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Time to Freeze Standing Water in Steel Pipe

  1. Dec 3, 2008 #1
    I'm at little rusty on my heat transfer and could use some help.

    I'm trying to calculate the approximate time to freeze standing water in a 18inch steel pipe. I have some parameters and made some assumptions and they are:

    The pipe is 18inch carbon steel
    The standing water is initially around 68oF
    The outside temp is around 0oF
    There is a constant breeze around 5mph
    No insulation around pipe
    The pipe is exsposed to the air and not buried
    The pipe is completely filled with water

    I've calculated this two different ways and came up with two completely different time values. One was around 8.7hrs and the other was around 4.8hrs.

    For the first method, I calculated the surface heat transfer coefficient (h) at the wind/pipe interface by:

    First calculating the Reynolds number (Re) of the wind/pipe interface from Re=VD/v(kinematic viscosity of air)
    Then I found the Prandtl number (Pr) for air at 0oF
    Then I calculated the Nusselt number (Nu) by Nu=0.023 x (Re)4/3 x (Pr)1/3
    Then I found the thermal conductivity of the air kf
    Then I calculated the surface heat transfer coefficient(h) by h=(Nu x kf)/Dpipe

    I came up with h~11 W/(m2 K)

    Once I found the surface heat transfer coefficient, I calculated the Biot number (Bi) by Bi=hD/ks , where ks is the thermal conductivity of the carbon steel pipe.

    I found Bi=0.003364666 , and with Bi<0.1 I figured could use the lumped capacitance method.

    Note: density=rho=p
    ps=density of steel
    pw=density of water
    cs=heat capacitance of steel
    cw=heat capacitance of water
    Ti=initial water temp
    Tinf=air temp
    T=water at 32oF or 273.15 K

    Using the lumped capacitance method, time (t) in secs can be found from t=[(pVc)tot/(hAs)]*ln[(Ti-Tinf)/(T-Tinf)]

    Where (pVc)tot=[((pscs(Do-Di))/4)s+((pwcw(Di))/4)w]

    and so ((pVc)tot)*(1/h)*ln[(Ti-Tinf)/(T-Tinf)]=t

    This method is how I came up with 8.7hrs. I came up with 4.8hrs using a method out of an ASHRAE handbook.

    Is the method okay for a good approximation? Is 8.7hrs a good approximation? If there's something I'm doing wrong or a better approach, please let me know.

    Thanks for the help.
    Last edited: Dec 3, 2008
  2. jcsd
  3. Dec 4, 2008 #2


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    Did you neglect thermal conduction between the ambient temperature pipe and the water? I think there's actually a decent effect from the thickness of the pipe. I remember writing a fortran program that calculated the "critical" thickness of a pipe/insulation in order to keep heat loss at a minimum. Make the insulation too large and convection outside the pipe increases more than the resistance of conduction through the pipe.

    You might want to take that into account, which it seems the standards do, which would decrease the freezing time.
  4. Dec 4, 2008 #3


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    I'm interested in getting a little more detail on the tube's geometry, specifically its wall thickness. Right now, you're calculating how long it takes to get the water down to freezing temp but not forming ice, but not how long it takes to actually freeze, is this what you're actually after?

    Just looking briefly at your convection calculation it seems to be about right. I think you may be off a slight bit on your Biot number calculation, but it does seem to be much less than 0.1 in any case (for a tube wall of about 0.375").

    Your result seems to be off by about a factor of 2, did you accidentally drop a 2 or a 0.5 anywhere in your calcs?
    Last edited: Dec 4, 2008
  5. Dec 4, 2008 #4
    I did the calc. again and took into consideration the conduction through the pipe wall and found a new "overall" heat transfer value of around ~[tex]11.2 \frac{W}{m^2 K}[/tex]

    So it went from around ~[tex]10.5 \frac{W}{m^2 K}[/tex] to ~[tex]11.2 \frac{W}{m^2 K}[/tex] and the new time is 8.2 hrs...

    Any idea if this is the correct approach? I've worked it a couple times and keep coming up with the same result. The 8.2hrs just seems a bit high to me.

    I looked over the other method out of the ASHRAE handbook and found that I made a mistake in the calculation and now I'm getting around ~ 11hrs... I'm not familiar at all with that method so it may not be correct.

    Thanks in advance for any help you can lend!

    Oh, and I found a new [tex]B_i[/tex] number and it was [tex]B_i[/tex]=0.084 , so it's a little higher than earlier.
    Last edited: Dec 4, 2008
  6. Dec 5, 2008 #5


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    Are you assuming an initial temperature for the pipe? No, I guess you're doing a simple steady-state analysis with the pipe at ambient temperature.

    Also, now that I think of it, adding in the conduction term was probably erroneous. At steady state, the pipe itself will remain at constant temperature. This means that heat in = heat out. In other words, the heat transfer due to conduction from the water to the pipe will equal the heat transfer due to convection from the pipe to the air.

    Can you tell us a little more about the ASHRAE method?
  7. Dec 5, 2008 #6
    I've attached an ASHRAE example where they are calculating the time to freeze orange juice in a 1ft diameter container. I used a similar calculation but with using the properties of water. Now that I look at their result for the orange juice it is ~17hrs... and the thermal conductiviy and heat capacity of the water isn't too far off from the orange juice so the ~11hrs might be close... though it seems high.

    Using the first method, the ~8.2hrs might be a good approximation for the time for the center of the pipe to reach the freezing piont???

    Attached Files:

    Last edited: Dec 5, 2008
  8. Dec 9, 2008 #7
    I don't think that can use the lumped capacitance method when dealing with a phase change so I've tried another approach... Is this the correct method?

    I calculated the surface heat transfer coefficient (h) by [tex]h=\frac{N_u * k_f}{D_o}[/tex], where [tex]k_f[/tex]= thermal conductivity of air

    Then I calculated the overall heat transfer coefficient (U) by [tex]U=\frac{1}{\frac{1}{h*2*PI*r_o}+{\frac{ln\frac{r_i}{r_o}}{2*PI*k_s}}}[/tex], where [tex]k_s[/tex]=thermal conductivity of the steel pipe.

    Then I calculated the heat loss per length of pipe by [tex]q = U*A_s*(T_i-T_\infty )[/tex], where [tex]A_s[/tex]=suface area of pipe per length of pipe, [tex]T_i[/tex]= initial water temp and [tex]T_\infty[/tex]=outside air temp.

    The I calculated the time for the water to reach the freezing temperature by:

    [tex]t_c = \frac{C_{pw} * M_{lw} *(T_i - T_f )}{q}[/tex] , where [tex]C_{pw}[/tex]= specific heat of water, [tex]M_{lw}[/tex]= mass of water per length of pipe, and [tex]T_i[/tex]=intial temp of water, [tex]T_f[/tex]=freezing temp of water and [tex]t_c[/tex] is in seconds.

    I calculated [tex]t_c=14485.17 s [/tex] or 4.02 hrs

    Then I calculated the time for the water to actually freeze once it reaches the freezing temp by:

    [tex]t_f = \frac{h_{fs} * M_{li} }{q}[/tex], where [tex]h_{fs}[/tex]=latent heat of fusion for water, [tex]M_{li}[/tex]=mass of ice per length of pipe.

    I calculated [tex]t_f= 58098.86 s [/tex] or 16.14 hrs

    So the total time to freeze would be [tex]t_{tot}= t_c + t_f = 20.16 hrs[/tex]

    I've attached a word file of my entire calculation if you would like to check my numbers..

    Thanks for the help

    Attached Files:

  9. Dec 9, 2008 #8


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    I think you're probably right about that.

    The problem with this method is that you're assuming the heat flux stays constant as the water cools, however as the water and pipe are cooling the heat flux goes down because the temperature difference is getting smaller. Additionally, the thermal properties of water change as it cools and turns into ice.
  10. Dec 10, 2008 #9
    So you're saying that I should probably not use the same heat rate in the freezing time as I used in the cooling time because of the smaller delta T?... makes sense. I'll try that. I'm sure that will make the freezing time longer because of the smaller heat rate.

    Isn't the change in thermal properties, during the freezing time, accounted for in the latent heat of fusion term?
  11. Dec 10, 2008 #10


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    I think what you're going to want to do is set up a series of differential equations that describe the heat flux in the two separate regions (pipe wall and water), define a set of intial conditions and boundary conditions, and then solve the system of equations either analytically or numerically.

    Think about it this way, as the water is cooling, it will first freeze at the surface of the pipe and move its way in. The really sticky problem is, the thermal conductivity of ice is higher than that of liquid water, so how the ice forms is kind of indeterminite since we are assuming the water is perfectly still (which could be a poor assumption).

    I don't know, there's no simple solution to the problem you're solving, but I think differential equations are your best bet...
  12. Dec 12, 2008 #11
    Yeah, that would probably work but I think I can I use the lumped method to determine just the time it takes for the center of the (non flowing) water to reach the freezing temperature. Which would be my [tex]t_c [/tex] (time to cool the water to its freezing temp). I think the lumped method would account for the transient heat rate while the water is cooling.

    and then use

    [tex]t_f= \frac{(h_{fs} * M_i) }{ q} [/tex]

    where [tex]h_{fs}[/tex] is the latent heat of fusion for water, [tex]M_i[/tex] is the mass of water and q is the heat rate in [W/m] calcuated at the freezing temp for water.

    Then just add [tex]t_c[/tex] to [tex]t_f[/tex] to get the total freezing time?

    Any thoughts?

    Last edited: Dec 12, 2008
  13. Dec 15, 2008 #12


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    If you want a very rough estimate of how long it will take, maybe you can make something work with the lumped capacitance method. The lumped capacitance method is based on differential equations that descibe the heat flux in a part, and certain assumptions are made to simplify the solution.

    First, the lumped capcitance method is meant to be used on a single piece of material that is in direct contact with the environment. This is not the case since your water has a pipe surrounding it, which skews the results since the steel pipe acts as an insulator. The method assumes that all of the material is the same temperature, or at least that the temperature gradient is very small. This is what the Biot number accomplishes, it compares the thermal resitance within the part to thermal resistance outside of the part. In your case, you have to decide if the lumped capacitance method can apply to both the water and the pipe wall.

    One thing you might do is decide if you want to assume the pipe is the same temperature as the surrounding air. If so, you can make you life easier by "ignoring" the pipe.
  14. Dec 16, 2008 #13
    freezing a can of soda will cause its bottom and top to bulge so badly the can will not stand up. What's happened?
  15. Dec 18, 2008 #14


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    It's because as the water in the can freezes and becomes ice, it expands. The crystalline structure of ice is such that it is less dense than water (hence, ice has a larger volume per mass than water) and so the can has to expand out to contain the larger volume of ice.
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