Editing Bernoulli's principle (section)

== Derivations ==
{{math proof
| title = Bernoulli equation for incompressible fluids
| proof =
The Bernoulli equation for incompressible fluids can be derived by either [[integral|integrating]] [[Newton's second law of motion]] or by applying the law of [[conservation of energy]], ignoring [[viscosity]], compressibility, and thermal effects.
; Derivation by integrating Newton's second law of motion
The simplest derivation is to first ignore gravity and consider constrictions and expansions in pipes that are otherwise straight, as seen in [[Venturi effect]].  Let the {{mvar|x}} axis be directed down the axis of the pipe.

Define a parcel of fluid moving through a pipe with cross-sectional area {{mvar|A}}, the length of the parcel is {{math|d''x''}}, and the volume of the parcel {{math|''A'' d''x''}}. If [[mass density]] is {{math|ρ}}, the mass of the parcel is density multiplied by its volume {{math|1=''m'' = ''ρA'' d''x''}}. The change in pressure over distance {{math|d''x''}} is {{math|d''p''}} and [[flow velocity]] {{math|1=''v'' = {{sfrac|d''x''|d''t''}}}}.

Apply [[Newton's second law of motion]] (force&nbsp;= mass&nbsp;×&nbsp;acceleration) and recognizing that the effective force on the [[fluid parcel|parcel of fluid]] is {{math|−''A'' d''p''}}. If the pressure decreases along the length of the pipe, {{math|d''p''}} is negative but the force resulting in flow is positive along the {{mvar|x}} axis.

<math display="block">\begin{align}
m \frac{\mathrm{d}v}{\mathrm{d}t}&= F \\
\rho A \mathrm{d}x \frac{\mathrm{d}v}{\mathrm{d}t} &= -A \mathrm{d}p \\
\rho \frac{\mathrm{d}v}{\mathrm{d}t} &= -\frac{\mathrm{d}p}{\mathrm{d}x} 
\end{align}</math>

In steady flow the velocity field is constant with respect to time, {{math|1=''v'' = ''v''(''x'') = ''v''(''x''(''t''))}}, so {{mvar|v}} itself is not directly a function of time {{mvar|t}}. It is only when the parcel moves through {{mvar|x}} that the cross sectional area changes: {{mvar|v}} depends on {{mvar|t}} only through the cross-sectional position {{math|''x''(''t'')}}.
<math display="block">\frac{\mathrm{d}v}{\mathrm{d}t}
= \frac{\mathrm{d}v}{\mathrm{d}x}\frac{\mathrm{d}x}{\mathrm{d}t}
= \frac{\mathrm{d}v}{\mathrm{d}x}v
= \frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{v^2}{2} \right).</math>

With density {{mvar|ρ}} constant, the equation of motion can be written as
<math display="block">\frac{\mathrm{d}}{\mathrm{d}x} \left( \rho \frac{v^2}{2} + p \right) =0</math>
by integrating with respect to {{mvar|x}}
<math display="block"> \frac{v^2}{2} + \frac{p}{\rho}= C</math>
where {{mvar|C}} is a constant, sometimes referred to as the Bernoulli constant. It is not a [[universal constant]], but rather a constant of a particular fluid system. The deduction is: where the speed is large, pressure is low and vice versa.

In the above derivation, no external work–energy principle is invoked. Rather, Bernoulli's principle was derived by a simple manipulation of Newton's second law.
[[File:BernoullisLawDerivationDiagram.svg|thumb|center|600px|A streamtube of fluid moving to the right. Indicated are pressure, elevation, flow speed, distance ({{mvar|s}}), and cross-sectional area. Note that in this figure elevation is denoted as {{mvar|h}}, contrary to the text where it is given by {{mvar|z}}.]]
; Derivation by using conservation of energy
Another way to derive Bernoulli's principle for an incompressible flow is by applying conservation of energy.<ref name="FeynmanLeightonSands1963V2">{{cite book |first1=R.P. |last1=Feynman |author-link1=R.P. Feynman |first2=R.B. |last2=Leighton |author-link2=R.B. Leighton |first3=M. |last3=Sands |year=1963 |title=The Feynman Lectures on Physics |isbn=978-0-201-02116-5 |volume=2 |title-link=The Feynman Lectures on Physics|publisher=Addison-Wesley }}{{rp|at= §40–3|pp=40–6 to 40–9}}</ref> In the form of the [[work (physics)|work-energy theorem]], stating that<ref>{{cite book |last=Tipler |first=Paul |title=Physics for Scientists and Engineers: Mechanics |edition=3rd extended |publisher=W. H. Freeman |year=1991 |isbn=978-0-87901-432-2}}, p. 138.</ref>
{{block indent | em = 1.5 | text = ''the change in the kinetic energy ''{{math|''E''<sub>kin</sub>}}'' of the system equals the net work {{mvar|W}} done on the system''; <math display="block">W = \Delta E_\text{kin}.</math>}}
Therefore,
{{block indent | em = 1.5 | text = ''the [[Mechanical work|work]] done by the [[force]]s in the fluid equals increase in [[kinetic energy]].''}}
The system consists of the volume of fluid, initially between the cross-sections {{math|''A''<sub>1</sub>}} and {{math|''A''<sub>2</sub>}}. In the time interval {{math|Δ''t''}} fluid elements initially at the inflow cross-section {{math|''A''<sub>1</sub>}} move over a distance {{math|1=''s''<sub>1</sub> = ''v''<sub>1</sub> Δ''t''}}, while at the outflow cross-section the fluid moves away from cross-section {{math|''A''<sub>2</sub>}} over a distance {{math|1=''s''<sub>2</sub> = ''v''<sub>2</sub> Δ''t''}}. The displaced fluid volumes at the inflow and outflow are respectively {{math|''A''<sub>1</sub>''s''<sub>1</sub>}} and {{math|''A''<sub>2</sub>''s''<sub>2</sub>}}. The associated displaced fluid masses are – when {{mvar|ρ}} is the fluid's [[density|mass density]] – equal to density times volume, so {{math|''ρA''<sub>1</sub>''s''<sub>1</sub>}} and {{math|''ρA''<sub>2</sub>''s''<sub>2</sub>}}. By mass conservation, these two masses displaced in the time interval {{math|Δ''t''}} have to be equal, and this displaced mass is denoted by&nbsp;{{math|Δ''m''}}:
<math display="block">\begin{align}
\rho A_1 s_1 &= \rho A_1 v_1 \Delta t = \Delta m, \\
\rho A_2 s_2 &= \rho A_2 v_2 \Delta t = \Delta m.
\end{align}</math>

The work done by the forces consists of two parts:
* The ''work done by the pressure'' acting on the areas {{math|''A''<sub>1</sub>}} and {{math|''A''<sub>2</sub>}} <math display="block">W_\text{pressure}=F_{1,\text{pressure}} s_{1} - F_{2,\text{pressure}} s_2 =p_1 A_1 s_1 - p_2 A_2 s_2 = \Delta m \frac{p_1}{\rho} - \Delta m \frac{p_2}{\rho}.</math>
* The ''work done by gravity'': the gravitational potential energy in the volume {{math|''A''<sub>1</sub>''s''<sub>1</sub>}} is lost, and at the outflow in the volume {{math|''A''<sub>2</sub>''s''<sub>2</sub>}} is gained.  So, the change in gravitational potential energy {{math|Δ''E''<sub>pot,gravity</sub>}} in the time interval {{math|Δ''t''}} is
<math display="block">\Delta E_\text{pot,gravity} = \Delta m\, g z_2 - \Delta m\, g z_1. </math>
Now, the [[Energy#Potential energy|work by the force of gravity is opposite to the change in potential energy]], {{math|1=''W''<sub>gravity</sub> = −''ΔE''<sub>pot,gravity</sub>}}: while the force of gravity is in the negative {{mvar|z}}-direction, the work—gravity force times change in elevation—will be negative for a positive elevation change {{math|1=Δ''z'' = ''z''<sub>2</sub> − ''z''<sub>1</sub>}}, while the corresponding potential energy change is positive.<ref name="FeynmanLeightonSands1963V1">{{Cite book| first1=R.P. | last1=Feynman |author-link1=R. P. Feynman | first2=R.B. | last2=Leighton| author-link2=R. B. Leighton | first3=M. | last3=Sands | year=1963 | title=The Feynman Lectures on Physics | isbn=978-0-201-02116-5|volume= 1| title-link=The Feynman Lectures on Physics | publisher=Addison-Wesley }}</ref>{{rp|at= §14–3| p=14–4}} So: <math display="block">W_\text{gravity} = -\Delta E_\text{pot,gravity} = \Delta m\, g z_1 - \Delta m\, g z_2.</math>
And therefore the total work done in this time interval {{math|Δ''t''}} is
<math display="block">W = W_\text{pressure} + W_\text{gravity}.</math>
The ''increase in kinetic energy'' is
<math display="block">\Delta E_\text{kin} = \tfrac12 \Delta m\, v_2^2-\tfrac12 \Delta m\, v_1^2.</math>
Putting these together, the work-kinetic energy theorem {{math|1=''W'' = Δ''E''<sub>kin</sub>}} gives:<ref name="FeynmanLeightonSands1963V2" />
<math display="block">\Delta m \frac{p_1}{\rho} - \Delta m \frac{p_2}{\rho} + \Delta m\, g z_1 - \Delta m\, g z_2 = \tfrac12 \Delta m\, v_2^2 - \tfrac12 \Delta m\, v_1^2</math>
or
<math display="block">\tfrac12 \Delta m\, v_1^2 + \Delta m\, g z_1 + \Delta m \frac{p_1}{\rho} = \tfrac12 \Delta m\, v_2^2 + \Delta m\, g z_2 + \Delta m \frac{p_2}{\rho}.</math>
After dividing by the mass {{math|1=Δ''m'' = ''ρA''<sub>1</sub>''v''<sub>1</sub> Δ''t'' = ''ρA''<sub>2</sub>''v''<sub>2</sub> Δ''t''}} the result is:<ref name="FeynmanLeightonSands1963V2" />
<math display="block">\tfrac12 v_1^2 +g z_1 + \frac{p_1}{\rho}=\tfrac12 v_2^2 +g z_2 + \frac{p_2}{\rho}</math>
or, as stated in the first paragraph:
{{NumBlk||<math display="block">\frac{v^2}{2}+g z+\frac{p}{\rho} = C</math> |{{EquationRef|Eqn. 1}}, which is also Equation (A)}}
Further division by {{mvar|g}} produces the following equation. Note that each term can be described in the [[length]] dimension (such as meters). This is the head equation derived from Bernoulli's principle:
{{NumBlk||<math display="block">\frac{v^2}{2 g}+z+\frac{p}{\rho g}=C</math>|{{EquationRef|Eqn. 2a}}}}
The middle term, {{mvar|z}}, represents the potential energy of the fluid due to its elevation with respect to a reference plane. Now, {{math|z}} is called the elevation head and given the designation {{math|''z''<sub>elevation</sub>}}.

A [[free fall]]ing mass from an elevation {{math|''z'' > 0}} (in a [[vacuum]]) will reach a [[speed]]
<math display="block">v = \sqrt{ {2 g}{z} },</math>
when arriving at elevation {{math|1=''z'' = 0}}. Or when rearranged as ''head'':
<math display="block">h_v =\frac{v^2}{2 g}</math>
The term {{math|{{sfrac|''v''<sup>2</sup>|2''g''}}}} is called the ''velocity [[Hydraulic head|head]]'', expressed as a length measurement. It represents the internal energy of the fluid due to its motion.

The [[hydrostatic pressure]] ''p'' is defined as
<math display="block">p = p_0 - \rho g z ,</math>
with {{math|''p''<sub>0</sub>}} some reference pressure, or when rearranged as ''head'':
<math display="block">\psi = \frac{p}{\rho g}.</math>
The term {{math|{{sfrac|''p''|''ρg''}}}} is also called the ''[[pressure head]]'', expressed as a length measurement. It represents the internal energy of the fluid due to the pressure exerted on the container. The head due to the flow speed and the head due to static pressure combined with the elevation above a reference plane, a simple relationship useful for incompressible fluids using the velocity head, elevation head, and pressure head is obtained.
{{NumBlk||<math display="block">h_{v} + z_\text{elevation} + \psi = C</math>|{{EquationRef|Eqn. 2b}}}}
If Eqn. 1 is multiplied by the density of the fluid, an equation with three pressure terms is obtained:
{{NumBlk||<math display="block">\frac{\rho v^2}{2}+ \rho g z + p = C</math>|{{EquationRef|Eqn. 3}}}}
Note that the pressure of the system is constant in this form of the Bernoulli equation. If the static pressure of the system (the third term) increases, and if the pressure due to elevation (the middle term) is constant, then the dynamic pressure (the first term) must have decreased. In other words, if the speed of a fluid decreases and it is not due to an elevation difference, it must be due to an increase in the static pressure that is resisting the flow.

All three equations are merely simplified versions of an energy balance on a system.
}}

{{math proof
| title = Bernoulli equation for compressible fluids
| proof =
The derivation for compressible fluids is similar. Again, the derivation depends upon (1) conservation of mass, and (2) conservation of energy. Conservation of mass implies that in the above figure, in the interval of time {{math|Δ''t''}}, the amount of mass passing through the boundary defined by the area {{math|''A''<sub>1</sub>}} is equal to the amount of mass passing outwards through the boundary defined by the area {{math|''A''<sub>2</sub>}}:
<math display="block">0= \Delta M_1 - \Delta M_2 = \rho_1 A_1 v_1 \, \Delta t - \rho_2 A_2 v_2 \, \Delta t.</math>
Conservation of energy is applied in a similar manner: It is assumed that the change in energy of the volume
of the streamtube bounded by {{math|''A''<sub>1</sub>}} and {{math|''A''<sub>2</sub>}} is due entirely to energy entering or leaving through one or the other of these two boundaries. Clearly, in a more complicated situation such as a fluid flow coupled with radiation, such conditions are not met. Nevertheless, assuming this to be the case and assuming the flow is steady so that the net change in the energy is zero,
<math display="block">\Delta E_1 - \Delta E_2 = 0</math>
where {{math|Δ''E''<sub>1</sub>}} and {{math|Δ''E''<sub>2</sub>}} are the energy entering through {{math|''A''<sub>1</sub>}} and leaving through {{math|''A''<sub>2</sub>}}, respectively. The energy entering through {{math|''A''<sub>1</sub>}} is the sum of the kinetic energy entering, the energy entering in the form of potential gravitational energy of the fluid, the fluid thermodynamic internal energy per unit of mass ({{math|''ε''<sub>1</sub>}}) entering, and the energy entering in the form of mechanical {{math|''p'' d''V''}} work:
<math display="block">\Delta E_1 = \left(\tfrac12 \rho_1 v_1^2 + \Psi_1 \rho_1 + \varepsilon_1 \rho_1 + p_1 \right) A_1 v_1 \, \Delta t</math>
where {{math|1=''Ψ'' = ''gz''}} is a [[conservative force|force potential]] due to the [[Earth's gravity]], {{mvar|g}} is acceleration due to gravity, and {{mvar|z}} is elevation above a reference plane. A similar expression for {{math|Δ''E''<sub>2</sub>}} may easily be constructed.
So now setting {{math|1=0 = Δ''E''<sub>1</sub> − Δ''E''<sub>2</sub>}}:
<math display="block">0 = \left(\tfrac12 \rho_1 v_1^2+ \Psi_1 \rho_1 + \varepsilon_1 \rho_1 + p_1 \right) A_1 v_1 \, \Delta t - \left(\tfrac12 \rho_2 v_2^2 + \Psi_2 \rho_2 + \varepsilon_2 \rho_2 + p_2 \right) A_2 v_2 \, \Delta t</math>
which can be rewritten as:
<math display="block"> 0 = \left(\tfrac12 v_1^2 + \Psi_1 + \varepsilon_1 + \frac{p_1}{\rho_1} \right) \rho_1 A_1 v_1 \, \Delta t - \left(\tfrac12 v_2^2 + \Psi_2 + \varepsilon_2 + \frac{p_2}{\rho_2} \right) \rho_2 A_2 v_2 \, \Delta t </math>
Now, using the previously-obtained result from conservation of mass, this may be simplified to obtain
<math display="block"> \tfrac12 v^2 + \Psi + \varepsilon + \frac{p}{\rho} = \text{constant} \equiv b </math>
which is the Bernoulli equation for compressible flow.

An equivalent expression can be written in terms of fluid enthalpy ({{mvar|h}}):
<math display="block"> \tfrac{1}{2} v^2 + \Psi + h = \text{constant} \equiv b </math>
}}