Problem 167

(Ⅰ) By the Law of Sines, $(2\sin A+\sin C)\cos B + \sin B\cos C = 0$, which can be written as:
$$2\sin A\cos B + \sin C \cos B + \sin B\cos C = 0$$
$$2\sin A \cos B + \sin(B+C) = 0$$
Since $A + B + C = \pi$, we have $\sin(B+C) = \sin A$. Thus, the equation becomes $2\sin A \cos B + \sin A = 0$. Because $\sin A \neq 0$, we can factor out $\sin A$:
$$\sin A(2\cos B + 1) = 0$$
From this, we get $\cos B = -\frac{1}{2}$. As $B$ is an interior angle of a triangle, the value of $B$ is $\boxed{\frac{2\pi}{3}}$.

(Ⅱ) The area $S$ of the triangle can be expressed using the formula $S = \frac{1}{2}ac\sin B$. By substituting $B = \frac{2\pi}{3}$ and $a+c=4$, we have:
$$S = \frac{1}{2}a(4-a)\sin\left(\frac{2\pi}{3}\right)$$
$$S = \frac{\sqrt{3}}{4}(4a - a^2)$$
$$S = \frac{\sqrt{3}}{4}[4 - (a-2)^2]$$
To maximize $S$, we want to minimize $(a-2)^2$, which occurs when $a=2$ because $0 < a < 4$. Thus, the maximum value of $S$ is $\boxed{\sqrt{3}}$.