There's a bit more to this subject than what's found with a Google search. Because of expressed interest here I'll offer some further information which additionally validates my former statements.

The Navier-Stokes equations are fundamental partial differential equations which describe an incompressible fluid's flow. Laplace, Euler's pressure to inertial differential & Bernoulli equations are related, as is atmospheric pressure differential & applied vacuum in relation to capillarity. Colebrook equation may be invoved to some extent for overall system behaviour, D'arcy-Weisbach speaks of friction effects or major loss in pipe & Moody diagram helps explain. Hydraulic diameter must be calculated in relation to laminar, transient or turbulent flow with kinematic viscosity effects of fluid dynamics in system relevant to friction coefficient.

Thanks very much to The Engineering Toolbox for the following, used with permission:

Pressure and Pressure Loss

According the Energy Equation for a fluid the total energy can be summarized as elevation energy, velocity energy and pressure energy.

The Energy Equation can then be expressed as:
p1 + ρ v12 / 2 + ρ g h1 = p2 + ρ v22 / 2 + ρ g h2 + ploss (1)

where:
p = pressure in fluid (Pa (N/m2), psi (lb/ft2))
ploss = pressure loss (Pa (N/m2), psi (lb/ft2))
ρ = density of the fluid (kg/m3, slugs/ft3)
v = flow velocity (m/s, ft/s)
g = acceleration of gravity (m/s2, ft/s2)
h = elevation (m, ft)
For horizontal steady state flow v1 = v2 and h1 = h2, - (1) can be transformed to:
ploss = p1 - p2 (2)

The pressure loss is divided in major loss due to friction and minor loss due to change of velocity in bends, valves and similar.

The pressure loss in pipes and tubes depends on the flow velocity, pipe or duct length, pipe or duct diameter, and a friction factor based on the roughness of the pipe or duct, and whether the flow is turbulent or laminar - the Reynolds Number of the flow. The pressure loss in a tube or duct due to friction, major loss, can be expressed as:

ploss = λ (l / dh) (ρ v2 / 2) (3)

where:
ploss = pressure loss (Pa, N/m2)
λ = friction coefficient
l = length of duct or pipe (m)
dh = hydraulic diameter (m)
(3) is also called the D'Arcy-Weisbach Equation. (3) is valid for fully developed, steady, incompressible flow.

Head and Head Loss
The Energy equation can be expressed in terms of head and head loss by dividing each term by the specific weight of the fluid. The total head in a fluid flow in a tube or a duct can be expressed as the sum of elevation head, velocity head and pressure head.
p1 / γ + v12 / 2 g + h1 = p2 / γ + v22 / 2 g + h2 + hloss (4)

where:
hloss = head loss (m, ft)
γ = ρ g = specific weight (N/m3, lb/ft3)
For horizontal steady state flow v1 = v2 and h1 = h2, - (4) can be transformed to:
hloss = h1 - h2 (5)

where:
h = p / γ = head (m, ft)
The head loss in a tube or duct due to friction, major loss, can be expressed as:
hloss = λ (l / dh) (v2 / 2 g) (6)
where
hloss = head loss (m, ft)
Friction Coefficient - λ

The friction coefficient depends on the flow - if it is laminar, transient or turbulent and the roughness of the tube or duct. To determine the friction coefficient we first have to determine if the flow is laminar, transient or turbulent - then use the proper formula or diagram.

Friction Coefficient for Laminar Flow:
For fully developed laminar flow the roughness of the duct or pipe can be neglected. The friction coefficient depends only the Reynolds Number - Re - and can be expressed as:
λ= 64 / Re (7)
where
Re = the dimensionless Reynolds number
The flow is
laminar when Re < 2300
transient when 2300 < Re < 4000
turbulent when Re > 4000

Friction Coefficient for Transient Flow
If the flow is transient - 2300 < Re < 4000 - the flow varies between laminar and turbulent flow and the friction coefficient is not possible to determine.

Friction Coefficient for Turbulent Flow
For turbulent flow the friction coefficient depends on the Reynolds Number and the roughness of the duct or pipe wall. On functional form this can be expressed as:
λ = f( Re, k / dh ) (8)

where:
k = relative roughness of tube or duct wall (mm, ft)
k / dh = the roughness ratio
The friction coefficient - λ - can be calculated by the Colebrooke Equation:
1 / λ1/2 = -2,0 log10 [ (2,51 / (Re λ1/2)) + (k / dh) / 3,72 ] (9)

Since the friction coefficient - λ - is on both sides of the equation, it must be solved by iteration. If we know the Reynolds number and the roughness - the friction coefficient - λ - in the particular flow can be calculated.

A graphical representation of the Colebrooke Equation is the Moody Diagram. With the Moody diagram we can find the friction coefficient if we know the Reynolds Number - Re - and the
Roughness Ratio - k / dh. In the Moody diagram we can see how the friction coefficient depends on the Reynolds number for laminar flow - how the friction coefficient is undefined for transient flow - and how the friction coefficient depends on the roughness ratio for turbulent flow.

For hydraulic smooth pipes - the roughness ratio limits zero - and the friction coefficient depends more or less on the Reynolds number only.

For a fully developed turbulent flow the friction coefficient depends on the roughness ratio only.
Example - Pressure Loss in Air Ducts
Air at 0 oC is flows in a 10 m galvanized duct - 315 mm diameter - with velocity 15 m/s.
Reynolds number are expressed as:
Re = dh v ρ / μ (10)

where:
Re = Reynolds number
v = velocity
ρ = density
μ = dynamic or absolute viscosity

Reynolds number calculated:
Re = (15 m/s) (315 mm) (10-3 m/mm ) (1.23 kg/m3) / (1.79 10-5 Ns/m2)
= 324679 (kgm/s2)/N
= 324679 ~ Turbulent flow

Turbulent flow indicates that Colebrooks equation (9) must be used to determine the friction coefficient - λ -.
With roughness - ε - for galvanized steel 0.15 mm, the roughness ratio can be calculated:
Roughness Ratio = ε / dh
= (0.15 mm) / (315 mm)
= 4.76 10-4
Using the graphical representation of the Colebrooks equation - the Moody Diagram - the friction coefficient - λ - can be determined to:

λ = 0.015
The major loss for the 10 m duct can be calculated with the Darcy-Weisbach Equation:
hloss = λ ( l / dh ) ( ρ v2 / 2 )
= 0.015 ((10 m) / (0.315 m)) ( (1.23 kg/m3) (15 m/s)2 / 2 )
= 65 Pa (N/m2)

There now guys, hopefully this information above will reveal some of the mysteries of hydraulic systems & how to arrive at solutions, such as Surge Loops to influence fluid's motion.