# Radiation heat transfer - mistake in attached solution?

• pyroknife
In summary, the conversation discusses a problem where the answer is off by a magnitude of 10 compared to the solutions provided. The equation and values are given, and it is suggested that there may be a mistake in the calculation.
pyroknife
I attached the problem of interest (probably the only part you guys need to see is the equations at the bottom).

My answers are off by a magnitude of 10 compared to the solutions. (copy and pasted it below since the pic is a little hard to see at the bottom)
q = σε f*A(T^4 − Ts^4 )
= 5.67 Χ 10^−8 Χ ε Χ 1 Χ (π Χ 0.1^2 )(T^4 − 77^4 )

= 1.7813 Χ 10 −8 ε (T^4 − 77^4 )

When they did 5.67*10^-8 * 1 * pi * (.1)^2, they got 1.7813*10^-8.
I plugged that into my calculator, but I got 1.7813*10^-9.

Normally, I would think this was a typo, but in the next part (not included in image), they plotted values that were a magnitude of 10 different from mine. This has me thinking that I might have missed something.

What do you guys think?

#### Attachments

• Untitled.png
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Am I missing something or is it a typo? It's likely that you are missing something. The equation you posted can be written as q = σε f*A(T^4 − Ts^4 ), where σ is the Stefan-Boltzmann constant, ε is the emissivity, f is the view factor, A is the area of the surface, T is the temperature, and Ts is the temperature of the surroundings. By substituting in the appropriate values for each variable, one can arrive at the solution given. It's possible that you made an error when substituting in one of the values.

## 1. What is radiation heat transfer?

Radiation heat transfer is the transfer of thermal energy through electromagnetic waves. It does not require a medium to propagate, unlike conduction and convection heat transfer.

## 2. What are the factors that affect radiation heat transfer?

The factors that affect radiation heat transfer include the temperature difference between two objects, the emissivity of the objects, and the distance between them. Other factors such as surface properties and the presence of intervening materials can also affect radiation heat transfer.

## 3. What is the mistake in the attached solution for radiation heat transfer?

The mistake in the attached solution could be due to incorrect calculation of the Stefan-Boltzmann constant, emissivity, or temperature difference. It could also be caused by using the wrong formula or making a calculation error.

## 4. How can we accurately calculate radiation heat transfer?

To accurately calculate radiation heat transfer, we need to use the correct formula and ensure that all relevant parameters such as the Stefan-Boltzmann constant, emissivity, and temperature difference are accurately measured and accounted for. It is also important to double check calculations for any errors.

## 5. Is radiation heat transfer dangerous?

Radiation heat transfer is a natural process that occurs all around us. It is only dangerous if the radiation levels are extremely high, such as in the case of nuclear radiation. In most cases, the amount of radiation heat transfer is not harmful to humans.

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