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Max power plant output w/o killing river smelt?

  1. Jan 18, 2014 #1
    1. The problem statement, all variables and given/known data

    My simplified version of the equation:

    Power plant proposes output power increase from 188MW to 535 MW

    Excess heat is dumped into river with smelt. Smelt can withstand a 3.44C temp increase

    River = 222ft wide at dump site. Do not know depth or cross section here

    Further down the river, there is a concrete area that is 88.7ft wide and 36.2ft deep. The flow rate here is 10.4mph (4.649216 m/s)

    The planned efficiency of the plant is 37.30%.

    What is the max power output w/o endangering the smelt?

    ------------------------------------
    The Actual question:

    A battle has erupted over a local power plant owner\'s decision to raise the power delivered by the plant from 188 MW to 535 MW (The delivered power accounts for energy losses). The local population is increasing, so the area needs more power. However, the plant expends its heated waste water* into the Frustration River; local citizens are concerned that the heated water will raise water temperatures to the point where the endangered river smelt will die from oxygen depletion. Local biologists called into the fray report that the smelt can only withstand an average increase of 3.44°C. The river is 222 ft wide at the point where the waste water is dumped, but the depth and cross section is not well known. Farther downstream, however, the water flows through a narrow man-made concrete channel of rectangular cross-section, which has a width of 88.7 ft and depth of 36.2 ft. The water flows through this channel at 10.4 mph. The power plant claims a planned efficiency of 37.30% for its new power plant design. Based on this information, what is the maximum power that the power plant can output without endangering the river smelt?


    2. Relevant equations


    Related equations:
    Q= mc delta T
    C = 4186 J/ kg* K
    Work = F*d -- Power = work/time -- power = (f*d)/time -- Force = m*g -- P=m*g*Velocity -- Density=m/Volume - - mass = density/volume - - Power = (density/volume) * gravity * velocity - -Density of water = 1000 kg/m^3 - - Watt = Joule/second - - Joule = (kg m^2)/ s^2
    Efficiency of any engine = 1 – (Qc/Qh ) … Efficiency of carnot = 1 – (Tc/Th) - -celsius to Kelvin = C + 273.15

    3. The attempt at a solution

    my attempt at this problem was first finding all those related equations. I don't know how to piece the equations together to get to where I need to be.

    The problem gives you a width at the dump site and then the area of a rectangular cross section at a place farther down the river? Why do I care about what goes on at a different spot when the width isn't the same?

    Random guess:

    First use the 3 measurements to find volume. 222 * 88.7*36.2 = 7.1282868 x10^5 ft^3. converting this to SI units (m^3) we multiply this by 0.02831684659 to get 20185.06 m^3

    mass = density / volume

    Using Q = mcΔT I would guess to do Q = [itex]\frac{density}{volume}[/itex]*4186 * 3.44 +273.15 =>> Q= (1000/20185.06) * 4186 * 276.59 = 57359.538 J

    or: if using P = (density/volume) * gravity * velocity: P=.04954 * 9.8 * 0.44704 = 0.217 watts (really low...)

    For a cyclic process heat engine: efficiency = work/heat input. Solving for heat input in joules we would do Qh = work/efficiency or Qh = 0.217/0.373 = 0.58188 joules
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 18, 2014 #2

    TSny

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    Hello, Sorbik. Welcome to PF.

    Consider the region (or volume) of the river that lies between the cross section at the power plant and the cross section at the man-made channel. For steady flow of the river, the amount of water in this region does not change over time.

    So, if X m3 of water flow through the channel each second, how many m3 flow past the power plant each second?
     
    Last edited: Jan 18, 2014
  4. Jan 18, 2014 #3
    Thanks for the welcome. Help is scarce with online classes.

    Here's my work so far. the power I got at the end is pretty big... the original problem was only thinking about boosting power from 188MW to 535MW whereas I got 1.6MMW

    I'm still not sure how to apply the 222 ft portion of the problem. There's a width of the dump site and a width of the river at man made channel. I don't think it's 222*88.7*36.2 to get volume in ft^3 since it's not width*length*depth, the values are width*width*depth
     
    Last edited: Jan 18, 2014
  5. Jan 18, 2014 #4

    TSny

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    That looks good, except what does ΔT represent in Q = mcΔT?

    Of course, you still need to deal with the efficiency factor.
     
  6. Jan 18, 2014 #5
    delta T is max increase in temp the river smelt can survive which is 3.44C. This is converted to Kelvin to cancel with the Kelvin in the specific heat value of 4186 J/kg K

    For the efficiency, I was going to use 0.373 = 1 - Qc/Qh where Qc is the value I get from Q=mcΔT since it is the heat being dumped into the river
     
  7. Jan 18, 2014 #6

    TSny

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    Can you show that a temperature change on the Celsius scale is the same as the temperature change on the Kelvin scale?

    Your expressions for efficiency look good. Can you write Qh in terms of Qc and the useful energy output, or work W?
     
  8. Jan 18, 2014 #7
    Got the change part. 3.44 change in C is also 3.44 change in K

    I'm still getting some massive number for Qh though in comparison to the initial power range in the problem. Also not sure if that initial 222ft width specification means anything.

    QH is max plant output before the fishes die.
     
    Last edited: Jan 18, 2014
  9. Jan 18, 2014 #8

    SteamKing

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    I think you are getting hung up using the definition of thermal efficiency in your calculations.

    Mechanical efficiency is defined as the ratio of actual work obtained to the maximum amount of work which could be obtained from a perfect system with no losses.

    Let's say a theoretical power plant which has no losses could put out 1000 KW. A similar plant with an efficiency of 37.3% can output only 0.373*1000 KW = 373 KW. The extra 1000 - 373 = 627 KW is lost as heat.
     
  10. Jan 18, 2014 #9
    So, 31,851.651MW * 0.373 = 11880.66582MW is the power output. That's still like... 20 times the amount the problem is expecting with the 535MW desired power output.
     
    Last edited: Jan 18, 2014
  11. Jan 18, 2014 #10

    TSny

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    Your calculation of Qh looks correct to me. But note that Qh is not the power output.

    But I think you will get a very large answer for the power output. It appears that the answer will be much larger than 535 MW.

    Edit: [OK, I see you have now calculated the power output. Looks good to me (except you might want to round off to an appropriate number of significant figures).]
     
  12. Jan 18, 2014 #11
    awesome!
     
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