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

[Xyimon]

Homework Statement

Hi all,

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 be cooled by the cooling equipment, and will then 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 144 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 (how large of a chiller) will need to be applied to the tank, and how much water will the separate reservoir need to hold to keep the water at optimum temperature, enough to chill the bars.

Homework Equations

I believe that I need to evaluate whether the lumped capacitance method is applicable to the scenario by checking the value of the Biot number, but this is beyond my understanding, I have looked into it via wikipedia and some journals but I am struggling to understand which heat transfer coefficient (h) and thermal conductivity (ks) values to use for my scenario

The Attempt at a Solution

So far I have worked out the following:

(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 / 144secs = -4757.64 (J/s or W) (Why is it J/s or W?)

And then finally, -4757.64 x 6bars = -28.546 K/watts

This answer is not of much use, instead I would need to determine the optimal water temperature for the cooling purpose, and the the optimal temperature depends on the time available for the cooling process.

Any help would be greatly appreciated.

 

[SteamKing]

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 144 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 (how large of a chiller) will need to be applied to the tank, and how much water will the separate reservoir need to hold to keep the water at optimum temperature, enough to chill the bars.

Homework Equations

I believe that I need to evaluate whether the lumped capacitance method is applicable to the scenario by checking the value of the Biot number, but this is beyond my understanding, I have looked into it via wikipedia and some journals but I am struggling to understand which heat transfer coefficient (h) and thermal conductivity (ks) values to use for my scenario

The Attempt at a Solution

So far I have worked out the following:

(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

Always carry your units through your calculations. What are the units of -685100 above?

Next, average heat loss per second per bar = -685100 / 144secs = -4757.64 (J/s or W) (Why is it J/s or W?)

Again, if you carry your units through the calculation, you'll see why the answer comes out J/s.

And then finally, -4757.64 x 6bars = -28.546 K/watts

The result is -28.546 KW, not K/watt, which implies some sort of ratio.

This answer is not of much use, instead I would need to determine the optimal water temperature for the cooling purpose, and the the optimal temperature depends on the time available for the cooling process.

Any help would be greatly appreciated.

What is your definition of the 'optimal' cooling water temperature? The only constraints you have specified is that six bars must be cooled from 85 C to below 28 C in less than 144 sec.

 

[Xyimon]

The bars must be <=28deg after the 144 seconds, so I need to work out what temperature the water needs to be kept at in order to remove that amount of heat from the bars in that period of time

I am having issues trying to work out a value needed to formulate the Biot number which is needed to work out what temperature a bath of water would have to be kept at to cool a specific amount of mild steel in an allotted period of time. (Quite a mouthful, I know)

So far I have worked out the following:

K = 43w/(m.k)
Cp = 620 J/(Kg.K)
SpGr = 7.86ps OR 7860Kg/m3
Cylinder volume = 0.00216 m2
Cylinder Surface = 0.177 m2

Bi = h*Lc/k

h expressed in [W/m2/K] or equivalent units is heat transfer coefficient between the fluid (water) and the body (steel bar), but I cannot work this out for the life of me.

From this I am close to working out the Biot number, but to do so I need the unit surface conductance of my specific size of mild steel, which is worked out by the Thermal transmittance, Φ = A × U × (T1 - T2) ?

I believe I need to calculate the U-Value, which is where I get lost...

Regards,