A flat belt drives a pulley 500 mm in diameter at 450 rpm. The tight side tension is in the belt is three times the slack side tension. Calculate the width of the belt required to transmit 18 kW of power if the maximum allowable tension in the belt is 14 N/mm width.

The radius of the pulley is $r = \frac{D}{2} = \frac{0.5 \text{ m}}{2} = 0.25 \text{ m}$.

$$v = \omega \times r = 15 \pi \text{ rad/s} \times 0.25 \text{ m} = 3.75 \pi \text{ m/s}$$

The power transmitted by a belt drive is given by $P = (T_1 - T_2) \times v$.
We are given $T_1 = 3 T_2$, so $T_1 - T_2 = 3 T_2 - T_2 = 2 T_2$.
Substituting this into the power equation:

$$P = 2 T_2 \times v$$ $$18000 \text{ W} = 2 T_2 \times 3.75 \pi \text{ m/s}$$ $$18000 = 7.5 \pi T_2$$ $$T_2 = \frac{18000}{7.5 \pi} \text{ N}$$

Now, calculate $T_1$:

$$T_1 = 3 T_2 = 3 \times \frac{18000}{7.5 \pi} = \frac{54000}{7.5 \pi} = \frac{7200}{\pi} \text{ N}$$ $$T_1 \approx \frac{7200}{3.14159} \approx 2291.83 \text{ N}$$

The maximum allowable tension in the belt is $14 \text{ N/mm width}$. This means the total tight side tension $T_1$ must be equal to or less than the allowable tension per unit width multiplied by the belt width.

$$\text{Width} = \frac{T_1}{\text{Allowable tension per mm width}}$$ $$\text{Width} = \frac{2291.83 \text{ N}}{14 \text{ N/mm}} \approx 163.70 \text{ mm}$$

The width of the belt required is $\text{163.70 mm}$.