Question

A $0.2\mathrm{m}\times 0.2\mathrm{m}$ street sign surface has an absorptivity of $0.6$ and an emissivity of $0.7$, while the street sign is subjected to a cross flow wind at ${20}^{\circ }\mathrm{C}$ with a velocity of $1\mathrm{m}/\mathrm{s}$. Solar radiation is incident on the street sign at a rate of $1100\mathrm{W}/{\mathrm{m}}^2$, and the surrounding temperature is ${20}^{\circ }\mathrm{C}$. Determine the surface temperature of the street sign. Evaluate the air properties at ${30}^{\circ }\mathrm{C}$. Treat the sign surface as a vertical plate in cross flow.

Short answer

Question: Determine the temperature on the surface of a street sign subjected to solar radiation and cross-flow wind, given that the sign dimensions are $0.2m\times 0.2m$, sign surface absorptivity is $\alpha =0.6$, sign surface emissivity is $\epsilon =0.7$, wind temperature is $T_{wind}={20}^{\circ }C$, wind velocity is $1m/s$, solar radiation is $1100W/m^2$, surrounding temperature is ${20}^{\circ }$C, and air properties are evaluated at ${30}^{\circ }$C. Answer: To determine the temperature on the surface of the street sign, follow these steps: 1. Find the radiative heat transfer: $q_{rad}^{″}=660W/m^2$. 2. Find the convective heat transfer coefficient using the empirical correlation for a vertical plate in cross-flow. 3. Calculate the convective heat transfer: $q_{conv}^{″}=h(T_s-T_{wind})$. 4. Apply the energy balance equation: $q_{rad}^{″}=q_{conv}^{″}$. 5. Evaluate air properties at the specified temperature, ${30}^{\circ }$C. 6. Solve for the surface temperature, $T_s$, of the street sign.

Step-by-step solution

Gather given information

We are given following data - Sign dimensions: $0.2m\times 0.2m$ - Sign surface absorptivity: $\alpha =0.6$ - Sign surface emissivity: $\epsilon =0.7$ - Wind temperature: $T_{wind}={20}^{\circ }C$ - Wind velocity: $1m/s$ - Solar radiation: $1100W/m^2$ - Surrounding temperature: ${20}^{\circ }$C - Air properties to be evaluated at ${30}^{\circ }$C

Find the radiative heat transfer

The absorbed solar radiation (radiative heat gain) on the street sign can be represented as $q_{rad}^{″}=\alpha G_s$, where: - $q_{rad}^{″}$ - Radiative heat transfer per unit area - $G_s$ - Incident solar radiation $q_{rad}^{″}=0.6\times 1100=660W/m^2$

Find the convective heat transfer coefficient

As the surface is treated as a vertical plate in cross-flow, we can use the following empirical correlation (assuming turbulent flow) for calculating the convective heat transfer coefficient $h$: $h=0.664\sqrt{R{e}_x}\times P{r}^{(1/3)}\times \frac{{\left(\frac{\mu }{{\mu }_w}\right)}^a}{fR{e}_{x}Pr}$ where, - $R{e}_x$ - Reynold's number based on length x - $Pr$ - Prandtl number - $\mu$ and ${\mu }_w$ - Dynamic viscosities of fluid and near-wall fluid, respectively - $f$ - 0.037 to 0.047 in turbulent flow - $a$ - 0.14 to 0.17 in turbulent flow Note: We can use the heat transfer charts to determine the Reynolds number, Prandtl number, fluid, and wall dynamic viscosity at the specified temperature.

Calculate the convective heat transfer

The convective heat transfer per unit area between the wind and the sign surface can be represented as $q_{conv}^{″}=h(T_s-T_{wind})$, where: - $q_{conv}^{″}$ - Convective heat transfer per unit area - $T_s$ - Surface temperature of the sign

Apply the energy balance equation

Now, let's apply the energy balance equation for radiative and convective heat transfer: $q_{rad}^{″}=q_{conv}^{″}$ Plug in the expressions found above: $660W/m^2=h(T_s-20)$ Solve for $T_s$.

Evaluate air properties

Evaluate the air properties such as viscosity and Prandtl number or use the heat transfer charts at the specified temperature, ${30}^{\circ }$C.

Final Calculation

After plugging in the values of $h$ and the air properties, we can solve for the surface temperature $(T_s)$ of the street sign, which is the temperature we're trying to determine in this problem.

Key concepts

Radiative Heat Transfer

Radiative heat transfer occurs when energy is transferred through electromagnetic waves, such as sunlight. This form of heat transfer can happen even if there is a vacuum between objects. In the given exercise, the street sign absorbs solar radiation. This is calculated using the formula:

• $q_{rad}^{″}=\alpha G_s$

where:

• $\alpha$ is the absorptivity of the sign (0.6 in this example), indicating 60% of the solar energy is absorbed.
• $G_s$ is the incident solar radiation $1100W/m^2$.

Plugging in the values, we find $q_{rad}^{″}=0.6\times 1100=660W/m^2$.
This value represents the amount of radiant energy absorbed per unit area. Understanding absorptivity is essential because it determines how much heat the object gains from radiation.

Convective Heat Transfer

Convective heat transfer involves the movement of heat between a solid surface and a fluid (like air or water) moving past it.
It depends on the properties of the fluid and the conditions of flow. In this context, the street sign is treated as a vertical plate experiencing cross-flow wind.The convective heat transfer coefficient $h$ is crucial and can be determined using empirical correlations which include factors like Reynolds and Prandtl numbers:

• $R{e}_x$ is the Reynolds number, which characterizes the flow regime (laminar or turbulent).
• $Pr$ is the Prandtl number, relating the fluid’s viscosity and thermal conductivity.
  

The equation to estimate convective heat transfer is:

• $q_{conv}^{″}=h(T_s-T_{wind})$

where:

• $T_s$ is the surface temperature of the sign.
• $T_{wind}$ is the wind temperature, given as ${20}^{\circ }C$.

By solving for $h$ and knowing $q_{rad}^{″}$, you can balance this with $q_{conv}^{″}$ to find the surface temperature.

Energy Balance Equation

The Energy Balance Equation is a powerful tool for solving heat transfer problems.
It allows us to set the incoming energy equal to the outgoing energy, thus maintaining a thermal equilibrium.In this exercise, the energy absorbed by the street sign through radiation must equal the energy lost through convection—since they are not heated or cooled otherwise:

• $q_{rad}^{″}=q_{conv}^{″}$

Substituting the known formulas gives:

• $660W/m^2=h(T_s-20)$

Solving this equation requires determining $h$ using air properties evaluated at ${30}^{\circ }C$.
The equation illustrates the balance, ensuring that the energy absorbed by the sign equals the energy it releases.
This foundational principle helps engineers design systems that efficiently manage heat flows.