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 ramandithe plunger is .
Calculate the weight of the load that can be exerted by the ram of the hydraulic
press.
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 .
A pressure pump has an accumulator to keep a constant pressure in a closed
system. The diameterof the accumulator is 200 mm and the mass of the ballast
is 900 kg .

Ask by Salazar Goodwin. in South Africa

Mar 30,2025

Question

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 ramandithe plunger is .
Calculate the weight of the load that can be exerted by the ram of the hydraulic
press.
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 .
A pressure pump has an accumulator to keep a constant pressure in a closed
system. The diameterof the accumulator is 200 mm and the mass of the ballast
is 900 kg .

Ask by Salazar Goodwin. in South Africa

Mar 30,2025

UpStudy AI · Accepted Answer

**(a) Weight of the Load Exerted by the Ram**

1. **Force at the Plunger:**  
   An effort of $$185\,\text{N}$$ is applied at the end of a lever with a mechanical advantage of $$12$$. Thus the nominal force at the plunger is  
   $$
   F_{\text{nominal}} = 12 \times 185\,\text{N} = 2220\,\text{N}.
   $$

2. **Accounting for Efficiency:**  
   With a hydraulic efficiency of $$82\%$$ (or 0.82), the effective force transmitted to the hydraulic fluid is  
   $$
   F_{\text{effective}} = 0.82 \times 2220\,\text{N} \approx 1820.4\,\text{N}.
   $$

3. **Hydraulic Multiplication:**  
   The ratio of the diameters of the ram to the plunger is $$8:1$$. Since the area scales with the square of the diameter, the area (and thus force) multiplication factor is  
   $$
   \frac{A_{\text{ram}}}{A_{\text{plunger}}} = 8^2 = 64.
   $$
   Therefore, the force (or weight) that can be exerted by the ram is  
   $$
   F_{\text{load}} = 64 \times 1820.4\,\text{N} \approx 116505.6\,\text{N}.
   $$

4. **Result:**  
   The ram can exert a load of approximately  
   $$
   1.17 \times 10^5\,\text{N}.
   $$
   If desired, this force can be converted to an equivalent mass by dividing by the acceleration due to gravity (assuming $$g \approx 9.81\,\text{m/s}^2$$),  
   $$
   m \approx \frac{116505.6\,\text{N}}{9.81\,\text{m/s}^2} \approx 11875\,\text{kg}.
   $$

---

**(b) Power Required to Drive the Pump**

1. **Plunger Area Calculation:**  
   The plunger diameter is $$125\,\text{mm} = 0.125\,\text{m}$$. The radius is  
   $$
   r = \frac{0.125}{2} = 0.0625\,\text{m}.
   $$
   Thus, the area is  
   $$
   A = \pi r^2 = \pi (0.0625)^2 \approx \pi (0.00390625) \approx 0.01227\,\text{m}^2.
   $$

2. **Volume per Stroke for One Cylinder:**  
   With a stroke length of $$400\,\text{mm} = 0.400\,\text{m}$$, the volume displaced by one cylinder per stroke is  
   $$
   V_{\text{stroke}} = A \times L = 0.01227\,\text{m}^2 \times 0.400\,\text{m} \approx 0.004908\,\text{m}^3.
   $$

3. **Total Displaced Volume per Revolution:**  
   The pump has three cylinders, and assuming each cylinder delivers one stroke per revolution, the total volume per revolution is  
   $$
   V_{\text{rev}} = 3 \times 0.004908\,\text{m}^3 \approx 0.014724\,\text{m}^3.
   $$

4. **Flow Rate at 500 rpm:**  
   At $$500\,\text{rpm}$$, the flow per minute is  
   $$
   V_{\text{min}} = 500 \times 0.014724\,\text{m}^3 \approx 7.362\,\text{m}^3/\text{min}.
   $$
   Converting to cubic meters per second,  
   $$
   Q = \frac{7.362}{60} \approx 0.1227\,\text{m}^3/\text{s}.
   $$

5. **Hydraulic Power Delivered:**  
   With a delivery pressure of $$500\,\text{kPa} = 500\,000\,\text{Pa}$$, the hydraulic power is  
   $$
   P_{\text{hydraulic}} = \text{Pressure} \times Q = 500\,000\,\text{Pa} \times 0.1227\,\text{m}^3/\text{s} \approx 61350\,\text{W}.
   $$

6. **Accounting for Motor Efficiency:**  
   Given the motor efficiency of $$80\%$$ (or 0.80), the required input power is  
   $$
   P_{\text{input}} = \frac{P_{\text{hydraulic}}}{0.80} \approx \frac{61350}{0.80} \approx 76687.5\,\text{W}.
   $$

7. **Result:**  
   The pump requires approximately  
   $$
   77\,\text{kW}
   $$
   of power to operate at 500 rpm.

---

**(c) Pressure in the Closed System with an Accumulator**

1. **Accumulator Area:**  
   The accumulator has a diameter of $$200\,\text{mm} = 0.2\,\text{m}$$, so the radius is  
   $$
   r = \frac{0.2}{2} = 0.1\,\text{m}.
   $$
   The area is  
   $$
   A = \pi r^2 = \pi (0.1)^2 \approx \pi \times 0.01 = 0.031416\,\text{m}^2.
   $$

2. **Force Due to the Ballast:**  
   The ballast has a mass of $$900\,\text{kg}$$. The force exerted by the ballast (its weight) is  
   $$
   F = m \times g = 900\,\text{kg} \times 9.81\,\text{m/s}^2 \approx 8829\,\text{N}.
   $$

3. **System Pressure:**  
   The pressure maintained by the accumulator is given by  
   $$
   P = \frac{F}{A} = \frac{8829\,\text{N}}{0.031416\,\text{m}^2} \approx 280820\,\text{Pa}.
   $$
   Converting to kilopascals,  
   $$
   P \approx 281\,\text{kPa}.
   $$

---

**Final Answers:**

- (a) The ram can exert a load of approximately $$1.17 \times 10^5\,\text{N}$$ (or equivalently a mass of about $$11875\,\text{kg}$$).
- (b) The pump requires about $$77\,\text{kW}$$ of power at $$500\,\text{rpm}$$.
- (c) The constant pressure in the closed system is approximately $$281\,\text{kPa}$$.
- (a) The ram can exert a load of approximately $$1.17 \times 10^5\,\text{N}$$. - (b) The pump requires about $$77\,\text{kW}$$ of power at $$500\,\text{rpm}$$. - (c) The constant pressure in the closed system is approximately $$281\,\text{kPa}$$.

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