Example of a heptagonal polyhedron? I did a lot of googling but I'm unable to find an example of a convex polyhedron in 3-dimensional space, such that its faces are all congruent irregular heptagons. 
Is there a reason such a shape can't exist?
Also in parallel what is the word for a polyhedron, such that all of its faces are congruent but not necessarily face-transitive. 
 A: $$ V - E + F = 2  $$
let's see, $$  E = 7 F / 2  $$
Each vertex meets at least three faces, $$ V \leq 7F / 3. $$
$$ V - E + F \leq \frac{7F}{3} - \frac{5F}{2} = \frac{-F}{6}  $$
$$ V - E + F \leq  \frac{-F}{6}  $$
$$ 2 \leq  \frac{-F}{6}  $$
which is bad
A: The Euler characteristic of a polyhedron  $F + V - E = 2$
If we glue $n$ heptagons together we have 
$F = n$
Since two faces meet at each edge
$E = \frac {7n}{2}$
And we must have at least 3 faces meeting at a vertex (unless you want to include degenerate heptagons with straight angles, and are really something with fewer sides)
$V \le \frac {7n}{3}$
and for any $n$
$F+V - E < 0$
You might be able to make some sort of torus, though.
Alternatively,
At each vertex the sum of the angles must be less than $360^\circ$ if the shape is convex.
And if we look at the differential between the sum of the angles and $360$ and sum it across all of the vertices, the sum equals $720^\circ$ if the surface is closed (and simply connected).
The average angle in a heptagon is $\frac {5}{7} 180^\circ$
The average vertex is concave. i.e. $ (1+ \frac {1}{7}) 360^\circ$
If you tile with heptagons you will either get "swiss cheese" i.e. a multi-holed torus, or you will get a model of a hyperbolic plane.
