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Calculate the following: 11.1.1 The fluid pressure in the hydraulic system in MPa 11.1.2 The magnitude of the force that will be exerted onto the bearing by the ram 11.2 State ONE function of a hydraulic reservoir - NSC Mechanical Technology Fitting and Machining - Question 11 - 2021 - Paper 1

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Question 11

Calculate-the-following:--11.1.1-The-fluid-pressure-in-the-hydraulic-system-in-MPa--11.1.2-The-magnitude-of-the-force-that-will-be-exerted-onto-the-bearing-by-the-ram--11.2-State-ONE-function-of-a-hydraulic-reservoir-NSC Mechanical Technology Fitting and Machining-Question 11-2021-Paper 1.png

Calculate the following: 11.1.1 The fluid pressure in the hydraulic system in MPa 11.1.2 The magnitude of the force that will be exerted onto the bearing by the ra... show full transcript

Worked Solution & Example Answer:Calculate the following: 11.1.1 The fluid pressure in the hydraulic system in MPa 11.1.2 The magnitude of the force that will be exerted onto the bearing by the ram 11.2 State ONE function of a hydraulic reservoir - NSC Mechanical Technology Fitting and Machining - Question 11 - 2021 - Paper 1

Step 1

11.1.1 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:

  • FF is the force in Newtons (25 kN or 25000 N), and
  • AA is the area of the plunger given by the formula A=πd24A = \frac{\pi d^2}{4}.

First, we calculate the area of the plunger:

  • Diameter d=35mm=0.035md = 35 mm = 0.035 m
  • Area: A=π(0.035)24=9.62×104m2A = \frac{\pi (0.035)^2}{4} = 9.62 \times 10^{-4} m^2

Now substituting into the pressure formula:

P=250009.62×104=25958.1Pa=25.96MPaP = \frac{25000}{9.62 \times 10^{-4}} = 25958.1 Pa = 25.96 MPa

Step 2

11.1.2 The magnitude of the force that will be exerted onto the bearing by the ram

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Answer

To find the force exerted by the ram, we can rearrange the initial formula where the area will be calculated with the diameter of the ram:

  • Diameter D=120mm=0.120mD = 120 mm = 0.120 m
  • Area of the ram: A=πD24=π(0.120)24=1.131×102m2A = \frac{\pi D^2}{4} = \frac{\pi (0.120)^2}{4} = 1.131 \times 10^{-2} m^2

Using the pressure calculated in step 11.1.1: F=P×A=25958.1×1.131×102=293,88NF = P \times A = 25958.1 \times 1.131 \times 10^{-2} = 293,88 N

Step 3

11.2 State ONE function of a hydraulic reservoir.

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Answer

One function of a hydraulic reservoir is to act as a fluid storage tank, which helps in maintaining a sufficient supply of hydraulic fluid for the system.

Step 4

11.3 Give TWO reasons why pneumatic systems are very efficient.

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Answer

  1. Pneumatic tools are environmentally friendly, making them suitable for various applications.
  2. They are easy to use and require less physical effort from the operator, thereby enhancing productivity.

Step 5

11.4 State TWO uses of pneumatic systems.

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Answer

  1. Drills
  2. Nail guns

Step 6

11.5.1 The rotational frequency of the pulley on the washing machine in r/s

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Answer

To find the rotational frequency of the pulley: Using the relationship between the diameters and speeds:

N1D1=N2D2N_1 D_1 = N_2 D_2

Let N1N_1 be the speed of the driver pulley and D1=600mm=0.6mD_1 = 600 mm = 0.6 m, D2=800mm=0.8mD_2 = 800 mm = 0.8 m. The speed of the driver pulley is given as N1=7.2r/sN_1 = 7.2 r/s.

Now substituting: N2=N1D1D2=7.2×0.60.8=5.4r/sN_2 = \frac{N_1 D_1}{D_2} = \frac{7.2 \times 0.6}{0.8} = 5.4 r/s

Step 7

11.5.2 The power that can be transmitted in kW

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Answer

The power transmitted can be calculated using:

P=T×(T1T2)DN60P = T \times \frac{(T_1 - T_2) D N}{60}

Where:

  • T1=300NT_1 = 300 N (tight side),
  • T2=120NT_2 = 120 N,
  • D=0.8mD = 0.8 m, and
  • N=5.4r/sN = 5.4 r/s.

Now substituting:

P=(300120)×0.8×5.4=2.44kWP = (300 - 120) \times 0.8 \times 5.4 = 2.44 kW

Step 8

11.6.1 The rotation frequency of the output shaft if the motor rotates at 2300 r/min

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Answer

To find the output frequency:

Using: Noutput=TA×TC×NATB×TDN_{output} = \frac{T_A \times T_C \times N_A}{T_B \times T_D}

Where gears have:

  • TA=30T_A = 30, TB=40T_B = 40, NA=2300N_A = 2300, TC=20T_C = 20, and TD=60T_D = 60.

Therefore:

Noutput=30×20×230040×60=575r/minN_{output} = \frac{30 \times 20 \times 2300}{40 \times 60} = 575 r/min

Step 9

11.6.2 The ratio between the input shaft and the output shaft of the system.

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Answer

The gear ratio can be calculated as:

Gear Ratio=Product of teeth on driven gearsProduct of teeth on driver gears=40×6030×20=4:1Gear\ Ratio = \frac{Product\ of\ teeth\ on\ driven\ gears}{Product\ of\ teeth\ on\ driver\ gears} = \frac{40 \times 60}{30 \times 20} = 4:1

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