termo 2.modeli5.arjaniti.

8/9/2019 termo 2.modeli5.arjaniti.

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Fizik Teknike (Detyre kursi 2)

I. - Percaktimi i parametrave, per pikat karakteristike te ciklit, sipas diagrames <h-s>

• Per piken 1.

p1 = 24 (bar)

v1 = 0.08 (m3/kg)t1 = 220 oC

h1 = 2630 (kj/kg)

s1 = 5.9 (kj/kgK)

u1 = h1 – p1v1 = 2630 – 2400 * 0.08 = 2438 (kj/kg)

• Per piken 2.

p2 = 24 (bar)

v2 = 0.1 (m3/kg)

t2 = 250 oC

h2 = 2880 (kj/kg)

s1 = 6.4 (kj/kgK)

u2 = h2 – p2v2 = 2880 – 2400 * 0.1 = 2640 (kj/kg)

• Per piken 3

p3 = 5 (bar)

v3 = 0.49 (m3/kg)

t3 = 250 oC

h3 = 2960 (kj/kg)

s3 = 7.27 (kj/kgK)

u3 = h3 – p3v3 = 2960 – 500 * 0.49 = 2615 (kj/kg)

1




• Per piken 4.

p4 = 5 (bar)

v4 = 0.35 (m3/kg)

t4 = 150 oC

h4 = 2540 (kj/kg)s4 =6.37 (kj/kgK)

u4 = h4 – p4v4 = 2540 – 500 * 0.35 = 2365 (kj/kg)

Tabela nr.1

Gjendja p

(kpa)

v

(m

3

/kg)

T

(K)

h

(kj/kg)

s

(kj/kgK)

u

(kj/kg)1 2400 0.08 493 2630 5.9 2438

2 2400 0.1 523 2880 6.4 2640

3 500 0.49 523 2960 7.27 2715

4 500 0.35 423 2540 6.37 2365

2




II . – Paraqitja skematike e ciklit ne diagramat <p-v>, <T-s>, <h-s>.

p

<p – v > K

p= cost

1 2 t=cost

4 3

x= 0 p=cost x= 1

x= 0.9

v

TK t= cost

<T – s >

p= cost 2 3

2

1

p=cos

4

x= 0 x= 1

x= 0.9

s

h

t= cost

<h – s > p= cost3

2

K 1

p=cost

x= 0 4 x= 1

x= 0.9

s

3




- Per cdo process te ciklit te llogariten:

• Procesi (1-2), proces izobarik.

• # i energjise se brendeshme.

Δu = ? Δu = u2 – u1 = 2640 – 2438 = 202 (kj/kg)

• # i entalpise.

Δh = ? Δh = h2 – h1 = 2880 – 2630 = 250 (kj/kg)

• # i entropies

Δs = ? Δs = s2 – s1 = 6.4 – 5.9 = 0.5 (kj/kgK)

• Puna e procesit.

?=l )/(48)08.01.0(2400)( 12 kg kjvv p pdvl =−⋅=−⋅== ∫

• Puna teknike.

?=t

l

∫ =−= 0vdpl t

• Sasia e nxetesise

q = ? q = Δh = h2 – h1 = 2630 – 2880 = 250 (kj/kg)

• Procesi (2-3), proces izotermik.


Δu = ? Δu = u3 – u2 = 2715 - 2640 = 75 (kj/kg)

• # i entalpise.

Δh = ? Δh = h3 – h2 = 2960 – 2880 = 80 (kj/kg)

• # i entropies

Δs = ? Δs = s3 – s2 = 7.27 – 6.4= 0.87 (kj/kgK)

4





?=l )/(01.3807501.455 kg kjuql =−=∆−=

• Puna teknike.

?=t l )/(01.3758001.455 kg kjhql t =−=∆−=


q = ? q = T Δs = 523 *0.87= 455.01 (kj/kg)

• Procesi (3-4), proces izobarik


Δu = ? Δu = u4 – u3 = 2365 – 2715 = -350 (kj/kg)

• # i entalpise.

Δh = ? Δh = h4 – h3 = 2540 – 2960 = - 420 (kj/kg)

• # i entropies

Δs = ? Δs = s4 – s3 = 6.37 – 7.27 = - 0.87


?=l )/(70)49.035.0(500)( 12 kg kjvv p pdvl −=−⋅=−⋅== ∫

• Puna teknike.

?=t l ∫ =−= 0vdpl t


q = ? q = Δh = h4 – h3 = 2540 – 2960 = - 420 (kj/kg)

• Procesi (4-1), proces “izotermik”.

5





Δu = ? Δu = u1 – u4 = 2438 - 2365 = 73 (kj/kg)

• # i entalpise.

Δh = ? Δh = h1 – h4 = 2630 – 2540 = 90 (kj/kg)

• # i entropies

Δs = ? Δs = s1 – s4 = 5.9 – 6.37= - 0.47 (kj/kgK)


?=l )/(26.28873)26.215( kg kjuql −=−−=∆−=

• Puna teknike.

?=t l )/(26.30590)26.215( kg kjhql t −=−−=∆−=


q = ? q = T Δs = 458 *(- 0.47) = -215.26 (kj/kg)

Tabela nr.2

Procesi Δu

(kj/kg)

Δh

(kj/kg)

Δs

(kj/kgK) )/( kg kj

l

)/( kg kj

l t q

(kj/kg)

(1-2’) 202 250 0.5 48 0 250

(2’-2) 75 80 0.87 380.01 375.01 455.01

(2-3) -350 -420 -0.9 -70 0 -420(3-1) 73 90 -0.47 -288.26 -305.26 -215.26

cikli 0 0 0 69.75 69.75 69.75

IV. -Te llogariten:

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• Puna e cikit

?=cl ( ) ( ) ( ) )/(75.6926.2887001.380481443)32(21 kg kjl l l l l c =−−+=+++= −−−−

• Rendimenti temik i cklit.

?=c

t η %9.9099.001.705

75.69≈===

N

cc

t

q

l η

• Rendimenti termik i ciklit Karno qe punon ne temp ekstreme.

?=k

t η %2.19192.0523

423523≈=

−

=

−

=

N

F N k

t T

T T η

• Paraqitja grafike e ciklit Karno ne <T – s >

T K

1’ 2 3

1

4’ 4 3’

x=0 x=1

x=0.85

s

• Eksergjia e nxetesise e fituar nga burimi i nxete.

?= xqN e ( ) ( )∑∑ =⋅−=>∆−>= )/(33137.127301.70500 0 kg kj sT qe xqN

7




• Paraqitja e eksergjise ne <T – s >.

T

K 2

T 1

exq

To

anergjia

s

V. - Te krahasohen te dhenat e nje pike te nxierra nga tabela me ato te nxierra nga

diagrama entalpi – entropi.

Sipas tabeles.

x = 0.92

h’ = 192 (kj/kg) hx = h’ + x (h” – h’) = 192 + 0.92 (2584 – 192 ) = 2393 (kj/kg)

h” = 2584 (kj/kg)

hx = ?

s’ = 0.6492 (kj/kgK) sx = s’ + x (8.1502 – 0.6492) = 7.51 (kj/kgK)

s” = 8.1502 (kj/kgK)

sx = ?

v’ = 0.0010103 (m3/kg) vx = v’ + x (v” – v’) = 0.0010103 + 0.92 (14.68 – 0.0010103)

v” = 14.68 (m3/kg) vx = 13.5 (m3/kg)

vx = ?

u’ = 192 (kj/kg)

u” = 2438(kj/kg) ux = u’ + x (u” – u’) = 192 + 0.92 ( 2468 – 192) = 2300 (kj/kg)

ux = ?

Sipas didagrames <h – s >.

8




x =0.92

hx = 2400 (kj/kg)

sx = 7.51 (kj/kgK)

vx = 13 (m3/kg)

ux = hx – pxvx = 2400 – 10 * 13= 2270 (kj/kg)

Tabela nr.3

hx

(kj/kg)

sx

(kj/kgK)

vx

(kj/kg)

ux

(kj/kg)

2400 7.51 13 2270 Sipas diagrames <h-s>

2393 7.51 12.5 2.30 Sipas tabeles

VI. - Ndertimi i vijave karakteristike te proceseve per piken x.

p

<p-v> K

9




v = cost

δq =0

p = cost

xt = cost

x=1

x=0 x= 0.92

v

T

<T-s>

K

p = cost

x

t = cost

v = cost

x=0.92

δq =0

p = cost

h

v = cost

<h-s>

t = cost

K

δq =0

s

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