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Number of V-belts for a motor drive

Short answer

Larger pulley D=420 mm. Belt speed v=10.56 m/s. Trial length ≈2504 mm; standard pitch length chosen 2500 mm giving exact centre distance 798 mm. Design power 18 kW, corrected belt rating 3.36 kW → number of belts = 18/3.36 = 5.36 → 6 B-section belts.

A 15 kW, 1440 rpm motor drives a machine at 480 rpm through B-section V-belts. The pitch diameter of the smaller pulley is 140 mm and the centre distance is about 800 mm. The service factor is 1.2. The rated power of one B belt at this speed and pulley size is 3.4 kW; the correction factors are 0.95 for the arc of contact (about 160 degrees) and 1.04 for belt length 2500 mm. Standard B-section pitch lengths are 2240, 2360, 2500, 2650 and 2800 mm. Find the larger pulley diameter, the belt speed, the belt length and actual centre distance, and the number of belts.

Pulley sizes and speed

D=d n1n2=140×1440480=420 mmD = d\,\frac{n_1}{n_2} = 140 \times \frac{1440}{480} = 420\ \text{mm} v=πdn160=π×0.14×144060=10.56 m/sv = \frac{\pi d n_1}{60} = \frac{\pi \times 0.14 \times 1440}{60} = 10.56\ \text{m/s}

(The speed is within the usual limit of 25 m/s for V-belts.)

Belt length

Trial centre distance C=800C = 800 mm:

L=2C+π2(D+d)+(D−d)24C=1600+879.6+24.5=2504 mmL = 2C + \frac{\pi}{2}(D + d) + \frac{(D - d)^2}{4C} = 1600 + 879.6 + 24.5 = 2504\ \text{mm}

Nearest standard pitch length: 2500 mm.

Actual centre distance

C=B+B2−(D−d)28,B=L4−π(D+d)8C = B + \sqrt{B^2 - \frac{(D - d)^2}{8}}, \qquad B = \frac{L}{4} - \frac{\pi (D + d)}{8} B=625−219.9=405.1 mmB = 625 - 219.9 = 405.1\ \text{mm} C=405.1+405.12−28028=405.1+164 106−9800=405.1+392.8=797.9 mmC = 405.1 + \sqrt{405.1^2 - \frac{280^2}{8}} = 405.1 + \sqrt{164\,106 - 9800} = 405.1 + 392.8 = 797.9\ \text{mm}

Contact angle on the small pulley:

θ=180∘−2sin⁡−1D−d2C=180∘−2sin⁡−12801596=159.8∘\theta = 180^\circ - 2\sin^{-1}\frac{D - d}{2C} = 180^\circ - 2\sin^{-1}\frac{280}{1596} = 159.8^\circ

This agrees with the 160 degrees assumed for the factor 0.95.

Number of belts

Design power=1.2×15=18 kW\text{Design power} = 1.2 \times 15 = 18\ \text{kW} Corrected rating per belt=3.4×0.95×1.04=3.36 kW\text{Corrected rating per belt} = 3.4 \times 0.95 \times 1.04 = 3.36\ \text{kW} Number of belts=183.36=5.36  ⇒  6 belts\text{Number of belts} = \frac{18}{3.36} = 5.36 \;\Rightarrow\; 6\ \text{belts}

Answer: D = 420 mm; v = 10.56 m/s; L = 2500 mm; C = 798 mm; 6 B-section belts.

✓ Answer checked by Claude Opus 5.5 (high effort)

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