QUESTION 5: HYDRAULICS
5.1 An effort of 185 N is applied at the end of a lever operating a plunger.
The efficiency of the hydraulic press is , while the mechanical advantage
is 12 . The ratio of the diameters of the rafnandtite plunger is .
Calculate the weight of the load that can be exerted by the ram of the hydraulic
press.
5.2 A three-cylinder pump with plunger diameters of 125 mm each and stroke
lengths of 400 mm has a delivery pressure of 500 kPa .
Calculate the power required to drive the pump at 500 rpm if the efficiency of
the motor is .

Ask by Howell Daniel. in South Africa

Mar 30,2025

Question

QUESTION 5: HYDRAULICS
5.1 An effort of 185 N is applied at the end of a lever operating a plunger.
The efficiency of the hydraulic press is , while the mechanical advantage
is 12 . The ratio of the diameters of the rafnandtite plunger is .
Calculate the weight of the load that can be exerted by the ram of the hydraulic
press.
5.2 A three-cylinder pump with plunger diameters of 125 mm each and stroke
lengths of 400 mm has a delivery pressure of 500 kPa .
Calculate the power required to drive the pump at 500 rpm if the efficiency of
the motor is .

Ask by Howell Daniel. in South Africa

Mar 30,2025

UpStudy AI · Accepted Answer

**Problem 5.1**

1. The applied effort is  
   $$
   F_{\text{effort}} = 185\,\text{N}.
   $$

2. The lever provides a mechanical advantage of  
   $$
   \text{MA} = 12,
   $$
   so the force transmitted to the master plunger is  
   $$
   F_{\text{master}} = 185\,\text{N} \times 12 = 2220\,\text{N}.
   $$

3. The diameters of the master and the ram plungers are in the ratio  
   $$
   8:1.
   $$
   Since the area is proportional to the square of the diameter, the area ratio is  
   $$
   \left(\frac{8}{1}\right)^2 = 64.
   $$
   This means that for a given pressure, the ram (slave) will exert a force 64 times that on the master plunger.

4. In an ideal situation the ram force would be  
   $$
   F_{\text{ram, ideal}} = F_{\text{master}} \times 64 = 2220\,\text{N} \times 64 = 142080\,\text{N}.
   $$

5. However, the press operates at an efficiency of  
   $$
   \eta = 82\% = 0.82.
   $$
   Therefore the actual load force is  
   $$
   F_{\text{load}} = 142080\,\text{N} \times 0.82 \approx 116506\,\text{N}.
   $$
   
Thus, the weight of the load that can be exerted by the ram is approximately  
$$
1.17 \times 10^5\,\text{N}.
$$

---

**Problem 5.2**

1. Each cylinder has a plunger with diameter  
   $$
   d = 125\,\text{mm} = 0.125\,\text{m}.
   $$
   The cross‐sectional area of a plunger is  
   $$
   A = \frac{\pi}{4}d^2 = \frac{\pi}{4}(0.125)^2 \approx 0.01227\,\text{m}^2.
   $$

2. The stroke length is  
   $$
   L = 400\,\text{mm} = 0.4\,\text{m},
   $$
   so the volume displaced per stroke per cylinder is  
   $$
   V_{\text{stroke}} = A \times L \approx 0.01227\,\text{m}^2 \times 0.4\,\text{m} \approx 0.004908\,\text{m}^3.
   $$

3. The pump has three cylinders, and at 500 revolutions per minute (rpm) each cylinder completes one stroke per revolution. Thus, the total volume delivered per revolution is  
   $$
   V_{\text{rev}} = 3 \times 0.004908\,\text{m}^3 \approx 0.014724\,\text{m}^3.
   $$
   At 500 rpm the volume delivered per minute is  
   $$
   V_{\text{min}} = 500 \times 0.014724\,\text{m}^3 \approx 7.362\,\text{m}^3/\text{min}.
   $$
   Converting this to cubic meters per second:  
   $$
   Q = \frac{7.362}{60} \approx 0.1227\,\text{m}^3/\text{s}.
   $$

4. The hydraulic pump delivers at a pressure of  
   $$
   p = 500\,\text{kPa} = 500\,000\,\text{Pa}.
   $$
   The hydraulic power output is given by  
   $$
   P_{\text{hydraulic}} = p \times Q = 500\,000\,\text{Pa} \times 0.1227\,\text{m}^3/\text{s} \approx 61350\,\text{W}.
   $$

5. The motor driving the pump has an efficiency of  
   $$
   \eta = 80\% = 0.8.
   $$
   Therefore, the required input power to the motor is  
   $$
   P_{\text{motor}} = \frac{P_{\text{hydraulic}}}{\eta} = \frac{61350\,\text{W}}{0.8} \approx 76600\,\text{W}.
   $$

Thus, the power required to drive the pump at 500 rpm is approximately  
$$
7.66 \times 10^4\,\text{W} \quad \text{or} \quad 76.6\,\text{kW}.
$$
**Problem 5.1** - **Given:** - Applied effort: 185 N - Mechanical advantage (MA): 12 - Diameter ratio of plungers: 8:1 - Efficiency of hydraulic press: 82% - **Calculations:** - Force transmitted to master plunger: 185 N × 12 = 2220 N - Area ratio: (8/1)² = 64 - Ideal ram force: 2220 N × 64 = 142,080 N - Actual load force: 142,080 N × 0.82 ≈ 116,506 N - **Answer:** The weight of the load is approximately 116,506 N. --- **Problem 5.2** - **Given:** - Plunger diameter per cylinder: 125 mm (0.125 m) - Stroke length: 400 mm (0.4 m) - Number of cylinders: 3 - Delivery pressure: 500 kPa (500,000 Pa) - Motor efficiency: 80% - Pump speed: 500 rpm - **Calculations:** - Cross-sectional area per cylinder: π/4 × (0.125)² ≈ 0.01227 m² - Volume per stroke per cylinder: 0.01227 m² × 0.4 m ≈ 0.004908 m³ - Total volume per revolution: 3 × 0.004908 m³ ≈ 0.014724 m³ - Volume per minute: 500 × 0.014724 m³ ≈ 7.362 m³/min - Volume per second: 7.362 / 60 ≈ 0.1227 m³/s - Hydraulic power: 500,000 Pa × 0.1227 m³/s ≈ 61,350 W - Motor power: 61,350 W / 0.8 ≈ 76,600 W - **Answer:** The power required to drive the pump is approximately 76,600 W or 76.6 kW.

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