5.4 Energy analysis of unsteady-flow processes

Energy Balance for Unsteady Flow

General Energy Balance Equation

The general energy balance for any system undergoing any process is:

$E_{in} - E_{out} = \Delta E_{system}$

• $E_{in}$ = total energy entering the system
• $E_{out}$ = total energy leaving the system
• $\Delta E_{system}$ = change in total energy of the system

In steady-flow analysis, properties at any point in the control volume stay constant over time, so $\Delta E_{system} = 0$. Unsteady-flow processes break that assumption. The fluid properties at a fixed point inside the control volume do change with time, which means mass and energy can accumulate or deplete within the control volume. That's the entire reason this topic gets its own treatment.

Unsteady-Flow Energy Balance Equation

The rate form of the energy balance for an unsteady-flow control volume is:

$\dot{Q}_{net,in} + \dot{W}_{net,in} + \dot{m}_{in}\left(h_{in} + \frac{1}{2}V_{in}^2 + gz_{in}\right) = \frac{dE_{CV}}{dt} + \dot{m}_{out}\left(h_{out} + \frac{1}{2}V_{out}^2 + gz_{out}\right)$

where:

• $\dot{Q}_{net,in}$ = net rate of heat transfer into the CV
• $\dot{W}_{net,in}$ = net rate of work input (excluding flow work, which is already wrapped into enthalpy)
• $\dot{m}_{in}, \dot{m}_{out}$ = mass flow rates at inlets and outlets
• $h$ = specific enthalpy
• $V$ = velocity
• $z$ = elevation
• $\frac{dE_{CV}}{dt}$ = rate of change of energy stored within the control volume

The stored energy term $\frac{dE_{CV}}{dt}$ includes internal, kinetic, and potential energy of the mass inside the CV:

$\frac{dE_{CV}}{dt} = \frac{d}{dt}\left(m_{CV} u_{CV}\right) + \frac{d}{dt}\left(\frac{m_{CV} V_{CV}^2}{2}\right) + \frac{d}{dt}\left(m_{CV} g z_{CV}\right)$

In most textbook problems, the kinetic and potential energy changes of the mass inside the CV are negligible, so this simplifies to $\frac{d}{dt}(m_{CV} u_{CV})$.

Integrated (Uniform-Flow) Form

Many unsteady-flow problems use the uniform-flow approximation: the state of the fluid at each inlet and outlet is assumed uniform and constant over the entire process duration, even though conditions inside the CV change. Under this assumption you can integrate the rate equation over the time interval from state 1 to state 2:

$Q_{net,in} + W_{net,in} + \sum_{in} m_i\left(h_i + \frac{V_i^2}{2} + gz_i\right) = \sum_{out} m_e\left(h_e + \frac{V_e^2}{2} + gz_e\right) + (m_2 u_2 - m_1 u_1)_{CV}$

• $m_i, m_e$ = total mass that entered or exited during the process
• $m_1, m_2$ = mass inside the CV at the initial and final states
• $u_1, u_2$ = specific internal energy of the CV contents at the initial and final states

This integrated form is the one you'll use most often for problems like filling a tank or discharging a vessel.

Mass Balance (Conservation of Mass)

Mass accumulation or depletion happens whenever the incoming and outgoing mass flow rates don't match. The rate form is:

$\dot{m}_{in} - \dot{m}_{out} = \frac{dm_{CV}}{dt}$

Integrated over the process duration:

$\sum m_i - \sum m_e = m_2 - m_1$

This equation is solved alongside the energy balance. You typically need both to close the problem.

Work in Unsteady Flow

Flow Work vs. Shaft Work

Work in unsteady-flow processes has two components:

1. Flow work ($Pv$ work): the work required to push mass into or out of the control volume. This is already accounted for when you use enthalpy ($h = u + Pv$) in the energy balance, so you don't add it separately.
2. Shaft work ($W_{shaft}$): work associated with rotating machinery like turbines, compressors, and pumps. This is what $W_{net,in}$ represents in the energy balance once flow work has been absorbed into the enthalpy terms.

A common mistake is to double-count flow work by including both $Pv$ terms and enthalpy in the same equation. If you're using $h$ for the flow streams, the flow work is already in there.

Boundary Work

For some unsteady-flow problems the CV boundary itself moves (e.g., a piston-cylinder device with an open valve). In that case, boundary work $\int P \, dV$ also appears in $W_{net,in}$. Be clear about whether your control volume has a fixed or moving boundary before writing the energy balance.

Energy Analysis of Unsteady Systems

Problem-Solving Steps

1. Sketch the system and define the control volume. Identify all inlets, outlets, and whether the CV boundary is fixed or moving.

2. Identify the initial and final states. List known properties ($P, T, v, u, h$) for the CV contents at states 1 and 2, and for each inlet/outlet stream.

3. Write the mass balance. Use $\sum m_i - \sum m_e = m_2 - m_1$ to relate the unknown masses.

4. Write the energy balance. Use the integrated (uniform-flow) form. Decide which terms are negligible:

   • Kinetic and potential energy changes are usually dropped unless the problem states high velocities or large elevation differences.
   • If the process is adiabatic, set $Q = 0$. If there's no shaft, set $W_{shaft} = 0$.

5. Look up or calculate properties. Use steam tables, ideal-gas relations, or other property data to find $h, u, v$ at each state.

6. Solve the system of equations. You typically have two equations (mass and energy) and two unknowns.

7. Check units and signs. Heat in and work in are positive on the left side of the balance as written above. Make sure your sign convention is consistent.

Common Unsteady-Flow Scenarios

• Filling a rigid tank (charging): No outlet, no shaft work, CV volume is constant. The energy balance simplifies significantly because $m_1$ may be zero (initially evacuated) and $W = 0$.
• Discharging a pressurized vessel: No inlet. Mass and energy inside the CV decrease over time.
• Filling or draining a piston-cylinder device: The boundary moves, so boundary work appears. The pressure inside may stay constant if the piston is loaded by a fixed weight.

That last point trips up a lot of students. The supply line fluid carries its flow work ($Pv$) with it, so you must use $h_i$, not $u_i$, for the incoming stream.