Q20Basic Mechanical Engineering

Question

Q.3. Find the power transmitted by a belt running over a pulley of 500 mm diameter at 300 rpm. The coefficient of friction between the belt and pulley is 0.24, angle of lap is 150° and maximum tension in the belt is 2.45 kN.

Answer 1

1. Identify the Given Data: - Diameter of the pulley ( $D$ ) = $500 mm = 0.5 m$ - Rotational Speed ( $N$ ) = $300 rpm$ - Coefficient of friction ( $μ$ ) = $0.24$ - Angle of lap/contact ( $θ$ ) = $15 0^{\circ}$ - Maximum tension in the belt ( $T_{1}$ , tight side) = $2.45 kN = 2450 N$

2. Convert Angle to Radians: The tension ratio formula requires the angle to be strictly in radians. $θ (in radians) = 150 \times \frac{π}{180} = 150 \times \frac{3.1416}{180} = 2.618 radians$

3. Calculate the Linear Velocity of the Belt ( $v$ ): The belt velocity is determined by the circumference and RPM of the pulley. $v = \frac{π \cdot D \cdot N}{60}$

v = \frac{3.1416 \times 0.5 \times 300}{60} = \frac{471.24}{60} = 7.854 m/s

4. Calculate the Tension in the Slack Side ( $T_{2}$ ): Using the fundamental flat belt tension ratio formula: $\frac{T _{1}}{T _{2}} = e^{μ θ}$

μ θ = 0.24 \times 2.618 = 0.62832

e^{0.62832} \approx 1.874

Now, substitute $T_{1}$ to find $T_{2}$ : $2450/ T_{2} = 1.874$ $T_{2} = \frac{2450}{1.874} \approx 1307.36 N$

5. Calculate the Power Transmitted ( $P$ ): The power transmitted is the product of the net driving force (tension difference) and the velocity.

P = (T_{1} - T_{2}) \times v

P = (2450 - 1307.36) \times 7.854

P = 1142.64 \times 7.854 = 8974.29 Watts

Converting to kilowatts: Power Transmitted ( $P$ ) $\approx 8.97 kW$

Q20Basic Mechanical Engineering

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