Calculate the belt length in meters, needed for the drive. Calculate the power transmitted.

Power transmitted, $P = 24 \text{ kW} = 24000 \text{ W}$.
Diameter of the driving pulley, $D = 130 \text{ mm} = 0.13 \text{ m}$.
Rotational speed of the driving pulley, $N = 240 \text{ r/min}$.
Contact angle, $\theta = 170^\circ$.
Coefficient of friction, $\mu = 0.3$.

The power transmitted is given by:

So, $T_1 - T_2 = \frac{24000}{1.6336} \approx 14691.48 \text{ N} \quad (Equation \ 1)$

The relationship between tight side tension ($T_1$) and slack side tension ($T_2$) is:

First, calculate the exponent:

Now, calculate $e^{\mu \theta_{rad}}$:

So, $\frac{T_1}{T_2} = 2.4354$, which means $T_1 = 2.4354 T_2 \quad (Equation \ 2)$

Substitute $Equation \ 2$ into $Equation \ 1$:

Now, find $T_1$ using $Equation \ 2$:

The Tight-side tension is $\boxed{\text{24.93 kN}}$ and the Slack-side tension is $\boxed{\text{10.24 kN}}$.