EGR 100 Take-Home Exam: Analyzing Thermistor Resistance & Temperature

This assignment solution focuses on analyzing the relationship between resistance and temperature in a thermistor, as part of an EGR 100 Introduction to Engineering Design course. The solution includes calculating temperature values in Celsius for given resistance values using a specific equation, finding the average and standard deviation of experimental resistance and temperature values, and creating scatter plots to visualize the relationship between average temperature and resistance over time. Error bars representing standard deviation are added to the data sets, along with trendlines, regression equations, and R-squared values. The calculations are presented in tables, and the analysis extends to discussing the factors determining the power output of water flow, including velocity, cross-section area, and conversion efficiency, along with a brief overview of Kaplan turbines. This document is available on Desklib, a platform offering a wide range of study resources and solved assignments for students.

Solution
( )
a) Calculate the temperature values in Celsius - C for the resistance values below using the
above equation.
Experiment 1
1/T0 1/B R for
Experiment
1
R0 (R/R0) ln(R/R0) 1/T T (K) T (0C)
0.00335402 0.00025316 800 10000 0.08 -2.5257 0.002715 368.3786 95.22857
0.00335402 0.00025316 1000 10000 0.1 -2.3026 0.002771 360.8702 87.72017
0.00335402 0.00025316 1200 10000 0.12 -2.1203 0.002817 354.9584 81.80839
0.00335402 0.00025316 1400 10000 0.14 -1.9661 0.002856 350.107 76.95701
0.00335402 0.00025316 1450 10000 0.145 -1.931 0.002865 349.0212 75.87117
0.00335402 0.00025316 1450 10000 0.145 -1.931 0.002865 349.0212 75.87117
0.00335402 0.00025316 1450 10000 0.145 -1.931 0.002865 349.0212 75.87117
0.00335402 0.00025316 1470 10000 0.147 -1.9173 0.002869 348.5992 75.44919
0.00335402 0.00025316 1500 10000 0.15 -1.8971 0.002874 347.9788 74.82884
0.00335402 0.00025316 1500 10000 0.15 -1.8971 0.002874 347.9788 74.82884
Experiment 2
1/T0 1/B R for
Experiment
2
R0 (R/R0) ln(R/R0) 1/T T (K) T (0C)
0.00335402 0.00025316 850 10000 0.085 -2.4651 0.00273 366.3084 93.15835
0.00335402 0.00025316 1100 10000 0.11 -2.2073 0.002795 357.7553 84.60535
0.00335402 0.00025316 1280 10000 0.128 -2.0557 0.002834 352.9097 79.7597
0.00335402 0.00025316 1450 10000 0.145 -1.931 0.002865 349.0212 75.87117
0.00335402 0.00025316 1520 10000 0.152 -1.8839 0.002877 347.5747 74.42466
0.00335402 0.00025316 1550 10000 0.155 -1.8643 0.002882 346.9762 73.82624
0.00335402 0.00025316 1550 10000 0.155 -1.8643 0.002882 346.9762 73.82624
0.00335402 0.00025316 1570 10000 0.157 -1.8515 0.002885 346.5865 73.43654
0.00335402 0.00025316 1600 10000 0.16 -1.8326 0.00289 346.0127 72.86273
0.00335402 0.00025316 1600 10000 0.16 -1.8326 0.00289 346.0127 72.86273

b) Find the average and standard deviation of the experimental resistance values and the
calculated temperature values.
Table of average value of resistance
R for
Experiment
1
R for
Experiment
2
Average
R
values
800 850 825
1000 1100 1050
1200 1280 1240
1400 1450 1425
1450 1520 1485
1450 1550 1500
1450 1550 1500
1470 1570 1520
1500 1600 1550
1500 1600 1550
Table showing the calculation of standard deviation
Average R
values̅ ( ̅ Variance s.d
825 -539.5 291060.3 61485.83 247.9634
1050 -314.5 98910.25
1240 -124.5 15500.25
1425 60.5 3660.25
1485 120.5 14520.25
1500 135.5 18360.25
1500 135.5 18360.25
1520 155.5 24180.25
1550 185.5 34410.25
1550 185.5 34410.25
Total 13645 553372.5
mean 1364.5
n 10

Table of average temperatures in degree Celsius
T (0C) T (0C) Average
temperature
values (0C)
95.22857 93.15835 94.19346
87.72017 84.60535 86.16276
81.80839 79.7597 80.78405
76.95701 75.87117 76.41409
75.87117 74.42466 75.14792
75.87117 73.82624 74.84871
75.87117 73.82624 74.84871
75.44919 73.43654 74.44287
74.82884 72.86273 73.84579
74.82884 72.86273 73.84579
Table showing the calculation of standard deviation
Average temperature values
(0C) ( ̅ ( ̅ Variance s.d
94.19346 15.74005 247.749 45.80363 6.767838
86.16276 7.709345 59.434
80.78405 2.330635 5.43186
76.41409 -2.03932 4.158846
75.14792 -3.30549 10.9263
74.84871 -3.6047 12.9939
74.84871 -3.6047 12.9939
74.44287 -4.01054 16.08447
73.84579 -4.60762 21.23021
73.84579 -4.60762 21.23021
Total 784.53415 412.2327
mean 78.453415
n 10

c) Plot the average temperature values in a scatter plot in triangles (y-axis), and the average
resistance values on a secondary y-axis in a scatter plot as squares. Feel free to change the
color of the data points, but make sure to graph the time on the x-axis.
Graph of average temperature – average resistance against time
d) The standard deviation of the each set of values should be used as error bars in each data set,
and add a trendline as well for each data set. The regression equations and R-squared values
for the trendlines should be displayed on the figure.
0
200
400
600
800
1000
1200
1400
1600
1800
0
10
20
30
40
50
60
70
80
90
100
0 2 4 6
average temperature and average
resistance
Time in minutes
Average temperature
values (0C)
Average R values

Graph of average temperature – average resistance against time
e) Create a table of all of your calculations.
1/T0 1/B R for
Experiment
1
R0 (R/R0) ln(R/R0) 1/T T (K) T (0C)
0.00335
402
0.00025
316
800 100
00
0.08 -2.5257 0.0027
15
368.3
786
95.22
857
0.00335
402
0.00025
316
1000 100
00
0.1 -2.3026 0.0027
71
360.8
702
87.72
017
0.00335
402
0.00025
316
1200 100
00
0.12 -2.1203 0.0028
17
354.9
584
81.80
839
0.00335
402
0.00025
316
1400 100
00
0.14 -1.9661 0.0028
56
350.1
07
76.95
701
0.00335
402
0.00025
316
1450 100
00
0.145 -1.931 0.0028
65
349.0
212
75.87
117
0.00335
402
0.00025
316
1450 100
00
0.145 -1.931 0.0028
65
349.0
212
75.87
117
0.00335
402
0.00025
316
1450 100
00
0.145 -1.931 0.0028
65
349.0
212
75.87
117
0.00335
402
0.00025
316
1470 100
00
0.147 -1.9173 0.0028
69
348.5
992
75.44
919
0.00335
402
0.00025
316
1500 100
00
0.15 -1.8971 0.0028
74
347.9
788
74.82
884
0.00335
402
0.00025
316
1500 100
00
0.15 -1.8971 0.0028
74
347.9
788
74.82
884
1/T0 1/B R for R0 (R/R0) ln(R/R0) 1/T T (K) T (0C)
y = -3.7097x + 88.655
R² = 0.6885
y = 141.39x + 975.67
R² = 0.7451
-80000
-60000
-40000
-20000
0
20000
40000
60000
80000
0
20
40
60
80
100
120
140
-1 0 1 2 3 4 5 6 7
average temperature and average
resistance
Time in minutes

Experiment
2
0.00335
402
0.00025
316
850 100
00
0.085 -2.4651 0.0027
3
366.3
084
93.15
835
0.00335
402
0.00025
316
1100 100
00
0.11 -2.2073 0.0027
95
357.7
553
84.60
535
0.00335
402
0.00025
316
1280 100
00
0.128 -2.0557 0.0028
34
352.9
097
79.75
97
0.00335
402
0.00025
316
1450 100
00
0.145 -1.931 0.0028
65
349.0
212
75.87
117
0.00335
402
0.00025
316
1520 100
00
0.152 -1.8839 0.0028
77
347.5
747
74.42
466
0.00335
402
0.00025
316
1550 100
00
0.155 -1.8643 0.0028
82
346.9
762
73.82
624
0.00335
402
0.00025
316
1550 100
00
0.155 -1.8643 0.0028
82
346.9
762
73.82
624
0.00335
402
0.00025
316
1570 100
00
0.157 -1.8515 0.0028
85
346.5
865
73.43
654
0.00335
402
0.00025
316
1600 100
00
0.16 -1.8326 0.0028
9
346.0
127
72.86
273
0.00335
402
0.00025
316
1600 100
00
0.16 -1.8326 0.0028
9
346.0
127
72.86
273
R for
Experim
ent 1
R for
Experim
ent 2
Average R
values
Average R
values
Varian
ce
s.d
800 850 825 825 -539.5 291060
.3
61485
.83
247.9634
1000 1100 1050 1050 -314.5 98910.
25
1200 1280 1240 1240 -124.5 15500.
25
1400 1450 1425 1425 60.5 3660.2
5
1450 1520 1485 1485 120.5 14520.
25
1450 1550 1500 1500 135.5 18360.
25
1450 1550 1500 1500 135.5 18360.
25
1470 1570 1520 1520 155.5 24180.
25
1500 1600 1550 1550 185.5 34410.
25
1500 1600 1550 1550 185.5 34410.
25

Tot
al
13645 553372
.5
mea
n
1364.5
n 10
T (0C) T (0C) Average
temperatur
e values
(0C)
Average
temperat
ure values
(0C)
(
Varian
ce
s.d
95.2285
7
93.1583
5
94.19346
87.7201
7
84.6053
5
86.16276 94.19346 15.740
05
247.74
9
45.80
363
6.767838
81.8083
9
79.7597 80.78405 86.16276 7.7093
45
59.434
76.9570
1
75.8711
7
76.41409 80.78405 2.3306
35
5.4318
6
75.8711
7
74.4246
6
75.14792 76.41409 -
2.0393
2
4.1588
46
75.8711
7
73.8262
4
74.84871 75.14792 -
3.3054
9
10.926
3
75.8711
7
73.8262
4
74.84871 74.84871 -3.6047 12.993
9
75.4491
9
73.4365
4
74.44287 74.84871 -3.6047 12.993
9
74.8288
4
72.8627
3
73.84579 74.44287 -
4.0105
4
16.084
47
74.8288
4
72.8627
3
73.84579 73.84579 -
4.6076
2
21.230
21
73.84579 -
4.6076
2
21.230
21
Tot
al
784.5342 412.23
27
mea
n
78.45342
n 10

Q2)
a) Water possess energy by the virtue of its flow
The factors that determine the power output depend on
a) The velocity V (m/s)
b) The cross-section area A
c) The overall conversion efficiency (ŋ)
The principle of converting kinetic energy to mechanical energy and then electrical energy
K.E = ½ *m*v2……………………………………..1
M = mass of water
M = A * ………….2
Where,
A = cross-section area
Combining equation 1 and 2
K.E = (
K.E energy per second is equivalent to power
Power in water =
Power developed =
Representation of units
Power unit =
If we divide like terms

But, Energy =
Time = s
Power = Energy / time
Power =
b)
KAPLAN TURBINE