The top of an insulated cylindrical container is covered by a disc having emissivity 0.6 and conduct

Q: Why do we have to convert temperatures to Kelvin for the radiation formula, but we can use Celsius for the conduction formula?

In the Stefan-Boltzmann law, the temperature is raised to the fourth power ($T^4$). The ratio or difference of fourth powers depends on the absolute zero scale, so Kelvin is mandatory. However, in the conduction formula, we only use the temperature difference ($\Delta T$). A difference of $1^\circ\text{C}$ is exactly the same as a difference of $1 \text{ K}$, so using Celsius is perfectly fine and often easier.

Q: What does 'steady state' actually mean in this context?

Steady state means that the temperatures at any given point in the system are no longer changing with time. For this to happen, the energy entering a region must exactly equal the energy leaving it. Here, the heat entering the top surface via conduction equals the heat leaving it via radiation.

Q: Why did we ignore convection?

The problem explicitly states to 'Neglect the heat loss due to convection.' In reality, a hot surface exposed to air would also lose heat by warming the air molecules that touch it, which then rise (convection). Ignoring it simplifies the problem to just conduction and radiation.

Q: What if the disc was made of a material with higher thermal conductivity?

If $K$ was higher, the disc would conduct heat more easily. To maintain the same radiation loss (and thus the same top surface temperature), the required temperature difference across the disc would be smaller. This means the oil wouldn't need to be as hot.

Q: Is the temperature of the oil uniform?

We assume it is. The problem mentions 'circulating oil', which implies that the fluid is constantly moving and mixing, keeping its temperature relatively uniform throughout the container.

Question

The top of an insulated cylindrical container is covered by a disc having emissivity 0.6 and conductivity 0.167 W/Km and thickness 1 cm. The temperature is maintained by circulating oil as shown.
(a) Find the radiation loss to the surroundings in $	ext{W/m}^2$ if temperature of the upper surface of disc is $127^\circ	ext{C}$ and temperature of surroundings is $27^\circ	ext{C}$.
(b) Also find the temperature of the circulating oil. Neglect the heat loss due to convection.
(Given, $\sigma = \frac{17}{3} 	imes 10^{-8} 	ext{ Wm}^{-2}	ext{K}^{-4}$)

Sigma Percentile · Accepted Answer

This problem is a beautiful demonstration of energy balance in a steady-state thermal system. We have a container of hot oil, and heat is escaping through a disc at the top. The heat must first conduct through the solid disc and then radiate away from its surface into the cooler surroundings. Let's break down this journey of heat step-by-step.

### Part (a): Calculating the Radiation Loss

The top surface of the disc is exposed to the surroundings. Because it's hot, it radiates energy. The net rate of heat radiated per unit area (which we can call the intensity $I$) is governed by the **Stefan-Boltzmann Law**:

$$I = e\sigma (T^4 - T_0^4)$$

Here, $e$ is the emissivity of the surface, $\sigma$ is the Stefan-Boltzmann constant, $T$ is the absolute temperature of the radiating surface, and $T_0$ is the absolute temperature of the surroundings.

Before we plug in the numbers, we must convert our temperatures from Celsius to Kelvin. This is a common trap!

$$T = 127^\circ	ext{C} + 273 = 400 	ext{ K}$$
$$T_0 = 27^\circ	ext{C} + 273 = 300 	ext{ K}$$

Now, let's substitute the given values into our radiation equation:

$$I = 0.6 	imes \left(\frac{17}{3} 	imes 10^{-8}ight) 	imes (400^4 - 300^4)$$

Calculating the fourth powers:

$$400^4 - 300^4 = (256 	imes 10^8) - (81 	imes 10^8) = 175 	imes 10^8 	ext{ K}^4$$

Substituting this back:

$$I = 0.6 	imes \frac{17}{3} 	imes 10^{-8} 	imes 175 	imes 10^8$$

The $10^{-8}$ and $10^8$ elegantly cancel each other out.

$$I = 0.2 	imes 17 	imes 175 = 595 	ext{ W/m}^2$$

So, the top surface is losing energy at a rate of **$595 	ext{ W/m}^2$**.

### Part (b): Finding the Oil Temperature

Now, where is this radiated heat coming from? It's coming from the hot oil below the disc. For the top surface to maintain a constant temperature of $127^\circ	ext{C}$, the heat it radiates away must be exactly replenished by the heat conducted up from the oil. This is the essence of a **steady state**.

The rate of heat conduction per unit area through the disc is given by Fourier's Law of Heat Conduction:

$$\frac{Q}{At} = \frac{K(	heta - T)}{l}$$

Where $K$ is the thermal conductivity, $	heta$ is the temperature of the oil (the hotter side), $T$ is the temperature of the top surface (the cooler side), and $l$ is the thickness of the disc.

Equating the conduction rate to the radiation rate we just found:

$$\frac{K(	heta - T)}{l} = I$$

Let's plug in the known values. Remember to convert the thickness $l = 1 	ext{ cm}$ to meters ($0.01 	ext{ m}$).

$$\frac{0.167 	imes (	heta - 127)}{0.01} = 595$$

Notice that we can keep the temperatures in Celsius here because a temperature difference in Celsius is numerically equal to a temperature difference in Kelvin.

Rearranging to solve for $	heta$:

$$	heta - 127 = \frac{595 	imes 0.01}{0.167}$$

$$	heta - 127 = \frac{5.95}{0.167} \approx 35.63^\circ	ext{C}$$

$$	heta = 127 + 35.63 = 162.63^\circ	ext{C}$$

The temperature of the circulating oil must be maintained at approximately **$162.6^\circ	ext{C}$** to keep the system in this steady state.

Visualized Solution

Visualizing the Setup

Stefan-Boltzmann Law for Radiation

Substituting Values for Radiation

Calculating Radiation Loss

Steady State Heat Conduction

Setting up the Conduction Equation

Solving for Oil Temperature

Final Conclusion

The Sigma Insight: Heat Transfer

Part (a)

Part (b)

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