Queueing Theory: How to estimate steady-state queue length for single queue, N servers?
You need to apply Little's law: 


  The long-term average number of customers in a stable queue L is equal to the long-term average arrival rate, λ, multiplied by the long-term average time a customer spends in the queue, W; or expressed algebraically: L = λW.


The beauty of this law is that it does not depend on the distribution of arrivals or the service time (whether it is Markovian or not, etc.). More technically, and in Kendall's notation, it is true for the general GI/G/m queues.

We now assume that service time follows an exponential distribution (with parameter μ), and the arrivals follow a Poisson distribution (with parameter λ). In addition, we assume there's only one server. That is, our queue is modeled as M/M/1. 

Using Little's Law, it can be shown (see formula (6.15) on page 247 of this book) that:

$W = \frac{\lambda/\mu^2}{1-\lambda/\mu}$

Note that the book uses different notations than here. It also states the formula holds for M/G/1-PS and M/G/1-LCFS queues.

Using Little's Law, we have $L = {\lambda^2 \over \mu^2-\lambda\mu}$.

In your case, λ = 1/(KT), and μ = 1/T. Hence L = 1/K(K-1).



PS: Little's Law has 3 variants. It can be applied to the whole system, to the queue itself, or to the service center. See pages 259-260 of this book for more info.



Edit: The case for M/M/c queues is much trickier. Here, you need to apply Erlang C formula. 

To derive the following formulas, you can take a look at Section 5.2.3 of this book.

Let $a := \lambda/\mu$ and $\rho := a/c$, where c denote the number of servers. Then Erlang C formula, $C[c,a]$ is obtained by:

$C[c,a] = \frac{a^c}{c!} / ((1-\rho)\sum\limits_{n=0}^{c-1}\frac{a^n}{n!}+\frac{a^c}{c!})$

Now we have: $L = \rho C[c,a] / (1- \rho)$.

As a sanity check, note that for $c=1$ we derive the previous answer: $L = {\lambda^2 \over \mu^2-\lambda\mu}$.