termo 1' modeli 4

8/9/2019 termo 1' modeli 4

1/12

2

II.

Percaktimi i parametrave, kur dihet:

cp= 0.903 (kj/kgK)

cv= 0.669 (kj/kgK)

R = 0.234 (kj/kgK)

3.1=v

p

c

c5

cp cv = R

Per piken 1.

p1 = 30 (bar)= 30 * 105

v1= 0.026 (m3/kg)

T1= ? )(732234

026.01030 5111111 K

R

vpTRTvp =

===

Per piken 2.

p2 = ?

v2 = ? - procesi (1 2) politropik, si i tille na jep:

T2 = ?

n = -10

s(1-2) = 0.690 (kj/kgK) )(870293ln 690.0690.0

12

1

2

1

2 KeeTTeT

T

T

Tcs nn c

s

c

s

n =====

( )paT

Tpp

p

T

p

T nn

n

n

n

n

349.7447354320

8701030

110

10

51

1

2

121

2

1

1

2

2 =

=

==

027.0349.7447354

870234

2

2

2222=

=

==

p

TRvRTvp (m3/kg)

3


2/12


3/12

p

p1 1P1 = 30 (bar) q = 0

T1 = 320 (K)

T = 0

Po = 1 (bar)

Px xTo = 273 (K) po o

v

e x1 = ? v1 vx vo

e x1 = lmax = l(1-x) + l(x-o)

(1-x) process adiabatic

q = l(1-x) + h l(1-x) =-h = cp( T1 Tx)

0

(x 1) process izotermik

q = l(x-o)+ h l(x-o) = Tx s Tx = To

0

s = R

== kgKkj

p

p

x

732.0348.2283979

1ln234.0ln 1

( )paT

Tpp

p

T

p

T kk

oxk

x

k

x

k

k

348.2283979320

2731030

135..1

35.1

51

1

111

1

1 =

=

==

e x1= 0.907 (320 273) + 273*0.732 = 236.705 (kj/kg)

5

III.


4/12

Procesi (1-2) politropik.

Treguesi i politropes.

n = ? 10=

= ncc

ccn

vn

pn

Nxetesia specifike.

cn= ? )/(690.013.1

35.13.1672.0

1kgKkj

n

kncc vn =

=

=

i energjise se brendeshme.

u = ? u = cv (T2 T1) = 0.669 (870 320 ) = 362.95 (kj/kg)

i entropies.

s = ? s = )/(690.0320

870ln669.0ln

1

2 kgkjT

Tcn ==

i entalpise.

h = ? h = cp ( T2 T1) = 0.903 (870 320 ) = 416.65 (kj/kg)]

Sasia e nxetesis.

q = ? q = cn(T2 T1) = 0.690 (870 320) = 379.5 (kj/kg)

Puna e kryer.

?=l )/(7.11)870320(234.0110

1)(

1

112 kgkjTTR

nl =

=

=

Puna teknike.

?=tl )/(117)7.11(10 kgkjlnlt ===

6

Eksergjia e nxetsis.


5/12

exq = ? exq = q To s = 379.5 273*0.690 = 191.13 (kj/kgK)

T

T22

T11

eksergjia

To anergjia

Ss1

s2

Procesi (2 3), politropik.


n = ? 3.1

027.0

140.0log

900

736.7670log

log

log

4

3

3

2

3322====

v

v

p

p

nvpvp nn

Nxetesia specifike.

cn= ? )/(112.01

kgKkjcn

kncc pvn ==

=


u = ? u = cv (T3 T2) = 0.669 (539 - 870) = -214.439 (kj/kg)

i entropies.

s = ? s =)/(054.0

870

539ln112.0ln

2

3 kgKkjT

Tcp ==

i entalpise.

h = ? h = cn ( T3 T2) = 0.903 (539 - 870) = -278.893 (kj/kg)

7

Sasia e nxetesis.


6/12

q = ? q = cn(T3 T2) = -0.112(539 870 ) = 37.072 (kj/kg)

Puna e kryer.

?=l )/(18.258)539870(234.013.1

1)(

1

123 kgkjTTR

nl =

=

=

Puna teknike.

?=tl )/(634.33818.7583.1 kgkjlnlt ===


exq = ? exq = q To s = 37.072 273*0.112 = 22.33 (kj/kgK)

T

T22

T33

eksergjia

To anergjia

Ss3

s2

8

Procesi (3 4) adiabatik.


7/12


n = ? 35.1==

= kncc

ccn

vn

pn

Nxetesia specifike.

cn= ? 01=

=

n

kn

cc vn


u = ? u = cv (T4 T3) = 0.667 (320 -539) = - 148.54 (kj/kg)

i entropies.

s = ? s = 0

539

320ln0ln

3

4==

T

Tcn

i entalpise.

h = ? h = cp ( T4 T3) = 0.903 (320 - 539) = - 197.757 (kj/kg)

Sasia e nxetesis.

q = ? q = T s = 0

Puna e kryer.

?=l )/(417.146)320539(234.0135.1

1)(

1

143 kgkjTTR

kl =

=

=

Puna teknike.

?=tl )/(342.194417.14635.1 kgkjlklt ===


exq = ? exq = q To s = 0

9

Procesi (4-1) izotermik.


8/12


9/12

exq = ? exq = q To s = 2240.64 273 * 0.752 = 35.344 (kj/kg)

T

1 2

eksergjia

anergjia

s1 s4 S

Tabela nr. 2

Procesi n u(kj/kg) s(kj/kgK) h(kj/kg) q(kj/kg)

l

(kj/kg)t

l

(kj/kg) e

xq

(kj/kg)

(1-2) -10 362.95 0.690 466.65 379.5 11.7 -117 191.13

(2-3) 1.3 -214.439 0.054 -278.893 37.072 258.18 0 22.33

(3-4) 1.35 -148.54 0 -197.757 0 146.417 194.342 0

(4-1) 1 0 -0.752 0 -240.64 -240.64 -240.64 35.344

cikli - 0 0.008 0 175.932 175.657 175.606 -

Rendimenti termik i ciklit.

%2.42422.0573.416

657.175===

N

cc

tq

l

11

IV.

Eksergjia e dhene nga niveli i siperm


10/12

?=xqNe ( ) ( ) ==>>= )/(276.211752.0273572.41600 0 kgkjsTqexqN

Eksergjia e dhene nga niveli poshtem.

?=xqFe ( ) ( ) ==


11/12

Ndertimi i ciklit te barasvlefshem Karno.

( )

( ))(554

752.0

657.416

0

01 K

s

qTbv ==>

>=

( )

( ))(319

752.0

64.240

0

02 K

s

qTbv

==


12/12

p2=p3 2

pvn=cost

1

3

p4 4

q=0

dT=0

p1 4

v

v1 v3 v4

T

T2 2Pv-10=cost pv1.3=cost

3

q=0

T4 1 4

s2 s3 s1 s

h

pv-10=cost 2

h2 1 pv1.3=cost

h3 3

q=0

dT=0

h1 4

s1 s2 s3 s

Documents

termo 1' modeli 4