# Automation: steady state error calculation block diagram

Hello I have a problem determining the value of K in this block diagram, to calculate this value, I applied this method, but the value at the end of K I is different from 216, you could explain to me where I'm wrong  So, for this problem, you first need to find the transfer function from U to E. An easy way to do this to use the forward gain over one plus the loop gain. This simply mean $\frac{E(s)}{U(s)} = \frac{forward~gain}{1+loop~gain}$.
The forward gain in this problem is 1 (U to E has no gains between it) while the loop gain is $\frac{K}{(s+2)(s+3)}*\frac{1}{s+4}$ (all gains around loop from E back to U).
The steady state error is $e_{ss} = \lim_{s\to0} s\frac{E(s)}{U(s)} U(s)$. This simplifies to what you had above. If you substitute everything is, you have $0.1 = \lim_{s\to0}\frac{1}{1+\frac{K}{(s+2)(s+3)(s+4)}}$.
Multiply both top and bottom by $(s+2)(s+3)(s+4)$ and set s equal to 0. Multiply it all out and you should get 216.