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11.1 FIGURE 11.1 below shows a hydraulic lift - NSC Mechanical Technology Fitting and Machining - Question 11 - 2022 - Paper 1

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Question 11

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11.1 FIGURE 11.1 below shows a hydraulic lift. The lift needs to raise a maximum load of 4 500 kg at the ram with a diameter of 120 mm. The diameter of the plunger i... show full transcript

Worked Solution & Example Answer:11.1 FIGURE 11.1 below shows a hydraulic lift - NSC Mechanical Technology Fitting and Machining - Question 11 - 2022 - Paper 1

Step 1

The fluid pressure in the hydraulic system in MPa

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Answer

To calculate the fluid pressure in the hydraulic system, we use the formula:

P=FAP = \frac{F}{A}

Where:

  • F=4500 kg×10 m/s2=45000 NF = 4500 \text{ kg} \times 10 \text{ m/s}^2 = 45000 \text{ N}
  • The area AA of the ram can be calculated using:

A=πd24A = \frac{\pi d^2}{4}

Substituting the diameter of the ram, d=120 mm=0.12 md = 120 \text{ mm} = 0.12 \text{ m}:

A=π(0.12)24=0.01131 m2A = \frac{\pi (0.12)^2}{4} = 0.01131 \text{ m}^2

Now, substituting this value into the pressure formula:

P=450000.011313978.5 N/m2=3.98 MPaP = \frac{45000}{0.01131} \approx 3978.5 \text{ N/m}^2 = 3.98 \text{ MPa}

Step 2

The force to be applied on the plunger

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Answer

To find the force to be applied on the plunger, we can use the same principle but for the plunger area:

Aplunger=π(0.032)24A_{plunger} = \frac{\pi (0.032)^2}{4}

Calculating this:

Aplunger=π(0.032)240.000804 m2A_{plunger} = \frac{\pi (0.032)^2}{4} \approx 0.000804 \text{ m}^2

Now we can calculate:

F=P×Aplunger3.98×106N/m2×0.000804m23184 NF = P \times A_{plunger} \approx 3.98 \times 10^6 \, \text{N/m}^2 \times 0.000804 \, \text{m}^2 \approx 3184 \text{ N}

Step 3

State ONE function of a hydraulic non-return valve.

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Answer

A hydraulic non-return valve is used to provide one-directional oil flow and prevent the fluid from flowing back.

Step 4

Give TWO reasons why a spring-loaded double-action control valve is used in a water line.

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Answer

  1. To protect water supplies from back flow.
  2. To prevent water contamination in case one valve gets jammed.

Step 5

Name the TWO different types of pressure gauges.

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Answer

  1. Bourdon tube gauge.
  2. Schrader gauge.

Step 6

The rotational frequency in r/s of the pulley on the concrete mixer

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Answer

To find the rotational frequency, we can use:

N=Nmotor×DmotorDpulleyN = \frac{N_{motor} \times D_{motor}}{D_{pulley}}

Given:

  • Nmotor=1320 r/minN_{motor} = 1320 \text{ r/min}
  • Dmotor=0.085 mD_{motor} = 0.085 \text{ m}
  • Dpulley=0.375 mD_{pulley} = 0.375 \text{ m}

Substituting, we first convert NmotorN_{motor} to r/s:

N=13206022 r/sN = \frac{1320}{60} \approx 22 \text{ r/s}

So,

N=22×0.0850.3754.99 r/sN = \frac{22 \times 0.085}{0.375} \approx 4.99 \text{ r/s}

Step 7

The power that can be transmitted in watt

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Answer

The power transmitted can be calculated using:

P=(TtightTslack)×πD×N60P = \frac{(T_{tight} - T_{slack}) \times \pi D \times N}{60}

Where:

  • Ttight=275 NT_{tight} = 275 \text{ N}
  • Tslack=120 NT_{slack} = 120 \text{ N}
  • D=0.085 mD = 0.085 \text{ m}
  • N=22 r/sN = 22 \text{ r/s}

Calculating gives:

P=(275120)×π×0.085×132060911.59 WP = \frac{(275-120) \times \pi \times 0.085 \times 1320}{60} \approx 911.59 \text{ W}

Step 8

The number of teeth on the driver gear TA

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Answer

To find the number of teeth on driver gear A (TAT_A), we can set up the equation:

Ninput=TA×TETB×TFN_{input} = \frac{T_A \times T_E}{T_B \times T_F}

Given:

  • NA=320 r/minN_A = 320 \text{ r/min}
  • NF=720 r/minN_F = 720 \text{ r/min}
  • TB=20,TE=50,TF=20,TC=45T_B = 20, T_E = 50, T_F = 20, T_C = 45

We solve for TAT_A:

TA=720×20×25320×45=10T_A = \frac{720 \times 20 \times 25}{320 \times 45} = 10

Step 9

The gear ratio of the system

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Answer

Gear ratio can be calculated as:

Gear ratio=Product of the number of teeth on driven gearsProduct of the number of teeth on driving gears\text{Gear ratio} = \frac{\text{Product of the number of teeth on driven gears}}{\text{Product of the number of teeth on driving gears}}

In our case: Gear ratio=20×25×5010×45×20=0.44\text{Gear ratio} = \frac{20 \times 25 \times 50}{10 \times 45 \times 20} = 0.44

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