Mass Flow Rate:

d mcvdt

=Σ mi−Σ me – time rate of change of mass within a control volume.

m=ρAV= AVv ---- (AV) represents the volumetric flow rate (m3/s)

If it is an ideal gas:

pv = RT v=RTp

m= AVv m=(AV ) p

RT

Steady state for a control volume:

Σ mi=Σ me

Steady State - Circumstance in which there is no accumulation of mass or energy within a control volume. It is independent of time.

Energy Rate Balance:

Derivation:

One dimensional flow-control volume Energy rate balance:

W – is the work associated with fluid pressure – what was used in chapter 3 for internal specific energy

Wcv – Accounts for all other work effects – mass flow rate for a control volume.

Enthalpy:

Simplified Energy Rate Balance:

Set this equal to zero. Read the problem statement careful to determine if Q, KE, or PE are negligible.

If the problem reads a turbine or compressor operate adiabatically, Q = 0

Components in a System:

Nozzles and Diffusers:

There is no heat transfer that can be done within a nozzle nor a diffuser. They also do not create power nor require power to operate, thus W=0 The only terms in the energy balance for either a nozzle or a diffuser should be enthalpy and

velocity.

Turbines:

Power is always developed in a Turbine, thus will always have a positive W. The mass in a turbine equals the mass out. Most problems will ask to solve for the Power, or you will need to use the power to solve for the

mass flow rate and then go from there in the problem.

Compressors and Pumps:

Compressors and Pumps both require power to operate, thus Work will be done onto the system – Work and Power will be negative.

Compressor is used when the substance is a gas (vapor) Pump is used for when the substance is liquid

Heat Exchanger:

Heat exchanger does not require power nor produce power, thus W = 0 Sum the mass and enthalpies in and out for the heat exchanger as shown in the equation below:

Valves/ Nozzles:

No power is developed or necessary to operate, heat transfer is negligible. Enthalpy going into a nozzle equals enthalpy at its exit:

Transient Operation:

The state will change with time until it reaches steady state. Ex: the startup or shutdown of turbines, compressors, and motors.

Second law of Thermodynamics:

Clausius Statemento It is impossible for any system to operate in such a way that the sole result

would be an energy transfer by heat from a cooler body to a hotter body.

o This could only be possible with an input of work into the system (Heat pumps/Refrigeration cycles from chapter 2

Kelvin -Planck Statemento It is impossible for any system to operate in a thermodynamic cycle and deliver

a net amount of energy by work to its surroundings while receiving energy by heat transfer from a single reservoir.

o There must be some Qout for a system (exhaust).

The second law states:

It is impossible for any system to operate in a way that entropy is conserved.o Entropy can be zero or positive, but never negative.

Entropy

Entropy is produced, unlike mass and energy which is conserved.o It is produced whenever nonidealities or irreversibilities are present, such as friction.

Measured in kJkg∗K

∨ BTUlb∗R

Irreversible Process

True if the system and all parts of its surroundings cannot be exactly restored to their respective initial states after the process has occurred.

Will include one or more of the following:o Heat transfer through a finite temperature differenceo Unrestrained expansion of a gas or liquid to a lower pressureo Spontaneous chemical reactiono Friction – sliding or in the flow of fluidso Electric current through a resistance o Inelastic deformation

Reversible Process - if both the system and its surroundings can return to their initial states then a process is said to be reversible.

Cycles:

Power Cycle:

Heat Pumps and Refrigeration Cycles

CoP for a refrigeration cycle:

CoP for a Heat Pump:

To find if a system is able to work, theoretically:

Maximum CoP or Efficiency:

----- Must be absolute Temperature that is used (K or R)

If it asks for minimum work required to operate, use maximum CoP or Efficiency.