24v Power supplies - input and output power
100 – 240 VAC 2.7A. The output is 24 – 28 V DC 10 – 8.6A.


The input current will be maximum at the minimum input voltage. So \$ P = VI = 100 \times 2.7 = 270\ \text W \$. 

Output is \$ 24 \times 10 = 240 \ \text W \$ or \$ 28 \times 8.6 = 240.8 \ \text W \$.

Efficiency is \$ \frac {P_{OUT}}{P_{IN}} = \frac {240}{270} = 89% \$ which is OK.

From this information we can work backwards and see what the input current will be at 240 V: \$ I = \frac {P}{V} = \frac {270}{240} = 1.1\ \text A \$.


  My first question is where is the missing 408W?


It's still back at the power station. Everything is OK.


  If the input is rated at 2.7 A x 8 power packs then I will need an RCB in the consumer unit rated at over 21.6 A.


You have omitted location and supply voltage details from both your question and your user profile. If you are in 110 V land your max current will be probably 2.5 A per unit. In 230 V land it will be about 1.2 A per unit.