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HORIZONTAL SHELL AND TUBE CONDENSERSKari Saari26. maaliskuuta 20101
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7172019 Ene-39 4024 Shell and Tube Condensers 2
httpslidepdfcomreaderfullene-39-4024-shell-and-tube-condensers-2 112
HORIZONTAL SHELL AND TUBE CONDENSERS
Kari Saari
26 maaliskuuta 2010
1
7172019 Ene-39 4024 Shell and Tube Condensers 2
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2
CONDUCTANCE OF TUBE CONDENSERS
Conductance of a horizontal tube condenser can be calculated based on the heat transfercoefficients inside and outside of the tube and on the heat conductivity of the metallictube wall Conductance per unit length of tube is defined as
Gprime
= G
l (1)
where l is length of the tube and G is conductance of the tube
Conductance per unit length of the tube can be calculated using the following equation
1Gprime
= Rprime
= 1πds
983080 1hs
+ Rdirt
1048617+ 1
πdu
1hu
+ ln852008duds983081
2πλ (2)
where Rprime
is the heat transfer resistance per unit length of the tube ds is inside diameterof the tube hs is heat transfer coefficient inside of the tube (cooling water side) Rdirt isheat transfer resistance of the dirt layer caused by cooling water du is outside diameterof the tube hu is heat transfer coefficient outside of the tube (condensing wapour) and λis the heat conductivity of metallic tube wall
The extra heat transfer resistance caused by sedimentation from normal cooling water is
approximately Rdirt = 0 05m2
K kW When the cooling water is rather dirty the extra
resistance is approcimately Rdirt = 0 10 2m2K kW
The total conductance of the condenser is
G = LGprime
(3)
where L = N l is the total length of all tubes N is total number of tubes and l is length
of one single tube
The heat flow Φ in the condenser is calculated similarly as in all heat exchangers
Φ = Gθ = mh∆hL = mvc p∆T (4)
where θ is the mean temperature difference between condensing vapour and coolingwater mh is mass flow of the condensing vapour ∆hL is latent heat of vapourization
mv is mass flow of cooling water c p is the specific heat of cooling water and ∆T is thetemperature rise of cooling water Logarithmic mean tremperature difference can beused as the mean temperature difference
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θ = θln = (T h minus T v1)minus (T h minus T v2)
lnT hminusT v1T hminusT v2
(5)
where T h is the condensing temperature of the vapour T v1 is the inlet temperature of the cooling water and T
v2
is outlet temperature of the cooling water If you want to useeffectivity of the condenser it can be expressed as
ε = ∆T
θ0= 1minus eminusz (6)
where ∆T is temperature rise of cooling water θ0is temperature difference between con-densing vapour temperature and inlet cooling water temperature z = G C min is dimen-sionless conduvtance of condenser which can be calculated using the following equation
z = GC min
= minusln(1minus ε) (7)
COOLING WATER SIDE
The heat transfer coefficient inside of the tubes can be calculated based on classicalcorrelations of Nusselt number
N u = hsds
λ (8)
One of the most used correlations is Dittus-Boelter correlation
Nu = 0023Re08P r04 (9)
and the other more precise correlation especially for short tubes is Hausen-correlationwhich takes into account the enhanced heat transfer at the inlet of the tube (the thin
thermal boundary layer) According to Hausen the mean Nusselt number is
N u =f 8
(Reminus 1000)P r
1 + 127radic
f 8
(P r23 minus 1)
1 +
1048616dsl
85200923 (10)
where l is length of the tubeds is inside diameter of the tube and the coefficient f canbe calculated using the following equation
f = (182log10Reminus 164)minus2 (11)
The frictional pressure drop inside of the tubes is
7172019 Ene-39 4024 Shell and Tube Condensers 2
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4
∆ p = ξ L
ds
1
2ρw2 (12)
where w is the mean velocity of the flow ρ is density of the fluid L is length of the tubeand ds is inside diameter of the tube The friction coefficient of the flow ξ is dependent
on the Reynolds number of the flow and on the roughness of the tube For smooth tubesthe friction coefficient can be calculated using the correlation of Blasius
ξ = 031644radic
Re(13)
which is valid when Re lt 105
If the tubes are not smooth but have have a roughness k the friction coefficient can be
approximated as
1radic ξ
= 2lg(dsk
) + 114 (14)
This is strictly valid only if Re gt 10000000 For smaller Reunolds numbers the Moodydiagram can be used to calculate the friction coefficient as a function of Reynoldsnumber
Figure 1 Moody diagram for friction coefficient
This diagram is based on the correlation of Colebrookin and White
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1radic ξ
= minus20lg
1048667 252
Reradic
ξ +
k
371ds
1048669 (15)
which can be used recursively to solve the friction coefficient The roughness of thetubes is given in the following table
Table 1 Roughness of different tube materials
Material k [mm]steel 003
mildly rusty steal 025
aluminium 0025copper 0025rubber 0025
glass 00025galvanized metals 015
smooth plastic 00025
CONDENSATION OF VAPOUR
In condensers the effective temperature of condensing vapour is affected by non-condensablegases and by pressure drop of the vapour flow If the vapour velocity in the condenseris too small the non-condensable gases are gathered on the heat transfer surfaces Thepartial pressure of water vapour is decreased in the near region of heat transfer tubeswhen also the saturation temperature is decreased In this way the available tempera-ture difference between vapour and cooling water is decreased which can be seen as adiminished effective heat transfer coefficient of condensing vapour On the other hand if the vapour velocity is too large the pressure drop of the vapour decreases the saturation
temperature of the condensing vapour and the available temperature difference betweenvapour and cooling water is again decreased There exists experimentally specified opti-mum point (optimum velocity) for different kind of shell and tube condensers where thevapour velocity is high enough to prevent the accumulation of non-condensable gases andsmall enough not to increase the pressure drop This optimum is dependent on the tem-perature level where the condensation is happening In the following graph the optimum
point is locating on top of the curves which are plotted using the product852008
mprimeprime
h
9830812sum
as a
parameter [Tekniikan kaumlsikirja 4 page 487]
7172019 Ene-39 4024 Shell and Tube Condensers 2
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Figure 2 Optimal operation point of shell and tube condensers heat transfer coefficient
as a function of condensation temperature and product 852008 mprimeprime
h9830812
sumThe parameter
852008 m
primeprime
h
9830812sum
which is used for dimensioning of the condenser is calculated
based on the specific vapour flow of the condenser mprimeprime
h and on the pressure dropcoefficient of the condenser
sum The specific vapour flow of the condenser is defined as
mprimeprime
h = m
S (16)
where m is the mass flow of the condensing vapour and S is the heat transfer surfacearea where the condensation is happening (outside area of the tubes)
If the vapour velocity is too high the operation point is moving to the left from thetop and the effective heat transfer is decreased If the velocity is too small the non-condensable gases are gathering on the tube surfaces and the operation point is movingto the right from the top thus decreasing again the effective heat transfer coefficient
Lets consider next how the pressure drop coefficient sum
is calculated The total outsideheat transfer area of condenser tubes is S tot and the tubes are positioned triangularly insuch a way that the number of the tube rows in the direction of the flow is n At theinlet (the first row of the tube bundle) the cross sectional flow area for the vapour flow
is A(1) The heat transfer surface area (outside surface area of the tubes) where thevapour is condensing after the cross section A(1) is marked as S (1) Analogically at thetube row x the cross sectional flow area for the vapour flow is A(x) and the heattransfer surface area following this cross section is S (x)
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Figure 3 Cross sectional area A and condensing surface area S of the condenser
The mass flow of vapour which is flowing through the cross section A(x) and iscondensing on the surface area S (x) is mh(x) when the specific mass flow of thecondenser after the row x is
mprimeprime
h(x) = mh(x)
S (x) (17)
The vapour velocity at the tube row x is then
u(x) = mh(x)
ρA(x) =
mprimeprime
h(x)S (x)
ρA(x) (18)
The frictional pressure drop of one tube row x is
∆ p(x) = 1
2ρξ [u(x)]2 (19)
where ξ (x) is friction coefficient of tube row x If it is assumed that density of thevapour ρ is constant the frictional pressure drop of all n tube rows is
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∆ p = 1
2ρ
x=n9917611
ξ (x)
1048616 m
primeprime
h(x)S (x)
A(x)
8520092 (20)
When it is also assumed that the friction coefficient ξ (x) is the same for all rows
ξ (x) = ξ = constant and that the specific mass flow is constant all over the condenserthe pressure drop of the condenser can be written as
∆ p = ξ
2ρ
852008m
primeprime
h
9830812 n9917611
1048616S (x)
A(x)
8520092
= ξ
2ρ
852008m
primeprime
h
9830812991761
(21)
where
n9917611
1048616S (x)A(x)
8520092
=991761
(22)
which is pressure drop coefficient of the condenser
In normal condensers the geometrical positioning of the tubes is triangular in such away that the ratio sd = 13
s
d
Fig 4 Tube arrangement
Normally in shell and tube condensers the vapour velocity is tried to keep approximately
constant when the accumulation of non-condensable gases is not a problem if thevelocity is high enough The pressure drop coefficient for this kind of condenser can becalculated based on the first row values A(1) and S (1) because in this case
7172019 Ene-39 4024 Shell and Tube Condensers 2
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S (x)
A(x) =
S (1)
A(1) = constant
and the pressure drop coefficient Σ is
991761 = n
1048616S (1)
A(1)
8520092
(23)
At different pressure levels the shell and tube condensers have very different specificmass flows This is mainly due to the fact that the specific volume of vater wapour isvery stronly dependent on the pressure level At low pressures the volume flows areincreasing dramatically which is restricting the specific mass flow of the condenser The
following table gives approximate dimensioning values for the product 852008mprimeprime
h9830812
sum as a
function of water vapour saturation pressure In the table is also given the frictioncoefficient for one tube row which is needed in pressure drop calculations
Table 2 Approximate dimensioning values for vapour side
p0 t0 ξ 852008
mprimeprime
h
9830812sum
kN m2
C - 852008 kgm2s9830812
0 0 12 012 20 098 17 40 083 10
20 60 070 102
70 90 059 103
250 130 050 104
800 170 043 105
2800 230 037 106
8000 300 032 107
More precisely the dimensioning value for the product852008
mprimeprime
h
9830812sum
can be read from Fig 2which gives it as a function of saturation temperature The values for the product852008
mprimeprime
h
9830812sum
can be manipulated a little further Based on the eq 16 you can write for therow x
S (x)
A(x)
= u(x)
vh(x)m
primeprime
h(x)
(24)
where vh(x) is the specific volume of the vapour at row x Pressure drop coefficient sum
can then be written as
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991761 =
n991761i=1
1048667 u(x)
vh(x)mprimeprime
h(x)
10486692
(25)
If it is assumed that the specific vapour flow and specific volume of the vapour areconstant all over the condenser this can be written as
(vhmprimeprime
h)2991761
=n991761i=1
[u(x)]2 (26)
If the cross sectional flow area of the condenser is decreasing in such a way that thevelocity within the condenser with n rows is constant u(x) = u(1) = u(n) = u and
n991761i=1
[u(x)]2
= nu2
(27)
The constant velocity within the condenser is then
u =
991770 (vhm
primeprime
h)2sum
n (28)
Examining the top values of the curves in Fig 2 and making some calculations it ispossible to notice that the following condition is approximately true for the top points of the curves ( valid only when T lt 200oC )
(vhmprimeprime
h)2991761
= 75
1048667m2
s2
1048669 (29)
Now you can substitute this equation into Eq 26 and solve the velocity as a function of total row number n
u = u(1) = u(n) =
991770 752
n =
75radic n
983131m
s
983133 (30)
which is the dimensioning target velocity for the shell and tube condenser with constantvapour velocity It is independent of the pressure or temperature level Using thisvelocity the condenser is working approximately on top of the curves in Fig 2 whichmeans that the pressure drop is reasonable and on the other hand the velocity is highenough to sweep away the accumulating non condensable gases
The following figure gives the tube arrangement for a 4-row shell and tube condenserThe vapour is flowing inwards and condenses on each tube row In the middle of thecondenser is the suction channel which is used to pump away the non condensable gaseswhich are flowing through the condenser together with a small amount of passage vapour
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Figure 5 Condenser tube locations number of tubes and areast
Based on the equations presented in the fig 5 the total outside surface area of the con-
denser tubes is
S tot = Ltotπdu = NLπdu (31)
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where L is length of one tube = LtotN and N is number of tubes
Energy balance (Eq 4) gives the total mass flow of vapour mtot = φhL when thespecific vapour flow is
mprimeprime
h = mtot
S tot=
φhL
NLπdu(32)
The condensing surface after the cross section area A1 is S 1 when the mass flow at thecross section A1 is
m1 = S 1S tot
mtot = (N minus 3n)Lπdu
NLπdumtot =
N minus 3n
N mtot (33)
When the specific volume of vapour is vh this mass flow corresponds to the volume flow
˙V 1 = m1vh =
N
minus3n
N mtotvh (34)
Based on this volume flow the velocity at the cross section A1 is then
u1 =V 1A1
=N minus3nN
mtotvh1 8Lndu
(35)
When the vapour velocity within the condenser is approximately constant u1 = u = unand the dimensioning can be done based on the velocity u1 The condenser must bedimensioned in such a way that this vapour velocity u1 at the cross section A1 is the same(or a little larger) than the target velocity u (Eq 30)
DIMENSIONING OF HORIZONTAL TUBE BANK (STEP BYSTEP)
The dimensioning can begin by choosing the cooling water velocity to be about 2ms inorder to get large heat transfer coefficient of cooling water Then choose the row numbern (when also the number of tubes N is defined see Fig 5) Now you can calculate themass flow and temperature rise of cooling water and the total needed conductance andvapour mass flow using equations 4-7 The heat transfer coefficient of cooling water can
be calculated using equations 8-10 and the heat transfer coefficient for vapour side isgiven in Fig 2 by using the top values of curves The conductance per unit length of thetube is then calculated using equations 1-3 after which the lentgth of one tube L can becalculated because N (number of tubes) is already fixed Finally the vapour velocity u1
at cross section A1 can be calculated using the equation 35 and the known mass flow of vapour This velocity should be the same as the targer velocity u in equation 30 If thisis not the case choose a new row number n until the velocity u1 is the same or a littlelarger than target velocity
You can calculate the frictional pressure drop of cooling water using Moody diagramThe pressure drop of vapour side can be calculated using the equation 20 after you first
calculate the pressure drop coefficient sum
using the equation 23 or 25
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CONDUCTANCE OF TUBE CONDENSERS
Conductance of a horizontal tube condenser can be calculated based on the heat transfercoefficients inside and outside of the tube and on the heat conductivity of the metallictube wall Conductance per unit length of tube is defined as
Gprime
= G
l (1)
where l is length of the tube and G is conductance of the tube
Conductance per unit length of the tube can be calculated using the following equation
1Gprime
= Rprime
= 1πds
983080 1hs
+ Rdirt
1048617+ 1
πdu
1hu
+ ln852008duds983081
2πλ (2)
where Rprime
is the heat transfer resistance per unit length of the tube ds is inside diameterof the tube hs is heat transfer coefficient inside of the tube (cooling water side) Rdirt isheat transfer resistance of the dirt layer caused by cooling water du is outside diameterof the tube hu is heat transfer coefficient outside of the tube (condensing wapour) and λis the heat conductivity of metallic tube wall
The extra heat transfer resistance caused by sedimentation from normal cooling water is
approximately Rdirt = 0 05m2
K kW When the cooling water is rather dirty the extra
resistance is approcimately Rdirt = 0 10 2m2K kW
The total conductance of the condenser is
G = LGprime
(3)
where L = N l is the total length of all tubes N is total number of tubes and l is length
of one single tube
The heat flow Φ in the condenser is calculated similarly as in all heat exchangers
Φ = Gθ = mh∆hL = mvc p∆T (4)
where θ is the mean temperature difference between condensing vapour and coolingwater mh is mass flow of the condensing vapour ∆hL is latent heat of vapourization
mv is mass flow of cooling water c p is the specific heat of cooling water and ∆T is thetemperature rise of cooling water Logarithmic mean tremperature difference can beused as the mean temperature difference
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θ = θln = (T h minus T v1)minus (T h minus T v2)
lnT hminusT v1T hminusT v2
(5)
where T h is the condensing temperature of the vapour T v1 is the inlet temperature of the cooling water and T
v2
is outlet temperature of the cooling water If you want to useeffectivity of the condenser it can be expressed as
ε = ∆T
θ0= 1minus eminusz (6)
where ∆T is temperature rise of cooling water θ0is temperature difference between con-densing vapour temperature and inlet cooling water temperature z = G C min is dimen-sionless conduvtance of condenser which can be calculated using the following equation
z = GC min
= minusln(1minus ε) (7)
COOLING WATER SIDE
The heat transfer coefficient inside of the tubes can be calculated based on classicalcorrelations of Nusselt number
N u = hsds
λ (8)
One of the most used correlations is Dittus-Boelter correlation
Nu = 0023Re08P r04 (9)
and the other more precise correlation especially for short tubes is Hausen-correlationwhich takes into account the enhanced heat transfer at the inlet of the tube (the thin
thermal boundary layer) According to Hausen the mean Nusselt number is
N u =f 8
(Reminus 1000)P r
1 + 127radic
f 8
(P r23 minus 1)
1 +
1048616dsl
85200923 (10)
where l is length of the tubeds is inside diameter of the tube and the coefficient f canbe calculated using the following equation
f = (182log10Reminus 164)minus2 (11)
The frictional pressure drop inside of the tubes is
7172019 Ene-39 4024 Shell and Tube Condensers 2
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4
∆ p = ξ L
ds
1
2ρw2 (12)
where w is the mean velocity of the flow ρ is density of the fluid L is length of the tubeand ds is inside diameter of the tube The friction coefficient of the flow ξ is dependent
on the Reynolds number of the flow and on the roughness of the tube For smooth tubesthe friction coefficient can be calculated using the correlation of Blasius
ξ = 031644radic
Re(13)
which is valid when Re lt 105
If the tubes are not smooth but have have a roughness k the friction coefficient can be
approximated as
1radic ξ
= 2lg(dsk
) + 114 (14)
This is strictly valid only if Re gt 10000000 For smaller Reunolds numbers the Moodydiagram can be used to calculate the friction coefficient as a function of Reynoldsnumber
Figure 1 Moody diagram for friction coefficient
This diagram is based on the correlation of Colebrookin and White
7172019 Ene-39 4024 Shell and Tube Condensers 2
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5
1radic ξ
= minus20lg
1048667 252
Reradic
ξ +
k
371ds
1048669 (15)
which can be used recursively to solve the friction coefficient The roughness of thetubes is given in the following table
Table 1 Roughness of different tube materials
Material k [mm]steel 003
mildly rusty steal 025
aluminium 0025copper 0025rubber 0025
glass 00025galvanized metals 015
smooth plastic 00025
CONDENSATION OF VAPOUR
In condensers the effective temperature of condensing vapour is affected by non-condensablegases and by pressure drop of the vapour flow If the vapour velocity in the condenseris too small the non-condensable gases are gathered on the heat transfer surfaces Thepartial pressure of water vapour is decreased in the near region of heat transfer tubeswhen also the saturation temperature is decreased In this way the available tempera-ture difference between vapour and cooling water is decreased which can be seen as adiminished effective heat transfer coefficient of condensing vapour On the other hand if the vapour velocity is too large the pressure drop of the vapour decreases the saturation
temperature of the condensing vapour and the available temperature difference betweenvapour and cooling water is again decreased There exists experimentally specified opti-mum point (optimum velocity) for different kind of shell and tube condensers where thevapour velocity is high enough to prevent the accumulation of non-condensable gases andsmall enough not to increase the pressure drop This optimum is dependent on the tem-perature level where the condensation is happening In the following graph the optimum
point is locating on top of the curves which are plotted using the product852008
mprimeprime
h
9830812sum
as a
parameter [Tekniikan kaumlsikirja 4 page 487]
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6
Figure 2 Optimal operation point of shell and tube condensers heat transfer coefficient
as a function of condensation temperature and product 852008 mprimeprime
h9830812
sumThe parameter
852008 m
primeprime
h
9830812sum
which is used for dimensioning of the condenser is calculated
based on the specific vapour flow of the condenser mprimeprime
h and on the pressure dropcoefficient of the condenser
sum The specific vapour flow of the condenser is defined as
mprimeprime
h = m
S (16)
where m is the mass flow of the condensing vapour and S is the heat transfer surfacearea where the condensation is happening (outside area of the tubes)
If the vapour velocity is too high the operation point is moving to the left from thetop and the effective heat transfer is decreased If the velocity is too small the non-condensable gases are gathering on the tube surfaces and the operation point is movingto the right from the top thus decreasing again the effective heat transfer coefficient
Lets consider next how the pressure drop coefficient sum
is calculated The total outsideheat transfer area of condenser tubes is S tot and the tubes are positioned triangularly insuch a way that the number of the tube rows in the direction of the flow is n At theinlet (the first row of the tube bundle) the cross sectional flow area for the vapour flow
is A(1) The heat transfer surface area (outside surface area of the tubes) where thevapour is condensing after the cross section A(1) is marked as S (1) Analogically at thetube row x the cross sectional flow area for the vapour flow is A(x) and the heattransfer surface area following this cross section is S (x)
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Figure 3 Cross sectional area A and condensing surface area S of the condenser
The mass flow of vapour which is flowing through the cross section A(x) and iscondensing on the surface area S (x) is mh(x) when the specific mass flow of thecondenser after the row x is
mprimeprime
h(x) = mh(x)
S (x) (17)
The vapour velocity at the tube row x is then
u(x) = mh(x)
ρA(x) =
mprimeprime
h(x)S (x)
ρA(x) (18)
The frictional pressure drop of one tube row x is
∆ p(x) = 1
2ρξ [u(x)]2 (19)
where ξ (x) is friction coefficient of tube row x If it is assumed that density of thevapour ρ is constant the frictional pressure drop of all n tube rows is
7172019 Ene-39 4024 Shell and Tube Condensers 2
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∆ p = 1
2ρ
x=n9917611
ξ (x)
1048616 m
primeprime
h(x)S (x)
A(x)
8520092 (20)
When it is also assumed that the friction coefficient ξ (x) is the same for all rows
ξ (x) = ξ = constant and that the specific mass flow is constant all over the condenserthe pressure drop of the condenser can be written as
∆ p = ξ
2ρ
852008m
primeprime
h
9830812 n9917611
1048616S (x)
A(x)
8520092
= ξ
2ρ
852008m
primeprime
h
9830812991761
(21)
where
n9917611
1048616S (x)A(x)
8520092
=991761
(22)
which is pressure drop coefficient of the condenser
In normal condensers the geometrical positioning of the tubes is triangular in such away that the ratio sd = 13
s
d
Fig 4 Tube arrangement
Normally in shell and tube condensers the vapour velocity is tried to keep approximately
constant when the accumulation of non-condensable gases is not a problem if thevelocity is high enough The pressure drop coefficient for this kind of condenser can becalculated based on the first row values A(1) and S (1) because in this case
7172019 Ene-39 4024 Shell and Tube Condensers 2
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S (x)
A(x) =
S (1)
A(1) = constant
and the pressure drop coefficient Σ is
991761 = n
1048616S (1)
A(1)
8520092
(23)
At different pressure levels the shell and tube condensers have very different specificmass flows This is mainly due to the fact that the specific volume of vater wapour isvery stronly dependent on the pressure level At low pressures the volume flows areincreasing dramatically which is restricting the specific mass flow of the condenser The
following table gives approximate dimensioning values for the product 852008mprimeprime
h9830812
sum as a
function of water vapour saturation pressure In the table is also given the frictioncoefficient for one tube row which is needed in pressure drop calculations
Table 2 Approximate dimensioning values for vapour side
p0 t0 ξ 852008
mprimeprime
h
9830812sum
kN m2
C - 852008 kgm2s9830812
0 0 12 012 20 098 17 40 083 10
20 60 070 102
70 90 059 103
250 130 050 104
800 170 043 105
2800 230 037 106
8000 300 032 107
More precisely the dimensioning value for the product852008
mprimeprime
h
9830812sum
can be read from Fig 2which gives it as a function of saturation temperature The values for the product852008
mprimeprime
h
9830812sum
can be manipulated a little further Based on the eq 16 you can write for therow x
S (x)
A(x)
= u(x)
vh(x)m
primeprime
h(x)
(24)
where vh(x) is the specific volume of the vapour at row x Pressure drop coefficient sum
can then be written as
7172019 Ene-39 4024 Shell and Tube Condensers 2
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10
991761 =
n991761i=1
1048667 u(x)
vh(x)mprimeprime
h(x)
10486692
(25)
If it is assumed that the specific vapour flow and specific volume of the vapour areconstant all over the condenser this can be written as
(vhmprimeprime
h)2991761
=n991761i=1
[u(x)]2 (26)
If the cross sectional flow area of the condenser is decreasing in such a way that thevelocity within the condenser with n rows is constant u(x) = u(1) = u(n) = u and
n991761i=1
[u(x)]2
= nu2
(27)
The constant velocity within the condenser is then
u =
991770 (vhm
primeprime
h)2sum
n (28)
Examining the top values of the curves in Fig 2 and making some calculations it ispossible to notice that the following condition is approximately true for the top points of the curves ( valid only when T lt 200oC )
(vhmprimeprime
h)2991761
= 75
1048667m2
s2
1048669 (29)
Now you can substitute this equation into Eq 26 and solve the velocity as a function of total row number n
u = u(1) = u(n) =
991770 752
n =
75radic n
983131m
s
983133 (30)
which is the dimensioning target velocity for the shell and tube condenser with constantvapour velocity It is independent of the pressure or temperature level Using thisvelocity the condenser is working approximately on top of the curves in Fig 2 whichmeans that the pressure drop is reasonable and on the other hand the velocity is highenough to sweep away the accumulating non condensable gases
The following figure gives the tube arrangement for a 4-row shell and tube condenserThe vapour is flowing inwards and condenses on each tube row In the middle of thecondenser is the suction channel which is used to pump away the non condensable gaseswhich are flowing through the condenser together with a small amount of passage vapour
7172019 Ene-39 4024 Shell and Tube Condensers 2
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Figure 5 Condenser tube locations number of tubes and areast
Based on the equations presented in the fig 5 the total outside surface area of the con-
denser tubes is
S tot = Ltotπdu = NLπdu (31)
7172019 Ene-39 4024 Shell and Tube Condensers 2
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12
where L is length of one tube = LtotN and N is number of tubes
Energy balance (Eq 4) gives the total mass flow of vapour mtot = φhL when thespecific vapour flow is
mprimeprime
h = mtot
S tot=
φhL
NLπdu(32)
The condensing surface after the cross section area A1 is S 1 when the mass flow at thecross section A1 is
m1 = S 1S tot
mtot = (N minus 3n)Lπdu
NLπdumtot =
N minus 3n
N mtot (33)
When the specific volume of vapour is vh this mass flow corresponds to the volume flow
˙V 1 = m1vh =
N
minus3n
N mtotvh (34)
Based on this volume flow the velocity at the cross section A1 is then
u1 =V 1A1
=N minus3nN
mtotvh1 8Lndu
(35)
When the vapour velocity within the condenser is approximately constant u1 = u = unand the dimensioning can be done based on the velocity u1 The condenser must bedimensioned in such a way that this vapour velocity u1 at the cross section A1 is the same(or a little larger) than the target velocity u (Eq 30)
DIMENSIONING OF HORIZONTAL TUBE BANK (STEP BYSTEP)
The dimensioning can begin by choosing the cooling water velocity to be about 2ms inorder to get large heat transfer coefficient of cooling water Then choose the row numbern (when also the number of tubes N is defined see Fig 5) Now you can calculate themass flow and temperature rise of cooling water and the total needed conductance andvapour mass flow using equations 4-7 The heat transfer coefficient of cooling water can
be calculated using equations 8-10 and the heat transfer coefficient for vapour side isgiven in Fig 2 by using the top values of curves The conductance per unit length of thetube is then calculated using equations 1-3 after which the lentgth of one tube L can becalculated because N (number of tubes) is already fixed Finally the vapour velocity u1
at cross section A1 can be calculated using the equation 35 and the known mass flow of vapour This velocity should be the same as the targer velocity u in equation 30 If thisis not the case choose a new row number n until the velocity u1 is the same or a littlelarger than target velocity
You can calculate the frictional pressure drop of cooling water using Moody diagramThe pressure drop of vapour side can be calculated using the equation 20 after you first
calculate the pressure drop coefficient sum
using the equation 23 or 25
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3
θ = θln = (T h minus T v1)minus (T h minus T v2)
lnT hminusT v1T hminusT v2
(5)
where T h is the condensing temperature of the vapour T v1 is the inlet temperature of the cooling water and T
v2
is outlet temperature of the cooling water If you want to useeffectivity of the condenser it can be expressed as
ε = ∆T
θ0= 1minus eminusz (6)
where ∆T is temperature rise of cooling water θ0is temperature difference between con-densing vapour temperature and inlet cooling water temperature z = G C min is dimen-sionless conduvtance of condenser which can be calculated using the following equation
z = GC min
= minusln(1minus ε) (7)
COOLING WATER SIDE
The heat transfer coefficient inside of the tubes can be calculated based on classicalcorrelations of Nusselt number
N u = hsds
λ (8)
One of the most used correlations is Dittus-Boelter correlation
Nu = 0023Re08P r04 (9)
and the other more precise correlation especially for short tubes is Hausen-correlationwhich takes into account the enhanced heat transfer at the inlet of the tube (the thin
thermal boundary layer) According to Hausen the mean Nusselt number is
N u =f 8
(Reminus 1000)P r
1 + 127radic
f 8
(P r23 minus 1)
1 +
1048616dsl
85200923 (10)
where l is length of the tubeds is inside diameter of the tube and the coefficient f canbe calculated using the following equation
f = (182log10Reminus 164)minus2 (11)
The frictional pressure drop inside of the tubes is
7172019 Ene-39 4024 Shell and Tube Condensers 2
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4
∆ p = ξ L
ds
1
2ρw2 (12)
where w is the mean velocity of the flow ρ is density of the fluid L is length of the tubeand ds is inside diameter of the tube The friction coefficient of the flow ξ is dependent
on the Reynolds number of the flow and on the roughness of the tube For smooth tubesthe friction coefficient can be calculated using the correlation of Blasius
ξ = 031644radic
Re(13)
which is valid when Re lt 105
If the tubes are not smooth but have have a roughness k the friction coefficient can be
approximated as
1radic ξ
= 2lg(dsk
) + 114 (14)
This is strictly valid only if Re gt 10000000 For smaller Reunolds numbers the Moodydiagram can be used to calculate the friction coefficient as a function of Reynoldsnumber
Figure 1 Moody diagram for friction coefficient
This diagram is based on the correlation of Colebrookin and White
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5
1radic ξ
= minus20lg
1048667 252
Reradic
ξ +
k
371ds
1048669 (15)
which can be used recursively to solve the friction coefficient The roughness of thetubes is given in the following table
Table 1 Roughness of different tube materials
Material k [mm]steel 003
mildly rusty steal 025
aluminium 0025copper 0025rubber 0025
glass 00025galvanized metals 015
smooth plastic 00025
CONDENSATION OF VAPOUR
In condensers the effective temperature of condensing vapour is affected by non-condensablegases and by pressure drop of the vapour flow If the vapour velocity in the condenseris too small the non-condensable gases are gathered on the heat transfer surfaces Thepartial pressure of water vapour is decreased in the near region of heat transfer tubeswhen also the saturation temperature is decreased In this way the available tempera-ture difference between vapour and cooling water is decreased which can be seen as adiminished effective heat transfer coefficient of condensing vapour On the other hand if the vapour velocity is too large the pressure drop of the vapour decreases the saturation
temperature of the condensing vapour and the available temperature difference betweenvapour and cooling water is again decreased There exists experimentally specified opti-mum point (optimum velocity) for different kind of shell and tube condensers where thevapour velocity is high enough to prevent the accumulation of non-condensable gases andsmall enough not to increase the pressure drop This optimum is dependent on the tem-perature level where the condensation is happening In the following graph the optimum
point is locating on top of the curves which are plotted using the product852008
mprimeprime
h
9830812sum
as a
parameter [Tekniikan kaumlsikirja 4 page 487]
7172019 Ene-39 4024 Shell and Tube Condensers 2
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6
Figure 2 Optimal operation point of shell and tube condensers heat transfer coefficient
as a function of condensation temperature and product 852008 mprimeprime
h9830812
sumThe parameter
852008 m
primeprime
h
9830812sum
which is used for dimensioning of the condenser is calculated
based on the specific vapour flow of the condenser mprimeprime
h and on the pressure dropcoefficient of the condenser
sum The specific vapour flow of the condenser is defined as
mprimeprime
h = m
S (16)
where m is the mass flow of the condensing vapour and S is the heat transfer surfacearea where the condensation is happening (outside area of the tubes)
If the vapour velocity is too high the operation point is moving to the left from thetop and the effective heat transfer is decreased If the velocity is too small the non-condensable gases are gathering on the tube surfaces and the operation point is movingto the right from the top thus decreasing again the effective heat transfer coefficient
Lets consider next how the pressure drop coefficient sum
is calculated The total outsideheat transfer area of condenser tubes is S tot and the tubes are positioned triangularly insuch a way that the number of the tube rows in the direction of the flow is n At theinlet (the first row of the tube bundle) the cross sectional flow area for the vapour flow
is A(1) The heat transfer surface area (outside surface area of the tubes) where thevapour is condensing after the cross section A(1) is marked as S (1) Analogically at thetube row x the cross sectional flow area for the vapour flow is A(x) and the heattransfer surface area following this cross section is S (x)
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Figure 3 Cross sectional area A and condensing surface area S of the condenser
The mass flow of vapour which is flowing through the cross section A(x) and iscondensing on the surface area S (x) is mh(x) when the specific mass flow of thecondenser after the row x is
mprimeprime
h(x) = mh(x)
S (x) (17)
The vapour velocity at the tube row x is then
u(x) = mh(x)
ρA(x) =
mprimeprime
h(x)S (x)
ρA(x) (18)
The frictional pressure drop of one tube row x is
∆ p(x) = 1
2ρξ [u(x)]2 (19)
where ξ (x) is friction coefficient of tube row x If it is assumed that density of thevapour ρ is constant the frictional pressure drop of all n tube rows is
7172019 Ene-39 4024 Shell and Tube Condensers 2
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∆ p = 1
2ρ
x=n9917611
ξ (x)
1048616 m
primeprime
h(x)S (x)
A(x)
8520092 (20)
When it is also assumed that the friction coefficient ξ (x) is the same for all rows
ξ (x) = ξ = constant and that the specific mass flow is constant all over the condenserthe pressure drop of the condenser can be written as
∆ p = ξ
2ρ
852008m
primeprime
h
9830812 n9917611
1048616S (x)
A(x)
8520092
= ξ
2ρ
852008m
primeprime
h
9830812991761
(21)
where
n9917611
1048616S (x)A(x)
8520092
=991761
(22)
which is pressure drop coefficient of the condenser
In normal condensers the geometrical positioning of the tubes is triangular in such away that the ratio sd = 13
s
d
Fig 4 Tube arrangement
Normally in shell and tube condensers the vapour velocity is tried to keep approximately
constant when the accumulation of non-condensable gases is not a problem if thevelocity is high enough The pressure drop coefficient for this kind of condenser can becalculated based on the first row values A(1) and S (1) because in this case
7172019 Ene-39 4024 Shell and Tube Condensers 2
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S (x)
A(x) =
S (1)
A(1) = constant
and the pressure drop coefficient Σ is
991761 = n
1048616S (1)
A(1)
8520092
(23)
At different pressure levels the shell and tube condensers have very different specificmass flows This is mainly due to the fact that the specific volume of vater wapour isvery stronly dependent on the pressure level At low pressures the volume flows areincreasing dramatically which is restricting the specific mass flow of the condenser The
following table gives approximate dimensioning values for the product 852008mprimeprime
h9830812
sum as a
function of water vapour saturation pressure In the table is also given the frictioncoefficient for one tube row which is needed in pressure drop calculations
Table 2 Approximate dimensioning values for vapour side
p0 t0 ξ 852008
mprimeprime
h
9830812sum
kN m2
C - 852008 kgm2s9830812
0 0 12 012 20 098 17 40 083 10
20 60 070 102
70 90 059 103
250 130 050 104
800 170 043 105
2800 230 037 106
8000 300 032 107
More precisely the dimensioning value for the product852008
mprimeprime
h
9830812sum
can be read from Fig 2which gives it as a function of saturation temperature The values for the product852008
mprimeprime
h
9830812sum
can be manipulated a little further Based on the eq 16 you can write for therow x
S (x)
A(x)
= u(x)
vh(x)m
primeprime
h(x)
(24)
where vh(x) is the specific volume of the vapour at row x Pressure drop coefficient sum
can then be written as
7172019 Ene-39 4024 Shell and Tube Condensers 2
httpslidepdfcomreaderfullene-39-4024-shell-and-tube-condensers-2 1012
10
991761 =
n991761i=1
1048667 u(x)
vh(x)mprimeprime
h(x)
10486692
(25)
If it is assumed that the specific vapour flow and specific volume of the vapour areconstant all over the condenser this can be written as
(vhmprimeprime
h)2991761
=n991761i=1
[u(x)]2 (26)
If the cross sectional flow area of the condenser is decreasing in such a way that thevelocity within the condenser with n rows is constant u(x) = u(1) = u(n) = u and
n991761i=1
[u(x)]2
= nu2
(27)
The constant velocity within the condenser is then
u =
991770 (vhm
primeprime
h)2sum
n (28)
Examining the top values of the curves in Fig 2 and making some calculations it ispossible to notice that the following condition is approximately true for the top points of the curves ( valid only when T lt 200oC )
(vhmprimeprime
h)2991761
= 75
1048667m2
s2
1048669 (29)
Now you can substitute this equation into Eq 26 and solve the velocity as a function of total row number n
u = u(1) = u(n) =
991770 752
n =
75radic n
983131m
s
983133 (30)
which is the dimensioning target velocity for the shell and tube condenser with constantvapour velocity It is independent of the pressure or temperature level Using thisvelocity the condenser is working approximately on top of the curves in Fig 2 whichmeans that the pressure drop is reasonable and on the other hand the velocity is highenough to sweep away the accumulating non condensable gases
The following figure gives the tube arrangement for a 4-row shell and tube condenserThe vapour is flowing inwards and condenses on each tube row In the middle of thecondenser is the suction channel which is used to pump away the non condensable gaseswhich are flowing through the condenser together with a small amount of passage vapour
7172019 Ene-39 4024 Shell and Tube Condensers 2
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Figure 5 Condenser tube locations number of tubes and areast
Based on the equations presented in the fig 5 the total outside surface area of the con-
denser tubes is
S tot = Ltotπdu = NLπdu (31)
7172019 Ene-39 4024 Shell and Tube Condensers 2
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where L is length of one tube = LtotN and N is number of tubes
Energy balance (Eq 4) gives the total mass flow of vapour mtot = φhL when thespecific vapour flow is
mprimeprime
h = mtot
S tot=
φhL
NLπdu(32)
The condensing surface after the cross section area A1 is S 1 when the mass flow at thecross section A1 is
m1 = S 1S tot
mtot = (N minus 3n)Lπdu
NLπdumtot =
N minus 3n
N mtot (33)
When the specific volume of vapour is vh this mass flow corresponds to the volume flow
˙V 1 = m1vh =
N
minus3n
N mtotvh (34)
Based on this volume flow the velocity at the cross section A1 is then
u1 =V 1A1
=N minus3nN
mtotvh1 8Lndu
(35)
When the vapour velocity within the condenser is approximately constant u1 = u = unand the dimensioning can be done based on the velocity u1 The condenser must bedimensioned in such a way that this vapour velocity u1 at the cross section A1 is the same(or a little larger) than the target velocity u (Eq 30)
DIMENSIONING OF HORIZONTAL TUBE BANK (STEP BYSTEP)
The dimensioning can begin by choosing the cooling water velocity to be about 2ms inorder to get large heat transfer coefficient of cooling water Then choose the row numbern (when also the number of tubes N is defined see Fig 5) Now you can calculate themass flow and temperature rise of cooling water and the total needed conductance andvapour mass flow using equations 4-7 The heat transfer coefficient of cooling water can
be calculated using equations 8-10 and the heat transfer coefficient for vapour side isgiven in Fig 2 by using the top values of curves The conductance per unit length of thetube is then calculated using equations 1-3 after which the lentgth of one tube L can becalculated because N (number of tubes) is already fixed Finally the vapour velocity u1
at cross section A1 can be calculated using the equation 35 and the known mass flow of vapour This velocity should be the same as the targer velocity u in equation 30 If thisis not the case choose a new row number n until the velocity u1 is the same or a littlelarger than target velocity
You can calculate the frictional pressure drop of cooling water using Moody diagramThe pressure drop of vapour side can be calculated using the equation 20 after you first
calculate the pressure drop coefficient sum
using the equation 23 or 25
7172019 Ene-39 4024 Shell and Tube Condensers 2
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4
∆ p = ξ L
ds
1
2ρw2 (12)
where w is the mean velocity of the flow ρ is density of the fluid L is length of the tubeand ds is inside diameter of the tube The friction coefficient of the flow ξ is dependent
on the Reynolds number of the flow and on the roughness of the tube For smooth tubesthe friction coefficient can be calculated using the correlation of Blasius
ξ = 031644radic
Re(13)
which is valid when Re lt 105
If the tubes are not smooth but have have a roughness k the friction coefficient can be
approximated as
1radic ξ
= 2lg(dsk
) + 114 (14)
This is strictly valid only if Re gt 10000000 For smaller Reunolds numbers the Moodydiagram can be used to calculate the friction coefficient as a function of Reynoldsnumber
Figure 1 Moody diagram for friction coefficient
This diagram is based on the correlation of Colebrookin and White
7172019 Ene-39 4024 Shell and Tube Condensers 2
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5
1radic ξ
= minus20lg
1048667 252
Reradic
ξ +
k
371ds
1048669 (15)
which can be used recursively to solve the friction coefficient The roughness of thetubes is given in the following table
Table 1 Roughness of different tube materials
Material k [mm]steel 003
mildly rusty steal 025
aluminium 0025copper 0025rubber 0025
glass 00025galvanized metals 015
smooth plastic 00025
CONDENSATION OF VAPOUR
In condensers the effective temperature of condensing vapour is affected by non-condensablegases and by pressure drop of the vapour flow If the vapour velocity in the condenseris too small the non-condensable gases are gathered on the heat transfer surfaces Thepartial pressure of water vapour is decreased in the near region of heat transfer tubeswhen also the saturation temperature is decreased In this way the available tempera-ture difference between vapour and cooling water is decreased which can be seen as adiminished effective heat transfer coefficient of condensing vapour On the other hand if the vapour velocity is too large the pressure drop of the vapour decreases the saturation
temperature of the condensing vapour and the available temperature difference betweenvapour and cooling water is again decreased There exists experimentally specified opti-mum point (optimum velocity) for different kind of shell and tube condensers where thevapour velocity is high enough to prevent the accumulation of non-condensable gases andsmall enough not to increase the pressure drop This optimum is dependent on the tem-perature level where the condensation is happening In the following graph the optimum
point is locating on top of the curves which are plotted using the product852008
mprimeprime
h
9830812sum
as a
parameter [Tekniikan kaumlsikirja 4 page 487]
7172019 Ene-39 4024 Shell and Tube Condensers 2
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6
Figure 2 Optimal operation point of shell and tube condensers heat transfer coefficient
as a function of condensation temperature and product 852008 mprimeprime
h9830812
sumThe parameter
852008 m
primeprime
h
9830812sum
which is used for dimensioning of the condenser is calculated
based on the specific vapour flow of the condenser mprimeprime
h and on the pressure dropcoefficient of the condenser
sum The specific vapour flow of the condenser is defined as
mprimeprime
h = m
S (16)
where m is the mass flow of the condensing vapour and S is the heat transfer surfacearea where the condensation is happening (outside area of the tubes)
If the vapour velocity is too high the operation point is moving to the left from thetop and the effective heat transfer is decreased If the velocity is too small the non-condensable gases are gathering on the tube surfaces and the operation point is movingto the right from the top thus decreasing again the effective heat transfer coefficient
Lets consider next how the pressure drop coefficient sum
is calculated The total outsideheat transfer area of condenser tubes is S tot and the tubes are positioned triangularly insuch a way that the number of the tube rows in the direction of the flow is n At theinlet (the first row of the tube bundle) the cross sectional flow area for the vapour flow
is A(1) The heat transfer surface area (outside surface area of the tubes) where thevapour is condensing after the cross section A(1) is marked as S (1) Analogically at thetube row x the cross sectional flow area for the vapour flow is A(x) and the heattransfer surface area following this cross section is S (x)
7172019 Ene-39 4024 Shell and Tube Condensers 2
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Figure 3 Cross sectional area A and condensing surface area S of the condenser
The mass flow of vapour which is flowing through the cross section A(x) and iscondensing on the surface area S (x) is mh(x) when the specific mass flow of thecondenser after the row x is
mprimeprime
h(x) = mh(x)
S (x) (17)
The vapour velocity at the tube row x is then
u(x) = mh(x)
ρA(x) =
mprimeprime
h(x)S (x)
ρA(x) (18)
The frictional pressure drop of one tube row x is
∆ p(x) = 1
2ρξ [u(x)]2 (19)
where ξ (x) is friction coefficient of tube row x If it is assumed that density of thevapour ρ is constant the frictional pressure drop of all n tube rows is
7172019 Ene-39 4024 Shell and Tube Condensers 2
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∆ p = 1
2ρ
x=n9917611
ξ (x)
1048616 m
primeprime
h(x)S (x)
A(x)
8520092 (20)
When it is also assumed that the friction coefficient ξ (x) is the same for all rows
ξ (x) = ξ = constant and that the specific mass flow is constant all over the condenserthe pressure drop of the condenser can be written as
∆ p = ξ
2ρ
852008m
primeprime
h
9830812 n9917611
1048616S (x)
A(x)
8520092
= ξ
2ρ
852008m
primeprime
h
9830812991761
(21)
where
n9917611
1048616S (x)A(x)
8520092
=991761
(22)
which is pressure drop coefficient of the condenser
In normal condensers the geometrical positioning of the tubes is triangular in such away that the ratio sd = 13
s
d
Fig 4 Tube arrangement
Normally in shell and tube condensers the vapour velocity is tried to keep approximately
constant when the accumulation of non-condensable gases is not a problem if thevelocity is high enough The pressure drop coefficient for this kind of condenser can becalculated based on the first row values A(1) and S (1) because in this case
7172019 Ene-39 4024 Shell and Tube Condensers 2
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S (x)
A(x) =
S (1)
A(1) = constant
and the pressure drop coefficient Σ is
991761 = n
1048616S (1)
A(1)
8520092
(23)
At different pressure levels the shell and tube condensers have very different specificmass flows This is mainly due to the fact that the specific volume of vater wapour isvery stronly dependent on the pressure level At low pressures the volume flows areincreasing dramatically which is restricting the specific mass flow of the condenser The
following table gives approximate dimensioning values for the product 852008mprimeprime
h9830812
sum as a
function of water vapour saturation pressure In the table is also given the frictioncoefficient for one tube row which is needed in pressure drop calculations
Table 2 Approximate dimensioning values for vapour side
p0 t0 ξ 852008
mprimeprime
h
9830812sum
kN m2
C - 852008 kgm2s9830812
0 0 12 012 20 098 17 40 083 10
20 60 070 102
70 90 059 103
250 130 050 104
800 170 043 105
2800 230 037 106
8000 300 032 107
More precisely the dimensioning value for the product852008
mprimeprime
h
9830812sum
can be read from Fig 2which gives it as a function of saturation temperature The values for the product852008
mprimeprime
h
9830812sum
can be manipulated a little further Based on the eq 16 you can write for therow x
S (x)
A(x)
= u(x)
vh(x)m
primeprime
h(x)
(24)
where vh(x) is the specific volume of the vapour at row x Pressure drop coefficient sum
can then be written as
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991761 =
n991761i=1
1048667 u(x)
vh(x)mprimeprime
h(x)
10486692
(25)
If it is assumed that the specific vapour flow and specific volume of the vapour areconstant all over the condenser this can be written as
(vhmprimeprime
h)2991761
=n991761i=1
[u(x)]2 (26)
If the cross sectional flow area of the condenser is decreasing in such a way that thevelocity within the condenser with n rows is constant u(x) = u(1) = u(n) = u and
n991761i=1
[u(x)]2
= nu2
(27)
The constant velocity within the condenser is then
u =
991770 (vhm
primeprime
h)2sum
n (28)
Examining the top values of the curves in Fig 2 and making some calculations it ispossible to notice that the following condition is approximately true for the top points of the curves ( valid only when T lt 200oC )
(vhmprimeprime
h)2991761
= 75
1048667m2
s2
1048669 (29)
Now you can substitute this equation into Eq 26 and solve the velocity as a function of total row number n
u = u(1) = u(n) =
991770 752
n =
75radic n
983131m
s
983133 (30)
which is the dimensioning target velocity for the shell and tube condenser with constantvapour velocity It is independent of the pressure or temperature level Using thisvelocity the condenser is working approximately on top of the curves in Fig 2 whichmeans that the pressure drop is reasonable and on the other hand the velocity is highenough to sweep away the accumulating non condensable gases
The following figure gives the tube arrangement for a 4-row shell and tube condenserThe vapour is flowing inwards and condenses on each tube row In the middle of thecondenser is the suction channel which is used to pump away the non condensable gaseswhich are flowing through the condenser together with a small amount of passage vapour
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Figure 5 Condenser tube locations number of tubes and areast
Based on the equations presented in the fig 5 the total outside surface area of the con-
denser tubes is
S tot = Ltotπdu = NLπdu (31)
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where L is length of one tube = LtotN and N is number of tubes
Energy balance (Eq 4) gives the total mass flow of vapour mtot = φhL when thespecific vapour flow is
mprimeprime
h = mtot
S tot=
φhL
NLπdu(32)
The condensing surface after the cross section area A1 is S 1 when the mass flow at thecross section A1 is
m1 = S 1S tot
mtot = (N minus 3n)Lπdu
NLπdumtot =
N minus 3n
N mtot (33)
When the specific volume of vapour is vh this mass flow corresponds to the volume flow
˙V 1 = m1vh =
N
minus3n
N mtotvh (34)
Based on this volume flow the velocity at the cross section A1 is then
u1 =V 1A1
=N minus3nN
mtotvh1 8Lndu
(35)
When the vapour velocity within the condenser is approximately constant u1 = u = unand the dimensioning can be done based on the velocity u1 The condenser must bedimensioned in such a way that this vapour velocity u1 at the cross section A1 is the same(or a little larger) than the target velocity u (Eq 30)
DIMENSIONING OF HORIZONTAL TUBE BANK (STEP BYSTEP)
The dimensioning can begin by choosing the cooling water velocity to be about 2ms inorder to get large heat transfer coefficient of cooling water Then choose the row numbern (when also the number of tubes N is defined see Fig 5) Now you can calculate themass flow and temperature rise of cooling water and the total needed conductance andvapour mass flow using equations 4-7 The heat transfer coefficient of cooling water can
be calculated using equations 8-10 and the heat transfer coefficient for vapour side isgiven in Fig 2 by using the top values of curves The conductance per unit length of thetube is then calculated using equations 1-3 after which the lentgth of one tube L can becalculated because N (number of tubes) is already fixed Finally the vapour velocity u1
at cross section A1 can be calculated using the equation 35 and the known mass flow of vapour This velocity should be the same as the targer velocity u in equation 30 If thisis not the case choose a new row number n until the velocity u1 is the same or a littlelarger than target velocity
You can calculate the frictional pressure drop of cooling water using Moody diagramThe pressure drop of vapour side can be calculated using the equation 20 after you first
calculate the pressure drop coefficient sum
using the equation 23 or 25
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5
1radic ξ
= minus20lg
1048667 252
Reradic
ξ +
k
371ds
1048669 (15)
which can be used recursively to solve the friction coefficient The roughness of thetubes is given in the following table
Table 1 Roughness of different tube materials
Material k [mm]steel 003
mildly rusty steal 025
aluminium 0025copper 0025rubber 0025
glass 00025galvanized metals 015
smooth plastic 00025
CONDENSATION OF VAPOUR
In condensers the effective temperature of condensing vapour is affected by non-condensablegases and by pressure drop of the vapour flow If the vapour velocity in the condenseris too small the non-condensable gases are gathered on the heat transfer surfaces Thepartial pressure of water vapour is decreased in the near region of heat transfer tubeswhen also the saturation temperature is decreased In this way the available tempera-ture difference between vapour and cooling water is decreased which can be seen as adiminished effective heat transfer coefficient of condensing vapour On the other hand if the vapour velocity is too large the pressure drop of the vapour decreases the saturation
temperature of the condensing vapour and the available temperature difference betweenvapour and cooling water is again decreased There exists experimentally specified opti-mum point (optimum velocity) for different kind of shell and tube condensers where thevapour velocity is high enough to prevent the accumulation of non-condensable gases andsmall enough not to increase the pressure drop This optimum is dependent on the tem-perature level where the condensation is happening In the following graph the optimum
point is locating on top of the curves which are plotted using the product852008
mprimeprime
h
9830812sum
as a
parameter [Tekniikan kaumlsikirja 4 page 487]
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6
Figure 2 Optimal operation point of shell and tube condensers heat transfer coefficient
as a function of condensation temperature and product 852008 mprimeprime
h9830812
sumThe parameter
852008 m
primeprime
h
9830812sum
which is used for dimensioning of the condenser is calculated
based on the specific vapour flow of the condenser mprimeprime
h and on the pressure dropcoefficient of the condenser
sum The specific vapour flow of the condenser is defined as
mprimeprime
h = m
S (16)
where m is the mass flow of the condensing vapour and S is the heat transfer surfacearea where the condensation is happening (outside area of the tubes)
If the vapour velocity is too high the operation point is moving to the left from thetop and the effective heat transfer is decreased If the velocity is too small the non-condensable gases are gathering on the tube surfaces and the operation point is movingto the right from the top thus decreasing again the effective heat transfer coefficient
Lets consider next how the pressure drop coefficient sum
is calculated The total outsideheat transfer area of condenser tubes is S tot and the tubes are positioned triangularly insuch a way that the number of the tube rows in the direction of the flow is n At theinlet (the first row of the tube bundle) the cross sectional flow area for the vapour flow
is A(1) The heat transfer surface area (outside surface area of the tubes) where thevapour is condensing after the cross section A(1) is marked as S (1) Analogically at thetube row x the cross sectional flow area for the vapour flow is A(x) and the heattransfer surface area following this cross section is S (x)
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Figure 3 Cross sectional area A and condensing surface area S of the condenser
The mass flow of vapour which is flowing through the cross section A(x) and iscondensing on the surface area S (x) is mh(x) when the specific mass flow of thecondenser after the row x is
mprimeprime
h(x) = mh(x)
S (x) (17)
The vapour velocity at the tube row x is then
u(x) = mh(x)
ρA(x) =
mprimeprime
h(x)S (x)
ρA(x) (18)
The frictional pressure drop of one tube row x is
∆ p(x) = 1
2ρξ [u(x)]2 (19)
where ξ (x) is friction coefficient of tube row x If it is assumed that density of thevapour ρ is constant the frictional pressure drop of all n tube rows is
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∆ p = 1
2ρ
x=n9917611
ξ (x)
1048616 m
primeprime
h(x)S (x)
A(x)
8520092 (20)
When it is also assumed that the friction coefficient ξ (x) is the same for all rows
ξ (x) = ξ = constant and that the specific mass flow is constant all over the condenserthe pressure drop of the condenser can be written as
∆ p = ξ
2ρ
852008m
primeprime
h
9830812 n9917611
1048616S (x)
A(x)
8520092
= ξ
2ρ
852008m
primeprime
h
9830812991761
(21)
where
n9917611
1048616S (x)A(x)
8520092
=991761
(22)
which is pressure drop coefficient of the condenser
In normal condensers the geometrical positioning of the tubes is triangular in such away that the ratio sd = 13
s
d
Fig 4 Tube arrangement
Normally in shell and tube condensers the vapour velocity is tried to keep approximately
constant when the accumulation of non-condensable gases is not a problem if thevelocity is high enough The pressure drop coefficient for this kind of condenser can becalculated based on the first row values A(1) and S (1) because in this case
7172019 Ene-39 4024 Shell and Tube Condensers 2
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S (x)
A(x) =
S (1)
A(1) = constant
and the pressure drop coefficient Σ is
991761 = n
1048616S (1)
A(1)
8520092
(23)
At different pressure levels the shell and tube condensers have very different specificmass flows This is mainly due to the fact that the specific volume of vater wapour isvery stronly dependent on the pressure level At low pressures the volume flows areincreasing dramatically which is restricting the specific mass flow of the condenser The
following table gives approximate dimensioning values for the product 852008mprimeprime
h9830812
sum as a
function of water vapour saturation pressure In the table is also given the frictioncoefficient for one tube row which is needed in pressure drop calculations
Table 2 Approximate dimensioning values for vapour side
p0 t0 ξ 852008
mprimeprime
h
9830812sum
kN m2
C - 852008 kgm2s9830812
0 0 12 012 20 098 17 40 083 10
20 60 070 102
70 90 059 103
250 130 050 104
800 170 043 105
2800 230 037 106
8000 300 032 107
More precisely the dimensioning value for the product852008
mprimeprime
h
9830812sum
can be read from Fig 2which gives it as a function of saturation temperature The values for the product852008
mprimeprime
h
9830812sum
can be manipulated a little further Based on the eq 16 you can write for therow x
S (x)
A(x)
= u(x)
vh(x)m
primeprime
h(x)
(24)
where vh(x) is the specific volume of the vapour at row x Pressure drop coefficient sum
can then be written as
7172019 Ene-39 4024 Shell and Tube Condensers 2
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991761 =
n991761i=1
1048667 u(x)
vh(x)mprimeprime
h(x)
10486692
(25)
If it is assumed that the specific vapour flow and specific volume of the vapour areconstant all over the condenser this can be written as
(vhmprimeprime
h)2991761
=n991761i=1
[u(x)]2 (26)
If the cross sectional flow area of the condenser is decreasing in such a way that thevelocity within the condenser with n rows is constant u(x) = u(1) = u(n) = u and
n991761i=1
[u(x)]2
= nu2
(27)
The constant velocity within the condenser is then
u =
991770 (vhm
primeprime
h)2sum
n (28)
Examining the top values of the curves in Fig 2 and making some calculations it ispossible to notice that the following condition is approximately true for the top points of the curves ( valid only when T lt 200oC )
(vhmprimeprime
h)2991761
= 75
1048667m2
s2
1048669 (29)
Now you can substitute this equation into Eq 26 and solve the velocity as a function of total row number n
u = u(1) = u(n) =
991770 752
n =
75radic n
983131m
s
983133 (30)
which is the dimensioning target velocity for the shell and tube condenser with constantvapour velocity It is independent of the pressure or temperature level Using thisvelocity the condenser is working approximately on top of the curves in Fig 2 whichmeans that the pressure drop is reasonable and on the other hand the velocity is highenough to sweep away the accumulating non condensable gases
The following figure gives the tube arrangement for a 4-row shell and tube condenserThe vapour is flowing inwards and condenses on each tube row In the middle of thecondenser is the suction channel which is used to pump away the non condensable gaseswhich are flowing through the condenser together with a small amount of passage vapour
7172019 Ene-39 4024 Shell and Tube Condensers 2
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Figure 5 Condenser tube locations number of tubes and areast
Based on the equations presented in the fig 5 the total outside surface area of the con-
denser tubes is
S tot = Ltotπdu = NLπdu (31)
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where L is length of one tube = LtotN and N is number of tubes
Energy balance (Eq 4) gives the total mass flow of vapour mtot = φhL when thespecific vapour flow is
mprimeprime
h = mtot
S tot=
φhL
NLπdu(32)
The condensing surface after the cross section area A1 is S 1 when the mass flow at thecross section A1 is
m1 = S 1S tot
mtot = (N minus 3n)Lπdu
NLπdumtot =
N minus 3n
N mtot (33)
When the specific volume of vapour is vh this mass flow corresponds to the volume flow
˙V 1 = m1vh =
N
minus3n
N mtotvh (34)
Based on this volume flow the velocity at the cross section A1 is then
u1 =V 1A1
=N minus3nN
mtotvh1 8Lndu
(35)
When the vapour velocity within the condenser is approximately constant u1 = u = unand the dimensioning can be done based on the velocity u1 The condenser must bedimensioned in such a way that this vapour velocity u1 at the cross section A1 is the same(or a little larger) than the target velocity u (Eq 30)
DIMENSIONING OF HORIZONTAL TUBE BANK (STEP BYSTEP)
The dimensioning can begin by choosing the cooling water velocity to be about 2ms inorder to get large heat transfer coefficient of cooling water Then choose the row numbern (when also the number of tubes N is defined see Fig 5) Now you can calculate themass flow and temperature rise of cooling water and the total needed conductance andvapour mass flow using equations 4-7 The heat transfer coefficient of cooling water can
be calculated using equations 8-10 and the heat transfer coefficient for vapour side isgiven in Fig 2 by using the top values of curves The conductance per unit length of thetube is then calculated using equations 1-3 after which the lentgth of one tube L can becalculated because N (number of tubes) is already fixed Finally the vapour velocity u1
at cross section A1 can be calculated using the equation 35 and the known mass flow of vapour This velocity should be the same as the targer velocity u in equation 30 If thisis not the case choose a new row number n until the velocity u1 is the same or a littlelarger than target velocity
You can calculate the frictional pressure drop of cooling water using Moody diagramThe pressure drop of vapour side can be calculated using the equation 20 after you first
calculate the pressure drop coefficient sum
using the equation 23 or 25
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6
Figure 2 Optimal operation point of shell and tube condensers heat transfer coefficient
as a function of condensation temperature and product 852008 mprimeprime
h9830812
sumThe parameter
852008 m
primeprime
h
9830812sum
which is used for dimensioning of the condenser is calculated
based on the specific vapour flow of the condenser mprimeprime
h and on the pressure dropcoefficient of the condenser
sum The specific vapour flow of the condenser is defined as
mprimeprime
h = m
S (16)
where m is the mass flow of the condensing vapour and S is the heat transfer surfacearea where the condensation is happening (outside area of the tubes)
If the vapour velocity is too high the operation point is moving to the left from thetop and the effective heat transfer is decreased If the velocity is too small the non-condensable gases are gathering on the tube surfaces and the operation point is movingto the right from the top thus decreasing again the effective heat transfer coefficient
Lets consider next how the pressure drop coefficient sum
is calculated The total outsideheat transfer area of condenser tubes is S tot and the tubes are positioned triangularly insuch a way that the number of the tube rows in the direction of the flow is n At theinlet (the first row of the tube bundle) the cross sectional flow area for the vapour flow
is A(1) The heat transfer surface area (outside surface area of the tubes) where thevapour is condensing after the cross section A(1) is marked as S (1) Analogically at thetube row x the cross sectional flow area for the vapour flow is A(x) and the heattransfer surface area following this cross section is S (x)
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Figure 3 Cross sectional area A and condensing surface area S of the condenser
The mass flow of vapour which is flowing through the cross section A(x) and iscondensing on the surface area S (x) is mh(x) when the specific mass flow of thecondenser after the row x is
mprimeprime
h(x) = mh(x)
S (x) (17)
The vapour velocity at the tube row x is then
u(x) = mh(x)
ρA(x) =
mprimeprime
h(x)S (x)
ρA(x) (18)
The frictional pressure drop of one tube row x is
∆ p(x) = 1
2ρξ [u(x)]2 (19)
where ξ (x) is friction coefficient of tube row x If it is assumed that density of thevapour ρ is constant the frictional pressure drop of all n tube rows is
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∆ p = 1
2ρ
x=n9917611
ξ (x)
1048616 m
primeprime
h(x)S (x)
A(x)
8520092 (20)
When it is also assumed that the friction coefficient ξ (x) is the same for all rows
ξ (x) = ξ = constant and that the specific mass flow is constant all over the condenserthe pressure drop of the condenser can be written as
∆ p = ξ
2ρ
852008m
primeprime
h
9830812 n9917611
1048616S (x)
A(x)
8520092
= ξ
2ρ
852008m
primeprime
h
9830812991761
(21)
where
n9917611
1048616S (x)A(x)
8520092
=991761
(22)
which is pressure drop coefficient of the condenser
In normal condensers the geometrical positioning of the tubes is triangular in such away that the ratio sd = 13
s
d
Fig 4 Tube arrangement
Normally in shell and tube condensers the vapour velocity is tried to keep approximately
constant when the accumulation of non-condensable gases is not a problem if thevelocity is high enough The pressure drop coefficient for this kind of condenser can becalculated based on the first row values A(1) and S (1) because in this case
7172019 Ene-39 4024 Shell and Tube Condensers 2
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S (x)
A(x) =
S (1)
A(1) = constant
and the pressure drop coefficient Σ is
991761 = n
1048616S (1)
A(1)
8520092
(23)
At different pressure levels the shell and tube condensers have very different specificmass flows This is mainly due to the fact that the specific volume of vater wapour isvery stronly dependent on the pressure level At low pressures the volume flows areincreasing dramatically which is restricting the specific mass flow of the condenser The
following table gives approximate dimensioning values for the product 852008mprimeprime
h9830812
sum as a
function of water vapour saturation pressure In the table is also given the frictioncoefficient for one tube row which is needed in pressure drop calculations
Table 2 Approximate dimensioning values for vapour side
p0 t0 ξ 852008
mprimeprime
h
9830812sum
kN m2
C - 852008 kgm2s9830812
0 0 12 012 20 098 17 40 083 10
20 60 070 102
70 90 059 103
250 130 050 104
800 170 043 105
2800 230 037 106
8000 300 032 107
More precisely the dimensioning value for the product852008
mprimeprime
h
9830812sum
can be read from Fig 2which gives it as a function of saturation temperature The values for the product852008
mprimeprime
h
9830812sum
can be manipulated a little further Based on the eq 16 you can write for therow x
S (x)
A(x)
= u(x)
vh(x)m
primeprime
h(x)
(24)
where vh(x) is the specific volume of the vapour at row x Pressure drop coefficient sum
can then be written as
7172019 Ene-39 4024 Shell and Tube Condensers 2
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10
991761 =
n991761i=1
1048667 u(x)
vh(x)mprimeprime
h(x)
10486692
(25)
If it is assumed that the specific vapour flow and specific volume of the vapour areconstant all over the condenser this can be written as
(vhmprimeprime
h)2991761
=n991761i=1
[u(x)]2 (26)
If the cross sectional flow area of the condenser is decreasing in such a way that thevelocity within the condenser with n rows is constant u(x) = u(1) = u(n) = u and
n991761i=1
[u(x)]2
= nu2
(27)
The constant velocity within the condenser is then
u =
991770 (vhm
primeprime
h)2sum
n (28)
Examining the top values of the curves in Fig 2 and making some calculations it ispossible to notice that the following condition is approximately true for the top points of the curves ( valid only when T lt 200oC )
(vhmprimeprime
h)2991761
= 75
1048667m2
s2
1048669 (29)
Now you can substitute this equation into Eq 26 and solve the velocity as a function of total row number n
u = u(1) = u(n) =
991770 752
n =
75radic n
983131m
s
983133 (30)
which is the dimensioning target velocity for the shell and tube condenser with constantvapour velocity It is independent of the pressure or temperature level Using thisvelocity the condenser is working approximately on top of the curves in Fig 2 whichmeans that the pressure drop is reasonable and on the other hand the velocity is highenough to sweep away the accumulating non condensable gases
The following figure gives the tube arrangement for a 4-row shell and tube condenserThe vapour is flowing inwards and condenses on each tube row In the middle of thecondenser is the suction channel which is used to pump away the non condensable gaseswhich are flowing through the condenser together with a small amount of passage vapour
7172019 Ene-39 4024 Shell and Tube Condensers 2
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Figure 5 Condenser tube locations number of tubes and areast
Based on the equations presented in the fig 5 the total outside surface area of the con-
denser tubes is
S tot = Ltotπdu = NLπdu (31)
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where L is length of one tube = LtotN and N is number of tubes
Energy balance (Eq 4) gives the total mass flow of vapour mtot = φhL when thespecific vapour flow is
mprimeprime
h = mtot
S tot=
φhL
NLπdu(32)
The condensing surface after the cross section area A1 is S 1 when the mass flow at thecross section A1 is
m1 = S 1S tot
mtot = (N minus 3n)Lπdu
NLπdumtot =
N minus 3n
N mtot (33)
When the specific volume of vapour is vh this mass flow corresponds to the volume flow
˙V 1 = m1vh =
N
minus3n
N mtotvh (34)
Based on this volume flow the velocity at the cross section A1 is then
u1 =V 1A1
=N minus3nN
mtotvh1 8Lndu
(35)
When the vapour velocity within the condenser is approximately constant u1 = u = unand the dimensioning can be done based on the velocity u1 The condenser must bedimensioned in such a way that this vapour velocity u1 at the cross section A1 is the same(or a little larger) than the target velocity u (Eq 30)
DIMENSIONING OF HORIZONTAL TUBE BANK (STEP BYSTEP)
The dimensioning can begin by choosing the cooling water velocity to be about 2ms inorder to get large heat transfer coefficient of cooling water Then choose the row numbern (when also the number of tubes N is defined see Fig 5) Now you can calculate themass flow and temperature rise of cooling water and the total needed conductance andvapour mass flow using equations 4-7 The heat transfer coefficient of cooling water can
be calculated using equations 8-10 and the heat transfer coefficient for vapour side isgiven in Fig 2 by using the top values of curves The conductance per unit length of thetube is then calculated using equations 1-3 after which the lentgth of one tube L can becalculated because N (number of tubes) is already fixed Finally the vapour velocity u1
at cross section A1 can be calculated using the equation 35 and the known mass flow of vapour This velocity should be the same as the targer velocity u in equation 30 If thisis not the case choose a new row number n until the velocity u1 is the same or a littlelarger than target velocity
You can calculate the frictional pressure drop of cooling water using Moody diagramThe pressure drop of vapour side can be calculated using the equation 20 after you first
calculate the pressure drop coefficient sum
using the equation 23 or 25
7172019 Ene-39 4024 Shell and Tube Condensers 2
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Figure 3 Cross sectional area A and condensing surface area S of the condenser
The mass flow of vapour which is flowing through the cross section A(x) and iscondensing on the surface area S (x) is mh(x) when the specific mass flow of thecondenser after the row x is
mprimeprime
h(x) = mh(x)
S (x) (17)
The vapour velocity at the tube row x is then
u(x) = mh(x)
ρA(x) =
mprimeprime
h(x)S (x)
ρA(x) (18)
The frictional pressure drop of one tube row x is
∆ p(x) = 1
2ρξ [u(x)]2 (19)
where ξ (x) is friction coefficient of tube row x If it is assumed that density of thevapour ρ is constant the frictional pressure drop of all n tube rows is
7172019 Ene-39 4024 Shell and Tube Condensers 2
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∆ p = 1
2ρ
x=n9917611
ξ (x)
1048616 m
primeprime
h(x)S (x)
A(x)
8520092 (20)
When it is also assumed that the friction coefficient ξ (x) is the same for all rows
ξ (x) = ξ = constant and that the specific mass flow is constant all over the condenserthe pressure drop of the condenser can be written as
∆ p = ξ
2ρ
852008m
primeprime
h
9830812 n9917611
1048616S (x)
A(x)
8520092
= ξ
2ρ
852008m
primeprime
h
9830812991761
(21)
where
n9917611
1048616S (x)A(x)
8520092
=991761
(22)
which is pressure drop coefficient of the condenser
In normal condensers the geometrical positioning of the tubes is triangular in such away that the ratio sd = 13
s
d
Fig 4 Tube arrangement
Normally in shell and tube condensers the vapour velocity is tried to keep approximately
constant when the accumulation of non-condensable gases is not a problem if thevelocity is high enough The pressure drop coefficient for this kind of condenser can becalculated based on the first row values A(1) and S (1) because in this case
7172019 Ene-39 4024 Shell and Tube Condensers 2
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9
S (x)
A(x) =
S (1)
A(1) = constant
and the pressure drop coefficient Σ is
991761 = n
1048616S (1)
A(1)
8520092
(23)
At different pressure levels the shell and tube condensers have very different specificmass flows This is mainly due to the fact that the specific volume of vater wapour isvery stronly dependent on the pressure level At low pressures the volume flows areincreasing dramatically which is restricting the specific mass flow of the condenser The
following table gives approximate dimensioning values for the product 852008mprimeprime
h9830812
sum as a
function of water vapour saturation pressure In the table is also given the frictioncoefficient for one tube row which is needed in pressure drop calculations
Table 2 Approximate dimensioning values for vapour side
p0 t0 ξ 852008
mprimeprime
h
9830812sum
kN m2
C - 852008 kgm2s9830812
0 0 12 012 20 098 17 40 083 10
20 60 070 102
70 90 059 103
250 130 050 104
800 170 043 105
2800 230 037 106
8000 300 032 107
More precisely the dimensioning value for the product852008
mprimeprime
h
9830812sum
can be read from Fig 2which gives it as a function of saturation temperature The values for the product852008
mprimeprime
h
9830812sum
can be manipulated a little further Based on the eq 16 you can write for therow x
S (x)
A(x)
= u(x)
vh(x)m
primeprime
h(x)
(24)
where vh(x) is the specific volume of the vapour at row x Pressure drop coefficient sum
can then be written as
7172019 Ene-39 4024 Shell and Tube Condensers 2
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991761 =
n991761i=1
1048667 u(x)
vh(x)mprimeprime
h(x)
10486692
(25)
If it is assumed that the specific vapour flow and specific volume of the vapour areconstant all over the condenser this can be written as
(vhmprimeprime
h)2991761
=n991761i=1
[u(x)]2 (26)
If the cross sectional flow area of the condenser is decreasing in such a way that thevelocity within the condenser with n rows is constant u(x) = u(1) = u(n) = u and
n991761i=1
[u(x)]2
= nu2
(27)
The constant velocity within the condenser is then
u =
991770 (vhm
primeprime
h)2sum
n (28)
Examining the top values of the curves in Fig 2 and making some calculations it ispossible to notice that the following condition is approximately true for the top points of the curves ( valid only when T lt 200oC )
(vhmprimeprime
h)2991761
= 75
1048667m2
s2
1048669 (29)
Now you can substitute this equation into Eq 26 and solve the velocity as a function of total row number n
u = u(1) = u(n) =
991770 752
n =
75radic n
983131m
s
983133 (30)
which is the dimensioning target velocity for the shell and tube condenser with constantvapour velocity It is independent of the pressure or temperature level Using thisvelocity the condenser is working approximately on top of the curves in Fig 2 whichmeans that the pressure drop is reasonable and on the other hand the velocity is highenough to sweep away the accumulating non condensable gases
The following figure gives the tube arrangement for a 4-row shell and tube condenserThe vapour is flowing inwards and condenses on each tube row In the middle of thecondenser is the suction channel which is used to pump away the non condensable gaseswhich are flowing through the condenser together with a small amount of passage vapour
7172019 Ene-39 4024 Shell and Tube Condensers 2
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11
Figure 5 Condenser tube locations number of tubes and areast
Based on the equations presented in the fig 5 the total outside surface area of the con-
denser tubes is
S tot = Ltotπdu = NLπdu (31)
7172019 Ene-39 4024 Shell and Tube Condensers 2
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12
where L is length of one tube = LtotN and N is number of tubes
Energy balance (Eq 4) gives the total mass flow of vapour mtot = φhL when thespecific vapour flow is
mprimeprime
h = mtot
S tot=
φhL
NLπdu(32)
The condensing surface after the cross section area A1 is S 1 when the mass flow at thecross section A1 is
m1 = S 1S tot
mtot = (N minus 3n)Lπdu
NLπdumtot =
N minus 3n
N mtot (33)
When the specific volume of vapour is vh this mass flow corresponds to the volume flow
˙V 1 = m1vh =
N
minus3n
N mtotvh (34)
Based on this volume flow the velocity at the cross section A1 is then
u1 =V 1A1
=N minus3nN
mtotvh1 8Lndu
(35)
When the vapour velocity within the condenser is approximately constant u1 = u = unand the dimensioning can be done based on the velocity u1 The condenser must bedimensioned in such a way that this vapour velocity u1 at the cross section A1 is the same(or a little larger) than the target velocity u (Eq 30)
DIMENSIONING OF HORIZONTAL TUBE BANK (STEP BYSTEP)
The dimensioning can begin by choosing the cooling water velocity to be about 2ms inorder to get large heat transfer coefficient of cooling water Then choose the row numbern (when also the number of tubes N is defined see Fig 5) Now you can calculate themass flow and temperature rise of cooling water and the total needed conductance andvapour mass flow using equations 4-7 The heat transfer coefficient of cooling water can
be calculated using equations 8-10 and the heat transfer coefficient for vapour side isgiven in Fig 2 by using the top values of curves The conductance per unit length of thetube is then calculated using equations 1-3 after which the lentgth of one tube L can becalculated because N (number of tubes) is already fixed Finally the vapour velocity u1
at cross section A1 can be calculated using the equation 35 and the known mass flow of vapour This velocity should be the same as the targer velocity u in equation 30 If thisis not the case choose a new row number n until the velocity u1 is the same or a littlelarger than target velocity
You can calculate the frictional pressure drop of cooling water using Moody diagramThe pressure drop of vapour side can be calculated using the equation 20 after you first
calculate the pressure drop coefficient sum
using the equation 23 or 25
7172019 Ene-39 4024 Shell and Tube Condensers 2
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∆ p = 1
2ρ
x=n9917611
ξ (x)
1048616 m
primeprime
h(x)S (x)
A(x)
8520092 (20)
When it is also assumed that the friction coefficient ξ (x) is the same for all rows
ξ (x) = ξ = constant and that the specific mass flow is constant all over the condenserthe pressure drop of the condenser can be written as
∆ p = ξ
2ρ
852008m
primeprime
h
9830812 n9917611
1048616S (x)
A(x)
8520092
= ξ
2ρ
852008m
primeprime
h
9830812991761
(21)
where
n9917611
1048616S (x)A(x)
8520092
=991761
(22)
which is pressure drop coefficient of the condenser
In normal condensers the geometrical positioning of the tubes is triangular in such away that the ratio sd = 13
s
d
Fig 4 Tube arrangement
Normally in shell and tube condensers the vapour velocity is tried to keep approximately
constant when the accumulation of non-condensable gases is not a problem if thevelocity is high enough The pressure drop coefficient for this kind of condenser can becalculated based on the first row values A(1) and S (1) because in this case
7172019 Ene-39 4024 Shell and Tube Condensers 2
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9
S (x)
A(x) =
S (1)
A(1) = constant
and the pressure drop coefficient Σ is
991761 = n
1048616S (1)
A(1)
8520092
(23)
At different pressure levels the shell and tube condensers have very different specificmass flows This is mainly due to the fact that the specific volume of vater wapour isvery stronly dependent on the pressure level At low pressures the volume flows areincreasing dramatically which is restricting the specific mass flow of the condenser The
following table gives approximate dimensioning values for the product 852008mprimeprime
h9830812
sum as a
function of water vapour saturation pressure In the table is also given the frictioncoefficient for one tube row which is needed in pressure drop calculations
Table 2 Approximate dimensioning values for vapour side
p0 t0 ξ 852008
mprimeprime
h
9830812sum
kN m2
C - 852008 kgm2s9830812
0 0 12 012 20 098 17 40 083 10
20 60 070 102
70 90 059 103
250 130 050 104
800 170 043 105
2800 230 037 106
8000 300 032 107
More precisely the dimensioning value for the product852008
mprimeprime
h
9830812sum
can be read from Fig 2which gives it as a function of saturation temperature The values for the product852008
mprimeprime
h
9830812sum
can be manipulated a little further Based on the eq 16 you can write for therow x
S (x)
A(x)
= u(x)
vh(x)m
primeprime
h(x)
(24)
where vh(x) is the specific volume of the vapour at row x Pressure drop coefficient sum
can then be written as
7172019 Ene-39 4024 Shell and Tube Condensers 2
httpslidepdfcomreaderfullene-39-4024-shell-and-tube-condensers-2 1012
10
991761 =
n991761i=1
1048667 u(x)
vh(x)mprimeprime
h(x)
10486692
(25)
If it is assumed that the specific vapour flow and specific volume of the vapour areconstant all over the condenser this can be written as
(vhmprimeprime
h)2991761
=n991761i=1
[u(x)]2 (26)
If the cross sectional flow area of the condenser is decreasing in such a way that thevelocity within the condenser with n rows is constant u(x) = u(1) = u(n) = u and
n991761i=1
[u(x)]2
= nu2
(27)
The constant velocity within the condenser is then
u =
991770 (vhm
primeprime
h)2sum
n (28)
Examining the top values of the curves in Fig 2 and making some calculations it ispossible to notice that the following condition is approximately true for the top points of the curves ( valid only when T lt 200oC )
(vhmprimeprime
h)2991761
= 75
1048667m2
s2
1048669 (29)
Now you can substitute this equation into Eq 26 and solve the velocity as a function of total row number n
u = u(1) = u(n) =
991770 752
n =
75radic n
983131m
s
983133 (30)
which is the dimensioning target velocity for the shell and tube condenser with constantvapour velocity It is independent of the pressure or temperature level Using thisvelocity the condenser is working approximately on top of the curves in Fig 2 whichmeans that the pressure drop is reasonable and on the other hand the velocity is highenough to sweep away the accumulating non condensable gases
The following figure gives the tube arrangement for a 4-row shell and tube condenserThe vapour is flowing inwards and condenses on each tube row In the middle of thecondenser is the suction channel which is used to pump away the non condensable gaseswhich are flowing through the condenser together with a small amount of passage vapour
7172019 Ene-39 4024 Shell and Tube Condensers 2
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Figure 5 Condenser tube locations number of tubes and areast
Based on the equations presented in the fig 5 the total outside surface area of the con-
denser tubes is
S tot = Ltotπdu = NLπdu (31)
7172019 Ene-39 4024 Shell and Tube Condensers 2
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where L is length of one tube = LtotN and N is number of tubes
Energy balance (Eq 4) gives the total mass flow of vapour mtot = φhL when thespecific vapour flow is
mprimeprime
h = mtot
S tot=
φhL
NLπdu(32)
The condensing surface after the cross section area A1 is S 1 when the mass flow at thecross section A1 is
m1 = S 1S tot
mtot = (N minus 3n)Lπdu
NLπdumtot =
N minus 3n
N mtot (33)
When the specific volume of vapour is vh this mass flow corresponds to the volume flow
˙V 1 = m1vh =
N
minus3n
N mtotvh (34)
Based on this volume flow the velocity at the cross section A1 is then
u1 =V 1A1
=N minus3nN
mtotvh1 8Lndu
(35)
When the vapour velocity within the condenser is approximately constant u1 = u = unand the dimensioning can be done based on the velocity u1 The condenser must bedimensioned in such a way that this vapour velocity u1 at the cross section A1 is the same(or a little larger) than the target velocity u (Eq 30)
DIMENSIONING OF HORIZONTAL TUBE BANK (STEP BYSTEP)
The dimensioning can begin by choosing the cooling water velocity to be about 2ms inorder to get large heat transfer coefficient of cooling water Then choose the row numbern (when also the number of tubes N is defined see Fig 5) Now you can calculate themass flow and temperature rise of cooling water and the total needed conductance andvapour mass flow using equations 4-7 The heat transfer coefficient of cooling water can
be calculated using equations 8-10 and the heat transfer coefficient for vapour side isgiven in Fig 2 by using the top values of curves The conductance per unit length of thetube is then calculated using equations 1-3 after which the lentgth of one tube L can becalculated because N (number of tubes) is already fixed Finally the vapour velocity u1
at cross section A1 can be calculated using the equation 35 and the known mass flow of vapour This velocity should be the same as the targer velocity u in equation 30 If thisis not the case choose a new row number n until the velocity u1 is the same or a littlelarger than target velocity
You can calculate the frictional pressure drop of cooling water using Moody diagramThe pressure drop of vapour side can be calculated using the equation 20 after you first
calculate the pressure drop coefficient sum
using the equation 23 or 25
7172019 Ene-39 4024 Shell and Tube Condensers 2
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9
S (x)
A(x) =
S (1)
A(1) = constant
and the pressure drop coefficient Σ is
991761 = n
1048616S (1)
A(1)
8520092
(23)
At different pressure levels the shell and tube condensers have very different specificmass flows This is mainly due to the fact that the specific volume of vater wapour isvery stronly dependent on the pressure level At low pressures the volume flows areincreasing dramatically which is restricting the specific mass flow of the condenser The
following table gives approximate dimensioning values for the product 852008mprimeprime
h9830812
sum as a
function of water vapour saturation pressure In the table is also given the frictioncoefficient for one tube row which is needed in pressure drop calculations
Table 2 Approximate dimensioning values for vapour side
p0 t0 ξ 852008
mprimeprime
h
9830812sum
kN m2
C - 852008 kgm2s9830812
0 0 12 012 20 098 17 40 083 10
20 60 070 102
70 90 059 103
250 130 050 104
800 170 043 105
2800 230 037 106
8000 300 032 107
More precisely the dimensioning value for the product852008
mprimeprime
h
9830812sum
can be read from Fig 2which gives it as a function of saturation temperature The values for the product852008
mprimeprime
h
9830812sum
can be manipulated a little further Based on the eq 16 you can write for therow x
S (x)
A(x)
= u(x)
vh(x)m
primeprime
h(x)
(24)
where vh(x) is the specific volume of the vapour at row x Pressure drop coefficient sum
can then be written as
7172019 Ene-39 4024 Shell and Tube Condensers 2
httpslidepdfcomreaderfullene-39-4024-shell-and-tube-condensers-2 1012
10
991761 =
n991761i=1
1048667 u(x)
vh(x)mprimeprime
h(x)
10486692
(25)
If it is assumed that the specific vapour flow and specific volume of the vapour areconstant all over the condenser this can be written as
(vhmprimeprime
h)2991761
=n991761i=1
[u(x)]2 (26)
If the cross sectional flow area of the condenser is decreasing in such a way that thevelocity within the condenser with n rows is constant u(x) = u(1) = u(n) = u and
n991761i=1
[u(x)]2
= nu2
(27)
The constant velocity within the condenser is then
u =
991770 (vhm
primeprime
h)2sum
n (28)
Examining the top values of the curves in Fig 2 and making some calculations it ispossible to notice that the following condition is approximately true for the top points of the curves ( valid only when T lt 200oC )
(vhmprimeprime
h)2991761
= 75
1048667m2
s2
1048669 (29)
Now you can substitute this equation into Eq 26 and solve the velocity as a function of total row number n
u = u(1) = u(n) =
991770 752
n =
75radic n
983131m
s
983133 (30)
which is the dimensioning target velocity for the shell and tube condenser with constantvapour velocity It is independent of the pressure or temperature level Using thisvelocity the condenser is working approximately on top of the curves in Fig 2 whichmeans that the pressure drop is reasonable and on the other hand the velocity is highenough to sweep away the accumulating non condensable gases
The following figure gives the tube arrangement for a 4-row shell and tube condenserThe vapour is flowing inwards and condenses on each tube row In the middle of thecondenser is the suction channel which is used to pump away the non condensable gaseswhich are flowing through the condenser together with a small amount of passage vapour
7172019 Ene-39 4024 Shell and Tube Condensers 2
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11
Figure 5 Condenser tube locations number of tubes and areast
Based on the equations presented in the fig 5 the total outside surface area of the con-
denser tubes is
S tot = Ltotπdu = NLπdu (31)
7172019 Ene-39 4024 Shell and Tube Condensers 2
httpslidepdfcomreaderfullene-39-4024-shell-and-tube-condensers-2 1212
12
where L is length of one tube = LtotN and N is number of tubes
Energy balance (Eq 4) gives the total mass flow of vapour mtot = φhL when thespecific vapour flow is
mprimeprime
h = mtot
S tot=
φhL
NLπdu(32)
The condensing surface after the cross section area A1 is S 1 when the mass flow at thecross section A1 is
m1 = S 1S tot
mtot = (N minus 3n)Lπdu
NLπdumtot =
N minus 3n
N mtot (33)
When the specific volume of vapour is vh this mass flow corresponds to the volume flow
˙V 1 = m1vh =
N
minus3n
N mtotvh (34)
Based on this volume flow the velocity at the cross section A1 is then
u1 =V 1A1
=N minus3nN
mtotvh1 8Lndu
(35)
When the vapour velocity within the condenser is approximately constant u1 = u = unand the dimensioning can be done based on the velocity u1 The condenser must bedimensioned in such a way that this vapour velocity u1 at the cross section A1 is the same(or a little larger) than the target velocity u (Eq 30)
DIMENSIONING OF HORIZONTAL TUBE BANK (STEP BYSTEP)
The dimensioning can begin by choosing the cooling water velocity to be about 2ms inorder to get large heat transfer coefficient of cooling water Then choose the row numbern (when also the number of tubes N is defined see Fig 5) Now you can calculate themass flow and temperature rise of cooling water and the total needed conductance andvapour mass flow using equations 4-7 The heat transfer coefficient of cooling water can
be calculated using equations 8-10 and the heat transfer coefficient for vapour side isgiven in Fig 2 by using the top values of curves The conductance per unit length of thetube is then calculated using equations 1-3 after which the lentgth of one tube L can becalculated because N (number of tubes) is already fixed Finally the vapour velocity u1
at cross section A1 can be calculated using the equation 35 and the known mass flow of vapour This velocity should be the same as the targer velocity u in equation 30 If thisis not the case choose a new row number n until the velocity u1 is the same or a littlelarger than target velocity
You can calculate the frictional pressure drop of cooling water using Moody diagramThe pressure drop of vapour side can be calculated using the equation 20 after you first
calculate the pressure drop coefficient sum
using the equation 23 or 25
7172019 Ene-39 4024 Shell and Tube Condensers 2
httpslidepdfcomreaderfullene-39-4024-shell-and-tube-condensers-2 1012
10
991761 =
n991761i=1
1048667 u(x)
vh(x)mprimeprime
h(x)
10486692
(25)
If it is assumed that the specific vapour flow and specific volume of the vapour areconstant all over the condenser this can be written as
(vhmprimeprime
h)2991761
=n991761i=1
[u(x)]2 (26)
If the cross sectional flow area of the condenser is decreasing in such a way that thevelocity within the condenser with n rows is constant u(x) = u(1) = u(n) = u and
n991761i=1
[u(x)]2
= nu2
(27)
The constant velocity within the condenser is then
u =
991770 (vhm
primeprime
h)2sum
n (28)
Examining the top values of the curves in Fig 2 and making some calculations it ispossible to notice that the following condition is approximately true for the top points of the curves ( valid only when T lt 200oC )
(vhmprimeprime
h)2991761
= 75
1048667m2
s2
1048669 (29)
Now you can substitute this equation into Eq 26 and solve the velocity as a function of total row number n
u = u(1) = u(n) =
991770 752
n =
75radic n
983131m
s
983133 (30)
which is the dimensioning target velocity for the shell and tube condenser with constantvapour velocity It is independent of the pressure or temperature level Using thisvelocity the condenser is working approximately on top of the curves in Fig 2 whichmeans that the pressure drop is reasonable and on the other hand the velocity is highenough to sweep away the accumulating non condensable gases
The following figure gives the tube arrangement for a 4-row shell and tube condenserThe vapour is flowing inwards and condenses on each tube row In the middle of thecondenser is the suction channel which is used to pump away the non condensable gaseswhich are flowing through the condenser together with a small amount of passage vapour
7172019 Ene-39 4024 Shell and Tube Condensers 2
httpslidepdfcomreaderfullene-39-4024-shell-and-tube-condensers-2 1112
11
Figure 5 Condenser tube locations number of tubes and areast
Based on the equations presented in the fig 5 the total outside surface area of the con-
denser tubes is
S tot = Ltotπdu = NLπdu (31)
7172019 Ene-39 4024 Shell and Tube Condensers 2
httpslidepdfcomreaderfullene-39-4024-shell-and-tube-condensers-2 1212
12
where L is length of one tube = LtotN and N is number of tubes
Energy balance (Eq 4) gives the total mass flow of vapour mtot = φhL when thespecific vapour flow is
mprimeprime
h = mtot
S tot=
φhL
NLπdu(32)
The condensing surface after the cross section area A1 is S 1 when the mass flow at thecross section A1 is
m1 = S 1S tot
mtot = (N minus 3n)Lπdu
NLπdumtot =
N minus 3n
N mtot (33)
When the specific volume of vapour is vh this mass flow corresponds to the volume flow
˙V 1 = m1vh =
N
minus3n
N mtotvh (34)
Based on this volume flow the velocity at the cross section A1 is then
u1 =V 1A1
=N minus3nN
mtotvh1 8Lndu
(35)
When the vapour velocity within the condenser is approximately constant u1 = u = unand the dimensioning can be done based on the velocity u1 The condenser must bedimensioned in such a way that this vapour velocity u1 at the cross section A1 is the same(or a little larger) than the target velocity u (Eq 30)
DIMENSIONING OF HORIZONTAL TUBE BANK (STEP BYSTEP)
The dimensioning can begin by choosing the cooling water velocity to be about 2ms inorder to get large heat transfer coefficient of cooling water Then choose the row numbern (when also the number of tubes N is defined see Fig 5) Now you can calculate themass flow and temperature rise of cooling water and the total needed conductance andvapour mass flow using equations 4-7 The heat transfer coefficient of cooling water can
be calculated using equations 8-10 and the heat transfer coefficient for vapour side isgiven in Fig 2 by using the top values of curves The conductance per unit length of thetube is then calculated using equations 1-3 after which the lentgth of one tube L can becalculated because N (number of tubes) is already fixed Finally the vapour velocity u1
at cross section A1 can be calculated using the equation 35 and the known mass flow of vapour This velocity should be the same as the targer velocity u in equation 30 If thisis not the case choose a new row number n until the velocity u1 is the same or a littlelarger than target velocity
You can calculate the frictional pressure drop of cooling water using Moody diagramThe pressure drop of vapour side can be calculated using the equation 20 after you first
calculate the pressure drop coefficient sum
using the equation 23 or 25
7172019 Ene-39 4024 Shell and Tube Condensers 2
httpslidepdfcomreaderfullene-39-4024-shell-and-tube-condensers-2 1112
11
Figure 5 Condenser tube locations number of tubes and areast
Based on the equations presented in the fig 5 the total outside surface area of the con-
denser tubes is
S tot = Ltotπdu = NLπdu (31)
7172019 Ene-39 4024 Shell and Tube Condensers 2
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12
where L is length of one tube = LtotN and N is number of tubes
Energy balance (Eq 4) gives the total mass flow of vapour mtot = φhL when thespecific vapour flow is
mprimeprime
h = mtot
S tot=
φhL
NLπdu(32)
The condensing surface after the cross section area A1 is S 1 when the mass flow at thecross section A1 is
m1 = S 1S tot
mtot = (N minus 3n)Lπdu
NLπdumtot =
N minus 3n
N mtot (33)
When the specific volume of vapour is vh this mass flow corresponds to the volume flow
˙V 1 = m1vh =
N
minus3n
N mtotvh (34)
Based on this volume flow the velocity at the cross section A1 is then
u1 =V 1A1
=N minus3nN
mtotvh1 8Lndu
(35)
When the vapour velocity within the condenser is approximately constant u1 = u = unand the dimensioning can be done based on the velocity u1 The condenser must bedimensioned in such a way that this vapour velocity u1 at the cross section A1 is the same(or a little larger) than the target velocity u (Eq 30)
DIMENSIONING OF HORIZONTAL TUBE BANK (STEP BYSTEP)
The dimensioning can begin by choosing the cooling water velocity to be about 2ms inorder to get large heat transfer coefficient of cooling water Then choose the row numbern (when also the number of tubes N is defined see Fig 5) Now you can calculate themass flow and temperature rise of cooling water and the total needed conductance andvapour mass flow using equations 4-7 The heat transfer coefficient of cooling water can
be calculated using equations 8-10 and the heat transfer coefficient for vapour side isgiven in Fig 2 by using the top values of curves The conductance per unit length of thetube is then calculated using equations 1-3 after which the lentgth of one tube L can becalculated because N (number of tubes) is already fixed Finally the vapour velocity u1
at cross section A1 can be calculated using the equation 35 and the known mass flow of vapour This velocity should be the same as the targer velocity u in equation 30 If thisis not the case choose a new row number n until the velocity u1 is the same or a littlelarger than target velocity
You can calculate the frictional pressure drop of cooling water using Moody diagramThe pressure drop of vapour side can be calculated using the equation 20 after you first
calculate the pressure drop coefficient sum
using the equation 23 or 25
7172019 Ene-39 4024 Shell and Tube Condensers 2
httpslidepdfcomreaderfullene-39-4024-shell-and-tube-condensers-2 1212
12
where L is length of one tube = LtotN and N is number of tubes
Energy balance (Eq 4) gives the total mass flow of vapour mtot = φhL when thespecific vapour flow is
mprimeprime
h = mtot
S tot=
φhL
NLπdu(32)
The condensing surface after the cross section area A1 is S 1 when the mass flow at thecross section A1 is
m1 = S 1S tot
mtot = (N minus 3n)Lπdu
NLπdumtot =
N minus 3n
N mtot (33)
When the specific volume of vapour is vh this mass flow corresponds to the volume flow
˙V 1 = m1vh =
N
minus3n
N mtotvh (34)
Based on this volume flow the velocity at the cross section A1 is then
u1 =V 1A1
=N minus3nN
mtotvh1 8Lndu
(35)
When the vapour velocity within the condenser is approximately constant u1 = u = unand the dimensioning can be done based on the velocity u1 The condenser must bedimensioned in such a way that this vapour velocity u1 at the cross section A1 is the same(or a little larger) than the target velocity u (Eq 30)
DIMENSIONING OF HORIZONTAL TUBE BANK (STEP BYSTEP)
The dimensioning can begin by choosing the cooling water velocity to be about 2ms inorder to get large heat transfer coefficient of cooling water Then choose the row numbern (when also the number of tubes N is defined see Fig 5) Now you can calculate themass flow and temperature rise of cooling water and the total needed conductance andvapour mass flow using equations 4-7 The heat transfer coefficient of cooling water can
be calculated using equations 8-10 and the heat transfer coefficient for vapour side isgiven in Fig 2 by using the top values of curves The conductance per unit length of thetube is then calculated using equations 1-3 after which the lentgth of one tube L can becalculated because N (number of tubes) is already fixed Finally the vapour velocity u1
at cross section A1 can be calculated using the equation 35 and the known mass flow of vapour This velocity should be the same as the targer velocity u in equation 30 If thisis not the case choose a new row number n until the velocity u1 is the same or a littlelarger than target velocity
You can calculate the frictional pressure drop of cooling water using Moody diagramThe pressure drop of vapour side can be calculated using the equation 20 after you first
calculate the pressure drop coefficient sum
using the equation 23 or 25