A1: 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 s2, when the fist piston has a stroke of s1= 400 mm.
- The pressure within this pressure system in bar and Pascal.
A2: 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 * s1 = F2 * s2 = 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 * sPedal/2 = FPiston* scontact point (one side level)
=> FKolben = 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