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To cool a steel bar with water.

  1. Apr 3, 2009 #1

    Suppose an infinitely long round bar made of steel of 25 mm diameter, which is traveling at a speed of 10 m/sec. The bar is at a temperature of 950 degrees Celsius. I need to find out how much water (liters per minute or liters per second) do I need to put in contact with the bar in order to cool its surface to 900 degrees Celsius.

    I would like to know about formula(s) in order to have a theoretical approach to the solution of this problem.

    Thank you.
  2. jcsd
  3. Apr 3, 2009 #2
    Is this homework?
  4. Apr 3, 2009 #3
    No, it isn´t homework.


  5. Apr 4, 2009 #4
    OK. In one second, the steel bar moves 10 meters or 1000 cm. The diameter is 2.5 cm, so 2.5 cm dia x 1000 cm is 4909 cubic cm, which has a mass of about 38,780 grams. How much water (liters) would it take to cool this mass frrom 950 deg C to 900 deg C?
  6. Apr 4, 2009 #5
    If the bar is infinitely long, you will need an infinite amount of water to cool it. I guess you'll need 50 times that infinity to cool it 50 degrees.

    And if it is infinite, how can you tell that it is moving?

    Finally, if it is infinite, this isnt a 'real' problem to be solved (hence the homework inference above).
  7. Apr 5, 2009 #6
    The flow is infinitely large, that´s true. But not the flow rate. At any given instant of time, if I apply a sufficient amount of water (water at ambient temperature, for example 20 °C) I would be able to decrease the surface temperature of the bar 50 °C. And that´s the solution I´m looking for.

    Of course, the real life problem is not about an infinitely long bar; it´s about bars that move along a continuous rolling mill train, a steel mill for rolling hot steel bars. At the point were the bar is of 25 mm diameter, it´s lengh is more than 200 meters.
  8. Apr 5, 2009 #7
    To get efficient cooling, you will need a long counterflow heat exchanger design, with cold water entering at the downstram end and flowing upstream for several (maybe even 10) meters. You need a high Reynold's number to get good turbulence and mixing in the water.
  9. Apr 5, 2009 #8
    You need to figure out the mass of steel you want to cool, per second. To do that, from the geometry of the bar and the speed its going figure out how many kg per second passes by a given point. Then multiply that times the heat capacity of the material and times the change in temp (50 C) to get the energy that must be removed per second. Then figure out how much water you have to boil away to take away that energy. You do that by checking the enthalpy difference between the water temp (you said 20 C) and the steam enthalpy.

    Sorry, I don't think in metric units but when I did this just now I get about 20 gallons (US) per minute. I think that's around 1.25 liter/second.

    That's alot of steam being made, you will need to do something about that.
  10. Apr 6, 2009 #9
    Can you please help me a little more by showing me the formulas that were applied here to get the 20 gallons per minute result?
  11. Apr 6, 2009 #10
    Using my posts #s 4 and 7, You want to cool about 38.8 Kg of steel from 950 deg C to 900 deg C per second. You cannot cool the surface only (unless you want to hard temper just the surface) because of the high thermal conductivity to the center of the rod. So with a specific heat of 0.12 kcal per Kg per deg C for iron, and a say 50 deg C rise for water with a specific heat capacity of 1.0 kcal per Kg per deg. C, you will need (0.12/1) x 38.8 kG = 4.6 Kg (4.6 liters) of water per second. If you want to heat the water from 20 deg C to 900 deg C with a counterflow heat exchanger, it will be a lot less because the heat of vaporization of water is much higher (540 kcal per kg at 100 deg C). Designing good seals for the heat exchanger might be a problem, so you have to use a low pressure water loop. I don't have enthalpy tables handy.
    If you go from 20 deg C to steam at 100 deg C, that is 80 plus 540 = 620 kcal per Kg, rather than the 50 kcal (50 deg C rise) I used above, then the water use is (50/620) times 4.6 liters/sec = 0.37 liters per sec, or 22.2 liters per minute (about 6 gallons per minute).
    Last edited: Apr 6, 2009
  12. Apr 6, 2009 #11
    Bob S,

    OK..Thank you.
  13. Apr 7, 2009 #12
    I can't find my scribbling to come up with 20 gpm, so I just re-did it and I came up with 7 gpm (for the case where you are simply boiling the water). I take that as a verification of Bob S's 6 gpm result. There must have been a mistake in my 20 gpm calc.

    It is still a lot of steam, do they really do it this way in a production facility? You'd need big hoods and a ventilation system to do something with all that steam.
  14. Apr 8, 2009 #13
    There is steam generation; but I don´t think it´s very abundant. I will get a picture for you tomorrow, so you can see this. Meanwhile, take a look at the one I´m attaching. Here you see the conduct and the holes for the water.

    Attached Files:

  15. Apr 9, 2009 #14
    Here are the pictures showing the device in operation.

    Attached Files:

  16. Apr 9, 2009 #15
    thanks for sharing the photos, capterdi
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