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.

Here are the calculations for Question 6: Hydraulics.

6.1 A mass loaded accumulator

Given: Ram diameter $D = 400 mm = 0.4 \text{ m}$ Initial mass $m_0 = 580 \text{ kg}$ Hydraulic pressure $P = 1.0 MPa = 1.0 \times 10^6 \text{ Pa}$ Ram moves distance $s = 250 mm = 0.25 \text{ m}$ in time $t = 4 \text{ s}$ Assume acceleration due to gravity $g = 9.81 m/s^2$.

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

Step 1: Calculate the area of the ram. $A = \frac{\pi D^2}{4} = \frac{\pi (0.4 m)^2}{4} = 0.04\pi m^2$

Step 2: Calculate the total force required to generate the pressure. $F = P \times A = (1.0 \times 10^6 Pa) \times (0.04\pi m^2) = 40000\pi \text{ N}$

Step 3: Calculate the total mass required to produce this force. $M = \frac{F}{g} = \frac{40000\pi N}{9.81 m/s^2} \approx 12809.77 \text{ kg}$

Step 4: Calculate the additional mass required. $m_{add} = M - m_0 = 12809.77 kg - 580 kg = \text{12229.77 kg}$

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

Step 1: Use the force calculated in 6.1.1. $F = 40000\pi \text{ N}$

Step 2: Calculate the work done. $W = F \times s = (40000\pi N) \times (0.25 m) = 10000\pi J = \text{31415.93 J}$

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

Step 1: Use the work done calculated in 6.1.2. $W = 10000\pi \text{ J}$

Step 2: Calculate the power transmitted. $P_{transmitted} = \frac{W}{t} = \frac{10000\pi J}{4 s} = 2500\pi W = \text{7853.98 W}$

6.2 The plunger of a three-cylinder water pump

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

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

Step 1: Calculate the area of one plunger. $A_{plunger} = \frac{\pi d^2}{4} = \frac{\pi (0.08 m)^2}{4} = 0.0016\pi m^2$

Step 2: Calculate the theoretical flow rate of the pump. Assuming a single-acting pump, the number of strokes per second for each cylinder is $180 r/min / 60 s/min = 3 s^{-1}$. $Q_{theoretical} = N \times A_{plunger} \times L \times \frac{N_{rpm}}{60}$ $Q_{theoretical} = 3 \times (0.0016\pi m^2) \times (0.2 m) \times (3 s^{-1}) = 0.00288\pi m^3/\text{s}$

Step 3: Calculate the output power of the pump. $P_{out} = P \times Q_{theoretical} = (820 \times 10^3 Pa) \times (0.00288\pi m^3/s) = 2361.6\pi \text{ W}$

Step 4: Calculate the input power required, considering the efficiency. $P_{in} = \frac{P_{out}}{\eta} = \frac{2361.6\pi W}{0.88} = \text{8430.68 W}$

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

Step 1: Calculate the theoretical flow rate in $\text{m}^3/\text{min}$. $Q_{theoretical, min} = Q_{theoretical} \times 60 s/min = (0.00288\pi m^3/s) \times 60 = 0.1728\pi m^3/\text{min}$

Step 2: Calculate the actual flow rate considering a slip of $5\%$. $Q_{actual, min} = Q_{theoretical, min} \times (1 - slip)$ $Q_{actual, min} = (0.1728\pi m^3/min) \times (1 - 0.05) = 0.16416\pi m^3/\text{min}$

Step 3: Convert the actual flow rate to litres per minute. $Q_{actual, litres/min} = Q_{actual, min} \times 1000 litres/m^3$ $Q_{actual, litres/min} = (0.16416\pi m^3/min) \times 1000 = \text{515.69 litres/min}$

6.3 A hydraulic press

Given: Ram diameter $D_{ram} = 90 mm = 0.09 \text{ m}$ Plunger diameter $d_{plunger} = 18 mm = 0.018 \text{ m}$ Plunger stroke $L_{plunger} = 35 mm = 0.035 \text{ m}$ 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 $80\%$.

Step 1: Calculate the force exerted by the 4-ton load. $F_{load} = 4 tons \times 1000 kg/ton \times 9.81 m/s^2 = 4000 kg \times 9.81 m/s^2 = 39240 \text{ N}$

Step 2: Calculate the area of the ram and the plunger. $A_{ram} = \frac{\pi D_{ram}^2}{4} = \frac{\pi (0.09 m)^2}{4} = 0.002025\pi m^2$ $A_{plunger} = \frac{\pi d_{plunger}^2}{4} = \frac{\pi (0.018 m)^2}{4} = 0.000081\pi m^2$

Step 3: Calculate the theoretical force on the plunger. The pressure in the system is $P = \frac{F_{load}}{A_{ram}}$. The theoretical force on the plunger is $F_{plunger, theoretical} = P \times A_{plunger}$. $F_{plunger, theoretical} = F_{load} \times \frac{A_{plunger}}{A_{ram}} = 39240 N \times \frac{0.000081\pi m^2}{0.002025\pi m^2}$ $F_{plunger, theoretical} = 39240 N \times \frac{0.000081}{0.002025} = 39240 N \times \frac{1}{25} = 1569.6 \text{ N}$

Step 4: Calculate the actual force on the plunger, considering the efficiency of $80\%$. $F_{plunger, actual} = \frac{F_{plunger, theoretical}}{\eta} = \frac{1569.6 N}{0.80} = 1962 \text{ N}$

Step 5: Calculate the input force required on the lever, considering its mechanical advantage of 10. $F_{input} = \frac{F_{plunger, actual}}{MA_{lever}} = \frac{1962 N}{10} = \text{196.2 N}$

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

Step 1: Calculate the volume displaced by one theoretical plunger stroke. $V_{plunger, theoretical} = A_{plunger} \times L_{plunger} = (0.000081\pi m^2) \times (0.035 m) = 0.000002835\pi m^3$

Step 2: Calculate the volume required to lift the ram by $180 \text{ mm}$. $H_{ram} = 180 mm = 0.18 \text{ m}$ $V_{ram} = A_{ram} \times H_{ram} = (0.002025\pi m^2) \times (0.18 m) = 0.0003645\pi m^3$

Step 3: Calculate the actual volume delivered per plunger stroke, considering a slip of $4\%$. $V_{plunger, actual} = V_{plunger, theoretical} \times (1 - slip)$ $V_{plunger, actual} = (0.000002835\pi m^3) \times (1 - 0.04) = 0.0000027216\pi m^3$

Step 4: Calculate the number of strokes required. $n_{actual} = \frac{V_{ram}}{V_{plunger, actual}} = \frac{0.0003645\pi m^3}{0.0000027216\pi m^3} = 133.928...$ Since the number of strokes must be a whole number to achieve the desired lift, we round up. $n_{actual} = \text{134 strokes}$

Drop the next question.