P_{1 }+ ρgh_{1 }+ ½ρv_{1}^{2 }
= P_{2 }+ ρgh_{2 }+ ½ρv_{2}^{2 }

or

P + ρgh + ½ρv^{2 }= constant.

If a liquid is at rest, then

P_{1 }+ ρgh_{1
}^{ }= P_{2 }+ ρgh_{2},

P_{bottom} + ρgh_{bottom} = P_{top} + ρgh_{top},

P_{bottom} = P_{top} + ρg(h_{top} - h_{bottom}).

The pressure in the fluid increases linearly with depth.

If a liquid is at rest in a horizontal pipe, P_{1 }
^{ }= P_{2}, the pressure is the same everywhere.

P_{1 }+ ½ρv_{1}^{2 }= P_{2
}+ ½ρv_{2}^{2}.

If a liquid (or a gas which is not being compressed) is flowing frictionless in a steady state through a horizontal pipe with a varying cross-sectional area, then the pressure depends on the speed of the fluid.

The faster the fluid is flowing, the lower is the pressure at the same height.

This may seem counterintuitive to you, but it is a consequence of conservation of energy. The molecules of a fluid at room temperature are always in motion, even if the fluid as a whole is at rest. This disordered motion is responsible for the pressure exerted by the fluid, even in gravity-free space. In a pipe, it results in collisions with the walls. If a fluid is flowing trough a horizontal pipe at a steady rate, then the molecules also have ordered motion. In a narrow section of the pipe the fluid is flowing faster, and more of its energy goes into the ordered motion. This leaves less energy for the random motion and therefore results in softer collisions and lower pressure. Phenomena which can be understood with the help of Bernoulli's equation include the Pitot tube, the Venturi effect, atomizers, hurricanes, flapping flags, etc.

A Venturi tube may be used as a fluid flow meter. If the difference in
pressure P_{1 }- P_{2 }= 21 kPa, find the fluid flow rate in m^{3}/s
given that the radius of the outlet tube is 1 cm, the radius of the inlet tube is
2 cm, and the fluid is gasoline (ρ = 700 kg/m^{3}).

Solution:

P_{1 }+ ρgh_{1 }+ ½ρv_{1}^{2
}= P_{2
}+ ρgh_{2 }+ ½ρv_{2}^{2},

h is constant, so P_{1
}+ ½ρv_{1}^{2 }= P_{2 }+ ½ρv_{2}^{2}.

P_{1 }- P_{2 }= ½ρv_{2}^{2 }- ½ρv_{1}^{2}.

21 kPa = 350 kg/m^{3
}(v_{2}^{2 }- v_{1}^{2}).

From the equation of continuity we have

Area 1*v_{1} = Area 2*v_{2}. v_{1 }= (A_{2}/A_{1})v_{2}.

Inserting this into the above equation we have

(1-(A_{2}/A_{1})^{2})v_{2}^{2}) =
(21000/350)(m/s)^{2}.

(A_{2}/A_{1})^{2 }= (1/4)^{2
}= 1/16. v_{2}^{2
}= (21000/350)(16/15)(m/s)^{2 }= 64(m/s)^{2.}.

v_{2
}= 8 m/s.

Volume flow rate: A_{2}v_{2} =
π(0.01 m)^{2}*8 m/s = 0.0025 m^{3}/s.

Real fluids are not ideal fluids. But in a short enough section, the laminar flow of a real fluid may be approximately treated as ideal, if the energy loss in this section is very small compared to the ordered kinetic energy of the fluid in the section. Then Bernoulli's equation is approximately valid for this section of the real fluid.