A v-belt drive is used to transmit 38 kW power at 1440 r.p.m. from a three phase induction motor to a centrifugal pump, required to be operated at 360 r.p.m. The motor pulley pitch diameter is 225 mm and the groove angle is $38^0$. The central distance between the pulleys is 1 m. The coefficient of friction for the belt pulley combination is 0.2 and the density of the belt material is 0.97 gm/cc. If the allowable tension in the belt is 800 N, determine:

i) the number of belts required; and ii) the pitch length of the belt.

Assume suitable cross-section for the belt based on the power to be transmitted.

Mumbai University > Mechanical Engineering > Sem 7 > Machine Design 2

Marks: 10M

Year: Dec 2016

$d = 225 mm\\ N_1 = 1440\\ N_2 = 360\\ i = 4\\ c = 1m$

For P = 38 and d = 225 mm

Select cross section (PSG 7.58)

For c - 8

Top width W = 22 mm

Nominal thickness T = 14 mm

D = idn D = 4 x 225 x 0.98 (assuming slip less than 2 %) D = 882 mm $D \approx 890 mm$

Length of belt

$L = 2c + \frac{\pi}{2} (D+d)+ \frac{(P - d)^2}{4C}\\ L = 3846 mm$

sts length available = 3713 mm for cc/s

corrected center distance = 0.92 m

No. of belt $n = \frac{P \times Fa}{KW \times F_c \times F_d}$

$F_c = 1$ for selected belt (PSG 7.60)

$F_d$ = correction factor to a

Arc of contact

$d = 180 - 60 \frac{D - d}{C}\\ d = 136.63\\ F_d = 0.88 \text{(for d = 136)}$

$F_a = 1.2$ for induction motor and centrifugal pump

$KW = \bigg[ 1.47s^{-0.09} - \frac{142.7}{de} - n2.34 \times 10^{-4} s^2 \bigg]s\\ \text{where} \hspace{0.2cm} d_c = d_p \times F_b\\ \hspace{1cm} = 225 \times F_b\\ F_b = 1.14 (for i = 4)\\ d_e = 256.5\\ s = 16..964\\ kw = 8.644\\ n = 5.97\\ n \approx 6 \text{belts}$