# How long to freeze standing water in steel pipe

1. Dec 3, 2008

### jt316

I posted this in another section and haven't receive any feedback so I thought I might have it in the wrong section. Thanks in advance for any help.

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.

2. Dec 9, 2008

### jt316

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 $$h=\frac{N_u * k_f}{D_o}$$, where $$k_f$$= thermal conductivity of air

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

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

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

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

I calculated $$t_c=14485.17 s$$ or 4.02 hrs

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

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

I calculated $$t_f= 58098.86 s$$ or 16.14 hrs

So the total time to freeze would be $$t_{tot}= t_c + t_f = 20.16 hrs$$

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

Thanks for the help

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3. Dec 9, 2008

### Staff: Mentor

Boy, I hated my heat transfer class: it's a lot of work for something that in real life is never actually done and isn't all that accurate anyway. Yeah, it's good to learn the theory, but I don't know that the theory is really all that useful to actual engineers. Anway, that's why I didn't respond before....

I think the flaw may be in the calculation of the pip-air heat transfer coefficient. I don't see anything about the roughness of the pipe in there. I can't remember if the calculation assumes it is smooth, but in any case, if it isn't smooth, you get much more turbulent flow and much more convection. I'm not certain of this, so consider it just food for thought.

4. Dec 9, 2008

### jt316

Ah... so you think this is a homework problem? I was wondering why no one was responding to the problem...

I am a mechanical engineer and this is actually a design related problem. We are in the design phase of a project that requires the installation of a cooling tower and 18 inch water lines. We are trying to determine that if there was ever failure of the pump, how long we would have before the water in the pipe would freeze. We have a contractor working for us who thinks this pipe will never freeze and we have another mechanical engineer that think it might, under these conditions.

I've been given the task to determine when it would actually freeze. Well, at least a good approximation.

It's been several years since I've had heat transfer as well.

Any and all help is greatly appreciated.

Thanks