Question

Liquid mercury at \(250^{\circ} \mathrm{C}\) is flowing in parallel over a flat plate at a velocity of \(0.3 \mathrm{~m} / \mathrm{s}\). Surface temperature of the \(0.1-\mathrm{m}\)-long flat plate is constant at \(50^{\circ} \mathrm{C}\). Determine \((a)\) the local convection heat transfer coefficient at \(5 \mathrm{~cm}\) from the leading edge and \((b)\) the average convection heat transfer coefficient over the entire plate.

Short answer

Question: Determine the local convection heat transfer coefficient at a specific position 5 cm from the leading edge and the average convection heat transfer coefficient over the entire plate. Answer: (a) The local convection heat transfer coefficient at 5 cm from the leading edge is approximately 241048 W/(m^2 ⋅ K). (b) The average convection heat transfer coefficient over the entire plate is approximately 1712950 W/(m^2 ⋅ K).

Step-by-step solution

Obtain given information

We are given the following information: Temperature of liquid mercury, \(T_L = 250^\circ C\) Velocity of liquid mercury, \(u = 0.3 m/s\) Surface temperature of the flat plate, \(T_P = 50^\circ C\) Length of the flat plate, \(L = 0.1 m\) Position we would like to compute local (convection) heat transfer coefficient, \(x = 0.05 m\)

Calculate Reynolds number

To make use of the Nusselt number and heat transfer coefficient formulas, first, we need to calculate the Reynolds number at the position \(x = 5 cm\). Reynolds number, \(Re_x = \frac{u x}{\nu}\) where, \(\nu\) is the kinematic viscosity of liquid mercury. Based on the provided information, you can look up the kinematic viscosity of liquid mercury at \(250^\circ C\) in a reference table. For instance, \(\nu \approx 1.125 \times 10^{-7} m^2/s\). Now we can calculate the Reynolds number at \(x = 0.05 m\): \(Re_x = \frac{0.3 \times 0.05}{1.125 \times 10^{-7}} \approx 133333\)

Calculate Prandtl number

Next, we need to calculate the Prandtl number, which characterizes the relative thickness of the momentum and thermal boundary layers. Prandtl number, \(Pr = \frac{\nu}{\alpha}\) where, \(\alpha\) is the thermal diffusivity of liquid mercury. You can look up the thermal diffusivity of liquid mercury at \(250^\circ C\) in a reference table. For instance, \(\alpha \approx 1.0 \times 10^{-7} m^2/s\). Now we can calculate the Prandtl number: \(Pr = \frac{1.125 \times 10^{-7}}{1.0 \times 10^{-7}} \approx 1.125\)

Calculate local Nusselt number and Local convection heat transfer coefficient

Now that we have the Reynolds and Prandtl numbers, we can determine the local Nusselt number using the following formula: Local Nusselt number, \(Nu_x = C \cdot Re_x^m \cdot Pr^n\) Here, \(C\), \(m\), and \(n\) are empirical constants, typical values are \(C = 0.332\), \(m = 0.5\), and \(n = 1/3\) for laminar flow over a flat plate. Therefore, \(Nu_x = 0.332 \cdot (133333)^{0.5} \cdot (1.125)^{1/3} \approx 4598\) Now, we can determine the local convection heat transfer coefficient at \(x = 0.05 m\) by using the following formula: Local convection heat transfer coefficient, \(h_x = \frac{k \cdot Nu_x}{x}\) where, \(k\) is the thermal conductivity of liquid mercury. You can look up the thermal conductivity of liquid mercury at \(250^\circ C\) in a reference table. For instance, \(k \approx 26.3 W/(m \cdot K)\). Now we can calculate the local convection heat transfer coefficient: \(h_x = \frac{26.3 \times 4598}{0.05} \approx 241048 W/(m^2 \cdot K)\) The local convection heat transfer coefficient at \(5 cm\) from the leading edge is approximately \(241048 W/(m^2 \cdot K)\)

Calculate the average convection heat transfer coefficient

Next, we will determine the average convection heat transfer coefficient over the entire plate. First, we need to calculate the average Nusselt number with the following formula: Average Nusselt number, \( \overline{Nu}_L = C \cdot Re_L^m \cdot Pr^n\) Here, \(Re_L\) is the Reynolds number at the end of the plate, \(L = 0.1 m\): \(Re_L = \frac{0.3 \times 0.1}{1.125 \times 10^{-7}} \approx 266667\) Now, we can calculate the average Nusselt number: \( \overline{Nu}_L = 0.332 \cdot (266667)^{0.5} \cdot (1.125)^{1/3} \approx 6515\) Finally, we can determine the average convection heat transfer coefficient over the entire plate using the following formula: Average convection heat transfer coefficient, \(\overline{h}_L = \frac{k \cdot \overline{Nu}_L}{L}\) Now we can calculate the average convection heat transfer coefficient: \(\overline{h}_L = \frac{26.3 \times 6515}{0.1} \approx 1712950 W/(m^2 \cdot K)\) The average convection heat transfer coefficient over the entire plate is approximately \(1712950 W/(m^2 \cdot K)\). Answer: (a) The local convection heat transfer coefficient at \(5 cm\) from the leading edge is approximately \(241048 W/(m^2 \cdot K)\). (b) The average convection heat transfer coefficient over the entire plate is approximately \(1712950 W/(m^2 \cdot K)\).

Key concepts

Convection Heat Transfer Coefficient

The convection heat transfer coefficient, denoted as \( h \), is a crucial parameter in heat transfer processes that involve fluid motion. It essentially quantifies the rate of heat transfer between a surface and a moving fluid. More specifically, \( h \) represents the amount of heat transferred per unit area per unit temperature difference between the fluid and the solid surface. To visualize this, imagine hot mercury flowing over a flat, cooler plate. The convection heat transfer coefficient would indicate how effectively heat is transferred from the mercury to the plate. High values of \( h \) suggest efficient heat transfer, while lower values imply resistance between the liquid and the solid surface. The coefficient is vital in engineering applications like heat exchangers, where understanding this rate helps design more effective systems for heating or cooling.In our example, we calculated \( h \) both locally at a specific point and averagely over the entire plate, illustrating how surface and flow characteristics affect heat exchange efficiency.

Nusselt Number

The Nusselt number, \( Nu \), is a dimensionless quantity that plays a pivotal role in convective heat transfer problems. It is defined as the ratio of convective to conductive heat transfer across a boundary. In simpler terms, \( Nu \) helps us understand how well a fluid over a surface enhances heat transfer compared to if the heat were conducted through a stationary fluid. A high \( Nu \) indicates that convection significantly dominates over conduction in the heat transfer process, leading to efficient heat movement.In the given problem, we used an empirical formula to determine the Nusselt number locally and averagely. The formula for \( Nu \) includes factors like the Reynolds number and Prandtl number, which describe flow and fluid properties. Overall, the Nusselt number helps in assessing the effectiveness of thermal systems.

Reynolds Number

The Reynolds number, \( Re \), is an essential dimensionless parameter in fluid dynamics that characterizes the nature of flow. It is determined by the ratio of inertial forces to viscous forces within the fluid. In practical terms, \( Re \) provides insight into whether a flow will be laminar or turbulent. Laminar flow occurs at low \( Re \) and is smooth and orderly, whereas turbulent flow appears at high \( Re \), with eddies and chaotic changes in pressure and velocity. For our exercise, we calculated \( Re \) for mercury flowing over a plate to classify the flow type and determine appropriate conditions for calculating the Nusselt number. It directly affects how heat and momentum are transferred between the fluid and the surface, influencing design decisions in various engineering applications.

Prandtl Number

The Prandtl number, \( Pr \), is another important dimensionless number in the field of heat transfer. It expresses the ratio of momentum diffusivity (viscous diffusion) to thermal diffusivity. In essence, \( Pr \) provides insights into the relative thickness of thermal and velocity boundary layers in a fluid. A high \( Pr \) means the velocity boundary layer is much thicker than the thermal boundary layer, which happens in oils. Conversely, a low \( Pr \) reflects a thicker thermal boundary layer common in liquid metals like mercury.In our solution, \( Pr \) was used in conjunction with \( Re \) to determine the local and average Nusselt numbers. These calculations are crucial for applications like cooling systems and understanding how efficiently heat is carried by the fluid.

Thermal Conductivity

Thermal conductivity, \( k \), is a fundamental property of a material that indicates its ability to conduct heat. It quantifies the amount of heat passed in a material by conduction. High thermal conductivity means that heat moves quickly through a material, while low values suggest the opposite. Think of \( k \) as how fast or slow heat energy is transferred through a substance. It is necessary to accurately determine the convection heat transfer coefficient in the given exercise, as the thermal conductivity of mercury was needed to calculate \( h \) using the Nusselt number.Thermal conductivity impacts how well materials can manage heat, which is vital for engineering materials used in insulation, electronic devices, and heat exchangers, ensuring efficient and safe thermal management.