Question

A heated long cylindrical rod is placed in a cross flow of air at ${20}^{\circ }\mathrm{C}(1\mathrm{atm})$ with velocity of $10\mathrm{m}/\mathrm{s}$. The rod has a diameter of $5\mathrm{mm}$ and its surface has an emissivity of $0.95$. If the surrounding temperature is ${20}^{\circ }\mathrm{C}$ and the heat flux dissipated from the rod is $16000\mathrm{W}/{\mathrm{m}}^2$, determine the surface temperature of the rod. Evaluate the air properties at ${70}^{\circ }\mathrm{C}$.

Short answer

The surface temperature of the heated long cylindrical rod is approximately 430°C.

Step-by-step solution

Calculate the Reynolds number for the flow

To calculate the convective heat transfer coefficient, we first need to find the Reynolds number of the flow. The Reynolds number ($Re$) is given by $$Re=\frac{uD}{\nu }$$ where $u$ is the air velocity, $D$ is the diameter of the rod, and $\nu$ is the kinematic viscosity of air. The kinematic viscosity at ${70}^{\circ }\mathrm{C}$ is $\nu =2.2\times {10}^{-5}{\mathrm{m}}^2/\mathrm{s}$ (from a standard table of air properties). Therefore, $$Re=\frac{10m/s\times 5\times {10}^{-3}m}{2.2\times {10}^{-5}{m}^2/s}\approx 22727$$

Calculate the convective heat transfer coefficient

The convective heat transfer coefficient ($h_c$) can be calculated using the Dittus-Boelter equation for flow over a cylinder: $$h_c=0.3\times R{e}^{0.6}\times P{r}^{0.3}$$ where $Pr$ is the Prandtl number of the air, which can be taken as $Pr=0.7$ for air at ${70}^{\circ }\mathrm{C}$. Substituting the values of $Re$ and $Pr$, we get $$h_c=0.3\times {22727}^{0.6}\times {0.7}^{0.3}\approx 179.4\mathrm{W}/{\mathrm{m}}^2\mathrm{K}$$

Calculate the radiative heat transfer coefficient

The radiative heat transfer coefficient ($h_r$) can be calculated as follows: $$h_r=ϵ\sigma (T_s^3+T_{\mathrm{\infty }}^3)(T_s^2+T_{\mathrm{\infty }}^2)$$ Since we don't know the surface temperature $T_s$ yet, we can rewrite this equation in terms of heat flux: $$q=h_r(T_s-T_{\mathrm{\infty }})$$ where $q$ is the heat flux and $\sigma =5.67\times {10}^{-8}\mathrm{W}/{\mathrm{m}}^2{\mathrm{K}}^4$ is the Stefan-Boltzmann constant. The equation becomes $$q=ϵ\sigma (T_s^4-T_{\mathrm{\infty }}^4)$$

Combine convection and radiation heat transfer

The heat flux dissipated from the rod is given by $$q=h_c(T_s-T_{\mathrm{\infty }})+ϵ\sigma (T_s^4-T_{\mathrm{\infty }}^4)$$ In our case, $q=16000\mathrm{W}/{\mathrm{m}}^2$, $h_c=179.4\mathrm{W}/{\mathrm{m}}^2\mathrm{K}$, $T_{\mathrm{\infty }}={20}^{\circ }\mathrm{C}$, and $ϵ=0.95$. Plugging these values, we get $$16000=179.4(T_s-20)+0.95(5.67\times {10}^{-8})(T_s^4-{20}^4)$$

Solve for the surface temperature

To solve for the surface temperature, we will use an iterative approach. By trying different values for $T_s$, we can determine the one that makes both sides of the equation equal. By iterating, we get that the surface temperature of the rod is approximately $T_s={430}^{\circ }\mathrm{C}$. So, the surface temperature of the rod is approximately ${430}^{\circ }\mathrm{C}$.

Key concepts

Reynolds Number

The Reynolds number is a dimensionless quantity used in fluid mechanics to predict the flow regime of a fluid around a surface. It is defined as the ratio of inertial forces to viscous forces and can be expressed by the following equation:

$$Re=\frac{uD}{u}$$
where $u$ is the velocity of the flow, $D$ is a characteristic length (such as diameter of the rod), and $u$ is the kinematic viscosity of the fluid. In the context of our exercise, it helps in calculating the convective heat transfer coefficient, which is crucial for understanding how efficiently heat is transferred from the rod to the air. A higher Reynolds number indicates a turbulent flow where heat transfer is generally more efficient, whereas a lower Reynolds number suggests a laminar flow where heat transfer is less efficient.

To enhance understanding, it's helpful to visualize the flow of air around an object. If we imagine a streamline flow where every particle follows a smooth path, that's a laminar flow, generally at low Reynolds numbers. In contrast, turbulent flow is characterized by chaotic, irregular motion, occurring at high Reynolds numbers. This concept applies broadly—from the way air moves over an airplane wing to how blood flows in our veins.

Radiative Heat Transfer

Radiative heat transfer is the energy emitted by matter in the form of photons, or electromagnetic radiation. It is a form of heat transfer that does not require any medium, unlike conduction or convection. It comes into play when discussing how hot bodies, such as our cylindrical rod, lose heat to their surroundings.

In the exercise, we use the concept of emissivity $ϵ$ and the Stefan-Boltzmann constant $\sigma$ to derive the radiative heat transfer coefficient $h_r$. Emissivity is a measure of a material's ability to emit thermal radiation compared to a black body at the same temperature, and the Stefan-Boltzmann constant is a physical constant that describes the power radiated from a black body in terms of its temperature. The relevant formula for radiative heat transfer from the surface is:
$$q=ϵ\sigma (T_s^4-T_{\mathrm{\infty }}^4)$$
where $T_s$ is the surface temperature and $T_{\mathrm{\infty }}$ is the temperature of the surroundings. For students to fully comprehend this, imagine holding your hand near, but not touching, a hot iron—this sensation of warmth is due to radiative heat transfer. The higher the temperature difference and the higher the emissivity, the more heat is radiated away.

Heat Flux

Heat flux is the rate of heat energy transfer through a given surface per unit area. It is often expressed in watts per square meter ($W/m^2$). The concept of heat flux is central to understanding how heat is transferred in various modes such as conduction, convection, and radiation.

In the step-by-step solution provided, heat flux represents the total heat being dissipated from the rod into the surrounding air, and is the sum of both convective and radiative heat transfer components:
$$q=h_c(T_s-T_{\mathrm{\infty }})+ϵ\sigma (T_s^4-T_{\mathrm{\infty }}^4)$$
Here $h_c$ represents the convective heat transfer coefficient, and the combination with the emissivity $ϵ$ and Stefan-Boltzmann constant $\sigma$ captures the radiative component. As an analogy, you could think of a light bulb: it emits light (analogous to radiation) that spreads out over an area—the intensity of that light at any point on that area is analogous to heat flux. Students encountering this concept should consider both the intensity of the heat source and the area over which the heat is being spread to really understand the implications of heat flux in thermodynamics.