Bitumen tank - Heat transfer rate 3

[MarcAndre]

Hi

I am working on an energy efficiency project in an asphalt making company in Quebec, Canada. I want to calculate the Heat transfer rate inside the Bitumen tanks from the thermail oil pipes to the bitumen. I've got all the data for the thermal oil, but I am still looking for the Thermal conductivity and the Specific heat of the bitumen.

I also want to calculate how much heat is loss through the walls of the tank (insulation: 2x 2'' mineral wool blanket 850F + aluminium cover). The outside wind is about 7mph.

I have downloaded the 3E Plus 3.2 insulation software, but I am not quite sure if I can consider the tank (108'' diam.) as a large pipe.

The desired bitumen temperature is 320F which is also the approximated temperature of the thermail oil flowing in the coils (2'' pipes) inside the tank.

Thermal oil specs:
Kinematic Viscosity = 10,78 cst (or mm^2/s)
Density: 822,1 kg/m^3
mass flow rate= 100 GPM pump --> 5.18 kg/s
Specific heat (Cp) = 2,15 kJ/kg/K
Thermal Conductivity = 0,126 W/m/K
Temp. thermal oil = 160C or 320F
Ambient Temp = 20C or 68F

I Used 18,9 W/m^2/K for the h convection of air

So, how do I calculate the effective heat transfer from the coils to the bitumen inside the tanks ? I would like to know how long it would take to increase the temperature of the bitumen, if we wuld shut off the thermal oil boiler on weekends and restart it before the production starts on Mondays.

thanks

 

[MarcAndre]

Also, I forgot to mention that the bitumen inside the tank doesn't move that much; we could consider it as almost static. What is its mass flow then ?? I think I am supposed to use that to calculate the h of bitumen.

Regards

Marc-Andre

 

[StoneCold]

MarcAndre
If you can measure the hot oil temp in and out of the tank and you know the flow rate then that will give you the current steady state heat loss by the tank. If you stop the flow of thermal oil and let the tank cool down a little you can back calculate a bulk heat capacity for the system by taking the (steady state heat loss {btu/min})*(minutes you let it cool down)/(temp difference).
It wont' be exact but it will give you a ball park. Obviously the rate of cooling in reality is asymptotic because the material near the wall will cool down and slow the cooling of the bulk of the material substantially.

All that being said I would say that cooling the asphalt off over the weekend is a bad idea. Maybe you could just lower the temperature to 250F for the weekend but shutting off the heater all together will probably get you into trouble. What if you can't get it started again on Monday? Startup's and shutdowns are the most dangerous part of any process, do you really want to add two of them per week to your process for a small amount of cash?

Things to think about, sometimes saving money can really bite you in the ***.

Good luck

StoneCold

 

[MarcAndre]

Thanks stoneCold

We are aware of the fact that if temp of the bitumen gets too low, it could be really bad. That is why I am in charge of calculating those heat transfer rates, to figure out if shutting down the thermal oil boiler would be too drastic. The next step is actually, as you said, to simply just lower its temperature.

But two weeks ago, we tried in one of our plant, to close simply close the boiler and see how the temperature would go down. My boss was going everyday to check (we had a 3 days break actually). And the temperature of the bitumen have in fact decreased but really slowly and not that much. Example from 320 to around 280F, if I remember. So we were very surprised and happy to see that.

So, in your opinion, the best way to do such calculations, is to do it in practice ?

Thanks for your help

Marc-Andre

 

[StoneCold]

MarcAndre
If you do the experiments I explained above it take a lot of other factors out of the equation. Like , what if you have gaps in your insulation, or wet insulation? What if the convective currents in your bitumen tank are different than the standard model due to product buildup on the walls. The experimental results will take all of those factors and lump them into a real heat loss number.
If your tank took three days to cool down then I would guess that if you determined your steady state heat loss it is small and therefore your cost savings are going to be small. But check it out, I could be wrong.

Goodluck
StoneCold

 

[MarcAndre]

The thing is that the boiler's firing rate is modulated according to the thermal oil's temperature, and therefore never really gets turn off. In fact, it works 24hrs a day, at different rates, but still uses fuel. It seems like heat is always transferred to the bitumen in the tank, even though the bitumen does not cool down very fast. That is why we think we could save a lot of energy if we could intentionally turn off or reduce the boiler's consumption at time where it is not really necessary.

This is what I think, and I want to verify it. That's why I am trying to find all these heat transfer rates, to get a good idea of what is really going on. We might be wrong, but we want to quantify the process and put $ to it

Thanks

M-A

 

[TD2K]

For the heat loss, as Stone said, I'd look at how much heat you are putting into the bitumen from the hot oil? Q = mCpdT. Rent a portable ultrasonic flow meter if you need to.

I'd then compare that to the heat losses from the tank assuming that the bitumen at the inside wall is about 320F and that the inside film coefficient is much less than the thermal resistance from the 2" of mineral wool (seems light to me for this temperature). If the two compare reasonably well, then you know the inside bitumen is close to 320F AND its heat transfer coefficient is insignificant to the insulation. If not, round 2 of the calculations.

I'm not surprised the bitumen didn't cool off very much, there's a huge thermal mass in that tank. I did some calculations on how much a rail car full of 50% caustic would cool down and got only about a deg F a day.

Do you have any properties for the bitumen as a baseline? I may have some references at the office but will have to look tomorrow.

 

[MarcAndre]

Hi TD2K

In fact, the insulation is made of two layers of 2'' mineral wool, which gives 4''.

As for your 2 paragraph, I'm not sure what I am supposed to compare. What do you mean by:

" If the two compare reasonably well, then you know the inside bitumen is close to 320F AND its heat transfer coefficient is insignificant to the insulation. If not, round 2 of the calculations."

For the bitumen specs, I am supposed to get some infos today by the mail, but I still didn't get it yet. I found on the web that Thermal conductivity is around 0,17W/m/K.
I still need to get the value for Cp (specfic heat) and viscosity is around 45cP (should get real data soon).

thanks

 

[TD2K]

Sorry for the confusing language.

If you go through the heat loss from the hot oil system you know how many kJ are going into the bitumen. Ignorning the effect of adding 'fresh' bitumen to the tank (if it's a different temperature than what is already there), those kJs have to be replacing those lost to the ambient surroundings.

Tank losses are through the wall, roof and floor. Heat losses through the wall through the wetted portion are first through the inside film coefficient from the bulk bitumen to the wall, through the steel, through the insulation and then to air (including air gaps if you want to) though the steel resistance is trivial and can be ignored. For the walls above the liquid level you need to include the air inside film coeficient from the bulk air to the inside wall. You have similar losses through the floor and the roof. For the roof losses, you also have the bitumen to the air and then air to the roof heat transfer coefficients to include. When I've done tank heat loss calculations those air film coefficients are significant in reducing the heat losses.

If you think about what is happening in the tank, there are convection cells bringing 'hot' bitumen to the wall where it cools due to the heat losses, it sinks and flows backwards the middle of the tank. Depending on the magnitude of this heat transfer coefficient compared to the resistance from the mineral wool insulation, the inside wall temperature could be low enough to have a significant effect on your heat losses. For example, the inside wall temperature isn't 320F, it's 250F to pull a number out of the air so the dT for heat losses through the wall is 250F - 68F, not 320F - 68F. Worse, you could have a layer of bitumen 'solidifying' on the walls for some thickness providing another 'insulation' layer you aren't accounting for.

If you make the assumption that the inside film coefficient is insignificant compared to the thermal resistance of the 4" of insulation then the inside wall temperature is going to be the bulk temperature or about 320F. If so, your heat losses based on this temperature should be in the range of what you calculate you are inputting into the tank from the hot oil system. If this assumption though gives you a much higher heat loss than what you believe is happening from the hot oil input, then you need to get more precise on including the inside bitumen effect.

 

[MarcAndre]

Wow, that's a very good explanation of what is going on. Thank you very much. I should be able to do something good out of all this useful information.

I'll give you guys some feedback of what are my results when I get a better idea.

Thanks again

 

[TD2K]

There are lots of ways to 'tweak' the tank heat loss numbers. Couple of the electrical heat tracing supply companies have free calculators that do heat loss calculations. Their numbers seem reasonable to what I calculate manually for heat losses through the insulated shell and floor. I don't believe their numbers for heat losses through an uninsulated roof, it appears that they don't include properly the air film coefficients but they can't tell me exactly what their program is doing, it's just a black box program to them.

As long as I come up with a number that is reasonably close to what I know/believe is going into the tank I'd be satisified I know what the heat losses are and that I don't have a huge heat sink that I'm not accounting for. You can then use that to see how long you can operate without the heater running before you could expect to see problems or what are the energy savings from different/additional insulation.

 

[MarcAndre]

Hi guys

I have talked with an engineer at the company that designs and makes our bitumen tanks. He sent me two charts that give a good approximation of "Average Overall Heat Transfer Coefficients" for different heating and cooling applications. Also, he gave me an approximate heat loss coefficient for storage tanks.

I have:
U= 10 to 20 BTU/(hr ft^2 F) (For heat tr. oil to bitumen)
U= 0.11 BTU/(hr ft^2 F) (for storage tank 2'' ins. for delta T about 200F)

Now, I don't understand something that should be trivial; i.e. how come I don't get the same kind of answer when comparing the following two equations (even when putting in the same units):

Q = mass flow * Cp * delta T (in kW)
Q = U * Area * delta T (in Btu/hr)

In my case, I have:

mass flow = 5.18 kg/s
Cp= 2.15 kJ/(kg* K)
Area= 350 ft of 2'' pipe = 183 ft^2


So, I don't see what I am doing wrong. Can someone help me please ? I'm still having trouble finding the amount of Energy or heat transferred to the Bitumen (per delta T)

Regards

Marc-Andre

 

[TD2K]

Q = m*Cp*dT is the heat transfer from the hot oil to the bitumen. The dT is the temperature difference of the hot oil (your 5.18 kg/s) between its temperature entering the coil and leaving the tank due to heat losses to the bitumen. I don't see those temperatures here (in your first post and third post you imply both the hot oil and the bitumen is 320F which would mean zero heat transfer to the bitumen).

OR

You can calculate the heat transfer from the hot oil coil to the bitumen. In this case, the equation is

Q = UAdTlm. The U is the 10 to 20 BTU/hrft2F figure.
A is the pipe external area for heat transfer and
dT is the log mean temperature difference between the hot oil and the bitumen (I'd use dT log mean since you have an inlet and outlet oil temperature. If the oil temperature doesn't vary much, you can just use the average oil temperature and the bitumen temperature).

OR

you can calculate the heat losses from the tank which has to be replaced by the hot oil coil. That equation is

Q = UAdT

U is the 0.11 BTU/hrft2F (if this is really for 2" of insulation, it would be 1/2 of this for the 4" of insulaton you have)
A is the tank external area for the wetted area (as I mentioned, you can then estimate the heat loss for the portion of the tank containing air and then through the floor)
dT is the temperature difference between the bitumen and the ambient air.

Those 3 numbers 'should' be comparible.

 

[MarcAndre]

Thank you very much TD2K.

This summarize very well what are my options. I'll check that out to see if the numbers make sense

Now, I would like to check about the log mean dT.
For example, let's say that the inlet temp is 320F and the outlet temp is 315F, and that the Bitumen temp is not changing a lot and is 310F. Are we saying that the log mean dT gives:

dT = (320-310) - (315-310)
--------------------------
ln ( (320-310)/(315-310) )

And do both the inlet and oulet hot oil temp should be higher than the Bitumen temp ? Could it be equal?

Thanks

Marc-Andre

 

[MarcAndre]

I was thinking about something. If for some reason, in another kind of application, the temperature of the oil (or maybe it would be water) is lower than the substance to be cooled, would we calculate the heat transfer rate using Q = m * Cp * dT ? I mean, would we use the Cp of the oil (or water) or the Cp of the substance being cooled. I am wondering if the Specific Heat is always for the "hot" side of a heat transfer process or is it associated with the media that is "flowing" ?

What happens when both media are flowing? Do we take the relative flow ?

Thanks

 

[TD2K]

Your LMTD calc is right. I personally find it's helpful to draw it on a piece of paper. Put a box where the heat transfer occurs and the hot side in and out and the cold side in and out (usually going different ways as most exchangers are counter current flow). then take the dT at each end. LMTD, and your formula is right, is (dT1 - dT2)/ln(dT1/dT2).

No, they can't be the same temperature. Try the math, one term goes to zero and the formula blows up (they can't be exactly the same temperature or you have no driving force for heat). The formula also blows up if you have one fluid condensing at a constant temperature and the the fluid boiling at a constant temperature but in that case, the dT is trival to see what it is.

"What happens when both media are flowing? Do we take the relative flow ?"


You can use either side (or both if you know the flows and dTs for each side). Since energy is conserved, the duty of the 'cold' side is exactly the same as the duty of the 'hot' side (I'm ignoring heat losses here).

Q on both sides, except for a sign change, is m*Cp*dT for a single phase fluid. If you have vaporization or condensation, you need to take into account the latent heat.

For your case, you very likely have better data for the thermal oil rather than knowing what is the temperature of the bitumen throughout the tank relative to the hot oil coil's surface. There will be a temperature variation for the bitument throughout the tank quantifying that difference makes the calculation of the heat transfer for your coil based on Uo external harder to get a reasonable value.

 

[25362]

There are two types of heat transfer, steady-state and unsteady-state heat transfer, and three common mechanisms: conduction, convection, and radiation.

In steady-state heat transfer temperatures do not change with time. In unsteady-state heat transfer, they are a function of both space and time and the heat flux per unit time changes, ie, heat may accumulate in the stagnant bitumen.

If there is a stage of equilibrium in which the heat lost by the hot oil equals the heat losses from the tank to its surroundings, one may assume a steady-state heat transfer by which the inlet and oulet temperatures of the circulating hot oil don't change with time.

Let's remember that Q in the formula mentioned by MarcAndre is a rate expression, namely kW, or kcal/h, or Btu/h.

In this case the formula measures the sensible heat lost by the warm "moving" side where one can measure the flow rate, and, under equilibrium conditions as above, this fluid experiences a steady rate of heat exchange.

A heat exchanger where both fluids, 1 and 2, are running without change of phase, ie, they interchange their sensible heat, one can use the formula on each side for comparison, as in:

Q=(mCp.[Δ]T)₁=(mCp.[Δ]T)₂​

[Δ]T refers to the temperature change each fluid undergoes.

When one uses Q = UA (LMTD), LMTD is the logarithmic mean temperature obtained from the temperature differences between both fluids at each end of the exchanger.
In this case it is U (the overall heat transfer coefficient) that incorporates the rate element.

In case there is also phase change in one or in both streams, the heat exchange includes, besides any sensible heat as above, also the latent heat of phase change.

Unsteady-state heat transfer is a t more complicated and analytical solutions are obtained from graphs or by compurized programs if available.

Have I complicated matters ?