Question

Engine oil at \(105^{\circ} \mathrm{C}\) flows over the surface of a flat plate whose temperature is \(15^{\circ} \mathrm{C}\) with a velocity of $1.5 \mathrm{~m} / \mathrm{s}\(. The local drag force per unit surface area \)0.8 \mathrm{~m}$ from the leading edge of the plate is (a) \(21.8 \mathrm{~N} / \mathrm{m}^{2}\) (b) \(14.3 \mathrm{~N} / \mathrm{m}^{2}\) (c) \(10.9 \mathrm{~N} / \mathrm{m}^{2}\) (d) \(8.5 \mathrm{~N} / \mathrm{m}^{2}\) (e) \(5.5 \mathrm{~N} / \mathrm{m}^{2}\) (For oil, use $\nu=8.565 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}, \rho=864 \mathrm{~kg} / \mathrm{m}^{3}$ )

Short answer

$$ R_{e x}=\approx 14050 $$ #tag_title#Step 2: Determine the skin friction coefficient Cf# #tag_content#Since the calculated Reynolds number is less than 500,000, we can assume the flow is laminar. For laminar flow over a flat plate, we can use the Blasius equation to determine the skin friction coefficient: $$ C_{f}=\frac{1.328}{\sqrt{R_{e x}}} $$ Calculating \(C_{f}\): $$ C_{f}=\frac{1.328}{\sqrt{14050}} $$

Step-by-step solution

Calculate Reynold's number x#

To calculate the Reynold's number, use the formula: $$ R_{e x}=\frac{U x}{\nu} $$ Where \(R_{e x}\) is the Reynold's number at a distance x from the leading edge, \(U\) is the velocity of the oil (1.5 m/s), x is the distance from the leading edge (0.8 m), and \(\nu\) is the kinematic viscosity of the oil (\(8.565 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\)). Calculating \(R_{e x}\): $$ R_{e x}=\frac{1.5 \cdot 0.8}{8.565 \times 10^{-5}} $$