6.4 Conservation of Energy

Conservation of Energy in Fluid Mechanics

Energy conservation in fluid mechanics extends the first law of thermodynamics to flowing systems. It lets you track how kinetic, potential, and internal energy change as fluid moves through a control volume, and it accounts for any heat added or work extracted along the way. This is the foundation for analyzing turbines, pumps, heat exchangers, and piping systems.

Conservation of Energy Equation

The energy equation for a control volume comes directly from the first law of thermodynamics: energy can't be created or destroyed, only converted between forms.

General integral form:

$\frac{\partial}{\partial t} \int_{CV} e \rho \, dV + \int_{CS} e \rho \vec{V} \cdot d\vec{A} = \dot{Q} - \dot{W}_{s} + \int_{CS} \frac{P}{\rho} \rho \vec{V} \cdot d\vec{A}$

Here's what each piece represents:

• $e$ = total specific energy (kinetic + potential + internal, per unit mass)
• $\rho$ = fluid density
• $\vec{V}$ = velocity vector
• $\dot{Q}$ = rate of heat transfer into the control volume (e.g., heat added by combustion)
• $\dot{W}_{s}$ = rate of shaft work done by the system (e.g., work extracted by a turbine). The subscript $s$ distinguishes shaft work from flow work, which is already captured by the pressure integral on the right side.
• $P$ = pressure

The first term on the left is the rate of energy storage inside the control volume. The second term is the net energy flux leaving through the control surface. The right side accounts for heat transfer, shaft work, and flow work (the pressure term that pushes fluid across the control surface).

Simplified form for steady flow with one inlet (1) and one outlet (2):

$\frac{V_1^2}{2} + gz_1 + \frac{P_1}{\rho} + u_1 + \frac{\dot{Q}}{\dot{m}} = \frac{V_2^2}{2} + gz_2 + \frac{P_2}{\rho} + u_2 + \frac{\dot{W}_{s}}{\dot{m}}$

• $V$ = fluid velocity
• $g$ = gravitational acceleration
• $z$ = elevation above a chosen reference datum
• $u$ = internal (thermal) energy per unit mass
• $\dot{m}$ = mass flow rate

The $P/\rho$ terms are the flow work (sometimes called pressure energy). They represent the work done by pressure forces to push fluid into and out of the control volume. Combined with internal energy, they form enthalpy: $h = u + P/\rho$. You'll often see the steady-flow energy equation written in terms of enthalpy instead.

Forms of Energy

Three forms of energy appear in the control volume energy equation. Understanding each one helps you decide which terms matter in a given problem.

Kinetic energy is energy due to the fluid's bulk motion:

$KE = \frac{1}{2}mV^2 \quad \text{(or per unit mass: } \frac{V^2}{2}\text{)}$

This term becomes significant when velocity changes are large, such as flow accelerating through a nozzle or decelerating into a diffuser.

Potential energy is energy due to elevation in a gravitational field:

$PE = mgz \quad \text{(or per unit mass: } gz\text{)}$

This matters when there's a meaningful height difference between inlet and outlet. Think of water falling through a hydroelectric turbine or being pumped up to a rooftop tank.

Internal energy ($u$) is the microscopic energy associated with molecular motion and intermolecular forces. It depends on the fluid's temperature and thermodynamic state. For example, steam entering a turbine at high temperature carries significant internal energy. When friction or viscous effects convert mechanical energy into heat, that energy shows up as an increase in $u$.

Energy Transfer and Problem-Solving

Energy enters or leaves a control volume through two mechanisms beyond the fluid flow itself: work and heat transfer.

Work is energy transfer due to a force acting over a distance:

$W = \int F \, ds$

In fluid systems, shaft work is the most common type. Pumps do work on the fluid (adding energy), while turbines extract work from the fluid (removing energy). The sign convention matters: $\dot{W}_s$ is positive when the system does work on the surroundings (turbine), and $\dot{Q}$ is positive when heat flows into the system.

Power is the rate of energy transfer:

$\dot{W} = \frac{dW}{dt}$

For a pump or turbine with a known mass flow rate, the shaft power relates to the energy change per unit mass by $\dot{W}_s = \dot{m} \cdot w_s$.

Steps for solving energy equation problems:

1. Define the control volume. Draw boundaries and label all inlets and outlets.
2. List your assumptions. Is the flow steady? Incompressible? Is there one inlet and one outlet? Can you neglect elevation changes or heat transfer?
3. Choose the right form of the energy equation. For steady, single-inlet/single-outlet flow, use the simplified form above. For unsteady or multi-port problems, go back to the integral form.
4. Identify known and unknown quantities. Write down pressures, velocities, elevations, temperatures, and any given pump/turbine power.
5. Substitute and solve. Plug in known values, keeping units consistent, and solve for the unknown.

Mechanical Energy Conservation

Mechanical energy is the sum of kinetic and potential energy:

$ME = \frac{1}{2}mV^2 + mgz$

When a flow is steady, incompressible, and inviscid (no friction or viscosity), and there's no shaft work or heat transfer, mechanical energy is conserved along a streamline. This gives you the Bernoulli equation:

$\frac{V_1^2}{2} + gz_1 + \frac{P_1}{\rho} = \frac{V_2^2}{2} + gz_2 + \frac{P_2}{\rho}$

Bernoulli is really a special case of the energy equation with all the "messy" terms (internal energy changes, heat, work, friction) set to zero. It's powerful for quick analysis of things like flow through a converging nozzle or a Venturi meter, but it breaks down whenever viscous effects are significant.

What happens with friction? Dissipative forces like wall friction and viscous shearing convert mechanical energy into internal energy (heat). This shows up as a head loss term, often written $h_L$, which you add to the downstream side of the energy equation:

$\frac{V_1^2}{2} + gz_1 + \frac{P_1}{\rho} = \frac{V_2^2}{2} + gz_2 + \frac{P_2}{\rho} + h_L$

In a long pipe, for instance, wall friction causes a pressure drop along the length. That "lost" pressure energy hasn't disappeared; it's been converted to thermal energy in the fluid. The extended Bernoulli equation (with $h_L$ and optional pump/turbine work terms) bridges the gap between the idealized Bernoulli equation and the full energy equation.