*Bernoulli’s equation* describes the relationship between pressure and velocity in fluids quantitatively. Hence, this equation is named after its discoverer, the Swiss scientist Daniel Bernoulli (1700–1782). **Bernoulli’s equation states that** for an incompressible, frictionless fluid, the following sum is constant: **P+(1/2)ρv ^{2}+ρgh=constant**, where P

*is the absolute pressure,*

*ρ*is the fluid density,

*v*is the velocity of the fluid,

*h*is the height above some reference point, and

*g*is the acceleration due to gravity.

If we follow a small volume of fluid along its path, various quantities in the sum of the above equation may change, but the total remains constant.

Let the subscripts 1 and 2 refer to any two points along the path that the bit of fluid follows; **Bernoulli’s equation** becomes **P _{1}+(1/2)ρv_{1}^{2}+ρgh_{1}=P_{2}+(1/2)ρv_{2}^{2}+ρgh_{2}**

## Bernoulli’s equation & the conservation of energy principle

Bernoulli’s equation is a form of the conservation of energy principle. Note that the second and third terms are the kinetic and potential energy with *m *replaced by *ρ*. In fact, each term in the equation has units of energy per unit volume.

Here, (1/2)ρv^{2}^{ }is the kinetic energy per unit volume.

*And, **ρgh* is the gravitational potential energy per unit volume.

Note that pressure P has units of energy per unit volume, too. Since *P = F/A*, its units are N/m^{2}. If we multiply these by m/m, we obtain N ⋅ m/m^{3} = J/m^{3}, or energy per unit volume.

Bernoulli’s equation is, in fact, just a convenient statement of conservation of energy for an incompressible fluid in the absence of friction.

## Derivation of Bernoulli’s equation

We will apply the conservation of energy principle to derive *Bernoulli’s equation.*

*At any point along the path that the bit of fluid follows, the total energy can be obtained by adding the pressure energy, kinetic energy, and gravitational potential energy at that point.*

So at point 1 in the above diagram (figure 1), the total energy E_{1} = pressure energy at point 1 + kinetic energy at point 1 + gravitational potential energy at point 2.

Hence, E_{1} = P_{1} V + (½) m v_{1}^{2} + mgh_{1}

Similarly, at point 2, the total energy E_{2} = P_{2} V + (½) m v_{2}^{2} + mgh_{2}

**By applying conservation of energy principle:**

E_{1} = E_{2}

=> P_{1} V + (½) m v_{1}^{2} + mgh_{1} = P_{2} V + (½) m v_{2}^{2} + mgh_{2}

=> V [ P_{1 }+ (½) ρ v_{1}^{2} + ρgh_{1} ] = V [ P_{2 }+ (½) ρ v_{2}^{2} + ρgh_{2}]

**=> P_{1 }+ (½) ρ v_{1}^{2} + ρgh_{1}= P_{2 }+ (½) ρ v_{2}^{2} + ρgh_{2}**

=> **P+(1/2)ρv ^{2}+ρgh=constant**, where P is the absolute pressure, ρ is the fluid density, v is the velocity of the fluid, h is the height above some reference point, and g is the acceleration due to gravity.

**Bernoulli’s Equation for Static Fluids**

Let us first consider the very simple situation where the fluid is static—that is, *v*_{1} = *v*_{2} = 0. Bernoulli’s equation in that case is

*P*_{1} + *ρgh*_{1} = *P*_{2} + *ρgh*_{2}.

**Bernoulli’s Principle—Bernoulli’s Equation at Constant Depth**

Another important situation is one in which the fluid moves but its depth is constant—that is, *h*_{1} = *h*_{2}. Under that condition, Bernoulli’s equation becomes:

** P _{1}+(1/2)ρv_{1}^{2}=P_{2}+(1/2)ρv_{2}^{2}**

## Sample Numerical problem with solution

**Question**: The speed of water in a hose increased from 1.96 m/s to 25.5 m/s going from the hose to the nozzle. Calculate the pressure in the hose, given that the absolute pressure in the nozzle is 1.0 × 10^{5} N/m^{2} (atmospheric, as it must be) and assuming level, frictionless flow.

**How to approach?**

Level flow means constant depth, so Bernoulli’s principle applies. We use the subscript 1 for values in the hose and 2 for those in the nozzle. We are thus asked to find *P*1.

**Solution**

** P _{1}+(1/2)ρv_{1}^{2}=P_{2}+(1/2)ρv_{2}^{2}**

Solving Bernoulli’s principle for *P*1 yields

P_{1} = P_{2} + (1/2)ρv_{2}^{2}− (1/2)ρv_{1}^{2} = P_{2} + (1/2)ρ(v_{2}^{2}−v_{1}^{2})

Substituting known values,

P_{1}=71.01×10^{5} N/m^{2}+(1/2)(10^{3} kg/m^{3})[(25.5 m/s)^{2}−(1.96 m/s)^{2}]=4.24×10^{5} N/m^{2}

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