Contents Exercise A - Steady-State Heat Conduction..................................................................1 OBJECTIVE..............................................................................................................1 METHOD...................................................................................................................1 THEORY....................................................................................................................1 APPARATUS/EQUIPMENT.....................................................................................2 PROCEDURE............................................................................................................3 RESULTS AND DISCUSSIONS..............................................................................3 CONCLUSION..........................................................................................................7 REFERENCE.............................................................................................................7 Exercise B – The Fourier Rate Equation.......................................................................8 OBJECTIVE..............................................................................................................8 METHOD...................................................................................................................8 THEORY....................................................................................................................8 APPARATUS/EQUIPMENT...................................................................................10 PROCEDURE..........................................................................................................12 RESULTS AND DISCUSSIONS............................................................................12 CONCLUSION........................................................................................................18 REFERENCE...........................................................................................................18
HT11 Issue 11: Linear Heat Conduction Exercise A - Steady-State Heat Conduction OBJECTIVE To measure the temperature distribution for steady-state conduction of energy through a uniform plane wall and demonstrate the effect of a change in heat flow. METHOD By measuring the change in temperature with distance resulting from the linear conduction of heat along a simple bar at different rates of heat flow through the bar. THEORY Provided that the heated and cooled sections are clamped tightly together, so that the two end faces are in good thermal contact, the two sections can be considered to be one continuous wall of uniform cross section and material. According to Fourier’s law of heat conduction: If a plane wall of thickness (Δχ) and area (A) supports a temperature difference then the heat transfer rate per unit time (Q) by conduction through the wall is found to be: Q∝ A
∆T ∆x
Q=C
∆T ∆x
where C is a constant which will be investigated in a later exercise. The object of this exercise is to show that for a simple plane wall where the material and cross section are constant: Q∝
∆T ∆x
1
ΔT : Change in temperature Q
: Heat Flow
Δχ : Change in position
Figure 1: Heat transfer process through conduction.
APPARATUS/EQUIPMENT
Heater
Cold Water Supply
Thermocouples
HT11 Linear Heat Conduction Accessory HT10XC Computer Compatible Heat Transfer Service Unit Figure 2: Set-up of Apparatus 2
PROCEDURE 1. The front mains switch is switched on. (The RCD and any circuit breakers at the rear of the service unit are checked in case the panel meters do not illuminate; all switches at the rear should be up.) 2. The selector switch is set to ‘MANUAL’ for manual operation. The cooling water is turned on and the flow control valve (not the pressure regulator) is adjusted to give approximately a constant flow rate. 3. The heater voltage is set to 4 volts by adjusting the voltage control potentiometer to give a reading of 4 volts on the top panel meter with the selector switch set to position V. 4. The HT11 is allowed to stabilize and the temperatures is monitored using the lower selector switch/meter. 5. The temperatures are recorded after they are stable. The temperature values are noted down by hand from the front panel display, using the selector switch to select each required value from T1 to T8 in turn (except T4 and T5). 6. Steps 3 to 5 are repeated by setting different heater voltages of 7V, 10V and 14V accordingly. The results are tabulated.
RESULTS AND DISCUSSIONS For this exercise, the distance between each thermocouples are constant as follow: i. ii. iii. iv.
Distance between thermocouples T1 and T3, Δχ13 = 0.03m Distance between thermocouples T6 and T8, Δχ68 = 0.03m Distance between each thermocouple = 0.015m Distance between thermocouple T3 or T6 and the end face = 0.0075m
The raw data obtained in this exercise is shown in Table 1 below. Table 1:
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Heater Voltage, V (Volts) Heater Current, I (Amps) Heated section of high temperature, T1 (°C) Heated section of mid temperature, T2 (°C) Heated section of low temperature, T3 (°C) Cooled section of high temperature, T6 (°C) Cooled section of mid temperature, T7 (°C) Cooled section of low temperature, T8 (°C)
4 0.42 30.8 30.4 30.0 28.8 28.6 28.3
7 0.72 37.6 36.5 35.2 31.2 30.2 29.2
10 1.03 48.2 46.0 43.5 34.8 32.7 30.6
14 1.44 67.1 62.7 58.0 41.4 37.2 33.1
For each set of readings, the derived results are tabulated in Table 2 below. Table 2: Heater Voltage Heater Current Heat Flow (power to heater) Temperature difference in heated section Temperature difference in cooled section
V (Volts) I (Amps) Q=VI (Watts) ΔThot=T1-T3 (°C) ΔTcold=T6-T8 (°C)
4 0.42 1.68
7 0.72 5.04
10 1.03 10.3
14 1.44 20.16
0.8
2.4
4.7
9.1
0.5
2
4.2
8.3
Comparison between ΔThot and ΔTcold in the two sections at the same/different heat flow:
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Graph of Temperature Difference against Heat Flow 10 f(x) = 0.45x + 0.09 f(x) = 0.42x - 0.16
8 6
ΔT (°C)
4 2 0
0
5
10
15
20
25
Q (Watts) ΔThot
Linear (ΔThot)
Linear (ΔThot)
Linear (ΔThot)
Linear (ΔThot)
ΔTcold
Linear (ΔTcold)
From the graph, it can be deduced that ΔT hot is greater than ΔTcold at the same heat flow. Meanwhile, ΔThot and ΔTcold also increases with the increase of heat flow. When the heat flow increases, the difference between ΔThot and ΔTcold also widens.
Graph of T aginst χ 80 70 f(x) = - 501.14x + 76.23
60 50
T (°C)
f(x) = - 260.19x + 52.96
40
f(x) = - 123.62x + 39.81 f(x) = - 36.38x + 31.39
30 20 10 0 0
0.02
0.03
0.05
0.06
0.08
χ (m) Q=1.68Watts
Linear (Q=1.68Watts)
Q=5.04Watts
Linear (Q=5.04Watts)
Q=10.3Watts
Linear (Q=10.3Watts)
Q=20.16Watts
Linear (Q=20.16Watts)
5
0.09
0.11
From the graph, it can be observed that each temperature profile is a straight line and the magnitude of the gradient of each profile increases linearly with the increase of heat flow (Note that the negative sign of the gradient is not significant and being ignored in analysis because it only indicates that ΔT falls with the increase of Δχ). The gradients are tabulated in Table 3 corresponding with heat flow. Table 3: Q (Watts)
Gradient (°C/m)
1.68 5.04 10.3 20.16
36.381 123.62 260.19 501.14
Q (Watts ∙m/℃) Gradient 0.0462 0.0408 0.0396 0.0402
From Table 3, it can be deduced that
Q Gradient
is a constant, which is
approximately equal to 0.04 Watts ∙m/℃ . Experimental Errors and Cumulative Influence on Calculated Values for a) Heat Flow, Q The second law of thermodynamics stated that the entropy of an isolated system always increases over time, which means energy will dissipate or lose during energy transferring or transformation. This explain why our calculated values for heat flow (Q=IV) is not the exact power supplied to the heater. The system has encountered loss of heat energy in energy transfer. b) ΔThot and ΔTcold Calculation of ΔThot and ΔTcold might encounter two errors: i.
Insufficient Time for HT11 to Stabilize Due to time constraint to carry out the experiment, we only allocated 10 minutes for the equipment to stabilize for each heat flow. In fact, 10 minutes is not long enough as we still observed some fluctuation of the temperature reading at the moment of
6
recording result. 20-30 minutes will be more appropriate for HT11 to stabilize in order to obtain a more accurate result. ii.
Absence of Sensor SFT2 to Control the Cooling Water Flowrate Without sensor SFT2, we were unable to monitor the flowrate of cooling water during the experiment. Therefore, the flowrate of cooling water might not be fixed throughout the experiment as we could hardly monitor the flowrate simply by controlling the valve.
CONCLUSION Fouriers law of heat conduction is proven, which for a wall of constant conductivity and constant cross sectional area, the temperature gradient is directly proportional to the rate of heat flow. The heat flow through a plane wall will create temperature ∆T difference which changes linearly with distance. The equation, Q=C ∆ x
is valid.
ΔThot and ΔTcold increases linearly with the heat flow. When the heat flow increases, the gap between ΔThot and ΔTcold widens. The measurement accuracy of the equipment is high and the heat loss from the equipment is minimal. This can be proven as we managed to get a constant for Q Gradient
of each temperature profile, which is approximately equal to 0.04
Watts ∙m/℃ . It means that the error encountered is very small and hence we can deduce that the accuracy of the equipment is high, REFERENCE J.Moran, M. (2011). Fundamentals of Engineering Thermodynamics. United States of America: Don Fowley.
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Exercise B – The Fourier Rate Equation OBJECTIVE To understand the use of the Fourier Rate Equation in determining rate of heat flow through solid materials for one-dimensional steady flow of heat. METHOD To demonstrate Fourier’s Law for the linear conduction of heat along a simple bar by measuring the change in temperature with distance at different rates of heat flow through the bar and using the measurements to calculate the conductivity of the bar. THEORY Provided that the heated and cooled sections are clamped tightly together, so that the two end faces are in good thermal contact, the three sections can be considered to be one continuous wall of uniform cross section and material. According to Fourier’s law of heat conduction: If a plane wall of thickness (Δχ) and area (A) supports a temperature difference then the heat transfer rate per unit time (Q) by conduction through the wall is found to be: Q∝ A
∆T where ∆ x =(x b−x a ) ∆x
ΔT : Change in temperature Q
: Heat Flow
Δχ : Change in position
8
If the material of the wall is homogeneous and has a thermal conductivity k (the constant of proportionality) then: Q=−kA
∆T where ∆T =(T a−T b) ∆x
It should be noted that heat flow is positive in the direction of temperature fall hence the negative sign in the equation. For convenience the equation can be rearranged to avoid the negative sign as follows: Q=kA
∆T where ∆ T =(T a−T b ) ∆x
Note: In this exercise the HT11 is configured as a simple plane wall where k and A are constant therefore the object is to show that the Fourier equation can be used to relate Q, ΔT and Δχ.
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APPARATUS/EQUIPMENT
Cold Water Supply HT11 Linear Heat Conduction Accessory HT10XC Computer Compatible Heat Transfer Service Unit Heater
25mm Brass Heater
Thermocouples
Figure 3: Set-up of Apparatus
10
Voltage Display Unit
Voltage Potentiometer Current Display Unit
11
PROCEDURE 1. The front mains switch is switched on. (The RCD and any circuit breakers at the rear of the service unit are checked in case the panel meters do not illuminate; all switches at the rear should be up.) 2. The selector switch is set to ‘MANUAL’ for manual operation. The cooling water is turned on and the flow control valve (not the pressure regulator) is adjusted to give approximately a constant flow rate. 3. The heater voltage is set to 4 volts by adjusting the voltage control potentiometer to give a reading of 4 volts on the top panel meter with the selector switch set to position V. 4. The HT11 is allowed to stabilize and the temperatures is monitored using the lower selector switch/meter. 5. The temperatures are recorded after they are stable. The temperature values are noted down by hand from the front panel display, using the selector switch to select each required value from T1 to T8 in turn. 6. Steps 3 to 5 are repeated by setting different heater voltages of 7V, 10V and 14V accordingly. The results are tabulated. RESULTS AND DISCUSSIONS For this exercise the raw data is tabulated under the following headings in Table 4.
Heater Voltage Heater Current Heated section of high temperature Heated section of mid temperature Heated section of low temperature Intermediate section of high temperature Intermediate section of low temperature Cooled section of high temperature Cooled section of mid temperature Cooled section of low temperature
V (Volts) I (Amps) T1 (°C) T2 (°C) T3 (°C)
4
7
10
14
0.42 32 31.7 31.1
0.72 38.4 37.3 36
1.03 49 46.8 44.3
1.44 69 64.8 60
T4 (°C)
30.2
33.9
40.2
52
T5 (°C) T6 (°C) T7 (°C) T8 (°C)
29.6 29.2 28.8 28.5
32.7 30.9 30 29.1
37.8 33.9 32 30.2
47.7 39.8 36 32.4
Table 4: For this exercise the applicable constants are shown in Table 5 below. 12
Table 5: Distance between thermocouples T1 and T3, χ13 (m) Distance between thermocouples T4 and T5, χ45 (m) Distance between thermocouples T6 and T8, χ68 (m) Diameter of bar, D (m) Note: The distance between each thermocouple is 0.015 m. The
0.03 0.015 0.03 0.025 distance between
thermocouple T3, T4, T5 or T6 and the end face is 0.0075 m. For each set of readings the derived results are tabulated in Table 6 below. Table 6: Heater V (Volts) Voltage Heater I (Amps) Current Heat Flow (Power to Q=VI (Watts) heater) 2 CrossπD A= (m2 ) sectional area 4 of bar Temperature difference in ΔThot =T1-T3 (°C) heated section x13 Q k hot = Conductivity ∆ T hot A hot in heated section (W / m ℃) Temperature difference in intermediate section
ΔTint =T4-T5 (°C)
4
7
10
14
0.42
0.72
1.03
1.44
1.68
5.04
10.3
20.16
0.000491
0.000491
0.000491
0.000491
0.9
2.4
4.7
9
114.0823
128.3425
133.934
136.8987
0.6
1.2
2.4
4.3
85.5617
128.3425
131.1437
143.2661
0.7
1.8
3.7
7.4
∫ ¿ A∫ ¿ Conductivity in intermediate section
∆T¿ x Q ∫ ¿= 45¿ k¿ (W / m ℃)
Temperature difference in
ΔTcold =T6-T8 (°C)
13
cooled section Conductivity in cooled section
k cold =
x 68 Q ∆ T cold Acold 146.6772
171.1234
170.1323
166.4984
(W / m ℃)
Comparison between the calculated values for the thermal conductivity of Brass in the three sections at the same/different heat flows is shown in Table 7. Table 7: Q (Watts) khot (W/m°C) kint (W/m°C) kcold (W/m°C)
1.68 114.0823 85.5617 146.6772
5.04 128.3425 128.3425 171.1234
10.3 133.934 131.1437 170.1323
20.16 136.8987 143.2661 166.4984
Graph of Thermal Conductivity against Heat Flow 200 150
k (W/m°C) 100 50 0
0
5
10
15
20
25
Q (Watts) khot (W/m°C)
Linear (khot (W/m°C))
kint (W/m°C)
Linear (kint (W/m°C))
kcold (W/m°C)
Linear (kcold (W/m°C))
At the same heat flow, kcold is always the highest as compared to kint and khot. However, fluctuation is observed between kint and khot. Initially, khot is higher than kint. However as heat flow increases, kint and khot are almost on par and finally kint has become higher than khot when the heat flow is the highest. 14
At different heat flow, both k int and khot increases linearly with heat flow. However, kcold has encountered a fluctuation trend, which it increases at first and drops later. Although kcold falls slightly as heat flow increases, it is still higher than kint and khot. Hence, it can be deduced that thermal conductivity of Brass is always the best in cooled section. For intermediate section and hot section, its thermal conductivity is dependent on the rate of heat flow. The thermal conductivity in intermediate section is better when heat flow is high and vice-versa.
Graph of T against χ 80 70 Q=1.68Watts
Linear (Q=1.68Watts) f(x) = - 369.13x + 75.13
Q=5.04Watts
60 50 f(x) = - 189.84x + 52.09 40
Q=10.3Watts f(x) = - 93.73x + 39.86
30
f(x) = - 35.95x + 32.56
Linear (Q=5.04Watts) T (°C)
Linear (Q=10.3Watts)
20 10 Q=20.16Watts
Linear (Q=20.16Watts)
0 0
0.02
0.03
0.05
0.06
0.08
0.09
0.11
0.12
0.14
x (m)
From the graph, it can be observed that each temperature profile is a straight line and the magnitude of the gradient of each profile increases linearly with the increase of heat flow (Note: The negative sign is not significant because for convenience in this exercise, the equation of Fourier’s Law of heat conduction has been rearranged). From Fourier’s Equation of heat conduction, 15
Q=kA
∆T ∆x
Rearrange the equation to find k, k=
Q ∆T ÷ A ∆x
where
∆T is the gradients of each temperature profile ∆x
Thus, the gradients are tabulated in Table 8 corresponding with heat flow in order to calculate thermal conductivity of Brass. Table 8: Q (Watts)
A (m2)
Gradient (°C/m)
k ( W /m℃ ¿
1.68 5.04 10.3 20.16
0.000491 0.000491 0.000491 0.000491
35.952 93.73 189.84 369.13
95.20 109.54 110.53 111.26
Average conductivity of the brass ¯ ¿
¿
( 95.20+ 109.54+110.53+111.26 ) 4
¿ 106.6 3
W /m℃
As compared with the values previously obtained for each individual section of the bar in Table 7, it is noticeable that the average conductivity is very different from the conductivity of each individual section. The difference between them is getting greater and greater when the heat flow increases. Given the real thermal conductivity of Brass should be in the range 110 - 128 W/m°C. Thus, our result, 106.63 W /m℃
has encountered error and falls out of range.
16
|110−106.63 |×100 =3.06 110
Percentage Error=
Experimental Errors and Cumulative Influence on Calculated Values for a) Heat Flow, Q The second law of thermodynamics stated that the entropy of an isolated system always increases over time, which means energy will dissipate or lose during energy transferring or transformation. This explain why our calculated values for heat flow (Q=IV) is not the exact power supplied to the heater. The system has encountered loss of heat energy in energy transfer. b) ΔThot, ΔTint and ΔTcold Calculation of ΔThot, ΔTint and ΔTcold might encounter two errors: i.
Insufficient Time for HT11 to Stabilize
Due to time constraint to carry out the experiment, we only allocated 10 minutes for the equipment to stabilize for each heat flow. In fact, 10 minutes is not long enough as we still observed some fluctuation of the temperature reading at the moment of recording result. 20-30 minutes will be more appropriate for HT11 to stabilize in order to obtain a more accurate result. ii.
Absence of Sensor SFT2 to Control the Cooling Water Flowrate Without sensor SFT2, we were unable to monitor the flowrate of cooling water during the experiment. Therefore, the flowrate of cooling water might not be fixed throughout the experiment as we could hardly monitor the flowrate simply by controlling the valve. c) khot, kint and kcold The experimental errors in calculating khot, kint and kcold is most probably due to heat loss from the equipment. The reason is that a small amount of heat loss is inevitable as the temperature of the bar increases and hence the calculated value for the conductivity will increase at higher operating temperatures. 17
d) Average Thermal Conductivity of Brass Minor error (3.06%) of the average thermal conductivity of brass is due to the cumulative influence of experimental errors on calculated values for Q, ΔT hot, ΔTint and ΔTcold. Moreover, the given range of 110 - 128 W/m°C for average thermal conduCtivity of Brass is based on the assumption that no heat loss from the equipment. However in actual experiment, heat loss is inevitable as stated by second law of thermodynamics.
CONCLUSION The Fourier rate equation,
Q=kA
∆T ∆x
is proven and can be used to relate
temperature difference, heat flow and distance in a solid material of constant crosssectional area. The equation can also be used to calculate the constant, which is the thermal conductivity of a solid material. The temperature gradient increases linearly with the heat flow. Although the average thermal conductivity obtained in this exercise does not fulfill the range of real thermal conductivity of Brass, the measurement accuracy of the equipment is still high because only 3.06% of percentage error is encountered. In addition, the error in this exercise is higher than exercise A because the heat loss from the equipment is higher. This is the factor which contributes the most to our experiment error in this exercise. The reason is that the presence of brass bar has increased the heat loss of the equipment.
REFERENCE J.Moran, M. (2011). Fundamentals of Engineering Thermodynamics. United States of America: Don Fowley.
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