Question

A hydraulic circuit uses 25 litres of fluid per min. The fluid is supplied by a pump having a fixed displacement of 12.5 cm^3 per revolution driven at 3000 rpm. The pump has a volumetric efficiency of 0.85 and torque efficiency of 0.88. If the system pressure is set at 18 MPa by the relief valve. What will be the power required to drive the pump?

(a) 13.48 KW

(b) 12.20 KW

(c) 11.09 KW

(d) 12.78 KW

This question was posed to me in an interview for job.

Asked question is from Numericals on Estimation of Tractor Power topic in portion Tillage, Disc Plough & Numericals of Farm Machinery

Answer 1

Correct option is (d) 12.78 KW

Easy explanation: np = \(\frac{Dp* ∆p}{2πTp} \)

Tp = \(\frac{12.6*10^{-6}*18*10^6}{2π*0.88}\)=40.69 Nm

P required = 2π*Tp*Np

P = 2π*40.69*3000/60

P = 12.78 KW.