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MP2303/AE2006: THERMODYNAMICS REVIEW OF FUNDAMENTALS
A. PROPERTIES OF PURE SUBSTANCES (a) Water/Steam or Refrigerants
Use Property Tables for the appropriate substance
To determine in which region the state is: Given T and P, using the saturation pressure table For the given P, if T < Tsat → subcooled liquid For the given P, if T > Tsat → superheated vapour If other properties are given, e.g, v and T, using the saturation temperature table, for given T
fvv ≤ → subcooled liquid v v vf ≤ ≤ g → saturated liquid-vapour mixture
gvv ≥ → superheated vapour Can use u, h or s in similar manner
(i) Subcooled Liquid - approximate using sat. liquid properties at
given T v ≈ vf(T), u ≈ uf(T), s ≈ sf(T) h ≈ hf(T) + vf(P-Psat) (second term can be neglected for low values of P)
(ii) Sat. Liquid-Vapour Mixture - given P, from sat. pressure table, T = Tsat - given T, from sat. temperature table, P = Psat Using sat. pressure or sat. temperature table (depending on whether P or T is given) - obtain gf vv ,
- given v, obtain dryness fraction, xv vv v
f
g f=
−
−
- given x, obtain v x v xf gv= − +( )1 - other properties may be similarly obtained,
e.g. fgfgf xhhxhhxh +=+−= )1(
(iii) Superheated Vapour - use Superheated Vapour Tables
given P, T determine v, u, h, s by interpolation
(b) Ideal Gas (e.g. Air) Pv RT= u u T= ( ) du C dTv=h h T= ( ) dh C dTp=
Assuming constant , Cv CP
u u C T Tv2 1 2 1− = −( ) h h C T TP2 1 2 1− = −( ) s s C T T R P PP2 1 2 1 2 1− = −ln( / ) ln( / ) or
)/ln()/ln( 121212 vvRTTCss v +=−
(c) Incompressible Substance (Liquids/Solids) constant=v
C;
C Cp v= = Assuming constant C u u C T T2 1 2 1− = −( )
)()( 121212 PPvTTChh f −+−=− )/ln( 1212 TTCss =−
B. CLOSED SYSTEM ANALYSIS
- Fixed Mass or Non-Flow System (a) First Law (Energy Balance):
Q W E U KE PE− = = + +∆ ∆ ∆ ∆ - For stationary systems: ∆ ∆KE PE, = 0
)( 12 uumUWQ −=∆=− - (in); Q = +ve Q = −ve (out);
W = +ve (out); W = −ve (in);
- Moving boundary work, ∫= PdVWb Need the process equation linking P and V, e.g. for polytropic process, [ ] constant2211 == nn vPvP
nvPvPPdv
mWb
−−
== ∫ 111222
1 for work done against a spring
( )12212
1 2vvPPPdv
mWb −⎟
⎠⎞
⎜⎝⎛ +
== ∫
- Other types of work, W , W may be present. shaft elec
(b) Second Law (Entropy Balance):
)( 12 ssmSSTQ
genb
−=∆=+∫δ
- ∫bTQδ
is entropy transfer by heat transfer; +ve
for heat in and –ve for heat out. - is the absolute temperature at which the
heat is transferred bT
- is the entropy generation due to irreversibilities; and is always +ve
genS
For “isolated” systems (e.g. system and surroundings), a special type of closed system:
0 ,0 == WQ 0=∆U
0)( ≥=∆+∆=∆ gensurrsys SSSS
C. STEADY STATE CONTROL VOLUME (SSCV) - a region in space, with mass flow across
boundary; also called Open or Flow System - deals mostly with rate processes - no changes inside the CV (steady state) - changes is between inlet state and exit state
Typical SSCVs: Turbine, Compressors, Pumps, Nozzles, Valves, Heat Exchangers (includes Boilers, Condensors and Evaporators)
SSCV with single inlet, single exit
(a) Mass flow rate: ( )
& & &m m mvi e= = =
AV
(b) First Law (Energy Rate Balance):
( ) ⎥⎦
⎤⎢⎣
⎡−+⎟⎟
⎠
⎞⎜⎜⎝
⎛−+−=− )(
22
22
ieie
ie zzghhmWQ VV&&&
- here is usually shaft work or electrical work, NOT moving boundary work ( ) because volume of SSCV does not change
W&
∫PdV
(c) Second Law (Entropy Rate Balance):
)( iegenb
ssmSTQ
−=+∫ &&&δ
Reversible steady flow work (with negligible KE, PE changes between inlet and outlet)
∫−= vdPmW && e.g. for a polytropic flow process, n
iinee vPvP =
)(1 iiee
e
ivPvP
nnvdP
mW
−−
−=−= ∫&
&
for an incompressible liquid flow (pump)
)( ie
e
iPPvvdP
mW
−−=−= ∫&
&
D. SPECIAL PROCESSES
(a) Isentropic Process
When a process is adiabatic ( ) and reversible ( ), then
0or =QQ &
0or =gengen SS &
or (constant entropy) 12 ss = ie ss =
For an Ideal Gas, the isentropic relation is
(i) assuming constant , , Cv CP vP CCk /=
1
1
2
2
1
1
2−
⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
kkk
TT
vv
PP
This is equivalent to a polytropic process with n = k.
(b) Isothermal Process
12 TT = or (constant temperature) ie TT =
For an Ideal Gas constant2211 === RTvPvP
; 12 uu = 12 hh =
1
22
1ln
vvRTPdv
mW
== ∫ ; i
ee
i PPRTvdP
mW ln−=−= ∫&
&
E. ANALYSIS OF CYCLES (a) 1st Law for cycles:
0=∆ cyU outnetoutin WQQ ,=−
(b) 2nd Law for cycles:
0=∆ cyS 0≤∫ TQδ
(Clausius Ineq.)
(c) Performance of Cycles: (i) Heat engines:
in
out
in
outin
in
outnet
QQQ
QW
−=−
== 1,η
H
Lrev T
T−=≤ 1ηη
(ii) Refrigerators:
LH
L
innet
Lref QQ
QW
QCOP
−==
,
LH
Lrefrevref TT
TCOPCOP−
=≤ ,
(iii) Heat Pumps:
LH
H
innet
Hhp QQ
QW
QCOP−
==,
LH
Hhprevhp TT
TCOPCOP−
=≤ ,