Heat Transfer Help: Estimating Cooling Capability w/ Q=mcDT

In summary: I did not expect any direct answeres but this is a very nice site.In summary, the conversation discusses a material that caught on fire during the filtration process, causing the deluge system to activate. The speaker wants to know if there are any basic ways to prove that their current fire suppression system is adequate. They mention calculating the heat needed to cool the material and considering other theories and formulas. The material has known properties such as mass, volume, and density. The conversation ends with the speaker seeking help in calculating the mass of water required to cool the material to a safe temperature and mentioning they have not yet consulted the MSDS for more information.
  • #1
William12
8
0
OK so I have a little issue I'm dealing with.. here are the basics..

There is a material that was filtered and came out of the filtration process at about 330 degrees (it comes out as a solid filter cake). It somehow caught on fire which caused the deluge system to activate.

I just want to know if there are any basic ways of proving that our current system is adequate.

Here is what I thought about doing. Calculating the heat that would need to be "lost" in cooling the cake from 330 degrees to some safe temperature by using Q=mcDT and then somehow seeing if our current volumetric flow rate of water can provide that capability.

Are there any other BASIC (im not trying to make it too difficult) theories or formulas I could use?

The filtered material has known properties such as mass, volume, density... etc
 
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  • #2
hBasically I want Help with calculating mass of water required to drop the temperature of the cake from 330 to another "safe" temperature
 
  • #3
William12 said:
OK so I have a little issue I'm dealing with.. here are the basics..

There is a material that was filtered and came out of the filtration process at about 330 degrees (it comes out as a solid filter cake). It somehow caught on fire which caused the deluge system to activate.

I just want to know if there are any basic ways of proving that our current system is adequate.

Here is what I thought about doing. Calculating the heat that would need to be "lost" in cooling the cake from 330 degrees to some safe temperature by using Q=mcDT and then somehow seeing if our current volumetric flow rate of water can provide that capability.

Are there any other BASIC (im not trying to make it too difficult) theories or formulas I could use?

The filtered material has known properties such as mass, volume, density... etc

Welcome to the PF.

What do you mean by "somehow caught on fire"? Did something else catch it on fire, or was the material above its combustion temperature? Can you say what the material is? What is listed in Section 5 of the MSDS for this material (that is the Fire Fighting section of the MSDS)?

When you say you want to "prove that your current system is adequate", which system? The filtration system? The fire suppression system? Some new system that you want to incorporate that will cool the material below its combustion temperature before exposure to air?
 
  • #4
I want to prove the fire suppression system is adequate. To be honest, we are not sure why it caught on fire, some say static accumilation, some say autoignition. We are currently investigating. I have not had a chance to view the msds yet so I couldn't really tell you what the filtered material is besides that it contains celite
 
  • #5
berkeman said:
Welcome to the PF.

What do you mean by "somehow caught on fire"? Did something else catch it on fire, or was the material above its combustion temperature? Can you say what the material is? What is listed in Section 5 of the MSDS for this material (that is the Fire Fighting section of the MSDS)?

When you say you want to "prove that your current system is adequate", which system? The filtration system? The fire suppression system? Some new system that you want to incorporate that will cool the material below its combustion temperature before exposure to air?
See above
 
  • #6
I'm not sure we can help you prove anything about the fire suppression system. We certainly are not experts on HAZMAT fires and their suppression. Do you have any HAZMAT qualified experts accessible to you at your work? You may be able to get some help through your local OSHA office, or your local fire department's HAZMAT team.
 
  • #7
berkeman said:
I'm not sure we can help you prove anything about the fire suppression system. We certainly are not experts on HAZMAT fires and their suppression. Do you have any HAZMAT qualified experts accessible to you at your work? You may be able to get some help through your local OSHA office, or your local fire department's HAZMAT team.
Thanks for the support
 

1. How do you calculate the cooling capability with the equation Q=mcDT?

To calculate the cooling capability, you will need to know the mass of the substance (m), its specific heat capacity (c), and the change in temperature (DT). Simply plug these values into the equation Q=mcDT and solve for Q, which represents the amount of heat transferred.

2. What does each variable in the Q=mcDT equation represent?

The variable Q represents the amount of heat transferred, measured in joules (J). The variable m represents the mass of the substance in kilograms (kg). The variable c represents the specific heat capacity of the substance in joules per kilogram per degree Celsius (J/kg°C). The variable DT represents the change in temperature, measured in degrees Celsius (°C).

3. How does heat transfer help with estimating cooling capability?

The equation Q=mcDT is a useful tool for estimating the cooling capability of a substance because it allows you to calculate the amount of heat that will be transferred based on the properties of the substance and the change in temperature. This can help you determine the most efficient way to cool a substance, such as choosing the appropriate cooling equipment or adjusting the rate of cooling.

4. Can the Q=mcDT equation be used for all types of substances?

Yes, the Q=mcDT equation can be used for all types of substances as long as you have the necessary information for each variable. However, it is important to note that the specific heat capacity (c) may vary for different substances, so it is important to use the correct value for the substance you are working with.

5. How does the Q=mcDT equation relate to the laws of thermodynamics?

The Q=mcDT equation is based on the first law of thermodynamics, which states that energy cannot be created or destroyed, only transferred or converted from one form to another. In this case, the equation helps us understand how heat energy is transferred between substances and how it can be used for cooling or other purposes.

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