What is the Cooling Time for Steel Tubes with Ambient Nitrogen Gas Flow?

In summary: I don't know what to do. I've been trying to do the math and it just doesn't seem to work.In summary, Dave is trying to calculate the cooling rate of Steel using Newton's Law of Cooling. He has a load of Steel Tubes and is trying to figure out how long it will take for the Load to cool to a certain temperature while flowing 150 SCFH of ambient temperature nitrogen gas. He is also trying to calculate the KW lost in the process. Unfortunately, the calculations don't seem to work.
  • #1
davejc
4
0
TL;DR Summary
Trying to calculate cooling time for a load of steel tubes from 1100ºF using ambient temperature Nitrogen Gas.
I am working on a new application and am trying to figure out how I can calculate the cooling rate of Steel.

I have a load of Steel Tubes. 1200lb total mass. Tubes have some variance on size depending on what is being run on a given day, but nominally, we are looking at 1" OD x .1875" Wall x 17" Long Steel Tube.

Max Temp of the Load is 1100ºF.

What I am trying to come up with a calculation for is how long it will take for that load to cool to <500ºF while flowing 150 SCFH of ambient temperature nitrogen gas. I cannot just open the door and blow ambient air as it will cause the parts to oxidize.I'm certain this is not as easy as just calculating heat loss based on radiation and I'm also certain that I haven't given enough information. We typically only deal with the heatup portion of these projects, but working with a new customer who wants to calculate cooling time.

Thank you

- Dave
 
Engineering news on Phys.org
  • #2
What you've provided isn't a bad start. The way I'd approach it is to numerically model Newton's Law of Cooling using a spreadsheet. If that lights the bulb let me know how it goes, otherwise I can talk you through that. A couple of points:

1. Theoretically speaking you never get to ambient temperature, but you can decide how close and calculate how long it takes to get there.
2. Is this in a closed, insulated container? In other words, are there other avenues of heat loss? Radiation?
 
  • Like
Likes berkeman
  • #3
It is inside a heat treating furnace. So, I will have heat radiating off the Load and off the Furnace Wall.

I know I will never get the load to ambient temperature. The next step of the process requires that the parts be below 500ºF.

I am looking over Newtons Law of Cooling. Calculus was quite a few years ago, I will gladly accept any tips you have.
 
  • #4
davejc said:
I am looking over Newtons Law of Cooling. Calculus was quite a few years ago, I will gladly accept any tips you have.
Newton's Law of Cooling states that the heat transfer rate is proportional to temperature difference. So as the tube cools, the heat transfer drops and the rate of cooling slows. That's a differential equation that makes calculating the temperature at a time (t) in the future difficult.

But if you use a spreadsheet you can approximate it numerically.
  • First column: time
  • Column 2: Sample temp
  • Column 3: Heat transfer rate (m*Cp*ΔT) [that's mass flow rate of air]
  • Column 4: Temperature change in the time interval
Pick an interval that's small enough for a relatively smooth curve. Probably should also pick an effectiveness fraction for the heat transfer.

It is inside a heat treating furnace. So, I will have heat radiating off the Load and off the Furnace Wall.
You mean the air is flowing through the furnace, cooling both the sample and the furnace? If the sample and furnace are radiating against each other at the same temp, but if the furnace wall has to be cooled, that will impact the calcs...
 
  • #5
Yes, the part as well as the Furnace Wall will be radiating heat. The Heating Elements are not giving off any energy, but the insulation is holding some heat in it.

Trying to wrap my head around the Excel Setup, and I probably did this wrong.

Colum 1 = Time
Column 2 = Current Temp
Column 3 = Heat Transfer
Column 4 = KW Change
Column 5 = New Temp

I am using 1 Minute as my Time Base
Start Temp is 1100ºF.
I calculated that I put 43.47kw into the material to heat it to 1100. kw = (1200lb * 0.12 [Specific Heat of Steel] * 1030 [Temperature Rise]) / 3412 [BTU > kw Conversion]

Heat Transfer I am calculating as Q = ((M X Cp x dT) * 60) / 3412. M = 0.182lb/min of Nitrogen. Cp = 0.248btu/lbºF. dT = Current Temp - 70. This gives me btu/min. X 60 = BTU/hr. /3412 = KW.

KW Change = 43.47 - the Heat Transfer

If I interpret this as the KW lost in 1 minute, then plug it back into the initial KW calculation but solve for DeltaT. New Temperature = (KW Change * 3412) / (1200 * 0.12). This becomes the Sample Temp for the next row.

So row 1
Sample Temp = 1100. Heat Transfer = 0.818kw. KW Change = 43.47 - 0.818 = 42.653. New Temp = 1010.629.
Row 2:
Sample Temp - 1010.629. Heat Transfer = 0.747. KW change = 41.906. New Temp = 992.939.

This tells me that it'll take 42 minutes to get below 500ºF, which can't be correct.
 
  • #6
davejc said:
This tells me that it'll take 42 minutes to get below 500ºF, which can't be correct.
I got 47min. I think you may have done some extra steps to find some of the constants instead of googling them? Otherwise, maybe some rounding errors between us. But close enough for a first pass.

What strikes me is the airflow is really low. Did you really mean CFH? CFM?

CoolingSteel.jpg
 
Last edited:
  • #7
It may be some rounding errors, we have close enough answers.

Where I think we are not correct is that there is no account for the fact that the Furnace Wall is radiating heat as well. The heat capacity of the Nitrogen is not exclusively pulling heat from the Steel, it is also pulling from the fire brick.

The Nitrogen Flow rate is 150 CFH. The Furnace Chamber is only 27cuft. We flowing ~5 furnace volumes per hour.

I am trying to get some more information from my customer on this. I am expecting the cooling time to be somewhere in the 5hr range. Even if I reduce the Gas Flow rate to 25% its current value (simulating that the only 25% of the capacity of the Nitrogen is going to the steel), I am only at a little over 3hrs.

I do thank you for the help thus far, Russ.
 
  • #8
@russ_watters are you sure that your spreadsheet is right? I get very similar numbers and 46 minutes to 500F, but only if I assume 150 cfm (9000 ft3/hour). With the given 150 cfh I get a first-minute heat removal of 46.3 Btu, for a steel temperature change of 0.32F. Carrying it to 500F takes 2800 minutes (47 hours). Maybe I have a stray factor of 60 in my spreadsheet.
 

Related to What is the Cooling Time for Steel Tubes with Ambient Nitrogen Gas Flow?

1. What is the cooling time for steel tubes?

The cooling time for steel tubes varies depending on the size and thickness of the tube, as well as the cooling method used. Generally, it can take anywhere from a few hours to several days for steel tubes to cool completely.

2. How does the cooling time for steel tubes affect their strength?

The cooling time for steel tubes can have a significant impact on their strength. If the tubes are cooled too quickly, they may become brittle and prone to cracking. On the other hand, if they are cooled too slowly, they may not reach their full strength potential.

3. What factors influence the cooling time for steel tubes?

The cooling time for steel tubes is influenced by several factors, including the initial temperature of the tube, the ambient temperature, the type of cooling method used, and the composition of the steel itself. Thicker and larger tubes will also take longer to cool compared to thinner and smaller tubes.

4. Can the cooling time for steel tubes be accelerated?

Yes, the cooling time for steel tubes can be accelerated through the use of cooling methods such as quenching or air cooling. These methods involve rapidly lowering the temperature of the tubes, which can reduce the cooling time significantly.

5. Why is the cooling time for steel tubes important?

The cooling time for steel tubes is important because it can affect the overall quality and properties of the tubes. If the tubes are not cooled properly, they may not meet the required strength and durability standards, which can lead to potential safety hazards and issues in their intended use.

Similar threads

  • Mechanical Engineering
Replies
6
Views
1K
Replies
7
Views
1K
  • Other Physics Topics
Replies
3
Views
2K
  • Mechanical Engineering
Replies
3
Views
1K
Replies
3
Views
2K
Replies
7
Views
1K
Replies
3
Views
1K
Replies
3
Views
1K
Replies
9
Views
6K
Replies
3
Views
3K
Back
Top