May I ask you something? If we assume that is the Area (30m^2) of the outside wall of one room. Then every second we need 1035 Watt to keep the room temperature constant. Then in one hour the required energy of one heater to keep the temperature constant will be 1035 times 3600 seconds which is 3726000 Watts or 3726kW. This is an extremely high number. I would appreciate if you explain me where am i wrong.
Where did you determine that 1035 Watts per second was necessary? That is not the calculation that is being done in this video. You may be confusing this with a J/s which is a Watt (not the same as a Watt/s). You would rarely ever need 1035 Watts per second to hold a constant temperature. Keep in mind the average space heater runs on 1500 Watts, so 1035 is not far off. The correct calculation you are looking for is 1035W * Hours used (let's assume it's so cold you run it for the entire day, 24 hours) = 24,840 Watt Hours or 24.84 kWh, the average household in America uses around 30 kWh a day, which makes sense considering you are powering a heater for an entire day. If we assume something more reasonable like 16 hours of run time, the value becomes 16.56 kWh for that day which is more realistic. Remember, Watt is a unit of *power*, Kilowatt-hour is a unit of *energy*.
the first way should give you same answer, check your temperature. For example, q=-(.69)(5*6)*((5-20)/(.30)), there's a second negative when you do 5-20.
Area = A = (5 m * 6 m). I think there is confusion over the fact that my multiplication sign (*) looks like a plus sign in the video. If you compute the area A as 5*6 = 30 m2 you will get the correct answer and this will also resolve the earlier question about units.
I feel the area wasn't calculated properly. Bearing in mind we are dealing with a 3 dimensional object, having a thickness of 0.3m, length of 6m and height of 5m. The surface area from my point of view should be 2[(5*6)+(5*0.3)+(6*0.3)] = approx 46m^2 Therefore the heat transfer rate, q° = - 0. 69W/m°C * 46m^2 * (-50°C/m) = 1587W Doing the unit analysis, this works just fine.
@@tammyallison8735 the direction of the heat is 1-d in the assumption so you calculate the area which the heat will cross only and if you imagine that in real life, it is not true because heat transfer is randomly distributed ^_^
so easy to understand! thanks again!
Hello What if you add multiple layers with air in between
in real life situation, how do i calculate the thermal conductivity?
Hi professor! Do you recommend a book concerning heat transfer?
May I ask you something? If we assume that is the Area (30m^2) of the outside wall of one room. Then every second we need 1035 Watt to keep the room temperature constant. Then in one hour the required energy of one heater to keep the temperature constant will be 1035 times 3600 seconds which is 3726000 Watts or 3726kW. This is an extremely high number. I would appreciate if you explain me where am i wrong.
Where did you determine that 1035 Watts per second was necessary? That is not the calculation that is being done in this video. You may be confusing this with a J/s which is a Watt (not the same as a Watt/s). You would rarely ever need 1035 Watts per second to hold a constant temperature. Keep in mind the average space heater runs on 1500 Watts, so 1035 is not far off. The correct calculation you are looking for is 1035W * Hours used (let's assume it's so cold you run it for the entire day, 24 hours) = 24,840 Watt Hours or 24.84 kWh, the average household in America uses around 30 kWh a day, which makes sense considering you are powering a heater for an entire day. If we assume something more reasonable like 16 hours of run time, the value becomes 16.56 kWh for that day which is more realistic. Remember, Watt is a unit of *power*, Kilowatt-hour is a unit of *energy*.
I want to know about which book are you using . Can you tell me the name of book?
Could you provide an online certificate for passing through your courses?
Doing unit analysis I get W/m or J/ms.
k: W/(mK), A: m^2, dT/dx: K/m ... W/(mK) * m^2 * K/m = W.
20-5/0-0.3=50 so when i time that positive 50 with the rest include -k it will be - answer ? correct me
if you did that, it would be 15/(-.3), which would give you -50, and if you multiple with rest with -k you would still get a positive.
Should the temperatures be in Kelvin?
For temperature difference, it is doesn't matter which system you're dealing with.
i.e. T1 - T2 = (10-5)degC = (283-278)K
at any units you can calculate the conduction rate and so
I get - 1035 when I use the first way
the first way should give you same answer, check your temperature. For example, q=-(.69)(5*6)*((5-20)/(.30)), there's a second negative when you do 5-20.
how did the answer is 1035w? my answer is 379.5w can you explain why 1035w?
Area = A = (5 m * 6 m). I think there is confusion over the fact that my multiplication sign (*) looks like a plus sign in the video. If you compute the area A as 5*6 = 30 m2 you will get the correct answer and this will also resolve the earlier question about units.
exactly, since when did the area of a rectangle become = (L+B) xD
I feel the area wasn't calculated properly.
Bearing in mind we are dealing with a 3 dimensional object, having a thickness of 0.3m, length of 6m and height of 5m.
The surface area from my point of view should be 2[(5*6)+(5*0.3)+(6*0.3)] = approx 46m^2
Therefore the heat transfer rate,
q° = - 0. 69W/m°C * 46m^2 * (-50°C/m)
= 1587W
Doing the unit analysis, this works just fine.
@@tammyallison8735 the direction of the heat is 1-d in the assumption so you calculate the area which the heat will cross only and if you imagine that in real life, it is not true because heat transfer is randomly distributed ^_^
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