This paper studies the fundamental relations among integer flows, modulo orientations, integer-valued and real-valued circular flows, and monotonicity of flows in signed graphs. A (signed) graph is modulo-$(2p+1)$-orientable if it has an orientation such that the indegree is congruent to the outdegree modulo $2p+1$ at each vertex. An integer-valued $\frac{2p+1}{p}$-flow is a flow taking integer values in $\{\pm p, \pm (p+1)\}$. Extending a fundamental result of Jaeger to signed graphs, we show that a bridgeless signed graph is modulo-$(2p+1)$-orientable if and only if it admits an integer-valued $\frac{2p+1}{p}$-flow. It was conjectured by Raspaud and Zhu that, for any signed graph, the admission of a circular $r$-flow implies the admission of an integer-valued $\lceil r \rceil$-flow. Although this conjecture has been disproved in general, it is confirmed in this paper for bridgeless signed graphs if $r=\frac{2p+1}{p}$ and $p \geq 3$.

A mod $(2p+1)$-orientation $D$ is an orientation of $G$ such that $d_D^+(v)-d_D^-(v)\equiv 0 \pmod {2p+1}$ for any vertex $v \in V(G)$. Jaeger conjectured that every $4p$-edge-connected graph has a mod $(2p+1)$-orientation. A graph $G$ is strongly ${\mathbb Z}_{2p+1}$-connected if for every mapping $b: V(G) \mapsto {\mathbb Z}_{2p+1}$ with $\sum_{v\in V(G)}b(v)=0$, there exists an orientation $D$ of $G$ such that $d_D^+(v)-d_D^-(v)= b(v)$ in ${\mathbb Z}_{2p+1}$ for any $v \in V(G)$. A strongly ${\mathbb Z}_{2p+1}$-connected graph admits a mod $(2p+1)$-orientation, and it is a contractible configuration for mod $(2p+1)$-orientation. We prove Jaeger's module orientation conjecture is equivalent to its restriction to bipartite simple graphs and investigate strongly ${\mathbb Z}_{2p+1}$-connectedness of certain bipartite graphs, particularly for $p=2$. We also show that if $G$ is a simple graph with $|V(G)|\ge N(p)= 1152p^4$ and $\min\{\delta(G),\delta(G^c)\}\ge 4p$, then either $G$ or $G^c$ is strongly ${\mathbb Z}_{2p+1}$-connected. When $p=2$, the value of $N(2)$ can be reduced to $N(2) = 80$.