Heat transfer over time in a rubber pipe 2

[enrjdean]

Hi

I haven't had any hits on the search regarding this particular topic, if they are there, I appologise.

I'm having trouble calculating the following:

20" rubber pipe submerged in sea water at 27-28C has had crude running through it at 6000m^3. The crude, has been pumped at around 65C. The fluid has now come to rest. The heat transfer coeeficient of the rubber is 8.3 W/m2K and wall thickness is 25mm

How long till the crude reaches the temp of the sea water?

I've had a look at Newtons law of cooling, but am having trouble finding an equation which takes into account the insulation offered through the pipe wall.

Can anyone give me pointers on how best to approach this problem?

Thanks

 

[IRstuff]

Your heat transfer coefficient is for what? The units do not correspond to thermal conductivity, which is W/m-K.

Normally, you calculate thermal_cond*delta_T/thickness as the heat loss, and you make assumptions about how the heat is distributed within the oil. You can assume that it's all located at the pipewall for a fastest possible cooldown, or assume it's all in the center for a slowest cooldown. Naturally, your delta_T will change as a function of the amount of heat you pull out, so it'll be an iterative calculation.

TTFN

FAQ731-376

 

[enrjdean]

Hi

Sorry for the mix up in units, will post the correct ones tomorrow. Is also 6000m^3/hr.

I assume that the fluid is left in the pipe and that the pipe is full.

So using (k) thermal_cond*delta_T/thickness I get a heat transfer rate, Watts. What I really need is to get to the answer in seconds.

Can I plug back the heat transfer rate into the equation to get seconds?

Delta'Q'/Dela't' = kA * ( Delta'T' / x)

Thanks

 

[IRstuff]

You have "x" joules of heat in the oil. However, as I discussed earlier, you need to model how the heat is distributed within the oil and what thermal resistance it sees within the oil to get a more meaningful transient time. If you're worried about the oil cooling down too fast, then you assume the worst-case with all the heat up against the pipe wall.

TTFN

FAQ731-376

 

[ione]

Just for a first estimate, assume the crude mass is all at the same temperature and consider:

Q = (k*deltaT)/d

Where:

Q = thermal power per square metre of pipe [J/(m^2*s)]
k = pipe material thermal conductivity [W/(m*°C)]
deltaT = temperature difference [°C]
d = pipe thickness [m]

you can also write:

Q = (M*cp*deltaT)/t

Where:

M = crude oil mass in a length of pipe having 1 square external area [kg/m^2]
cp = crude specific heat at the reference temperature [J/(kg*°C)]
t = time

equalling the two expression of Q and solving for t you get your answer

t = d*(M*cp*deltaT)/ (k*deltaT)

 

[enrjdean]

Hi guys

attahced is the spreadhseet I have made up with the above equations, thanks!

One question tho, I have calculated the mass of the contents sitting in the pipe and assume that the pipe is full. What is meant by "crude oil mass in a length of pipe having 1 square external area"?

Do you mean mass per sqm over the surface area of the pipe?

 

[ione]

You know the external pipe diameter so, for your given pipe, you can calculate the length of pipe which has got an external area of 1 square metre. Once you have calculated the length of pipe above mentioned, and given the internal diameter of your pipe you can calculate the oil volume and the density of oil you calculate the mass corresponding to the control volume.

L = 1/(pi*D)

V = [pi*d^2/4]*L = [pi*d^2/4]* 1/(pi*D)

M = rho*V

Where

L = length of pipe with an external surface of 1 m^2
V = control volume of oil
d = pipe internal diameter
D = pipe external diameter
M = oil mass
rho = oil density

 

[enrjdean]

Hi Ione

Thanks for your response.

I assume by using the control volume this will give me the rate in sec's per sqm due to temp drop. Multiplying this over the length of the pipe will give me the total time over the whole length?

 

[prex]

There is a mistake in your spreadsheet (the second deltat lacks a parenthesis), but also you didn't understand that there are two deltaT's to be accounted for(OK ione's explanation was not very clear).
So let's restart.
The change in temperature of the oil for a power (heat per unit time) output to pipe wall Q and for a unit length of pipe (W/m) is:
[Δ]T=Q/([ρ]AC_p)
where:
T=oil temperature (°C)
[ρ]=oil density (kg/m³)
A=area of pipe section=[π]D_i²/4 (m²)
C_p=specific heat of oil (J/kg°C)
The heat lost through pipe wall in the unit time and per unit length is
Q(W/m)=k[π]D_avg(T-T_out)/d
where:
k=pipe wall thermal conductivity (W/m°C)
D_avg=some average pipe wall diameter (m) (may be taken as (D_o+D_i)/2)
T_out=sea temperature (°C)
d=pipe wall thickness=(D_o-D_i)/2
From the above you can easily calculate [Δ]T at time increments of 1 sec (or larger), easy to do with a spreadsheet, but, for such a simple model, there is also an exponential decay formula in closed form.

All the above assumes a uniform temperature throughout the oil. As convection could be a minor phenomenon in your crude, especially when temperature goes down, the time to cool down can be grossly underestimated with this procedure. If you need an idea of how much, you should adopt the model of heat conducting through a solid cylinder having the heat conductivity of the oil. It's more complex, but can still be solved by formula.
Also to be noted that, as your pipe is quite thick, a non negligible contribution to the cool down time may be offered by the heat present in the wall (the mass of the pipe is of the order of two times the mass of the oil).

prex

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[ione]

I have not interpreted the problem this way, anyway I could be wrong.
In your opening post you’ve said at a certain time the flow stops (“the fluid has now come to rest”), so the overall length of the pipe is in the same condition as that of the control volume you are considering (the control volume is representative of the overall volume of fluid stored in the pipe).

 

[enrjdean]

Prex

ok, so the oil disaptes heat (in this case power, Watts) through the wall of the pipe which has some conductivity (k). 1 watt = Joules/sec. So through iteration I can find what the total time it would take to drop to sea temp.

I hope I have followed your notes correctly. I have attahced the spreadsheet, if you could have a look over, I'd be greatful.

 

[IRstuff]

To expand on Prex's comments w.r.t. underestimation.

You can somewhat bracket the time by assuming ALL the heat is in the middle of the oil, and you would simply add the thermal resistance of the oil to that of the pipe. This would give you a closer approximation than assuming that all the heat is immediately available on the inside of the pipe.

TTFN

FAQ731-376

 

[ione]

Rubber thermal conductivity = 85 W/mK ????? I do not think so.


Heat loss for conduction through the wall pipe (a cylinder)

Q = (2*pi*L*k)*deltaT/ln(D-d) (1)

Where:

Q = heat loss [W]
L = pipe length [m]
k = pipe thermal conductivity [W/mK]
D = pipe outer diameter [m]
d = pipe inside diameter [m]
deltaT = temperature difference [Tamb-Tcrude] assuming the overall mass is at the same temperature

Q = M*cp*deltaT*t (2)

M = crude oil mass [kg]
cp = crude oil specific heat [J/(kgK)]
t = time required for cooling [second]

Equal expression (1) and (2) and solve for t

This should give you a rough estimate. As already pointed out by IRstuff you can take into account the thermal resistance of the oil and consider the oil thermal resistance and the pipe thermal resistance as two resistances in series.

 

[prex]

No, enrjdean, you didn't take the equations correctly. What you should obtain is an asymptotic decay of temperature, so an infinite time to equilibrium (but of course you'll assume that the equilibrium is reached when the temperature differential is less than a given delta).
Also, as noted above, the conductivity of pipe wall is unrealistic, and your data on ID, OD and wall thickness do not match.
Come back with some more correct data and perhaps with a better description of your goal, to help us suggest how to determine a reasonable (not infinite of course) cool down time.
On that basis I'll see if I can return a corrected spreadsheet.
You seem to be far out of your field of expertise, though, if you could ask for help to someone closer to you, that would be best for you.

prex

http://www.xcalcs.com

: Online engineering calculations

http://www.megamag.it

: Magnetic brakes and launchers for fun rides

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: Air bearing pads

 

[ione]

Prex,

An asymptotic decay is what one would get using Newton’s cooling law

T(t) = Tamb + (Tin –Tamb)*e^(-c*t)

This equation leads to T(0) = Tin and T(?) = Tamb

Tamb = outer ambient temperature
Tin = starting temperature of oil
c = constant

Were you making reference to this?

 

[prex]

Yes, sure. You can call it Newton's law or otherwise, but of course it is an exponential decay law.

prex

http://www.xcalcs.com

: Online engineering calculations

http://www.megamag.it

: Magnetic brakes and launchers for fun rides

http://www.levitans.com

: Air bearing pads

 

[ione]

enrjdean,

Try this equation

T(t) = Tamb + (Ti – Tamb)*e^[-2*pi*L*k*t/(ln(D/d)*M*cp)]

Tamb = 65 °C
Ti = 27 °C
L = 320 m
k = 0.14 W/m°C (rubber thermal conductivity)
D = 0.87 m
d = 0,5 m
M = 59670 kg
Cp = 1900 J/(kg °C)
t = time for cooling (seconds)

It takes 1000000 seconds to reach T = 27.36 °C approx

 

[enrjdean]

Hi Ione

I had looked at Newton's law of cooling, but couldn't put the steps together that you guys were trying to get across. I've plugged some numbers into the equation and am starting to make a bit more sense of it; thanks for sticking with me!!

The large value ok K I can only assume is due to the flanges at each end of the pipe. But using 85 W/mK as a value of K makes the assumption that this value is constant across the length of the pipe?

 

[ione]

Thermal conductivity is a property of the material and the coefficient k is a measure of the rate at which heat flows through the material. It is an intrinsic property of the material and it depends on the chemical composition, porosity, density and crystal lattice of the material. We assume the material has an isotropic behaviour and consider thermal conductivity as a constant.

85 W/mK is a value that could be typical of a metal, not of rubber.

Meanwhile have you grabbed the meaning of the equation above?