The Instrumentation And Measurement Engineering Essay

The Instrumentation And Measurement Engineering EssayA single strain gauge having a exemption of 500, gauge cistron of 2 and a temperature coefficient of 1 - 10-5 per C at room temperature is mounted on the beam and connected in the arm AB of the bridge shown in figure Q4 for measuring a strain in cantilaver beam. The other three arms BC, CD and DA of the bridge have electrical opposite of coulomb, 100 and 500 respectively. The detector connected crossways A and C of the bridge has a opponent (Rg) of 100 and sensitivity 5 mm per A. The potential put out to the bridge is 12 V.Deter seconde the detector deflection for a gauge strain of 0.002.Given-R1 = 500 R2 = 100 R3 = 500 R4 = 100 Resistance across A and C of the bridge, Rg = 100 = temperature coefficient of 1 x 10-5 CGauge strain, = 0.002Voltage supply = 12 VSensitivity 5 mm / AGauge factor of 2 dissolvent-When a strain is introduced, the strain sensitivity, which is also called the gage factor (GF), and also the strain i s set upd as the tally of deformation per unit length of an object when a load is applied. Strain is calculated by dividing the total deformation of the skipper length by the original length (L).Substitute all the value that given and find out change in resistance,After get the change in resistance, R1. Total resistance measured is equal to R1 and R3 in parallel and R2 and R4 in parallel. If strain gauge is changes even a little bit in value can sweat the bridge unbalanced and can define that R1 = R1 + R1.Wheatstone resistive bridge sensors can be analyzed using Thevenins Theorem, where the go is reduced to potency sources with series resistanceVoltage across the bridge, VAC varies change as strain gauge, R1So we can deter minute of arce the voltage across the bridge, VAC terminals by applying the Ohms Law.R is the resistance that across the A and C of the bridgeLastly, determine the detector deflection for a gauge strain and the deflection are known as below,Deflection = Sen sitivity x Current, Ig flow at VACDeflection = (5 mm / A) x 29.71 ADeflection = 148.55 mmDetermine the change in strain indicated for an increase of 20 C in room temperature.Substitute the value that to the equivalent change in strainQUESTION 2A telephone line result be lend oneselfd to carry measurement data as a frequency-modulated signal from 5kHz to 6kHz. The line is shared with unwanted voice data below 500Hz, and replacement disruption spends above 500kHz. Design a band-pass RC filter that reduces the unwanted voice by 80% and reduces the switching mental disturbance by 90%. Assume CH is 0.05F, and use a resistance ratio r of 0.02. What is the Vout/Vin on the passband frequency of 5.5kHz?Given-Frequency modulated signal from 5 kHz to 6 kHz unwished-for voice data below 500HzSwitching noise above 500 kHzReduced unwanted voice 80%Reduced switching noise 90%CH = 0.05 FResistance ratio, r = 0.02First need to find out the low pass filterPassive RC Low Pass perkAlso know tha t the capacitive reactance of a capacitor in an AC overlap is given as belowThe High Pass Filter is the guide opposite to the low pass filter. This filter has no proceeds voltage from DC (0Hz), up to a specified cut-off frequency (c) baksheesh. This lower cut-off frequency point is 70.7% or -3dB (dB = -20log Vout/Vin) of the voltage gain allowed to pass.Passive RC High Pass FilterAlso know that the capacitive reactance of a capacitor in an AC circuit is given as belowAfter getting value of then substituting to theSubstitute the R1 value to the law to get the R2 value as headUsing the FCL find out the C2Band Pass Filter CircuitPassive RC Band Pass FilterBand Pass Filters passes signals within a certain band or spread of frequencies without distorting the foreplay signal or introducing extra noise. This band of frequencies can be any width and is known as the filters BandwidthBand Pass Filter Bode PlotQUESTION 3Describe how the sensor control works in cycle with relay in fillin g and draining weewee from the tank. Find the value of amplifier gain, K, required to open the valve when the level chance oned 1.5 m.Description-Input flow Q1 and Q2 fill the tank without controlled. When the level of weewee in tank reaches the height h = 1.5m, the level sensor sends signals voltage, Vh to the amplifier to affix the voltage to relays voltage Vr = KVh with a gain of K, which the voltage of relay will be large enough to drive the relay closes.As the relays voltage reaches Vr = 6V, the relay is closed and activates the valve to open and water in the tank is started to drain out. After whatever time, the water level drops to 1.1m and the level sensor will read the signal. Again, the voltage Vh is amplified to Vr = 4.8V and latch the relay to open. The open relay is then instructed the valve to close.Even though the water stops draining out, the water tank is still filling with water. The water level will increase to h = 1.5m again. The same cycle is expected to oc cur continuously.Given-Level sensors voltage, Vh = 0.8h + 0.4VRelays voltage, Vr = KVhRelays voltage closes, Vr = 6Vh = 1.5mFind the amplifier gain, K antecedent-Level sensor linear static operating characteristics which is given, Vh = 0.8h + 0.4V. Apply this formula to the voltage relay to get the value of amplifier gain, K. Substitute all the info that given to the relays voltage, Vr = KVhAt what level does the valve close?Given-Relays voltage closes, Vr = 4.8VLevel sensors voltage, Vh = 0.8h + 0.4VAmplifier gain, K = 3.75Solution-When the valve close, the voltage of relay, Vr = 4.8V and given that K = 3.75 and rest to the formula relays voltage to get the level, h of valve close.1.28 = 0.8h + 0.40.8h = 0.88h = 1.1 mSuppose Q1 = 5 m3/min, Q2 = 2 m3/min, and Qout = 9 m3/min (when open). Determine the time for water level to rise from 1.1 to 1.5 meters and the time to drain out. Find the total time of cycle.Given-Input flow rate (velocity), Q1 = 5m3/min and Q2 = 2m3/min make flow r ate (velocity), Qout = 9m3/min (when valve open)Solution-Time for water level to rise from 1.1m to 1.5mVelocity shows how fast an object is moving to which direction. total velocity can be calculated by dividing displacement over time.Where the t1 is the time when water start rise at height 1.1m. Assume the t1 = 0 (Initial time)Time for water level to drain out from 1.5m to 1.1mOutput flow rate (velocity), Qout = 9m3/min (when valve open)t2 is the time when water reaches 1.1m, water drains out is stopped. Assume t2 = 0QUESTION 4A measurement of temperature using a sensor that creates 6.5 mV/C must measure to 100C. A 6-bit ADC with a 10V extension service is used. Develop a circuit to interface the sensor and the ADC. Find the temperature resolution.Given-Output of the sensor = 6.5 mV/ C measure to 100C6-bit ADC = 10VrefSolution-Find the output sensor during 100C where the output sensor 6.5 mV/ C measure 1C is given.Resolution can define electrically, and expressed in volts. The minimum change in voltage required to guarantee a change in the output code level is called the LSB (least authoritative bit, since this is the voltage represented by a change in the LSB).The resolution Q of the ADC is equal to the LSB voltage. The voltage resolution of an ADC is equal to its overall voltage measurement identify divided by the number of discrete voltage intervalsN is the number of voltage intervals,EFSR is the full scale voltage range, 10 VNormally, the number of voltage intervals is given by,Where the M is the ADCs resolution in bits.Solution-Develop a circuit to interface the sensorBlock DiagramFigure 4.1 Interfacing an Analog Output Temperature Sensor to an ADCAt first sensor consists of a band gap reference circuit that produces a voltage.A switched capacitor op amp amplifier is used to amplify the temperature coefficient to a voltage mV/C because of the ease of building capacitors that are a ratio of each other.Lowpass filter is used to remove the switching n oise of the amplified signal. The output signal is then driven by a buffer amplifier.The temperature sensors output pin is driven by an op amp that has output opposition (ROUT). The input of the ADC consists of a simple sample and hold circuit.A switch is used to connect the signal source with a sampling capacitor, while the ADC measures the CSAMPLE capacitors voltage in order to determine the temperature. The ROUT and RSWITCH resistances and the CSAMPLE capacitor form a time constant that must be less than the sampling rate (TSAMPLE) of the ADC as shown.An foreign capacitor can be added to the output pin to provide additional filtering and to form an anti-aliasing filter for the ADC. This capacitor may impact the time response of the sensor and the designer must allow time for the capacitor to charge sufficiently between ADC conversions.Also, the sensor amplifier may oscillate if the filter capacitor is too large. A small resistor of approximately 10 to 100 can be added between t he output pin of the sensor and CFILTER to isolate the sensors amplifier from the capacitive load.The output impedance of the sensor (ROUT) varies as a function of frequency. Thus, a series resistor should be added to the effective ROUT resistance if CFILTER is intended to serve as the ADCs anti-aliasing filter. The output impedance of the TC1047A is less than 1 because operational amplifier A2 functions as a voltage buffer.The output impedance of the sensor is low due to the negative feedback of the buffer circuit topology. The negative feedback results in an output impedance that is equal to the impedance of the amplifier divided by the open-loop gain of the amplifier. The open-loop gain of the op amp is relatively large which, in turn, forces the output impedance to be small.QUESTION 5A pressure sensor has a resistance that changes with pressure according to R = (0.15 k/psi)p + 2.5 k. This resistance is then converted to a voltage with the transfer function,The sensor time consta nt is 350 ms. At t = 0, the pressure changes suddenly from 40 psi to cl psi.What is the voltage output at 0.5 s? What is the indicated pressure at this time?Given- drag sensor has resistance changes with pressure, R = (0.15 k/psi)P + 2.5 k take function of voltage,Sensor time constant, = 350 msAt t = 0, Pressure, P change suddenly from 40 psi 150 psiSolution-Pressure changes suddenly from 40 psi to 150 psi and we can assume that the initial pressure, Po = 150 psiVoltage output after t = 0.5 sec, find pressure after 0.5 sec firstBasic formula a quantity pressure depends exponentially on time t if.During the resistance 0.5 sec, P = 35.95 psi. Substitute the indicated pressure at 0.5s, P in resistance changes with pressure according to to find out the resistance during 0.5 secVoltage output at 0.5 s,2. At what time does the output reach 5.0 V?Find out the resistance by substitute the output voltage at the formula that givingIndicated pressure at output 5.0 V50 psiA quantity pressure depends exponentially on time t if, substitute the value of initial pressure, indicated pressure output at 5.0 V and time constant is giving 350 ms. Finally we can get the time does the output reach 5.0 V.,