Simplest Machine Model for a langauge $ L = \{(w\# )^{k}\;|\; w\in \Sigma^{*},\;k\in \mathbb{N}\}$ What is the simplest machine model accepting the following language? 

$$ L = \{(w\# )^{k}\;|\; w\in \Sigma^{*},\;k\in \mathbb{N}\}$$

In other words, $L$ is obtained by taking each string $w$ in $\Sigma^*$, 
and then creating the strings $w\#\;,\; w\#w\#\;,\; w\#w\#w\#\;,...$

Here $\#$ is a separator symbol which is not in $\Sigma$.

The simplest I could think of was queue automata with $2$-queues. But these are already Turing complete. Is there some well studied class of machines that is not Turing complete and accepts $L$?
Just to refine Abuzer's answer, the language can also be recognized by a
Stateless 2-head 1-way DFA which is usually defined in this way [Ibarra et al., On stateless multihead automata: hierarchies and the emptiness problem (2010)]:

For an input string $w \in \Sigma$, machines work on a tape containing $\text{^} w \$$ and
start with all heads on the left end marker. At every step of the computation,
the symbols $a_1,..., a_k$ currently scanned by all $k$ heads are considered, any
corresponding transition $a_1,...,a_k \to d_1,...,d_k \in \delta, a_i \in \Sigma \cup \{\text{^}, \$\}$ is chosen, and each $i$-th heads is moved according to $d_i \in \{ Stay, Right \}$. If no such transition exists, the automaton rejects. If any of the heads falls off the tape, the automaton rejects as well.
If the transition instructs all heads to Stay, the automaton halts and accepts. The string is accepted if there exists a computation resulting in acceptance.

If $\Sigma = \{0,1,\#\}$, then your language is recognized by the following Stateless 2-head 1-way DFA:

   ^   ,  ^   -&gt; S , R
   ^   ,{0,1} -&gt; S , R
   ^   ,  #   -&gt; R , R
   0   ,  0   -&gt; R , R
   1   ,  1   -&gt; R , R
   #   ,  #   -&gt; R , R
{0,1,#},  $   -&gt; S , S