91Ó°ÊÓ

First, let's convert the dimensions of the plate and cylinder into meters, as we will be working in SI units: - Plate thickness, \(t = 1mm = 0.001m\) - Plate length, \(L = 25mm = 0.025m\) - Plate width, \(w = 50mm = 0.05m\)

For a laminar flow over a flat plate, we use the Nusselt number correlation for parallel flow over isothermal surfaces: \[Nu = 0.664 \, Re^{1/2} Pr^{1/3}\] Where, \(Nu\) is the Nusselt number, \(Re\) is the Reynolds number and \(Pr\) is the Prandtl number. The Reynolds number is defined as: \[Re = \frac{u_{\infty} L}{\nu}\] Where, \(\nu\) is the kinematic viscosity of air. We can find the value of kinematic viscosity from air property tables at \(25^{\circ}C\): \(\nu = 15.69 \times 10^{-6} \,m^2/s\) Now, let's calculate the Reynolds number: \[Re = \frac{3 * 0.025}{15.69 \times 10^{-6}} = 4795\] Considering the convective conditions associated with the flat plate, here we assume the Prandtl number \(Pr = 0.71\), which is a standard assumption for air at room temperature. Now, let's calculate the Nusselt number: \[Nu = 0.664 * (4795)^{1/2} * (0.71)^{1/3} = 55.26\] The heat transfer coefficient (h) can then be calculated as: \[h = \frac{Nu * k}{L}\] Where, k is the thermal conductivity of air. We can find the value of thermal conductivity from air property tables at \(25^{\circ}C\): \(k = 0.02624 \, W/mK\) \[h = \frac{55.26 * 0.02624}{0.025} = 61.24 \, W/m^2K\]

For the flat plate, we have to consider both the top and the bottom surfaces. The total convection heat loss (\(Q\)) can be calculated using the formula: \[Q = 2 * h * A * (T_s - T_{\infty})\] Where, \(A\) is the area of one surface of the plate. \[A = L * w = 0.025 * 0.05 = 0.00125 \, m^2\] Now, let's calculate the convection heat loss: \[Q = 2 * 61.24 * 0.00125 * (80 - 25) = 8.48 \, W\]

We can calculate the volume of the plate using the formula: \[V_{plate} = t * L * w = 0.001 * 0.025 * 0.05 = 1.25 \times 10^{-6} \, m^3\]

Given that the cylinder has the same volume as the plate, we can calculate the cylinder's radius (r) and length (\(L_c\)) using the following equations: \[V_{cylinder} = V_{plate}\] \[\pi * r^2 * L_c = 1.25 \times 10^{-6}\] Given that \(L_c = 50mm = 0.05m\), we can now solve for the radius: \[r = \sqrt{\frac{1.25 \times 10^{-6}}{3.14 * 0.05}} = 0.001599 \, m\]

To find the heat transfer coefficient for the cylinder, we will use the Churchill-Bernstein correlation for the Nusselt number (appropriate for isothermal laminar flow over cylinders): \[Nu = 0.3 + \frac{0.62 * Re^{1/2} * Pr^{1/3}}{\left[1 + (0.4 / Pr)^{2/3}\right]^{1/4} * \left[1 + (Re / 282000)^{5/8}\right]^{4/5}}\] Now, let's calculate the Reynolds number for the cylinder: \[Re_c = \frac{u_{\infty} L_c}{\nu} = \frac{3 * 0.05}{15.69 \times 10^{-6}} = 9590\] Now, calculate the Nusselt number using the Churchill-Bernstein correlation: \[Nu_c = 0.3 + \frac{0.62 * (9590)^{1/2} * (0.71)^{1/3}}{\left[1 + (0.4 / 0.71)^{2/3}\right]^{1/4} * \left[1 + (9590 / 282000)^{5/8}\right]^{4/5}} = 58.17\] The heat transfer coefficient for the cylinder can then be calculated as: \[h_c = \frac{Nu_c * k}{L_c}\] \[h_c = \frac{58.17 * 0.02624}{0.05} = 32.23 \, W/m^2K\]

Now, let’s calculate the convection heat loss for the cylinder: \[Q_c = h_c * A_c * (T_s - T_{\infty})\] The surface area of the cylinder (\(A_c\)) is: \[A_c = 2 * \pi * r * L_c = 2 * 3.14 * 0.001599 * 0.05 = 0.0016 \, m^2\] Now, calculate the convection heat loss: \[Q_c = 32.23 * 0.0016 * (80 - 25) = 2.84 \, W\]

We have calculated the convection heat losses for both the flat plate and the equivalent cylinder. Now let's compare these values: For the flat plate: \(Q = 8.48 \, W\) For the cylinder: \(Q_c = 2.84 \, W\) In this case, the convection heat loss from the flat plate is significantly higher than that from the cylinder, with the plate experiencing approximately 2.98 times more heat loss than the cylinder.