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34. The speed of a hydraulic cylinder is controlled by means of a series pressure-compensated flow control valve, as shown in the given circuit. Given: Pump: V, - 25 cm'/rev п 1000 грm n, - 0.95 n. - 0.93 7, - 1 Directional control valve: Pa, Pla Q= 6×107JAP Relief valve: Cracking pressure = 6 MPa Override pressure 0 Hydraulic cylinder: P, f= 2000 Ns/m v = 0.1 m/s F = 9000 N Piston diameter = 60 mm Rod diameter = 25 mm No inner leakage Calculate: P=0 P, P, P, P Qu Qy Power losses in the DCV Power losses in the FCV (M- Pump real flow rate

34. The speed of a hydraulic cylinder is controlled by means of a series pressure-compensated flow control valve, as shown in the given circuit. Given: Pump: V, - 25 cm'/rev п 1000 грm n, - 0.95 n. - 0.93 7, - 1 Directional control valve: Pa, Pla Q= 6×107JAP Relief valve: Cracking pressure = 6 MPa Override pressure 0 Hydraulic cylinder: P, f= 2000 Ns/m v = 0.1 m/s F = 9000 N Piston diameter = 60 mm Rod diameter = 25 mm No inner leakage Calculate: P=0 P, P, P, P Qu Qy Power losses in the DCV Power losses in the FCV (M- Pump real flow rate

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
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34. The speed of a hydraulic cylinder is controlled by means of a series pressure-compensated flow control valve, as shown in the given circuit. Given: Pump: V, = 25 cm'/rev n = 1000 rpm n, = 0.95 n. = 0.93 n. = 1 P,Q, Pa, Directional control valve: Q= 6x10-7 VAP Relief valve: Cracking pressure = 6 MPa P. Override pressure = 0 P=0 Hydraulic cylinder: f= 2000 Ns/m v = 0.1 m/s F= 9000 N Piston diameter = 60 mm Rod diameter = 25 mm No inner leakage Calculate: P=0 P, P, P, Pp Qu Qy Power losses in the DCV M) Power losses in the FCV Pump real flow rate
Transcribed Image Text:34. The speed of a hydraulic cylinder is controlled by means of a series pressure-compensated flow control valve, as shown in the given circuit. Given: Pump: V, = 25 cm'/rev n = 1000 rpm n, = 0.95 n. = 0.93 n. = 1 P,Q, Pa, Directional control valve: Q= 6x10-7 VAP Relief valve: Cracking pressure = 6 MPa P. Override pressure = 0 P=0 Hydraulic cylinder: f= 2000 Ns/m v = 0.1 m/s F= 9000 N Piston diameter = 60 mm Rod diameter = 25 mm No inner leakage Calculate: P=0 P, P, P, Pp Qu Qy Power losses in the DCV M) Power losses in the FCV Pump real flow rate
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I confused following this picture why (3.85-1.033)*10^-6 

6. Power loss in DCV Q=6 × 10-7√√Ap =6 × 10-7√(3.85 – 1.033) × 106 -3 =1.007 × 10-³ m³/s
Transcribed Image Text:6. Power loss in DCV Q=6 × 10-7√√Ap =6 × 10-7√(3.85 – 1.033) × 106 -3 =1.007 × 10-³ m³/s
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