Question

Consider a tractor that delivers a total of 50 gpm through the remote outlets at rated speed of 2200 rpm. The pump operates at engine speed. Neglect efficiencies in this problem.

a) What is the displacement of the pump?

b) At maximum outlet pressure of 3000 psi, how much power is going into the hydraulic pump and system?

c) At maximum outlet pressure of 3000 psi, how much torque is the pump requiring?

d) Sketch the total flow versus pressure curve for this tractor.

e) Sketch the total flow versus engine speed curve for this tractor.

f) Sketch the maximum pressure versus engine speed curve for this tractor.

Answer 1

A)

Answer:-

Pump Displacement = 231 x GPM ÷ RPM

= 131 x 50/2200

= 5.25 Gallon/Revolution

= 2.10 Cubic Inches

B)

Answer:-

If Outlet Pressure = 3000 Psi

Power = GPM X PSI X .0007

= 50 X 3000 X .0007

= 1050 horsepower

= 782.984 KW

C)

Answer:-

1) Motor Torque from GPM, PSI and RPM:-

Torque = GPM X PSI X 36.77 / RPM

= 50 * 3000 * 36.77 / 2200

= 20507.045 inch pounds second

2) Motor Torque from Pressure and Displacement:

Torque = Horsepower X 63025 / RPM

   = 1050 * 63025 / 2200

= 30080.113 inch pound

answered by: anonymous