Question

Consider an opaque plate that is well insulated on the edges and it is heated at the bottom with an electric heater. The plate has an emissivity of \(0.67\), and is situated in an ambient surrounding temperature of \(7^{\circ} \mathrm{C}\) where the natural convection heat transfer coefficient is \(7 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). To maintain a surface temperature of \(80^{\circ} \mathrm{C}\), the electric heater supplies \(1000 \mathrm{~W} / \mathrm{m}^{2}\) of uniform heat flux to the plate. Determine the radiosity of the plate under these conditions.

Short answer

Answer: To find the radiosity of the plate, follow these steps: 1. Calculate the heat transfer by convection (q_conv) using the formula: q_conv = h * (T_s - T_inf) 2. Apply the energy balance equation: q_heater = q_conv + q_rad 3. Find the emissive power (E_b): q_rad = ε * σ * (T_s^4 - T_inf^4) 4. Calculate the radiosity (J): J = E_b + (q_conv / emissivity) By plugging in the given values and performing calculations according to the solution steps above, the radiosity of the plate can be determined.

Step-by-step solution

Calculate the heat transfer by convection

Calculate the heat transfer by convection between the plate and ambient surroundings using the natural convection heat transfer coefficient: \(q_{conv} = h \cdot (T_s - T_\infty)\) where \(h = 7 \: \mathrm{W/m^2 \cdot K}\) (given natural convection heat transfer coefficient) \(T_s = 80^{\circ} \mathrm{C}\) (desired surface temperature) \(T_\infty = 7^{\circ} \mathrm{C}\) (ambient surrounding temperature) Plugging in the values, we will find the heat transfer by convection.

Apply the energy balance equation

Use the energy balance equation to find the relationship between heat losses from the surface by convection and radiation. Let \(q_{rad}\) be the heat transfer by radiation. \(q_{heater} = q_{conv} + q_{rad}\) where \(q_{heater} = 1000 \: \mathrm{W/m^2}\) (heat supplied by the heater) Use the calculated value of \(q_{conv}\) from the previous step and plug it into the equation to find the heat transfer by radiation, \(q_{rad}\).

Find the emissive power

To find the emissive power for the given plate, we need to use the relationship: \(q_{rad} = \epsilon \cdot \sigma \cdot (T^4_s - T^4_\infty)\) where \(\epsilon = 0.67\) (emissivity of the plate) \(\sigma = 5.67 \times 10^{-8} \: \mathrm{W/m^2 \cdot K^4}\) (Stefan-Boltzmann constant) We know \(q_{rad}\) from the energy balance equation. We can rearrange the equation to directly calculate the emissive power \(E_b = \sigma \cdot T^4_s\).

Calculate the radiosity

Finally, we will calculate the radiosity using the following equation: \(J = E_b + \cfrac{q_{conv}}{emissivity} = \sigma \cdot T_s^4 + \cfrac{q_{conv}}{0.67}\) Plug in the values of emissive power \(E_b\) and the calculated heat transfer by convection, \(q_{conv}\), for the given system, to find the radiosity of the plate.

Key concepts

Convection Heat Transfer

Convection heat transfer refers to the process of heat being transferred between the surface of an object and a fluid (like air or water) surrounding it through the motion of the fluid. In the given problem, the opaque plate is influenced by natural convection, where warmer air rises and cooler air comes into contact with the plate in a cyclic manner. This process is quantified using the convection heat transfer coefficient, denoted by the symbol \(h\). The value for this specific situation is \(7 \, \text{W/m}^2 \cdot \text{K}\).
To calculate the amount of heat transferred by convection, we use the formula:\[q_{conv} = h \cdot (T_s - T_\infty)\]where:\\(q_{conv}\) is the convection heat transfer rate (in watts per square meter), \\(T_s\) is the surface temperature of the plate (80°C), and \\(T_\infty\) is the ambient air temperature (7°C).
This relationship tells us how much heat is being lost from the plate due to cold air flowing around it.

Emissivity

Emissivity is a key concept when dealing with thermal radiation. It is a measure of an object's ability to emit energy as thermal radiation, expressed on a scale from 0 to 1. An emissivity of 0 means that the object is a perfect reflector, while an emissivity of 1 indicates that it is a perfect emitter of radiation.
For the opaque plate in the exercise, the emissivity is given as 0.67. This indicates that the plate emits 67% of the radiation compared to an ideal blackbody (a perfect emitter).
Emissivity influences the amount of radiative heat transfer from the plate to the surroundings. When you calculate the radiative heat transfer \(q_{rad}\) using the formula:\[q_{rad} = \epsilon \cdot \sigma \cdot (T^4_s - T^4_\infty)\]\(\epsilon\) represents emissivity, and \(\sigma = 5.67 \times 10^{-8} \, \text{W/m}^2 \cdot \text{K}^4\) is the Stefan-Boltzmann constant, representing the potential power output per unit area for a blackbody at a certain temperature.
Understanding emissivity helps us comprehend how efficiently the plate radiates heat compared to the theoretical maximum.

Energy Balance

The principle of energy balance is crucial for understanding how energy conservation plays out in thermal systems. In the context of the heated plate, energy balance involves equating the energy supplied to the energy lost.
The equation used is:\[q_{heater} = q_{conv} + q_{rad}\]Here, \(q_{heater}\) is the energy input from the electric heater, which is \(1000 \, \text{W/m}^2\). It is balanced by the convection heat transfer \(q_{conv}\) and the radiation heat transfer \(q_{rad}\), collectively making up the total heat loss from the surface of the plate.
Achieving an energy balance ensures that the temperature remains steady at 80°C, as the energy supplied equals the energy dissipated through convection and radiation. This balance is vital in systems engineering to predict and maintain desired thermal conditions.

Radiosity

Radiosity is a fundamental concept in radiative heat transfer, representing the total radiation leaving a surface per unit area. It is a combination of emitted radiation and reflected radiation from a surface.
To calculate the radiosity \(J\) of the plate, the following formula is used:\[J = E_b + \cfrac{q_{conv}}{\text{emissivity}}\]Where \(E_b\) is the emissive power given by \(\sigma \cdot T_s^4\) and \(\text{emissivity}\) is the same emissivity factor affecting how much of \(q_{conv}\) is reflected back as part of the radiosity.
By understanding radiosity, students can appreciate how radiation and convection interact, offering a comprehensive view of the energy transfer from a surface under given conditions. This approach is vital for calculating the radiative characteristics of systems in thermal management applications.