Oil at \(60^{\circ} \mathrm{C}\) flows at a velocity of \(20 \mathrm{~cm} / \mathrm{s}\) over a \(5.0\)-m-long and \(1.0-\mathrm{m}\)-wide flat plate maintained at a constant temperature of \(20^{\circ} \mathrm{C}\). Determine the rate of heat transfer from the oil to the plate if the average oil properties are: \(\rho=880 \mathrm{~kg} / \mathrm{m}^{3}, \mu=0.005 \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}, k=0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(c_{p}=2.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\).

The Reynolds number (Re) is important in determining the flow regime (laminar or turbulent) of the fluid and can be calculated using the formula: \(Re = \frac{\rho vL}{\mu}\) where \(ρ\) is the fluid density, \(v\) is the flow velocity, \(L\) is the characteristic length, and \(μ\) is the dynamic viscosity of the fluid. Here, the characteristic length would be the length of the flat plate. Using the given values, \(Re = \frac{880 \mathrm{~kg/m^{3}} \times 0.2 \mathrm{~m/s} \times 5.0 \mathrm{~m}}{0.005 \mathrm{~kg/m \cdot s}} = 176000\)

The Prandtl number (Pr) is calculated using the following formula: \(Pr = \frac{c_{p} \mu}{k}\) where \(c_{p}\) is the specific heat at constant pressure, and \(k\) is the thermal conductivity of the fluid. Using the given values, \(Pr = \frac{2000 \mathrm{~J/kg \cdot K} \times 0.005 \mathrm{~kg/m \cdot s}}{0.15 \mathrm{~W/m \cdot K}} = 66.67\)

Since the flow is turbulent (Re > 10000), we will use the Dittus-Boelter equation for the Nusselt number (Nu) for turbulent flow over a flat plate: \(Nu = 0.0296 Re^{4 / 5} Pr^{n}\) For cooling cases, \(n = 1 / 3\). So, \(Nu = 0.0296 \times (176000)^{4 / 5} \times (66.67)^{1 / 3} = 4250.91\)

The convective heat transfer coefficient (h) can be calculated using the Nusselt number and the thermal conductivity: \(h = \frac{Nu \times k}{L}\) \(h = \frac{4250.91 \times 0.15 \mathrm{~W/m \cdot K}}{5.0 \mathrm{~m}} = 127.53 \mathrm{~W/m^{2} \cdot K}\)

Now that we have the convective heat transfer coefficient, we can use it to find the rate of heat transfer from the oil to the plate using the following formula: \(q = h \times A \times (T_{\mathrm{oil}} - T_{\mathrm{plate}})\) where \(A\) is the surface area of the flat plate, and \(T_{\mathrm{oil}}\) and \(T_{\mathrm{plate}}\) are the temperatures of the oil and the plate, respectively. Since the plate is 5.0 m long and 1.0 m wide, its surface area is: \(A = 5.0 \mathrm{~m} \times 1.0 \mathrm{~m} = 5.0 \mathrm{~m^{2}}\) Given temperatures are in Celsius, we have: \(T_{\mathrm{oil}} = 60^{\circ} \mathrm{C} = 333.15 \mathrm{~K}\) \(T_{\mathrm{plate}} = 20^{\circ} \mathrm{C} = 293.15 \mathrm{~K}\) So, the rate of heat transfer is: \(q = 127.53 \mathrm{~W/m^{2} \cdot K} \times 5.0 \mathrm{~m^{2}} \times (333.15 \mathrm{~K} - 293.15 \mathrm{~K}) = 25403.9 \mathrm{~W}\)