Measuring Power output of DC generator

  • #1
Deathz
26
1
Hello everyone,

I would like to ask some questions regarding my thesis. I have a problem identifying which load I should use to calculate the power output of the small generator.

I use an Archimedes screw turbine as the prime mover of my generator, and the water flow is low in my locality, which is why the maximum voltage it could produce is about 4.7 volts.

What I need to do is compare the output efficiency of the modified and non-modified generator.

The modified generator would produce about 4.7 volts, while the non-modified one would produce 2.5 volts.

I want to measure the current to differentiate the efficiency of the two.

I used a 12V and 1-watt load, but the problem I encountered is that it will only draw 0.036 amps, resulting in an output power of 0.036×4.7, which is very small, given that my mechanical input is about 128 watts, as per the p=ygh equation.

What specific load should I use to ensure at least one centiampere to achieve a bigger power output?
 
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  • #2
Welcome to PF.

First, we need to know the energy available. That comes from the height the water falls, and the flow volume per unit time.
 
  • #3
If a DC generator produces an unloaded open-circuit voltage, then it is probable that you should load the generator until the voltage falls to half that voltage.

What is your generator that turns so slowly and produces so little voltage?
Does it have a commutator, or is it an alternator with a rectifier?
 
  • #4
Baluncore said:
If a DC generator produces an unloaded open-circuit voltage, then it is probable that you should load the generator until the voltage falls to half that voltage.

What is your generator that turns so slowly and produces so little voltage?
Does it have a commutator, or is it an alternator with a rectifier?
It has a commutator.
 
  • #5
Baluncore said:
Welcome to PF.

First, we need to know the energy available. That comes from the height the water falls, and the flow volume per unit time.
The water us about 0.0259 m³/s
 
  • #6
I need to check your arithmetic and units. How high is the drop?
 
  • #7
I also have a problem in computing the efficiency. for example I use a load with a resistance of 150 ohm, I would get the power output and and input (using the p=ygQh or the power produce by the water). But when I increase the resistance of the load under the same flowrate, of course it will decrease the power output and if I use a the same power input since the flowrate is the same as the 150 ohm it will decrease the efficiency.

To understand better.
Input at flowrate of 0.03 m3/s = 70 watts
Load Resistance 150 ohm = 3 watts
Load resistance 220 ohm = 2 watts

It changes the efficiency if I will use the same input power of 70 watts.
 
  • #8
Baluncore said:
I need to check your arithmetic and units. How high is the drop?
The drop us 0.56 m
 
  • #9
You are confusing yourself more.

From how high does the water fall ?
Or what is the difference in height of the Archimedes screw axis, from water entry to the water exit?
 
  • #10
I think I might be wrong in calculating the efficiency. Could you suggest how to calculate the efficiency if the prime mover is an Archimedes turbine?
Baluncore said:
You are confusing yourself more.

From how high does the water fall ?
Or what is the difference in height of the Archimedes screw axis, from water entry to the water exit?
 
  • #11
Baluncore said:
You are confusing yourself more.

From how high does the water fall ?
Or what is the difference in height of the Archimedes screw axis, from water entry to the water exit?
Sorry. I am using a small Archimedes, and the height of the water fall is only 0.56 meter
 
  • #12
0.56 metre drop.
0.0259 m³/s = 25.9 kg/sec
g = 9.8 m/s2
Power = m*g*h = 25.9 * 9.8 * 0.56 = 142. watts.

What are you using as the DC generator ?
Make and model, or a link to the generator ?
 
  • #13
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Baluncore said:
0.56 metre drop.
0.0259 m³/s = 25.9 kg/sec
g = 9.8 m/s2
Power = m*g*h = 25.9 * 9.8 * 0.56 = 142. watts.

What are you using as the DC generator ?
Make and model, or a link to the generator ?
 
  • #14
How do you speed up the turbine shaft to turn that motor ?
Is it a chain drive ?
What is the ratio ?
 
  • #15
Baluncore said:
How do you speed up the turbine shaft to turn that motor ?
Is it a chain drive ?
What is the ratio ?
It is chain drive and 8 teeth in the motor, 32 teeth in the prime mover. Which gives, 0.25
 
  • #16
It is chain drive and 8 teeth in the motor, 32 teeth in the prime mover. Which gives, 0.25
 
  • #17
Deathz said:
The modified generator would produce about 4.7 volts, while the non-modified one would produce 2.5 volts.
What is the modification ?
 
  • #18
Hel
Baluncore said:
What is the modification ?
I change the motor's magnet into a neodymium magnet. I also changed the turns of the motor. The original turns is 30 and the diameter of wire is 25. I changed it into 36 and a diameter of 26.
 
  • #19
I tested it using a drill as a prime mover and the output current and voltage increased. My problem now is to answer the objectives of my thesis, which is to compare the efficiency of the two generators.
 
  • #20
When I had my Data gathering, I use a flashlight as a Load in order to get the current but the voltage output would only give as what I've said in the thread. That is why I'm asking the correct way of getting the power output by asking of the correct load to use or the correct computation of the power input.
 
  • #21
The LED of the flashlight gets busted when I manually rotate the turbine and the voltage reaches above 5V, which makes me think that I am using an incorrect load.

Honestly, the maximum voltage when I use a drill is 14V+ for the modified version and 10V+ for the unmodified one. Unfortunately, when using the turbine, the low flow rate of the water limits its output voltage to only 4.7V.
 
  • #22
I think it's okay to have a low volts since I am just going to compare the two. The problem is I don't know how to correctly get the input and output power in order to compare them
 
  • #23
Deathz said:
I change the motor's magnet into a neodymium magnet.
A stronger magnet will induce a greater voltage, but I wonder how well it fits mechanically. There may be a bigger air gap in the magnetic path that could limit the power.
Is there only one magnet ?
Did you check the pole pattern before fitting it ?
How similar were the dimensions?

Deathz said:
The original turns is 30 and the diameter of wire is 25. I changed it into 36 and a diameter of 26.
That should give an increase in voltage of 36/30 = 1.2 times.


The picture you posted shows an auger (blue) in a semi-cylindrical housing (black). There will be problems with friction, or the seal leakage, between the auger and the outer tube. That may be one cause of low efficiency.

An Archimedes screw could have the helix attached to the outer tube, so the outer tube then rotates with the helix, so there is no leakage, and no seal friction. An Archimedes screw can also be made from a spiral of pipe without any seals.
Google search, images; Archimedes screw
 
  • #24
There is an air gap, the magnet have the same size with the orginal magnet since I bought a customize one.
 
  • #25
The motor have 4 poles
 
  • #26
So far i have learned that in order

To calculate the power input, I will use the P=mgh, right?

I still don't know what load should I used with 4.7 volts opej circuit output
 
  • #27
Deathz said:
To calculate the power input, I will use the P=mgh, right?
Yes, but there is friction and leakage in the screw turbine.

Deathz said:
I still don't know what load should I used with 4.7 volts opej circuit output
For maximum power output, you will need to load the generator until the voltage falls to 2.35 volts. Another way is to measure the internal resistance of the motor when it is not turning, then load it with that resistance.
 
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  • #28
Baluncore said:
Yes, but there is friction and leakage in the screw turbine.


For maximum power output, you will need to load the generator until the voltage falls to 2.35 volts. Another way is to measure the internal resistance of the motor when it is not turning, then load it with that resistance.
Can I use a resistor as a load?
 
  • #29
But if I will use a higher resistance load to get the voltage into half, the turbine will stop rotating.
 
  • #30
Do you have a link, a book or a tutorial on how to get the efficiency of a generator having a hydro prime mover?
 
  • #31
Baluncore said:
Yes, but there is friction and leakage in the screw turbine.


For maximum power output, you will need to load the generator until the voltage falls to 2.35 volts. Another way is to measure the internal resistance of the motor when it is not turning, then load it with that resistance.
Okay, once the voltage decreases, the current also increases right? So it is important to have a constant RPM between different load?
 
  • #32
If I am going to compare the efficiency of the two? What should be the constant parameter sir? Is it okay if i will base on the flow rate?
 
  • #33
Deathz said:
Can I use a resistor as a load?
Yes. You will need to adjust the load, and it will need to dissipate the power generated.
Deathz said:
But if I will use a higher resistance load to get the voltage into half, the turbine will stop rotating.
It will slow down, but it will not stop.
If it stops, the voltage will be zero, not half.

If you want to compare the efficiency of the original and modified generators, you should run them at a specified RPM, from some standard motor, not from an Archimedes turbine with unknown flow, leakage and friction parameters.
 
  • #34
Baluncore said:
Yes. You will need to adjust the load, and it will need to dissipate the power generated.

It will slow down, but it will not stop.
If it stops, the voltage will be zero, not half.

If you want to compare the efficiency of the original and modified generators, you should run them at a specified RPM, from some standard motor, not from an Archimedes turbine with unknown flow, leakage and friction parameters.
If I'm not using Archimedes as a reference, how can I determine the power input based on the input RPM?
 
  • #35
Like for example, I will use a drill in order to control na Input RPM, how should I compute the power input?
 

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