Naked Science Forum
General Science => General Science => Topic started by: Happy Mark on 16/11/2012 20:37:19
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I think I am getting confused regarding the kinetic energy formula regarding mechanical energy.
I am trying to fine out how much power/work my little turbine will do. Do I need to add together the rotary KE and the PE to get the power. If my little turbine can continually lift a mass of 225grams to a height of 3 Mtr every second, I calculate that to be 6.64 watts/joules of power. the question is do I add the PE that it has lifted to that height as if it was to drop it would also be 6.64 joules of energy this would make it 13.28 joules of work done. I am sorry for asking this but I just get very confused with all the kinetic energy stuff.
Many thanks for any help
Happy Mark
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A few points here to clear up Mark
(1) when your little little turbine is lifting the mass it is putting in some work to get that mass to a height, and the mass is gaining some energy; when the mass expresses its increased potential energy in its return to the bottom, it is losing that potential energy. In other words, the two energy terms you are referring to have opposite sign, so the overall sum of both of them is zero. However, that does not mean that your little turbine is not expressing any power; all it really means is that the 225 gram mass moving up and down is a "middle man" that does not really enter into energy calculations (until frictional dissipation and efficiency come into play)
(2) You seem unsure about the terms "power" and "energy" and the units "watt" and "joule". To clarify things a bit (I hope) "power" is the "energy" produced/dissipated per unit time.
Watt is a power unit, while joule is an energy unit. The watt is exactly the same as a joule per second.
(3) It is unclear exactly what setup you have there, but if we suppose that your little turbine is operating an ideal winch that is reeling in a suspended mass on a frictionless rope or pulley towards a skyhook, then the power is given by the increase in potential energy per second, which will in turn be determined by the mass suspended and its vertical velocity. In other words both of the energy-type terms you mention are really just the same energy looked at in two different ways.
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I am trying to fine out how much power/work my little turbine will do. Do I need to add together the rotary KE and the PE to get the power.
I don't think the useful work done by an energy "generating" machine includes the overhead costs of moving its component parts, ( e.g. frictional forces at the turbine bearing ).
...In practice, however, some work is always wasted in overcoming friction and raising moving parts, and therefore the useful work done by a machine is always less than the work done by the effort.
http://scienceuniverse101.blogspot.co.uk/2012/03/work-done-by-machine-efficiency.html
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Hi Damocles and RD and many thanks for your replies.
Just to confirm a few of your questions, The turbine is lifting a weight vertical at a constant speed. I expect I am getting mixed up with work and power. I was looking at the power in watts that the turbine is generating or could be used.
I have attached a small generator to the turbine and it generates 6.5 watts but then I need to work out the efficiency of the generator, looking at the technical details of the generator I think it is about 50% efficient. I fully understand that there will be bearing and friction losses. I have looked at many different ways for calculating power. Such as Nm x radians/s = power watts . or .5 x I^2 x V^2 for power
I know where the moment is (I) and I know the rpm or rotary speed the turbine at the center of mass. I think the think that is confusing me is where does the Half in the EK calculation from. Looking at this another way if the little generator was 50% efficient the turbine would need to generate 13 watts of power to turn it.
It all comes down to the drag coefficient of the turbine blade or lift coefficient that I am playing with.
Many thanks, I hope that my answers may be of help so I can confirm the correct way in calculating the turbines power.
Happy Mark
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... I have attached a small generator to the turbine and it generates 6.5 watts but then I need to work out the efficiency of the generator, looking at the technical details of the generator I think it is about 50% efficient ...
To work out the efficiency you'd have to know the wind power input , which depends on the wind speed (see attached), then compare that with the useful work output, ( lifting your weight).
https://www.google.com/search?q=home+made+anemometer
http://www.ocgi.okstate.edu/owpi/EducOutreach/Library/Lesson1_windenergycalc.pdf
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Hi RD and many thanks for the reply.
Yes I know the wind speed and the power in the wind at the wind speed, I am also aware of the Betz 59% limit.But what I have read under kinetic and potential energy of motion is.
KE stands for kinetic energy and I under stand that. (Note that a change in the velocity will have a much greater effect on the amount of kinetic energy because that term is squared. The total amount of mechanical energy in a system is the sum of both potential and kinetic energy, also measured in Joules (J).
Total Mechanical Energy = Potential Energy + Kinetic Energy If I know the kinetic energy as rotation where is the potential energy is it the mass you have lifted waiting to fall.
Many thanks
Happy Mark
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Hi RD and many thanks for the reply.
Yes I know the wind speed and the power in the wind at the wind speed, I am also aware of the Betz 59% limit.But what I have read under kinetic and potential energy of motion is.
KE stands for kinetic energy and I under stand that. (Note that a change in the velocity will have a much greater effect on the amount of kinetic energy because that term is squared. The total amount of mechanical energy in a system is the sum of both potential and kinetic energy, also measured in Joules (J).
Total Mechanical Energy = Potential Energy + Kinetic Energy If I know the kinetic energy as rotation where is the potential energy is it the mass you have lifted waiting to fall.
Many thanks
Happy Mark
In this particular instance there is no potential energy involved -- you are only "using" the kinetic energy of the air motion to drive your turbine. If you were to speed up your turbine wirh a heavy rock lowered from a windlass driving the axle, then you would have to add the kinetic and potential terms, but there is no "exploited" potential energy in this setup
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Hi Damocles
Sorry for the delay in replying but I have been away for the week. I have been doing a lot of reading on the subject and as you have said and explained it is correct.
I think my problem is on turning the generator and the power I get out of it. I purchased another wind turbine that has the same sweep area as mine it was a three bladed drag VAWT turbine. They claim a 20 % efficiency for this turbine. if you get the turbine to do some work such as lifting a weight it would not be anything like it ,maybe 2% but when you connected it up to a 12v battery. This is what is my problem it does not do any work until it gets up to the battery voltage, the battery voltage goes to 13v with .5 amp going into the battery. this equals 6.5watts now the big question is this correct or should it be 1v gain times.5 amps for it rated power .5 watts. now Betz theory states 59.3% is the max power in the wind. is it for work done such as lifting a weight or power into a battery. I have tried many turbines and they are all the same. They all count the total voltage and amps for their rating not the voltage gain times amps. what should be the correct method for the power of a turbine. I would like to know this before I put my design in the public domain
Many thanks for your time,
Happy Mark
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Energy is "lost" (usually via heat) whenever energy is converted from one type to another...
converting motion into electricity (the dynamo)
converting electricity into chemical energy (charging the battery)
converting chemical energy into electrical energy (discharging the battery)
...They all count the total voltage and amps for their rating not the voltage gain times amps. what should be the correct method for the power of a turbine.
The "voltage gain" would depend on the charge state of the battery.
A fair comparison between models of turbine would be comparing them under the same typical electrical load,
e.g. have the turbines each power an identical light-bulb (~10W 12volt), whilst monitoring the windspeed , voltage and current. The voltage across the output terminals of the turbine, multiplied by the current anywhere in the light-bulb circuit will give you the electrical power output of the turbine at that wind-speed.
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Hi RD thanks for the reply.
I have tried that with both turbines, if I use a 12v 6w bulb, when you put the other turbine on it it will nearly stall the turbine. I have been told this is the residence of the bulb. hence he turbine turns very slowly and hardly any voltage is generated so you might only get 3v into the bulb @.5 amps = 1.5w. if I run the bulb off the battery and connect the turbine to the battery I get 12v into the battery @.5 amps 6w.
So how do they calculate the efficiency of a wind turbine. firstly it can't be by lifting a weight as they can't lift the weight at the speed continuously. you cant run a bulb or a load directly from a turbine. You could charge a battery due to it's low internal residence especially when the battery has little charge in it.
So me thinking how do they raty a turbine for power generation ? my turbine will do far more work than the other one i have tested it against. I would be only too happy to but my turbine in a wind tunnel facility and prove it.
Happy Mark
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To obtain the maximum power transfer, the load resistance (bulb/battery) must match the resistance of the generator (turbine)... http://en.wikipedia.org/wiki/Maximum_power_theorem
Presumably in this case that would be the internal resistance of the battery the turbine is designed to charge.
However the internal resistance of a real battery is variable ...
Internal resistance of a battery is dependent on the specific battery's size, chemical properties, age, temperature and the discharge current. Measurement of the internal resistance of a battery is a guide to its condition
http://en.wikipedia.org/wiki/Internal_resistance
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The maximum power theorem suggests that you select the optimum balance between current and voltage from the generator:
- If the load resistance is too low, the current will be high, but the voltage into the load will be low, resulting in low power output. (Plus, for high-power wind turbines in the Megawatt range, this means you will be dissipating up to a Megawatt inside the generator, which is likely to overheat and burn out.)
- If the load resistance is too high, the voltage will be high, but the current into the load will be low, resulting in less than the theoretical maximum power output. (However, the power dissipated inside the generator will be lower, leading to improved electrical efficiency, reduced temperatures and longer lifetime, so this is the way they are normally operated.)
- In practice, the output voltage (and frequency) of the generator varies with wind speed, and this must feed into a grid with approximately constant voltage and frequency. This adaptation is sometimes done with an inverter circuit, which can take whatever voltage is produced by the turbine/generator, and produce the right output voltage and current to feed the maximum amount of power into the power grid, without overheating.
- A simpler circuit would have a variable transformer, which can dynamically adjust the output voltage, but not change the frequency.
Similarly, the maximum power theorem suggests that you select the optimum balance between speed and torque exerted by the generator on the turbine shaft:
- If the generator's torque is too low, the turbine speed will be high, but the power into the load will be low. (Plus, in high wind conditions, "centrifugal force" could tear the blades apart.)
- If the generator's torque is too high, the turbine will stall, and will generate no output power.
- In practice, the optimum torque and speed of the turbine varies with wind speed. If it is to feed into a constant-frequency grid without an inverter, then the turbine must rotate at a fixed speed; the speed may be adjusted by adjusting the blade angle.
Generating maximum output power involves dynamically balancing several variables, usually under computer control.
This level of sophistication is not available on a fixed-blade turbine feeding into a generator with no inverter. All you can do is produce a fixed wind speed (eg with a domestic fan, placed in different positions), and measure the generator output power into different electrical resistance loads.