Instrumentation Measurement & Lab
Temperature Alarm Project
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Part 1
Question 1. A temperature sensor is exposed to a sudden change of 20^{o}C to 80^{o}C. The sensor outputs 0.02 volts for every^{ o}C of temperature and has a 2.3 second time constant.
a. What is the sensor output voltage at 1.5 seconds?
Solution: The transient component of the voltage is given by
v(t)=V_{s} (1e^{t/τ} )u(t)
Where
V_{s} = 0.02*(8020)=1.2 V
v(t=1.5)=1.2(1e^{1.5/2.3})u(1.5)=1.2(10.52091)=0.5749 V
Therefore, total Voltage is steady state voltage + transient voltage
V(t=1.5)=V_{0} + v_{t} = 0.02*20+0.5749=0.9749 V
b. At what time, t, does the sensor output become 1.0 volt?
Solution:
V=V_{0}+v(t)=V_{0}+V_{s} (1e^{t/τ} )u(t)
0.4+1.2(1e^{t/2.3} )=1
1.2(1e^{t/2.3} )=10.4=0.6
(1e^{t/2.3} )=0.6/1.2=0.5
e^{t/2.3}=0.5
t/2.3=ln?(0.5)
t=2.3*0.69315=1.59424 sec
Question 2. A sensor, R, that changes resistance is used in a bridge circuit as shown below.
a. What sensor resistance, R, will null the bridge?
Solution:
The bridge will be null if the resistance network is balanced
R_{1}/R_{2} = R_{3}/R
458/677=344/R
R=344*677/458=508.489083 ?
b. What is the offnull voltage if R changes by 0.5 O?
Solution:
We have
ΔV=E(R_{2}/(R_{2}+R_{3} )R/(R_{1}+R))=5(677/(677+458)508.989083/(344+508.989083))
ΔV=5(677/(677+458)508.989083/(344+508.989083))=1.1826 mV
Question 3. A measurement system has noise above 20 kHz and the data signal frequency is between 100 and 500 Hz. Design an RC filter that reduces the noise by 95%, i.e., only 5% is left. Show the schematic and the component values.
Solution:
Transfer function:
G(s) = 121951.2195122/s+121951.2195122
R = 8.2 Ω
C = 1 uF
Cut off frequency
fc = 19409.139401451[Hz]
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Question 4. Given a Type J TC with a 25^{o}C reference.
a. What is the temperature if the TC voltage is 22.87 mV?
Solution: From the Type J Thermocouple we have
dV/dT=0.055321 mV/°C
Therefore V=is 22.87 mV
T=441.639 °C
b. What voltage would result from a temperature of 225oC?
Solution:
dV/dT=0.055321 mV/°C
Therefore, for T=225
ΔV= 11.0642 mV
Question 5. A 10bit ADC has a 5.00volt reference.
a. What binary output is produced by an input of 3.04 volts?
Solution:
V

Vref

V Vref

ADC output


3.04

2.5

0.54

1

MSB

0.54

1.25

0.71

0


0.54

0.625

0.085

0


0.54

0.3125

0.2275

1


0.2275

0.15625

0.07125

1


0.07125

0.078125

0.00687

0


0.07125

0.039063

0.032188

1


0.032188

0.019531

0.012656

1


0.012656

0.009766

0.002891

1


0.002891

0.004883

0.00199

0

LSB

Therefore, Binary output is 1001101110
b. Suppose the output is found to be 1F4h. What is the possible input voltage?
Solution:
Binary 
0 
1 
1 
1 
1 
1 
0 
1 
0 
0 
v 
2.5 
1.25 
0.625 
0.3125 
0.15625 
0.078125 
0.039063 
0.019531 
0.009766 
0.004883 
Therefore, input Voltage = 1.25+0.625+0.3125+0.15625+0.078125+0.019531= 2.441406 V
Question 6. A 12 bit bipolar DAC has a 10volt reference.
a. If the hex input is 5D7h what is the DAC output voltage?
Solution:
Binary 
0 
1 
0 
1 
1 
1 
0 
1 
0 
1 
1 
1 
v 
10 
5 
2.5 
1.25 
0.625 
0.3125 
0.15625 
0.078125 
0.039063 
0.019531 
0.009766 
0.004883 
Output Voltage = 10 + (5+1.25+0.625+0.3125+0.078125+0.019531+0.009766+0.004883)
DAC output Voltage = 2.7002 V
b. What input is required to get a zero volt output?
Solution: The zero output voltage will be available for DAC input (100000000000)_{2}
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Part 2
Question 1. A golfer estimates that his golfball is 250 yards to the hole. At the hole, another golfer has dropped a water bottle that is reflecting light back to the golfer and his ball. What is the approximate time difference based on golfers distance to the hole/water bottle?
Solution:
Time=(total distnace)/(speed of light)=457.2/299792458=1.525 μs
Question 2. A CdS photocell has a resistance versus light intensity as shown in the following graph. The cell has a time constant of 25 ms. At t=0, the cell is exposed to a sudden change of light intensity 20 W/m2 to 80 W/m^{2}. What is the sensor resistance after 20 ms?
Figure 1: CdS Resistance Versus Light Intensity
Solution:
R(t)=R_{0} (e^{t/τ} )u(t)
R(t=20m)=8.5k(e^{20m/25m} )u(t)=3.82k?
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Question 3: A force of 120 poundsforce is used to open a valve. What is the area of the diaphragm if the pressure difference is 90 kPa (~ 13 psi) that must provide this force?
Solution:
Area=Force/Pressure=(533.787 N)/(90 kPa)=0.005930967m^{2}
Question 4: A processing plant is being proposed and will need to use of the city's water system to be sucessful. The plant will need a minimum of 2500 gallons of water per minute (gpm) for its operation. The city's current water supply pipe to the proposed plant is 12 inches in diameter with a water velocity of 1.5 m/s. Will the city's current water supply meet the plant's needs or will it need to be upgraded? Show calculations to verify your answer.
Solution:
Volume of water out of pipe in 1 min=π*(diameter of pipe/2)^{2}*velocity of water*time
Volume of water out of pipe in 1 min=π*(0.1524/2)^{2}*1.5*60=1.641732232m^{3}
City water Need=2500 gallons=9.463529m^{3}
Therefore, city's current water supply does not meet the plant's needs. Yes it need to be updated.
Question 5. From the figure below, construct the truth table for the logic diagram. What function does this represent?
Solution:
This represents NOR Gate.
Truth table
A

B

Output

0

0

1

0

1

0

1

0

0

1

1

0

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