1. ChemE 260 Entropy Balances On Open and Closed Systems
Dr. William Baratuci
Senior Lecturer
Chemical Engineering Department
University of Washington
TCD 8: A & B CB 6: 10 + Supplement
May 10, 2005

2. Entropy Balance: Closed System
1st Law:
2nd Law, Internally Reversible Processes:
Boundary Work, Internally Reversible Processes:
(Usually assume Wtot = Wb.)
Gibbs 1st Equation:
Entropy Balance Equation
Integral Form:
Differential Form:
Rate Form:
We start with closed systems, just to be thorough. The interesting application of the 2nd Law is for open systems.
This is a quick run down of all the equations that apply to closed systems.
The new part is the entropy balance equation.
The integral form is the one we will use most for closed systems.
It is essentially the definition of entropy generation.
It is not really new.
The rate form is a nice lead in to entropy balances on open systems.
Baratuci
ChemE 260
May 10, 2005

3. Entropy Balance: Open System
General:
Steady-state, SISO:
Entropy generation within the system:
The general form of the entropy balance equation states that the entropy of the system changes for four different reasons
Mass entering and leaving the system carries entropy with it.
The the rate at which the entropy of the system changes as a result is equal to the rate at which mass flow carries entropy into the system minus the rate at which mass flow carries entropy out of the system
Heat transfer to and from the system results in a change in the entropy of the system
Each place in the system where heat transfer occurs changes the entropy of the system at a rate of Qdot / T, where T is the temperature of the system at the particular location where heat transfer occurs.
The net rate at which heat transfer changes the entropy of the system is the sum of all the individual Qdot / T terms. For example: QH / Tsys,H – QC / Tsys,C for a heat engine with an evaporator operating at Tsys,H and a condenser operating at Tsys,C .
Note: These T’s are the temperatures within the system and NOT the temperatures of the reservoirs !
The last term is the rate of entropy generation within the system.
The general entropy balance equation can be simplified a great deal for a SISO system operating at steady-state.
Often, our objective is to determine the rate of entropy generation in a system.
You will see why this is a key quantity later in this chapter.
It is very important to realize that this is ONLY the entropy generation within the system.
What about the reservoirs and the surroundings ??
Baratuci
ChemE 260
May 10, 2005

4. The 1st Law and Entropy From the definition of entropy:
1st Law, Steady-state, Internally Reversible, SISO:
From the definition of entropy:
Gibbs 2nd Equation:
When we apply the 1st Law to a SISO, internally reversible, steady-state process, the work term includes all forms of work OTHER THAN boundary work.
In this course, that usually this means shaft work.
Our goal here is to derive an equation that will let us determine the shaft work for the process, so we begin by solving for Wnot b.
We divide by m-dot to get the specific work on the left-hand side of the equation.
The heat transfer is the key that connects the 1st Law to the 2nd Law.
We can manipulate the definition of entropy and use it to eliminate Q from the 1st Law.
Then, we can use Gibbs 2nd Equation to express T dS as dH – V dP.
This helps us get S out of the equation. This is a good thing because is notoriously difficult to measure.
On the next slide, we put all of these equations together to eliminate Q and then S.
Integrating for an open system:
Baratuci
ChemE 260
May 10, 2005

5. Mechanical Energy Balance Equation
Combine all the equations from the previous slide:
The MEBE:
Bernoulli Equation
If Wnot b = 0 :
Bernoulli Equation
(incompressible fluid)
The first cool thing that happens is that H drops out.
This is good because H is also a bit difficult to measure.
The result is the Mechanical Energy Balance Equation or MEBE.
This is a very useful equation in fluid mechanics
But it does not take friction into account because we started this analysis by assuming that the process was internally reversible.
Another famous equation is the Bernoulli Equation.
It is just the MEBE with no work at all.
Bernoulli is especially useful for analyzing flow in pipes.
If the fluid flowing through the pipes is an incompressible liquid, then Bernoulli can be simplified even further.
In this case it is customary to use  = 1 / Vhat.
The result that we will use most frequently in this course applies when changes in kinetic and potential energies are negligible.
This is a surprising and all too often confusing result.
Look at it VERY carefully.
It says that the specific non-boundary (usually SHAFT) work is equal to the integral of Vhat dP from the inlet to the outlet pressure.
This is CONFUSING sometimes because we found a similar looking result earlier in this course:
Specific BOUNDARY work is the integral of P dVhat
This is NOT a typo !
What does this result look like on a PV Diagram ?
Usually, Wnot b = Wsh
If Ekin = Epot = 0 :
Baratuci
ChemE 260
May 10, 2005

6. Shaft Work & PV Diagrams
Polytropic Processes
or:
 1 :
On a PV diagram, shaft work is equal to the area to the LEFT of the process path FOR
Steady-state
Internally reversible
SISO
Shaft work and flow work only
Ekin = Epot = 0
For a known process path, such as a polytropic process path, we can evaluate the shaft work by direct integration.
Results are shown here for  = 1 and  1
Keep in mind that  = 1 does not necessarily imply an isothermal process here !
The results shown here apply when the assumptions listed above are valid: SISO, SS, Int. Rev, Ekin = Epot = 0 & polytropic
The equations apply for all fluids
Liquids, real gases, and ideal gases
If the fluid is an ideal gas and  = 1, then the process is isothermal.
 = 1 :
Baratuci
ChemE 260
May 10, 2005

7. PV Diagram: Polytropic
Assumptions
Steady-state
Internally reversible
SISO
Shaft work and flow work only
Ekin = Epot = 0
This diagram shows how the value of  effects the shape of the polytropic process path on a PV Diagram.
As  increases, the process path becomes more steep and the specific volume changes by a smaller and smaller amount for the same change in pressure.
Observations
I have shown  = 1 following the red isotherm.
This is ONLY the case for an ideal gas with constant heat capacities.
The path for  =  is only isentropic if the fluid is an ideal gas and its heat capacities are constant.
The isochoric process requires the most shaft work to raise the pressure from P1 to P2.
With all of the assumptions we have made, no shaft work can be done in an isobaric process.
Keep in mind that boundary work and flow work can still be done.
In the next slide, I present equations for specific Ws for various polytropic processes for real fluids and ideal gases.
Baratuci
ChemE 260
May 10, 2005

8. Summary: Wsh Polytropic Processes
Process Type
Real Fluids
Ideal Gases
= 0
 = 1
Isothermal
  1: Polytropic
 =  = constant
Isentropic
 = : Isochoric
Here are all of the most useful equations for calculating shaft work for polytropic processes.
Once again, the assumptions we made are:
Steady-state
Internally reversible
SISO
Shaft work and flow work only
Ekin = Epot = 0
The ideal gas column adds the assumption that the fluid behaves as an ideal gas.
In order for  =  to mean that an IG is undergoing an isentropic process, the heat capacities ( and therefore ) must be constant.
Baratuci
ChemE 260
May 10, 2005
Wsh is actially all work other than flow work.

9. Next Class … Problem Session After that… Isentropic Efficiency
Define an efficiency for a process by comparing actual performance to the performance of an isentropic process
Nozzles, compressors and turbines
New diagram: HS Diagram
Multi-Stage Compressors
Intercooler HEX’s reduce work input requirement
Baratuci
ChemE 260
May 10, 2005

10. Example #1
Air is compressed from 1 bar and 310 K to 8 bar. Calculate the specific work and heat transfer if the air follows a polytropic process path with d = Assume air is an ideal gas in this process.
Baratuci
ChemE 260
May 10, 2005

11. Example #2
A turbine lets down steam from 5 MPa and 500oC to saturated vapor at 100 kPa while producing 720 kJ/kg of shaft work. The outer surface of the turbine is at an average temperature of 200oC. Determine the heat losses from the turbine and the entropy generation in the turbine in kJ/kg-K.
Baratuci
ChemE 260
May 10, 2005