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1 Give TWO examples where the individual production method could be applied. 1.2 State TEN principles of good workshop layout.

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QUESTION 1

1.1 Give TWO examples where the individual production method could be applied.

  • Shipbuilding
  • Custom-made furniture

1.2 State TEN principles of a good workshop layout.

  1. Flow of work: Materials and products should move smoothly with minimal backtracking.
  2. Space utilization: Efficient use of all available space.
  3. Accessibility: Easy access to machines, tools, and work areas.
  4. Safety: Minimize hazards and provide clear emergency routes.
  5. Flexibility: Ability to adapt to changes in production.
  6. Communication: Facilitate easy interaction among workers.
  7. Supervision: Allow for effective monitoring of work.
  8. Comfort and environment: Good lighting, ventilation, and temperature control.
  9. Minimization of material handling: Reduce movement of materials.
  10. Orderliness and cleanliness: Promote a tidy and organized workspace.

QUESTION 2

2.1 The following are the particulars of an open flat-belt drive between a motor and a machine:

  • Belt speed = 900 m/min900 \text{ m/min}
  • Centre distance between shafts = 4.5 m4.5 \text{ m}
  • Speed of motor = 1400 r/min1400 \text{ r/min}
  • Contact angle on motor pulley = 182∘182^\circ
  • Power = 14.5 kW14.5 \text{ kW}
  • Speed of machine = 2300 r/min2300 \text{ r/min}
  • Contact angle on machine pulley = 178∘178^\circ
  • Thickness of belt = 5 mm5 \text{ mm}
  • Maximum allowable tension = 160 N per cm width of belt160 \text{ N per cm width of belt}
  • Coefficient of friction = 0.3

2.1.1 Calculate the diameter of the pulleys.

  1. Convert belt speed to meters per second.

    v=900 m/min=90060 m/s=15 m/sv = 900 \text{ m/min} = \frac{900}{60} \text{ m/s} = 15 \text{ m/s}

  2. Calculate the diameter of the motor pulley (DmD_m).

    The formula for belt velocity is v=πDN60v = \frac{\pi D N}{60}. Rearranging for diameter:

    Dm=60vπNmD_m = \frac{60v}{\pi N_m} Dm=60×15 m/sπ×1400 r/min=9001400π m≈0.2047 mD_m = \frac{60 \times 15 \text{ m/s}}{\pi \times 1400 \text{ r/min}} = \frac{900}{1400\pi} \text{ m} \approx 0.2047 \text{ m} Dm≈204.7 mmD_m \approx 204.7 \text{ mm}

  3. Calculate the diameter of the machine pulley (DmachD_{mach}).

    Dmach=60vπNmachD_{mach} = \frac{60v}{\pi N_{mach}} Dmach=60×15 m/sπ×2300 r/min=9002300π m≈0.1244 mD_{mach} = \frac{60 \times 15 \text{ m/s}}{\pi \times 2300 \text{ r/min}} = \frac{900}{2300\pi} \text{ m} \approx 0.1244 \text{ m} Dmach≈124.4 mmD_{mach} \approx 124.4 \text{ mm}

    The diameters of the pulleys are 204.7 mm\boxed{\text{204.7 mm}} and 124.4 mm\boxed{\text{124.4 mm}}.

    2.1.2 Calculate the width of the belt in mm.

  4. Calculate the effective tension (F1−F2F_1 - F_2).

    P=(F1−F2)vP = (F_1 - F_2)v F1−F2=Pv=14.5×103 W15 m/s≈966.67 NF_1 - F_2 = \frac{P}{v} = \frac{14.5 \times 10^3 \text{ W}}{15 \text{ m/s}} \approx 966.67 \text{ N}

  5. Determine the smaller contact angle and convert it to radians.

    The smaller contact angle is θmach=178∘\theta_{mach} = 178^\circ.

    θrad=178∘×π180∘≈3.1067 rad\theta_{\text{rad}} = 178^\circ \times \frac{\pi}{180^\circ} \approx 3.1067 \text{ rad}

  6. Calculate the ratio of tensions F1F2\frac{F_1}{F_2}.

    F1F2=eμθrad=e0.3×3.1067=e0.93201≈2.5397\frac{F_1}{F_2} = e^{\mu \theta_{\text{rad}}} = e^{0.3 \times 3.1067} = e^{0.93201} \approx 2.5397

  7. Solve for F1F_1 (tight side tension).

    We have F1−F2=966.67F_1 - F_2 = 966.67 and F1=2.5397F2F_1 = 2.5397 F_2.
    Substitute F1F_1 into the first equation:

    2.5397F2−F2=966.672.5397 F_2 - F_2 = 966.67 1.5397F2=966.671.5397 F_2 = 966.67 F2=966.671.5397≈627.83 NF_2 = \frac{966.67}{1.5397} \approx 627.83 \text{ N} F1=2.5397×627.83≈1594.5 NF_1 = 2.5397 \times 627.83 \approx 1594.5 \text{ N}

  8. Calculate the width of the belt.

    The maximum allowable tension is 160 N/cm160 \text{ N/cm}. The tight side tension F1F_1 is the maximum tension.

    Width=F1Max allowable tension per cm=1594.5 N160 N/cm≈9.966 cm\text{Width} = \frac{F_1}{\text{Max allowable tension per cm}} = \frac{1594.5 \text{ N}}{160 \text{ N/cm}} \approx 9.966 \text{ cm}

    Convert to mm:

    Width=9.966 cm×10 mm/cm≈99.66 mm\text{Width} = 9.966 \text{ cm} \times 10 \text{ mm/cm} \approx 99.66 \text{ mm}

    The width of the belt is 99.66 mm\boxed{\text{99.66 mm}}.

    2.1.3 Calculate the length of the belt in metres.

  9. Use the formula for the length of an open flat belt drive.

    L=π(Dm+Dmach2)+2C+(Dm−Dmach)24CL = \pi \left( \frac{D_m + D_{mach}}{2} \right) + 2C + \frac{(D_m - D_{mach})^2}{4C}

    Using Dm=0.2047 mD_m = 0.2047 \text{ m}, Dmach=0.1244 mD_{mach} = 0.1244 \text{ m}, and C=4.5 mC = 4.5 \text{ m}:

    L=π(0.2047+0.12442)+2(4.5)+(0.2047−0.1244)24(4.5)L = \pi \left( \frac{0.2047 + 0.1244}{2} \right) + 2(4.5) + \frac{(0.2047 - 0.1244)^2}{4(4.5)}

  10. Perform the calculations.

    L=π(0.32912)+9+(0.0803)218L = \pi \left( \frac{0.3291}{2} \right) + 9 + \frac{(0.0803)^2}{18} L=π(0.16455)+9+0.0064480918L = \pi (0.16455) + 9 + \frac{0.00644809}{18} L≈0.5170+9+0.000358L \approx 0.5170 + 9 + 0.000358 L≈9.517358 mL \approx 9.517358 \text{ m}

    The length of the belt is 9.52 m\boxed{\text{9.52 m}}.

    2.2 Calculate the number of V belts required for this drive if the maximum allowable tension per belt is 900 N and the belt has to transmit 30 kW.

  11. Calculate the belt velocity (vv).

    v=πDN60=π×0.25 m×1800 r/min60 s/min=7.5π m/s≈23.56 m/sv = \frac{\pi D N}{60} = \frac{\pi \times 0.25 \text{ m} \times 1800 \text{ r/min}}{60 \text{ s/min}} = 7.5\pi \text{ m/s} \approx 23.56 \text{ m/s}

  12. Calculate the centrifugal tension (FcF_c).

    Fc=mv2=0.5 kg/m×(23.56 m/s)2≈277.58 NF_c = m v^2 = 0.5 \text{ kg/m} \times (23.56 \text{ m/s})^2 \approx 277.58 \text{ N}

  13. Calculate the effective coefficient of friction (μe\mu_e) for a V-belt.

    The angle of the V-groove is 2α=40∘2\alpha = 40^\circ, so α=20∘\alpha = 20^\circ.

    μe=μsin⁡α=0.25sin⁡20∘≈0.250.3420≈0.7310\mu_e = \frac{\mu}{\sin \alpha} = \frac{0.25}{\sin 20^\circ} \approx \frac{0.25}{0.3420} \approx 0.7310

  14. Convert the contact angle to radians and calculate μeθ\mu_e \theta.

    θrad=140∘×π180∘≈2.4435 rad\theta_{\text{rad}} = 140^\circ \times \frac{\pi}{180^\circ} \approx 2.4435 \text{ rad} μeθrad=0.7310×2.4435≈1.7860\mu_e \theta_{\text{rad}} = 0.7310 \times 2.4435 \approx 1.7860

  15. Calculate the ratio of tensions F1−FcF2−Fc\frac{F_1 - F_c}{F_2 - F_c}.

    F1−FcF2−Fc=eμeθrad=e1.7860≈5.965\frac{F_1 - F_c}{F_2 - F_c} = e^{\mu_e \theta_{\text{rad}}} = e^{1.7860} \approx 5.965

  16. Calculate the tensions F1F_1 and F2F_2 per belt.

    Given maximum allowable tension per belt F1=900 NF_1 = 900 \text{ N}.

    900 N−277.58 NF2−277.58 N=5.965\frac{900 \text{ N} - 277.58 \text{ N}}{F_2 - 277.58 \text{ N}} = 5.965 622.42 NF2−277.58 N=5.965\frac{622.42 \text{ N}}{F_2 - 277.58 \text{ N}} = 5.965 F2−277.58 N=622.42 N5.965≈104.34 NF_2 - 277.58 \text{ N} = \frac{622.42 \text{ N}}{5.965} \approx 104.34 \text{ N} F2=104.34 N+277.58 N=381.92 NF_2 = 104.34 \text{ N} + 277.58 \text{ N} = 381.92 \text{ N}

  17. Calculate the power transmitted per belt (PbeltP_{belt}).

    Pbelt=(F1−F2)v=(900 N−381.92 N)×23.56 m/sP_{belt} = (F_1 - F_2)v = (900 \text{ N} - 381.92 \text{ N}) \times 23.56 \text{ m/s} Pbelt=518.08 N×23.56 m/s≈12204.6 W≈12.20 kWP_{belt} = 518.08 \text{ N} \times 23.56 \text{ m/s} \approx 12204.6 \text{ W} \approx 12.20 \text{ kW}

  18. Calculate the number of belts required.

    Number of belts=Total power to transmitPower per belt=30 kW12.20 kW≈2.459\text{Number of belts} = \frac{\text{Total power to transmit}}{\text{Power per belt}} = \frac{30 \text{ kW}}{12.20 \text{ kW}} \approx 2.459

    Since the number of belts must be a whole number, round up.
    The number of V belts required is 3\boxed{\text{3}}.

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