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Calculate the following: 11.1.1 The fluid pressure in the hydraulic system in MPa 11.1.2 The diameter in millimetres of piston B 11.1.3 State TWO advantages of a chain drive when compared to a belt drive - 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-diameter-in-millimetres-of-piston-B--11.1.3-State-TWO-advantages-of-a-chain-drive-when-compared-to-a-belt-drive-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 diameter in millimetres of piston B 11.1.3 State TWO advantages of a ... 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 diameter in millimetres of piston B 11.1.3 State TWO advantages of a chain drive when compared to a belt drive - 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, we use the formula:

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

where:

  • F=1.32 kN=1320 NF = 1.32\ kN = 1320\ N (force on piston A)
  • AA=πDA24A_A = \frac{\pi D_A^2}{4} (cross-sectional area of piston A)

Substituting DA=0.025 mD_A = 0.025\ m:

AA=π(0.025)24=4.91×104 m2A_A = \frac{\pi (0.025)^2}{4} = 4.91 \times 10^{-4}\ m^2

Now, calculating the pressure:

P=1320 N4.91×104 m2=2.69×106 Pa=2.69 MPaP = \frac{1320\ N}{4.91 \times 10^{-4}\ m^2} = 2.69 \times 10^6 \ Pa = 2.69\ MPa

Step 2

11.1.2 The diameter in millimetres of piston B

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Answer

For piston B, we need to find its diameter using the pressure equalization condition:

Using the formula for pressure and area: PA=PBFAAA=FBABP_A = P_B \Rightarrow \frac{F_A}{A_A} = \frac{F_B}{A_B}

Rearranging: AB=FBAAFAA_B = \frac{F_B \cdot A_A}{F_A}

Substituting values:

  • FB=6.45 kN=6450 NF_B = 6.45\ kN = 6450\ N
  • AA=4.91×104 m2A_A = 4.91 \times 10^{-4}\ m^2
  • FA=1320 NF_A = 1320\ N

Plugging in: AB=6450 N4.91×104 m21320 N=2.40×103 m2A_B = \frac{6450\ N \cdot 4.91 \times 10^{-4}\ m^2}{1320\ N} = 2.40 \times 10^{-3}\ m^2

Now, calculate diameter: DB=4AB/π=42.40×103/π=0.05528 m=55.28 mmD_B = \sqrt{4A_B/\pi} = \sqrt{4 \cdot 2.40 \times 10^{-3}/\pi} = 0.05528\ m = 55.28\ mm

Step 3

11.1.3 State TWO advantages of a chain drive when compared to a belt drive.

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Answer

  • Higher efficiency and less energy loss due to slippage.
  • Greater durability and longer operational life under load.

Step 4

11.1.4 State ONE application of a hydraulic system.

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Answer

Hydraulic systems are commonly used in construction machinery, such as excavators and cranes, to provide powerful lifting and moving capabilities.

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