Determine the Flow Regime and Calculate Drag Force and Heat Transfer Rate
Since the Reynolds number is \(750,000\), which is greater than \(5 \times 10^5\), the flow is considered to be turbulent. For a turbulent flow over a flat plate, we can use the following correlations to calculate the drag force and heat transfer rate:
For drag force, the formula is:
\(F_D = \frac{1}{2} \rho u^2 C_D A\)
where \(F_D\) is the drag force, \(C_D\) is the drag coefficient, and \(A\) is the area of the plate. The drag coefficient can be found using the formula:
\(C_D = \frac{0.074}{Re_x^{1/5}}\)
For the heat transfer rate, the formula is:
\(Q = h A \Delta T\)
where \(Q\) is the heat transfer rate, \(h\) is the convective heat transfer coefficient, and \(\Delta T\) is the temperature difference between the plate and the fluid. The heat transfer coefficient can be calculated using the formula:
\(h = \frac{k}{x} * \frac{0.0296 * Re_x^{4/5} Pr^{1/3}}{(1 + (0.4 / Pr)^{2/3})^{1/4}}\)
where \(k\) is the thermal conductivity of the fluid, and \(Pr\) is the Prandtl number. For engine oil, assume the standard values: \(k = 0.145 \, W/m*K\) and \(Pr = 1200\).
Now, we can calculate the drag coefficient, drag force, and heat transfer rate.
\(C_D = \frac{0.074}{(750,000)^{1/5}} = 0.00445\)
The area of the plate per unit width is \(A = 10 \, m\) (since the width is given as 10 meters).
\(F_D = \frac{1}{2} (900 \, kg/m^3) (2.5 \, m/s)^2 (0.00445) (10 \, m) = 313.1 \, N\)
\(h = \frac{0.145 \, W/m*K}{10 \, m} * \frac{0.0296 * (750,000)^{4/5} (1200)^{1/3}}{(1 + (0.4 / 1200)^{2/3})^{1/4}} = 341.2 \, W/m^2*K\)
\(\Delta T = 85^{\circ}C - 35^{\circ}C = 50^{\circ}C\)
\(Q = (341.2 \, W/m^2*K) (10 \, m) (50 \, K) = 170,600 \, W\)
In conclusion, the total drag force over the entire plate per unit width is \(313.1 \, N\), and the rate of heat transfer over the entire plate per unit width is \(170,600 \, W\).
