Q20Basic Mechanical Engineering

Question

Q.3 A belt is running over a pulley of 600 mm diameter at 200 rpm. The coefficient of friction is 0.25 and angle of lap is 160°. If maximum tension is 2.5 kN, find the power transmitted.

Answer 1

In mechanical power transmission systems, a flat belt drive systematically transfers rotary kinetic energy completely from a driving pulley to a driven pulley strictly via surface friction. The total mechanical power transmitted is mathematically fundamentally dependent entirely on the absolute difference in tension strictly between the tight side ( $T_{1}$ ) and the slack side ( $T_{2}$ ) of the belt, multiplied directly by the linear velocity ( $v$ ) of the belt itself.

Step 1: Given Geometric and Physical Parameters

Maximum allowable belt tension (Tight side, $T_{1}$ ): $2500 N$
Pulley rotational speed ( $N$ ): $200 RPM$
Pulley physical diameter ( $D$ ): $0.6 m$
Coefficient of static friction ( $μ$ ): $0.25$
Angle of contact ( $θ$ ): $16 0^{\circ}$

Step 2: Calculating Linear Belt Velocity ( $v$ )

The absolute linear velocity of the belt physically traveling over the pulley surface is rigorously defined by the mathematical product of the pulley's exact circumference and its rotational frequency in revolutions per second.

v = \frac{π D N}{60}

v = \frac{π \times 0.6 \times 200}{60} = \frac{120 π}{60} = 2 π \approx 6.283 m/s

Step 3: Calculating Slack Side Tension ( $T_{2}$ )

The limiting tension ratio strictly between the tight side and slack side of a flat belt actively on the verge of physical slipping is rigorously governed universally by the exponential friction equation. Before applying this, the contact angle MUST be mathematically converted strictly from degrees to radians.

θ_{rad} = 16 0^{\circ} \times \frac{π}{18 0 ^{\circ}} = \frac{16 π}{18} = \frac{8 π}{9} \approx 2.792 radians

Now, we rigorously apply the fundamental belt tension ratio formula:

\frac{T _{1}}{T _{2}} = e^{μ θ}

\frac{2500}{T _{2}} = e^{(0.25 \times 2.792)} = e^{0.698} \approx 2.01

By systematically solving for the unknown slack side tension $T_{2}$ :

T_{2} = \frac{2500}{2.01} \approx 1243.78 N

Step 4: Final Power Calculation ( $P$ )

The total effective mechanical driving force actively turning the pulley is exactly the mathematical difference in tensions ( $T_{1} - T_{2}$ ). The total physical power ( $P$ ) transmitted is strictly the product of this net driving force and the linear velocity.

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

P = (2500 - 1243.78) \times 6.283 = 1256.22 \times 6.283 \approx 7892.8 Watts

Converting strictly to kilowatts (kW) precisely yields:

P \approx 7.89 kW

Therefore, the maximum safe continuous mechanical power transmitted by the belt drive is rigorously mathematically calculated to be exactly $7.89 kW$ .

Q20Basic Mechanical Engineering

Question

Answer

Step 1: Given Geometric and Physical Parameters

Step 2: Calculating Linear Belt Velocity (v)

Step 3: Calculating Slack Side Tension (T2​)

Step 4: Final Power Calculation (P)

Step 2: Calculating Linear Belt Velocity ( $v$ )

Step 3: Calculating Slack Side Tension ( $T_{2}$ )

Step 4: Final Power Calculation ( $P$ )