Lect22-23 - Lecture notes 22-23 PDF

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Isentropic Processes For an isentropic process s1= s2 P  s 2 − s1 = s 2o − s1o − R ln 2  = 0  P1  P  s 2o − s1o = ln 2  R  P1   s 2o − s1o  exp( s 2o / R ) P2  = = exp o P1 R exp( s   1 / R) Define relative pressure Pr ≡ exp(s o / R ) Pr 2  P2    = P  1  s = const Pr ...

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Isentropic Processes For an isentropic process s1= s2

P s 2 − s1 = s 2o − s1o − R ln 2  P1 P s2o − s1o = ln 2 R  P1

  = 0 

  

 s 2o − s1o  exp(s 2o / R ) P2  = = exp o P1 R   exp(s1 / R ) Define relative pressure Pr ≡ exp(s o / R ) Pr 2  P2  =   P  1  s =const Pr 1

For ideal gas

v2 T2  P1  T2  Pr1  (T2 / Pr 2 ) =  =  = v1 T1  P2  T1  Pr 2  (T1 / Pr1 )

Define vr ≡ T / Pr , so

v v2  = r2    v1  s = const vr 1

Values of Pr and Tr as a function of temperature for air are tabulated in Table A-22 144

Isentropic Process for ideal gas with constant cv and cP

T  v s 2 − s1 = cV ln 2  + R ln 2  T1   v1 note cV = R /(k − 1)  1 T 0 = R ln 2  k −1  T1 T 0 = ln 2  T1

  

1 k −1

  = 0 

  v   + ln 2    v1 

v  + ln 2   v1 

1   k −1  v    T 0 = ln  2  ⋅  2    T1   v1     Take exponential of both sides

T  exp(0) = 1 =  2   T1 

1 k −1

v  ⋅  2   v1 

 T2  v    =  1   T1 sc==const  v2  const

k −1

p

v  Pv T Pv but 2 = 2 2 substituting 2 2 =  1  T1 P1v1 P1v1  v 2 

k −1

145

this yields

 P2  v    =  1   P1  sc==const  v2  const

k

or P1v1k = P2 v 2k

p

Recall, for a polytropic compression or expansion process Pvn= const, for the special case of an isentropic process (adiabatic and reversible) Æ n= k Combining the two equations yields  v1   v2  T2   T1

1 k

  P2  T   =   =  2    P1   T1   P   =  2  cs==const  P1  const

1 k −1

k −1 k

p

146

Control volume entropy rate balance

Similar approach to that used to derive conservation of energy m& 2 &1 m

CONTROL VOLUME

&3 m

Q& j dS CV =∑ +∑m & i si − ∑ m & e se + S& gen j T i e dt j Rate of Entropy change

Rate of entropy transfer

Rate of Entropy production

If temperature in CV is not uniform Tj corresponds to the temperature at different points on the control surface where heat is transferred For steady-state, one inlet and one outlet, isothermal CV S&gen 1  Q& CV  0=   + s in − s out + m&  T  m& 147

Isentropic efficiencies of Turbines and Compressors

Recall, for a turbine First Law (steady-state, neglecting KE and PE effects and heat losses) yields P1

Expansion (P2 < P1)

W

T

W&CV = h1 − h2 > 0 P2 m& S& gen An entropy balance yields s 2 − s1 = ≥0 m&

(Wout)

For an actual turbine, irreversibilities are present, so accessible states are such that s2 > s1 1 T P1

2 P2

2s

s

The state labeled 2s on the T-s and h-s diagrams would be attained only in the limit of no irreversibilities, i.e., internally reversible expansion ( S& gen = 0 ) and thus s2 = s1 148

The maximum theoretical amount of turbine work output is obtained for an isentropic expansion W& CV = h1 − h2s m&

Since h1 - h2 < h1 - h2s the actual work produced is less than the ideal isentropic turbine produces The difference is gauged by the isentropic turbine efficiency defined by W&CV / m& h −h = 1 2 ηt = (W& CV / m& )s h1 − h2 s Note, ηt < 1

149

Recall, for a compressor First Law (steady-state, neglecting KE and PE effects and heat losses) yields

P1

C

Compression (P2 > P1) W& CV = h1 − h2 < 0 (Win) m&

W

P2

An entropy balance yields

s2 − s1 =

S& gen m&

≥0

For an actual compressor irreversibilities are always present so s2 > s1

T 2

P2

2s P1 1 s

The state labeled 2s on the T-s and h-s diagrams would be attained only in the limit of no irreversibilities, i.e., internally reversible compression where S& gen = 0 and thus s2 = s1 150

The minimum theoretical amount of compressor work required corresponds to isentropic compression W& CV = − ( h2s − h1 ) m&

Since h2 – h1 > h2s – h1 the actual work input is more than the ideal isentropic compressor requires The difference is gauged by the isentropic compressor efficiency defined by

ηc =

(W&CV / m& )s W& CV / m&

=

h1 − h2s h1 − h2

Note, ηc < 1

151

Internally Reversible Steady-State Flow Work

For a single inlet and exit (1-inlet, 2-exit) CV at steadystate neglecting KE and PE effects conservation of energy

 V12 − V22  W&CV Q& CV  + g ( z1 − z2 ) = + (h1 − h 2 ) +  2  m& m&  For an internally reversible process Q& / m& = ∫ Tds W& CV = ∫12 Tds + ( h1 − h2 ) m&

Recall:

Tds = dh − vdp → ∫12 Tds = (h2 − h1 ) − ∫12 vdp W& CV = ( h2 − h1 ) − ∫12 vdP + ( h1 − h2 ) m&

For pumps, turbines, compressors when ∆KE= ∆PE= 0  W&CV    = −∫12 vdP  m&  int rev

Pumps and compressors dP > 0 Æ work done on system Turbines dP < 0 Æ work done by system 152

Total Heat Transferred

Total Work

Liquids – liquid are incompressible, so v1=v2= v 2  W&CV    = − ∫ vdP = − v( P2 − P1 ) &  int 1  m rev

Gases - when each unit of gas through the CV undergoes a polytropic process Pvn= const 1 2 dP  W&CV    = − ∫ vdP = − ( const ) n ∫12 1 int &  rev 1  m Pn

For the special case of an ideal gas where Pv = RT nRT1  T2  W&CV    =−  − 1 int &  rev n − 1  T1   m

n ≠ 1,

153

recall, for polytropic process

 W& CV    & int  m rev

T2  P2  =   T1  P1 

n −1 n

n −1    nRT1   P2  n 1 =−   −  n − 1   P1    

so

n≠ 1

Recall: If the process is internally reversible and adiabatic (isentropic) for constant cp and cv Æ Pvk= const Substitute n= k in above equations to get work per unit mass for isentropic process (implies k = const ≠ f (T ) )

For the case of n=1: P1v1 = P2v2 Æ T1=T2 (isothermal) ∫ vdP gives:

 W&CV    = − RT ln( P2 P1 ) & int  m rev

n=1

154...