Answer according to engineering science n4 level

Ram piston diameter $D = 220 \text{ mm} = 0.22 \text{ m}$
Plunger diameter $d = 30 \text{ mm} = 0.03 \text{ m}$
Mechanical advantage of the lever $MA_{lever} = 15$
Load mass $m = 2 \text{ Mg} = 2000 \text{ kg}$
Acceleration due to gravity $g = 9.81 \text{ m/s}^2$
Load force $F_{ram} = m \times g = 2000 \text{ kg} \times 9.81 \text{ m/s}^2 = 19620 \text{ N}$

The ideal mechanical advantage of the jack is the ratio of the ram area to the plunger area.

A slip of 20% means the efficiency $\eta = 100\% - 20\% = 80\% = 0.80$.
The actual mechanical advantage is $MA_{actual} = \eta \times IMA_{total}$.
The force applied to the lever (effort) is $F_{effort} = \frac{F_{ram}}{MA_{actual}}$.

The force that must be applied to the lever is approximately 30.40 N.

5.1.2. Number of strokes needed to lift the load by 180 mm if there is an efficiency of 95%.

Load lift height $H = 180 \text{ mm} = 0.18 \text{ m}$
Plunger stroke length $L_s = 60 \text{ mm} = 0.06 \text{ m}$
Efficiency $\eta = 95\% = 0.95$

Volume lifted by ram $V_{ram} = A_{ram} \times H = \frac{\pi D^2}{4} \times H = \frac{\pi (0.22 \text{ m})^2}{4} \times 0.18 \text{ m} = 0.006842 \text{ m}^3$
Volume displaced per plunger stroke $V_{plunger} = A_{plunger} \times L_s = \frac{\pi d^2}{4} \times L_s = \frac{\pi (0.03 \text{ m})^2}{4} \times 0.06 \text{ m} = 0.00004241 \text{ m}^3$

Since there is an efficiency of 95%, more strokes are needed.

Since the number of strokes must be a whole number, we round up.
The number of strokes needed is 170 strokes.

5.2. A two-cylinder water pump is running at 120 r/min. The diameter of the plunger is 150 mm and the stroke length is 250 mm.

5.2.1. The volume of water delivered in l/s.

Number of cylinders $n_c = 2$ (assuming single-acting cylinders)
Running speed $N = 120 \text{ r/min} = \frac{120}{60} \text{ r/s} = 2 \text{ r/s}$
Plunger diameter $d = 150 \text{ mm} = 0.15 \text{ m}$
Stroke length $L_s = 250 \text{ mm} = 0.25 \text{ m}$

For a two-cylinder single-acting pump, there are two delivery strokes per revolution.
Volume delivered per stroke $V_{stroke} = A_{plunger} \times L_s = 0.017671 \text{ m}^2 \times 0.25 \text{ m} = 0.0044178 \text{ m}^3$
Volume delivered per second $Q = n_c \times V_{stroke} \times N = 2 \times 0.0044178 \text{ m}^3 \times 2 \text{ r/s} = 0.017671 \text{ m}^3/\text{s}$

$1 \text{ m}^3 = 1000 \text{ l}$

The volume of water delivered is approximately 17.67 l/s.

5.2.2. The pressure needed to pump the water through a height of 30 m.

Height $h = 30 \text{ m}$
Density of water $\rho = 1000 \text{ kg/m}^3$
Acceleration due to gravity $g = 9.81 \text{ m/s}^2$

The pressure needed is 294.3 kPa.

5.3. An irrigation system fills an empty circular dam with a diameter of 50 m and a height of 20 m with a pressure of 250 MPa. Determine the amount of work needed to fill the dam to capacity.

Dam diameter $D = 50 \text{ m}$
Dam height $H = 20 \text{ m}$
Density of water $\rho = 1000 \text{ kg/m}^3$
Acceleration due to gravity $g = 9.81 \text{ m/s}^2$
The pressure of 250 MPa is the pump's capability and is not directly used to calculate the work done to fill the dam (potential energy).

Volume of the dam $V = \frac{\pi D^2}{4} \times H$

When filling an empty dam, the water is lifted from the bottom to an average height of half the dam's height.

The amount of work needed to fill the dam to capacity is approximately $3.85 \times 10^9 \text{ J}$ (or 3.85 GJ).