# Diffusion across surface with constant temperature

• Nikstykal
In summary, to increase the carbon content of a Ti Rod from 0.2%wt to 0.5%, the temperature of the furnace would need to be around 763 Kelvin. This can be calculated using equations 1 and 3, and by solving for the diffusion coefficient (D) and temperature (T). The final carbon content at a depth of 44 mm and after 48 hours can be verified using equation 2. It is important to use consistent units and the correct values for x and t in the calculations.
Nikstykal

## Homework Statement

A Ti Rod is to be put into a furnace to try and increase the carbon content of the rod. Initially, the carbon content of the rod is 0.2%wt. The carbon content of the furnace is 1.0%wt. What would the temperature have to be in order to get a final carbon content in the Ti rod to be 0.5%, using a depth of 44 mm over 48 hours.

Given: erf(k)=z value table, R = 8.314 J/mol*k, Di = 3.2x10^-3 cm2/s, Q = 83.862 KJ

## Homework Equations

1: J=-D dC/dx
2: (C(x,t) - Ci)/(Cs - Ci) = 1 - erf(x/2√(DT))
3: D = Di*e^(-Q/RT)

## The Attempt at a Solution

My thought process to solve this was to use equation 2 to solve for a value of D. I set Cx= 0.5, Ci=0.2, and Cs=1.0. Then I rearranged the equation so erf(x/2√(Dt)) was on its own side and looked at the erf value table to match my erf value with the table's to find out what x/(2√(Dt)) has to equal. It wasn't given so I had to interpolate, which I ended up with 0.627.

Once found I could then plug in 0.04cm for x and 1.728*10^5 seconds for t to solve for D. I think it was around 5.92*10^-9 cm2/s.

Rearranging equation 3 to solve for T, I get T = Q/R * (1 /( ln Di - ln D) which I got 763 Kelvin.

The one thing I am really unsure of is if I can leave the carbon content in %wt and if my order of evaluation is correct. Thanks.

Hello,

Thank you for your response. Your approach is correct, and your calculations for D and T seem to be accurate. However, there are a few things to keep in mind:

1. The carbon content in %wt can be used as long as all other units are consistent. In this case, since the diffusion coefficient is given in cm2/s, the depth should also be in cm and the time should be in seconds.

2. You have solved for the diffusion coefficient at a specific depth (x = 0.04 cm) and time (t = 1.728*10^5 seconds). However, the final carbon content of 0.5% is desired at a depth of 44 mm (4.4 cm) after 48 hours (1.728*10^5 seconds). So, you will need to use the appropriate values for x and t in equation 2 to solve for D.

3. The order of evaluation seems to be correct, but just to be sure, you can double-check your calculations by plugging in the values of D and T into equation 2 and solving for the final carbon content at a depth of 44 mm and after 48 hours. The result should be close to 0.5%.

Overall, your approach and calculations seem to be correct. Keep in mind the units and make sure to use the appropriate values for x and t in equation 2. Good luck!

## What is diffusion?

Diffusion is the process of particles, such as molecules, moving from an area of higher concentration to an area of lower concentration. This is a result of random motion and does not require an external force.

## What is surface diffusion?

Surface diffusion is a type of diffusion that occurs only on the surface of a material. It is the movement of particles along the surface, rather than through the bulk of the material.

## How does temperature affect diffusion?

Increasing the temperature will generally increase the rate of diffusion. This is because higher temperatures provide more energy to the particles, causing them to move faster and therefore diffuse more quickly.

## What is the role of a constant temperature in diffusion?

A constant temperature provides a stable environment for diffusion to occur. This allows for accurate measurements and observations of the diffusion process without the interference of temperature fluctuations.

## Why is diffusion across a surface with constant temperature important?

Diffusion across a surface with constant temperature is important for understanding and controlling various processes in different industries, such as in materials science and chemical engineering. It is also relevant in biological systems, such as the diffusion of oxygen and nutrients in cells.

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