Question

The local atmospheric pressure in Denver, Colorado (elevation \(1610 \mathrm{~m}\) ), is \(83.4 \mathrm{kPa}\). Air at this pressure and \(20^{\circ} \mathrm{C}\) flows with a velocity of \(8 \mathrm{~m} / \mathrm{s}\) over a \(1.5 \mathrm{~m} \times 6 \mathrm{~m}\) flat plate whose temperature is \(140^{\circ} \mathrm{C}\). Determine the rate of heat transfer from the plate if the air flows parallel to the \((a)\)-m-long side and \((b)\) the \(1.5 \mathrm{~m}\) side.

Short answer

Question: Calculate the rate of heat transfer from a flat plate to the air for both cases: (a) when the air flows parallel to the 6-meter long side, and (b) when it flows parallel to the 1.5-meter long side. The plate has dimensions of 1.5 m x 6 m. The velocity of the air is 8 m/s, the air is at 20°C, and the atmospheric pressure is 83.4 kPa. The surface temperature of the plate is 140°C. Answer: The rate of heat transfer from the plate to the air is 6409 W when the air flows parallel to the 6-meter long side (case a), and 15802 W when it flows parallel to the 1.5-meter long side (case b).

Step-by-step solution

Find the properties of air at the given pressure and temperature

First, we need to find the properties of the air, such as the kinematic viscosity (\(\nu\)), thermal conductivity (\(k\)), and Prandtl number (Pr) at the given atmospheric pressure of 83.4 kPa and the temperature of 20°C. Here are the properties of air at 20°C: - \(\nu = 15.1 \times 10^{-6} \frac{m^2}{s}\) - \(k = 0.026 \frac{W}{m \cdot K}\) - \(Pr = 0.707\)

Compute the Reynolds number for both cases

Now we need to calculate the Reynolds numbers for both cases (a) and (b) by using the given dimensions and velocity of the air. The Reynolds number can be calculated using the formula: $$ Re = \frac{UL}{\nu} $$ where \(U\) is the air velocity, \(L\) is the characteristic length (m-long side for case a, and 1.5 m side for case b), and \(\nu\) is the kinematic viscosity. Case (a): $$ Re_a = \frac{8 \times 6}{15.1 \times 10^{-6}} \approx 3.16 \times 10^{6} $$ Case (b): $$ Re_b = \frac{8 \times 1.5}{15.1 \times 10^{-6}} \approx 7.93 \times 10^{5} $$

Calculate the Nusselt number for both cases

The Nusselt number (Nu) can be estimated using the Gnielinski correlation for turbulent flow over a flat plate: $$ Nu = \frac{(0.037 Re^{\frac{4}{5}} - 871) Pr}{(1 + (0.220 Pr^{\frac{1}{5}} - 1) Re^{-1}Pr^{-1})^{2}} $$ Case (a): $$ Nu_a = \frac{(0.037 (3.16 \times 10^6)^{\frac{4}{5}} - 871) (0.707)}{(1 + (0.220 (0.707)^{\frac{1}{5}} - 1) (3.16 \times 10^6)^{-1}(0.707)^{-1})^{2}} \approx 1370 $$ Case (b): $$ Nu_b = \frac{(0.037 (7.93 \times 10^5)^{\frac{4}{5}} - 871) (0.707)}{(1 + (0.220 (0.707)^{\frac{1}{5}} - 1) (7.93 \times 10^5)^{-1}(0.707)^{-1})^{2}} \approx 847 $$

Calculate the heat transfer coefficient for both cases

We can now calculate the heat transfer coefficient (h) for both cases using the Nusselt number and thermal conductivity: $$ h = \frac{Nu \cdot k}{L} $$ Case (a): $$ h_a = \frac{1370 \times 0.026}{6} \approx 5.92 \frac{W}{m^2 \cdot K} $$ Case (b): $$ h_b =\frac{847 \times 0.026}{1.5} \approx 14.8 \frac{W}{m^2 \cdot K} $$

Calculate the rate of heat transfer for both cases

Finally, we can determine the rate of heat transfer (Q) for both cases using the heat transfer coefficient (h), the surface area (A), and the temperature difference (T_difference) between the plate and the air: $$ Q = hA(T_{plate} - T_{air}) $$ Case (a): $$ Q_a = 5.92 \times (1.5 \times 6) \times (140 - 20) = 5.92 \times 9 \times 120 \approx 6409 W $$ Case (b): $$ Q_b = 14.8 \times (1.5 \times 6) \times (140 - 20) = 14.8 \times 9 \times 120 \approx 15802 W $$ The rate of heat transfer from the plate to the air is 6409 W when the air flows parallel to the 6-meter long side (a), and 15802 W when the air flows parallel to the 1.5-meter long side (b).

Key concepts

Reynolds Number

Understanding the Reynolds number is critical when approaching problems related to fluid dynamics and heat transfer. It's a dimensionless quantity that offers insight into the flow regime of a fluid over a surface, indicating whether the flow is laminar or turbulent. Calculated with the formula \( Re = \frac{UL}{u} \), where \( U \) represents the velocity of the fluid, \( L \) is a characteristic linear dimension (such as the length of a plate over which fluid flows), and \( u \) is the kinematic viscosity of the fluid.

The example in the exercise shows two different Reynolds numbers for airflow over different sides of a flat plate. A larger \( Re \) often denotes turbulent flow, which impacts how heat is transferred from the plate to the air, as turbulence enhances the mixing of the fluid and thus the heat transfer rate.

In practical applications, knowing the Reynolds number helps engineers design systems that optimize the flow for effective cooling or heating, which is essential, for instance, in HVAC system designs or in the aerospace industry.

Nusselt Number

The Nusselt number, represented by \( Nu \), is another dimensionless figure crucial in heat transfer calculations. It correlates the convective to conductive heat transfer across a boundary layer and is used to determine the heat transfer coefficient.

In the given problem, we calculate the Nusselt number using the Gnielinski correlation, which is suitable for turbulent flow past a flat plate. This correlation takes into account the Reynolds number, Prandtl number (\( Pr \)), and some constants to estimate the heat transfer features of the flow. The formula \[ Nu = \frac{(0.037 Re^{\frac{4}{5}} - 871) Pr}{(1 + (0.220 Pr^{\frac{1}{5}} - 1) Re^{-1}Pr^{-1})^{2}} \] helps us predict how effectively the air flowing over the plate can remove heat from the surface.

As such, the Nusselt number connects the fluid properties and flow conditions to the heat transfer characteristics, making it an indispensable tool in thermal analysis and system design, for applications ranging from industrial heat exchangers to solar panels.

Heat Transfer Coefficient

The heat transfer coefficient \( h \) is a measure of a fluid's ability to transfer heat to or from a surface. Determined by the formula \( h = \frac{Nu \cdot k}{L} \), where \( k \) is the thermal conductivity of the fluid and \( L \) is the characteristic length, this coefficient is crucial in quantifying the rate of heat transfer.

In our problem, the heat transfer coefficient differs for both cases (a) and (b) due to the change in characteristic length. Since \( h \) is directly proportional to the Nusselt number, a higher \( Nu \) results in a higher heat transfer coefficient, which translates to more efficient heat transfer.

The heat transfer coefficient enables engineers and scientists to compare different heat exchange scenarios and optimize systems for cooling or heating efficiency. It's used extensively in designing radiators, heat sinks, and various types of heat exchanger equipment found in power plants, automotive industry, and electronics cooling systems.