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## Lesson 28. SOLVING NUMERICAL

Module 6. Power transmission
Lesson 28
SOLVING NUMERICAL
Design a V belt drive for the following: Drive: AC motor, operating speed is 1440 rpm and operates for over 10 hours. The equipment driven is a compressor, which runs at 900 rpm and the power transmission is 20kw
Since it is a V belt drive, let us consider belt speed, v = 25 m/sec. Design power P The value 1.3 is selected from design data book for the given service condition. Now, Hence, obvious choice for belt section is Now, V= π*d Where n=Rotational speed RPM d d V = Velocity of belt standard sizes are, d d First combination gives the speed ratio to be 1.68 Second combination gives the speed ratio to be 1.58. So, it is better to choose the second combination because it is very near to the given speed ratio. Therefore, selected pulley diameters are d Center distance, C should be such that, d Let us consider, C = 1500 mm, this value satisfies the above condition. Considering an open belt drive, the belt length, Inside length of belt = 4444 – 56 = 4388 mm The nearest value of belt length for C-section is 4394 mm (from design data book) Therefore, the belt designation is Power rating (kW) of one C-section belt Equivalent small pulley diameter is, d C For the belt speed of 23 m/sec, the given power rating (KW) =12.1 KW For the obtained belt length, the length correction factor C Determination of angle of wrap For the angle of wrap of 3.00 radian (smaller pulley), the angle of wrap factor, C Therefore, incorporating the correction factors, 2 numbers of C 4394\173 belts are required for the transmission of 20kW. |