Instrument types and performance characteristics

 

2.3.1 Zero order instrument

If all the coefficients a1 ...an other than a0 in equation (2.2) are assumed zero, then:

                                     a0q0 = b0qi or q0 = b0qi/a0 = Kqi

where K is a constant known as the instrument sensitivity as defined earlier. Any instrument that behaves according to equation (2.3) is said to be of zero order type. Following a step change in the measured quantity at time t, the instrument output moves immediately to a new value at the same time instant t, as shown in Figure 2.10. A potentiometer, which measures motion, is a good example of such an instrument, where the output voltage changes instantaneously as the slider is displaced along the potentiometer track.

 

2.3.2 First order instrument

If all the coefficients a2 ...an except for a0 and a1 are assumed zero in equation (2.2) then:

      a1 dq0/ dt + a0q0 = b0qi

Any instrument that behaves according to equation (2.4) is known as a first order instrument. If d/dt is replaced by the D operator in equation (2.4), we get:

a1Dq0 = a0q0 = b0qi and rearranging this then gives q0 = (b0/a0)qi/ [1 + (a1/a0)D]

Defining K = b0/a0 as the static sensitivity and  = a1/a0 as the time constant of the system, equation (2.5) becomes:

 

                                                                  q0 = Kqi/ 1 +  D

 

If equation (2.6) is solved analytically, the output quantity q0 in response to a step change in qi at time t varies with time in the manner shown in Figure 2.11. The time constant  of the step response is the time taken for the output quantity q0 to reach 63% of its final value

 

The liquid-in-glass thermometer (see Chapter 14) is a good example of a first order instrument. It is well known that, if a thermometer at room temperature is plunged into boiling water, the output e.m.f. does not rise instantaneously to a level indicating 100°C, but instead approaches a reading indicating 100°C in a manner similar to that shown in Figure 2.11.

 

 A large number of other instruments also belong to this first order class: this is of particular importance in control systems where it is necessary to take account of the time lag that occurs between a measured quantity changing in value and the measuring instrument indicating the change. Fortunately, the time constant of many first order instruments is small relative to the dynamics of the process being measured, and so no serious problems are created.

 

Example 2.3

A balloon is equipped with temperature and altitude measuring instruments and has radio equipment that can transmit the output readings of these instruments back to ground. The balloon is initially anchored to the ground with the instrument output readings in steady state. The altitude-measuring instrument is approximately zero order and the temperature transducer first order with a time constant of 15 seconds. The temperature on the ground, T0, is 10°C and the temperature Tx at an altitude of x metres is given by the relation: Tx = T0 - 0.01x

(a) If the balloon is released at time zero, and thereafter rises upwards at a velocity of 5 metres/second, draw a table showing the temperature and altitude measurements reported at intervals of 10 seconds over the first 50 seconds of travel. Show also in the table the error in each temperature reading.

(b) What temperature does the balloon report at an altitude of 5000 metres?

 

Solution In order to answer this question, it is assumed that the solution of a first order differ[1]ential equation has been presented to the reader in a mathematics course. If the reader is not so equipped, the following solution will be difficult to follow.

Let the temperature reported by the balloon at some general time t be Tr. Then Tx is related to Tr by the relation:

 

                                       Tr = Tx/1 + D = T0 - 0.01x/ 1 + D = 10 - 0.01x/ 1 + 15D

 

It is given that x = 5t, thus: Tr = 10 - 0.05t/ 1 + 15D

 

The transient or complementary function part of the solution (Tx = 0) is given by:

Trcf = Ce^-t/15

The particular integral part of the solution is given by: Trpi = 10 - 0.05(t – 15)

Thus, the whole solution is given by: Tr = Trcf + Trpi = Ce^-t/15 + 10 - 0.05(t – 15)

Applying initial conditions: At t = 0, Tr = 10, i.e. 10 = Ce^-0 + 10 - 0.05(-15)

Thus C = -0.75 and therefore: Tr = 10 - 0.75e^-t/15 - 0.05(t – 15)

Using the above expression to calculate Tr for various values of t, the following table can be constructed:

(b) At 5000 m, t = 1000 seconds. Calculating Tr from the above expression:

                                 Tr = 10 - 0.75e^-1000/15 - 0.05(1000 – 15)

The exponential term approximates to zero and so Tr can be written as:

                                           Tr ³ 10 - 0.05(985) = -39.25°C

 

This result might have been inferred from the table above where it can be seen that the error is converging towards a value of 0.75. For large values of t, the transducer reading lags the true temperature value by a period of time equal to the time constant of 15 seconds. In this time, the balloon travels a distance of 75 metres and the temperature falls by 0.75°. Thus for large values of t, the output reading is always 0.75° less than it should be.