PASCO TD-8564 User Manual

Includes
Teacher's Notes
and
Typical
Experiment Results
THERMAL EFFICIENCY
Instruction Manual and Experiment Guide for the PASCO scientific Model TD-8564
012-05443A
3/94
12.9
12.9
12.4
12.4
12.0
12.0
11.6
11.6
11.2
11.2
10.8
10.8
10.4
10.4
K
K
scientific
5±1%
5±1%
12 VDC MAX
12 VDC MAX
T
°C
°C
K
K
75
75
9.12
9.12
76
76
8.81
8.81
77
77
8.52
8.52
78
78
8.24
8.24
79
79
7.96
7.96
80
80
7.70
7.70
81
81
7.45
7.45
Cold
Reservoir
c
°C
°C
K
K
°C
°C
85
85
6.53
6.53
95
95
86
86
6.33
6.33
96
96
87
87
6.12
6.12
97
97
88
88
5.93
5.93
98
98
89
89
5.74
5.74
99
99
90
90
5.56
5.56
100
100
91
91
5.39
5.39
101
101
Hot
Reservoir
Q
c
W
Q
h
Heat
Engine
T
h
WATER
WATER
PUMP
PUMP
7.5 - 12 VDC
7.5 - 12 VDC
@500mA
@500mA
K
K
°C
°C
461
461
-5
-5
436
436
-4
-4
413
413
-3
-3
391
391
-2
-2
370
370
-1
-1
351
351
0
0
332
332
1
1
PASCO
Model TD-8564
Model TD-8564
THERMAL EFFICIENCY
THERMAL EFFICIENCY
APPARATUS
27.4
27.4
26.4
26.4
25.3
25.3
24.4
24.4
23.4
23.4
22.5
22.5
21.7
21.7
K
K
°C
°C
55
55
56
56
57
57
58
58
59
59
60
60
61
61
APPARATUS
HEATER
2.0Ω2.01.0Ω1.00.5Ω0.5
K
K
°C
°C
18.6
18.6
65
65
17.9
17.9
66
66
17.3
17.3
67
67
16.6
16.6
68
68
16.0
16.0
69
69
15.5
15.5
70
70
14.9
14.9
71
71
THERMISTOR
THERMISTOR
SELECT
SELECT
PELTIER
PELTIER
DEVICE
DEVICE
COOLING
COOLING
WATER
WATER
THERMISTOR TABLETHERMISTOR TABLE
K
K
°C
K
K
°C
°C
K
K
°C
°C
K
K
°C
°C
K
269
269
5
5
161
161
15
255
255
242
242
230
230
218
218
207
207
197
197
15
6
6
153
153
16
16
7
7
146
146
17
17
8
8
139
139
18
18
9
9
133
133
19
19
10
10
126
126
20
20
11
11
120
120
21
21
95.4
95.4
91.1
91.1
87.0
87.0
83.1
83.1
79.4
79.4
75.9
75.9
K
100
100
25
25
63.4
63.4
26
26
60.7
60.7
27
27
58.1
58.1
28
28
55.6
55.6
29
29
53.2
53.2
30
30
51.0
51.0
31
31
48.9
48.9
°C
°C
°C
41.2
41.2
45
45
35
35
39.6
39.6
46
46
36
36
37.9
37.9
47
47
37
37
36.4
36.4
48
48
38
38
34.9
34.9
49
49
39
39
33.5
33.5
50
50
40
40
32.2
32.2
51
51
41
41
© 1991 PASCO scientific $10.00
012-05443A Thermal Efficiency Apparatus
T able of Contents
Copyright, Warranty and Equipment Return...................................................ii
Introduction .....................................................................................................1
Quick Start.......................................................................................................2
Theory ............................................................................................................. 3
HEAT ENGINE:
Introduction ...............................................................................................3
Actual Efficiency.......................................................................................3
Carnot Efficiency.......................................................................................3
Adjusted Efficiency ...................................................................................3
HEAT PUMP (REFRIGERATOR):
Introduction ...............................................................................................4
Actual Coefficient of Performance............................................................ 4
Maximum Coefficient of Performance...................................................... 4
Adjusted Coefficient of Performance ........................................................4
MEASUREMENTS USING THE THERMAL EFFICIENCY APPARATUS:
Direct Measurements.................................................................................5
Temperatures .......................................................................................5
Power Delivered to the Hot Reservoir (PH) ......................................... 6
Power Dissipated by the Load Resistor (PW)....................................... 6
Indirect Measurements ..............................................................................6
Internal Resistance...............................................................................6
Heat Conduction and Radiation...........................................................6
Heat Pumped from the Cold Reservoir................................................7
EXPERIMENTS:
1 — Heat Engine and Temperature Difference ......................................... 9
2 — Heat Engine Efficiency (Detailed Study) .........................................13
3 — Heat Pump Coefficient of Performance............................................17
4 — Thermal Conductivity.......................................................................20
5 — Load for Optimum Performance.......................................................21
Teacher’s Guide..............................................................................................25
Technical Support................................................................. Inside Back Cover
i
Thermal Efficiency Apparatus 012-05443A
Copyright, Warranty and Equipment Return
Please—Feel free to duplicate this manual subject to the copyright restrictions below.
Copyright Notice
The PASCO scientific Model TD-8564 Thermal Effi­ciency Apparatus manual is copyrighted and all rights reserved. However, permission is granted to non-profit educational institutions for reproduction of any part of this manual providing the reproductions are used only for their laboratories and are not sold for profit. Reproduc­tion under any other circumstances, without the written consent of PASCO scientific, is prohibited.
Limited Warranty
PASCO scientific warrants this product to be free from defects in materials and workmanship for a period of one year from the date of shipment to the customer. PASCO will repair or replace, at its option, any part of the product which is deemed to be defective in material or workman­ship. This warranty does not cover damage to the product caused by abuse or improper use. Determination of whether a product failure is the result of a manufacturing defect or improper use by the customer shall be made solely by PASCO scientific. Responsibility for the return of equipment for warranty repair belongs to the customer. Equipment must be properly packed to prevent damage and shipped postage or freight prepaid. (Damage caused by improper packing of the equipment for return ship­ment will not be covered by the warranty.) Shipping costs for returning the equipment, after repair, will be paid by PASCO scientific.
Equipment Return
Should the product have to be returned to PASCO scientific for any reason, notify PASCO scientific by letter, phone, or fax BEFORE returning the product. Upon notification, the return authorization and shipping instructions will be promptly issued.
ä
NOTE: NO EQUIPMENT WILL BE
ACCEPTED FOR RETURN WITHOUT AN AUTHORIZATION FROM PASCO.
When returning equipment for repair, the units must be packed properly. Carriers will not accept responsibility for damage caused by improper packing. To be certain the unit will not be damaged in shipment, observe the following rules:
The packing carton must be strong enough for the
item shipped.
Make certain there are at least two inches of
packing material between any point on the apparatus and the inside walls of the carton.
Make certain that the packing material cannot shift
in the box or become compressed, allowing the instrument come in contact with the packing carton.
Credits
This manual authored by: Ann Hanks This manual edited by: Ann Hanks and Eric Ayars Teacher’s Guide written by: Eric Ayars
Address: PASCO scientific
10101 Foothills Blvd. Roseville, CA 95747-7100
Phone: (916) 786-3800 FAX: (916) 786-3292 email: techsupp@pasco.com web: www.pasco.com
ii
012-05443A Thermal Efficiency Apparatus
Introduction
The Thermal Efficiency Apparatus can be used as a heat engine or a heat pump. When used as a heat engine, heat from the hot reservoir is used to do work by running a current through a load resistor. The actual efficiency of this real heat engine can be ob­tained and compared to the theoretical maximum efficiency . When used as a heat pump to transfer heat from the cold reservoir to the hot reservoir, the actual coefficient of performance and the theoretical maxi­mum coefficient of performance can be obtained.
The apparatus is built around a thermoelectric con­verter called a Peltier device. To simulate the theoreti­cal heat engines found in textbooks which have infinite hot and cold reservoirs, one side of the Peltier device is maintained at a constant cold temperature by pump­ing ice water through the block and the other side of the Peltier device is maintained at a constant hot temperature using a heater resistor imbedded in the block. The temperatures are measured with ther­mistors which are imbedded in the hot and cold blocks.
Additional Equipment Needed
Then, in 1834, Jean-Charles-Athanase Peltier discov­ered the opposite of the Seebeck Effect, that a current flowing through a junction of dissimilar metals causes heat to be absorbed or freed, depending on the direc­tion in which the current is flowing.
2
Since the Ther­mal Efficiency Apparatus is operated in this manner the thermoelectric converter is called a Peltier device. However, the Thermal Efficiency Apparatus also exhibits the Seebeck Effect because the two sides of the device are maintained at different temperatures.
Today the Seebeck Effect is achieved using pn junc­tions. The arrangement of the dissimilar semiconduc­tors is as seen in Figure 1. If the left side of the device is maintained at a higher temperature than the right side, then holes generated near the junction drift across the junction into the p region and electrons drift into the n region. At the cold junction on the right side, the same process occurs but at a slower rate so the net effect is a flow of electrons in the n region from the hot side to the cold side. Thus there is a current from the cold side to hot side in the n region.
3
To perform the experiments in this manual, you will need the following equipment in addition to the Thermal Efficiency Apparatus.
• 1 DC power supply capable of 2.5A at 12V (SF-9584)
• 3 kg (7 lbs) ice and a bucket for the ice-water bath
• Ohmmeter (SB-9624)
• 1 Ammeter (up to 3A) (SB-9624A)
• 2 Voltmeters (SB-9624A)
• Patch Cords (SE-9750-51)
History
The principle upon which the Thermal Efficiency Apparatus operates has been known since the 1800’s but has only become practical since the recent devel­opment of semiconductors.
In 1821 the Russian-German physicist Thomas Johann Seebeck discovered that when a junction of dissimilar metals is heated, a current is produced. enon is now known as the Seebeck Effect and is the basis of the thermocouple.
1
This phenom-
Cold
)
(T
Hot (Th)
p
n
p
n
Copper
Figure 1: Arrangement of Thermocouples
1
Timetables of Science, by Alexander Hellemans and
c
I
I
I
I
I
Load resistor
Bryan Bunch, Simon & Schuster, NY, 1988, p.281.
2
IBID, p.301.
3
Circuits, Devices, and Systems, 3rd ed., by Ralph J. Smith, Wiley, 1976, p.543.
1
Thermal Efficiency Apparatus 012-05443A
Quick Start
The following sections of this manual are essential to operate the Thermal Efficiency Apparatus and will give the user the minimum amount of information necessary to get started quickly:
Theory
Heat Engine
• Introduction
• Actual Efficiency
• Carnot Efficiency
Measurements Using the Thermal Efficiency Apparatus
Direct Measurements
• Temperatures
• Power to the Hot Reservoir
• Power Used by the Load Resistor
Experiment — 1: Heat Engine Efficiency and Temperature Difference
The other portions of the manual provide a more detailed explanation of the operation of the Thermal Efficiency Apparatus in other modes as well as the heat engine mode.
2
012-05443A Thermal Efficiency Apparatus
Theory
Heat Engine
Introduction
A heat engine uses the temperature difference between a hot reservoir and a cold reservoir to do work. Usu­ally the reservoirs are assumed to be very large in size so the temperature of the reservoir remains constant regardless of the amount of heat extracted or delivered to the reservoir. This is accomplished in the Thermal Efficiency Apparatus by supplying heat to the hot side using a heating resistor and by extracting heat from the cold side using ice water.
In the case of the Thermal Efficiency Apparatus, the heat engine does work by running a current through a load resistor. The work is ultimately converted into heat which is dissipated by the load resistor (Joule heating).
A heat engine can be represented by a diagram (Figure
2). The law of Conservation of Energy (First Law of Thermodynamics) leads to the conclusion that Q
= W + QC, the heat input to the engine equals the
H
work done by the heat engine on its surroundings plus the heat exhausted to the cold reservoir.
Cold
Reservoir
Q
c
Q
Hot
Reservoir
h
NOTE: Since you will be measuring the rates at which energy is transferred or used by the Thermal Efficiency Apparatus all measurements will be power rather than energy. So P
= dQH/dt and then the equation
H
QH = W + QC becomes PH=PW+PC and the efficiency becomes
P
W
e =
P
H
Carnot Efficiency
Carnot showed that the maximum efficiency of a heat engine depends only on the temperatures between which the engine operates, not on the type of engine.
– T
T
H
e
=
Carnot
where the temperatures must be in Kelvin. The only engines which can be 100% efficient are ones which operate between TH and absolute zero. The Carnot efficiency is the best a heat engine can do for a given pair of temperatures, assuming there are no energy losses due to friction, heat conduction, heat radiation, and Joule heating of the internal resistance of the device.
C
T
H
T
c
W
Figure 2: Heat Engine
Heat
Engine
T
h
Actual Efficiency
The efficiency of the heat engine is defined to be the work done divided by the heat input
W
e =
Q
H
So if all the heat input was converted to useful work, the engine would have an efficiency of one (100% efficient). Thus, the efficiency is always less than one.
Adjusted Efficiency
Using the Thermal Efficiency Apparatus, you can account for the energy losses and add them back into the powers PW and PH. This shows that, as all losses are accounted for, the resulting adjusted efficiency approaches the Carnot efficiency, showing that the maximum efficiency possible is not 100%.
3
Thermal Efficiency Apparatus 012-05443A
W
Heat Pump (Refrigerator)
Introduction
A heat pump is a heat engine run in reverse. Normally, when left alone, heat will flow from hot to cold. But a heat pump does work to pump heat from the cold reser­voir to the hot reservoir, just as a refrigerator pumps heat out of its cold interior into the warmer room or a heat pump in a house in winter pumps heat from the cold outdoors into the warmer house.
In the case of the Thermal Efficiency Apparatus, heat is pumped from the cold reservoir to the hot reservoir by running a current into the Peltier device in the direction opposite to the direction in which the Peltier device will produce a current.
A heat pump is represented in a diagram such as Figure 3.
NOTE: The arrows are reversed compared to the heat in Figure 2. By conservation of energy,
Q
+ W = QH,
or in terms of power
C
PC+PW=PH.
This is similar to efficiency because it is the ratio of what is accomplished to how much energy was ex­pended to do it. Notice that although the efficiency is always less than one, the COP is always greater than one.
Maximum Coefficient of Performance
As with the maximum efficiency of a heat engine, the maximum COP of a heat pump is only dependent on the temperatures.
T
=
TH– T
C
C
κ
max
where the temperatures are in Kelvin.
Adjusted Coefficient of Performance
If all losses due to friction, heat conduction, radiation, and Joule heating are accounted for, the actual COP can be adjusted so it approaches the maximum COP.
Ohmmeter
h
Hot
Reservoir
T
h
Cold
Reservoir
T
c
Q
c
W
Q
Heat
Pump
Figure 3: Heat Pump
Actual Coefficient of Performance
Instead of defining an efficiency as is done for a heat engine, a coefficient of performance (COP) is defined for a heat pump. The COP is the heat pumped from the cold reservoir divided by the work required to pump it
P
κ
= COP =
C
.
P
9V Power Supply In
Rubber
Hoses
WATER
PUMP
7.5 - 12 VDC @500mA
In
COOLING
WATER
THERMISTOR
SELECT
PELTIER DEVICE
Out
THERMISTOR TABLE
K
°C
K
°C
K
°C
K
°C
K
°C
461
-5
269
5 436 413 391 370 351 332 315 298 283
161
-4
255
6
153
-3
242
7
146
-2
230
8
139
-1
218
9
133
0
207
10
126
1
197
11
120
2
187
12
115
3
178
13
109
4
169
14
104
K
15
100
25
63.4
16
95.4
26
60.7
17
91.1
27
58.1
18
87.0
28
55.6
19
83.1
29
53.2
20
79.4
30
51.0
21
75.9
31
48.9
22
72.5
32
46.8
23
69.3
33
44.9
24
66.3
34
43.0
K
°C
41.2
45
27.4
35
39.6
46
26.4
36
37.9
47
25.3
37
36.4
48
24.4
38
34.9
49
23.4
39
33.5
50
22.5
40
32.2
51
21.7
41
30.9
52
20.9
42
29.7
53
20.1
43
28.5
54
19.3
44
Figure 4: Thermal Efficiency Apparatus
scientific
PASCO
Model TD-8564 THERMAL EFFICIENCY APPARATUS
HEATE R
°C 55
56 57 58 59 60 61 62 63 64
5Ω±1%
12 VDC MAX
2.01.00.5
K
°C
K
°C
18.6
65
12.9
75
17.9
66
12.4
76
17.3
67
12.0
77
16.6
68
11.6
78
16.0
69
11.2
79
15.5
70
10.8
80
14.9
71
10.4
81
14.4
72
10.1
82
13.8
73
9.76
83
13.4
74
9.43
84
K
°C
K
°C
9.12
85
6.53
95
8.81
86
6.33
96
8.52
87
6.12
97
8.24
88
5.93
98
7.96
89
5.74
99
7.70
90
5.56
100
7.45
91
5.39
101
7.21
92
5.22
102
6.98
93
5.06
103
6.75
94
4.91
104
4
012-05443A Thermal Efficiency Apparatus
C
Measurements Using the Thermal Efficiency Apparatus
Direct Measurements
Three quantities may be directly measured with the Thermal Efficiency Apparatus: temperatures, the power delivered to the hot reservoir, and the power dissipated by the load resistors. The details of how these measurements are made follow.
Temperatures
The temperatures of the hot and cold reservoirs are determined by measuring the resistance of the thermistor imbedded in the hot or cold block. To do this, connect an ohmmeter to the terminals located as shown in Figure 4. The switch toggles between the hot side and the cold side. The thermistor reading can be converted to a temperature
Table 1: Resistance to Temperature Conversion Chart
k °C k °C k °C k °C k °C
461 -5 436 -4 413 -3
146 17 139 18 133 19
53.2 39
51.0 40
48.9 41
by using the chart located on the front of the Thermal Efficiency Apparatus and in Table 1. Notice that as the temperature increases, the thermistor resistance decreases (100 k is a higher temperature than 200 k).
NOTE: To get the exact temperature reading the user must interpolate between numbers on the chart. For example, suppose the ohmmeter reads
118.7 k. This reading lies between 120 k = 21°C and 115 k = 22°C. The reading is 120-118.7 = 1.3 k above 21°C which is
1.3kΩ×
1°C
120 – 115k
= 0.26°
Therefore 118.7 k is 21.26°C.
21.7 61
20.9 62
20.1 63
9.76 83
9.43 84
9.12 85 391 -2 370 -1 351 0 332 1 315 2 298 3 283 4 269 5 255 6 242 7 230 8 218 9 207 10 197 11 187 12 178 13 169 14 161 15
126 20 120 21 115 22 109 23 104 24 100 25
95.4 26
91.1 27
87.0 28
83.1 29
79.4 30
75.9 31
72.5 32
69.3 33
66.3 34
63.4 35
60.7 36
58.1 37
46.8 42
44.9 43
43.0 44
41.2 45
39.6 46
37.9 47
36.4 48
34.9 49
33.5 50
32.2 51
30.9 52
29.7 53
28.5 54
27.4 55
26.4 56
25.3 57
24.4 58
23.4 59
19.3 64
18.6 65
17.9 66
17.3 67
16.6 68
16.0 69
15.5 70
14.9 71
14.4 72
13.8 73
13.4 74
12.9 75
12.4 76
12.0 77
11.6 78
11.2 79
10.8 80
10.4 81
8.81 86
8.52 87
8.24 88
7.96 89
7.70 90
7.45 91
7.21 92
6.98 93
6.75 94
6.53 95
6.33 96
6.12 97
5.93 98
5.74 99
5.56 100
5.39 101
5.22 102
5.06 103 153 16
55.6 38
22.5 60
5
10.1 82
4.91 104
Thermal Efficiency Apparatus 012-05443A
Power Delivered to the Hot Reservoir (PH)
The hot reservoir is maintained at a constant temperature by running a current through a resistor. Since the resis­tance changes with temperature, it is necessary to mea­sure the current and the voltage to obtain the power input. Then P
= IHVH.
H
Power Dissipated by the Load Resistor (PW)
The power dissipated by the load resistor is determined by measuring the voltage drop across the known load resistance and using the formula
2
V
PW=
.
R
The load resistors have a tolerance of 1%.
2
V
NOTE: We may use the equation
PW=
for
R
measuring the power in the load resistor because the temperature (and therefore resistance) of this resistor does not change significantly. We may not use this equation to measure power in the heating resistor, since its temperature (and resistance) changes.
When the Thermal Efficiency Apparatus is operated as a heat pump rather than as a heat engine, the load resistors are not used so it is necessary to measure both the current and the voltage. So the current into the Peltier device is measured with an ammeter, and the voltage across the Peltier device is measured with a voltmeter and the power input is calculated with the formula P
= IWVW.
W
Indirect Measurements
It will be necessary to know three additional quantities in the experiments: device;
The amount of heat conducted through the
device and the amount radiated away;
Figure 5: Procedure for Finding Internal Resistance
The internal resistance of the Peltier
The amount of
Peltier Device
R
r
V
l
l
V
s
heat pumped from the cold reservoir. These quantities may be determined indirectly with the Thermal Effi­ciency Apparatus in the following ways.
Internal Resistance
Before the adjusted efficiency can be calculated, it is necessary to calculate the internal resistance. This is accomplished by measuring the voltage drop across the Peltier device when an external load is applied.
First run the Thermal Efficiency Apparatus with a load resistor (R) as in figure 6. The electrical equivalent of this setup is shown in figure 5. Kirchoff’s Loop Rule gives
VS– Ir – IR =0
Next, run the Thermal Efficiency Apparatus with no load, as in Figure 7. Since there is no current flowing through the internal resistance of the Peltier Device, the voltage drop across the internal resistance is zero and the voltage measured will just be VS.
Since we have measured V
rather than I in the heat
w
engine mode, the equation above becomes
V
w
Vs–
r – Vw=0
R
Solving this for the internal resistance gives us
– V
V
s
r =
w
R
V
.
w
You may also find the resistance by measuring the currents for two different load resistors and then solving the resulting loop rule equations simultaneously.
Heat Conduction and Radiation
The heat that leaves the hot reservoir goes two places: part of it is actually available to be used by the heat engine to do work while the other part bypasses the engine either by being radiated away from the hot reservoir or by being conducted through the Peltier device to the cold side. The portion of the heat which bypasses the engine by radiation and conduction would be trans­ferred in this same manner whether or not the device is connected to a load and the heat engine is doing work.
The Thermal Efficiency Apparatus is run with a load connected to measure P disconnected and the power input into the hot reservoir is adjusted to maintain the temperatures (less power is needed when there is no load since less heat is being drawn from the hot reservoir). See Figure 7. P
(Figure 6) and then the load is
H
is the power input
H(open)
6
012-05443A Thermal Efficiency Apparatus
to the hot reservoir when no load is present. Since, while there is no load, the hot reservoir is maintained at an equilibrium temperature, the heat put into the hot reser­voir by the heating resistor must equal the heat radiated and conducted away from the hot reservoir. So measuring the heat input when there is no load determines the heat loss due to radiation and conduction. It is assumed this loss is the same when there is a load and the heat engine is operating.
Heat Pumped from the Cold Reservoir
When the Thermal Efficiency Apparatus is operated as a heat pump, conservation of energy yields that the rate at which heat is pumped from the cold reservoir, PC, is equal to the rate at which heat is delivered to the hot reservoir, PH, minus the rate at which work is being done, P
W
(Figure 3).
The work can be measured directly but the heat delivered to the hot reservoir has to be measured indirectly. Notice that when the heat pump is operating, the temperature of the hot reservoir remains constant. Therefore, the hot reservoir must be in equilibrium and the heat delivered to it must equal the heat being conducted and radiated away. So a measurement of the heat conducted and radiated away at a given temperature difference will also be a measurement of the heat delivered to the hot reservoir. The heat conducted and radiated is measured by running the device with no load and measuring the heat input needed to maintain the temperature of the hot side (Figure 7).
T
H
Conducted
Power
V
Engine
A
Power
Supply
V
Figure 6: Heat Engine With A Load
V
A
Conducted
Power
PH (open)
Power Supply
P
W
T
C
T
H
V
Figure 7: No Load
7
T
C
Thermal Efficiency Apparatus 012-05443A
Copy-Ready Experiments
The following experiments are written in worksheet form.
Feel free to photocopy them for use in your lab.
NOTE: The first paragraph in each experiment lists all the equipment needed to perform the experiment. Be sure to read this equipment list first, as the require­ments vary with each experiment.
8
012-05443A Thermal Efficiency Apparatus
Experiment 1: Heat Engine and T emperature Difference
EQUIPMENT NEEDED:
— Thermal Efficiency Apparatus — DC power supply capable of 2.5 A at 12 V — ohmmeter — ammeter (up to 3 A) — patch cords — 2 voltmeters — 3 kg (7 lbs) ice and a bucket for the ice-
water bath
Introduction
In this experiment the user will determine the actual efficiency and the Carnot efficiency of the heat engine as a function of the operating temperatures.
Setup
Prepare the ice-water bath and immerse both rubber tubes from the Thermal Efficiency Appara-
tus into the bath (Figure 4).
Plug the 9V transformer into the wall socket and into the pump on the Thermal Efficiency
Apparatus. You should now hear the pump running and water should be coming out of the rubber hose marked “out”.
Plug the ohmmeter into the thermistor terminals.Connect a DC power supply and a voltmeter and ammeter to the heater block terminals. Adjust
the voltage to about 11 V.
NOTE: This is just a suggested value chosen to make the hot temperature nearly at the
maximum allowed. Any voltage less than 12 V is suitable. The Thermal Efficiency Apparatus should not be run for more than 5 minutes with the hot side above 80°C. A thermal switch will automatically shut off the current to the heater block if it exceeds 93°C to prevent damage to the device.
V
A
Power
Supply
0.5
1
2
Figure 1.1
V
9
Thermal Efficiency Apparatus 012-05443A
Connect the 2Ω load resistor with a short patch cord as shown in Figure 1.1. Connect a voltmeter
across the load resistor. The choice of the 2 load resistor is arbitrary. Any of the load resis­tances may be used.
Procedure
Allow the system to come to equilibrium so that the hot and cold temperatures are constant.
This may take 5 to 10 minutes, depending on the starting temperatures. To speed up the process, increase the voltage across the heating resistor momentarily and then return it to the original setting. If it is desired to cool the hot side, the voltage can be momentarily decreased. Remember that the thermistor resistance goes down as the temperature increases.
Measure the temperature resistances of the hot side and the cold side by using the toggle
switch to switch the ohmmeter to each side. Record the readings in Table 1.1. Convert the resistances to temperatures using the chart on the front of the device or Table 1 as explained in the Measurements section and record these temperatures in Table 1.2.
Record the voltage (V
the load resistor (V
) across the heating resistor, the current (IH), and the voltage across
H
) in Table 1.1.
W
Lower the voltage across the heating resistor by about 2 V.Repeat Steps 1 through 4 until data for five different hot temperatures have been taken.
Table 1.1 Data for Heat Engine
Trial TH (kΩ)Tc (kΩ)TH (°C) TH (°C) V
1
2
3
4
5
H
I
H
V
w
Calculations
For each of the data runs, calculate the power supplied to the hot reservoir, P
power used by the load resistor, P
, and record these in Table 1.2.
W
, and the
H
Calculate the temperature difference for each trial and record it in Table 1.2.Calculate the actual efficiencies from the powers and record in Table 1.2.Calculate the Carnot (maximum) efficiencies from the temperatures and record in Table 1.2.
10
012-05443A Thermal Efficiency Apparatus
Table 1.2 Calculated Values
Trial P
1
2
3
4
5
H
P
w
Analysis and Questions
To compare the actual efficiency to the Carnot efficiency, construct a graph. Plot the Carnot efficiency vs. T and also plot the actual efficiency vs. T. This may be
done on the same graph.
NOTE: We are assuming by doing this that T
TH (k) Tc (k) ∆T (k) e
was nearly constant.
c
actual
e
Carnot
The Carnot efficiency is the maximum efficiency possible for a given temperature differ-
ence. According to the graph, is the actual efficiency always less than the Carnot effi­ciency?
Does the Carnot efficiency increase or decrease as the temperature difference increases?Does the actual efficiency increase or decrease as the temperature difference increases?The Carnot efficiency represents the best that a perfect heat engine can do. Since this heat
engine is not perfect, the actual efficiency is a percentage of the Carnot efficiency. The overall (actual) efficiency of a real heat engine represents the combination of the engine’s ability to use the available energy and the maximum energy available for use. From the data taken, what is the percentage of available energy used by this heat engine?
The actual efficiency of this heat engine is very low and yet heat engines of this type are
used extensively in remote areas to run things. How can such an inefficient device be of practical use?
11
Thermal Efficiency Apparatus 012-05443A
Notes:
12
012-05443A Thermal Efficiency Apparatus
V
V
A
Power
Supply
0.5
1
2
Experiment 2: Heat Engine Efficiency (Detailed Study)
EQUIPMENT NEEDED:
— Thermal Efficiency Apparatus — 1 DC power supply capable of 2.5 A at 12 V — ohmmeter — patch cords — ammeter (up to 3 A) — 2 voltmeters — 3 kg — (7 lbs) ice and a bucket for the ice-
water bath
Introduction
In this experiment the user will determine the actual efficiency and the Carnot efficiency of the heat engine and then compensate for the energy losses to show that the compensated actual efficiency approaches the Carnot efficiency.
Initial Setup
Prepare the ice-water bath and immerse both rubber tubes from the Thermal Efficiency
Apparatus into the bath (Figure 4).
Plug the 9V transformer into the wall socket and into the pump on the Thermal Efficiency
Apparatus. You should now hear the pump running and water should be coming out of the rubber hose marked “out”.
Plug the ohmmeter into the thermistor terminals.
Modes of Operation:
To obtain all the necessary data for the heat engine it is necessary to run the Thermal Efficiency Apparatus in two different modes. The Heat Engine Mode determines the actual efficiency of the Peltier device. The Open Mode determines the losses due to conduction and radiation. Data from both modes is used to calculate internal resistance and the Carnot Efficiency.
Heat Engine
A. Connect a DC power supply and a voltmeter and ammeter to the heater block terminals.
Turn on the voltage to about 11 V.
NOTE: This is just a suggested value
chosen to make the hot temperature nearly at the maximum allowed. Any voltage less than 12 V is suitable. The Thermal Efficiency Apparatus should not be run for more than 5 minutes with the hot side above 80°C. A thermal switch will automatically shut off the current to the heater block if it exceeds 93°C to prevent damage to the device.
Figure 2.1
13
Thermal Efficiency Apparatus 012-05443A
W
B. Connect the 2 load resistor with a short patch cord as shown in Figure 2.1. Connect a
voltmeter across the load resistor.
C. Allow the system to come to equilibrium so that the hot and cold temperatures are
constant. This may take 5 to 10 minutes, depending on the starting temperatures. To speed up the process, increase the voltage across the heating resistor momentarily and then return it to 11 V. If it is desired to cool the hot side, the voltage can be momentarily decreased. Remember that the thermistor resistance goes down as the temperature increases.
D. Measure the temperature resistances of the hot side and the cold side by using the toggle
switch to switch the ohmmeter to each side. Record the readings in Table 3. Convert the resistances to temperatures using the chart on the front of the device or Table 1 as explained in the Measurements section.
E. Record the voltage (V
) across the heating resistor, the current (IH), and the voltage
H
across the load resistor (VW) in Table 2.1.
Open
A. Disconnect the patch cord from the load resistor so no current is flowing through the
load and thus no work is being done. Now all the power delivered to the heating resistor is either conducted to the cold side or radiated away. Leave the voltmeter attached so that the Seebeck voltage (Vs) can be measured. (see figure 7)
B. Decrease the voltage applied to the hot side so that the system comes to equilibrium at
the same hot temperature as in the Heat Engine Mode. Since the temperature difference is the same as when the heat engine was doing work, the same amount of heat is now being conducted through the device when there is no load as when there is a load. (It may not be possible to exactly match the previous cold temperature.)
C. Record the resistances in Table 2.1 and convert them to degrees.
Also record V
, IH and Vp.
H
Calculations for the Heat Engine
Actual Efficiency: Calculate the actual efficiency using
P
W
e =
,
P
H
2
PW=
V
and PH = IHVH.
R
Table 2.1 Data
where
Record the powers in Table 2.2 and the efficiency in Table 2.3.
Mode TH (kΩ)Tc (kΩ)TH (°C) Tc (°C) V
Engine
Open
H
I
H
V
w
V
S
14
012-05443A Thermal Efficiency Apparatus
r
I
Table 2.2 Calculated Values
Internal Resistance = r = ________________
Mode Th (K) Tc (K) P
Engine (2 load)
Open
Table 2.3 Results
Actual Adjusted % Difference
Efficiency
h
Maximum
(Carnot)
P
w
I
w
Maximum Efficiency: Convert the temperatures to Kelvin and record in Table 2.2. Calcu-
late the Carnot efficiency using the temperatures and record in Table 2.3.
Adjusted Efficiency: The purpose of the following calculations is to account for all the
energy losses and adjust the actual efficiency so that it matches the Carnot efficiency.
2
A. First, the work done in the actual efficiency calculation only includes
dissipated by the load resistor R but, to account for total work done by the device, it
2
should also include I
r for the power dissipated by the internal resistance, r, of the device. This Joule heating of the Peltier device is not counted in the actual efficiency because it is not useful work. Thus, in the adjusted efficiency, the total work done in terms of power is
V
for the power
R
2
V
P
where
= PW+ I
W
=
W
V
W
. Calculate IW for the 2 load and record in Table 4.
R
W
2
r =
W
2
+ I
W
R
B. Second, the heat input must be adjusted. The heat that leaves the hot reservoir goes two
places. Part of it is actually available to be used by the heat engine to do work while the other part bypasses the engine either by being radiated away from the hot reservoir or by being conducted through the Peltier device to the cold side. The portion of the heat which bypasses the engine by radiation and conduction would be transferred in this same manner whether or not the device is connected to a load and the heat engine is doing work. Therefore this heat can be considered to not be available to do work and should not be included in the heat input in the adjusted efficiency.
P
= available heat = PH– P
H
H open
15
Thermal Efficiency Apparatus 012-05443A
R
The Thermal Efficiency Apparatus is run with a load connected to measure PH (Figure
6) and then the load is disconnected and the power input into the hot reservoir is ad­justed to maintain the temperatures (less power is needed when there is no load since less heat is being drawn from the hot reservoir). See Figure 7. P
H(OPEN)
is the power input to the hot reservoir when no load is present. Since, while there is no load, the hot reservoir is maintained at an equilibrium temperature, the heat put into the hot reservoir by the heating resistor must equal the heat radiated and conducted away from the hot reservoir. So measuring the heat input when there is no load determines the heat loss due to radiation and conduction. It is assumed this loss is the same when there is a load and the heat engine is operating.
Having accounted for the obvious energy losses, the adjusted efficiency should match the Carnot efficiency which assumes no energy loss. The adjusted efficiency is
e
adjusted
P
W
=
=
P
P
H
H
P
W
P
2
+ I
W
H open
r
Calculate the internal resistance, r, using the equation
– V
V
P
r =
W
V
W
which is derived in the Indirect Measurement section. Record this resistance in Table
2.2. Then calculate the adjusted efficiency and record the result in Table 2.3. Calculate the percent difference between the adjusted efficiency and the Carnot (maxi-
mum) efficiency
e
e
max
% Difference =
e
max
adjusted
× 100%
and record in Table 2.3.
Questions
If the difference between the temperature of the hot side and the cold side was decreased,
would the maximum efficiency increase or decrease?
The actual efficiency of this heat engine is very low and yet heat engines of this type are
used extensively in remote areas to run things. How can such an inefficient device be of practical use?
Calculate the rate of change in entropy for the system which includes the hot and cold
reservoirs. Since the reservoirs are at constant temperature, the rate of change in entropy is
Q / t
S
=
t
T
for each reservoir. Is the total change in entropy positive or negative? Why?
P
=
T
16
012-05443A Thermal Efficiency Apparatus
Experiment 3: Heat Pump Coefficient of Performance
EQUIPMENT NEEDED:
— Thermal Efficiency Apparatus — 1 DC power supplies capable of 2.5 A at 12 V — patch cords — ohmmeter — ammeter (up to 3 A) — voltmeter — 3 kg — (7 lbs) ice and a bucket for the
ice-water bath
NOTE: Before doing this experiment, it is necessary to perform the HEAT ENGINE
EFFICIENCY experiment to get the data necessary to determine the internal resistance of the Peltier device.
To complete the measurements for this experiment, use the following instructions to run the apparatus as a heat pump (pumping heat from the cold side to the hot side):
Setup
Prepare the ice-water bath and immerse both rubber tubes from the Thermal Efficiency
Apparatus into the bath (Figure 4).
for measuring temperatures
to AC supply
to ice water tub
Power Supply
+
for driving the Peltier device
V
for measuring V
Figure 3.1 Heat Pump Mode
17
A
for measuring I
w
Pw = VwI
w
w
Thermal Efficiency Apparatus 012-05443A
W
W
Plug the 9V transformer into the wall socket and into the pump on the Thermal Efficiency
Apparatus. You should now hear the pump running and water should be coming out of the rubber hose marked “out”.
Disconnect the power supply to the hot side. Connect the power supply directly across the
Peltier device with no load resistance. See Figure 3.1
Connect an ammeter and a voltmeter to the power supply.
Procedure
Increase the voltage until equilibrium is reached at the same hot temperature as in the
previous experiment. The hot side is now being heated by heat pumped from the cold side rather than the heater resistor.
Record the resistances and convert them to degrees. Also record the voltage (V
current (I
) in Table 3.1.
W
Analysis
Actual Coefficient of Performance: Calculate the actual COP using the data taken in the
Heat Engine experiment.
P
P
C
W
=
H (OPEN)
κ
=
P
– P
P
W
Record this result in Table 3.1.
Maximum Coefficient of Performance: Calculate the maximum COP using
T
=
TH– T
C
C
κ
MAX
and record this result in Table 3.1.
Adjusted Coefficient of Performance: Part of the power being applied to the Peltier device
is being dissipated in the Joule heating of the internal resistance of the device rather than being used to pump the heat from the cold reservoir. Therefore, to adjust for this, I be subtracted from the power input to the Peltier device. Then the COP becomes the heat pumped from the cold reservoir divided by work done to pump the heat, rather than divid­ing by the work done to pump the heat and heat the internal resistance. In terms of the power,
) and the
W
2
r must
κ
ADJUSTED
P
=
H (OPEN)
PW– I
P
2
W
r
Record this result in Table 3.1. Calculate the percent difference between the adjusted COP and maximum COP:
% Difference =
MAX
ADJUSTED
κ
MAX
× 100%
κ
κ
and record in Table 3.1.
18
012-05443A Thermal Efficiency Apparatus
Table 3.1 Heat Pump Data and Results
COP COP COP
(k)TC(k)TH(K) T
T
H
C
(K) V
I
W
P
W
actual max adj % diff
W
Questions
If the difference between the temperature of the hot side and the cold side was decreased,
would the maximum COP increase or decrease?
Calculate the rate of change in entropy for the system which includes the hot and cold
reservoirs. Since the reservoirs are at constant temperature, the rate of change in entropy is
Q / t
S
t
=
T
P
=
T
for each reservoir. Is the total change in entropy positive or negative? Why?
19
Thermal Efficiency Apparatus 012-05443A
Experiment 4: Thermal Conductivity
Introduction
The rate at which heat is conducted through a material of thickness x and cross-sectional area A depends on the difference in temperature between the sides (T) and the thermal conductivity (k) of the material.
Power =
Heat Time
kA (T)
=
x
For the Thermal Efficiency Apparatus, the Peltier device has 71 couples and each couple consists of 2 elements, so there is a total of 142 elements which conduct heat (Figure 9).
Each element has a length to area ratio of 8.460 cm
-1
. So
x
=
A
8.460cm 142
–1
. Use the data taken
in Experiment 2 for the Open Mode to calculate the thermal conductivity of the Peltier device:
k =
P
H (OPEN)
(x / A)
T
Question
How does the thermal conductivity of the Peltier device compare with the thermal conduc-
tivity of copper?
Copper
P
N
Figure 4.1 One Couple Equals Two Elements
20
012-05443A Thermal Efficiency Apparatus
Experiment 5: Load for Optimum Performance
EQUIPMENT NEEDED:
Theory
— Thermal Efficiency Apparatus — DCpower supply capable of 2.5 A at 12 V
V
s
r
— 3 kg (7 lbs) ice and a bucket for the ice-water bath — ohmmeter — ammeter (up to 3 A)
R
— 2 voltmeters — patch cords
Figure 5.1 Peltier device connected
to a load resistor
This experiment finds the load resistor which maximizes the power output of the heat engine. The power delivered to the load resistor, R, is P = I2R. The amount of current that flows through the load resistor varies as the load is varied. From Figure 10, VS = I(r+R) where VS is the Seebeck voltage and r is the internal resistance of the Peltier device.
So the power can be expressed in terms of the Seebeck voltage, the internal resistance, and the load resistance:
2
V
P =
s
r + R
R
Assuming the Seebeck voltage remains constant if the temperatures of the hot and cold reservoirs are constant, the power can be maximized with respect to the load resistance by taking the derivative and setting it equal to zero:
V
l
dP dR
=
S
(r + R)
=0
3
2
(r R)
V
This shows that when the load resistance is equal to the internal resistance of the Peltier device, the power delivered to the load will be a maximum.
Connect to appropriate AC supply
(powers pump to circulate ice water)
V
Place ends of tubing in ice water tub
0.5
2 1
A
V
Figure 5.2 Connecting the 0.5 load resistor
Power supply
21
Thermal Efficiency Apparatus 012-05443A
Procedure
Connect a DC power supply and a voltmeter and ammeter to the heater block terminals.
Turn on the voltage to about 11 V.
NOTE: This is just a suggested value chosen to make the hot temperature nearly at the
maximum allowed. Any voltage less than 12 V is suitable. The Thermal Efficiency Apparatus should not be run for more than 5 minutes with the hot side above 80°C. A thermal switch will automatically shut off the current to the heater block if it exceeds 93°C to prevent damage to the device.
Connect the 0.5W load resistor with a short patch cord as shown in Figure 11. Connect a
voltmeter across the load resistor.
NOTE: Alternatively, a variable power resistor (rheostat) may be used in place of the load
resistors supplied with the Thermal Efficiency Apparatus. This has the advantage of being able to continuously vary the load resistance. However, it will be necessary to measure the resistance of the load.
Allow the system to come to equilibrium so that the hot and cold temperatures are constant.
This may take 5 to 10 minutes, depending on the starting temperatures. To speed up the process, increase the voltage across the heating resistor momentarily and then return it to 11 V. If it is desired to cool the hot side, the voltage can be momentarily decreased. Remember that the thermistor resistance goes down as the temperature increases.
Measure the temperature resistances of the hot side and the cold side by using the toggle
switch to switch the ohmmeter to each side. Record the readings in Table 5.1. Convert the resistances to temperatures using the chart on the front of the device or Table 1 as explained in the Measurements section.
Record the voltage (V
the load resistor (V
R(Ω)TH(k)TC(k)TH(°K) (°K) V
0.5
1.0
1.5
2.0
2.5
3.0
) across the heating resistor, the current (IH), and the voltage across
H
) in Table 5.1.
W
Table 5.1: Heat Engine Data and Results
I
H
V
H
P
W
P
H
L
e
3.5
22
012-05443A Thermal Efficiency Apparatus
Calculate the power input to the hot side, P
2
V
resistor,
PL=
W
. Calculate the efficiency,
R
= IHVH, and the power dissipated by the load
H
P
L
e =
. Record all these values in Table 5.1.
P
H
Adjust the power input to the hot side to keep the temperature of the hot reservoir at the
same temperature as it was for the 0.5 resistor while Steps 1 through 6 are repeated for the other possible load resistances: 1, 1.5, 2, 2.5, 3, and 3.5 ohms.
Questions
For which load resistor is the efficiency a maximum?If you have done experiment 2: How does the load resistance for optimum efficiency
compare with the internal resistance measured in that experiment?
23
Thermal Efficiency Apparatus 012-05443A
Notes:
24
012-05443A Thermal Efficiency Apparatus
H
T eacher’s Guide
Experiment 1: Heat Engine and Temperature Difference
Notes on Setup
It may be necessary to prime the pump by sucking
on the output line briefly.
Notes on Calculations
2
V
W
Use the equations
P
efficiency =
e
Carnot
W
P
H
– T
T
H
=
C
T
H
PH= VHI
and
PW=
R
Notes on Analysis and Questions
0.2
0.18
0.16
0.14
0.12
0.1
0.08
Efficiency (%)
0.06
0.04
0.02 0
0 10203040506070
Carnot Efficiency
B
Actual Efficiency
J
B
B
J
J
Temperature Difference (°C)
B
B
B
B
J
J
J
J
B
J
Yes. ➁,➂ Both Carnot and actual efficiency increase with
increasing temperature difference. (for a constant cold temperature)
In these trials, 11-12% of the available energy was
used.
Although the efficiency is low, the reliability is ex-
tremely high. (There are no moving parts in the Peltier device.) One practical application of these devices is in satellite power supplies. A small piece of radioactive material is used as a source of heat, and a radiation fin is used as a heat sink. Another similar application is to use the temperature differ­ence between a nuclear isotope and arctic weather to run a remote unmanned weather station. Any ap­plication where the thermal mass of the available sources is large, the power requirements are small, and the required reliability is high is good for the Peltier device.
25
Thermal Efficiency Apparatus 012-05443A
Experiment 2: Heat Engine Efficiency (Detailed Study)
Notes on Setup
It may be necessary to prime the pump by sucking
on the output line briefly.
Sample Data
Mode Th (°C) Tc (°C) V Engine 57.9 3.5 10.00 2.02 0.890 Open 57.9 3.3 8.99 1.815 1.495
h
I
h
V
w
V
Calculated Values
Mode Th (K) Tc (K) P Engine 330.9 276.5 20.2 0.40 0.45 Open 330.9 276.3 16.3
Internal Resistance: r = 1.36
h
P
w
I
w
Results
Actual Adjusted Maximum
(Carnot)
Efficiency 1.96% 17.13% 16.44% -4.23%
Note that these results were obtained using slightly lower initial voltage than recommended in the lab. In general, mid-range temperatures give better results than extremely large or small temperature differences.
% Difference
s
Answers to Questions
If the temperature difference was decreased, the ef-
ficiency would also decrease.
See experiment 1, question 5.For the hot reservoir, S/t was -0.061. For the
cold reservoir, it was 0.073. The total change in en­tropy is positive. In any non-reversible process, the entropy will increase.
26
012-05443A Thermal Efficiency Apparatus
Experiment 3: Heat Pump Coefficient of Performance
Typical Results
Note that values of Ph and r were taken from experiment 2.
Th (K) Tc (K) V
330.9 275.5 3.64 1.63 5.93 1.75 4.97 4.48 9.9%
w
I
w
P
w
COP COP
max
COP
adj
Answers to Questions
The COP increases when the difference in tempera-
ture decreases.
For the hot reservoir, ∆S/t = +0.018. For the cold
reservoir, it is – 0.0215. The net change in entropy is negative. Work is done by the heat pump to de­crease the entropy.
Experiment 4: Thermal Conductivity
Answer to Questions
The thermal conductivity, based on the data taken
in experiment 2 of this guide, is 1.79 Watt/mK. By comparison, the thermal conductivity of copper (at 273 K) is 401 Watt/mK.
% diff
The Peltier device is made of Bismuth Telluride, which has an accepted thermal conductivity of approximately 1.6 Watt/mK
27
Thermal Efficiency Apparatus 012-05443A
Experiment 5: Load for Optimum Performance
Notes on Sample Data
2.4
2.3
2.2
2.1
% Efficiency
2
1.9
1
00.511.522.533.5
1
1
Load Resistance ( )
Answer to Question
The efficiency is a maximum when the 1.5 resistance is used. This is close to the value of the internal
1
1
1
resistance determined in experiment 2, as well.
28
012-05443A Thermal Efficiency Apparatus
T echnical Support
Feed-Back
If you have any comments about this product or this manual please let us know. If you have any sugges­tions on alternate experiments or find a problem in the manual please tell us. PASCO appreciates any cus­tomer feed-back. Your input helps us evaluate and improve our product.
To Reach PASCO
For Technical Support call us at 1-800-772-8700 (toll­free within the U.S.) or (916) 786-3800.
Contacting Technical Support
Before you call the PASCO Technical Support staff it would be helpful to prepare the following information:
• If your problem is with the PASCO apparatus, note: Title and Model number (usually listed on the label).
Approximate age of apparatus.
A detailed description of the problem/sequence of events. (In case you can't call PASCO right away, you won't lose valuable data.)
If possible, have the apparatus within reach when calling. This makes descriptions of individual parts much easier.
• If your problem relates to the instruction manual,
note: Part number and Revision (listed by month and year
on the front cover).
Have the manual at hand to discuss your questions.
29
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