SELECTION OF A PROCESS-CONTROL SYSTEM
A continuous industrial process contains four process centers, each of which has two variables that must be controlled. If a fast process-reaction rate is required with only small to moderate dead time, select a suitable mode of control. The system contains more than two resistance-capacity pairs. What type of transmission system would be suitable for this process?
Calculation Procedure:
1. Compute the number of process capacities
The number of process capacities ⫽(number of process centers)(number of vari- ables per center), or, for this system, 4⫻2⫽8 process capacities. This is defined as amultiplenumber of process capacities because the number controlled is greater than unity.
2. Analyze the process-time lags
A small to moderate dead time is allowed in this process control system. With such a dead-time allowance and with two or more resistance-capacity pairs in the system, a mode of control that provides for any number of process-time lags is desirable.
3. Select a suitable mode of control
Table 39 summarizes the forms of control suited to processes having various char- acteristics. This table is ageneral guide—it provides, at best, anapproximateaid
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TABLE39ProcessCharacteristicsversusModeofControl*
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in selecting control modes. Hence it is suitable for tentative selection of the mode of control. Final selection must be based on actual experience with similar systems.
Inspection of Table 39 shows that for a multiple number of processes with small to moderate dead time and any number of resistance-capacity pairs, a proportional plus reset mode of control is probably suitable. Further, this mode of control pro- vides for any (i.e., fast or slow) reaction rate. Since a fast reaction rate is desired, the proportional plus reset method of control is suitable because it can handle any process-reaction rate.
4. Select the type of transmission system to use Four types of transmission systems are used for process control today: pneumatic, electric, electronic, and hydraulic. The first three types are by far the most common.
Pneumatic transmission systems use air at 3 to 20 lb / in2(gage) (20.7 to 137.9 kPa) to convey the control signal through small-bore metal tubing at distances ranging to several thousand feet. The air used in pneumatic systems must be clean and dry. To prevent a process from getting out of control, a constant supply of air is required. Pneumatic controllers, receivers, and valve positioners usually have small air-space volumes of 5 to 10 in3 (81.0 to 163.9 cm3). Air motors of the diaphragm or piston type have relatively large volumes: 100 to 5000 in3(1639 to 81,935 cm3).
Pneumatic control systems are generally considered to be spark-free. Hence, they find wide use in hazardous process areas. Also, control air is readily available, and it can be ‘‘dumped’’ to the atmosphere safely. The response time of pneumatic control systems may be slower than that of electric or hydraulic systems.
Electric and electronic control systems are fast-response with the signal con- veyed by a wire from the sensing point to the controller. In hazardous atmospheres the wire must be protected against abrasion and breakage.
Hydraulic control systems are also rapid-response. These systems are capable of high power actuation. Slower-acting hydraulic systems use fluid pressures in the 50 to 100 lb / in2(344.8 to 689.5 kPa) range; fast-acting systems use fluid pressures to 5000 lb / in2(34,475 kPa).
Dirt and fluid flammability are two factors that may be disadvantages in certain hydraulic-control-system applications. However, new manufacturing techniques and nonflammable fluids are overcoming these disadvantages.
Since a fast response is desired in this process-control system, electric, elec- tronic, or hydraulic transmission of the signals would be considered first. With long distances between the sensing points [say 1000 ft (305 m) or more], an electric or electronic system would probably be best.
Next, determine whether the systems being considered can provide the mode of control (step 2) required. If a system cannot provide the necessary mode of control, eliminate the system from consideration.
Before a final choice of a system is made, other factors must be considered.
Thus, the relative cost of each type of system must be determined. Should an electric system prove too costly, the slightly slower response time of the pneumatic system might be accepted to reduce the initial investment.
Other factors influencing the choice of the type of a control system include type of controls, if any, currently used in the installation, skill and experience of the operating and maintenance personnel, type of atmosphere in which, and type of process for which, the controls will be used. Any of these factors may alter the initial choice.
Related Calculations. Use this general method to make a preliminary choice of controls for continuous processes, intermittent processes, air-conditioning sys- tems, combustion-control systems, etc. Before making a final choice of any control
ENVIRONMENTAL CONTROL AND ENERGY CONSERVATION 18.117
FIGURE 48 Temperature control of a simple process.
system, be certain to weigh the cost, safety, operating, and maintenance factors listed above. Last, the system chosenmustbe able to provide the mode of control required.
PROCESS-TEMPERATURE-CONTROL ANALYSIS
A water storage tank (Fig. 48) contains 500 lb (226.8 kg) of water at 150⬚F (65.6⬚C) when full. Water is supplied to the tank at 50⬚F (10.0⬚C) and is withdrawn at the rate of 25 lb / min (0.19 kg / s). Determine the process-time constant and the zero- frequency process gain if the thermal sensing pipe contains 15 lb (6.8 kg) of water between the tank and thermal bulb and the maximum steam flow to the tank is 8 lb / min (0.060 kg / s). The steam flow to the tank is controlled by a standard linear regulating valve whose flow range is 0 to 10 lb / min (0 to 0.076 kg / s) when the valve operator pressure changes from 5 to 30 lb / in2(34.5 to 206.9 kPa).
Calculation Procedure:
1. Compute the distance-velocity lag
The time in minutes needed for the thermal element to detect a change in temper- ature in the storage tank is thedistance-velocity lag,which is also called thetrans- portation lag,ordead time.For this process, the distance-velocity lagdis the ratio of the quantity of water in the pipe between the tank and the thermal bulb—that is, 15 gal (57.01 L)—and the rate of flow of water out of the tank—that is, 25 lb /
min (0.114 kg / s)—ord⫽15 / 25⫽0.667 min.
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2. Compute the energy input to the tank
This is atransient-control process;i.e., the conditions in the process are undergoing constant change instead of remaining fixed, as insteady-state conditions.For tran- sient-process conditions the heat balance is Hin ⫽ hout ⫹ hstor, whereHin ⫽ heat input, Btu / min;Hout⫽heat output, Btu / min;Hstor⫽heat stored, Btu / min.
The heat input to this process is the enthalpy of vaporizationhƒgBtu / (lb䡠min) of the steam supplied to the process. Since the regulating valve is linear, its sen- sitivitysis (flow-rate change, lb / min) / (pressure change, lb / in2). Or, by using the known valve characteristics, s ⫽ (10 ⫺ 0) / (30 ⫺ 5) ⫽ 0.4 (lb / min) / (lb / in2) [0.00044 kg / (kPa䡠s)].
With a change in steam pressure ofplb / in2(p⬘kPa) in the valve operator, the change in the rate of energy supply to the process isHin ⫽ 0.4 (lb / min) / (lb / in2)
⫻ p ⫻ hƒg. Taking hƒg as 938 Btu / lb (2181 kJ / kg) gives hin ⫽ 375p Btu / min (6.6p⬘kW).
3. Compute the energy output from the system
The energy output Hout ⫽ lb / min of liquid outflow ⫻ liquid specific heat, Btu / (lb䡠⬚F) ⫻(Ta⫺ 150⬚F), whereTa⫽ tank temperature,⬚F, at any time. When the system is in a state of equilibrium, the temperature of the liquid in the tank is the same as that leaving the tank or, in this instance, 150⬚F (65.6⬚C). But when steam is supplied to the tank under equilibrium conditions, the liquid temperature will rise to 150⫹Tr, whereTr⫽temperature rise,⬚F (Tr⬘,⬚C), produced by introducing steam into the water. Thus, the above equation becomesHout ⫽ 25 lb / min⫻ 1.0 Btu / (lb䡠⬚F)⫻Tr⫽25TrBtu / min (0.44Tr⬘ kW).
4. Compute the energy stored in the system
With the rapid mixing of the steam and water, Hstor⫽ liquid storage, lb⫻liquid specific heat, Btu / (lb䡠⬚F)⫻ Trq ⫽500⫻1.0⫻ trq, whereq⫽ derivative of the tank outlet temperature with respect to time.
5. Determine the time constant and process gain
Write the process heat balance, substituting the computed values inHin ⫽Hout ⫹ Hstor, or 375p ⫽ 25Tr ⫹ 500Trq. Solving gives Tr/p ⫽ 375 / (25 ⫹ 500q) ⫽ 15 / (1⫹ 20q).
The denominator of this linear first-order differential equation gives the process- system time constant of 20 min in the expression 1⫹20q. Likewise, the numerator gives the zero-frequency process gain of 15⬚F / (lb / in2) (1.2⬚C / kPa).
Related Calculations. This general procedure is valid for any liquid using any gaseous heating medium for temperature control with a single linear lag. Likewise, this general procedure is also valid for temperature control with a double linear lag and pressure control with a single linear lag.
COMPUTER SELECTION FOR INDUSTRIAL PROCESS-CONTROL SYSTEMS
Select the type of computer and its speed of operation for use in an industrial control application. The computer will be used to monitor and control two continuous-flow process operations. Budget limitations restrict the investment in the computer to abut $100,000 with a typical execution time of 10 ms or better.
ENVIRONMENTAL CONTROL AND ENERGY CONSERVATION 18.119
FIGURE 49 Digital-computer elements used in process control.
Calculation Procedure:
1. Analyze the computers available; select a suitable computer
Four types of computers are available for consideration in any control problem:
analog; digital; hybrid—consisting of analog and digital; and special-purpose—
analog or digital computers for industrial control.
Digital computers (Fig. 49) find wide use for controlling continuous-flow pro- cesses. When used to control continuous-flow processes, the digital computer is connectedonline;i.e., information reflecting the activity in the process being con- trolled is introduced to the data-processing system as soon as it occurs, and action is immediately initiated by the system to make any needed adjustments. Since digital computers are proven machines for process control, this type will be tenta- tively chosen for this process and its suitability investigated further.
The usual digital computer used in process control is a general-purpose one that can receive and transmit analog signals. A magnetic-drum-type stored memory is often used in control applications.
2. Determine the computer operating time
The speed of a computer depends on the actions needed to perform a calculation.
Thus, a drum or disk memory may have a 150-s add time with a memory access time of 16,000s. In such a computer the memory-access time is the controlling speed factor.
Figure 50 shows the execution times for a variety of digital computers of dif- ferent makes in solving the same bench-markor testproblems. A typical bench- mark problem consists of a pair of simultaneous equations having two unknowns.
Study of Fig. 50 shows that three machines—A, B, and D—meet the general requirements for this process with respect to speed and cost. Since each of these computers is produced by a different manufacturer, a fairly wide choice of units is available. When a chart such as Fig. 50 is not available, prepare one after computing the costs of machines available from various manufacturers.
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FIGURE 50 Execution time for an arithmetic benchmark problem shows the solution time re- quired by different makes of computers.
FIGURE 51 Execution time for a benchmark problem shows no clear relationships between solution time and computer cost.
3. Check the computer performance
Computer performance is rated according to three factors: speed margin, i.e., how the computer copes with the worst-case time combination of events in the process;
memory-storage margin, that is, provisions for worst-case data storage, retrieval, and working space; and reliability or consistency of performance.
Word length affects computer speed. Although a computer handling 24-bit words may seem to have an operation rate identical to a 12-bit-word computer, there may be as much as a 2:1 variation in the speed to the same degree of precision. To compare two computers, use their relativeproblem-solving speedsas a guide.
Manufacturers publish the relative problem-solving speeds of each model of computer. List these speeds for each of the suitable makes—A,B, andD, step 2.
Select the computer giving the fastest speed for the smallest investment.
4. Investigate the computer logic function
In some control applications, the logical-problem-solving ability of the computer may be more important than the arithmetic ability. In the logic function, the com- puter uses information transfers, manipulations, and comparisons, all of which take time. Using manufacturer’s data, plot the logic speed against cost for each computer for a specific bench-mark or test problem. Usually the execution time for a logical bench-mark problem shows no clear relation to price (Fig. 51). However, the plot does indicate the price range for various operating speeds.
5. Evaluate the computer-memory size
An online-control computer cannot deal with a real-time problem unless all instruc- tions defining the action are stored and available when needed. Distribution of the storage between high-speed random-access memory sections and lower-cost cyclic- access sections, such as drums and disks, influences computer speed. Good practice stores an image of the high-speed working memory in one of the slower-speed backup memories.
In machines with 12- to 15-bit words, more than one memory word may be needed to store an instruction or data item. Compare bench-mark problems to de- termine the real-time need of memory addresses between different computers. In general, more process-control applications are better served by machines having expandable memories. The tendency of many control-system designers is to un- derestimate the memory capacity needed. So choose a machine having the largest memory possible within the prevailing financial constraints.
ENVIRONMENTAL CONTROL AND ENERGY CONSERVATION 18.121
6. Evaluate the computer reliability
The mean time between failure (MTBF) is a good measure of computer reliability.
For a typical control computer, the MTBF should be in the 1000⫹ -h category.
Compare the MTBF for each of the computers in step 2. Choose the machine having the highest MTBF at the desired speed. Further, the computer should be capable of operating in the 50 to 120⬚F (10 to 48.9⬚C) ambient temperature range.
Related Calculations. Use this general method to choose a control computer for any process-type application. Where high-speed operations are involved, such as in missile-guidance applications, computers operating at nanosecond speeds are generally required. The selection procedure for such machines is different from that for process-control computers where millisecond speeds are usually satisfactory.
Note that the computer prices cited here are relative; for actual prices consult the manufacturers concerned.
The program or software prepared for automatic process control can cost as much as the computer hardware. The first process applications of computers in an industry often show that the programs needed cost more than anticipated. Thus, computer programming to perform startup and shutdown sequency monitoring in a process system may require 4000 instructions. For automatic sequency control, as many as 20,000 instructions may be needed. Further, each plant is unique, and little of the programming work done for one process can be applied to another.
Programming still uses the most time of any phase of total computer operation for problem solving. So even if the computer operating time were cut to nearly zero, the saving in machine time would not be significant when compared with the programming time. On the other hand, saving in computer machine time is impor- tant in process-control applications because processes can be held more closely to optimum levels with reduced machine time.
Direct digital control (DDC) is replacing analog control in some process appli- cations. The major advantage of DDC is that it removes the need for digital-to- analog converters and gives the computer more direct control over plant equipment while removing a source of spurious signals.
In process control, certain aspects of programming warrant special consideration.
These aspects include real-time operation, memory capability, and operator misuse.
Because a process computer functions in a real-time environment, a certain amount of ‘‘free’’ time must be available to allow for emergency reactions and special operator requests. A good rule of thumb is to allow 40 percent of free time within the computer; any less may cause the computer program to fall out of step with the process situation.
Most process-control computers use one of the variants of IBM’s original For- tran. This language finds its principal applications in relatively infrequent proce- dures, such as plant startup and shutdown. Minute-by-minute scanning is usually handled by a standard ‘‘scan, monitor, and alarm’’ program prepared manually for the particular scanning sequence dictated by operating requirements.
A self-checking program is an important feature of a process-control computer system. Unlike a scientific program in which the results obtained are printed out for perusal by a scientist or engineer, a closed-loop control system utilizes the computer’s calculations to act directly on the process plant itself. Should either the input to the computer or the data handling within it be in error, the resulting cal- culated control points and output signals will be incorrect and could result in haz- ardous operation. Double and triple checks may be necessary to ensure that the operating data are valid. As more and more input signals are utilized, the program- ming necessary to provide such validity checks becomes increasingly more com-
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plicated. Continuous calculation of a process heat balance can, for example, be useful in determining the validity of the input information, since an extensive range of input data figures in the calculation.
CONTROL-VALVE SELECTION FOR PROCESS CONTROL
Select a steam control valve for a heat exchanger requiring a flow of 1500 lb / h (0.19 kg / s) of saturated steam at 80 lb / in2(gage) (551.6 kPa) at full load and 300 lb / h (0.038 kg / s) at 40 lb / in2(gage) (275.8 kPa) at minimum load. Steam at 100 lb / in2(gage) (689.5 kPa) is available for heating.
Calculation Procedure:
1. Compute the valve flow coefficient
The valve flow coefficientCvis a function of the maximum steam flow rate through the valve and the pressure drop that occurs at this flow rate. In choosing a control valve for a process-control system, the usual procedure is to assume a maximum flow rate for the valve based on a considered judgment of the overload the system may carry. Usual overloads to not exceed 25 percent of the maximum rated capacity of the system. Using this overload range as a guide, assume that the valve must handle a 20 percent overload, or 0.20 (1500)⫽300 lb / h (0.038 kg / s). Hence, the rated capacity of this valve should be 1500⫹300⫽1800 lb / h (0.23 kg / s).
The pressure drop across a steam control valve is a function of the valve design, size, and flow rate. The most accurate pressure-drop estimate usually available is that given in the valve manufacturer’s engineering data for a specific valve size, type, and steam-flow rate. Without such data, assume a pressure drop of 5 to 15 percent across the valve as a first approximation. This means that the pressure loss across this valve, assuming a 10 percent drop at the maximum steam-flow rate, would be 0.10⫻80⫽ 8 lb / in2(gage) (55.2 kPa).
With these data available, compute the valve flow coefficient fromCv⫽ WK/
3(⌬p P2) , where0.5 W ⫽ steam flow rate, lb / h; K ⫽ 1 ⫹ (0.007 ⫻⬚F superheat of the steam); p ⫽ pressure drop across the valve at the maximum steam flow rate, lb / in2;P2⫽control-valve outlet pressure at maximum steam flow rate, lb / in2 (abs). Since the steam is saturated, it is not superheated andK ⫽ 1. Then Cv ⫽ 1500 / 3(8⫻94.7)0.5⫽18.1.
2. Compute the low-load steam flow rate
Use the relationW⫽3(Cv⌬p P2)0.5/K, where all the symbols are as before. Thus, with a 40-lb / in2(gage) (275.8-kPa) low-load heater inlet pressure, the valve pres- sure drop is 80 ⫺ 40 ⫽ 40 lb / in2 (gage) (275.8 kPa). The flow rate through the valve is thenW⫽3(18.1⫻ 40⫻54.7)0.5/ 1⫽598 lb / h (0.75 kg / s).
Since the heater requires 300 lb / h (0.038 kg / s) of steam at the minimum load, the valve is suitable. Had the flow rate of the valve been insufficient for the min- imum flow rate, a different pressure drop, i.e., a larger valve, would have to be assumed and the calculation repeated until a flow rate of at least 300 lb / h (0.038 kg / s) was obtained.