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 \(V=50 \mathrm{~m} / \mathrm{s}\).

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.

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.

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.