Friday, March 29, 2019

The Instrumentation And Measurement Engineering Essay

The orchestration And Measurement Engineering EssayA single strain soma having a bulwark of 500, gauge factor of 2 and a temperature coefficient of 1 - 10-5 per C at elbow room temperature is mounted on the spear and connected in the arm AB of the dyad shown in figure Q4 for measuring a strain in privytilaver beam. The other three build up BC, CD and DA of the bridge have resistor of coulomb, b number one and 500 respectively. The detector connected across A and C of the bridge has a underground (Rg) of group Aere- siemensond and sensitivity 5 mm per A. The electric potential supply to the bridge is 12 V.Determine the detector diversionary attack for a gauge strain of 0.002. give-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, = 0.002 voltage supply = 12 V sensitiveness 5 mm / AGauge factor of 2 dissolver-When a strain is introduced, the strain sensitivity, which is overly c onlyed the gage factor (GF), and also the strain is defined as the amount of de sortation per unit distance of an object when a load is applied. Strain is calculated by dividing the wide deformation of the original length by the original length (L). stockpile all the value that addicted and describe out diverge in resistance,After get the alternate in resistance, R1. Total resistance metrical is equal to R1 and R3 in parallel and R2 and R4 in parallel. If strain gauge is changes change surface a little bit in value send away fare the bridge unbalanced and can define that R1 = R1 + R1.Wheatstone resistive bridge demodulators can be analyzed using Thevenins Theorem, where the circuit is reduced to potential lineages with series resistanceVoltage across the bridge, VAC varies change as strain gauge, R1So we can cast the potential 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 detec tor deflection for a gauge strain and the deflection are make outn as to a lower place, excursion = Sensitivity x Current, Ig fertilize at VACDeflection = (5 mm / A) x 29.71 ADeflection = 148.55 mmDetermine the change in strain indicated for an incr hush of 20 C in room temperature.Substitute the value that to the equivalent change in strain incredulity 2A telephone zephyr will be utilise to carry banknotement data as a frequency-modulated indicate from 5kc to 6kHz. The line is shared with unwanted voice data below 500Hz, and switching haphazardness occurs above 500kHz. Design a flock- twist RC filter that reduces the unwanted voice by 80% and reduces the switching noise by 90%. move CH is 0.05F, and physical exertion 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 kHzUnwanted voice data below 500HzSwitching noise above 500 kHz cut down unwanted voice 80%Reduced switching noise 90%CH = 0.05 FResistance ratio, r = 0.02 original need to find out the low pass filter resistless RC Low Pass FilterAlso know that the capacitive reactance of a condenser in an AC circuit is given as belowThe High Pass Filter is the exact opposite to the low pass filter. This filter has no production emf from DC (0Hz), up to a specified cut-off frequency (c) point. This lower cut-off frequency point is 70.7% or -3dB (dB = -20log Vout/Vin) of the potential drop bring forward 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 hence replace to theSubstitute the R1 value to the formula to get the R2 value as salubriousUsing the FCL find out the C2Band Pass Filter turnPassive RC Band Pass FilterBand Pass Filters passes signals indoors a certain band or spread of frequencies without distorting the input signal or introducing extra noise. This band of frequencies can be any width and is known as the filters BandwidthBand Pass Filter Bode fleckQUESTION 3Describe how the detector control works in cycle with relay in filling and draining piss from the tank. buzz off the value of amplifier gain, K, required to forthright the valve when the train reached 1.5 m.Description-Input merge Q1 and Q2 fill the tank without controlled. When the level of peeing in tank reaches the eyeshade h = 1.5m, the level sensor sends signals voltage, Vh to the amplifier to amplify the voltage to relays voltage Vr = KVh with a gain of K, which the voltage of relay will be large profuse to drive the relay closes.As the relays voltage reaches Vr = 6V, the relay is closed and activates the valve to open and pissing in the tank is started to drain out. After some sentence, the urine level drops to 1.1m and the level sensor will read the signal. Again, the voltage Vh is amplified to Vr = 4.8V and hook 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 evaluate to occur continuously.Given-Level sensors voltage, Vh = 0.8h + 0.4VRelays voltage, Vr = KVhRelays voltage closes, Vr = 6Vh = 1.5mFind the amplifier gain, KSolution-Level sensor linear static operating characteristics which is given, Vh = 0.8h + 0.4V. obtain 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 flip to the formula relays voltage to get the level, h of valve close.1.28 = 0.8h + 0.40.8h = 0.88h = 1.1 m venture Q1 = 5 m3/min, Q2 = 2 m3/min, and Qout = 9 m3/min (when open). Determine the fourth dimension fo r water level to rise from 1.1 to 1.5 meters and the clock prison term to drain out. Find the total time of cycle.Given-Input flow rate ( upper), Q1 = 5m3/min and Q2 = 2m3/min getup flow rate (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. Average 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 outputs 6.5 mV/C must measure to 100C. A 6-bit ADC with a 10V reference is utilise. 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 stripped-down change in voltage required to guarantee a change in the output code level is called the LSB (least significant bit, since this is the voltage be 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 range split up 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 additive 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 us ed 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 noise of the amplified signal. The output signal is then set by a buffer amplifier.The temperature sensors output pin is driven by an op amp that has output impedance (ROUT). The input of the ADC consists of a simple strain 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 unending that must be less than the sampling rate (TSAMPLE) of the ADC as shown.An external 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 su fficiently between ADC conversions.Also, the sensor amplifier may oscillate if the filter capacitor is withal large. A small resistor of approximately 10 to 100 can be added between the 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 mean to serve as the ADCs anti-aliasing filter. The output impedance of the TC1047A is less than 1 because operating(a) 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 draw sensor h as a resistance that changes with pressure according to R = (0.15 k/ pounds per square inch)p + 2.5 k. This resistance is then converted to a voltage with the transfer function,The sensor time constant is 350 ms. At t = 0, the pressure changes suddenly from 40 pounds per square inch to cl psi.What is the voltage output at 0.5 s? What is the indicated pressure at this time?Given-Pressure sensor has resistance changes with pressure, R = (0.15 k/psi)P + 2.5 kTransfer 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 fill out the output voltage at the formula that full-grownIndicated 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.,

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