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Calculate the following: 11.1.1 The area of the ram if the diameter of the cylinder is 110 mm - NSC Mechanical Technology Fitting and Machining - Question 11 - 2023 - Paper 1

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Calculate the following: 11.1.1 The area of the ram if the diameter of the cylinder is 110 mm. 11.1.2 Force applied on the small piston (plunger). 11.1.3 The disp... show full transcript

Worked Solution & Example Answer:Calculate the following: 11.1.1 The area of the ram if the diameter of the cylinder is 110 mm - NSC Mechanical Technology Fitting and Machining - Question 11 - 2023 - Paper 1

Step 1

11.1.1 The area of the ram if the diameter of the cylinder is 110 mm.

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Answer

To find the area of the ram, we use the formula:

A=πD24A = \frac{\pi D^2}{4}

Substituting the diameter (D = 0.110 m):

A=π(0.110)24=0.0095m2A = \frac{\pi (0.110)^2}{4} = 0.0095 m^2

Step 2

11.1.2 Force applied on the small piston (plunger).

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Answer

Using the formula for force:

f=FaAf = \frac{F \cdot a}{A}

Given F = 350 N and A (area of the plunger) = 0.005 m²:

f=350×0.0050.0095184.21Nf = \frac{350 \times 0.005}{0.0095} \approx 184.21 N

Step 3

11.1.3 The displacement h of the small piston (plunger) in mm.

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Using the relationship:

h=AHah = \frac{A \cdot H}{a}

Where H is the area of the ram, and substituting values:

h=0.0095m20.025m0.005m2=0.0475m=47.5mmh = \frac{0.0095 m^2 \cdot 0.025 m}{0.005 m^2} = 0.0475 m = 47.5 mm

Step 4

11.2 What is the purpose of using pressure gauges in a hydraulic system?

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Answer

Pressure gauges serve several purposes:

  • They help to adjust pressure control valves.
  • They allow for determining the pressure being exerted.
  • For safety, as they indicate whether pressure limits are being approached.
  • They can indicate any leaks present in the system.

Step 5

11.3 State ONE advantage of applying pneumatics in a system.

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One advantage of pneumatics is that pneumatic tools are typically very environmentally friendly due to their clean operation.

Step 6

11.4.1 The rotational frequency of the driven pulley in r/sec.

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Answer

To find the rotational frequency of the driven pulley, we use:

NDN=NDR×DDRDDNN_{DN} = \frac{N_{DR} \times D_{DR}}{D_{DN}}

Substituting values:

NDN=25×753505.36r/secN_{DN} = \frac{25 \times 75}{350} \approx 5.36 r/sec

Step 7

11.4.2 The belt speed in m.s$^{-1}$.

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Answer

Using the formula for belt speed:

V=DDR×NDRDDNV = \frac{D_{DR} \times N_{DR}}{D_{DN}}

Substituting values:

V=0.075m×250.350m5.89m/sV = \frac{0.075 m \times 25}{0.350 m} \approx 5.89 m/s

Step 8

11.5 What protects a V-belt against sudden loads so that it cannot get damaged?

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Answer

The slippage of the V-belt is what protects it against sudden loads, ensuring the belt does not sustain damage.

Step 9

11.6.1 The rotation frequency of the output shaft in r/min.

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Answer

Using the relationship:

NB=product of teeth on driven gearsproduct of teeth on driver gears×NAN_{B} = \frac{product \ of \ teeth \ on \ driven \ gears}{product \ of \ teeth \ on \ driver \ gears} \times N_{A}

Calculating:

NB=30×2555×50×9525.91r/minN_{B} = \frac{30 \times 25}{55 \times 50} \times 95 \approx 25.91 r/min

Step 10

11.6.2 The power transmitted if the torque in the system is 120 Nm.

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Answer

The power transmitted is calculated using:

P=2πNT60P = \frac{2 \pi N T}{60}

Where T = 120 Nm and substituting for the frequency:

P=2π×95×120601193.81W1.19kWP = \frac{2 \pi \times 95 \times 120}{60} \approx 1193.81 W \approx 1.19 kW

Step 11

11.7 A fitter applies a force of 300 N on a ring spanner. What must the length of the ring spanner be to generate a torque of 135 Nm?

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Answer

Using the torque formula:

T=FRT = F \cdot R

Rearranging to find radius R:

R=TF=135Nm300N=0.45m or 450mmR = \frac{T}{F} = \frac{135 Nm}{300 N} = 0.45 m \ or \ 450 mm

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