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Heat transfer help!

  1. Aug 16, 2013 #1
    I am not really a heat transfer guy so I hope someone here can help me out here. I would like to know the formula to get this.

    Here is what I ma facing:

    I have a mandrel (basically a piece of cylindrical H-11 steel). It has OD of 3.5 inches and 37” in length. It has a core hole of about 7/8” Dia and 37 “ length. Water pumped into the core hole at 10 GPM thru a 3/4” dia pipe. The H-11 steel has .110 BTU/lb specific heat capacity.
    I want to be able to lower the temp of the mandrel from 850 to 200 deg F. How much water do I need?

    Please help.
     
  2. jcsd
  3. Aug 16, 2013 #2
    Well from a physics point of view, you've given us a lot of information we don't need and haven't given us all the information we do need! One thing we would need is: what temperature does the water start at? (We'd hope it's less than 200F!)

    I guess we could treat the (pretty unrealistic) problem of having the mandrel at 850F come in contact with a volume V of water at some starting temperature Ti. The relevant equation that you can solve this problem with is
    C=dQ/dT, where C is the specific heat capacity, Q is the heat transferred, and T is the temperature. You will need to integrate this expression, using the heat capacity of the mandrel, to find the decrease in heat Q needed to lower the temperature to 200F. Then you will need to use the heat capacity of water and use the same expression to figure out what volume V of water would change temperature from Ti to 200F if the heat transfer is the Q you calculated for the mandrel cooling.

    But it's unrealistic because you don't have unlimited time to let the temperatures equillibrate, and you will probably have a running reserve of water. Moreover if the heat is being generated continuously or you have some time constraint on how quickly it must cool, then you will need information about the thermal conductivity.
     
  4. Aug 16, 2013 #3
    The water inlet temp is 70 deg. Outlet unknown.

    I calculate the heat loss of the mandrel using Q = m x specific heat x temp change
    m is about 94 lbs..so 94 x .11 x -650 = -6721 BTU. The water is constantly flowing
     
  5. Aug 16, 2013 #4
    If the water flows constantly, then it really doesn't make much sense to talk about "how much water" unless you have it in some sort of closed recirculating system. If it is just coming from, say, the faucet, then you pretty much have unlimited water and this reserve could cool any finite system to exactly 70F provided there is no continuing heat generation.

    It seems like the problem has more to do with the rate of heat transfer rather than the magnitude of heat transfer. In that case you will have to use other equations like the Heat equation, Fourier's Law, etc. Maybe this will help you: http://en.wikipedia.org/wiki/Heat_conductivity#Resistance. But the method I suggested in my last post relies on "equilibrium" which would only hold when the temperatures have all equilibrated and there is no further transfer of heat. Things are more complicated out of equilibrium.
     
    Last edited: Aug 16, 2013
  6. Aug 16, 2013 #5
    It is in a closed circulating system. The water come from a tank and the water flow thru the mandrel. I am looking for the min amount of water needed in this tank to cool the mandrel down.
     
  7. Aug 16, 2013 #6
    To be more clear, the water from the tank circulate in a closed pipe system into the mandrel and go back to the tank. The same water keep circulating in this closed system.
     
  8. Aug 16, 2013 #7
    Take a look at the link I sent you. It has an example about how to apply the concept of "thermal resistance" to solve a problem which is almost identical to yours. You will need to get some more information about your mandrel and the water cooling system to be able to calculate their thermal resistances, but it looks like once you have those the answer pops right out of the equation in that "Resistance" section on the wiki page.
     
  9. Aug 16, 2013 #8
    So do i used this formula per what u said Q of mandrel = 94 lbs x 0.110 btu/lb x 650 = 6721
    then use water = lbs x 1 btu/lb x (200-70) = 6721
    and give me lbs of water = 51.7 lbs = 6 gallons of water?

    that sounds really low to me...
     
  10. Aug 16, 2013 #9
    I read it but I am not sure how I would figure out volumn of water from that...please explain
     
  11. Aug 16, 2013 #10
    Well I explained that what I said above would give the answer for a specific problem that I'm not convinced has any relevance to your actual problem. So I wouldn't be surprised if the answer you get using that method is irrelevant. The problem that my method applies to is bringing the mandrel into contact with the water and then letting them equillibrate.

    But if we were doing that problem, It would not surprise me that you get 51.7 pounds of water since you are assuming that water has about 9 times the specific heat of the mandrel.

    Well in the calculation you just showed me, you figured out the heat capacity of mass M of water by multiplying M with the specific heat capacity expressed in units of btu/(mass * degree kelvin). (BTU's? Ugh!) The specific heat capacity is a material property you need to look up. You should be able to do a similar algebraic maneuver to get the thermal resistance in terms of a volume V and a specific thermal resistance.
     
    Last edited: Aug 16, 2013
  12. Aug 16, 2013 #11
    This is an interesting heat transfer problem, but it has some complexities to it. The mandrel temperature is way above the boiling point of water (I'm assuming the pressure in the system is on the order of 1 atm.) Therefore, there is going to be some water vaporization (boiling) near the wall, and maybe further in. So there is some aspect to this involving two phase flow in the core hole, and the presence of water vapor near the wall can also slow down the rate of heat transfer. Another complexity is that the temperature is going to be varying with both time and spatial position within the mandrel as well as within the water in the core.

    Solving a problem like this all at once by writing and numerically solving the complicated PDEs for the system can be difficult and time consuming. It is much better to simplify as much as possible, breaking it down into bite sized chunks, and bounding the answer (i.e., finding the minimum possible time to cool the mandrel, and the minimum possible amount of water needed). This might give you all you really need to understand your system quantitatively.

    First of all let's take a look at the water in the core. Calculate the volume of the core in ft3. What is the volumetric throughput rate in ft3/min. From these two results, you can calculate the mean residence time of the water passing through the core (sec). What is that value? From the volumetric throughput rate, you can also calculate the mass flow rate of water through the core. What is that value in lb/hr? Do you think that, with that short a residence time and high a throughput rate, the average temperature of the water is going to increase much in passing through the core? To get a quantitative idea of how much the average water temperature changes in passing through the core, you need to have a rough idea of the heat transfer coefficient at the outer diameter of the core. Are you familiar with the term "heat transfer coefficient" and its definition?

    After you answer these questions, I will help you further by discussing some reasonable approximations to bound the answer, and will show you how to set up the differential equations to describe the water temperature as a function of location through the core tube, the heat flux at the outer diameter between the water and the mandrel, and the temperature vs time of the mandrel.

    Chet
     
  13. Aug 16, 2013 #12

    Integral

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    What is happening to the outside of your cylinder? Can you cool it? A fan of cold air could remove a lot of heat.
     
  14. Aug 16, 2013 #13

    SteamKing

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    1 U.S. gallon = 231 cu. in.
     
  15. Aug 16, 2013 #14
    To solve this problem you need to divide it into two primary cases:

    1)Internal convection via the flowing water

    2)Conduction through the pipe


    If you have a copy of "Fundamentals of Heat and Mass Transfer" (Incorpera DeWitt Bergman Lavine), you need to check out chapter 8 section 3 part 2 (8.3.2). This deals with the case of internal convection with a constant heat flux boundary condition.

    In your problem the easiest way to solve it would be to determine the heat flux at the inner surface. This is your first boundary condition. The second will be the temperature of 200 degrees at the mandrel surface. This may give you enough information to solve a polar form of the conduction equation.

    From Incopera, etc here are the equations you will need:

    [itex]q_{convection} = m'* (T_{mo} - T_{mi})[/itex]

    [itex]T_{m}(x) = T_{mi} + q_{s}''xP /(m' * c_p) [/itex]

    the subscript mo denotes mean flow temperature at the exit
    the subscript mi denotes mean flow temperature at the inlet
    q''s is the surface flux, assumed a constant in the above equation
    x is the distance along the mandrel
    P is the pressure drop along the mandrel

    This should be enough to get you preliminary answers
     
  16. Aug 17, 2013 #15
    We can bound the answer (to get the shortest amount of time needed to cool the mandrel) by considering the case where the inner surface of the mandrel (outer surface of the core) is assumed to be at the inlet water temperature of 70 F for all times. The actual amount of time for the cooling to take place will have to be longer than this. This is a straightforward transient heat conduction problem that is undoubtedly solved in Carslaw and Jaeger, Conduction of Heat in Solids. I don't have this analysis in front of me, but I do have the solution for a solid cylinder that is cooled from the outside. This will give an even shorter time for the cooling to take place, because there is much more surface area on the outside than in the mandrel. So it will give a still lower bound to the cooling time. At long times, the solution to cooling a cylinder from the outside approaches:

    [tex]\frac{(T-T_w)}{(T_0-T_w)}=\exp(-4(5.772)\frac{αt}{D^2})[/tex]

    where T is the average temperature of the cylinder at time t, Tw is the temperature at the cooling surface (70 F), T0 is the initial temperature of the cylinder (800 F), α is the thermal diffusivity of the steel (0.0052 in2/s), and D is the diameter of the cylinder (3.5 in). If we plug numbers into this relationship, we obtain t = 176 sec, or about 3 minutes (for a final mandrel average temperature of 200 F). Now this is the absolute minimum amount of time. The actual amount of time is going to be higher than this for the reasons that we have discussed above. I suggest you look up the solution in Carslaw and Jaeger for conductive cooling of mandrel of annular cross section cooled from the inside to pin down the answer more accurately. But, realistically, the answer is going to be on the order of 10s of minutes.

    There is also going to be some heat transfer resistance on the outside that needs to be considered. And the water temperature is going to increase a little as the water passes through the mandrel. For the situation above, we can estimate the average temperature rise of the water by dividing the total amount of heat removed by the mass of water flowing through the mandrel during the 3 minutes, times the heat capacity of water. The mandrel weights about 100 lb, so the heat removed will be about 100(0.11)(600)=6600 BTU. During the 3 minutes, 30 gallons of water flows, or about 250 lb. Since the heat capacity of water is 1 BTU/lb-F, the average temperature rise of the water through the mandrel in this very limiting case would be about 26 F. But, in practice, average the heat transfer rate will be much less than this, and the average temperature rise of the water will be much less (although more water will be required).
     
  17. Aug 18, 2013 #16
    This is follow up to my previous post. I re-solved this problem for the case originally specified in the problem statement, in which the cooling takes place at the surface of the inner core, rather than at the outer surface of the cylinder. As expected, I found that the cooling takes place more slowly, and the amount of water required is greater. In particular,

    1. The mandrel cooling requires about 5 times as long (~15 minutes)
    2. The amount of water required is about 5 times as much (~150 gallons)
    3. The average temperature rise of the water is about 5 times smaller (~5 F)
    I hope this is helpful.

    Chet
     
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