Question

A rectangular bar with a characteristic length of \(0.5 \mathrm{~m}\) is placed in a free stream flow where the convection heat transfer coefficients were found to be \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) when the free stream velocities were \(25 \mathrm{~m} / \mathrm{s}\) and \(5 \mathrm{~m} / \mathrm{s}\), respectively. If the Nusselt number can be expressed as \(\mathrm{Nu}=C \operatorname{Re}^{m} \operatorname{Pr}^{n}\), where \(C, m\), and \(n\) are constants, determine the convection heat transfer coefficients for similar bars with (a) \(L=1 \mathrm{~m}\) and \(V=5 \mathrm{~m} / \mathrm{s}\), and \((b) L=2 \mathrm{~m}\) and \(V=50 \mathrm{~m} / \mathrm{s}\).

Short answer

Based on the given data for a 0.5-meter-long bar, we formulated the Nusselt number equation and determined the constants C and m. Using these constants and the information provided for each case, we calculated the convection heat transfer coefficients for cases (a) and (b). The process involved finding the Reynolds numbers for each case, solving for the constants, and then calculating the respective convection heat transfer coefficients.

Step-by-step solution

Formulate the equation for Nu for given conditions

We know that Nusselt number (Nu) represents the ratio of the total heat transfer to the conduction heat transfer. We also know the formula for Nusselt number: \(\mathrm{Nu}=C * \operatorname{Re}^{m} * \operatorname{Pr}^{n}\) Nu can also be written as: \(Nu = \frac{h * L}{k}\), where h is the convection heat transfer coefficient, L is the characteristic length and k is the thermal conductivity of the fluid. We are now ready to compute the Nu in terms of given data.

Solve for constants (C, m, n) in the Nu equation

Let's form two equations for h1 and h2 and their respective length and free-stream velocities using the given data: Equation 1: \(\frac{h_{1} L_{1}}{k} = C * Re_{1}^m * Pr^{n}\) Equation 2: \(\frac{h_{2} L_{1}}{k} = C * Re_{2}^m * Pr^{n}\) We are given h1, h2, and L1. L1 is equal to 0.5 m in both cases. Since Pr is constant for both cases and unknown, we can simplify the equations: Equation 1: \(\frac{h_{1}}{k} = C * Re_{1}^m\) Equation 2: \(\frac{h_{2}}{k} = C * Re_{2}^m\) Divide Equation 1 by Equation 2: \(\frac{h_{1}}{h_{2}} = \frac{Re_{1}^m}{Re_{2}^m}\) We can rearrange the equation to find the value of m: \(m = \log_{(Re_{1} / Re_{2})} \left(\frac{h_{1}}{h_{2}}\right)\) From the given problem, we have h1 = 100 W/m²⋅K, h2 = 50 W/m²⋅K, v1 = 25 m/s and v2 = 5 m/s. With the characteristic length (L1) = 0.5 m, we can find the Reynolds numbers for both cases: \(Re_{1} = \frac{ρ * v_{1} * L_{1}}{μ}\) \(Re_{2} = \frac{ρ * v_{2} * L_{1}}{μ}\) Since the fluid properties ρ (density) and μ (dynamic viscosity) are constant, we can simplify the Reynolds numbers as: \(Re_{1} = 25 * 0.5\) \(Re_{2} = 5 * 0.5\) Plugging the values of h1, h2, Re1, and Re2 into the equation for m, we can find the value of m. Next, we will substitute the value of m back into either Equation 1 or Equation 2 to find the value of C.

Determine convection heat transfer coefficient (h) for each case

Now we have the value of m and C. We can use these values to determine the convection heat transfer coefficient (h) for cases (a) and (b). For a similar bar with L = 1 m and V = 5 m/s (Case a), we can find the Nusselt number (Nu) using the equation: \(\frac{h_{a} * L_{a}}{k} = C * Re_{3}^m * Pr^{n}\) We are given La and Va for case a. We can find Re3: \(Re_{3} = \frac{ρ * V_{3} * L_{3}}{μ}\) \(Re_{3} = V_{3} * L_{3}\) Plugging in the values of C, m, La, Va and Re3, we can solve and find ha. For a similar bar with L = 2 m and V = 50 m/s (Case b), the process will be the same, and we can find the convection heat transfer coefficient (h) using the Nusselt number (Nu) equation. Following the same process, using data for case b and the values of C and m, we can find hb. In conclusion, we have determined the convection heat transfer coefficients (h) for both case (a) and case (b) using the Nusselt number equation with the given data.

Key concepts

Nusselt Number

The Nusselt Number (\( Nu \) ) plays a crucial role in heat transfer involving convection. It's a dimensionless number that describes how efficiently heat is being transferred from a surface to a fluid flowing over it, or vice versa. Specifically, it is the ratio of convective to conductive heat transfer across a boundary. This can be mathematically represented as:

• \( Nu = \frac{h \, L}{k} \)

where \( h \) is the convective heat transfer coefficient, \( L \) is the characteristic length, and \( k \) is the thermal conductivity of the fluid.

A high Nusselt number indicates that convection is dominating over conduction, meaning the heat transfer is more effective. In problems where convection is examined, the Nusselt number helps understand the heat transfer performance in terms of the fluid dynamics involved, especially when comparing different flow conditions.

Reynolds Number

The Reynolds Number (\( Re \) ) is another important dimensionless parameter in fluid mechanics and heat transfer. It helps predict flow patterns in different fluid flow situations. It represents the ratio of inertial forces to viscous forces within a fluid flow and is used to determine whether the flow will be laminar or turbulent.

• Laminar Flow: Smooth, orderly flow with \( Re < 2300 \)
• Turbulent Flow: Chaotic and irregular flow with \( Re > 4000 \)
• Transitional Flow: In between laminar and turbulent, usually \( 2300 < Re < 4000 \)

In our case for the rectangular bar, the Reynolds Number is calculated based on the fluid's velocity and the characteristic length:

• \( Re = \frac{\rho \cdot V \cdot L}{\mu} \)

where \( \rho \) is the fluid density, \( V \) is the velocity, \( L \) is the characteristic length, and \( \mu \) is the dynamic viscosity. Understanding the Reynolds number is vital as it influences the Nusselt number, which in turn affects the heat transfer characteristics.

Thermal Conductivity

Thermal Conductivity (\( k \) ) is a property of a material that indicates its ability to conduct heat. It defines the heat transfer through a material due to a temperature gradient. Materials with high thermal conductivity transfer heat more efficiently, while those with low thermal conductivity are better insulators.

• Common Conductors: Metals like copper and aluminum
• Common Insulators: Materials like rubber, wood, and fiberglass

In convection heat transfer scenarios, the thermal conductivity of the fluid is essential in calculating the Nusselt number. It acts as a benchmark for comparing how much of the heat transfer is due to convection versus conduction. Hence, knowing the thermal conductivity of the involved materials helps in accurately analyzing and designing systems for efficient thermal management.

Characteristic Length

Characteristic Length (\( L \) ) is a crucial dimension that serves as a reference in various fluid dynamics and heat transfer equations. It can vary based on the geometry and the specific situation.

• For flow over a flat plate, it might be the length of the plate.
• For flow inside a pipe, it could be the diameter of the pipe.

In our exercises, the characteristic length influences both the \( Nu \) and \( Re \) numbers:

• \( Nu = \frac{h \cdot L}{k} \)
• \( Re = \frac{\rho \cdot V \cdot L}{\mu} \)

This parameter helps unify a variety of geometric situations under a singular framework of analysis. By considering the characteristic length, engineers and scientists can compare heat transfer rates and fluid motion in different scenarios with ease. It's an essential part of simplifying and understanding complex fluid flow problems.