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

QQ

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−

=≤ ,