Two phase flow assignment

7/23/2019 Two phase flow assignment

http://slidepdf.com/reader/full/two-phase-flow-assignment 1/24

Q1. Derive Bowring’s correlation explaining the theory of dry out and DNB in flow

boiling in brief.

Critical heat flux is that value of heat flux at which boiling crisis occurs. Boiling crisis inflow boiling is of two types, as follows,

i) Departure from Nucleate boiling: - his phenomenon is observe! at high heat flux

an! low "uality. #ere !ue to high heat flux number of bubble nucleation sites

are high. $fter reaching a certain heat flux value this !ensity becomes so high that

a!%acent bubbles coalesce to form a continuous film on the wall, preventing

further nucleation. Due to this reason cooler water from surroun!ings cannot

come in contact with hot wall causing wall temperature to rise an! heat transfer

coefficient to fall.

ii) Dry out:- his phenomenon is observe! at relatively lower heat flux an! very high

"uality. his is because in higher vapour mass fluxes, the flow regime becomes

annular an! at a certain tube length water &!ries out' from the walls.

Bowring assume! C#( to be linear function of "uality, as follows

qc=C 1+C

2 x

¿

..i)

*hereC

1, C

2=f ( G , d , p )

+ mass flux ! tube !iameter p system pressure

$pplying energy balance at inlet an! outlet of the tube we get

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b) qc × d × Z × π =m× (h ( z )−h i)

m× (h f + x¿ ×h fg−hi )

qc × d × Z × π =m× C pl × ∆T ¿+ m× x

¿× hfg

is the tube length, hi is enthalpy of water at inlet

x¿

is "uality at , h ( z ) is local flui! enthalpy at

∆ T ¿=hf −hi=degreeof subcoolingat inlet

m= Mas s flow rate=G × π /4 ×d2

x¿= q c × Z ×4

G × d × h fg

− C pl × ∆ T ¿hfg

/ubstituting the value of x¿

in e"uation i) we get

qc=C 1+C

2(

qc × Z ×4

G × d × hfg

−C pl × ∆ T ¿

hfg

)

qc(1−C 2 × Z ×4

G × d × hfg )=C 1−C 2 ×

C pl × ∆ T ¿h fg

C

(¿¿1−C 2

× C pl × ∆ T ¿

h fg

)

(1−C 2

× Z × 4

G × d × hfg

)

qc=¿

qc=(C

1

C 2

× hfg−C pl × ∆ T ¿)( hfg

C 2

− Z ×4

G × d )

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qc=

( × hfg−C pl × ∆ T ¿)

(! × hfg− Z ×4

G ×d )

*here = C 1

C 2 != 1

C 2

$ an! B are given as follows

=2.317× " 1

1+0.0143× " 2×√ d ×G

!=0.308× " 3/( hfg)

1+0.347× " 4×( G1356 )

n

(0, (1, (2 an! (3 are constants !epen!ent on pressure.

Q2. Water is flowing through a uniforly heated vertical tube of 1 length and !D 2"

with ass flow rate of #.1 $g%s. aount of sub cooling is &#' and syste

pressure is 1# bar. (alculate

a) (ritical heat flux at exit andb) *ocal +uality x, at exit and hence deterine the type of boiling crisis.

(rom chart for Bowring4s correlation for 5 06 bar we get

(06.378, (16.991, (26.3, (36.6099,

n 1-.6.6671;5 1-.6.6671;06 0.<17

=ass flux

G= m

π 4

× d2

= 0.1

π ×0.25×0.025

2=203.72

#g

m2

s

(rom Bowring4s correlation we get C#( :-qc=

(h fg × +∆ h¿)

(hfg × !+Z × 4

G ×d)

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*here h fg=2013.6 $% /#g

,

∆ h¿=C pl × ∆ T ¿=4.45×103×30=133.56 $& /#g

=2.317× " 11+0.0143× " 2× √ d ×G

=2.317× 0.478

1+0.0143×0.662×√ 0.025×203.72=0.848

!=0.308× " 3 /( hfg)

1+0.347× " 4×( G

1356)

n=0.308×

0.4 /(2013.6×103)

1+0.347×0.0166×(203.72

1356)1.9275

=6.117×10−8

∴qc= (2013×10

3×0.848+133.56×10

3 )

(2013

×10

3

×6.117

×10

−8

+1

×

4

203.72×0.025 )

=2.03 M' /m2

a)

c) qc × d × Z × π =m× (h ( z )−h i)

m × (h f + x¿ ×h fg−hi )=m×C pl × ∆ T ¿+ m× x¿ ×h fg

x¿=qc × d× Z× π

m ×h fg

−C pl × ∆ T ¿

h fg

=2.03×10

6×0.025×1× π

0.1×2013.6×103

−133.56×10

3

2013.6×103

x¿=0.725( )ere C)" leads¿ *+, -.T

Q&. -uppose the tube in Q1 is aintained at constant wall teperature of 12# o( and

delivering sae exit +uality. (alculate whether heat flux now is ore or less than

that calculated in previous proble.

$s the flow is in annular region, convective heat transfer !ominates over nucleate boiling

heat transfer. (rom Chen4s correlation we get

h2∅

=hnb+hcon/ection

'here

hnb=0 × h "Z

)ereh "Z =0.00122× ∆ T sat

0.24× ∆ 1sat 0.75 × ∆ 1sat

0.75× C pl0.45× 2l

0.49×# l0.79

3 0.5×h fg0.24× 4 l

0.29× 2g0.24

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Now ℜ2∅=ℜl× " 1.25

*hereℜl=

G4l

× (1− x¿) × d= 203.72

28.3×10−5

× (1−0.725 )×0.025=4949

>oc?hart-=artinelli parameter is given by

5 tt =( 1− x¿

x¿ )

0.9

×( 4l

4g )0.1

×( 2 g

2l )

0.5

=(1−0.725

0.725 )0.9

×( 28.3×10−5

12.06×10−6 )

0.1

×( 0.6958 )0.5

=0.0143

(rom ( vs 1/ 5 tt graph we get (7

∴ℜ2∅=ℜl × " 1.25=4949×75

1.25=1.09×106

¿ 0 /s

ℜ2∅ graphwe get 060

/o for the given con!ition we can safely neglecthnb term from total heat transfer term.

∴

h2∅=hcon/ection= " ×hcon/

Convective coefficient can be !etermine! by Dittus-Boelter e"uation

hcon/=# l ×0.023ℜ0.8× 1r

0.4/d 0.68×0.023×4949

0.8× 1.75

0.4

0.025=706.52

'

m2− $

∴ h2∅= " × hcon/=75×706.52=52.989

$'

m2− $

)eat flux q=h2∅× (T wall−T sat )=52.989×103 × (150−100 )=1.06 M' /m2

∴ @n this case the heat flux will be less than the critical heat flux. hus to get the

same "uality 1n! arrangement is safer from !esign consi!erations.

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Q. / hori0ontal tube of 1 length and 2" !D contains flow of water at 1at

pressure with ass flux 1### $g%2s. !nitial sub cooling is "#'.

a) 3ind axiu x¿

at (43 below which stratification is inevitable. 3ind the

corresponding (43.b) 3ind iniu x

¿

at (43 above which stratification is avoided. 3ind the

corresponding (43.

a) /tratification is inevitable at (r A 2 2

2g ×(¿¿ l−¿ 2 g)× g × d

1

¿ "r= x¿

cr ×G ×√ ¿

2

(¿¿ l−¿ 2 g)× g × d= 3

1000×√ 0.6×(958−0.6)×9.81×0.025=0.036

2g ×¿

∴ x¿

cr= "r

G × √ ¿

qc=G × d

4× Z × (C pl × ∆ T ¿ + x

¿× h fg)=10

3×0.025

4×1× (4.18×50+0.036×2257 )×10

3=1.81 M' /m2

b) /tratification is avoi!e! at (r 7 2

2g ×(¿¿ l−¿ 2 g)× g × d

1

¿ "r= x¿

cr ×G ×√ ¿

2

(¿¿ l−¿ 2 g)× g × d= 7

1000×√ 0.6×(958−0.6)×9.81×0.025=0.083

2g ×¿∴ x¿

cr= "rG

× √ ¿

qc=G × d

4× Z × (C pl × ∆ T ¿ + x

¿× h fg)=10

3×0.025

4×1× (4.18×50+0.083×2257 )×10

3=2.48 M' /m2

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Q". Water is flowing through a uniforly heated vertical tube of 1 length and !D 1#

c with ass flux of 1#& $g%25s. 3ind the re+uired aount of inlet sub cooling

to avoid DNB at exit where the flow is saturated. -aturation teperature 2## o(.

5 0.3< bar at sat 166 oC

(rom chart for Bowring4s correlation for 5 0.3< bar we get

(06.378, (16.69, (26.3, (36.6239

n 1-.6.6671;5 1-.6.6671;0.3< 0.887

*here h fg=1938.5 $% /#g

,

=2.317× " 1

1+0.0143× " 2× √ d ×G=2.317×

0.478

1+0.0143×0.506×√ 0.1×103=0.1456

!=0.308× " 3 /( hfg)

1+0.347× " 4×( G1356

)n=0.308×

0.4 /(1938.5×103)

1+0.347×0.0346×(1000

1356)1.887

=6.313×10−8

q c × d × Z × π = m× C pl × ∆T ¿+ m× x¿

× hfg .here x 6

qc= m ×C pl × ∆ T ¿

d × Z × π

=7.85× ∆ h¿

Now,qc=

(h fg × +∆ h¿)

(hfg × !+Z × 4

G ×d)=7.85× ∆ h¿

∴qc= (1938.5×10

3× 0.1456+∆ h¿ )

(1938.5×103

×6.313× 10−8+1×

4

1000×0.1)=7.85× ∆ h¿

/olving the above e"uation we get :-∆ h¿=24029.08

∴∆ T ¿=∆ h¿

C pl

=24029.08

4.5×103 =5.34 $

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Q6. Water is flowing through a uniforly heated vertical tube of 1 length and !D 2"

with ass flux of 1#& $g%25s. Degree of inlet sub cooling 2#'. 3ind out the

heat flux re+uired to get outlet +uality of x, #.2. !f boiling crisis occurs then find

out the distance at which it starts. -yste pressure 1 bar. /lso find out +uality atboiling crisis 7if any).


(06.378, (10.781, (26.3, (36.6663

n 1-.6.6671;5 1-.6.6671;0 0.<<17

*here

h fg=2257 $% /#g,

=2.317× " 1

1+0.0143× " 2× √ d ×G=2.317×

0.478

1+0.0143×1.782×√ 0.025×103=0.22

!=0.308× " 3/( hfg)

1+0.347× " 4×( G1356

)n=0.308×

0.4/ (2257×103)

1+0.347×0.0004×(1000

1356)1.99275

=5.45×10−8

Critical heat fluxqc= (h fg × +∆ h¿)

(hfg × !+Z × 4

G ×d)

∴qc= (2257×10

3×0.22+4.18×10

3×20)

(2257×103

×5.45×10−8+1×

4

1000×0.025)=2.05 M' /m

2

#eat flux re"uire! to obtain given "uality at exit :-

q x=

G× d

4× Z × ( C pl ×∆ T ¿+ x

¿×h fg )

10

3×0.025

4×1× (4.18×10

3 ×20+0.2×2257×103 )=3.34 M' /m2

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q x >C)"( Therefore boilingcrisis occurs

>et # be the height at which DNB occurs for the heat fluxq x

(2257×103 ×0.22+4.18×103×20)(2257×10

3×5.45×10

−8+ ) × 4

1000×0.025)=3.34 M' /m

2

/olving the above e"uation we get # 6.209 m

herefore "uality at # x

¿=4×q x × )

G × d × hfg

−C pl × ∆ T ¿

hfg

4×3.34×10

3

×0.3156

1000×0.025×2257×103−4.18×10

3

×20

2257×103 =0.038

Q8. Water is flowing through a uniforly heated vertical tube of 1 length with ass

flux of 1#& $g%25s. Degree of inlet sub cooling "#'. 3ind out the tube diaeter

re+uired to avoid dry out if saturated stea is to be obtained at outlet. -yste

pressure " bar.

(rom chart for Bowring4s correlation for 5 bar we get

(06.378, (10.60<, (26.3, (36.662

n 1-.6.6671;5 1-.6.6671; 0.<927

*here h fg=2107 $% /#g

,

=2.317× " 1

1+0.0143× " 2× √ d ×G=2.317×

0.478

1+0.0143×1.019×√ d ×103=

1.107526

1+14.57× 7 d

ME11M037



!=0.308× " 3/( hfg)

1+0.347× " 4×( G

1356)

n=0.308×

0.4/(2107×103)

1+0.347×0.0053×(1000

1356)1.99275

=5.84×10−8

Critical heat fluxqc= (h fg × + ∆ h¿)

(hfg × !+Z × 4

G ×d)

$gain we ?now that,

q x=G × d

4× Z × ( C pl ×∆ T ¿+ x¿ ×h fg )=10

3 ×d

4×1× (4.316×50+2107 )×10

3

¿580.7×106

×d

∴qc=(2257×10

3×

1.107526

1+14.57×√ d+4.18×10

3×20)

(2257×103

×5.45×10−8+1×

4

1000× d ) =580.7×10

6× d

earranging the above e"uation we get

d i+1=((

0.033

0.0295+di −1

)/14.57

)

2

i d i d i+1

0 0 3.30;06E-2

1 3.30;06E-2 2.308;06E-9

2 2.308;06E-9 9.909;06E-

3 9.909;06E- 9.23;06E-

9.23;06E- 9.293;06E-9 9.293;06E- 9.293;06E-

he re"uire! !iameter is 6.6929 mm.

@n practical engineering purposes the tube !iameters are much higher. herefore

!ry out always occurs un!er given con!itions.

ME11M037



Q9. Water is flowing through a uniforly heated vertical brass tube 7 1 length: inlet

diaeter 2" ) with ass flux of 1#& $g%25s. Degree of inlet sub cooling "#'.

Deterine the axiu degree of sub cooling re+uired to avoid dry out if

saturated stea is to be obtained at outlet. -yste pressure 1# bar. 4eat transfer

coefficient 1#'W%2 and elting point of brass ;&# o(.

(or this we have to fin! C#( at 0 for wFG0) <26 ℃

sat 086 oC at 5 06 bar

∴ 8nlet temperature T i=180−50=130℃

(rom chart for Bowring4s correlation for 5 bar we get

(06.378, (10.60<, (26.3, (36.662

n 1-.6.6671;5 1-.6.6671; 0.<927

*here

h fg=2107 $% /#g

,

=2.317× " 1

1+0.0143× " 2× √ d ×G=2.317×

0.478

1+0.0143×1.019×√ 0.025×103=0.335

!=0.308× " 3/( hfg)

1+0.347× " 4×( G

1356)

n=0.308×

0.4/(2107×103)

1+0.347×0.0053×(1000

1356)1.96375

=5.84×10−8

Critical heat fluxqc= (h fg × + ∆ h¿)

(hfg × !+Z × 4

G ×d)

$gain minimum heat flux re"uire! to raise the wall temperature to <26 ℃ is

ME11M037



q x= (T w ( z )−T

1)

(1/h+Z × 4

G × d × C pl

)=

(∆T ¿ ( z )+∆ T ¿)

(1/h+ Z × 4

G × d × C pl

)

Now, qc=q x

(2107×103

×0.335+4.316×103

× ∆ T ¿)

(2107×103

×5.84×10−8+1×

4

1000×0.025 )=

(750+ ∆ T ¿ )

( 1104 +1×

4

1000×0.025×4.316×103 )

2107×0.335+4.316×∆ T ¿=2.063× ∆ T ¿+1547.98

∆ T ¿=373.8 $

his ma?esT i=−193.8 $

@n practiceT i≫−193.8 $

so the tube will never fail un!er given con!itions.

Q;. / 1 long and 2" diaeter uniforly heated vertical tube delivers water at 1

bar and ass flux of 1### $g%25s with outlet +uality is to be $ept at 1#<. Degree

of inlet sub cooling "#'. 'eeping other paraeters sae: find out up to what

pressure liit the syste can wor$ without causing burnout.

#eat flux re"uire! to pro!uce the given "uality of steam at the outlet at 0bar:-

q x=G × d4× Z

× ( C pl ×∆ T ¿+ x¿ ×h fg )=103 ×0.025

4×1× (4.18×50+2257 )×10

3=2.717 M' /m2

(or the given problem we have to fin! out the operating pressure at which

q x 1.707 =*HmE1 is the C@@C$> #I$ (>JK

ME11M037



=2.317× " 1

1+0.0143× " 2× √ d ×G

!=0.308× " 3/( hfg)

1+0.347× " 4×( G1356

)

n

n=2−0.00725× 1

Critical heat fluxqc=

(h fg × + ∆ h¿)

(hfg × !+Z × 4

G ×d)

∆ h¿=C pl × ∆ T ¿=4.18×10

3

×50=209 $% / #g

= 31 32 3& 3 / B>1#59 h fg

'?%$g

qc

@W%

2

0 6.378 0.781 6.3 6.6663 6.11 .3 117 1.3<

6.378 0.60< 6.3 6.662 6.22 .83 1067 2.19

1.82 6.378 0.322 6.3 6.66193

6.190 .97 1076 1.73

>inearly interpolating between 5 1.82 an! 5 0 bar we get qc=2.717 M' /m2

at 5 1.99 bar.

Q1#. -aturated water flows through a vertical tube of 1 length and !D of 2" . @ass

flux is 1###$g%25s. (alculate

a) Axit +uality at critical heat flux.

b) 4eat transfer coefficient at DNB.

ME11M037




(06.378, (10.781, (26.3, (36.6663

n 1-.6.6671;5 1-6.6671; 0.<<17

*here h fg=2257 $% /#g ,

=2.317× " 1

1+0.0143× " 2× √ d ×G=2.317×

0.478

1+0.0143×1.782×√ 0.025×103=0.22

!=0.308× " 3/( hfg)

1+0.347× " 4×( G

1356)

n=0.308×

0.4/ (2257×103)

1+0.347×0.0004×(1000

1356)1.99275

=5.46×10−8


(h fg × + ∆ h¿)

(hfg × !+Z × 4

G ×d)

qc= (2257×10

3×0.22+0 )

(2257×103 ×5.46×10

−8+1× 4

1000×0.025)=1.753×10

6' /m2

$gain, x¿=4×q

x

× )

G × d × hfg−

C pl

× ∆ T ¿

hfg=

4

×1.753

×10

6

×1

1000×0.025×2257×103 −0=0.124

(or saturate! boiling heat transfer coefficient is given by the sum of nucleate boiling an!

convection, as followsh2∅=hnb+hcon/

Convective coefficient can be !etermine! by Dittus-Boelter e"uation

9u=hcon/

# l×d =0.023ℜ0.8

× 1r0.4

'here ℜ=

G4l

× (1− x¿ )× d= 1000

28.3×10−5

× (1−0.124 ) ×0.025=77385.16

ME11M037



∴hcon/=# l ×0.023 ℜ0.8× 1r0.4

d =0.6804×0.023×77385.16

0.8× 1.75

0.4

0.025

9278.02 *HmE1-L

$gain for nucleate boilingh2∅

hnb

=29 ℜ−0.3× "r0.2

'here "r= G

2

2l

2× g × d

× (1− x¿)2= 10

6

9582

×9.81×0.025× (1−0.124 )2

2.36<

∴hnb= h

2∅

29× ℜ−0.3× "r

0.2=

h2∅

29×77385.16−0.3

×3.4090.2

=0.79h2∅

∴h2∅=0.79h

2∅+6378.13

¿>h2∅

=6378.13

1−0.79=30.384

$'

m2− $ .. Ans

Q11. Water is flowing through a uniforly heated hori0ontal tube 7 1 length: inlet

diaeter "# ) with ass flow rate of #.1 $g%s. Degree of inlet sub cooling &#'.

Deterine the condition of the flow at (43 & @W%2. -yste is at 1 bar.

$t 0 bar, sat 066 ℃ ,

2g=0.6#g /m3

2l=958#g/m3

h fg 117 LMH?g

$pplying energy balance at inlet an! outlet of the tube

qc × d × Z × π =m× C pl × ∆T ¿+ m× x¿

× hfg

ME11M037



x¿=qc × d× Z× π − m ×C pl × ∆ T ¿

m× hfg

=3×10

6×0.05×1× π −0.1×4.18×103×30

0.1×3×106

=0.153

$gain,

2 2g ×(¿¿ l−¿ 2g)× g × d

1

¿

"r= x¿

cr × m

π d

2

4

×√ ¿

¿0.153×

0.1

π

0.052

4

×√ 1

0.6×(958−0.6)×9.81×0.025

3.9

∴ 2 (r 7

i.e. Both inertia an buoyancy forces are comparable.

O Not very prominently stratifie!.

Q12. 11& is flowing through a uniforly heated vertical tube 7 1 length: inlet diaeter

2" ) with ass flux of 1## $g%25s. Degree of inlet sub cooling "#'. Deterine

(43. -yste pressure 1 bar.

Civen:

2g=7.38 #g /m3

2l=1511 #g/m3

h fg 18 '?%$g

4g=1.1×10

−5 #g

m−s 4l=5.14×10

−4 #gm−s

C pl ;9&.9 ?%$g

3 =15.9×10−3 9 /m

ME11M037



#ere we use $hma!4s scaling law. *e assume the tube !imensions to be same.

i) ( 2g

2l )

water

=( 2g

2l )

refr

=4.884×10−3

(rom steam tables we get this ratio ( 2g

2l ) for water occurs at 5 P 8.9 bar

F sat 072.3 ℃¿

∴ (rom Bowring4s table we get

(06.378, (16.980, (26.3, (36.601

n 1-.6.6671;5 1-6.6671;8.9 0.<279

h fg 1629.1 LMH?g

2gw=4.35 #g /m3

2lw=890#g / m3

4gw=7.55×10

−5 #g

m−s

4lw=1.54×10−4 #g

m−s 3 w =0.05 9 /m

ii) ( ∆ h¿

h fg )

water

=( ∆ h¿

hfg )

refr

¿>( ∆ h¿)water=( ∆ h¿

hfg )

refr

× (h fg )water=68.14×10

4 $% / #g

iii) ∅=

(

G × d

4l

×

[( 4l

2

3 × d × 2l )2 /3

×( 4g

4l )1 /5

])refr

=

(

G ×d

4 l

×

[( 4 l

2

3 × d × 2l )2 /3

×( 4 g

4l )

1/5

])water

(100×0.025

5.14×10−4

×[( (5.14×10−4 )2

15.9×10−3

×0.025×1511 )2

3

×( 1.151.4 )1

5 ])refr

ME11M037



¿( G ×0.025

1.54×10−4

× [( (1.54×10−4 )2

0.05×0.025×890 )2

3

×( 7.5515.4 )1

5])water

Gwater=134.23

#g

m2−s

iv)

=2.317× " 1

1+0.0143× " 2×√ d ×G=2.317×

0.478

1+0.0143×0.681×√ 0.025×134.23=0.918

!=0.308×

" 3/( hfg)

1+0.347× " 4×( G

1356)

n =0.308×

0.4 /(2257×103)

1+0.347×0.0125×(134.23

1356)1.99275 =6.05×10

−8


(h fg × +∆ h¿)

(hfg × !+Z × 4

G ×d)

(qc)water=

(2036.2×103

×0.918+0)

(2036.2×10

3

×6.05×10

−8

+1×

4

134.23×0.025 )

=5.20×105

' /m2

iv) (qc)refr=(qc)water

×[ (G× hfg)refr

(G ×hfg)water ]=5.20×10

5×[ (100×147)refr

(134.23×2036.2 )water ]

¿2.79×104

' /m2

...........................$ns

Q1&. Water is flowing through a uniforly heated vertical tube 7 1 length: inlet

diaeter 2" ) with ass flux of 1### $g%25s. Degree of inlet sub cooling "#'.

and syste pressure 1 bar. /ount of heat fluxq0 "##'W%2. Due to soe

failure in the control syste heat flux starts to increase uniforly by

ME11M037



q t =q0×(1+0.08× t )

:where t is tie elapsed fro the inception of failure in

inutes . (alculate

a) Ehe tie re+uired to reach the (43 and

b) < change in exit +uality over that period.

a) (rom chart for Bowring4s correlation for 5 0 bar we get

(06.378, (10.781, (26.3, (36.6663

n 1-.6.6671;5 1-6.6671; 0.<<17

*here h fg=2257 $% /#g

,

=2.317× " 1

1+0.0143× " 2× √ d ×G=2.317×

0.478

1+0.0143×1.782×√ 0.025×103=0.22

!=0.308× " 3/( hfg)

1+0.347× " 4×( G

1356)

n=0.308×

0.4/ (2257×103)

1+0.347×0.0004×(10001356

)1.99275

=5.46×10−8

Critical heat flux

qc= (h fg × +∆ h¿)

(hfg × !+Z × 4

G ×d )

qc= (2257×10

3×0.22+0 )

(2257×103

×5.46×10−8+1×

4

1000×0.025)=1.753×10

6' /m

2

∴t =

1

0.08×( qc

q0

−1)= 1

0.08×(1.7530.5

−1)=31.38min

b) x0

¿=( q0

× Z ×4

G × d −C pl ∆ T ¿)=( 5×10

5 ×1×4

1000×0.025−4.18×10

3×50)=−0.056

ME11M037



xc¿=( qc × Z ×4

G ×d −C pl ∆T ¿)=( 1.753×10

6×1×4

103 ×0.025

−4.18×103×50)=0.0315

∴ Change in exit "uality

| xc

¿− x0

¿

x0¿

|×100

=156

Q1. -tea5Water ixture is flowing through a uniforly heated vertical tube 7 1

length: inlet diaeter 2" ) with exit +uality #.;. Degree of inlet sub cooling &#'.

and syste pressure 1 bar. 3ind the ass flux at which the exit +uality is obtained

at (43. What happens to (43 if ass flux is doubledF (opare the result with

theoretical observation.

q x=G × d

4× Z × ( C pl ×∆ T ¿+ x

¿×h fg )=

G×0.025

4×1× (4.18×30+0.9×2257 ) ×10

3=¿13.48G


(06.378, (10.781, (26.3, (36.6663

n 1-.6.6671;5 1-6.6671; 0.<<17

*here h fg=2257 $ % /#g

,

=2.317× " 1

1+0.0143× " 2× √ d ×G=2.317×

0.478

1+0.0143×1.782×√ 0.025×G=

1.107526

1+4.03×10−3

h fg × !=0.308× " 3

1+0.347× " 4×( G

1356 )n=0.308×

0.4

1+0.347×0.0004×( G

1356 )1.99275

≅ 0.1232

1+7.94 ×10−11

×

Z × 4

G× d=1×

4

G ×0.025=160

G

ME11M037




(h fg × + ∆ h¿)

(hfg × !+Z × 4

G ×d)

∆ h¿=C pl × ∆ T ¿=4.18

×10

3

×30

=125400

qc=(2257×10

3×

1.107526

1+4.03×10−3

×G+125400)

( 0.1232

1+7.94×10−11

× G2

+160

G )

+iven,qc=q x

∴(2257×10

3×

1.107526

1+4.03×10−3

×G+125400)

( 0.1232

1+7.94×10−11

×G2

+160

G )

=13.48×103

×G

earranging an! neglecting very small terms we get

Gi+1=( 2.5×106

2031.4+1.66×Gi

−103)/4.03

i Gi Gi+1

0 103 -86.6<

1 -86.6< 78.98

2 78.98 28.8

3 28.8 37.8

37.8 3.7

9 3.7 39.12

7 39.12 39.01

8 39.01 39.0

∴ =ass flux is + 46.15

#g

m2−s

ME11M037



∴ qc=13.48× G ×103=13.48×46.15×10

3=6.22×105

*HmE1

*hen + 2× G=92.30

#g

m2−s

qc=

(2257×103

× 1.107526

1+4.03×10−3

×92.30

+125400)( 0.1232

1+7.94×10−11

× (92.30×103 )2

+ 160

92.30×103 )

1.04×106

*HmE1

xc

¿=( qc × Z ×4

G ×d −C pl ∆T ¿)=( 1.04×10

6×1×4

92.3×0.025−4.18×10

3×30)=0.74

Eherefore heat flux increases with decrease in exit +uality.

his in accor!ance with theory which states that C#( in high "uality Gone is inversely

proportional to exit "uality, other parameters being constant. his also shows that C#( is

a strong function of exit "uality in this Gone.

Q1". -tea5Water ixture is flowing through a uniforly heated vertical tube 7 1

length: inlet diaeter 2" ) with exit +uality #.#". Degree of inlet sub cooling &#'.

and syste pressure 1 bar. 3ind the ass flux at which the exit +uality is obtained

at (43. What happens to (43 if ass flux is doubledF (opare the result with

theoretical observation.

q x=G× d

4× Z

× ( C pl ×∆ T ¿+ x¿

×hfg )

¿G ×0.025

4 ×1× (4.18×30+0.05×2257 )×10

3=1.49×103

×G

ME11M037




(06.378, (10.781, (26.3, (36.6663

n 1-.6.6671;5 1-6.6671; 0.<<17

*here h fg=2257 $% /#g ,

=2.317× " 1

1+0.0143× " 2× √ d ×G=2.317×

0.478

1+0.0143×1.782×√ 0.025×G=

1.107526

1+4.03×10−3

h fg × !=0.308× " 3

1+0.347× " 4×( G

1356 )n=0.308×

0.4

1+0.347×0.0004×( G

1356 )1.99275

≅ 0.1232

1+7.94 ×10−11

×

Z × 4

G× d=1×

4

G ×0.025=160

G


(h fg × + ∆ h¿)

(hfg × !+Z × 4

G ×d)

∆ h¿=C pl × ∆ T ¿=4.18×103×30=125400

qc=(2257×10

3×

1.107526

1+4.03×10−3

×G+125400)

( 0.1232

1+7.94×10−11

× G2

+160

G )

+iven,qc=q x

∴ (2257×10

3×

1.107526

1+4.03×10−3

×G+125400

)( 0.1232

1+7.94×10−11

×G2

+160

G )

=1.49×103

×G

earranging an! neglecting very small terms we get

ME11M037



Gi+1=( 2.5×106

113+0.1836×Gi

−103)/4.03

/tarting withGi=1000:i=1

, after 26 iterations we get converge! value + 030

∴ =ass flux is + 1415

#g

m2−s

∴ qc=1.49× G×103=1.49×1415×10

3=2.11×106

*HmE1

*hen + 2× G=2830

#g

m2−s

qc=

(2257×103

× 1.107526

1+4.03×10−3

×2830

+125400)( 0.1232

1+7.94×10−11

× (2830×103 )2

+ 160

2830×103 )

1.82×106

*HmE1

xc¿=( qc × Z ×4

G ×d −C pl ∆T ¿)=( 1.82×10

6×1×4

2830×0.025−4.18×10

3 ×30)=−0.01

Eherefore heat flux decreases with decrease in exit +uality.

his in accor!ance with theory which states that C#( in low "uality Gone is !irectly

proportional to exit "uality, other parameters being constant. his also shows that C#( is

a wea? function of exit "uality in this Gone, change in C#( with change in "uality is not

so strong.

Two phase flow assignment

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Transcript of Two phase flow assignment