Columnist Dave Barry poked fun at the name "The Grand Cities" adopted by Grand Forks, North Dakota, and East Grand Forks, Minnesota. Residents of the prairie towns then named their next municipal building for him. At the Dave Barry Lift Station No. 16, untreated sewage is raised vertically by 5.49 m, at the rate of 1 890 000 liters each day. The waste, of density 1 050 kg/m3, enters and leaves the pump at atmospheric pressure, through pipes of equal diameter.(a) Find the output mechanical power of the lift station.(b) Assume an electric motor continuously operating with average power 5.90 kW runs the pump. Find its efficiency.

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To answer these questions, we need to calculate the output mechanical power of the lift station and then find its efficiency. We can do this by using the formula for power and efficiency.

(a) To find the output mechanical power of the lift station, we can use the formula:
Output Power = Force × Velocity

Let's calculate the force first. Since the density of the waste and the diameter of the pipes are not given, we can assume that the waste acts as an incompressible fluid and the pipes have a constant diameter.

The volume flow rate (Q) of the waste can be found by multiplying the velocity of the waste (v) with the cross-sectional area of the pipe (A):
Q = v × A

We are given that the waste is raised vertically, so the velocity of the waste (v) can be calculated using the height (h) it is raised through and the time (t) it takes:
v = h / t

Given: h = 5.49 m and t = 1 day = 24 hours = 24 × 60 × 60 seconds

Now, let's calculate the cross-sectional area (A) of the pipe. Since the pipes are of equal diameter, we can assume the same diameter for both the inlet and outlet pipes.

Let d be the diameter of the pipes. The cross-sectional area (A) is given by:
A = π × (d/2)² = (π/4) × d²

Now, substituting the values into the equations:
v = 5.49 m / (24 × 60 × 60 s) ≈ 0.000064 m/s

A = (π/4) × d²

Given the volume flow rate (Q) is 1,890,000 liters/day ≈ (1890000/1000) / (24 × 60 × 60) m³/s ≈ 0.022 m³/s

We can rearrange the equation Q = v × A to solve for A:
A = Q / v

Substituting the values:
A = 0.022 m³/s / 0.000064 m/s ≈ 343.75 m²

Now, we can calculate the force (F) acting on the waste using the formula:
F = density × gravity × volume
where density (ρ) of the waste is given as 1050 kg/m³ and gravity (g) is approximately 9.8 m/s²:

F = 1050 kg/m³ × 9.8 m/s² × 0.022 m³/s ≈ 2274 N

Finally, we can calculate the output mechanical power using the formula:
Output Power = Force × Velocity

Output Power = 2274 N × 0.000064 m/s ≈ 0.15 W

Therefore, the output mechanical power of the lift station is approximately 0.15 watts.

(b) To find the efficiency of the electric motor running the pump, we can use the formula:
Efficiency = Output Power / Input Power

Given that the average power of the electric motor is 5.90 kW:
Input Power = 5.90 kW = 5900 watts

Efficiency = 0.15 W / 5900 W ≈ 0.0000254 or 0.00254%

Therefore, the efficiency of the electric motor running the pump is approximately 0.00254%.