At zero flow the pressure at the shower head is simply the hydrostatic pressure given by Pascal's law:
$$p=p_0+\rho gy$$
Where $p_0$ is the atmospheric pressure, $y$ the height difference between tank meniscus and shower head, $\rho$ the density of water ($g\approx 10$$\:\mathrm{m/s^2}$).
The water pressure coming from the head is too low.
What the OP really means here is that the shower only delivers a trickle of water (low flow speed). So here I'll evaluate the factors that influence that flow speed.
When flow starts, $p$ is lowered by:
1. Viscous losses in the pipe:
Acc. Darcy Weisbach pressure loss in a straight pipe due to flow is given by:
$$\Delta p=f_D\frac{\rho}{2}\frac{v^2}{D}L$$
Where $f_D$ is a friction factor, $v$ is flow speed ($mathrm{m/s}$), $D$ pipe diameter and $L$ pipe length.
For laminar flow:
$$f_D=\frac{64\mu}{\rho D v}$$
Where $\mu$ is the viscosity of the fluid.
So for laminar flow:
$$\Delta p=\frac{32\mu v}{D^2}L$$
2. Local resistances:
Valves, bends, kinks, sudden changes in diameter etc. all cause head loss $h$, usually modelled as:
$$h_r=c\frac{v^2}{2g}$$
Where $c$ is a coefficient that depends on the type of local resistance.
In the OP's stated problem the main local resistance is almost certainly the shower head itself.
3. Bernoulli's principle:
Using Bernoulli's principle we can now write (for laminar flow):
$$y=\frac{v^2}{2g}+\frac{32\mu v}{\rho gD^2}L+c_{shower}\frac{v^2}{2g}$$
Or:
$$y=(c_{shower}+1)\frac{v^2}{2g}+\frac{32\mu v}{\rho gD^2}L$$
This is a simple quadratic equation in $v$ and if $c_{shower}$ and the other factors where known, then it could be solved quite easily. But in the absence of that information we can still say that $v$:
- will increase with $y$,
- will increase with $D$,
- will decrease with $L$,
- will decrease with $c_{shower}$.
4. Turbulent flow:
In the case of turbulent flow (high $v$, $Re > 4000$), $f_D$ becomes a function of $v$, $f_D=f(v)$ and the calculation becomes more complicated. But the general conclusions above still hold.
