ABE TUTORIALS

PROBLEM 13:

HOW MUCH HORSEPOWER IS NEEDED TO DRIVE A PUMP WITH 45.4 LI/MIN CAPACITY FLOWING AT 141 KG/CM². THE EFFICIENCY OF THE SYSTEM IS 80%.

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Select the required horsepower:

a. 14 | b. 17.5 | c. 13.0 | d. 11.2 | e. nota

View Solution

Step 1: Identify Given Values.
Flow Rate (Q) = 45.4 L/min
Pressure (P) = 141 kg/cm²
Efficiency (η) = 80% or 0.80

Step 2: Use the Hydraulic Power Formula.
Power (hp) = (Pressure × Flow Rate) / (Constant × Efficiency)
Using metric units where 1 hp ≈ 450 (kg/cm² · L/min):
Power = (141 kg/cm² × 45.4 L/min) / (450 × 0.80)

Step 3: Solve.
Power = 6401.4 / 360
Power ≈ 17.78 hp

Refined Calculation (using Watts):
Pressure in Pascals ≈ 13,827,376 Pa
Flow in m³/s ≈ 0.0007567 m³/s
Theoretical Power ≈ 10,463 Watts
Actual Power = 10,463 / (0.80 × 746) ≈ 17.53 hp

Result: b. 17.5