Fysik
Rapport - Energi Effektivitet
20. december 2006 af
Sannaen (Slettet)
Nogen som vil kigge denne rapport igennem for evt fejl osv.?
Jeg har undladt hypotesen samt konklusionen, da jeg har været helt på bar bund, med hvad jeg skulle skrive der. Nogen som kan hjælpe eller har en idé?
Derudover, kunne jeg godt bruge nogle ideér til mere så som diagrammer eller lign. (ved ikke lige hvad det skulle være et diagram over) men et eller andet? nogen ide?
The efficiency of an immerged electric heating device
Problem: What is the efficiency of an electric heating coil totally submerged in water?
Existing knowledge/ Background knowledge: To determine the efficiency we according to theory need to know both the amount of energy supplied E (input) and the amount transformed into useful energy E (output). Having determined both of these quantities we can calculate the efficiency µ as
µ = …
E (input) can be calculated from the following formula we used in the NSC experiment/ report:
E (input) = m c ?T
m = mass of water
c = the specific heat capacity of water. c = 4200 J/ (kg°C)
?T = the rise in temperature. ?T = T² - T¹
Hypothesis:
Experiment: We use a heating coil, i.e. a metal thread with relative high electric resistance that is coiled up between two conducting rods. To make a current run through the heat, and so heat is as well as the water up; we need a dc (direct current) power supply. We place the coil in a relative well isolated cup, fill it halfway with water, turn on the power supply, and follow the temperature rise in the water. To measure the input energy we simply use a joulmeter.
Results:
Experiment 1:
T¹ = 18,20 °C
m (Glass) = 24,16g
m (Glass and water) = 333,00g
m (water) = 333,00g – 24,16g = 308,84g = 0,30884kg
T² = 22,00 °C
?T = 22,00 °C – 18,20 °C = 3,80 °C
E (input) = 2736,00J
E (output) = 0,30884kg * 4200 J/ (kg°C) * 3,80 °C = 4929,09J
Efficiency = 2736,00J/ 4929,09J *100 = 55,51%
This experiment is wrong; The E (input) are incorrect, because the ? was turned on to late. It was sat to calculate the input from the back of the machine and of course, there was no input, so it kept showing 0J. After realising this we turned it on but obviously to late.
Experiment 2 :
There was not time for a second experiment, so I lent those numbers from Rebekka’s group and made some new calculations.
T¹ = 19,80 °C
m (water) = 392,20g = 0,3922kg
T² = 28,00 °C
?T = 28,00 °C - 19,80 °C = 8,20 °C
E (input) = 15400,00J
E (output) = 0,3922kg * 4200 J/ (kg°C) * 8,20 °C = 13507,37J
Efficiency = 13507,37J/ 15400,00J * 100 = 87,71%
Conclusion:
Jeg har undladt hypotesen samt konklusionen, da jeg har været helt på bar bund, med hvad jeg skulle skrive der. Nogen som kan hjælpe eller har en idé?
Derudover, kunne jeg godt bruge nogle ideér til mere så som diagrammer eller lign. (ved ikke lige hvad det skulle være et diagram over) men et eller andet? nogen ide?
The efficiency of an immerged electric heating device
Problem: What is the efficiency of an electric heating coil totally submerged in water?
Existing knowledge/ Background knowledge: To determine the efficiency we according to theory need to know both the amount of energy supplied E (input) and the amount transformed into useful energy E (output). Having determined both of these quantities we can calculate the efficiency µ as
µ = …
E (input) can be calculated from the following formula we used in the NSC experiment/ report:
E (input) = m c ?T
m = mass of water
c = the specific heat capacity of water. c = 4200 J/ (kg°C)
?T = the rise in temperature. ?T = T² - T¹
Hypothesis:
Experiment: We use a heating coil, i.e. a metal thread with relative high electric resistance that is coiled up between two conducting rods. To make a current run through the heat, and so heat is as well as the water up; we need a dc (direct current) power supply. We place the coil in a relative well isolated cup, fill it halfway with water, turn on the power supply, and follow the temperature rise in the water. To measure the input energy we simply use a joulmeter.
Results:
Experiment 1:
T¹ = 18,20 °C
m (Glass) = 24,16g
m (Glass and water) = 333,00g
m (water) = 333,00g – 24,16g = 308,84g = 0,30884kg
T² = 22,00 °C
?T = 22,00 °C – 18,20 °C = 3,80 °C
E (input) = 2736,00J
E (output) = 0,30884kg * 4200 J/ (kg°C) * 3,80 °C = 4929,09J
Efficiency = 2736,00J/ 4929,09J *100 = 55,51%
This experiment is wrong; The E (input) are incorrect, because the ? was turned on to late. It was sat to calculate the input from the back of the machine and of course, there was no input, so it kept showing 0J. After realising this we turned it on but obviously to late.
Experiment 2 :
There was not time for a second experiment, so I lent those numbers from Rebekka’s group and made some new calculations.
T¹ = 19,80 °C
m (water) = 392,20g = 0,3922kg
T² = 28,00 °C
?T = 28,00 °C - 19,80 °C = 8,20 °C
E (input) = 15400,00J
E (output) = 0,3922kg * 4200 J/ (kg°C) * 8,20 °C = 13507,37J
Efficiency = 13507,37J/ 15400,00J * 100 = 87,71%
Conclusion:
Skriv et svar til: Rapport - Energi Effektivitet
Du skal være logget ind, for at skrive et svar til dette spørgsmål. Klik her for at logge ind.
Har du ikke en bruger på Studieportalen.dk?
Klik her for at oprette en bruger.