Question

Engine oil flows in a 15 -cm-diameter horizontal tube with a velocity of $1.3 \mathrm{~m} / \mathrm{s}\(, experiencing a pressure drop of \)12 \mathrm{kPa}$. The pumping power requirement to overcome this pressure drop is (a) \(190 \mathrm{~W}\) (b) \(276 \mathrm{~W}\) (c) \(407 \mathrm{~W}\) (d) \(655 \mathrm{~W}\) (e) \(900 \mathrm{~W}\)

Short answer

Answer: (b) 276 W

Step-by-step solution

Calculate the area of the tube

We are first going to find the area of the cross-section of the tube, which will be used to calculate the flow rate. Area (m^2) = π × (Diameter/2)^2 We convert diameter from cm to m: Diameter = 15 cm = 0.15 m Area (m^2) = π × (0.15 / 2)^2 = 0.0177 m^2

Calculate the flow rate of the engine oil

Now we will find the flow rate of the oil in the tube using the area we calculated in step 1 and the given velocity of the oil. Flow Rate (m^3/s) = Area of the tube (m^2) × Velocity of the oil (m/s) Flow Rate (m^3/s) = 0.0177 m^2 × 1.3 m/s = 0.02301 m^3/s

Convert pressure drop from kPa to Pa

We will need to convert the given pressure drop from kPa to Pa for use in the pumping power formula. 1 kPa = 1000 Pa Pressure Drop = 12 kPa = 12 × 1000 = 12,000 Pa

Calculate the pumping power requirement

Finally, we will calculate the pumping power requirement using the flow rate and the pressure drop. Pumping Power (W) = Flow Rate (m^3/s) × Pressure Drop (Pa) Pumping Power (W) = 0.02301 m^3/s × 12,000 Pa = 276.12 W The correct answer to this problem is (b) 276 W (rounded to the nearest whole number).