Steady Flow Processes

Chief Noter

July 28, 2021
8 min read
1st Law Control Volumes Processes SFEE Thermo Thermodynamics

This notes sheet covers the fundamentals of thermodynamic steady flow processes: how to model flow through a control volume with the steady flow energy equation (SFEE), specific applications of steady flow, mass & volume flow rates and the Rankine cycle.

Modelling steady flow through a control volume
The Steady Flow Energy Equation (SFEE)
Applications of the SFEE
Mass & volume flow rates
The Rankine Cycle
Summary

The basic form of the first law, ΔQ – ΔW = ΔE, only applies to closed systems – no mass can transfer across the system boundary, only energy in the form of heat and work.

In reality, perfectly closed systems are quite rare (take a turbine, for example: air flows in as well as heat, shaft work and hot air flow out), and as such a different model is required:

COntrol volume analysis, steady flow processes, thermodynamics, steady flow through a control volume

The energy transfers across the control volume surface are:

Shaft work
Heat transfer
Energy in the working fluid (kinetic, potential, and internal energies)

To simplify the process, we only look at the energy inputs and outputs: what goes on inside the control volume is irrelevant.

We call where the working fluid enters and exits ports.

The control volume above has two ports: one inflow and one outflow port.

The Steady Flow Energy Equation (SFEE)

In order to solve problems involving steady flow through a control volume, we use the Steady Flow Energy Equation (SFEE) instead of the simple first law equation:

$\dot Q - \dot W_sh = \Sigma _{out} [\dot m (h+\frac{c^2}{2}+gZ)] - \Sigma_{in} [\dot m (h+\frac{c^2}{2}+gZ)]$

In a less mathematically intimidating form:

Left hand side: rate of energy transfer – rate of shaft work
Right hand side: sum of output mass energy flow rates – sum of input mass energy flow rates

Note: The dot above the letters means it is a rate of change with respect to time:

$\dot Q = \frac{dQ}{dt} \quad \dot W_{sh} = \frac{dW_{sh}}{dt} \quad \dot m = \frac{dm}{dt}$

Thankfully, kinetic and potential energies can often be ignored, simplifying the equation immensely. However, it is important you start with the above form when solving problems, else you will likely forget parts.

See the derivation of the SFEE here

Mass Continuity Equation

We can use the SFEE in conjunction with the principle of conservation of mass. This means that the mass in the control volume must stay constant, so the total input mass flow rate = total output mass flow rate:

$\Sigma_{in} \dot m = \Sigma_{out} \dot m$

Therefore, for a two-port control volume where potential energies are negligible, the SFEE reduces to:

$\dot Q - \dot W_{sh} = \dot m(h_2 - h_1 + \frac{1}{2}(c_2^2-c_1^2))$

When kinetic energy can also be ignored, this becomes:

$\dot Q - \dot W_{sh} = \dot m (h_2-h_1)$

The point is, the SFEE in its full form may look difficult, but it really isn’t. Just be neat and methodical in your workings, and always start from the full form and then see which terms you can set to zero.

Applications of the SFEE

There are a number of very common examples where the SFEE is applied and simplified. These are worth knowing, as crop up in almost all problems involving steady flow.

This is also where some actual real-life engineering gets involved. Which is fun.

Each example has a standard block diagram symbol. These should be learned as they are used to describe multi-stage processes like the Rankine Cycle (see below).

Partly Closed Valve (Throttling)

Partly closed valve, thermodynamic throttling, throttling, steady flow processes through a control volume, SFEE, steady flow energy equation

This is the simplest form of the SFEE: every term in the SFEE other than the enthalpies are negligible:

Heat transfer is negligible when the valve is well insulated
There is no shaft work
Kinetic energy change is negligible
Potential energy change is negligible

There are two ports, so the SFEE reduces to:

$0 = h_2 - h_1$

From this you can see that throttling (the process that takes place in a partly-closed valve) is isenthalpic.

Nozzles

A simple nozzle can be modelled as a duct of reducing cross-sectional area, used to increase the kinetic energy of the working fluid at the expense of its enthalpy.

There is no heat transfer to or from the working fluid
There is no shaft work
There is a massive change in kinetic energy
Potential energy is negligible

Again, there are two ports, so the SFEE becomes:

$0 = h_2 - h_1 +\frac{1}{2} (c_2^2 - c_1^2)$

Boilers

Heat is transferred to the working fluid to the point of saturation, and the fluid evaporates.

Boilers are isobaric, meaning there is no work done.

Kinetic energy is negligible
Potential energy is negligible

There are two ports, so the SFEE reduces to:

$\dot Q = \dot m (h_2-h_1)$

Pumps & Compressors

pump block diagram, compressor block diagram, steady flow through a pump, steady flow through a compressor, SFEE, steady flow energy equation

Pumps and compressors are exactly the same thing: the only difference is that a pump refers to a liquid and a compressor refers to a gas (as liquids are generally incompressible).

There is typically no heat transfer – the process is adiabatic
There is a pressure change, so there is shaft work
Kinetic energy and potential energy can sometimes be ignored, but not always

When KE and PE are negligible, the SFEE reduces to:

$- \dot W = \dot m (h_2-h_1)$

If KE and PE are not negligible (e.g. pumping water up a hill), the SFEE becomes:

$- \dot W = \dot m v(p_2-p_1)$

Where $v$ is the specific volume. In this instance, specific work is given by $v \Delta P$ , not $P \Delta v$ .

Turbines

The opposite of a pump or compressor, the working fluid expands in a turbine to provide shaft work.

Heat transfer is generally negligible
There is a lot of output shaft work (h₁ > h₂, so W > 0)
Kinetic energy is sometimes negligible, but often not
Potential energy can usually be ignored

Two ports, so the SFEE becomes:

$- \dot W = \dot m (h_2-h_1 + \frac{1}{2} (_2^2-c_1^2))$

Heat Exchanger

A heat exchanger (as its name suggests) transfers heat from one working fluid to another. This means that if we make the control volume the whole exchanger with four ports, there is no net heat transfer.

Obviously, there is also no shaft work or change in kinetic and potential energy. Therefore:

$0 = \dot m_1 h_1 - \dot m_2 h_2 + \dot m_3 h_3 - \dot m_4 h_4$

Using mass continuity equations for ports 1 & 2 and 3 & 4, we can simplify the SFEE even further to:

$0 = \dot m_1 (h_1-h_2) + \dot m_3 (h_3-h_4)$

Heat addition block diagram, heater block diagram, steady flow through a heater, steady flow processes through a control volume, SFEE, steady flow energy equation

However, if we make the control volume just one of the two fluids with a positive or negative heat transfer from the other fluid, there are two ports and there is a heat transfer:

$\dot Q_34 = \dot m_3 (h_4-h_3)$

Where the heat transfer comes from the hot stream:

$\dot Q_34 = \dot Q_12 = \dot m_1(h_2-h_1)$

Mixing Chamber

A typical mixing chamber has two inflow ports and one outflow port, though there could be many more inflow ports.

There is no net heat transfer between the control volume and its surroundings (there may be a heat transfer between the mixing fluids, but this is inside the boundaries so irrelevant)
There is no shaft work
KE and PE are both negligible

The only remaining factors are enthalpy and flow rates:

$0 = \dot m_3h_3 - \dot m_2h_2$

Using mass continuity, we know that:

$\dot m_1 + \dot m_2 = \dot m_3$

Mass & Volume Flow Rates

The mass flow rate can be calculated from the velocity and density of the working fluid, and the size of the duct/pipe it is flowing through:

$\dot m = \rho CA$

$\rho$ is the density of the working fluid
$C$ is the velocity
$A$ is the area of the duct/pipe

From this, we can calculate the volume flow rate:

$\dot V = \frac{\dot m}{\rho} = AC$

Combining these, and using specific volume gives us the most useful form:

$\dot m = \frac{CA}{v}$

The Rankine Cycle

First put forward by William Rankine in 1859, the Rankine Cycle models the performance of steam turbine systems:

rankine cycle, rankine cycle diagram, rankine cycle block diagram, SFEE, steady flow energy equation

Cold liquid is pressurised in the pump
Pressurised liquid is heated in the boiler (isobaric process) and vaporises
Vapour expands in the turbine as temperature and pressure reduce; dryness fraction decreases from 1 to inside the vapour-dome
Wet vapour condenses at constant pressure to become a saturated liquid

We cannot use the basic first law, $Q-W=\Delta E$ for control volumes, as this does not take into account the mass transfer across the CV boundary
Instead, we look at the energy at the inputs and outputs using the steady flow energy equation (SFEE):
- $\dot Q - \dot W_{sh} = \Sigma [\dot m (h+ \frac{c^2}{2}+gZ)]_{out} - \Sigma [\dot m (h+\frac{c^2}{2}+gZ)]_{in}$
We also use mass continuity: $\Sigma \dot m_{in} = \Sigma \dot m_{out}$
The mass flow rate can be calculated from density, velocity or volume and duct area:
- $\dot m = \rho CA$
- $\dot m = \frac{CA}{v}$
The Rankine Cycle models the performance of a turbine engine, and consists of four stages:
1. Cold liquid is pressurised in the pump
2. The liquid is vaporised in the boiler
3. The vapour expands in the turbine
4. The vapour condenses in the condenser

Steady Flow Processes

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