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:

QUESTION 6: HYDRAULICS

6.1

• Given: Ram diameter $D = 400 mm = 0.4 \text{ m}$. Initial mass $m_{initial} = 580 \text{ kg}$. Required hydraulic pressure $P = 1.0 MPa = 1.0 \times 10^6 \text{ Pa}$. Ram moves distance $s = 250 mm = 0.25 \text{ m}$ in $t = 4 \text{ s}$. Assume $g = 9.81 m/s^2$.

Step 1: Calculate the area of the ram. $A = \frac{\pi D^2}{4} = \frac{\pi (0.4 m)^2}{4} = \frac{\pi \times 0.16}{4} = 0.04\pi m^2 \approx 0.12566 m^2$

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

Step 2: Calculate the total mass required to achieve the desired pressure. The pressure is given by $P = \frac{F}{A} = \frac{m_{total} \cdot g}{A}$. $m_{total} = \frac{P \cdot A}{g} = \frac{(1.0 \times 10^6 Pa) \times (0.04\pi m^2)}{9.81 m/s^2}$ $m_{total} = \frac{125663.7}{9.81} \approx 12809.76 kg$ Step 3: Calculate the additional mass. $m_{additional} = m_{total} - m_{initial} = 12809.76 kg - 580 kg = 12229.76 kg$

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

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

Step 1: Calculate the force exerted by the ram. $F = P \cdot A = (1.0 \times 10^6 Pa) \times (0.04\pi m^2) = 125663.7 N$ Step 2: Calculate the work done. $W = F \cdot s = (125663.7 N) \times (0.25 m) = 31415.925 J$

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

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

Step 1: Calculate the power. $P_{power} = \frac{W}{t} = \frac{31415.925 J}{4 s} = 7853.98 W$

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

6.2

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

Step 1: Calculate the area of one plunger. $A_p = \frac{\pi d^2}{4} = \frac{\pi (0.08 m)^2}{4} = \frac{\pi \times 0.0064}{4} = 0.0016\pi m^2 \approx 0.0050265 m^2$ Step 2: Convert pump speed to strokes per minute. 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%.

Step 3: Calculate the theoretical volume delivered per minute. $Q_{theoretical} = N_c \cdot A_p \cdot L \cdot N$ $Q_{theoretical} = 3 \times (0.0016\pi m^2) \times (0.2 m) \times (180 min^{-1})$ $Q_{theoretical} = 0.542867 m^3/min$ Step 4: Calculate the theoretical power output of the pump. $P_{out} = P \cdot Q_{theoretical} = (820 \times 10^3 Pa) \times \left(\frac{0.542867}{60} m^3/s\right)$ $P_{out} = 820 \times 10^3 \times 0.00904778 \approx 7419.18 W$ Step 5: Calculate the power required to drive the pump (input power). $P_{in} = \frac{P_{out}}{\eta_{overall}} = \frac{7419.18 W}{0.88} \approx 8430.89 W$

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

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

Step 1: Calculate the actual volume delivered per minute. $Q_{actual} = Q_{theoretical} \times (1 - slip)$ $Q_{actual} = (0.542867 m^3/min) \times (1 - 0.05)$ $Q_{actual} = 0.542867 \times 0.95 = 0.51572365 m^3/min$ Step 2: Convert the volume to litres per minute. $Q_{actual\_litres} = 0.51572365 m^3/min \times 1000 litres/m^3 = 515.72 litres/min$

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

6.3

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

Step 1: Calculate the areas of the ram and plunger. $A_{ram} = \frac{\pi D_{ram}^2}{4} = \frac{\pi (0.09 m)^2}{4} = \frac{\pi \times 0.0081}{4} = 0.002025\pi m^2 \approx 0.0063617 m^2$ $A_{plunger} = \frac{\pi d_{plunger}^2}{4} = \frac{\pi (0.018 m)^2}{4} = \frac{\pi \times 0.000324}{4} = 0.000081\pi m^2 \approx 0.00025447 m^2$

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

Step 2: Calculate the force of the load on the ram. $F_{load} = W \cdot g = 4000 kg \times 9.81 m/s^2 = 39240 N$ Step 3: Calculate the theoretical force required on the plunger. The pressure is constant throughout the hydraulic system: $P = \frac{F_{plunger\_theoretical}}{A_{plunger}} = \frac{F_{load}}{A_{ram}}$. $F_{plunger\_theoretical} = F_{load} \frac{A_{plunger}}{A_{ram}} = (39240 N) \times \frac{0.000081\pi m^2}{0.002025\pi m^2}$ $F_{plunger\_theoretical} = 39240 \times \frac{0.000081}{0.002025} = 39240 \times 0.04 = 1569.6 N$ Step 4: Calculate the actual force applied to the plunger, considering efficiency. $F_{applied\_plunger} = \frac{F_{plunger\_theoretical}}{\eta} = \frac{1569.6 N}{0.80} = 1962 N$ Step 5: Calculate the force required on the lever. $F_{lever} = \frac{F_{applied\_plunger}}{MA} = \frac{1962 N}{10} = 196.2 N$

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

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

Step 1: Calculate the volume of water required to lift the ram by $H_{ram}$. $V_{ram\_lift} = A_{ram} \cdot H_{ram} = (0.002025\pi m^2) \times (0.18 m) = 0.0003645\pi m^3 \approx 0.0011451 m^3$ Step 2: Calculate the theoretical volume displaced by the plunger per stroke. $V_{plunger\_stroke} = A_{plunger} \cdot L_{plunger} = (0.000081\pi m^2) \times (0.035 m) = 0.000002835\pi m^3 \approx 0.000008908 m^3$ Step 3: Calculate the actual volume delivered by the plunger per stroke, considering slip. $V_{actual\_stroke} = V_{plunger\_stroke} \times (1 - slip) = (0.000002835\pi m^3) \times (1 - 0.04)$ $V_{actual\_stroke} = 0.000002835\pi \times 0.96 m^3 \approx 0.000008551 m^3$ Step 4: Calculate the number of strokes required. $N_{strokes} = \frac{V_{ram\_lift}}{V_{actual\_stroke}} = \frac{0.0011451 m^3}{0.000008551 m^3} \approx 133.91$ Since the number of strokes must be an integer, we round up.

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

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