Why no interaction in two-way ANOVA without replication This may seem like a silly question, but I can't figure out (or find anywhere) why it's not possible to calculate interaction with a 2-way ANOVA without replication?
Has it something to do with the d.f for the interaction terms being zero?  
 A: It's not clear what you mean by 'not possible' and 'without replication'
Suppose (for concreteness) you have two two-level treatments $A$ and $B$, and you observe one value of $Y$ for each of the four combinations (not $A$, not $B$), (not $A$, $B$), ($A$, not $B$), ($A$, $B$).   Call them $Y_{00}$, $Y_{01}$, $Y_{10}$, $Y_{11}$
You have four numbers, and there are five parameters you want to estimate: the grand mean, the two main effects, the interaction, and the error variance. That won't work.
You can either estimate the interaction: $(Y_{11}-Y_{10})- (Y_{01}-Y_{00})$, or estimate the error variance. Mathematically, that's your choice.
If you estimate the interaction, your model will fit the data perfectly, and you will have no information left to estimate the error variance (the sum of squared residuals will be zero and the degrees of freedom will be zero, and the variance is the ratio of the two).
Alternatively, if don't estimate an interaction, you have one degree of freedom left to estimate the error variance. 
Neither of these is ideal: in the former case you'll be treating any real error variance as if it was part of the interaction; in the latter case you'll be treating any real interaction as if it was part of the error variance.  
It usually makes more sense to push the interactions into the error variance than vice versa -- the error variance then becomes a generalised category of "stuff we didn't model" -- and your software might well insist that's the only way to do it. 
A: I believe it is something to do with the fact that ANOVA only takes a look at averages, so having no replications won't afford that information. To be specific, the ANOVA chart will not produce P-values for you to make decisions with unless you have at least one replication. 
