With the help of a hydraulic press, a load of 80 kN is to be lifted. A force of F1 = 200 N is applied to the first piston with the diameter of 30 mm. Calculate:
- The required diameter of the load piston in mm
- The piston stroke d2, when the fist piston has a stroke of d1= 400 mm.
- The pressure within this pressure system in bar and Pascal.
In the hydraulic brake system shown below, the brake pedal is 30 cm long. The piston on the brake pedal has an area of 5 cm2. The two brake pistons have a diameter of 50 mm. Calculate the total force which brakes the wheel!
Solution A1: hydraulic press:
1. W1 = W2 ⇒ F1 * d1 = F2 * d2 = 30 Nm ⇒ F2 = 30 Nm / 0,0011 m = 27,27 kN
F2 stands for the weight of the mass, which is lifted by the load piston:
FG = F2 = m * g ⇒ m = F2 / g = 27,27 kN / 9,81 m/s2 = 2779,82 kg
2. Solution: d2 = 200 mm
3. p2 = F2 / A2 = 27270 N / 314,15 cm2 = 136,01 bar = 1,3601 * 107 Pa = 13,601 MPa
Solution A2: hydraulic brake system
FPedal * dPedal/2 = FPiston* scontact point (one side level)
=> Fpiston = FPedal * sPedal/2 = 60 N * 15 cm = 90 N
. scontact point 10 cm
Pressure within the system: p = FPiston / APiston = 90 N /5 cm2 = 18 N / cm2 = 1,8 bar
FBrake pad = p * ABrake pad , ABrake pad = (ᴫ d2Brake pad) / 4= ᴫ 52 cm2 / 4= 19,63 cm2
FBrake pad = 18 N / cm2 * 19,63 cm2 = 353,34 N
Total brake force = 2 * FBrake pad = 2* 353,34 N = 706,68 N