resistance, Rh
Figure 2-10. Flow restricting
Two main types of flow take place in liquid systems: laminar and turbulent flow. Laminar flow occurs when the fluid velocity is relatively low and the liquid flows in layers, so the flow is directly proportional to differential pressure or head on the liquid. Unfortunately, laminar flow is seldom encountered in actual practice and usually occurs only with very viscous fluids at low velocities.
Turbulent flow occurs when the fluid velocity is relatively high and the velocity of the liquid at any point varies irregularly. When turbulent flow
40 Measurement and Control Basics
occurs from a tank discharging under its own head or pressure, the flow is found by the following equation:
q = K A 2 g h |
(2-21) |
Where q is the flow rate (ft3/s), K is a flow coefficient, A is the area of the discharge orifice (ft2), g is gravitation constant (ft/s2), and h is pressure head of liquid (ft). We can define hydraulic resistance (Rh) to flow as follows:
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Potential |
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h |
(2-22) |
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Therefore, the instantaneous rate of change of hydraulic resistance to flow is
Rh = dh dq
Rearranging Equation 2-21, we arrive at:
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KA 2g |
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and differentiating Equation 2-24 with respect to q gives us,
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d q |
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(2-23)
(2-24)
(2-25)
Multiplying the top and bottom of Equation 2-25 by the square root of h yields
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(2-26) |
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d q |
K A 2 g h |
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Chapter 2 – Process Control Loops |
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According to Equation 2-21, the denominator of Equation 2-26 is q, so substituting q into Equation 2-26 gives us,
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(2-27) |
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Therefore, instantaneous hydraulic resistance is
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(2-28) |
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The hydraulic resistance is analogous to electrical resistance in that it is inversely proportional to flow q but directly proportional to two times the differential pressure, h, or the driving potential. The difference lies in the fact that turbulent flow involves the square root of the driving potential or head h.
However, we will now show that liquid capacitance is directly analogous to electrical capacitance. Figure 2-11 shows a tank being filled with a liquid. The equation for the volume (V) of the liquid in the tank is given by the following equation:
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V(t) = Ah(t) |
(2-29) |
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= the volume of liquid as a function of time |
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height of liquid |
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A |
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Note that the volume V of the tank and the liquid height or head are a function of time. The flow of liquid into the tank, qi, and the flow liquid out of the tank, qo, vary with time.
If we solve Equation 2-29 for A, we obtain the following: |
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quantity |
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h(t) potential |
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Comparing this equation to the equation for electrical capacitance
(i.e., C = q/V) clearly shows that liquid capacitance Cl is simply the surface area of the liquid in the tank, or Cl = A. Furthermore, taking the derivative of Equation 2-29 with respect to time yields
dV (t) |
= A |
h(t) |
(2-31) |
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dt dt
42 Measurement and Control Basics
Flow in, qi
Liquid Level
h
Flow out, qo
Figure 2-11. Liquid storage tank
We know that the instantaneous rate of change of volume, dV/dt, is given by the flow in (qi), minus the flow out (qo), or
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dV |
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(2-32) |
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Using Equations 2-31 and 2-32, we obtain |
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qi − qo = A |
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If we assume turbulent flow from the tank, then qo = 2h/R, or |
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qi = A |
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Note that Equation 2-34 is a first-order linear differential equation that expresses liquid level as a function of time, with the fluid flow in (qi) as the forcing function.
The example that follows shows how to determine the differential equation for a typical fluid flow system.
Chapter 2 – Process Control Loops |
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EXAMPLE 2-2
Problem: Determine the differential equation for flow (qo) out of the process tank shown in Figure 2-11. Find the system time constant if the operating head is 5 m, the steady state flow is 0.2m3/s, and the surface area of the liquid is 10 m2.
Solution: Equation 2-33 states that
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= A |
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To obtain the differential equation in terms of variable qo, dh/dt must be expressed in terms of dqo/dt. We know from the definition of hydraulic resistance (Equation 2-23) that
Rh = dh(t) dq(t)
or dh(t) = Rh dq(t). Taking the derivative of this equation with respect to time yields
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= Rh |
dqo |
dt |
dt |
Therefore, the system equation becomes
ARh d qo + qo = qi dt
To calculate the system time constant, remember that for our system, the liquid capacitance Cl is equal to the surface area of liquid in the tank; therefore, the system time constant τ is given by RhCl or RhA. The hydraulic resistance is given by
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2 h |
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Since h = 5 m and q = 0.2 m3/s in our example, then
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0 .2 m 3 |
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/ s |
The system time constant τ is given by
τ = Rh A = (50 s / m 2)(10 m 2) = 500 s.
44 Measurement and Control Basics
Thermal Systems
The final physical process we will examine to understand process time constants is a thermal system. The basic thermal processes encountered in the process industries are the mixing of hot and cold fluids, the exchange of heat through adjoining bodies, and the generation of heat by combustion or chemical reaction.
Two laws of thermodynamics are used in the study of thermal systems. The first governs the way in which heat energy is produced and determines the amount generated. The second governs the flow of heat.
Temperature changes in an isolated body conform to the first law of thermodynamics. For a given body, heat input raises the internal energy, and the rate of change of body temperature will be proportional to the heat flow to the body. The constant that relates temperature change and heat flow is called the thermal capacity of the body:
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dT |
= q |
(2-35) |
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where
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= thermal capacitance (cal/°C) |
dT/dt = the rate of change of temperature °C/s
q |
= heat flow (cal/s) |
The thermal capacitance of a body is found by multiplying the specific heat of the material by the weight of the body:
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C = MS |
(2-36) |
where |
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M |
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the mass of the body (gm) |
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S |
= |
the specific heat of the material (cal/gm)(°C) |
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Thermal capacitance is analogous to electric capacitance. For example, in Figure 2-12, heat flowing into a body with thermal capacitance C causes the temperature (T) to rise above the ambient value To.
Heat flow and charge flow as well as temperature and voltage are analogous quantities. Heat transmission takes place by conduction, convection, or radiation. Conduction involves transmission through adjoining bodies, convection involves transmission and mixing, and radiation uses electromagnetic waves to transfer heat.