Heat transfer rate from Mild steel bars to a chilled water tank issue

[Xyimon]

Hi all,

This is my first post here (hopefully of many), and there is an issue I am struggling with for my work.

I am helping in designing a conveyor that takes freshly milled mild steel bars and submerges them in a tank of cooled water to reduce their temperature enough to to not damage the proceeding equipment.

The tank of water will be connected to a separate reservoir of water which will contain the cooling equipment, and will be pumped around in a cycle as the cooling equipment is not allowed near the milling machines.

The question is as follows:

Bars of mild steel with a radius of 25mm and a length of 1100mm, as transported into a tank of water, with a constant flow meaning that there will be 6 bars submerged in the water at any time, and will take a total of roughly 48 seconds to pass through the tank.

The tank of water holds 275L.

The bars will enter the water at roughly 85deg C, and need to be reduced to below 28deg C by the time they leave.

How much cooling will need to be applied to the tank to keep the water at optimum temperature, enough to chill the bars.

Any information or help with this problem would be greatly appreciated, I will be checking this forum regularly so please don't hesitate to ask if any further information is required.

Regards,

 

[LittleInch]

Why don't you try and work it out yourself?

You know the mass of metal per second / minute going into the water. you know how much temperature you want to loose so can equate that to energy (Joules)/sec / minute.

All that heat needs to be taken away by the water so this heat energy is what your heat load rejection (cooling) needs to be designed for.

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[IRstuff]

Do you care whether the rods get below that threshold temperature?

The minimum cooling is the amount of Joule heat removed from the bars

TTFN
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[Xyimon]

IRstuff, the 28deg C is the maximum temperature that they are allowed to be, so anywhere between say 15 and 28 would be fine.

LittleInch, I have been trying but am struggling with the formula required.

Regards,

 

[LittleInch]

Mass of metal in one bar x heat capacity (Kj/kg/K) x temp change ( Tin - T out) = amount of heat loss per bar (a)

Average heat loss per second per bar = (a)/48secs = (b) (J/s) or W

total heat loss per second required in your bath = (b) X number of bars in your bath at any one time (6) in Watts

To maintain a steady water temperature you need to remove this amount of heat per second as a minimum

This is pretty basic stuff??



My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[ione]

If you already have determined the batch temperature compatible with your process, then, as stated above, the minimum cooling is the amount of heat removed from the bars. You just need to know the mass of the steel bars to be cooled, the starting and the final temperature, and the specific heat of steel.

Things would be a bit different if you should have to determine the optimal water temperature for cooling purpose. The optimal temperature depends on the time available for your cooling process. The lower is the water temperature, the faster is the cooling process.
In this case you should evaluate weather the lumped capacitance method is applicable to your scenario, checking the value of the Biot number. If lumped capacitance method is not applicable then you should perform a more complicated transient analysis.

 

[Xyimon]

Ok right, pretty sure I have got it but I need to confirm I am using the correct units so I will type it out below:

(Mass of metal in 1 bar = 17kg), (Specific Heat Capacity of Mild steel = 0.62j/kg/k or 620j/kg/k SO I use 620j/kg/k as its in Kg, correct?), (Temp change (T Final - T Initial) so 20deg - 85 deg = -65deg)

So we have 17kg x 620j/kg/k x -65 = -685100

Next, average heat loss per second per bar = -685100 / 48secs = -14272.9 (J/s or W) (Why is it J/s or W?)

And then finally, -14272.9 x 6 = -85637.5 watts

Is this correct? I understand I know very little about this but it is far from my area of expertise so I am rather lost.

All help is greatly appreciated,

Regards,

 

[Xyimon]

If i use:

17kg x 0.62j/kg/k x -65 = -685.1

-685.1 / 48 seconds = -14.273 J/s or W

-14.273 Js/w x 6 bars = -85.6375 watts,

This seems much more realistic, but I am still unsure. Also, what is the units for the restful of the initial calculation 17kg x 0.62j/kg/k x -65 = -685.1?

Regards,

 

[LittleInch]

I looked and found a variety of specific heat from 420 to 620 J/kg/C.

Never mind, your first one is correct. J/s = Watts

85 kW of heat rejection sounds about right to me. It's an approximate number and works on the lumped capacitance method which may not fully apply

think of it the other way around - If you had 102kg of steel (your 6 bars) and you needed to raise its temperature 65 Degrees C in only 48 seconds, you would need some pretty big burners / electric immersion heaters to maintain the water temperature yes? Well it works the other way around.

I don't know if this is a working system ,but you may find 48 seconds isn't long enough to cool the bars down as assuming you keep the water at say 10C, the heat loss rate will constantly fall as the temperature of the bar falls


My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[Xyimon]

LittleInch,

That's very helpful thank you, the overall aim of this was to try and work out the following:

Connected to the tank of water will be another reservoir that is actually cooled, and the resulting water will be pumped into the tank that the bars are dipped into, as the cooling machine is not allowed to be near the milling machinery.

It was my aim to find out how large the separate reservoir would need to be, and how much cooling/what type of cooler would be required.

Any points in the right direction would be very useful.

Regards,

 

[Xyimon]

Also, what is the lumped capacitance method and why may it not fully apply? Is there a specific article that I could read as I can only find rather complex formula documenting it...

Regards,

 

[LittleInch]

If you can get a sample or even half a sample, I would try doing some tests in a BIG bath of water at say 10C so the water doesn't heat up much and measure temperature versus time for your rod to see if its going to work in your timescale. Measuring your before and after water temp will also serve to see if this simple calculation is correct. After that I would guess it's an industrial chiller unit for you - 85 kw is a decent size and you'll need a place for the condenser to reject the heat to atmosphere.

I would go for a 100kW at least...

Good luck.

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[Xyimon]

ione,

It turns out that the amount of power/cooling required is far too much to make this viable, and you stated that "Things would be a bit different if you should have to determine the optimal water temperature for cooling purpose. The optimal temperature depends on the time available for your cooling process. The lower is the water temperature, the faster is the cooling process.
In this case you should evaluate weather the lumped capacitance method is applicable to your scenario, checking the value of the Biot number. If lumped capacitance method is not applicable then you should perform a more complicated transient analysis."

Due to this, I now have to work out what you preemptively stated above.

So, how do I go about checking the value of the Biot number in order to determine whether the lumped capacitance method is applicable?

As always, any help on this will be greatly appreciated

Regards,

 

[ione]

Xymon,

A lumped system is the one where the temperature varies with time but remains uniform throughout the body interested in the process.
The parameter to be checked in order to establish whether lumped capacitance can give you an accurate and reliable answer is the Biot number (Bi). Biot number basically expresses the ratio between the convection at the surface of a body and the conduction within the body. If conduction within the body is neatly predominant over convection at surface of the body then lumped capacitance is a good way to go, and the threshold which discriminates the approach to follow is Bi = 0.1.

Bi = h*Lc/k

Where:
h expressed in [W/m2/K] or equivalent units is heat transfer coefficient between the fluid (water in your case) and the body (steel bar in your case). *
Lc expressed in [m] or equivalent units is characteristic length of the body (generally Volume/Transverse area).
k expressed in [W/m/K] or equivalent units is thermal conductivity of the body.

* Note that there are empirical correlations (best fit of experimental data) which give the Nusselt number (Nu = h*D/k) as a function of dimensionless groups (in the attachment you can find a collection of those correlations).

 

[Xyimon]

Right,

I have Lc and k, but have been struggling for hours to work out h.

Regards,

 

[Xyimon]

ione,

From the data sheet that you provided me with, I am lost when it comes to the Nusselt number; any help?

 

[ione]

You have to identify your heat transfer scenario (free convection or forced convection further to heat transfer mechanisms + geometry involved). Use the relevant correlation to compute the Nusselt number. Then from the Nusselt number definition Nu = h*L/k and rearranging for h you can find the heat transfer coefficient. Please take it the right way and not as an offence, but it seems to me things are turning from "please give me some hints" to "please do it on my behalf". If you don't have the right background or knowledge to face the problem, ask your supervisor or mentor or whoever has done this before in your company to give you some guidance.

 

[Xyimon]

I am trying to work it out myself but I am at the point where I don't understand whats what and need a nudge in the right direction :/