Question

Consider a hot automotive engine, which can be approximated as a \(0.5-\mathrm{m}\)-high, \(0.40\)-m-wide, and \(0.8-\mathrm{m}\)-long rectangular block. The bottom surface of the block is at a temperature of \(100^{\circ} \mathrm{C}\) and has an emissivity of \(0.95\). The ambient air is at \(20^{\circ} \mathrm{C}\), and the road surface is at \(25^{\circ} \mathrm{C}\). Determine the rate of heat transfer from the bottom surface of the engine block by convection and radiation as the car travels at a velocity of \(80 \mathrm{~km} / \mathrm{h}\). Assume the flow to be turbulent over the entire surface because of the constant agitation of the engine block.

Short answer

Answer: The rates of heat transfer from the bottom surface of an engine block are 3571.04 W by convection and 341.74 W by radiation.

Step-by-step solution

Calculate the surface area of the bottom surface of the engine block

First, we need to find the area of the bottom surface of the engine block, which is a rectangle with a width of 0.4 m and a length of 0.8 m. We can calculate the area as: Area = width × length Area = 0.4 m × 0.8 m Area = 0.32 m²

Convert the car's velocity to m/s

We are given the car's velocity in km/h, but we need it to be in m/s. To do this, we can use the following conversion: 1 km/h = 1000 m / 3600 s So, the car's velocity in m/s would be: 80 km/h × (1000 m / 3600 s) = 22.22 m/s

Calculate the Reynolds number

We'll use the Reynolds number to determine the proper correlation for the heat transfer coefficient. The Reynolds number can be calculated as: Re = (ρ × V × L) / µ For air at 20°C, we will use the following properties: ρ = 1.204 kg/m³ (density) µ = 1.825 × 10⁻⁵ kg/(m.s) (dynamic viscosity) The Reynolds number for this case is: Re = (1.204 kg/m³ × 22.22 m/s × 0.8 m) / (1.825 × 10⁻⁵ kg/(m.s)) Re = 106657 Since Re > 4000, the flow is turbulent over the bottom surface of the engine block.

Calculate the heat transfer coefficient by convection

Using the Dittus-Boelter equation, we can find the heat transfer coefficient for turbulent flow as: h = 0.023 × Re^0.8 × Pr^1/3 × k/L For air at 20°C, we will use the following properties: k = 0.02624 W/(m.K) (thermal conductivity) Pr = 0.707 (Prandtl number) So the heat transfer coefficient can be calculated as: h = 0.023 × 106657^0.8 × 0.707^1/3 × (0.02624 W/(m.K)) / (0.8 m) h = 138.95 W/(m².K)

Calculate the heat transfer by convection

Now we can find the heat transfer by convection using the formula: q_conv = h × A × (T_s - T_inf) Where: T_s = 100°C (surface temperature of the engine block) T_inf = 20°C (ambient air temperature) q_conv = 138.95 W/(m².K) × 0.32 m² × (100 - 20) K q_conv = 3571.04 W

Calculate the heat transfer by radiation

To find the heat transfer by radiation, we will use the Stefan-Boltzmann law: q_rad = ε × σ × A × (T_s^4 - T_sur^4) Where: ε = 0.95 (emissivity of the surface) σ = 5.67 × 10⁻⁸ W/(m².K⁴) (Stefan-Boltzmann constant) T_s = 100°C + 273.15 K (converted to Kelvin) T_sur = 25°C + 273.15 K (temperature of the road surface converted to Kelvin) q_rad = 0.95 × 5.67 × 10⁻⁸ W/(m².K⁴) × 0.32 m² × ((373.15 K)^4 - (298.15 K)^4) q_rad = 341.74 W So the rate of heat transfer from the bottom surface of the engine block is: By convection: 3571.04 W By radiation: 341.74 W

Key concepts

Convection

Convection is a vital heat transfer mechanism that occurs within fluids such as gases and liquids. It is described by the movement of fluid layers, which results in the transfer of heat from one place to another. There are two main types of convection: natural and forced convection. In the case of our example with the car, we deal with forced convection, owing to the movement of air as the car travels.
In forced convection, the movement of the fluid is induced by an external force, such as a fan or, here, the motion of the car itself. The car accelerates the surrounding air over its surface, increasing the heat transfer rate significantly.
Convection is often quantified using the heat transfer coefficient, which can be determined from empirical correlations. For turbulent flow over a surface, the Dittus-Boelter equation is commonly used, as applied in this example. This equation gives us the value needed to calculate the rate of heat transfer by convection, which considers not just surface area and temperature difference, but also flow and fluid properties.

Radiation

Radiation is another critical mode of heat transfer, which occurs without the need for a medium to transfer energy between bodies. Instead, it involves the emission and absorption of electromagnetic waves, mainly in the infrared spectrum. All objects emit radiation, and the amount they emit is influenced by their temperature and surface properties.
A key factor in calculating radiation is the emissivity of the material, a measure of how effectively a surface emits energy as thermal radiation. In our engine block example, the block has an emissivity of 0.95, which indicates it is a very effective emitter of thermal energy.
The Stefan-Boltzmann law allows us to calculate the heat transfer rate by radiation, using the surface area and temperature of both the emitting and receiving bodies. It's important to convert temperatures to Kelvin for these calculations, as precise temperature-powered fourth-order computations come into play. Unlike convection, radiation doesn't require a moving fluid, and even occurs in a vacuum.

Turbulent Flow

Turbulent flow is a type of flow characterized by irregular swirling and eddies, contrasting with smooth and orderly laminar flow. This chaotic behavior enhances the mixing of the fluid, leading to better heat and momentum transfer.
In many practical situations, such as in our exercise where a car speeds through the air, turbulent flow is prevalent because the increased speed promotes irregular fluid motion. When designing systems or calculating heat transfer, assuming turbulent flow can often lead to more efficient designs, as it typically results in higher heat transfer coefficients.
Understanding whether flow is turbulent is key in choosing the appropriate correlations for calculations. For this, we often rely on the Reynolds number, a dimensionless quantity that predicts the nature of the flow.

Reynolds Number

The Reynolds number is a dimensionless value used in fluid mechanics to predict flow patterns in different fluid flow situations. It helps determine if the flow will be laminar or turbulent.
This number is calculated using the formula: \[ Re = \frac{\rho \times V \times L}{\mu} \]
where \( \rho \) is the density of the fluid, \( V \) is the velocity, \( L \) is a characteristic length, and \( \mu \) is the dynamic viscosity.
When the Reynolds number exceeds a critical range, like 4000, the flow is typically turbulent. In the engine block example, the car's speed contributed to a Reynolds number of over 106,000, indicating a highly turbulent flow.
The Reynolds number is crucial, as the type of flow it predicts directly impacts which heat transfer equations are suitable for use. High Reynolds numbers suggest the utilization of formulas and constants adapted to turbulent conditions, like the Dittus-Boelter equation employed in our solution.