Step 1: Determine the resistivity per unit length for the material of the conductor (you will probably have to look this up on a table of some sort - if you are using some type of cable instead of bus, chances are good the cable manufacturer has a good number to use somewhere on their website.)
Step 2: Convert the actual length of your conductor run to the same units as the resistivity measurement and multiply to obtain the actual resistance.
Step 3: Use Ohm's Law (E = I*R) and the current (in amperes) you're putting through the conductor to determine the voltage drop resulting from the conductor resistance.
You must get the cable resistance, usually in ohms/foot or ohms/meter, from the cable manufacturer. There are many different types of compounds used in conductors and their resistance will vary widely: hard or soft drawn copper, ACSR, direct bury hard copper...too many to generalize an easy answer.
Once you have Ohms/meter, calculate using Ohms law. Ohms/meter x meters x amps. I'd add a couple of ohms for terminations and splices just to be safe. Long overhead lines will also have distributed capacitance. That depends on the conductor orientation and number of rotations, pole height, etc. There is no easy way unless you make some assumptions on conductor resistance.
For "short" transmission runs (< 80 km), capacitance can be neglected. Usually, the only concern is the thermal loading on the transmission line (not the voltage drop, nor the system stability).
As mentioned earlier - get the "real" resistance data from the cable supplier to make your calculations.
In the interests of estimating, however, here is a quick-and-dirty approximation.
Assume aluminum conductor. Cross section is 50 mm2 (close enough to AWG 0).
Resistance per km at 25 C is 0.5511 ohm.
Line current = power / (volts * 1.732) = 70000 / (11000 * 1.732) = 3.67 Ampere
Therefore at 25 C for 22km run, voltage drop is: (22) * (0.5511) * (3.67) = 44.5 V
If we look at a hotter ambient (say 40 C) environment, the resistance per km will go up correspondingly.
This gives a 40 C drop as: (22) * (.5511 * (1+0.003952*(40-25))) * 3.67 = 47.1 V