Q.1. a) At t-0, a temperature sensor was suddenly changed from 25°C to 100°C. The sensor outpurs voltage given by the expression V=(0.06 V/°C)[T-20°C]. The following table gives the voltages measured and the times. Determine the average time constant of the sensor. [8]

t(seconds) 0 0.1 0.2 0.3 0.4 0.5V(volts) 0.3 1.8 2.8 3.4 3.9 4.2

b) Is driving an automobile best described as a servomechanism or a process-control system?Why? [2]

Q.2. a) A sensor with a nominal resistance of 60 Ω is used in a bridge with R1 = R2 = 120 Ω,

V = 12.0 V, and R3 = 150-Ω potentiometer shown in fig.1. It is necessary to resolve 0.1-Ω changes of the sensor resistance.

i) At what value of R3 will the bridge null?

ii) What voltage resolution must the null detector possess? [6]

b) A differential amplifier has R2 = 560 KΩ and R1 = 3.9 KΩ. When Va = Vb = 2.8 V the output is 69 mV. Find the CMR and CMRR.

[4]

Q.3. A) A sensor provides temperature data as 430 μV/°C. Develop a comparator circuit that goes high when the temperature reaches 620°C.

[3]

b) A 10-bit ADC has a 12.0-V reference.i. Find the output for inputs of 4.3 V and 8.2 V.ii. What range of inputs could have caused the output to become A5H?

[5]

c) An RTD has α (25 °C) = 0.006/°C. If R= 108 Ω at 25°C, find the resistance at 35°C.

[2]

Q.4. a) Calculate the intensity of a 12-watt source whose radius is 0.04 m at (a) the source in W/m2 and mW/cm2 and (b) 1.3 meters away if the divergence is 1.8º. [4]b) What are different types of Control Valve? Explain in detail [6]

Q.5. Simplify the following and implement the original and simplified equations with gates and ladder logic.

[10]

Q.6. a) A PI controller has Kp=4.5 and KI=7 s-1. Find the controller output for an error given by ep=3 sin(Пt). What is the phase shift between error and controller output? [4]

b) A proportional pneumatic controller has equal area bellows. If 3 to 15 psi signals are used on input and output, find the ratio of pivot distances that provides a 23% PB. [3]

c) A force transducer has a transfer function of 2.2mV/N. Design an alarm using a comparator that triggers at 1050 N. [3]

Q.1. Ans:

b) Ans:

Driving a car is a servomechanism because the purpose is to control the motion of the vehiclerather than to regulate a specific value. Therefore the objective is to cause the vehicle to follow aprescribed path. Of course keeping the speed constant during a trip could be considered processcontrol since the speed is being regulated.

Q.2. A)

(i) For a null condition: R1 R4 = R2 R3, therefore,R3 = R1 R4/R2 = (120)(60)/(120) = 60 Ω(ii) The detector resolution needed to resolve a resistance change of 0.1 Ω is found from the following equation when R4 has changed to 60.1 Ω (or 59.9 Ω).

For R4= 60.1Ω, ∆V= -4.41mV and For R4=59.9Ω, ∆V=-4.44mV

b) The amplifier gain is determined by Ad = R2/R1 = 560/3.9 = 144The common-mode gain is determined by Acm = Vo/Vin = 69 mV/2.8 = 0.025CMRR is the ratio between the differential gain (Ad) and the common-mode gain (Acm); thus,CMRR = Ad/Acm = 144/0.025 = 5760CMR is given in dB. The equation is:20 log 10 (CMRR) = 20 log 10 (5760) = 75.2 dB

Q.3. a) If a transfer function is 430 uV/ºC then a temperature of 620 ºC will result in an outputvoltage of,V = (430 x 10-6 V/ºC)(620 ºC)V = 0.2666 volts or 0.267 to three significant figures.We can construct a divider from a + 5 volt supply to obtain this required alarm voltage forthe comparator. One possible circuit then is shown below. Adjust R1 until Va = 0.267 V.

b) Given a 10-bit ADC with a 12.0 volt reference.(a) For an input of 4.3 volts we find the output as,N10 = Vin/ Vref* 2n=4.3/12*(210)=366.933=366=16EHFor an input of 8.2 volts the output is;N10 = (2n )*Vin/Vref=8.2/12*210=699.733=699=2BBH (b) For A5H we first find that A5H = 16510 Then,Vin=Vref( Input in decimal)/2n=12*165/210=1.934V

But the output will stay A5H until the input changes by the voltage of one LSB,

AV = Vref/2n =12/210=0.012V, so the range is 1.934 V to (1.934 + 0.012) = 1.946 V.

c) The equation to be used is: R(T) = R(T0)[1 + α0(T-T0)]WhereR(T) is the approximation of resistance at temperature TR(T0) is the resistance at temperature T0

α0 is the fractional change in resistance per degree of temperature at T0

Therefore: R(35°C) = 108[1 + 0.006(35 - 25)] = 114.48 Ω

Q.4. a)

b)

• Quick Opening: This type of valve is used predominantly for full On /full OFF control applications. The valve characteristic of fig. 7.47 shows that a relatively small motion of the valve stem results in maximum possible flow rate through the valve. Such a valve, for example, may allow 90% of maximum flow rate with only a 30% travel of the stem.

• Linear: This type of valve, as shown in fig. 7.47, has a flow rate that varies linearly with the stem position. It represents the ideal situation where the valve alone determines the pressure drop. The relationship is expressed as

where Q=flow rate(m3/s); S=Stem position(m) Qmax=maximum flow rate(m3/s) Smax=maximum stem position(m)

• Equal Percentage: A very important type of valve employed in flow control has a characteristic such that a given percentage change in stem position produces an equivalent change in flow-that is, an equal percentage.

• Generally, this type of valve does not shut off the flow completely in its limit of stem travel.

• Thus, Qmin represents the minimum flow when the stem is at one limit of its travel. At the other extreme, the valve allows a flow Qmax as its maximum, open-valve flow rate. For this type, we define rangeability,R, as the ratio

R=Qmax/Qmin Q.5.

Q.6. a)

b)

c)