Answer to Question #222818 in Mechanical Engineering for Satyam Yadav

Question

A flat belt drive is to be used to drive a reciprocating compressor running at 720 rpm by a 15 kw 1440 rpm electric motor the required centre distance is 2 m. select the flat belt for the drive from the manufacturer's catalogue

Answer 1

N= 720 rpm, P= 15 kW, Electric motor = 1440 rpm, center distance is 2 m

We will use belt velocity as 18 m/s .

$v_1= \frac{\pi d_1N}{60}$

$18 = \frac{\pi d 1440}{60}$

d= $\frac{18\times 60 }{\pi \times 1440}=0.238 m= 238 mm$

Load correction factor is 1.3 for compressor from table

So,

Maximum power = $1.3 \times 15 =$ 19.5 kW

Step 2: Arc of contact factor

$\alpha _s= 180 - 2sin^{-1}(\frac{(D-d)}{2C})$

$\alpha_s=175.23 ^o$

and the arc contact factor for this angle

F_d = 1.019

Step 3:

Corrected power = 1.019 $\times maximum [power$ = 19.87 kW

Step 4:

Corrected belt rating = $\frac{0.0118 \times 18.85 }{5.08}=0.0438 kW$

Step 5: Selection of belt

width $\times$ No of piles = corrected power/ correct belt rating

4 piles , w =$\frac{756.17}{4}=189.04 mm$

5 piles , w=$\frac{756.17}{5}=151 .2 mm$

We will select a HI-SPeed belt of 152 mm preferred width with 5 piles.