termo 3. modeli 4. arjaniti

8/9/2019 termo 3. modeli 4. arjaniti

1/12

Detyre kursi nr. 3 (Fizik Teknike) Arjanit

Ushrimi I.

- Te dhena:

tg = t1 = 800oC 1 = 35 (w/m

2K)

tu = t2 = 90oC 2 = 1750 (w/m

2K)

c = 8 mm c = 50 (w/mK)

d1 = 80 mm bl = 0.12 (w/mK)bl = 1.5 mm zm = 2 (w/mK)

zm = 1.5 mm

- Te percaktohen:

1.

a) Rezistenca termike.

t1

tm1tm2

tm3 tm4

t2

bl c zm

R = ? = iRR

)/(04255.02

105.1

50

108

12.0

105.1

35

111 2333

21

WKmRzm

zm

c

c

bl

bl

+

+

+=++++=

b) Koficientin e transmetimit te nxetesis.

?=k )/(5.2304255.0

11 2KmWR

k ===

1


2/12


c) Koficienti i barasvlefshem i perciellshmeris temike

?=bv )/(8203.0

25012.0

105.1108105.1

105.1108105.1333

333

mKWRRRR zmcbl

zmcbl

i

i

bv =

++

++

++=

++

++==

d) Rryma termike specifike

?=q )/(686.1604255.0

90800 221 mkWR

ttq

i

===

2. Temperaturat ndermjetese te murit.

CqRttR

ttq

o

m

m 32302857.016686800111

1

11 ===

=

CRRqtto

m 115)0125.002857.0(16686800)( 2112 =+=+=

CRRRqtt cblm 0113 112)00016.00125.002857.0(16686800)( =++=++=

CRqtto

m 10000057.01668690224 =+=+=

3. Paraqitja grafike e shperndarjes se vijave te temperatures.

t

t1

tm1

tm2 tm3

tm4

bl c zm t2

4. Realizimi i zgjidhjes per murin cilindrik.

2


3/12


t1

tm1

tm2tm3

tm4t2

dod1

d2

d3

- Percaktimi i diametrave.

do= d1 2zm = 80 2 * 1.5 = 77 mmd2= d1 + 2c = 80 + 2 * 8 = 96 mm

d3= d2 + 2bl = 96 + 2 * 1.5 = 99 mm

1. Te percaktohet:

a) Rezistenca termike.

?=R

)/(13876.0

17507714.3

1

77

80ln

214.32

1

80

96ln

5014.32

1

96

99ln

12.014.32

1

35107714.3

1

11

ln2

1

ln2

1

ln2

11

2

3

0201

2

2

3

13

WKmR

R

dd

d

d

d

d

d

dR zmcbl

=

+

+

+

+

=

=++++

=

b) Koficienti i transmetimit te nxetesise.

3


4/12


?=k )/(207.713876.0

11 2KmW

Rk ===

c) Koficenti i barasvlefshem i perciellshmeris termike.

?=bv

)/(21.0

77

80ln

2

1

80

96ln

50

1

96

99ln

12.0

177

99ln

ln1

ln1

ln1

ln

0

1

1

2

2

3

0

3

mKW

d

d

d

d

d

d

dd

bv

zmcbl

bv

=

++

=

++

=

d) Rryma termike specifike.

?=q )/(381.313876.0

90800 221 mkWR

ttq

i

=

=

=

2. Temperaturat ndermjetese te murit.

CqRttR

ttq

o

mm 48909191.03381800

111

1

11 ===

=

CRRqtto

blm 351)04083.009191.0(3381800)( 112 =+=+=

CRRRqtt cblm0

113 349)00058.004083.009191.0(3381800)( =++=++=

CRqtto

m 980024.0338190224 =+=+=

Ushtrimi 2.

4


5/12


- Te dhena:

Mu = 150 (kg/h) = 0.042 (kg/s) 1 = a = 800 (W/m2K)

tu = t2 = 24oC 2 = u = 1600 (W/m

2K)

tu = t2 = 34oC kk= 2000 (W/m

2K)

kf = 400 (W/m2K)

pav = 7 bar c = 59.3 (W/mK)

tav = 240 oC Rasti i pare.

70

80

1

2=

d

d

- Te percaktohet:

1. Sasin e avullit te nevojshem per ngrohjen e ujit.

Mav = ? Qu = Qav

Qav = Qk1 + Qf

2 = Qu

Mav [ (hA hB) + (hB hC ) = Mu cpu (t2 t2)

)(

)( '2"2

CA

cpuu

avhh

ttcMM

=

h

hA A

hB K B

hC C x=1 x=0

sC sA sA s

5


6/12


- Nga tabelat per presionin dhe temperaturen gjejme:

pav = 7 bar hA = 2958 (kj/kg)

tav = 240 oC hC = 697 (kj/kg)hB = 2764 (kj/kg)

tB = tC =165oC

)/(8.2)/(00078.0)6972958(

)2434(19.4042.0hkgskgMav ==

=

2. Siperfaqia kembysit te nxetesise per:

a) rryma me drejtime te njejta.

mestFkQ = dhe21

fk FFF +=

- ne zonen e kondesimit.

)(151.0)/(151.0)27642958(00078.0)(1 kWskjhhMQ BAk

av ====

CcM

hhMttttcMhhM o

puu

BAav

xxpuuBAav

2519.4042.0

)27642958(00078.024

)()()(

'22'22

=

+=

+==

C

tt

tt

ttttt

o

xB

A

xBA

mes177

25165

24240ln

)25165()24240(

ln

)()(

2

'2

2'2 =

=

=

)(00043.01772000

151 211 mtk

QF

mesk

k =

=

=

- ne zonen e ftohjes

)/(608.1)/(608.1)6972764(00078.0)(2 skWskjhhMQ CBav ====

6


7/12


tt

tt

ttttt

C

B

CxBmes

3165

2165ln

16()25165(

ln

()(

"2

"2

"22

=

=

)(0064.01262000

1608 222 mtk

QF

mesf

f =

=

=

)(00683.00064.000043.0221

mFFF fk =+=+=

b) per rryma me drejtime te kunderta.

mestFkQ = dhe21fk FFF +=

- ne zonen e kondesimit.

C

tt

tt

ttttt

o

B

xA

BxA

mes175

24165

25240ln

)24165()25240(

ln

)()(

'2

2

'22 =

=

=

)(00043.01752000

151 211

mtk

Q

Fmesk

k ===

- ne zonen e ftohjes

C

tt

tt

ttttt

o

C

xB

xCB

mes126

25165

34165ln

)25165()34165(

ln

)()(

"2

2

22" =

=

=

)(032.0126400

1608 222 mtk

QF

mesf

f =

=

=

)(03243.0032.000043.0221

mFFF fk =+=+=

3. paraqitja e skemes per shkembimin e nxetesise.

a) Per rryma me drejtime te njejta.

7


8/12


b) Per rryma me drejtime te kunderta.

4. Paraqitja grafike e shperndarjes se temperaturave ne diagramen < t F>

a) per rryma me drejtime te njejta.

t

tA

tB tC

t2t2x

t2

F


t

tA

8


9/12


tB tC

t2

t2x t2

F

Ushtrimi 3.

- Te dhena:

t1 = 120oC cpv =cp1 = 3.05 (kj/kgK)

t1 = 95oC cpu = cp2 = 4.19 (kj/kgK)

Mv = 120 (kg/h) = 0.033 (kg/s) k =1000 (W/m2K)

Mu = 85 (kg/h) = 0.024 (kg/s)

tu = t2 = 21

o

C

- Kerkohet:

1. Siperfaqia ftohese per rastet:

a. Rryma me drejtime te njejta.

F = ? mestk

QF

=

Qv = Qu

Mv cp1 (t1 t1) = Mu cp2 (t2 t2)

)/(5 3 9.2)1 2 09 5(0 5.30 3 3.0)('1"11

sk jttcMQpvv

====

CcM

QttttcMQQ

pu

v

puuv

0

3

3

2

'2"2'2"22 471019.4024.0

10539.221)( =

+=+===

C

tt

tt

ttttt

o

mes 71

4795

21120ln

)4795()21120(

ln

)()(

"2"1

'2'1

"2"1'2'1 =

=

=

9


10/12


)(0358.0711000

2539 2mtk

QF

mes

=

=

=

b. Per rryma me drejtime te kunderta.

tt

tt

tttttmes

9

1ln

47120(

ln

)()(

'2"1

''2'1

'2"1"2'1

=

=

)(0343.0741000

2539 2mtk

QF

mes

=

=

=

2. Paisjet per kembimin e nxetesise.

a) Per rryma me drejtime te njejta.

b) Per rryma me drejtime te kunderta.

10


11/12


3. Paraqitja grafike e shperndarjes se temperaturave ne diagramen < t F>

a) per rryma me drejtime te njejta.

t

t1

t1

t2

t2

F


t

t1

11


12/12


t2 t1

t2

F

4.

12

termo 3. modeli 4. arjaniti

Documents

Transcript of termo 3. modeli 4. arjaniti