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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).
(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 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
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!=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.
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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
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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
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=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.
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(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×(1000
1356)1.99275
=5.46×10−8
Critical heat fluxqc=
(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
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∴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
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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 Page 16
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#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 $% / #g
iii) ∅=
(
G × d
4l
×
[( 4l
2
3 × d × 2l )2 /3
×( 4g
4l )1 /5
])refr
=
(
G ×d
4 l
×
[( 4 l
2
3 × d × 2l )2 /3
×( 4 g
4l )
1/5
])water
(100×0.025
5.14×10−4
×[( (5.14×10−4 )2
15.9×10−3
×0.025×1511 )2
3
×( 1.151.4 )1
5 ])refr
ME11M037 Page 17
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¿( G ×0.025
1.54×10−4
× [( (1.54×10−4 )2
0.05×0.025×890 )2
3
×( 7.5515.4 )1
5])water
Gwater=134.23
#g
m2−s
iv)
=2.317× " 1
1+0.0143× " 2×√ d ×G=2.317×
0.478
1+0.0143×0.681×√ 0.025×134.23=0.918
!=0.308×
" 3/( hfg)
1+0.347× " 4×( G
1356)
n =0.308×
0.4 /(2257×103)
1+0.347×0.0125×(134.23
1356)1.99275 =6.05×10
−8
Critical heat fluxqc=
(h fg × +∆ h¿)
(hfg × !+Z × 4
G ×d)
(qc)water=
(2036.2×103
×0.918+0)
(2036.2×10
3
×6.05×10
−8
+1×
4
134.23×0.025 )
=5.20×105
' /m2
iv) (qc)refr=(qc)water
×[ (G× hfg)refr
(G ×hfg)water ]=5.20×10
5×[ (100×147)refr
(134.23×2036.2 )water ]
¿2.79×104
' /m2
...........................$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 Page 18
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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 Page 19
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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
(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×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 Page 20
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Critical heat fluxqc=
(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 Page 21
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∴ 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 Page 22
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(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×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
Critical heat fluxqc=
(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 Page 23
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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.