Subtopic: Steering and hydraulic control systems used in tractors
Question Statement:
A hydraulic system comprising of a pump and a single acting cylinder lifts 11 kN load. The pump flow rate is 25 L min⁻¹ and its overall efficiency is 80%. The cylinder diameter is 80 mm and its efficiency is 90%. If the pressure drop in the hydraulic circuit is 500 kPa, the power required to drive the pump in kW will be __________.
Official Answer Key:
1.4 to 1.7
Step-by-Step Solution & Explanation
• Cylinder area, A = (π/4)(0.08)² = 5.0265×10⁻³ m².
• Pressure needed at the cylinder to lift the load (accounting for cylinder efficiency): P_cyl = Load/(A×ηcyl) = 11,000/(5.0265×10⁻³ × 0.9) = 11,000/4.524×10⁻³ ≈ 2431 kPa.
• Pump delivery pressure must also overcome the circuit pressure drop: P_pump = 2431 + 500 = 2931 kPa.
• Flow, Q = 25 L/min = 4.1667×10⁻⁴ m³/s.
• Hydraulic output power of pump = P×Q = 2,931,000 × 4.1667×10⁻⁴ ≈ 1221 W.
• Pump input (shaft) power, accounting for 80% overall efficiency = 1221/0.8 ≈ 1527 W ≈ 1.53 kW, inside the accepted range 1.4–1.7.
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