The energy equation can be used to define the Bernoulli equation. This is due to the fact that the Bernoulli equation is a form of the energy equation.

**Energy Equation: No Work across the Control Surface**

Let’s first take a look at the energy equation with the following conditions. First, the flow will be one-dimensional. In addition, the flow will be steady. Third, fluid will be incompressible. Finally, work will be zero across the control surface of the control volume. As a result,

**(Eq 1) ** $\dot{m}\left[ \check{u}_{out}-\check{u}_{in}+\frac{p_{out}}{ρ}-\frac{p_{in}}{ρ}+ \frac{v^2_{out}-v^2_{in}}{2} +g(z_{out}-z_{in}) \right]=\dot{Q}_{net~in}$

$\check{u}$ = kinetic energy per unit mass

$p$ = pressure

$ρ$ = density

$v$ = velocity

$g$ = gravitational acceleration

$z$ = fluid height

$\dot{Q}$ = Heat Transfer across the control surface

$\dot{m}$ = mass flow rate

Next, we will dived equation 1 by the mass flow rate and rearranged it in the following order.

**(Eq 2) ** $\frac{p_{out}}{ρ}+\frac{v^2_{out}}{2}+gz_{out} = \frac{p_{in}}{ρ} + \frac{v^2_{in}}{2}+gz_{in} – (\check{u}_{ou}-\check{u}_{in}-q_{net~in})$

$q_{net~in} = \frac{\dot{Q}_{net~in}}{\dot{m}}$ = heat transfer per mass flow rate

Now, recall that Bernoulli equation is define by the following.

**(Eq 3) ** $p_{out} + \frac{ρv^2_{out}}{2} + γz_{out} = p_{in} + \frac{ρv^2_{in}}{2} + γz_{in}$

In the above equation the specific weight $γ = ρg$. Hence, if we were to divide Bernoulli equation by the density “$ρ$” equation 3 will become the following.

**(Eq 4)** $\frac{p_{out}}{ρ}+\frac{v^2_{out}}{2}+gz_{out} = \frac{p_{in}}{ρ} + \frac{v^2_{in}}{2}+gz_{in}$

Now if you were to compare equation 3 with equation 4 you should notice that they are basically the same equation where,

**(Eq 5) **$\check{u}_{out} – \check{u}_{in}-q_{net~in}=0$

for equation 2. However, for this to be the case the flow must frictionless. In reality flow is not frictionless. Due to this fact, by the second law of thermodynamics, the following would be true.

**(Eq 6) **$\check{u}_{out} – \check{u}_{in}-q_{net~in}>0$

As a result, this part of equation 2 will represent a loss. Hence equation 2 would become the following.

**(Eq 7) ** $\frac{p_{out}}{ρ}+\frac{v^2_{out}}{2}+gz_{out} = \frac{p_{in}}{ρ} + \frac{v^2_{in}}{2}+gz_{in} – loss$

**Energy Equation: With Work across the Control Surface**

In the previous section, an energy equation for a one-dimensional, incompressible, steady flow, with zero work being transferred over the control surface was related to Bernoulli equation. As a result we were able to define a head loss due to heat transfer. However, this is not the only form of energy that can cross the control surface. Work is also a form of energy that will cross the control surface. Hence we will also need to relate the Bernoulli equation to the energy equation when work is transferred over the control volume. For this example, I will say the work is be transferred through a shaft.

**(Eq 8) ** $\dot{m}\left[ \check{u}_{out}-\check{u}_{in}+\frac{p_{out}}{ρ}-\frac{p_{in}}{ρ}+ \frac{v^2_{out}-v^2_{in}}{2} +g(z_{out}-z_{in}) \right]=\dot{Q}_{net~in}+\dot{W}_{shaft~net~in}$

Next, we will dived equation 8 by the mass flow rate and rearranged it in the following order.

**(Eq 9)** $\frac{p_{out}}{ρ}+\frac{v^2_{out}}{2}+gz_{out} = \frac{p_{in}}{ρ} + \frac{v^2_{in}}{2}+gz_{in}+w_{shaft~net~in} – (\check{u}_{ou}-\check{u}_{in}-q_{net~in})$

$w_{shaft~net~in} = \frac{\dot{W}_{shaft~net~in}}{\dot{m}}$

or

**(Eq 10)** $\frac{p_{out}}{ρ}+\frac{v^2_{out}}{2}+gz_{out} = \frac{p_{in}}{ρ} + \frac{v^2_{in}}{2}+gz_{in}+w_{shaft~net~in} – loss$

Equation 10 represents a form of the energy equation that is used for a steady-in-the-mean flow for an incompressible fluid problem. Generally, it is called the mechanical energy equation or the extended Bernoulli equation. This equation is showing that as shaft work goes into the control volume a certain amount of loss will occur. As a result, in order to get the same head height that would have occurred without this loss, additional work will have to be put into the system.

Next, let’s multiply equation 10 by the fluid density “ρ”.

**(Eq 11)** $p_{out} + \frac{ρv^2_{out}}{2}+γz_{out} = p_{in}+\frac{ρv^2_{in}}{2}+γz_{in}+ρw_{shaft~net~in}-ρ(loss)$

Equation 11 will give us the energy per unit volume. In addition to multiplying equation 10 by the fluid density, we can also divide it by the gravitational constant.

**(Eq 12) **$\frac{p_{out}}{γ}+\frac{v^2_{out}}{2g}+z_{out} = \frac{p_{in}}{γ}+\frac{v^2_{in}}{2g}+z_{in}+h_s-h_L$

$h_s$ = shaft work head

$h_L$ = head loss

Equation 12 represents the energy per unit weight. By transforming the energy equation into this equation the notion of “head” can be defined, which in turn will allow us to properly define a head loss as well as a total head. A total head is the total height that a fluid can reach. If there is no head loss or shaft work present, than the total head can be found using the following equation.

**(Eq 13)** $H = \frac{p}{γ}+\frac{v^2}{2g}+z$

However, if a shaft work head and a head loss is applied, than the total head will become the following.

**(Eq 14) **$H_{out} = H_{in} + h_s-h_L$

In turn, equation 14 is another way to express equation 12.