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FIGURE 9.1 shows a belt-drive system with a 230 mm driver pulley - NSC Mechanical Technology Automotive - Question 9 - 2016 - Paper 1

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Question 9

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FIGURE 9.1 shows a belt-drive system with a 230 mm driver pulley. The belt speed in this system is 36 m.s<sup>-1</sup>. The tensile force in the slack side is 140 N ... show full transcript

Worked Solution & Example Answer:FIGURE 9.1 shows a belt-drive system with a 230 mm driver pulley - NSC Mechanical Technology Automotive - Question 9 - 2016 - Paper 1

Step 1

9.1.1 Rotation frequency of the driver pulley in r/min

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Answer

To calculate the rotation frequency of the driver pulley, we use the equation:

n=VπDn = \frac{V}{\pi D}

Where:

  • VV is the belt speed (36 m/s)
  • DD is the diameter of the driver pulley (0.23 m).

Substituting the values gives:

n=36π×0.23≈49.82 r/sn = \frac{36}{\pi \times 0.23} \approx 49.82 \text{ r/s}

To convert from r/s to r/min:

nmin=49.82 r/s×60≈2989.35 r/minn_{min} = 49.82 \text{ r/s} \times 60 \approx 2989.35 \text{ r/min}

Step 2

9.1.2 Power transmitted in this system

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Answer

The power transmitted can be calculated using the formula:

P=(TT−TS)⋅VP = (T_T - T_S) \cdot V

Where:

  • TTT_T is the tension in the tight side,
  • TST_S is the tension in the slack side (140extN140 ext{ N}), and
  • VV is the belt speed (36extm/s36 ext{ m/s}).

First, calculate the tension in the tight side using the ratio:

TTTS=2.5⇒TT=2.5×140=350extN\frac{T_T}{T_S} = 2.5 \Rightarrow T_T = 2.5 \times 140 = 350 ext{ N}

Now, substituting these values into the power equation:

P=(350−140)⋅36=7560extW≈7.56extkWP = (350 - 140) \cdot 36 = 7560 ext{ W} \approx 7.56 ext{ kW}

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