How Do I Calculate the Cooling Time for Nylon Pellets in a Heat Transfer System?

In summary, the speaker is working on a heat transfer design and has data on the materials and properties involved. They are trying to calculate how long it will take for one material to cool down using the lump mass approximation method and the assumption of conductive heat transfer. However, their calculations do not make sense and they are unsure what to do next. They are seeking assistance with solving the problem.
  • #1
JMorris
6
0
I am working on my first heat transfer design and I have almost all of my data, but I do not know what to do with it.

I know the following for the material on each side of a wall:
specific heat
viscosity
thermal conductivity
density
velocity

for the wall in between them I know the following:
thickness
mean surface area
thermal conductivity

i understand all about Reynalds, Nusselts, and Prandtl number as well.

i also know the temperature that both substances will initially be.

i am trying to figure out how long in time it will take to get one substance (nylon in solid pellet form) to drop from 350 F to 100 F. this nylon will be in a tube with an auger inside moving at a slow rate to transport through the system. On the other side of the wall in the tube will be cold water around 40 F that will be circulating in order to draw the heat out of the nylon.

if I can figure out the temperature of either side of the wall, I can solve for how long my tube needs to be through lump mass approx method.
Here are some values I have if I attempt to solve for the wall temp assuming it is convective heat transfer:

For Nylon
Cp specific heat capacity 1670 J/kg-K
k thermal conductivity 0.25 W/m-K
v velocity 1.524 m/hr (rate of auger)
p density 560.645 kg/m^3
R resistance 102317.0565
u viscosity ?? it is around for 1E+12 kg/m-s for solids but this gives me a "h" that doesn't make sense when i solve for it using the Nusselt number.

so that didnt seem to work so instead i assume it is conductive heat transfer and use the following:

k thermal conductivity 0.25 W/m-K
x thickness 0.889 m
A Surface Area 17.01674016 m^2
R resistance 0.208970694

but again this does not make sense cause i end up getting a temperature on the S.S. wall of less than 41 F when it is next to 350 F, and the water temp on the other side of the wall is 40 F.

what gives?

and thanx.
 
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  • #2
It sounds like you are trying to model the heat transfer between the two materials as a conduction process. The most direct way to approach this would be to use Fourier's Law of Heat Conduction, which states that the rate of heat transfer is proportional to the difference in temperature and the thermal conductivity of the material. To calculate the time it takes for the nylon to drop from 350 F to 100 F, you would need to calculate the rate of heat transfer between the two materials, and then divide the amount of energy needed to cool the nylon by the rate of heat transfer. You can then solve for the time it will take for the nylon to cool from 350 F to 100 F.
 
  • #3



Hi there,

It seems like you have a good understanding of the different parameters involved in heat transfer and have done some calculations already. However, it is important to note that heat transfer is a complex process and there are multiple factors that can affect the results. It is also important to have a clear understanding of the problem at hand and the assumptions made in your calculations.

To start, it would be helpful to have a clear diagram or description of your heat transfer system. From what you have described, it seems like you have a tube with an auger inside, where one side of the wall is in contact with hot nylon pellets and the other side is in contact with cold water. It would also be helpful to know the dimensions and materials of the tube and the auger, as well as the flow rate of the water and the rate at which the auger is moving the nylon pellets.

Additionally, it is important to have a clear understanding of the boundary conditions. Is the heat transfer considered steady-state or transient? Are there any external heat sources or sinks? Are there any assumptions made about the convective heat transfer coefficients on either side of the wall?

Once you have a clear understanding of the problem and the assumptions made, you can use appropriate equations and correlations to solve for the temperature at different points in the system and the time it takes for the nylon pellets to reach a certain temperature. It is also helpful to check your calculations using different methods, such as the lumped mass approximation method or finite difference methods.

In general, if you are still having trouble with your calculations, it would be best to consult with a heat transfer expert or a professor in your field for guidance. They may be able to provide valuable insights and suggestions for your design.

Good luck with your project!
 

1. What is heat transfer?

Heat transfer is the movement of thermal energy from one location to another. It occurs in three ways: conduction, convection, and radiation. Conduction is the transfer of heat through a solid material, convection is the transfer of heat through a fluid medium, and radiation is the transfer of heat through electromagnetic waves.

2. How does heat transfer affect everyday life?

Heat transfer plays a crucial role in our everyday lives. It is responsible for keeping our homes warm, cooking our food, and even regulating our body temperature. It also has a significant impact on industrial processes, such as power generation and manufacturing.

3. What are the factors that affect heat transfer?

The rate of heat transfer is affected by several factors, including the temperature difference between the two objects, the thermal conductivity of the materials, the surface area of contact, and the distance between the objects. Other factors, such as the type of material and the presence of insulation, can also influence heat transfer.

4. How is heat transfer calculated?

The rate of heat transfer is calculated using the equation Q = kA(T2 - T1)/d, where Q is the heat transferred per unit time, k is the thermal conductivity, A is the surface area of contact, T2 and T1 are the temperatures of the two objects, and d is the distance between them. This equation is known as Fourier's Law.

5. What are some real-world applications of heat transfer?

Heat transfer has many practical applications, such as heating and cooling systems, refrigeration, and cooking. It is also essential in industries such as metallurgy, chemical engineering, and aerospace, where precise control of heat transfer is necessary for various processes. In addition, heat transfer is crucial in fields like medicine and biology, where maintaining a stable body temperature is vital for human health.

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