$H(\kappa)$ a model of all the axioms of ZFC for $\kappa$ not inaccessible Is this possible?
For each cardinal $\kappa$, we can define $H(\kappa) = \{ x \mid |trclx| < \kappa\}$, where trcl is the transitive closure, and $|A|$ is the cardinality of $A$. For every regular cardinal, and any axiom $\phi$ of ZFC aside from the axiom of infinity and the power set axiom, we have $H(\kappa) \models \phi$.
Additionally, we have a Levy Reflection Principle to the effect that for any formula expressable in ZFC $\phi$, we have that:
$\forall \alpha \in ON \ \exists \beta \in ON [\beta \geq \alpha \text{ and } \phi \text{ is absolute for } R(\beta)]$
Where $R(\beta)$ is the cumulative hierarchy. 
As a homework exercise, I am to state a corresponding principle for $H(\beta)$, and to address the following question: is it possible that when we apply the principle to the power set axiom, that we have a model of ZFC $H(\kappa)$ for a $\kappa$ that is not inaccessible? 
Inaccessibility here is taken to mean regular plus unreachable via power sets (if $\alpha < \kappa$, then $|P(\alpha)| < \kappa$).
My thoughts: since $H(\kappa)$ is a model of the power set axiom, it will be closed under the power set operation and so we have $P(\alpha) < \kappa$ for every $\alpha < \kappa$. As for regularity, if we assume it's not regular then we have a cofinal subset of cardinality less than $\kappa$. Each element will be in $H(\kappa)$, so we have a set of subsets of $H(\kappa)$ indexed by a set in $H(\kappa)$. So I want to say that the union should be in $H(\kappa)$, which would be a contradiction (if it's in $H(\kappa)$ it must have cardinality less than $\kappa$, but the union of this set should have cardinality $\kappa$. But I'm not sure if it can be shown that the union really is in $H(\kappa)$. 
 A: Question: Is it consistent that $H(\kappa)\vDash ZFC$ even though $\kappa$ is not inaccessible?
Answer: yes. Let $\kappa$ be inaccessible. Within $V_\kappa$ we can adapt the usual argument for reflection to show:
Theorem: There is a club $C\subseteq \kappa$ such that $\forall\alpha\in C(H(\alpha)\prec V_\kappa)$ (that is, $H(\alpha)$ is an elementary substructure of $V_\kappa$). 
Clearly, for each such $\alpha$, $H(\alpha)\vDash ZFC$. But since the ordinals below $\kappa$ with co-finality $\omega$ are club, there will be club $\alpha$ such that $H(\alpha)\vDash ZFC$ which are not inaccessible.
Question: Are there $H(\kappa)$ which model $ZFC - Replacement + \neg Replacement$?
Answer: yes. We can define the beth function $\beth_\alpha$ such that $\beth_0 = 0$, $\beth_{\alpha+1} = 2^{\beth_\alpha}$, and $\beth_\lambda = \bigcup_{\alpha<\lambda}\beth_\alpha$. By recursion, we can define $\kappa_0 = \omega$, $\kappa_{n+1}$ to be the least ordinal such that for all $\beta<\kappa_\alpha$, $\beth_\beta< \kappa_{n+1}$, and $\kappa_\omega = \bigcup\kappa_n$. Then, $\kappa_\omega = \beth_{\kappa_\omega}$, and so $H(\kappa_\omega) \vDash ZFC - Replacement$. But since the $\kappa_n$ are definable, replacement will fail in $H(\kappa_\omega)$. 
