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Physics Assignment - Efficiency of an Electrical Appliance - Kettle.

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Dyl
#1
Jun1-12, 06:19 AM
P: 4
1. The problem statement, all variables and given/known data
Hey physicists. I have a problem. My assignment this term for physics is to test the efficiency of an electrical appliance, and to make suggestions on how to improve the efficiency for the future.

I conducted a number of experiments on the kettle, using 2 different kettles. Each kettle was tested 3 times for each volume of water, each testing 500mL, 1,000mL and 1,500mL volumes.

After all the data was recorded - Temperature before boil, time taken to boil, temp. after boil - I was able to work out the energy absorbed by the water, the average power taken to heat the water over the duration, and finally the efficiency.

From the data, the efficiencies of all three experiments per volume of water were averaged out, giving for kettle a:

500mL - 80%
1,000mL - 86%
1,500mL - 92%

This shows a 6% increase in efficiency per 500mL increase in volume. I graphed this to show the linear relationship, and then it came to explaining this... And I am stumped.

I have searched the internet for hours and hours on end, and have not been able to determine the reason for this increase in efficiency. PLEASE help me find out why? My teacher told me he can't tell me the reason, i have to find it myself.

He said two things will help me. The shape of the kettle, and the surface of the water compared to the volume.

From this I developed only 1 reason... The surface of the water was being reduced as more water was added in, due to the shape of the kettle (reducing in size as it got closer to the top of the kettle, thus the more water, the less surface area). This would explain that water heated by conduction between water and the element on the bottom of the kettle, which then became less dense and was displaced upwards by the more dense water; creating a convectional current. The heat lost would slowly decrease as more water was added due to the shape as there is a smaller surface area.

But the second kettle remained the same in shape from base to top. So this cannot be the full reason?


2. Relevant equations

The equations I used were viewed by my teacher and he told me they were fine, but I will show them anyway.

For Test 1, 500mL Kettle A:

Temp before boil: 19
Time taken (secs): 97s
Temp after boil: 98

This allowed me to determine the heat absorbed by the water using Q = m.c. Δt where m is mass of water, c is specific heat capacity of water, and Δt is the change in temp.

Q = 500 x 4.186 x (98-19)
Q = 2093 x (79)
Q = 165,347 J

This shows 165,347 J was absorbed by the water.

From this, we could find the average power to run the kettle with P = E/T where E is the energy absorbed and T is the time in sec.

P = E/T
P = 165,347/97
P= 1704.608247
∴ P ≈1705W

Finally the efficiency was determined using E = (Pout/Pin)x100% where Pout is the power output (1705W) and Pin is the power input (found on the kettle, 2100W).

E = (1705/2100)x 100
E = 0.8119047619 x 100
E ≈ 81%



3. The attempt at a solution

As stated earlier:

"I developed only 1 reason... The surface of the water was being reduced as more water was added in, due to the shape of the kettle (reducing in size as it got closer to the top of the kettle, thus the more water, the less surface area). This would explain that water heated by conduction between water and the element on the bottom of the kettle, which then became less dense and was displaced upwards by the more dense water; creating a convectional current. The heat lost would slowly decrease as more water was added due to the shape as there is a smaller surface area.

But the second kettle remained the same in shape from base to top. So this cannot be the full reason?



This is the only reason I can think of that would explain the efficiency of an electric kettle increasing as more water is added. The surface area in the kettle that changes shape, decreases the surface area of the water, and this reduces the availability of heat to escape through the water.

What other reasons could do this, and what could increase the efficiency of a kettle?
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gneill
#2
Jun1-12, 07:39 AM
Mentor
P: 11,689
Quote Quote by Dyl View Post
1. The problem statement, all variables and given/known data
Hey physicists. I have a problem. My assignment this term for physics is to test the efficiency of an electrical appliance, and to make suggestions on how to improve the efficiency for the future.

I conducted a number of experiments on the kettle, using 2 different kettles. Each kettle was tested 3 times for each volume of water, each testing 500mL, 1,000mL and 1,500mL volumes.

After all the data was recorded - Temperature before boil, time taken to boil, temp. after boil - I was able to work out the energy absorbed by the water, the average power taken to heat the water over the duration, and finally the efficiency.

From the data, the efficiencies of all three experiments per volume of water were averaged out, giving for kettle a:

500mL - 80%
1,000mL - 86%
1,500mL - 92%

This shows a 6% increase in efficiency per 500mL increase in volume. I graphed this to show the linear relationship, and then it came to explaining this... And I am stumped.

I have searched the internet for hours and hours on end, and have not been able to determine the reason for this increase in efficiency. PLEASE help me find out why? My teacher told me he can't tell me the reason, i have to find it myself.

He said two things will help me. The shape of the kettle, and the surface of the water compared to the volume.

From this I developed only 1 reason... The surface of the water was being reduced as more water was added in, due to the shape of the kettle (reducing in size as it got closer to the top of the kettle, thus the more water, the less surface area). This would explain that water heated by conduction between water and the element on the bottom of the kettle, which then became less dense and was displaced upwards by the more dense water; creating a convectional current. The heat lost would slowly decrease as more water was added due to the shape as there is a smaller surface area.

But the second kettle remained the same in shape from base to top. So this cannot be the full reason?


2. Relevant equations

The equations I used were viewed by my teacher and he told me they were fine, but I will show them anyway.

For Test 1, 500mL Kettle A:

Temp before boil: 19
Time taken (secs): 97s
Temp after boil: 98

This allowed me to determine the heat absorbed by the water using Q = m.c. Δt where m is mass of water, c is specific heat capacity of water, and Δt is the change in temp.

Q = 500 x 4.186 x (98-19)
Q = 2093 x (79)
Q = 165,347 J

This shows 165,347 J was absorbed by the water.

From this, we could find the average power to run the kettle with P = E/T where E is the energy absorbed and T is the time in sec.

P = E/T
P = 165,347/97
P= 1704.608247
∴ P ≈1705W

Finally the efficiency was determined using E = (Pout/Pin)x100% where Pout is the power output (1705W) and Pin is the power input (found on the kettle, 2100W).

E = (1705/2100)x 100
E = 0.8119047619 x 100
E ≈ 81%



3. The attempt at a solution

As stated earlier:

"I developed only 1 reason... The surface of the water was being reduced as more water was added in, due to the shape of the kettle (reducing in size as it got closer to the top of the kettle, thus the more water, the less surface area). This would explain that water heated by conduction between water and the element on the bottom of the kettle, which then became less dense and was displaced upwards by the more dense water; creating a convectional current. The heat lost would slowly decrease as more water was added due to the shape as there is a smaller surface area.

But the second kettle remained the same in shape from base to top. So this cannot be the full reason?



This is the only reason I can think of that would explain the efficiency of an electric kettle increasing as more water is added. The surface area in the kettle that changes shape, decreases the surface area of the water, and this reduces the availability of heat to escape through the water.

What other reasons could do this, and what could increase the efficiency of a kettle?
Hi Dyl, Welcome to Physics Forums.

Did you take note of the actual electrical power rating of the kettles (should be stamped on them somewhere)? How does your calculated power compare?

Did you remeasure the volume of water immediately after each trial?

Consider that the water doesn't only have a top surface...
Chestermiller
#3
Jun1-12, 09:19 AM
Sci Advisor
HW Helper
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Chestermiller's Avatar
P: 5,263
Quote Quote by Dyl View Post
1. The problem statement, all variables and given/known data
Hey physicists. I have a problem. My assignment this term for physics is to test the efficiency of an electrical appliance, and to make suggestions on how to improve the efficiency for the future.

I conducted a number of experiments on the kettle, using 2 different kettles. Each kettle was tested 3 times for each volume of water, each testing 500mL, 1,000mL and 1,500mL volumes.

After all the data was recorded - Temperature before boil, time taken to boil, temp. after boil - I was able to work out the energy absorbed by the water, the average power taken to heat the water over the duration, and finally the efficiency.

From the data, the efficiencies of all three experiments per volume of water were averaged out, giving for kettle a:

500mL - 80%
1,000mL - 86%
1,500mL - 92%

This shows a 6% increase in efficiency per 500mL increase in volume. I graphed this to show the linear relationship, and then it came to explaining this... And I am stumped.

I have searched the internet for hours and hours on end, and have not been able to determine the reason for this increase in efficiency. PLEASE help me find out why? My teacher told me he can't tell me the reason, i have to find it myself.

He said two things will help me. The shape of the kettle, and the surface of the water compared to the volume.

From this I developed only 1 reason... The surface of the water was being reduced as more water was added in, due to the shape of the kettle (reducing in size as it got closer to the top of the kettle, thus the more water, the less surface area). This would explain that water heated by conduction between water and the element on the bottom of the kettle, which then became less dense and was displaced upwards by the more dense water; creating a convectional current. The heat lost would slowly decrease as more water was added due to the shape as there is a smaller surface area.

But the second kettle remained the same in shape from base to top. So this cannot be the full reason?


2. Relevant equations

The equations I used were viewed by my teacher and he told me they were fine, but I will show them anyway.

For Test 1, 500mL Kettle A:

Temp before boil: 19
Time taken (secs): 97s
Temp after boil: 98

This allowed me to determine the heat absorbed by the water using Q = m.c. Δt where m is mass of water, c is specific heat capacity of water, and Δt is the change in temp.

Q = 500 x 4.186 x (98-19)
Q = 2093 x (79)
Q = 165,347 J

This shows 165,347 J was absorbed by the water.

From this, we could find the average power to run the kettle with P = E/T where E is the energy absorbed and T is the time in sec.

P = E/T
P = 165,347/97
P= 1704.608247
∴ P ≈1705W

Finally the efficiency was determined using E = (Pout/Pin)x100% where Pout is the power output (1705W) and Pin is the power input (found on the kettle, 2100W).

E = (1705/2100)x 100
E = 0.8119047619 x 100
E ≈ 81%



3. The attempt at a solution

As stated earlier:

"I developed only 1 reason... The surface of the water was being reduced as more water was added in, due to the shape of the kettle (reducing in size as it got closer to the top of the kettle, thus the more water, the less surface area). This would explain that water heated by conduction between water and the element on the bottom of the kettle, which then became less dense and was displaced upwards by the more dense water; creating a convectional current. The heat lost would slowly decrease as more water was added due to the shape as there is a smaller surface area.

But the second kettle remained the same in shape from base to top. So this cannot be the full reason?



This is the only reason I can think of that would explain the efficiency of an electric kettle increasing as more water is added. The surface area in the kettle that changes shape, decreases the surface area of the water, and this reduces the availability of heat to escape through the water.

What other reasons could do this, and what could increase the efficiency of a kettle?
Consider making a graph of the efficiency as a function of the surface-to-volume ratio of the water. Ask yourself, "where does the heat escape from the water?"

Chet

Dyl
#4
Jun1-12, 10:12 AM
P: 4
Physics Assignment - Efficiency of an Electrical Appliance - Kettle.

Thanks guys for replying!

I just realised, stupidly, that water inside the kettle will have more than one surface to escape from, and not just through latent heat loss through turning into water vapour; the water will actually be lost through convection with the sides of the kettle and radiation with the outside area of the kettle.

To put this into my assignment effectively, I began by stating the assumption that the kettle is a cylinder. From this I wanted to work out the surface area, so to find the height of the water in the kettle, I used V = pi r^2 h to get h = v/(pi x r^2) to get the height. Substituting in 500mL to find the height the water will reach in the kettle, we get h = 500/(pi x r^2) which gave me a height of approx. 2.1cm. Doing this to all volumes I receive the heights of 500mL, 1,000mL and 1,500mL as 2.1cm, 4.2cm and 6.3cm respectively.

I used this to then calculate the surface area of each using the formula, SA = (2 x pi x r x h) + (2 x pi x r^2) and got the following surface areas for volumes:

500mL = 596.51cm^2
1,000mL =711.96cm^2
1,500mL = 827.42cm^2

From this, we can compare the surface areas to the efficiencies that were calculated.

500mL = Efficiency of 80%, Surface Area: 596.51cm^2
1,000mL = Efficiency of 86%, Surface Area: 711.96cm^2
1,500mL = Efficiency of 92%, Surface Area: 827.42cm^2

The difference between the surface area of 500mL and 1,000mL, 115.45cm^2 (approx) represents a 6% increase in efficiency, further, the differences between 1,000mL and 1,500mL is also approx. 115.45cm^2, reflecting this same 6% increase in efficiency.

This proves that, the increase in volume does not increase the surface area at a rate that is proportional. Due to this, the surface area does not remain at a constant rate with respect to the increasing volume. As heat is lost through conduction with the sides of the kettle, through latent heat where the water is turning to vapour and escaping, and through radiation, this decrease in surface area that comes with the increase of volume results in less heat being lost and therefore more efficient heat distribution.

Is this correct??
Dyl
#5
Jun1-12, 10:24 AM
P: 4
Gneill, yes we did measure the volume afterwards, and there was an increase; this was due to thermal volume expansion and my teacher said to not include it in our assignment as the measurement.

Instead, just because I do believe it is slightly relevant, and it would demonstrate my understanding better, I was going to touch on it while I explain the components of the kettle, ie. the limit of water it can hold. The limit is set in place (normally 1800mL) to reduce the impact thermal volume expansion will have. If anything over 1.8L is put into the kettle the thermal expansion will cause the water to spill over and out of the spout.

I made use of the data that I had with surface area etc, and compared it to 1800mL. A volume of 1800mL corresponded to a surface area of 893.39cm^2. From this, I found the difference between this surface area and the surface area of 1500mL, 827.42cm^2. I got a difference of 65.97cm^2, and using this I worked out what the efficiency should be by dividing 115.45cm^2 (which we know gives us a difference of 6% efficiency) by 65.97cm^2 and got the answer of 1.75 which I then used to divide 6. 6/1.75 gave me 3.43 which was the difference in efficiency. Adding this 3.43% onto the efficiency of a 1,500mL boil and we get 99.43% efficiency.

Is there a better way to work out the difference in efficiency than the way I just did by the way?
Dyl
#6
Jun1-12, 10:27 AM
P: 4
Also, if 1800mL did correspond to an efficiency of 99.43%, what would happen if the kettle was to boil 2,000mL? The efficiency would then be 102% which cannot be possible. How can I explain this? I cannot create energy. :O
Chestermiller
#7
Jun1-12, 02:33 PM
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Quote Quote by Dyl View Post
Also, if 1800mL did correspond to an efficiency of 99.43%, what would happen if the kettle was to boil 2,000mL? The efficiency would then be 102% which cannot be possible. How can I explain this? I cannot create energy. :O
The behavior is much more complicated and detailed than that described simply by a change in surface area. The effect of surface area is just the first approximation to quantifying the behavior. To quantify things more accurately, you would need to consider and model many other effects. If the tea kettle allows venting (constant pressure), then there can be evaporation of water, and associated evaporative cooling. Even if the tea kettle does not allow venting, some water evaporation can occur into the head space. The heat transfer at the interface between the water and the kettle is going to involve both convection within liquid in the kettle, convective resistance from the kettle into the air, and radiative heat transfer into the room. The dominant effect will typically be convective resistance on the air side of the interface (primarily because the air has a very low thermal conductivity compared to the liquid water). There will typically be a thin air thermal boundary layer on the outside of the kettle that acts to insulate the kettle (but, of course, not completely). This needs to be taken into account in the modeling. I don't think that your teacher was considering modeling the behavior to this level of detail. But, it can be done.


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