QUESTION 3 HYDRAULICS AND PNEUMATICS 3.1 State THREE functions of an accumulator as it applies to hydraulic systems. 3.2 A hydraulic jack has a plunger diameter of 50 mm and a ram diameter of 200 mm . The work done by ram piston of 350 J is needed to lift a load of 2000 N . Calculate the following: 3.2.1 The load exerted on the plunger piston 3.2.2 The height moved by the ram 3.2.3 The volume displaced to the ram piston

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### 3.1 Functions of an Accumulator

1. **Energy Storage:** An accumulator stores hydraulic energy in the form of pressurized fluid, which can be released when demand increases.  
2. **Shock Absorption:** It dampens pressure fluctuations and sudden surges by absorbing shocks in the hydraulic system.  
3. **Maintaining Pressure:** It helps maintain a constant system pressure by compensating for leakage or temporary high flow demands.

---

### 3.2 Hydraulic Jack Calculations

Given parameters:  
- Plunger piston diameter: $$ d_p = 50\,\text{mm} $$  
- Ram piston diameter: $$ d_r = 200\,\text{mm} $$  
- Work done by the ram: $$ W = 350\,\text{J} $$  
- Load to be lifted: $$ F_r = 2000\,\text{N} $$

**Step 1. Calculate the Areas**

For any circular piston, the area is given by  
$$
A = \pi \left(\frac{d}{2}\right)^2
$$

- **Plunger piston area $$ A_p $$:**  
  Convert $$50\,\text{mm}$$ to meters: $$50\,\text{mm} = 0.05\,\text{m}$$.  
  $$
  A_p = \pi \left(\frac{0.05}{2}\right)^2 = \pi \left(0.025\right)^2 = \pi \times 0.000625 = 0.000625\pi\,\text{m}^2
  $$

- **Ram piston area $$ A_r $$:**  
  Convert $$200\,\text{mm}$$ to meters: $$200\,\text{mm} = 0.2\,\text{m}$$.  
  $$
  A_r = \pi \left(\frac{0.2}{2}\right)^2 = \pi \left(0.1\right)^2 = \pi \times 0.01 = 0.01\pi\,\text{m}^2
  $$

**Step 2. Use Pascal’s Principle**

Pascal’s principle tells us that the pressure $$ P $$ is the same on both pistons:  
$$
P = \frac{F_p}{A_p} = \frac{F_r}{A_r}
$$
Solving for the force on the plunger piston $$ F_p $$:
$$
F_p = \frac{A_p}{A_r} \times F_r
$$

Calculate the area ratio:
$$
\frac{A_p}{A_r} = \frac{0.000625\pi}{0.01\pi} = \frac{0.000625}{0.01} = 0.0625
$$

Thus,  
$$
F_p = 0.0625 \times 2000\,\text{N} = 125\,\text{N}
$$

**Answer to 3.2.1:** The load exerted on the plunger piston is $$125\,\text{N}$$.

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**Step 3. Calculate the Height Moved by the Ram**

The work done by the ram piston is given by:
$$
W = F_r \times h_r
$$
Solve for the height $$ h_r $$:
$$
h_r = \frac{W}{F_r} = \frac{350\,\text{J}}{2000\,\text{N}} = 0.175\,\text{m}
$$

**Answer to 3.2.2:** The height moved by the ram is $$0.175\,\text{m}$$.

---

**Step 4. Calculate the Volume Displaced to the Ram Piston**

The volume displaced is the product of the cross-sectional area of the ram piston and the displacement height:
$$
V = A_r \times h_r
$$
Substitute the values:
$$
A_r = 0.01\pi\,\text{m}^2 \quad \text{and} \quad h_r = 0.175\,\text{m}
$$
$$
V = 0.01\pi \times 0.175 = 0.00175\pi\,\text{m}^3
$$
Approximating using $$\pi \approx 3.1416$$:
$$
V \approx 0.00175 \times 3.1416 \approx 0.0055\,\text{m}^3
$$

**Answer to 3.2.3:** The volume displaced to the ram piston is approximately $$0.0055\,\text{m}^3$$.
**3.1 Functions of an Accumulator:** 1. **Energy Storage:** Stores hydraulic energy for later use. 2. **Shock Absorption:** Absorbs pressure fluctuations and shocks. 3. **Maintaining Pressure:** Keeps system pressure constant by compensating for leaks. **3.2 Hydraulic Jack Calculations:** 1. **Load on Plunger Piston:** 125 N 2. **Height Moved by Ram:** 0.175 m 3. **Volume Displaced to Ram Piston:** Approximately 0.0055 m³

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