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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)
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
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