Subtopic: Design and selection of machine elements – gears, pulleys, chains and sprockets and belts
Question Statement:
A flat belt drive is used to transmit power from a tractor PTO shaft to an irrigation pump. The belt has a linear mass density of $m = 0.5\text{ kg/m}$ and a maximum permissible tension of $T_{\text{max}} = 1500\text{ N}$. For maximum power transmission under centrifugal tension considerations, the optimum belt velocity is ________ $\text{m/s}$ (round off to 2 decimal places).
Answer Key:
31.62
Step-by-Step Solution & Explanation
```
[Driving Pulley] [Driven Pulley]
O===========( Flat Belt )===========O
T1 (Tight Side) --->
<--- T2 (Slack Side)
```
**Method 1: First Principles / Calculus of Max Power**
1. Centrifugal tension is given by:
$T_c = m v^2$
2. Effective driving tension capacity:
$T_1 - T_c = T_{\text{max}} - m v^2$
Since $\frac{T_1 - T_c}{T_2 - T_c} = e^{\mu \theta} = k$, power transmitted is:
$P = (T_1 - T_2) v = (T_1 - T_c)\left(1 - \frac{1}{k}\right) v = \left(1 - \frac{1}{k}\right)(T_{\text{max}} v - m v^3)$
3. Differentiating $P$ with respect to $v$ and equating to 0:
$\frac{dP}{dv} = 0 \implies T_{\text{max}} - 3 m v^2 = 0 \implies T_c = \frac{1}{3} T_{\text{max}}$
4. Solving for optimum velocity $v_{\text{opt}}$:
$v_{\text{opt}} = \sqrt{\frac{T_{\text{max}}}{3m}} = \sqrt{\frac{1500}{3 \times 0.5}} = \sqrt{\frac{1500}{1.5}} = \sqrt{1000} \approx 31.6228\text{ m/s}$
**Method 2: Direct Standard Design Formula**
$v = \sqrt{\frac{T_{\text{max}}}{3 m}} = \sqrt{1000} = 31.62\text{ m/s}$