Calculate the following: 1.1 The additional mass required to maintain the working hydraulic pressure. 1.2 The work done by the ram in the working stroke. 1.3 The power transmitted by the ram during the working stroke.

$$m_{add} = M - m_0 = 12809.77 \text{ kg} - 580 \text{ kg} = \boxed{\text{12229.77 kg}}$$

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

$$W = F \times s = (40000\pi \text{ N}) \times (0.25 \text{ m}) = 10000\pi \text{ J} = \boxed{\text{31415.93 J}}$$

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

$$P_{transmitted} = \frac{W}{t} = \frac{10000\pi \text{ J}}{4 \text{ s}} = 2500\pi \text{ W} = \boxed{\text{7853.98 W}}$$

6.2 The plunger of a three-cylinder water pump

Given:
Number of cylinders $N = 3$
Plunger diameter $d = 80 \text{ mm} = 0.08 \text{ m}$
Stroke length $L = 200 \text{ mm} = 0.2 \text{ m}$
Delivery pressure $P = 820 \text{ kPa} = 820 \times 10^3 \text{ Pa}$

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

Assuming a single-acting pump, the number of strokes per second for each cylinder is $180 \text{ r/min} / 60 \text{ s/min} = 3 \text{ s}^{-1}$.

$$Q_{theoretical} = N \times A_{plunger} \times L \times \frac{N_{rpm}}{60}$$ $$Q_{theoretical} = 3 \times (0.0016\pi \text{ m}^2) \times (0.2 \text{ m}) \times (3 \text{ s}^{-1}) = 0.00288\pi \text{ m}^3/\text{s}$$

$$P_{in} = \frac{P_{out}}{\eta} = \frac{2361.6\pi \text{ W}}{0.88} = \boxed{\text{8430.68 W}}$$

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

$$Q_{actual, min} = Q_{theoretical, min} \times (1 - \text{slip})$$ $$Q_{actual, min} = (0.1728\pi \text{ m}^3/\text{min}) \times (1 - 0.05) = 0.16416\pi \text{ m}^3/\text{min}$$

$$Q_{actual, litres/min} = Q_{actual, min} \times 1000 \text{ litres/m}^3$$ $$Q_{actual, litres/min} = (0.16416\pi \text{ m}^3/\text{min}) \times 1000 = \boxed{\text{515.69 litres/min}}$$

6.3 A hydraulic press

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 $L_{plunger} = 35 \text{ mm} = 0.035 \text{ m}$
Mechanical advantage of lever $MA_{lever} = 10$

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

$$A_{ram} = \frac{\pi D_{ram}^2}{4} = \frac{\pi (0.09 \text{ m})^2}{4} = 0.002025\pi \text{ m}^2$$ $$A_{plunger} = \frac{\pi d_{plunger}^2}{4} = \frac{\pi (0.018 \text{ m})^2}{4} = 0.000081\pi \text{ m}^2$$

The pressure in the system is $P = \frac{F_{load}}{A_{ram}}$. The theoretical force on the plunger is $F_{plunger, theoretical} = P \times A_{plunger}$.

$$F_{plunger, theoretical} = F_{load} \times \frac{A_{plunger}}{A_{ram}} = 39240 \text{ N} \times \frac{0.000081\pi \text{ m}^2}{0.002025\pi \text{ m}^2}$$ $$F_{plunger, theoretical} = 39240 \text{ N} \times \frac{0.000081}{0.002025} = 39240 \text{ N} \times \frac{1}{25} = 1569.6 \text{ N}$$

$$F_{input} = \frac{F_{plunger, actual}}{MA_{lever}} = \frac{1962 \text{ N}}{10} = \boxed{\text{196.2 N}}$$

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

$$H_{ram} = 180 \text{ mm} = 0.18 \text{ m}$$ $$V_{ram} = A_{ram} \times H_{ram} = (0.002025\pi \text{ m}^2) \times (0.18 \text{ m}) = 0.0003645\pi \text{ m}^3$$

$$V_{plunger, actual} = V_{plunger, theoretical} \times (1 - \text{slip})$$ $$V_{plunger, actual} = (0.000002835\pi \text{ m}^3) \times (1 - 0.04) = 0.0000027216\pi \text{ m}^3$$

$$n_{actual} = \frac{V_{ram}}{V_{plunger, actual}} = \frac{0.0003645\pi \text{ m}^3}{0.0000027216\pi \text{ m}^3} = 133.928...$$

Since the number of strokes must be a whole number to achieve the desired lift, we round up.

$$n_{actual} = \boxed{\text{134 strokes}}$$