A vented tanker is to be filled with fuel oil with density = 920 kg/m3 and viscosity = 0.045 Pa s from an underground reservoir using a 25-m-long, 4-cm-diameter smooth plastic hose. The elevation difference between the oil level in the reservoir and the top of the tanker where the hose is discharged is 5 m. The capacity of the tanker is 18 m3 and the filling time is 30 min. Assume the underground reservoir has its pressure maintained at atmospheric level. Determine the required power input to the pump if the mechanical efficiency of the pump is 82%. Express your answer in kW.

The first step is to calculate the flow rate of the oil. The capacity of the tanker is 18 m3 and it is filled in 30 minutes. Therefore, the flow rate (Q) is:

Q = Volume / Time Q = 18 m3 / 1800 s = 0.01 m3/s

Next, we calculate the velocity (v) of the oil in the hose using the flow rate and the cross-sectional area (A) of the hose. The diameter (d) of the hose is 4 cm = 0.04 m, so the radius (r) is d/2 = 0.02 m. The cross-sectional area is A = πr^2 = π(0.02 m)^2 = 0.001256 m2. Therefore, the velocity is:

v = Q / A v = 0.01 m3/s / 0.001256 m2 = 7.96 m/s

Next, we calculate the dynamic pressure (Pd) using the density (ρ) of the oil, the velocity of the oil, and the elevation difference (h). The dynamic pressure is:

Pd = ρgh + 0.5ρv^2 Pd = (920 kg/m3)(9.81 m/s2)(5 m) + 0.5(920 kg/m3)(7.96 m/s)^2 Pd = 45156 Pa + 27730 Pa = 72886 Pa

Next, we calculate the power (P) required to pump the oil using the dynamic pressure, the flow rate, and the mechanical efficiency (η) of the pump. The power is:

P = PdQ / η P = (72886 Pa)(0.01 m3/s) / 0.82 = 889.1 W = 0.889 kW

Therefore, the required power input to the pump is approximately 0.889 kW.