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Includes
Teacher's Notes and
Typical
Experiment Results
Instruction Manual and
Experiment Guide for the PASCO scientific
Model TD-8564
012-05443A
3/94
THERMAL EFFICIENCY
APPARATUS
PASCO scientific
PUMP
7.5 - 12 VDC
THERMISTOR
SELECT
COOLING
WATER
0.5
Ω
1.0
Ω
HEATER
2.0
Ω
5
Ω±
Cold
Reservoir
Q c
T c
K
Ω
461
436
° °
-5
-4
K
Ω
269
255
° °
5
6
5
6
K
Ω
161
153
° °
15
16
K
Ω
100
95.4
° °
25
26
THERMISTOR TABLE
K
Ω ° ° K Ω ° ° K Ω ° ° C
63.4
60.7
35
36
37.9
36.4
47
48
25.3
24.4
57
58
K Ω ° °
17.3
16.6
67
68
K Ω ° ° K Ω ° ° K Ω
12.0
11.6
77
78
8.52
8.24
87
88
6.12
5.93
° °
97
98
332
315
1
2
1
2
197
187
11
12
120
115
21
22
75.9
72 5
31
32
48.9
46 8
41
42
W
Q h
Hot
Reservoir
T h
Heat
Engine
© 1991 PASCO scientific $10.00
012-05443A Thermal Efficiency Apparatus
Table 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 (P
H
) ......................................... 6
Power Dissipated by the Load Resistor (P
W
) ....................................... 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
Copyright, Warranty and Equipment Return
012-05443A
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:
➀ The packing carton must be strong enough for the item shipped.
➁ Make certain there are at least two inches of packing material between any point on the apparatus and the inside walls of the carton.
➂ 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:
FAX: email: web:
(916) 786-3800
(916) 786-3292 [email protected]
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
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.
1
This phenomenon is now known as the Seebeck Effect and is the basis of the thermocouple.
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
Hot
(T h
) p n p n
Copper
I
I
I
I
Cold
(T c
)
I
Figure 1: Arrangement of Thermocouples
1
Timetables of Science, by Alexander Hellemans and
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
H
= W + Q
C
, the heat input to the engine equals the work done by the heat engine on its surroundings plus the heat exhausted to the cold reservoir.
Cold
Reservoir
Hot
Reservoir
Q c
Q h
T c
T h
W
Heat
Engine
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.
e
Carnot
=
T
H
– T
C
T
H where the temperatures must be in Kelvin. The only engines which can be 100% efficient are ones which operate between T
H
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.
Adjusted Efficiency
Using the Thermal Efficiency Apparatus, you can account for the energy losses and add them back into the powers P
W
and P
H
. 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%.
Figure 2: Heat Engine
Actual Efficiency
The efficiency of the heat engine is defined to be the work done divided by the heat input
e =
W
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.
➤ 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
Q
H
= dQ
H
H
/dt and then the equation
= W + Q
C
becomes P efficiency becomes
H
= P
W
+ P
C
and the
e =
P
W
P
H
3
Thermal Efficiency Apparatus
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
κ
= COP =
P
C
P
W
.
012-05443A
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
C or in terms of power
+ W = Q
H
,
P
C
+ P
W
= P
H
.
T c
Cold
Reservoir
Q c
Q h
Hot
Reservoir
T h
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.
κ
max
=
T
C
T
H
– T
C 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.
Ω
Ohmmeter
9V Power
Supply In
Rubber
Hoses
WATER
PUMP
7.5 - 12 VDC
@500mA
In
COOLING
WATER
THERMISTOR
SELECT
PELTIER
DEVICE
PASCO scientific
Model TD-8564
THERMAL EFFICIENCY
APPARATUS
HEATER
5 1%
12 VDC MAX
Out
0.5
Ω
1.0
Ω
2.0
Ω
W
Heat
Pump
K
Ω
461
436
413
391
370
351
332
315
298
283
°
C
-5
-4
-3
-2
-1
0
1
2
3
4
K
Ω
269
255
242
230
218
207
197
187
178
169
°
C
7
8
9
10
11
5
6
12
13
14
K
Ω
161
153
146
139
133
126
120
115
109
104
°
C
15
16
17
18
19
20
21
22
23
24
K
Ω
100
95.4
91.1
87.0
83.1
79.4
75.9
72.5
69.3
66.3
°
C
25
26
27
28
29
30
31
32
33
34
63.4
60.7
58.1
55.6
53.2
51.0
48.9
46.8
44.9
43.0
THERMISTOR TABLE
°
C C K
Ω °
C
35
36
37
38
39
40
41
42
43
44
41.2
39.6
37.9
36.4
34.9
33.5
32.2
30.9
29.7
28.5
45
46
47
48
49
50
51
52
53
54
27.4
26.4
25.3
24.4
23.4
22.5
21.7
20.9
20.1
19.3
55
56
57
58
59
60
61
62
63
64
K
Ω
18.6
17.9
17.3
16.6
16.0
15.5
14.9
14.4
13.8
13.4
°
C
65
66
67
68
69
70
71
72
73
74
K
Ω
12.9
12.4
12.0
11.6
11.2
10.8
10.4
10.1
9.76
9.43
°
C
75
76
77
78
79
80
81
82
83
84
K
Ω
9.12
8.81
8.52
8.24
7.96
7.70
7.45
7.21
6.98
6.75
°
C
85
86
87
88
89
90
91
92
93
94
K
Ω
6.53
6.33
6.12
5.93
5.74
5.56
5.39
5.22
5.06
4.91
°
C
95
96
97
98
99
100
101
102
103
104
Figure 4: Thermal Efficiency Apparatus
4
012-05443A Thermal Efficiency Apparatus
Measurements Using the Thermal Efficiency Apparatus
9
10
7
8
4
5
2
3
6
-1
0
1
-5
-4
-3
-2
14
15
16
11
12
13 k
Ω
242
230
218
207
315
298
283
269
255
461
436
413
391
370
351
332
197
187
178
169
161
153
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
°C 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
Ω
is a higher temperature than 200 k
Ω
).
➤ 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
Ω
. This reading lies between
120 k
Ω
= 21°C and 115 k
Ω
= 22°C. The reading is
120-118.7 = 1.3 k
Ω
above 21°C which is
1.3k
Ω ×
1
°
C
120 – 115k
Ω
= 0.26
Therefore 118.7 k
Ω
is 21.26°C.
°
C
Table 1: Resistance to Temperature Conversion Chart k
Ω
°C k
Ω
°C k
Ω
°C
146
139
133
126
120
115
109
104
100
24
25
95.4
26
91.1
27
87.0
28
83.1
29
79.4
30
75.9
31
72.5
32
21
22
23
17
18
19
20
69.3
33
66.3
34
63.4
35
60.7
36
58.1
37
55.6
38
53.2
39
51.0
40
48.9
41
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
22.5
60
21.7
61
20.9
62
20.1
63
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
10.1
82 k
Ω
°C
9.76
83
9.43
84
9.12
85
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
4.91
104
5
Thermal Efficiency Apparatus 012-05443A
Power Delivered to the Hot Reservoir (P
H
)
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
H
= I
H
V
H
.
Power Dissipated by the Load Resistor (P
W
)
The power dissipated by the load resistor is determined by measuring the voltage drop across the known load resistance and using the formula
P
W
=
V
2
.
R
The load resistors have a tolerance of 1%.
➤ NOTE: We may use the equation P
W
=
V
2
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
W
= I
W
V
W
.
Indirect Measurements
It will be necessary to know three additional quantities in the experiments:
➀
The internal resistance of the Peltier device; ➁ The amount of heat conducted through the device and the amount radiated away; ➂ The amount of
V s r
Peltier Device
V l
R l
Figure 5: Procedure for Finding Internal Resistance 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
V
S
– 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 V
S
.
Since we have measured V w rather than I in the heat engine mode, the equation above becomes
V s
–
V w
R
r – V w
= 0
Solving this for the internal resistance gives us
r =
V s
– V w
V w
R .
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
H
(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
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, P
C
, is equal to the rate at which heat is delivered to the hot reservoir,
P
H
, minus the rate at which work is being done, P
(Figure 3).
W
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).
Ω
T
H
Conducted
Power
Engine
V
A
Power
Supply P
W
Ω
V
Figure 6: Heat Engine With A Load
Conducted
Power
P
H
(open)
A
V
Power
Supply
V
Figure 7: No Load
7
T
C
T
H
T
C
Thermal Efficiency Apparatus
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.
012-05443A
8
012-05443A Thermal Efficiency Apparatus
Experiment 1: Heat Engine and Temperature Difference
EQUIPMENT NEEDED:
— Thermal Efficiency Apparatus
— ohmmeter
— patch cords
— 3 kg (7 lbs) ice and a bucket for the icewater bath
— DC power supply capable of 2.5 A at 12 V
— ammeter (up to 3 A)
— 2 voltmeters
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
➀
Prepare the ice-water bath and immerse both rubber tubes from the Thermal Efficiency Apparatus into the bath (Figure 4).
➁ 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”.
➂ Plug the ohmmeter into the thermistor terminals.
➃ 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.
Ω
A
V
Power
Supply
0.5
Ω
1
Ω
2
Ω
V
Figure 1.1
9
Thermal Efficiency Apparatus
➄ Connect the 2
Ω
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
Ω
load resistor is arbitrary. Any of the load resistances may be used.
Procedure
➀
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.
➁ 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.
➂ Record the voltage (V the load resistor (V
W
H
) across the heating resistor, the current (I
) in Table 1.1.
H
), and the voltage across
➃ Lower the voltage across the heating resistor by about 2 V.
➄
Repeat Steps 1 through 4 until data for five different hot temperatures have been taken.
Trial
1
2
3
4
5
Table 1.1 Data for Heat Engine
T
H
(k
Ω
) T c
(k
Ω
) T
H
(°C) T
H
(°C) V
H
I
H
V w
012-05443A
Calculations
➀ For each of the data runs, calculate the power supplied to the hot reservoir, P
H
, and the power used by the load resistor, P
W
, and record these in Table 1.2.
➁
Calculate the temperature difference for each trial and record it in Table 1.2.
➂ Calculate the actual efficiencies from the powers and record in Table 1.2.
➃ Calculate the Carnot (maximum) efficiencies from the temperatures and record in Table 1.2.
10
012-05443A Thermal Efficiency Apparatus
Trial
1
2
3
4
5
P
H
P w
Table 1.2 Calculated Values
T
H
(k) T c
(k)
∆
T (k) e actual e
Carnot
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 c
was nearly constant.
➀ The Carnot efficiency is the maximum efficiency possible for a given temperature difference. According to the graph, is the actual efficiency always less than the Carnot efficiency?
➁
Does the Carnot efficiency increase or decrease as the temperature difference increases?
➂
Does the actual efficiency increase or decrease as the temperature difference increases?
➃ 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?
➄ 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
Notes:
012-05443A
12
012-05443A Thermal Efficiency Apparatus
Experiment 2: Heat Engine Efficiency (Detailed Study)
EQUIPMENT NEEDED:
— Thermal Efficiency Apparatus
— ohmmeter
— 1 DC power supply capable of 2.5 A at 12 V
— patch cords
— ammeter (up to 3 A)
— 3 kg — (7 lbs) ice and a bucket for the icewater bath
— 2 voltmeters
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
➀ Prepare the ice-water bath and immerse both rubber tubes from the Thermal Efficiency
Apparatus into the bath (Figure 4).
➁ 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”.
➂
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.
➀ 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.
Ω ➤ 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.
0.5
Ω
1
Ω
2
Ω
V
A
V
Figure 2.1
Power
Supply
13
Thermal Efficiency Apparatus
B. Connect the 2
Ω
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
H
) across the heating resistor, the current (I
H across the load resistor (V
W
) in Table 2.1.
), and the voltage
➁ 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 (V s
) 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
H
, I
H
and V p
.
Calculations for the Heat Engine
➀ Actual Efficiency: Calculate the actual efficiency using
e =
P
W
P
H
, where P
W
=
V
2
W
R
and P
H
= I
H
V
H
.
Record the powers in Table 2.2 and the efficiency in Table 2.3.
012-05443A
Table 2.1 Data
T
H
(k
Ω
) T c
(k
Ω
) T
H
(°C) T c
(°C) V
H
I
H
V w
V
S
Mode
Engine
Open
14
012-05443A Thermal Efficiency Apparatus
Table 2.2 Calculated Values
Internal Resistance = r = ________________
T c
(K) P h
P w
Mode
Engine
(2
Ω
load)
Open
T h
(K) I w
Table 2.3 Results
Actual Adjusted
Maximum
(Carnot)
% Difference
Efficiency
➁
Maximum Efficiency: Convert the temperatures to Kelvin and record in Table 2.2. Calculate the Carnot efficiency using the temperatures and record in Table 2.3.
➂
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.
A. First, the work done in the actual efficiency calculation only includes
V
R
2
for the power dissipated by the load resistor R but, to account for total work done by the device, it should also include I
2 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
P
′
W
= P
W
+ I
2
W
r =
V
2
W
R
+ I
2
W r where I
W
=
V
W
R
. Calculate I
W
for the 2
Ω
load and record in Table 4.
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
′
H
= available heat = P
H
– P
H open
15
Thermal Efficiency Apparatus
The Thermal Efficiency Apparatus is run with a load connected to measure P
H
(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
′
H
=
P
W
+ I
2
W r
P
H
– P
H open
Calculate the internal resistance, r, using the equation
r =
V
P
– V
W
V
W
R 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 (maximum) efficiency
% Difference = e max
– e adjusted ×
100% e max and record in Table 2.3.
Questions
➀
If the difference between the temperature of the hot side and the cold side was decreased, would the maximum efficiency increase or decrease?
➁ 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?
➂ 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
∆
S
∆ t
=
∆
Q /
∆ t
T
=
P
T for each reservoir. Is the total change in entropy positive or negative? Why?
012-05443A
16
012-05443A Thermal Efficiency Apparatus
Experiment 3: Heat Pump Coefficient of Performance
EQUIPMENT NEEDED:
— Thermal Efficiency Apparatus
— patch cords
— ammeter (up to 3 A)
— 3 kg — (7 lbs) ice and a bucket for the ice-water bath
— 1 DC power supplies capable of 2.5 A at 12 V
— ohmmeter
— voltmeter
➤ 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
➀ Prepare the ice-water bath and immerse both rubber tubes from the Thermal Efficiency
Apparatus into the bath (Figure 4).
Ω for measuring temperatures
to AC supply to ice water tub for driving the Peltier device
–
+
Power
Supply
A
for measuring I w
P w
= V w
I w
V
for measuring V w
Figure 3.1 Heat Pump Mode
17
Thermal Efficiency Apparatus
➁ 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”.
➂
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
➃
Connect an ammeter and a voltmeter to the power supply.
Procedure
➀
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.
➁
Record the resistances and convert them to degrees. Also record the voltage (V
W current (I
W
) in Table 3.1.
) and the
Analysis
➀ Actual Coefficient of Performance: Calculate the actual COP using the data taken in the
Heat Engine experiment.
κ
=
P
C
P
W
=
P
H (OPEN)
– P
W
P
W
Record this result in Table 3.1.
➁ Maximum Coefficient of Performance: Calculate the maximum COP using
κ
MAX
=
T
C
T
H
– T
C and record this result in Table 3.1.
➂
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
2 r must 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,
κ
ADJUSTED
=
P
H (OPEN)
– P
W
P
W
– I
2
W r
Record this result in Table 3.1. Calculate the percent difference between the adjusted COP and maximum COP:
% Difference =
κ
MAX
–
κ
κ
ADJUSTED
MAX
×
100% and record in Table 3.1.
012-05443A
18
012-05443A Thermal Efficiency Apparatus
T
H
(k
Ω
) T
C
(k
Ω
) T
H
Table 3.1 Heat Pump Data and Results
(K) T
C
(K) V
W
I
W
P
W
COP COP COP actual max adj % diff
Questions
➀ If the difference between the temperature of the hot side and the cold side was decreased, would the maximum COP increase or decrease?
➁ 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
∆
S
∆ t
=
∆
Q /
∆ 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.
Heat
Power =
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).
x
Each element has a length to area ratio of 8.460 cm -1 . So
A
=
8.460cm
–1
. Use the data taken
142 in Experiment 2 for the Open Mode to calculate the thermal conductivity of the Peltier device:
k =
P
H (OPEN)
∆
T
(x / A)
Question
➀
How does the thermal conductivity of the Peltier device compare with the thermal conductivity 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:
— Thermal Efficiency Apparatus
— DCpower supply capable of 2.5 A at 12 V
— 3 kg (7 lbs) ice and a bucket for the ice-water bath
— ohmmeter
— ammeter (up to 3 A)
— 2 voltmeters
— patch cords
V s
R r
V l
Figure 5.1 Peltier device connected to a load resistor
Theory
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 = I
2
R. The amount of current that flows through the load resistor varies as the load is varied. From Figure
10, V
S
= I(r+R) where V
S
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:
P =
V s
r + R
2
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: dP dR
=
V
S
2
(r – R)
= 0
(r + R)
3
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)
Ω
Place ends of tubing in ice water tub
A
0.5
Ω
V
1
Ω
2
Ω
Figure 5.2 Connecting the 0.5
Ω
load resistor
21
V
Power supply
Thermal Efficiency Apparatus
Procedure
➀ 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.
➁
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.
➂ 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.
➃
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.
➄ Record the voltage (V the load resistor (V
W
H
) across the heating resistor, the current (I
) in Table 5.1.
H
), and the voltage across
012-05443A
1.5
2.0
2.5
R(
Ω
) T
H
0.5
1.0
3.0
3.5
(k
Ω
) T
C
Table 5.1: Heat Engine Data and Results
(k
Ω
) T
H
(°K) T° (°K) V
H
I
H
V
W
P
H
P
L e
22
012-05443A Thermal Efficiency Apparatus
➅ Calculate the power input to the hot side, P
H
= I
H
V
H
, and the power dissipated by the load resistor, P
L
=
V
2
W
. Calculate the efficiency, e =
R
P
L
P
H
. Record all these values in Table 5.1.
➆ 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
Ω
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
➀
For which load resistor is the efficiency a maximum?
➁ 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
Notes:
012-05443A
24
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