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Calculating Energy Loss (Kinetic Energy) and Output

  1. Apr 7, 2013 #1
    1. The problem statement, all variables and given/known data

    Hey everyone, trying to finish up a report and having a problem with getting to the final answer. It is a final report on hydro kinetic energy. We are planning to install a hydro kinetic turbine under a bridge and calculate what the energy output would be in KW. I am getting lost in the conversions and hoping someone can take a look at my math and tell me where there could be potential errors.


    2. Relevant equations

    Our Velocity is 1.65m/sec
    Flow rate past the turbine is 20.75m3/s

    I have calculated the flow rate loss past the turbine and plotted a graph from 50% to 3%

    V2 = V1x
    = x m/s


    The next step we took was to find Hr = (V1^2 - V2^2)/2g

    With Hr, we then solved for Pa = Hr*p*Q

    And finally, we are now at the step of our efficiency (35%) = Pout/Pa

    The issue is talking with our teacher we are getting some rather large numbers and it doesn't seem correct.
     
    Last edited: Apr 7, 2013
  2. jcsd
  3. Apr 7, 2013 #2
    Removed to clean this thread up (see the exel sheet below)
     
    Last edited: Apr 7, 2013
  4. Apr 7, 2013 #3
    attached is my excel sheet. That didn't print very well.

    I am not sure about all my units for all of these. Let me know if you guys can help me out.

    Thanks.
     

    Attached Files:

    Last edited: Apr 7, 2013
  5. Apr 7, 2013 #4

    haruspex

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    If you are getting unbelievable numbers, it is not going to depend on the gory details of the data. To make it easier on those reading your post, how about calculating a much simplified version. If you still get a surprising answer, post that.
     
  6. Apr 7, 2013 #5
    Perfect. I appreciate your response. I will try to make this explanation as near winded as possible but give as much info as possible on what we are doing.

    We are working on a final project for our civil engineering course. We decided to do our project on kinetic energy specifically kinetic fans in rivers. We chose a location with the following characteristics:

    Depth of water - ~5m
    Velocity of the water - ~1.65m/s

    Because we are not mechanical engineers we made assumptions as we are not designing a fan. More coming up with the expected results of if the turbine had certain characteristics.

    Assumptions we made (please let me know if some of these are out in left field). Data was collected from the internet which can be dodgy at times.

    Size of turbine chosen (4m diam)
    Efficiency - 35%

    In talking with our professor, he gave us a basic direction.

    First we figured expected loss at the turbine (Velocity):

    V2 = V1x

    *** EDIT*** At 5% loss, this will take our velocity after the kinetic fan to 1.57m/s (for review)


    As shown in exel, we did this from a expected loss of 50% to 3% to plot the chart.

    Our professor told us the next step we were missing was the following equation.

    Hr = (v1^2-v2^2) / 2g

    *** Edit**** Hr = (1.65^2 - 1.57^2) / (2(9.81))
    = .014

    **First question - What is Hr represent?***

    Next step is using the formula:

    PA = Hr*ρ*Q - where the flow being 20.75m3/s {12.56m2 * 1.65m/s}
    = .014 * 1000kg/m3 * 20.75m3/s
    = 280


    ***Second question - Our numbers we are getting are ending up high and I am unsure what this is suppose to be, flow in?***


    The next step is taking the efficiency of the pump (35%) and using the equation:

    Em = Pout / PA
    .35 = Pout / 280

    Pout = 98.26

    Knowing the Em (35%) and the PA from the above, we can solve for Pout. But I am unsure what to do from here. Is this my energy that we have? If so, what does this come out in? Joules?

    I guess the issue where we are coming up with, is I am not strong in physics and units to Joules/Watts etc. I am getting lost in the calculations.
     
    Last edited: Apr 7, 2013
  7. Apr 7, 2013 #6

    haruspex

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    That's certainly an improvement, but you use a lot of variables without defining them. E.g. what is PA?
    Hr appears to be some sort of "head equivalent", i.e. it's the height through which the water would drop to release the same amount of energy. No idea why it's interesting. Its units would be m, so PA has units kg m/s (momentum). I thought you'd be wanting power, kg m2/s3. Are you missing a factor g?
    Can you provide an online reference for these equations?
     
  8. Apr 8, 2013 #7
    I am trying to figure out what PA is. I am assuming that is Power in?

    I am wanting power, I am unsure what i need for units to get power.

    My teacher gave us some of these formula's so I cannot reference these unfortunately. They come from a hydraulics teacher.
     
  9. Apr 8, 2013 #8
    How would I go about getting what I am trying to acheive? is there another formula(s) I can use?

    There is another formula that I found on the internet, but I am unsure if it is what I am after.

    Can you confirm?
    Power = (1/2)(Turbine Efficiency) (Water Density) (Turbine Sweep Area) (Water Velocity)^3

    EDIT:

    https://www.physicsforums.com/showthread.php?t=451683&highlight=kinetic+energy+turbine

    I found this thread which is similar to what I am trying to do. The difference being I do not have any head.

    I have horizontal flow with an assumed 35% efficiency rating on my hydro-kinetic fan.
     
    Last edited: Apr 8, 2013
  10. Apr 8, 2013 #9

    haruspex

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    If PA is power then there's definitely something wrong with this pair of equations:
    Hr = (v1^2-v2^2) / 2g
    PA = Hr*ρ*Q​
    You shouldn't be losing that g. If we dispense with Hr and put the g back in we get:
    PA = (v1^2-v2^2) *ρ*Q/2
    which makes sense. It says the power = (energy yield/unit mass) * density * volumetric flow rate.
    That also makes sense, but I think the efficiency means something a bit different here. In your original approach you used the drop in water speed, whereas the above equation seems to be using the absolute speed, so I suspect the efficiency factor has to take into account that the water will not be completely halted. It may be a lot less than your 35%.

    The 'head-based' calculation doesn't have this problem. All of the lost PE can be assumed available. What you could do is use the Hr from your original calculation and plug it into a head-based calculation. That's equivalent to writing PA = Hr*ρ*Q*g instead of PA = Hr*ρ*Q.

    So, back to your original question. Having restored the lost g, you're going to get a numerical answer about 10 times what you had before, and you already thought that was too much. Having been consistent in using SI units (MKS, specifically), the answer is in Watts. You had Pout = 98.26, so that will become about 960W, or a bit under 1kW. Doesn't seem excessive to me.
     
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