01 วิชาที่1THERMODYNAMICS- มิย50 D(HIGH-LIGHT)

8/14/2019 01 1THERMODYNAMICS- 50 D(HIGH-LIGHT)

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1 (Thermodynamics)1

(Turbine)(Engine) (Electric Generator) 50-60

2 (SYSTEM/CLOSE SYSTEM/CONTROL MASS)

(OPEN SYSTEM /CONTROL VOLUME)-

(SURROUNDINGS)

(PROPERTY OF SUBSTANCE) 2 (INTENSIVE PROPERTY) (P) (T) () (v) (e) (EXTENSIVE PROPERTY) (m) (V) (E) (STATE)

1 30C 100 kPa

.


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(PHASE) 3 (SOLID) (LIQUID)

(VAPOUR OR GAS)

(PROCESS) 1 T

1, P

1 2 T

2, P

2

(PROCESS) (CYCLE)

P1 , T1 P2, T2 P3 , T3..P1 , T1

(UNITS )(SI UNIT)

10 18=a (atto), 15=f (femto), 12=p (pico), 9=n (nano), 6= (micro), 3=m(milli) 2= c (centi), 1=d (deci), +1=da (deka), +2=h (hecto), +3=k (kilo), +6=M (mega), +9=G (giga),

+12=T (Tera), +15=P (peta), +18=E (exa)

(MASS) m kg (.) (1 = 0.4536 kg)

1 kg 1 kg (TIME) t s () (LENGTH ) l L Z m (.)

(1 = 0.3048 m) (FORCE) F N, ()

F = ma

kg m/s2 kg.m/s2 N() 1 N = 1 kg.m/s2 1 (lb

f) = 4.448 N

1 . (kgf) = 9.81 N

(SPECIFIC VOLUME, v & DENSITY , ) v m3/kg ( litre/kg , 1 litre = 0.001 m3) m3/kmol 1 kmol = M kg M=v

kg/m3

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T O

C K 1 (Triple Point) 0.01OC 1 100OC K

K =OC + 273.15

K

(1OK = 1.8

OF F = Cx1.8 +32F C ) (PRESSURE) P

P = F/A

F N m2 P N/ m2 Pa 1 Pa = 1 N/m2 bar 1 bar = 105 Pa

( 1 lbf/in

2= 1 psi = 6.895 kPa)

1 (atm) = 101,325 Pa = 101.325 kPa = 14.7 psi.

= - = +

0 () (Static Pressure)

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P = L g (1.1 )

P= Pa , L= m() ,=kg/m

3, g = m/s2 9.81 m/s2

1.1 0.8 2 m

= Lg = (0.8x103 kg/m3) (2m) (9.81 m/s2)= 15.7x10

3kg.m/s

2.m

2

= 15.7x103

N/m2

= 15.7x103

Pa ()

. = 15.7x103+101,325 Pa=117025 Pa ( =101,325 Pa)

.2 m

1.1

70-120 mm 10.33 760 .

.

P

VAP= X L1. g + PATM

P

3 ( PURE SUBSTANCES) 3 2

1.2

PVAP

Y X

L1

ATM

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(Saturated liquid and Saturated Vapour)(Saturation)

(Saturation/Saturated

Point/State) ... (SATURATED....TABLE) 101.3 kPa 100

oC ()

Sat. Water(Steam) Table PS, kPa TS,

oC vf,m

3/kg vg,m

3/kg P -

5 32.88 0.001005 28.1900 (vapour pressure curve)

101.3 100 0.001044 1.6729 (PS

vs TS)

500 151.86 0.001093 0.3749

1000 179.91 0.001127 0.1944 T 1.3. (SATURATED TEMPERATURE), T

S

(SATURATED PRESSURE), PS (SATURATED LIQUID)

(SATURATED VAPOUR)

vf vg vfg = vg- vf (Quality) (Moisture)(Quality) , x = (, mv) / ( , ml+mv)(Moisture) , y = (, m

l) / ( , m

l+m

v)

100 x + y = 1 100% 101.3 kPa 30oC ? 101.3 kPa 3 kPa 30oC

&

.


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()

1. (Saturation) 2 30

oC vf = 0.001004 m

3/kg vg = 32.8932 m

3/kg

R-134a v

f= 0.000843 m

3/kg, v

g= 0.02671 m

3/kg

32 32

R-134a 30oC 771 kPa

8 R-22 1191.9 kPa 10%

10 kg 1 kg 1 g

2. -407C -32125 134a 23%, 25%52% -407C -17 1 MPa (Bubble) 19oC (Dew) 24 oC

(LPG)50%

( SUBCOOLED / COMPRESSED LIQUID ) :-

1 (TS) 2 (PS)

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3 (vf) 1.2 H2O 30

oC, 101.3 kPa ?

1 101 kPa TS = 100O

C TPS

1.3 H2O 30OC, v = 0.0009 m

3/kg ?

3 30OC vf= 0.001004 m3/kg v < vf

1.4 H2O 40

OC , 50,000 kPa ?

40OC vf

= 0.001008 m3/kg

(Compressed Liquid Water Table) v = 0.000987 m3/kg = (0.001008-0.000987)/0.000987x100% = 2% 1.5 H2O 30

OC , 0.001000 m3/kg ?

30OC vf= 0.001004 m3/kg 0.001000 m3/kg vf/ (SUPERHEATED VAPOUR)

1 (TS) 2 (PS) 3

1.6 (H2O) 200

O

C, 101 kPa ? 1 101 kPa TS = 100OC T > TS (Superheated

vapour)

2 200OC PS

= 1553.8 kPa P vg(Superheated vapour)

(CRITICAL POINT) (Final Saturated Point

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(CRITICAL POINT)

T, P v 0.1,1.0,.10, 20, 22.09 MPa 1.4

(Saturated liquid) (Saturated vapour)

P=10 MPa

P=1 MPa

T

CP (). P =22.09 MPa()P=20 MPa

P=0.1 MPa

1.4

()

v

, oC ,MPa

m3/kg

kg/ m

3

374.14 22.09 0.003155 317CO2 31.05 7.39 0.002143 467

O2

118.35 5.08 0.002438 410

N2

146.95 3.39 0.003215 311

H2 239.85 1.3 0.032192 31

CH4

82.75 4.6 0.00615 163

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( O2 )

118.35o

C 5.08 MPa 118.35 oC 5.08 MPa 30 oC 5.08 MPa 70 kg/ m3 15MPa 30 oC 200 kg/ m3

150 (NGV)(CH4) (TRIPLE POINT)

1.5 3 (TRIPLE POINT)

( Some Solid-Liquid-Vapor Triple Point Data ) , C ,kPa

CO2

-56.6 517

P

CP

T

P

CP

T

() () 1.5

-

-

-

-

-

-

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0.01 0.6113H

2(normal)

O2

N2

-259

-219

-210

-39

419

7.19

0.15

12.53

0.000000 13

5.066

CO

2 (101.0 kPa)

517 kPa

0.6113 kPa P,v,T

2 30oC 101 kPa 0.001004 m3/kg 0.001004 m3/kg 996.01594 kg/m3 (0.001004 =1/996.01594) 30

oC 101 kPa 60OC 30 MPa = 996.01594 m3/kg (v = 1/) 1 1 2 v = f(P,T) , P=f(v,T) ,

T = f(P,v), T = f( , P) X = f(Y,Z) 1.61.7

+

T=C

T1 T2 T3

T4P

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(Specific Internal Energy) (Specific Enthalpy) (Specific Entropy) u (Internal Energy) kJ/kg u

f ug

ufg = ug-uf 0 () 0 0.01C() (Absolute ) h (Enthalpy) kJ/kg

h = u + P.v u + P.v

v

1.6 P-v

1.7 T-v

P4

TP

P=C

3

2

P1

v

P

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ASHRAE 0 .0.01

oC() u u = h - P.v

R22,R134a NH3 0 kJ/kg -40oC

200 kJ/kg 0o

C h

f hg h

fg= h

g-h

f

s (Entropy) kJ/kg.K sfs

g s

fg= s

g-s

f

f (SUBSCRIPT f) g (SUBSCRIPT g) fg (SUBSCRIPT fg)

X = yXf+xX

g= X

f+ xX

fg= X

g- yX

fg.(1.2)

Xfg

= Xg

- Xf

(1.3)

X x y 1.8 50OC 20% 2 m3 v = vf + xvfg 50

OC vf = 0.001012 m

3/kg vfg = 12.0308 m

3/kg

v = 0.001012 +0.20x12.0308 = 2.41 m3/kg

m = V/v = 2 / 2.41 = 0.83 kg

1.9 0.4 m3 (H2O) 2 . 600 kPa

v = V/m = 0.4 /2 = 0.2 m3/kg (Saturated Water Table) P = 600 kPa vg=0.31567 m

3/kg vf= 0.001101 m

3/kg v = 0.2 m3/kg vf vg

-

v = vg y.vfg = vg - y (vg-vf) v = v

f+x(v

g- v

f) = v

f+x v

fg

0.2 = 0.001101 + x(0.31567-0.001101) = 0.001101 + x(0.31457)

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x = 0.6322 = 63.22% y = 0.3978 = 36.78% = mf=my = 2 x 0.3678 = 0.7356 kg = 1.2644 kg Vf= mfvf = 0.7356x0.001101 = 0.0008 m3 = 0.3992 m3

1.10 H2O P = 1000 kN/m2 T = 300

oC

P = 1000 kPa ,TS=179.91oC

= 300oC - v = 0.25794 m3/kg - P = 1MPa , T = 300

oC ( ) 4 kg/m3 v = = 0.25 m3/kg

1.11 20oC 40

oC

V = m = v = v1 = v2v1 = vg 20

oC = 0.14922 m

3/kg

v2

= v1

= 0.14922 m3/kg

T2

= 40oC

(Saturated Ammonia) (Superheated Ammonia) 40oC 0.14922 m3/kg 900 kPa ( 0.15582 m3/kg) 1000 kPa( 0.13868 m3/kg) 938 kPa (PERFECT / IDEAL GAS)

Pv = RT (1.4) P1v1/T1=P2v2/T2 (1.4a)Pv = RT (1.4b)

PV = mRT =nRT(1.4c)

P

1

V1

/T1

=P2

V2

/T2

(1.4d) P1

V1

/(m1

T1

)

=P2

V2

/(m2

T2

)(1.4e)

100 kPa100 m3101 kPam3?

P1/m

1=P

2/m

2

m2=(P

2/P1)m

1

m2=(101/100)m

1=1.01m

1

1%1 m3

R = () R = 0.287 kJ/kg.K R = 0.46152 kJ/kg.KR = = 8.31434 J/mole.K = 8.31434 kJ/kmol.K = MRM =(Molecular mass), m =, n =

:-- () ( 5 ) - () 2

1.12 6 x 10 x 4 m3

, P = 100 kPa , T = 25C m

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25)K(273.15g.K)0.287kJ/(k

4m106100kN/m

TR

PV=m

32

+

= = 280.5 kg

1.13 v P = 1000 kN/m2 (1MPa) T=300 C(573K) v = RT

PM

=8.3143 573

=1000 18

0 265m3

. / kg

P = 1 MPa , T = 300 C v = 0.25794 m3/kg 10 3%

4

W = Fdx W = W = F * x (1.5)2

1

Fdx

w = (F/m)dx w = w = (F/m) * x (1.5a)2

1

(F/m)dx

W = , w = , x = (+) ( )

N () m () N.m J ()

1 J = 1 N.m

(POWER) ,

W

t

W

dt

W=W .(1.6)=

W (watt) = J/s1.14 10 kg (g) 5 m/s2 20 m 40 s ??

W = F. x = (m.g).x= (10x5)20 = 1000 J = 1kJ 25W

40

1000

t

W=W ==

1.15 (CounterWeight) 1000kg 2 m/s ?

19.62kW19,620W2*9.81*1000m.g.Vt

m.g.x

t

F.x

t

W=W ======

1.8dL

P .

1.8 1.9

dL

P


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P 1.9 P P (Quasi-equilibrium) A

P*A dL W

W = P*A*dL = P*(A*dL) = PdV (1.7)

W = .(1.7a)2

1

PdV

P-V 1.10 V(X)

1.11

P

V W = P(V2

V1)

(Isothermal) T = PV =mRT= = C

W = )/VCln(V(C/V)dVPdV2

1

2

1

12 ==

E dC dt

W = EdC W/dt = EdC/dt= E.I

o

W

W E.I.cos o =o

W - , E - , I - , =Power Factor ( cos = 1)cos

2 1.10

1

P

1.11

1 2

V

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Q = q = =Q/m

Q dQ =

2

1

12 QQQ Q = Q1 21

2

2

1

Q=Q

Q

=Q

dt

J , W + ()

()

5 (Total Energy, E )

m kg

V m/s KE =mV

2/2 .(1.8)

ke = V2/2 .(1.8a)

KE kJ ke kJ/kg m kg

g m/s2

PE = mgZ .(1.9)

pe = gZ .(1.9a)

KE kJ pe kJ/kg u

f= 0

kJ/kg 0.01 oC -40 oC h

f= 0 kJ/kg 0 oC h

f= 200 kJ/kg

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2 u = f(P,v) , u = f(T,v) , u = f(P,T) uSUPERHEAT > ug > uf > uSUBCOOL LIQ.

u = y.u

f+ x.u

g= u

f+x.u

fg= u

g y.u

fg

ufg = ug-uf , x + y = 1

1.16 95oC 90% x = 0.9 y = 0.195 oC uf= 397.86 kJ/kg ufg = 2102.70 kJ/kg, ug= 2500.56 kJ/kg

u = 397.86+0.9x2102.70 = 2290.29 kJ/kg u = 2500.56-0.1x2102.70 =2290.29 kJ/kg1.17 -134a 100oC 4000 kPa

-134a 100oC PS= 3973.2 kPa

4000 kPa /( Subcooled/ compressed liquid) (Subcooled liquid ) u = u

f= 368.55 kJ/kg

6 1 1

() 2 12 1

() (-)

....(1.10)WQOO

=

= WQ (1.10a)1.18 200 W ?

= 200 = 200 W = WQ OO LQ

HQ

NETQ

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200 () 200 1.19 30 kW 10

35,000 kJ/ (Heat Engine)

Q H = 35,000x10/3,600 = +97.2 kW

= +30 kW

W

= WQ OO

Q L + 97.2 = +30

Q

L= 30 97.2 = 67.2 kW

67.2 kW1 (PROCESS) = WQ

() (Process)(E)

WQoo

=

(1.11) += EWQ += EWQ

E = U+KE+PE 1

Q = dU + d(mV2/2)+d(mgZ) +W (1.12) q = du + d(V2/2) + d(Zg) + w .(1.12a) 1Q 2 = (U2-U1) + (mV22/2 - mV21/2) + (mgZ2-mgZ1)+ 1W2(1.12b)

Q = U+KE+PE+W (1.12c)

1q

2= (u

2-u

1) + (V

2

2-V

2

1)/2+(Z

2-Z

1)g +

1w

2(1.12d)

1.20 1.13 ?

1(1.12c) Q = U+KE+PE+WKE & PE Q = U + WQ = -1500 kJ , W = -5000 kJ

1.13

U =?

W =5000 kJ

Q =1500 kJ

H

1.12

Q

W

QL

.


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-1500 = U -5000U = 5000 - 1500 = +3500 kJ Q U= 5000 kJ

1

(WE)(Wmech) 1.14 1

Q Q = U + KE + PE + WU1 Wmech U2 KE =0, PE = 0

WE 1.14 -Q =U -WE+ Wmech

U = WE-Q - Wmech U = 0 U WE=Q + Wmech = Wmech/ WE (P=)

Q = dU+dKE+dPE+W dKE dPE W = PdV

Q = dU+PdVQ = dU+dPV (P= )

Q = d(U+PV) = dH ..(1.13) q = dh (1.13a)

H = = U + PV h = H/m = u+Pv h u+Pv h u h u h Pv 0 () (h

f=0 kJ/kg)(Triple point) 0.01C

hf=0 kJ/kg 40C R-12,R-22 NH3

U = 0Q = W

E

WE

Q

Q = H2-H1 = m (h2-h1)

q = h2-h

1

W

dV

Q

1.15

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hf= 200 kJ/kg 0C R-134a

2 h = f(P,v) , h = f(T,v) , h = f(P,T) h

super heat> h

g>h

f> h

subcool

h = xhg+yhf= hf+xhfg = hgyhfg

1.21 1kg 100C )) 90% ?

q = dh) 1q2 = h2 h1 = hg hf = hfg = 2257 kJ / kg)

1q

2= h

2 h

1= (h

f+xh

fg) h

f= xh

fg= 0.90x2257 = 2031.3 kJ / kg

( SPECIFIC HEAT) 2

1 kg 1 1.16

1.17 1

q =dhq =du

20OC 21

OC 20

OC

21OC

C = q / dT (CONSTANT VOLUME SPECIFIC HEAT, C

V)

ol.K)....kJ/(kmT

u

T

uCg.K),.....kJ/(k

T

u

T

uC

vv

v

vv

v

=

=

(1.14)

(CONSTANT PRESSURE SPECIFIC HEAT, CP)

ol.K)....kJ/(kmT

h

T

h

Cg.K),.....kJ/(kT

h

T

h

CPP

P

PP

P

=

=

.(1.15)

1kg 1kg 1kg 1kg

1.16q q 1.17q = h2-h1q = u2-u1

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Cv

& Cp

C

du = CvdT = CdT

u = u2 u1 = C(T2 T1)dh = du + d(Pv)

v dh = du + v dP = CdT + v dP(1.16) P () v

dh = CdT h = h2 h1= C(T2 T1) = u .(1.17)(dP) dh = CdT + v dP C = 4.186 kJ/kg.K, C= 0.381 kJ/kg.K C=0.46 kJ/kg.K

Pv = RT, Pv RT= , PV = mRT = nRT = 8.3 nT

h = u + pv = u + RT = f(T)

CV CV =

VT

u

u CV

CV

=dT

du du = C

VdT

(1.18)==2

1

V12 dTCuuu

CV,av

u = u2-u1= CV,av(T2-T1) (1.18a)

==2

1

V12 dTCuuu kJ/kmol ....(1.19)

avV,C )T(TCuuu 12avV,12 == kJ/kmol ...(1.19a) kJ kg m kg n kmol C

P C

P=

PT

h

h C

P

CP = dTdh dh = CPdT

==

2

1

P12 dTChhh CP,av h = h2-h1= CP,av(T2-T1)(1.20)

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==2

1

P12 dTChhh kJ/kmol )T(TChhh 12P12 == kJ/kmol ..(1.21)

kJ

kg

m

kg n kmolC

po -3

298K(25oC),100kPa -2 (Mono atom) PC 20.8 kJ/(kmol.K)

P0C V0C ()

Cpo Cvo

Cpo - Cvo = R 8.314RCC VOPO == (1.22)

( Specific Heat Ratio ),V

P

C

Ck =

1.22P1 = 10 kPa, T1 = 200oC P2 = 50 kPa, T2 = 400oC h 1 h1 = 2879.52 kJ/kg , h2 = 3278.89 kJ/kg

h = 3278.89 - 2879.52 = 399.37 kJ/kg ()

2 25oC -2 Cpo = 1.872 kJ/kg.Kh = CpT = 1.872 x (400-200) = 374 kJ/kg.

[ (399.4-374)/399.4 = 6.4% ]

7 1 /

() 100 50 ()() 25 () 125 1.18 mi1, mi2 mi3 me1 me2

(1) mcv1(2) m

cv2

.


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mcv1

+( mi1+ m

i2+ m

i3) = m

cv2+( m

e1+ m

e2)

..(1.23) +=+ eCV2iCV1 mmmm

e

oo

iCV1CV2 mm

tmm

= (1.23a) 1.18 1 1 E

1 2 2

E2 m

i1, m

i2 m

i3 h

i1, h

i2 h

i3

me1 me2 he1, he2 Q W () 1

Q = ECV2

ECV1

+me1

(h+ke+pe)e1

+ me2

(h+ke+pe)e2[ m

i1(h+ke+pe )

i1+ m

i2(h+ke+pe)

i2+m

i3(h+ke+pe)

i3]+W

Q[(1) (2)]

1.18(1) (2)

()

W[(1) (2)]

()

mCV1

,ECV1 m

CV2,E

CV2

mi1,h

i1

mi2,h

i2

mi3,h

i3

me1

,he1

me2

,he2

Q = ECV2 ECV1 + me (h + ke+pe )e mi (h + ke + pe )i + W .(1.24)

Q = ECV2 ECV1 + m

e (h + V

2

/2+Zg )e mi (h + V

2

/2+Zg )i + W ...(1.24a)

=o

Qdt

dECV + (h + Voem 2/2+Zg )e (h + Vo

im2/2+Zg )i + .(1.24b)

o

W

ECV1=mCV1(u+V2/2+Zg)CV1(1.25) ECV2=mCV2(u+V

2/2+Zg)CV2(1.25a) t

EE

dt

dE CV1CV2CV = (1.25b)

(Steady Flow Process)

mi = me ..(1.26)

.


24/55

m i =

m e .(1.26a)

ECV1

= ECV2

Q = me (h + V

2/2+Zg )e

mi (h + V2/2+Zg )

i+ W kJ.(1.27)

= m e (h + V2/2+Zg )e m i (h + V

2/2+Zg )

i+

W kW .(1.27a)Q

Q = me he mi h i +W kJ ...(1.27b)Q =

m e he

m i h i + kW .(1.27c)

W

1.19 5 kg/s 5 kg/s

V

1= v

1.m / A

1, V

2= v

2.m / A

2

(V =) 1.19

m2=?m

1 1(1.26a)

m i =

m e = (1.27) (1.27a)

m

Q = m [(he-hi) + (V2

e/2 -V2

i/2) + g (Ze-Zi)] +W kJ ..(1.27d) Q = m [h + ke + pe] +W ..(1.27e)

= [(hQ

m e-hi) + (V

2

e/2 -V2

i/2) + g (Ze-Zi)] +W kW ..(1.27f)

= [h + ke + pe] + ..(1.27g)Qm

W

(1.27f)(1.27g) 1 .m

q = (he-hi) + (V2

e/2 -V2

i/2) + g (Ze-Zi)+w kJ/kg.. (1.27h)q = h + ke + pe +w .. (1.27i)

= h + kW (1.27j)m

W

Q

Q = h + W kJ (1.27k)q = h +w kJ/kg (1.27l)

- (Steady Flow)

1=5 kg/s

2

1

.


25/55

(ke = 0, pe = 0)

-P

e>>P

i (

W w = )

= h+ w = h+qW

m

Q

= (Pv + u)+ w = (Pv+u)+qW

m

Q

me=m

i=m

Ve=(PiVi/Ti) / (Pe/Te)

v u= CT

W = (vP + CT)+ w = ( vP+ CT)+q ..(1.28)

m

Q

() qQ 0u= CT = 0 ()

W = vP =

m

V P w = vP ...(1.29)

1.23 0.1 m3/s 101 kPa 303 kPa

() W =

V P

m .v = m

V

3/s 0.1 m3/s , P

i= 101 kPa P

e= 303 kPa

= 0.1 ( 303 101 ) = 20. 2W 20.2kW20.2kJ/s

m

kN

s

m2

3

==

20.2 kW 1.24 100 kPa 280 K 600 kPa 400 K

0.02 kg/s 16 kJ/kg

(1.27j) =Qm h+

W

=Q mCP(TeTi)+W

Q = .q = 0.02x16 =0.32 kW

m

CP 298 K 1.004 kJ/(kg.K)

0.32 = 0.02 x 1.004x(400280)+W

W = 2.410.32 = 2.73 kW

2.73 kW

(Turbine)(Engine)

.


26/55

Pe


27/55

1.27 (Hydraulic Turbine) 100 m 10,000 kg/s

(1.27f)Q =

m [(he-hi) + (V

2

e/2 -V2

i/2) + g (Ze-Zi)] +W

W

h = u +Pv

++++= W)g]Z-(Z/2)V-/2(V)vPv(P)u[(umQ ie2i

2eiieeie

Q = 0 , T

i= T

e= T ui = ue , vi = ve=v Pi = Pe = PATM

Vi=Ve

+= W)gZ{(Zm0 ie

W =

m (Z

i Z

e)g

m =10,000 kg/s v = 0.001 m3/kg = 1,000 kg/m3W = 10,000 x 100 x 9.81 W = 9,810,000 W = 9.81 MW.

70% , 85% =

9.81 x 0.7 x 0.85 = 5.84 MW

(Expansion Valve&Capillary Tube)

Pe


28/55

hef=165.8 kJ/kg , h

efg=216.36 kJ/kg , h

eg= 382.16 kJ/kg

40 (26.3) = 66.3 oC e he 256.54 kJ/kg

he = hef+ x hefg x =( he hef)/ hefg = (256.54 165.8)/216.36 = 0.42 = 42% 42%

i e 1(1.27h)

w = 0 q = he hi + (2

2

eV

2

2

iV

)+(zegzig)

i e

q = ue+ P

ev

e (u

i+ P

iv

i)+ (

2

2

eV

2

2

iV

)+(zegz

ig)

PiviPeve +( 2

2

iV

2

2

eV

)+ (zig zeg ) = (ueui) q ..(1.31)

(1.31) i e (u

eu

i)( q) kJ/kg Pa

(ve= vi= v)

Pi P

e+(

v

Vi

2

2

v

Ve

2

2

)+ (zig/v z

eg/v ) = [(u

eu

i) q]/v = P

F..(1.31a) Pa

Piv

i/g P

ev

e/g +(

g

Vi

2

2

g

Ve

2

2

)+ (zi z

e) = [(u

eu

i) q]/g = Z

F..(1.31b) m

PF Z

F

[(ueui) q]/v = PF [C(TeTi) q]/v = PF , TeTi= [v PF + q]/C [(u

eu

i) q]/g = Z

F [C(T

eT

i) q]/g = Z

F, T

eT

i= [g Z

F+ q]/C

(Bernoulli equation) P

iv

i- P

ev

e+(

2

2

iV

-2

2

eV

)+ (zig - z

eg ) = 0 ..(1.32)

Pi- Pe+(v

Vi

2

2

-v

Ve

2

2

)+ (zig/v - zeg/v ) = 0 ..(1.32a)

Piv

i/g - P

ev

e/g +(

g

Vi

2

2

-g

Ve

2

2

)+ (zi- z

e) = 0 ..(1.32b)

8

P1+V

1

2/(2v

1)+Z

1g/v

1=P

2+V

2

2/(2v

2)+Z

2g/v

2

2

.


29/55

(Steam Power Plant) (Gas Turbine

Plant) (Internal Combustion Engine)

= .(1.33)

1.29 35,000 kJ/litre 60 ./. 10 ./ 15 kW = (60 km/h)/(10 km/litre) x (35,000 kJ/litre) = 210000 kJ/h = 58.3 kW = /

= 15 / 58.3 = 0.26 =26% 1.30 25 kW 10 kW ?? = / = 25 / 10 = 2.5 = 250% ?

3 )

= (1.34)

1 (100%))

= .(1.35)

1 (100%))

...(1.36)

=

1 (100%)

.


30/55

1.31 20.2 kW 30 kW

= / = 20.2 / 30 = 0.67 = 67%

(Thermal Energy Reservoir or TemperatureReservoir)

2

(HEAT ENGINE)

1.24 1.25

TH

QH

(Boiler)

QL

(TH) (Thermal Efficiency)

th

H

th

Q

W = (1.37)

Wnet

(Condenser)

TL

1.24

(Turbine) P W

TH

QH

Wnet

TL

QL

1.25

.


31/55

1 W = QH Q

L

(1.38)H

L

H

LHth

Q

Q1

Q

QQ =

=

1 100%

(TH) (TL)(T

H)(T

L)

()) (REFRIGERATOR/HEAT PUMP)

(Low temperature reservoir)(Hightemperature reservoir) ()() 1.26 1.27

TH

QH

W

( ndense

(Evaporator)

L

Co r)

(Coefficient of Performance) COP

W

QCOP LR = ..(1.39)

1 W = QHQ

L

1/QQ

1

QQ

QCOP

LHLH

LR

=

= .(1.40)

W

TH

QH

TL

QL

1.27

Q TL

(Expansion Valve)

1.26

.


32/55

(Refrigerator) 1 (100%) 100%

QH

W

QCOP HHP = (1.41)

1 W = QH Q

L

.(1.42)HLLH

HHP

/QQ1

1

QQ

QCOP

=

=

(Heat Pump) 1 (100%) 100% 100% ( EletricResistance Heater) 100 kW 2 1 100 kW 2 100 kW 200 kW

(1.41) (1.40) COPHP- COPR= 1 (1.43)

1 (100%)

( Q

L)/.(Btu/h) (W)(W) (/

.) / Btu/h/W (EER=Energy EfficiencyRatio or Rating) COP = EER/3.412 5 10.6 10.6/3.412 = 3.11 (kW/TR) 1 12,000

/ 3.517 ? COP = 3.517/(kW/TR) 1.13 kW/TR COP = 3.517/1.13 = 3.11 1.32 (COP) 1.1 110% 0.20 kW 5 -20oC 1750 kJ 5 . 5 .

?

.


33/55

() 0.20 kW() = 1750 / (5*3600) = 0.10 kW

() = 0.20 + 0.1 = 0.30 kW

LQ

COP = QL/W = W/QL

= 0.30/1.1 = 0.27 kW/COPQW L

=

= 0.30 + 0.27 = 0.57 kW += WQQ LH 0.20 kW = 0.57- 0.20 =0.37 kW 5 .= 0.37 x 5 x 3600 = 6659 kJ

5 . = 0.20 kW

LQ

= 0.20/1.1 = 0.18 kW/COPQW L

=

= 0.20 + 0.18 = 0.38 kW += WQQ LH 0.20 kW = 0.38- 0.20 =0.18 kW

9 2

2 - (KELVIN - PLANCK STATEMENT)

1

1.28 1 W = QH th

H

W

Q= = 100% =100 %

1 100%

40%

THH

W 1.28

Q

.


34/55

2 (W)(QH)

QH= W

(CLAUSIUS STATEMENT)

THH

1.29

Q

W

W = 0 (QH = QL)

COP QRL= =

0

T

=

90%

10 (REVESIBLE&IRREVERSIBLE PROCESSES)

(Reversible Process)

W= 10 J W= -10 J W= 8 J W= -10 J

T1

, P1

T2

, P2

T1

, P1

T2

, P2

Q = 5 J 1.31 Q = -5 J Q = 5 J 1.32 Q = -7 J

1 + 5 7 = + 8 10 4

1. 2.

3. (dT0)

4. 2

H

HQ

TL

QL

1.30

.


35/55

11 (CARNOT CYCLE) (REVERSIBLE CYCLE)

(Totally Reversible Cycle)

TH

()

TL

()

()

THdTH

()

TLdTL

QH QH

QL QL

1.33

3

1 2

4

()

()Wnet

4 12 (- ) (dTH0) (Reversible isothermal process)23 (-) (Reversible adiabatic process34 (-) 12 (dT

L

0)

41 (-) 23 ()

2

HTLT

HQLQ = .(1.44)

H

H

L

L

T

Q

T

Q = .(1.44a)

.


36/55

.(1.44b)0T

Q

T

Q

L

L

H

H =

(1.44b) ..(1.44c) = 0TQ

(CARNOT HEAT ENGINE) th =

H

LH

H Q

QQ

Q

W =

QL / QH = TL / TH

th,CARNOT =H

LH

T

TT th,CARNOT =

H

L

T

T1 (1.45)

100% 0 (K) (R) 2

100% 1- T

L/ TH

1.33 30C 600C 70%

th,Carnot =273600

)27330()273600(

T

TT

H

LH

+

++=

= 65.29%

65.29% 70%

TH=600+273

Wne

WP

H=30+273

t

T

/ (CARNOT REFRIGERATOR/HEAT PUMP) COPR =

Q

Q Q

L

H L

QL

/ QH

= TL

/ TH

COPR,CARNOT =T

T T

L

H L..(1.46)

COPHP =LH

H

QQ

Q

COPHP,CARNOT = (1.47)

LH

H

TT

T

35% ?

.


37/55

1.34 53 kW (181,000 /.) 16 kW 35C 25C

COPR = 3.31653

W

Q

W

Q

o

o

LL ===

COPCARNOT

= 29.825)(27335)(273

25273

TT

T

LH

L =++

+=

= 2980 %

1.35 0C 20C 50 kW ? /

COPHP,CARNOT

= TH

/ (TH

TL) = (20+273)/(20-0) = 14.65 (1465%)

CARNOTHP,H /COPQW

= = 50 / 14.65 = 3.41 kW

12 (CLAUSIUS INEQUALITY)

0TQ

0T

QO

0TQO

= = 0TQ

< 0TQ

.(1.48)

0T

QO


38/55

S = 2

1T

Q. (1.49a)

m

Ss =

m

Q=q

T

q

ds = . .(1.49b)s2 - s1 = s =

2

1T

q..... .(1.49c)

T

q

kJ/K

(Specific entropy) s = S/m kJ/kg.K

(Absolute value) 0

sf = 0 kJ/kg.K 0.01C 134a(Refrigerant-134a) s

f= 0 kJ/kg.K 40C(40)

American Society of Heating, Refrigerating, Air Conditioning Engineers Inc.(ASHRAE) (International Convention) sf= 1kJ/kg.K 0C (Internally Reversible) sg

0.01C q=dh q = hg-hf= hfg = 2501.3 kJ/kg

ds = q/T =dh/T sg-sf= g

f

dh/T

sg-sf= sfg= (hg-hf)/T = hfg/T =2501.3/(0.01+273.15) = 9.1569 kJ/kg.K sg=sf+ sfg= 0 + 9.1569 = 9.1569kJ/kg.K (Saturated water table) R134a sg-sf=

sfg= hfg/T 0C hfg= 198.36 kJ/kg sg-sf= sfg= 198.36/(0+273.15) = 0.7262 kJ/kg.K sg= sf+ sfg=1 +0.7262 = 1.7262 kJ/kg.K ( -40C sf= 0 kJ/kg.K 0C sf= 0.197 kJ/kg.K sg=0.197 + 0.7262 = 0.9232 kJ/kg.K ) (Property)

ds = q/T

.


39/55

s SUPERHEAT > sg > sf> sSUBCOOL

s = xs

g+ y s

f=s

g-y s

fg= s

f+ xs

fg

1.36 10C, 0.2 MPa ? ( Saturated water:temperature table)10C 1.2276 kPa( 0.0012276MPa ) 0.2 MPa

() 10C s = sf 10C = 0.1510 kJ/kg.K

4

(Heat Engine)

1-2

1.34

L

4 3

a bs

TH

T

1 2

TH ()

TL

()

()

Q Q

T

THdTH

()

TLdTL

H H

QL QL

1.33

3

1 2

4

()

()

Wnet

.


40/55

S2-S

1=

2

1 REVT

Q T = TH

QH= TH(S2-S1 ) 1-2 2-3

S3-S

2=

3

2 REVT

Q Q = 0

S3-S2 = 0 S3= S2 Isentropic 3-4 1-2 Q

L= T

L(S

4-S

3) 3-4

4-1 2-3 S1= S4 Isentropic process

1 = WQ

Wnet = QH-QLW

net= 1-2 3-4

Wnet

= 1-2-3-4-1

th =H

LH

12H

43L12H

H

LH

T

TT

)S(ST

)S(ST)S(ST

Q

QQ =

=

(Refrigerator/Heat pump)

2-1 Q

H= T

H(S

1-S

2) 2-1

1-4 S

1= S

4 ( Isentropic ) 1.35

TH

T

1 2

TL

4 3

a bs

4-3 QL= TL(S3-S4 ) 4-3

3-2 S2= S3 Isentropic 1 Q=W

Wnet = QH-QLWnet = 2-1 . 4-3Wnet = 1-2-3-4-1

COP=LH

L

43L12H

43L

LH

L

TT

T

)S(ST)S(ST

)S(ST

QQ

Q

=

=

.


41/55

TH

QG TGREF/HP T

G T

L

TH T

L

1.36 Snet = Sgen = SG +SL +SH = 0 (a)

QH

QL

QG

SG = - QG/ TG, SL = - QL/ TL, SH = + QH/ TH (a) - QG/ TG- QL/ TL+ QH/ TH = 0 .(b)

QH = QG + QL (c)

(b)(c)

QL / QG = COPREF,CARNOT =

G

HG

LH

L

T

TT

TT

T(1.50)

QH

/ QG

= COPHP,CARNOT

=

G

LG

LH

H

T

TT

TT

T(1.51)

T

q

< 0T

Q

= 0dS 0 dS < dSTQ

> TQ

dS ..(1.52)

+= genSTQ

dS .(1.52a)

T

QdS > ..(1.53)

>2

1

12T

QSS ..(1.53a)

.


42/55

+=2

1

gen12 ST

QSS ..(1.53b)

(>)(EntropyGeneration = S

gen) 0

Sgen ( S

2S

1)

[ 2

1T

Q T

Q] ( S

gen)

+= genST

Q

dt

dS(1.54)

TdS = dU + PdV kJ(1.55)

Tds = du + Pdv kJ/kg..(1.55a) Td s = d u + Pd v kJ/kmol .....(1.55b)

TdS = dH VdP kJ....(1.56)

Tds = dh vdP kJ/kg (1.56a) Td s = d h v dP kJ/kmol ....(1.56b)

T

Pdv

T

duds += .(1.57)

T

vdP

T

dhds = .(1.58)

(Close system/Control mass orControl volume)

T

Pdv

T

duds +=

dv = 0

T

CdT

T

duds == =

2

1T

CdTs kJ/kg.K

C Cav

1

2av12

T

TlnCsss == ..(1.59)

.


43/55

1.37 500 K 50 kg 285 K 0.45 kJ/(kg.K) () () ()

() 12av12

TTlnmC)sm(sS == =(50kg)[0.45kJ/(kg.K)]ln[285K/500K]= 12.65 kJ/K

() 1

Q = U + W W = 0Q = U =mCav(T2-T1) = (50kg)[0.45 kJ/(kg.K)](285-500)K= - 4837.5 kJQ = - Q = +4837.5 kJ

S =Q/T = +4837.5/285 = 16.97 kJ/K() S =-12.65 + 16.97 = +4.32 kJ/K

T

Pdv

T

duds +=

T

vdP

T

dhds = du = CVOdT P/T = R/v

+=2

1

1

2VO12

v

vRln

T

dTCss (1.60)

dh =CPOdT v/T = R/P

=2

11

2PO12

P

PRln

T

dTCss .(1.61)

CVO

CPO

(T) -3 s C

VOC

PO

CPO= 20.8 kJ/(kmol.K), CVO= 12.5 kJ/(kmol.K) CVOCPO

1

2

1

2VO12

v

vRln

T

TlnCss += ..(1.60a)

1

2

1

2PO12

P

PRln

T

TlnCss = .(1.61a)

CPC

V

s2s

1= 0

1

2

1

2V12

v

vRln

T

TlnC0ss +==

.


44/55


45/55

v 1 k k =Cp/C

v

Pv

n= ..(1.65)

T/P

(n-1) / n= (1.66)

Tvn-1 = .(1.67)P (2) Pv Pv

nPv

k 1.37 T (2)Pvn (2) Pvk

(1) (2)Pv (1) (3)Pv

(3) Pvk Pvn Pv (3)Pvk (3)Pvn

v () s () P-v 1

2(Compression) 3 (Expansion) P-v X (1-2) ( 1-3)

T-s

X

T

W

()

()

si1,m

i1

si2,m

i2

si3

,mi3

CV1,m

CV1S

se1

,me1

e2

,me2

Q [(1)

s

(2)]TB

H (1) (2) 1.38

SCV2

,mCV2

.


46/55

-(Uniform State Uniform Flow / Transient Flow Process)

(SCV2

- SCV1

) +(mese -misi ) = T

Q+ S

gen(1.68)

(SCV2 - SCV1 ) +(mese -misi ) = 2

1T

Q+ Sgen (1.68a)

+=+ geniieeCV STQ

smsmdt

dS.(1.69)

=dt

dS CV

=

T

Q

=

ii

ee smsm

=

CVgen,S

-(Steady Flow Process) m

CV1= m

CV2 S

CV1= S

CV2(1.68)

(mese -misi ) = T

Q+ Sgen ..(1.70)

m(se si) = T

Q+ Sgen (1.70a)

genie s

T

qss += ....(1.71)

genie sT

dqss += .....(1.72)

genie sss = .(1.73)

se = si .(1.74)

Pvk= Tvk-1= T/P(k-1)/k=

.


47/55

1.40 1 MPa, 300oC 50 m/s 150 kPa 200 m/s 200 mm kW

1q = h+ke+pe+w q = h+ke+w(q ) q = 0, s

e= s

i w

w = hi-he+i2/2-e2/2P

i=1 MPa, T

i=300

oC, i=50m/s T i

we

Pe=150 kPa, e=200m/s ( ) s () 1.39

1 hi= 3051.15 kJ/kg, s

i= 7.1228 kJ/kg/K, v

i=0.25794 m

3/kg

2 Pe= 0.15 MPa,(Te=111.37oC) , se=si= 7.1228 kJ/kg/K

(quality=xe) ,

xe= (s

e-s

f)/s

fg s

f= 1.4336 kJ/kg.K, s

fg= 5.7897 kJ/kg.K

xe= (7.1228-1.4336)/5.7897=0.9827

he= h

f+ x

eh

fg h

f= 467.1 kJ/kg, h

fg= 2226.5 kJ/kg

he= 467.1+0.98x2226.5 =2655.0 kJ/kg

w = 3051.15-2665.0+50

2/(2x1000)-200

2/(2x1000)=377.5 kJ/kg

200mm= 0.2 m A =D2/4=(0.2)2/4=0.0314 m2

= A =0.0314x50 mV 3/s=1.57 m3/s

m = /v= 1.57/0.2579 = 6.09 kg/s

V

=

Wm w = 6.09 kg/s x377.5 kJ/kg = 2298 kW

1.41

R134a

.


48/55

1.017 MPa 40oC 0.1013 MPa ()

1 q = dh +w q w dh = 0 h

e= h

i -134a

i 40oC 1.017 MPa, si = 1.1909 kJ/kg.K hi = 256.54 kJ/kg e 0.1013 MPa he = hi =256.54 kJ/kg 26.3

oC he =

hfe

+x hefg

x = (he

- hfe)/ h

efg= (256.54-165.8)/216.36 =0.419

se = sfe +x sefg = 0.8690+0.419(0.8763) = 1.24 kJ/kg.K

ses

i= 1.24 1.1909 = 0.0453 kJ/kg.K

1.42 R-134a 1.017 MPa 40o C 0.1013 MPa se = si-134a i 40oC 1.017 MPa, si = 1.1909 kJ/kg.K hi = 256.54 kJ/kg e 0.1013 MPa s

e= s

i=1.1909 kJ/kg 26.3 oC

se

= sfe

+x sefg

x = (se - sfe)/ sefg = (1.1909-0.8690)/0.8763 = 0.37he = hfe +x hefg = 165.80+0.37(216.36) = 245.85 kJ/kg

1 w = h

i-h

e= 256.54-245.85 =10.69 kJ/kg

14 -

=2

1

Pdvw

-(Steady Flow Process)

w = - vdP- dke-dpe (1.75)

w = - vdP- ke-pe....(1.75a)e

i

.


49/55

w= - vdP ..(1.75b)

e

i

()=e

i

vdPmWoo

.. ..(1.75c)

(v)

()

ds = du/T + Pdv v dv ds = du/T = CdT/T ds = 0 dT 0 ds dT q = h +ke + pe + w q = h + w q = u +Pv + wu = q Pv w = q vP w ( v )T = [q vP w ] / c..(1.76) (w) w

vP (q) () 1.43 0.2 m3/s 100 kPa 600 kPa 75%

(1.75c) vdP = mWS e

i

.


50/55

)P(PV)Pv(PmW ieo

ieS ==

S

o

W = - 0.2(600-100) = -100 kW

= 100 / 0.75 = 133.3 kWo

W

T = [q vP w ] / cq , v = 0.001 m3/kg, P = 600-100 =500 kPa, c = 4.187 kJ/kg.Kw = = 133.3/(0.2/0.001) = 0.67 kJ/kg

oo

m/W

T = [q vP w ] / c =[ 0 0.001x500 (-0.67)] / 4.187 = 0.17OC()()()

q =

u + w

T = [q w ] / c

Pvk= C ()

P1/k

v = C1/k

=c v =1/kP

c w

= - vdP

e

i

w = -

e

i

)P

c( 1/k dP

)P(P

k

11

cw

)k

1(1

i)

k

1(1

e

=

)P(P1k

ckw

1)/k(ki

1)/k(ke

= ...(1.77)

1])/P[(PP1k

ckw 1)/k(kie

1)/k(ki

=

c = Pi1/k

vi1])/P[(PPvP

1k

kw

1)/k(kie

1)/k(kii

1/ki

=

1])/P[(PRT1k

k1])/P[(PPv

1k

kw

1)/k(kiei

1)/k(kieii

=

= (1.78)

(1.77) c = Pi

1/kv

i c = P

e

1/kv

e

)vPv(P1k

k)PvPPv(P

1k

kw iiee

1)/k(kii

1/ki

1)/k(kee

1/ke

=

= .. (1.79)

(1.79) Peve = RTe Pivi = RTi )T(TC)T(T

1-k

-kR)RT(RT

1k

kw iePieie ==

= ..(1.80)

()

.


51/55

Pv = RT = C () v = C/P w = - vdP

e

i

w = - Cln(Pe/Pi) = - RTi ln(Pe/Pi) .(1.81)

Pvn = C ()

Pvn= C () n 1 k ( Pv1 Pvn Pvk)

v = c / P1/n w= - vdP

e

i

1])/P[(PRT1n

n1])/P[(PPv

1n

nw 1)/n(niei

1)/n(nieii

=

= ..(1.82)

)vPv(P1n

nw iiee

= (1.83)

n = ln(P1/P

2) / ln(v

2/v

1) (1.84)

1.44 1 . 100 kPa 1 MPa )) w

= - vdP e

i

) vi = vf100 kPa= 0.001043 m3/kg

w = - vi(Pe-Pi) = vI(Pi-Pe) =(0.001043m3/kg)[(100-1000)kPa]= - 0.94 (m

3/kg)(kN/m

2)

w = - 0.94 kJ/kg

) 1(i) 100kPa ,vg=1.694m3/kg2(e) 1000kPa 1])/P[(PPv

1k

kw 1)/k(kieii

=

k-2 1.3271](1000/100)1.694x100[

11.327

1.327w

1)/1.327(1.327

= = -525 kJ/kg

5001 q = h+w =h

e

-hi

+w

.


52/55

q he

q = 0 s

e= s

i

hi =2675.5 kJ/kg si=7.3594 kJ/kg.K Pe=1 MPa

se = si = 7.3594 kJ/kg.K he = 3195.5 kJ/kgw = hi-he= 2675.5 3195.5 = 520 kJ/kg

(Air Compressor) w = - vdPe

i

dP (v)

()(Isothermal Process) () 1.45 27 oC 100 kPa () 450 kPa 227

oC ) ))

) Pv1.3

= )

) w = - vdP 1

q = h+w =h

e

i

e-hi+w

q 10%0.1w =he-hi+w w-0.1w = hi-he w =( hi-he)/0.9 = CP,av( Ti-Te)/0.9

1 CP -2 25

O

C CP,a= 1.004 kJ/kg.K

w = 1.004[(27+273)-(227+273)] / 0.9 = 1.004[(27-227] / 0.9 = - 223.11 kJ/kg

223.11 kJ/kg2-4.1 Ti = 27+273 = 300 K, hi =300.473 kJ/kg Te = 227+273 = 500K,h

e=503.360 kJ/kg

w = (300.473-503.360)/0.9 = -225.43 kJ/kg

225.43 kJ/kg

)27oC 100 kPa 450 kPa

.


53/55

1])/P[(PRT1k

kw

1)/k(kiei

=

-2 k = 1.4 w = -1.4/(1.4-1)x0.287x(27+273)[(450/100)

(1.4-1)/1.4-1]= -161.8 kJ/kg

161.8 kJ/kg) () w = - RTi ln(Pe/Pi)

w = -0.287x(27+273)ln(450/100) = -129.5 kJ/kg

129.5 kJ/kg) Pv1.3 = ]1)/P[(PRT

1n

n1])/P[(PPv

1n

nw 1)/n(niei

1)/n(nieii

=

=

n = 1.3 , R=0.287 kJ/kg.K, Ti = 300 K Pe =450 kPa, Pi =100 kPaw = -1.3/(1.3-1)x0.287x(27+273)[(450/100)

(1.3-1)/1.3-1]= -154.8 kJ/kg

154.8 kJ/kg 225.43 kJ/kg 161.87 kJ/kg 129.5 kJ/kg 154.8 kJ/kg)

= / 225.43 kJ/kg

= 161.87 / 225.43 = 0.72 (72%) = 129.5/225.43=0.57 (57%) = 154.8/225.43 =0.69 (69%)

k 1.4 n

P

Pi

Pe

(e) Pv Pvn

Pvk

(i)

v

1.40() ()

s

(i)

(e) Pvn

Pvk

(e) Pv

T

.


54/55

P-v 1.40 () - vdP (i)-(e) P

Pv = Pv

e

i

k

T-s 1.40() (i)-(e) s

(Gas Turbine) w = - vdPe

i

dP (v)() 1.46 1600K , 3 MPa () 100 kPa 850 K ) )) ) Pv1.3 = )

) w = - vdP 1

q = h+w =h

e

i

e-hi+w

(q= 0) w =( h

i-h

e) = C

P,av( T

i-T

e)

25OC CP,a= 1.004kJ/kg.K

w = 1.004(1600-850) = 753 kJ/kg) ()1600K 3 MPa 100 kPa

1])/P[(PRT1k

kw 1)/k(kiei

=

25OC k = 1.4 w = -1.4/(1.4-1)x0.287x1600[(100/3000)

(1.4-1)/1.4-1] = 999.02 kJ/kg

999.02 kJ/kg) ( T =)

w = - vdPe

i

.


55/55

Pv = RT = C () v = C/P w = - Cln(P

e/P

i) = - RT

iln(P

e/P

i) = -0.287x1600 ln(100/3000) = 1562 kJ/kg

1562 kJ/kg

) 1])/P[(PRT

1n

nw

1)/n(niei

=

w = -1.3/(1.3-1)x0.287x1600[(100/3000)(1.3-1)/1.3

-1] = 1082.15 kJ/kg

) / = /

= 753/999.02 = 0.75 (75%) = 753/1562 = 0.48 (48%)

= 753/1082.15 = 0.70 (70%)

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01 วิชาที่1THERMODYNAMICS- มิย50 D(HIGH-LIGHT)