OUTLINE OF CHAPTER 3 - ou.edu OF... · OUTLINE OF CHAPTER 3 Conservation of Energy Recall ... 12 1...

15
OUTLINE OF CHAPTER 3 Conservation of Energy Recall Θ = Energy(not specific energy) THUS (one phase, one componentÆ no generation) Surroundings System Mass in Mass out

Transcript of OUTLINE OF CHAPTER 3 - ou.edu OF... · OUTLINE OF CHAPTER 3 Conservation of Energy Recall ... 12 1...

OUTLINE OF CHAPTER 3

Conservation of Energy Recall

Θ = Energy(not specific energy)

THUS

(one phase, one component no generation)

Surroundings

System

Mass in

Mass out

Mechanisms for Energy to Enter/Leave With ingoing/outgoing fluid:

Note: Specific (per unit mass) internal energy is used. Heat Transfer (from various sources)

Q>0 if heat is added TO the system

Work (electrical)

sW >0 if work is added TO the system. Included in the shaft work term for convenience

F

Work of Fluid against pressure (open systems only) As fluid moves, it does work against the fluid that is ahead of it.

Work=F dL. But F=PA, and AdL=dV Work=P dV= ˆPV MΔ

THUS WE WRITE

But it could be any system

Specific MASS volume

Work

W >0 if work is added TO the system. In the case the system is behaving reversibly, then F=PA, where P is the pressure of the system, which is assumed uniform over A. Then

The negative value is added because W >0 if dVdt

<0

Energy Conservation (First Law of Thermodynamics)

Surroundings System

Mass in

Mass t

dA

Rate of change of Total Energy (Internal + External) of the system.

Energy added/subtracted associated to mass entering /leaving the system

Heat added

Shaft work

Expansion Work

Work done by fluid entering/leaving the system

Example: Turbine

Thus

1 1 1 2 2 2 1 1 1 2ˆ ˆ ˆ( ) ( )sW M PV M P V M V P P− = − = −

Work of a turbine/pump (no density changes in fluid)= Volumetric flowrate ×Pressure differential

Steady state

1 2

1 2

1 2

12

2 2

ˆ ˆ

0

M M

U U

v

M v neglegible

ψ ψ

=

==

=

Adiabatic

Turbine walls do not move (Volume remains constant)

ENTHALPY

where Molar Basis

= where H U PV= +

m: Mol. Weight

Commonly used forms:

DIFFERENTIAL FORMS

To obtain the equations in a molar basis:

KE and PE not important, no Shaft Work and only ONE Stream entering/leaving

and

1

Missing in book

1

1

CLOSED SYSTEM 1 0M =

or dU=δQ-PdV dH=δQ (for Constant P)

Very well know forms of FIRST LAW FOR CLOSED SYSTEMS

Equation for a Change of State 1 2 Integrate from t1 to t2

When fluid entering does not change in time

DIFFERENCE FORMS

To obtain the equations in a molar basis:

THERMODYNAMICS PROPERTIES OF MATTER IDEAL GAS

Real fluids need pressure or molar volume. But U and H are monotone with temperature (the higher the temperature, the higher is U or H). So add a known amount of heat and measure the change in temperature. For constant volume

Therefore

2 1{ ( ) ( )}VQC

N T t T t=

Known Measured

BUT dU=Q-PdV=Q (V is constant; Note that pressure will change) or Thus, one can correlate the change in temperature of a known amount of gas at constant volume to its change of internal energy (a zero needs to be picked) by making these measurements. Also, one obtains CV THEN…………….

A similar experiment, now at constant pressure

Then

Ideal Gases

But . Then . Then

* denotes functions of T only

We will use (in general)

Solids and Liquids PV U< .

Then

H U≈