A mass loaded accumulator has a ram diameter of 400 mm and a mass of 580 kg. A hydraulic pressure of 1,0 MPa is required as constant pressure in the hydraulic system. The ram moves through a distance of 250 mm in 4 s during a working stroke of the machine. Calculate the following:

$$A = \frac{\pi D^2}{4} = \frac{\pi (0.4 \text{ m})^2}{4} = \frac{\pi \times 0.16}{4} = 0.04\pi \text{ m}^2 \approx 0.12566 \text{ m}^2$$

6.1.1 The additional mass required to maintain the working hydraulic pressure.

The pressure is given by $P = \frac{F}{A} = \frac{m_{total} \cdot g}{A}$.

$$m_{total} = \frac{P \cdot A}{g} = \frac{(1.0 \times 10^6 \text{ Pa}) \times (0.04\pi \text{ m}^2)}{9.81 \text{ m/s}^2}$$ $$m_{total} = \frac{125663.7}{9.81} \approx 12809.76 \text{ kg}$$

$$m_{additional} = m_{total} - m_{initial} = 12809.76 \text{ kg} - 580 \text{ kg} = 12229.76 \text{ kg}$$

The additional mass required is $\boxed{\text{12229.76 kg}}$.

6.1.2 The work done by the ram in the working stroke.

$$W = F \cdot s = (125663.7 \text{ N}) \times (0.25 \text{ m}) = 31415.925 \text{ J}$$

The work done by the ram is $\boxed{\text{31415.93 J}}$.

6.1.3 The power transmitted by the ram during the working stroke.

$$P_{power} = \frac{W}{t} = \frac{31415.925 \text{ J}}{4 \text{ s}} = 7853.98 \text{ W}$$

The power transmitted by the ram is $\boxed{\text{7853.98 W}}$.

6.2

• Given: Plunger diameter $d = 80 \text{ mm} = 0.08 \text{ m}$. Stroke length $L = 200 \text{ mm} = 0.2 \text{ m}$. Number of cylinders $N_c = 3$. Delivery pressure $P = 820 \text{ kPa} = 820 \times 10^3 \text{ Pa}$. Pump speed $N = 180 \text{ r/min}$. Overall efficiency $\eta_{overall} = 88\% = 0.88$. Slip $= 5\% = 0.05$.

Since it's a three-cylinder pump, and assuming it's single-acting, $N = 180 \text{ strokes/min}$.

6.2.1 The power required to drive the pump at 180 r/min if the overall efficiency is 88%.

$$Q_{theoretical} = N_c \cdot A_p \cdot L \cdot N$$ $$Q_{theoretical} = 3 \times (0.0016\pi \text{ m}^2) \times (0.2 \text{ m}) \times (180 \text{ min}^{-1})$$ $$Q_{theoretical} = 0.542867 \text{ m}^3/\text{min}$$

$$P_{out} = P \cdot Q_{theoretical} = (820 \times 10^3 \text{ Pa}) \times \left(\frac{0.542867}{60} \text{ m}^3/\text{s}\right)$$ $$P_{out} = 820 \times 10^3 \times 0.00904778 \approx 7419.18 \text{ W}$$

$$P_{in} = \frac{P_{out}}{\eta_{overall}} = \frac{7419.18 \text{ W}}{0.88} \approx 8430.89 \text{ W}$$

The power required to drive the pump is $\boxed{\text{8430.89 W}}$.

6.2.2 The volume of water delivered per minute in litres, if the pump has a slip of 5%.

$$Q_{actual} = Q_{theoretical} \times (1 - \text{slip})$$ $$Q_{actual} = (0.542867 \text{ m}^3/\text{min}) \times (1 - 0.05)$$ $$Q_{actual} = 0.542867 \times 0.95 = 0.51572365 \text{ m}^3/\text{min}$$

$$Q_{actual\_litres} = 0.51572365 \text{ m}^3/\text{min} \times 1000 \text{ litres/m}^3 = 515.72 \text{ litres/min}$$

The volume of water delivered per minute is $\boxed{\text{515.72 litres/min}}$.

6.3

• Given: Ram diameter $D_{ram} = 90 \text{ mm} = 0.09 \text{ m}$. Plunger diameter $d_{plunger} = 18 \text{ mm} = 0.018 \text{ m}$. Plunger stroke length $L_{plunger} = 35 \text{ mm} = 0.035 \text{ m}$. Lever mechanical advantage $MA = 10$. Load $W = 4 \text{ tons} = 4000 \text{ kg}$. Efficiency $\eta = 80\% = 0.80$. Slip $= 4\% = 0.04$. Ram lift $H_{ram} = 180 \text{ mm} = 0.18 \text{ m}$. Assume $g = 9.81 \text{ m/s}^2$.

$$A_{ram} = \frac{\pi D_{ram}^2}{4} = \frac{\pi (0.09 \text{ m})^2}{4} = \frac{\pi \times 0.0081}{4} = 0.002025\pi \text{ m}^2 \approx 0.0063617 \text{ m}^2$$ $$A_{plunger} = \frac{\pi d_{plunger}^2}{4} = \frac{\pi (0.018 \text{ m})^2}{4} = \frac{\pi \times 0.000324}{4} = 0.000081\pi \text{ m}^2 \approx 0.00025447 \text{ m}^2$$

6.3.1 The force required to lift a 4-ton load if the efficiency of the press is 80%.

The pressure is constant throughout the hydraulic system: $P = \frac{F_{plunger\_theoretical}}{A_{plunger}} = \frac{F_{load}}{A_{ram}}$.

$$F_{plunger\_theoretical} = F_{load} \frac{A_{plunger}}{A_{ram}} = (39240 \text{ N}) \times \frac{0.000081\pi \text{ m}^2}{0.002025\pi \text{ m}^2}$$ $$F_{plunger\_theoretical} = 39240 \times \frac{0.000081}{0.002025} = 39240 \times 0.04 = 1569.6 \text{ N}$$

$$F_{lever} = \frac{F_{applied\_plunger}}{MA} = \frac{1962 \text{ N}}{10} = 196.2 \text{ N}$$

The force required to lift a 4-ton load is $\boxed{\text{196.2 N}}$.

6.3.2 The number of strokes required to raise the load 180 mm if the hydraulic system has a slip of 4%.

$$V_{actual\_stroke} = V_{plunger\_stroke} \times (1 - \text{slip}) = (0.000002835\pi \text{ m}^3) \times (1 - 0.04)$$ $$V_{actual\_stroke} = 0.000002835\pi \times 0.96 \text{ m}^3 \approx 0.000008551 \text{ m}^3$$

$$N_{strokes} = \frac{V_{ram\_lift}}{V_{actual\_stroke}} = \frac{0.0011451 \text{ m}^3}{0.000008551 \text{ m}^3} \approx 133.91$$

Since the number of strokes must be an integer, we round up.

The number of strokes required is $\boxed{\text{134 strokes}}$.