Ene-39 4024 Shell and Tube Condensers 2

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HORIZONTAL SHELL AND TUBE CONDENSERS

Kari Saari

26 maaliskuuta 2010

1



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



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



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



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]



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)



7

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



8

∆ 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



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



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



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)



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



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



3



(5)


v2


ε = ∆T



z = GC min


COOLING WATER SIDE


N u = hsds

λ (8)


Nu = 0023Re08P r04 (9)



N u =f 8

(Reminus 1000)P r

1 + 127radic

f 8

(P r23 minus 1)

1 +

1048616dsl

85200923 (10)






4

∆ p = ξ L

ds

1

2ρw2 (12)



ξ = 031644radic

Re(13)



approximated as

1radic ξ

= 2lg(dsk

) + 114 (14)






5

1radic ξ

= minus20lg

1048667 252

Reradic

ξ +

k

371ds

1048669 (15)












mprimeprime

h

9830812sum

as a




6



h9830812

sumThe parameter

852008 m

primeprime

h

9830812sum





mprimeprime

h = m

S (16)








7



mprimeprime

h(x) = mh(x)

S (x) (17)


u(x) = mh(x)

ρA(x) =

mprimeprime

h(x)S (x)

ρA(x) (18)


∆ p(x) = 1

2ρξ [u(x)]2 (19)




8

∆ p = 1

2ρ

x=n9917611

ξ (x)

1048616 m

primeprime

h(x)S (x)

A(x)

8520092 (20)



∆ 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)



s

d






9

S (x)

A(x) =

S (1)

A(1) = constant


991761 = n

1048616S (1)

A(1)

8520092

(23)



h9830812

sum as a



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


mprimeprime

h

9830812sum


mprimeprime

h

9830812sum


S (x)

A(x)

= u(x)

vh(x)m

primeprime

h(x)

(24)





10

991761 =

n991761i=1

1048667 u(x)

vh(x)mprimeprime

h(x)

10486692

(25)


(vhmprimeprime

h)2991761

=n991761i=1

[u(x)]2 (26)


n991761i=1

[u(x)]2

= nu2

(27)


u =

991770 (vhm

primeprime

h)2sum

n (28)


(vhmprimeprime

h)2991761

= 75

1048667m2

s2

1048669 (29)


u = u(1) = u(n) =

991770 752

n =

75radic n

983131m

s

983133 (30)





11



denser tubes is




12



mprimeprime

h = mtot

S tot=

φhL

NLπdu(32)


m1 = S 1S tot


NLπdumtot =

N minus 3n

N mtot (33)


˙V 1 = m1vh =

N

minus3n

N mtotvh (34)


u1 =V 1A1

=N minus3nN

mtotvh1 8Lndu

(35)











3



(5)


v2


ε = ∆T



z = GC min


COOLING WATER SIDE


N u = hsds

λ (8)


Nu = 0023Re08P r04 (9)



N u =f 8

(Reminus 1000)P r

1 + 127radic

f 8

(P r23 minus 1)

1 +

1048616dsl

85200923 (10)






4

∆ p = ξ L

ds

1

2ρw2 (12)



ξ = 031644radic

Re(13)



approximated as

1radic ξ

= 2lg(dsk

) + 114 (14)






5

1radic ξ

= minus20lg

1048667 252

Reradic

ξ +

k

371ds

1048669 (15)












mprimeprime

h

9830812sum

as a




6



h9830812

sumThe parameter

852008 m

primeprime

h

9830812sum





mprimeprime

h = m

S (16)








7



mprimeprime

h(x) = mh(x)

S (x) (17)


u(x) = mh(x)

ρA(x) =

mprimeprime

h(x)S (x)

ρA(x) (18)


∆ p(x) = 1

2ρξ [u(x)]2 (19)




8

∆ p = 1

2ρ

x=n9917611

ξ (x)

1048616 m

primeprime

h(x)S (x)

A(x)

8520092 (20)



∆ 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)



s

d






9

S (x)

A(x) =

S (1)

A(1) = constant


991761 = n

1048616S (1)

A(1)

8520092

(23)



h9830812

sum as a



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


mprimeprime

h

9830812sum


mprimeprime

h

9830812sum


S (x)

A(x)

= u(x)

vh(x)m

primeprime

h(x)

(24)





10

991761 =

n991761i=1

1048667 u(x)

vh(x)mprimeprime

h(x)

10486692

(25)


(vhmprimeprime

h)2991761

=n991761i=1

[u(x)]2 (26)


n991761i=1

[u(x)]2

= nu2

(27)


u =

991770 (vhm

primeprime

h)2sum

n (28)


(vhmprimeprime

h)2991761

= 75

1048667m2

s2

1048669 (29)


u = u(1) = u(n) =

991770 752

n =

75radic n

983131m

s

983133 (30)





11



denser tubes is




12



mprimeprime

h = mtot

S tot=

φhL

NLπdu(32)


m1 = S 1S tot


NLπdumtot =

N minus 3n

N mtot (33)


˙V 1 = m1vh =

N

minus3n

N mtotvh (34)


u1 =V 1A1

=N minus3nN

mtotvh1 8Lndu

(35)











4

∆ p = ξ L

ds

1

2ρw2 (12)



ξ = 031644radic

Re(13)



approximated as

1radic ξ

= 2lg(dsk

) + 114 (14)






5

1radic ξ

= minus20lg

1048667 252

Reradic

ξ +

k

371ds

1048669 (15)












mprimeprime

h

9830812sum

as a




6



h9830812

sumThe parameter

852008 m

primeprime

h

9830812sum





mprimeprime

h = m

S (16)








7



mprimeprime

h(x) = mh(x)

S (x) (17)


u(x) = mh(x)

ρA(x) =

mprimeprime

h(x)S (x)

ρA(x) (18)


∆ p(x) = 1

2ρξ [u(x)]2 (19)




8

∆ p = 1

2ρ

x=n9917611

ξ (x)

1048616 m

primeprime

h(x)S (x)

A(x)

8520092 (20)



∆ 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)



s

d






9

S (x)

A(x) =

S (1)

A(1) = constant


991761 = n

1048616S (1)

A(1)

8520092

(23)



h9830812

sum as a



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


mprimeprime

h

9830812sum


mprimeprime

h

9830812sum


S (x)

A(x)

= u(x)

vh(x)m

primeprime

h(x)

(24)





10

991761 =

n991761i=1

1048667 u(x)

vh(x)mprimeprime

h(x)

10486692

(25)


(vhmprimeprime

h)2991761

=n991761i=1

[u(x)]2 (26)


n991761i=1

[u(x)]2

= nu2

(27)


u =

991770 (vhm

primeprime

h)2sum

n (28)


(vhmprimeprime

h)2991761

= 75

1048667m2

s2

1048669 (29)


u = u(1) = u(n) =

991770 752

n =

75radic n

983131m

s

983133 (30)





11



denser tubes is




12



mprimeprime

h = mtot

S tot=

φhL

NLπdu(32)


m1 = S 1S tot


NLπdumtot =

N minus 3n

N mtot (33)


˙V 1 = m1vh =

N

minus3n

N mtotvh (34)


u1 =V 1A1

=N minus3nN

mtotvh1 8Lndu

(35)











5

1radic ξ

= minus20lg

1048667 252

Reradic

ξ +

k

371ds

1048669 (15)












mprimeprime

h

9830812sum

as a




6



h9830812

sumThe parameter

852008 m

primeprime

h

9830812sum





mprimeprime

h = m

S (16)








7



mprimeprime

h(x) = mh(x)

S (x) (17)


u(x) = mh(x)

ρA(x) =

mprimeprime

h(x)S (x)

ρA(x) (18)


∆ p(x) = 1

2ρξ [u(x)]2 (19)




8

∆ p = 1

2ρ

x=n9917611

ξ (x)

1048616 m

primeprime

h(x)S (x)

A(x)

8520092 (20)



∆ 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)



s

d






9

S (x)

A(x) =

S (1)

A(1) = constant


991761 = n

1048616S (1)

A(1)

8520092

(23)



h9830812

sum as a



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


mprimeprime

h

9830812sum


mprimeprime

h

9830812sum


S (x)

A(x)

= u(x)

vh(x)m

primeprime

h(x)

(24)





10

991761 =

n991761i=1

1048667 u(x)

vh(x)mprimeprime

h(x)

10486692

(25)


(vhmprimeprime

h)2991761

=n991761i=1

[u(x)]2 (26)


n991761i=1

[u(x)]2

= nu2

(27)


u =

991770 (vhm

primeprime

h)2sum

n (28)


(vhmprimeprime

h)2991761

= 75

1048667m2

s2

1048669 (29)


u = u(1) = u(n) =

991770 752

n =

75radic n

983131m

s

983133 (30)





11



denser tubes is




12



mprimeprime

h = mtot

S tot=

φhL

NLπdu(32)


m1 = S 1S tot


NLπdumtot =

N minus 3n

N mtot (33)


˙V 1 = m1vh =

N

minus3n

N mtotvh (34)


u1 =V 1A1

=N minus3nN

mtotvh1 8Lndu

(35)











6



h9830812

sumThe parameter

852008 m

primeprime

h

9830812sum





mprimeprime

h = m

S (16)








7



mprimeprime

h(x) = mh(x)

S (x) (17)


u(x) = mh(x)

ρA(x) =

mprimeprime

h(x)S (x)

ρA(x) (18)


∆ p(x) = 1

2ρξ [u(x)]2 (19)




8

∆ p = 1

2ρ

x=n9917611

ξ (x)

1048616 m

primeprime

h(x)S (x)

A(x)

8520092 (20)



∆ 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)



s

d






9

S (x)

A(x) =

S (1)

A(1) = constant


991761 = n

1048616S (1)

A(1)

8520092

(23)



h9830812

sum as a



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


mprimeprime

h

9830812sum


mprimeprime

h

9830812sum


S (x)

A(x)

= u(x)

vh(x)m

primeprime

h(x)

(24)





10

991761 =

n991761i=1

1048667 u(x)

vh(x)mprimeprime

h(x)

10486692

(25)


(vhmprimeprime

h)2991761

=n991761i=1

[u(x)]2 (26)


n991761i=1

[u(x)]2

= nu2

(27)


u =

991770 (vhm

primeprime

h)2sum

n (28)


(vhmprimeprime

h)2991761

= 75

1048667m2

s2

1048669 (29)


u = u(1) = u(n) =

991770 752

n =

75radic n

983131m

s

983133 (30)





11



denser tubes is




12



mprimeprime

h = mtot

S tot=

φhL

NLπdu(32)


m1 = S 1S tot


NLπdumtot =

N minus 3n

N mtot (33)


˙V 1 = m1vh =

N

minus3n

N mtotvh (34)


u1 =V 1A1

=N minus3nN

mtotvh1 8Lndu

(35)











7



mprimeprime

h(x) = mh(x)

S (x) (17)


u(x) = mh(x)

ρA(x) =

mprimeprime

h(x)S (x)

ρA(x) (18)


∆ p(x) = 1

2ρξ [u(x)]2 (19)




8

∆ p = 1

2ρ

x=n9917611

ξ (x)

1048616 m

primeprime

h(x)S (x)

A(x)

8520092 (20)



∆ 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)



s

d






9

S (x)

A(x) =

S (1)

A(1) = constant


991761 = n

1048616S (1)

A(1)

8520092

(23)



h9830812

sum as a



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


mprimeprime

h

9830812sum


mprimeprime

h

9830812sum


S (x)

A(x)

= u(x)

vh(x)m

primeprime

h(x)

(24)





10

991761 =

n991761i=1

1048667 u(x)

vh(x)mprimeprime

h(x)

10486692

(25)


(vhmprimeprime

h)2991761

=n991761i=1

[u(x)]2 (26)


n991761i=1

[u(x)]2

= nu2

(27)


u =

991770 (vhm

primeprime

h)2sum

n (28)


(vhmprimeprime

h)2991761

= 75

1048667m2

s2

1048669 (29)


u = u(1) = u(n) =

991770 752

n =

75radic n

983131m

s

983133 (30)





11



denser tubes is




12



mprimeprime

h = mtot

S tot=

φhL

NLπdu(32)


m1 = S 1S tot


NLπdumtot =

N minus 3n

N mtot (33)


˙V 1 = m1vh =

N

minus3n

N mtotvh (34)


u1 =V 1A1

=N minus3nN

mtotvh1 8Lndu

(35)











8

∆ p = 1

2ρ

x=n9917611

ξ (x)

1048616 m

primeprime

h(x)S (x)

A(x)

8520092 (20)



∆ 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)



s

d






9

S (x)

A(x) =

S (1)

A(1) = constant


991761 = n

1048616S (1)

A(1)

8520092

(23)



h9830812

sum as a



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


mprimeprime

h

9830812sum


mprimeprime

h

9830812sum


S (x)

A(x)

= u(x)

vh(x)m

primeprime

h(x)

(24)





10

991761 =

n991761i=1

1048667 u(x)

vh(x)mprimeprime

h(x)

10486692

(25)


(vhmprimeprime

h)2991761

=n991761i=1

[u(x)]2 (26)


n991761i=1

[u(x)]2

= nu2

(27)


u =

991770 (vhm

primeprime

h)2sum

n (28)


(vhmprimeprime

h)2991761

= 75

1048667m2

s2

1048669 (29)


u = u(1) = u(n) =

991770 752

n =

75radic n

983131m

s

983133 (30)





11



denser tubes is




12



mprimeprime

h = mtot

S tot=

φhL

NLπdu(32)


m1 = S 1S tot


NLπdumtot =

N minus 3n

N mtot (33)


˙V 1 = m1vh =

N

minus3n

N mtotvh (34)


u1 =V 1A1

=N minus3nN

mtotvh1 8Lndu

(35)











9

S (x)

A(x) =

S (1)

A(1) = constant


991761 = n

1048616S (1)

A(1)

8520092

(23)



h9830812

sum as a



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


mprimeprime

h

9830812sum


mprimeprime

h

9830812sum


S (x)

A(x)

= u(x)

vh(x)m

primeprime

h(x)

(24)





10

991761 =

n991761i=1

1048667 u(x)

vh(x)mprimeprime

h(x)

10486692

(25)


(vhmprimeprime

h)2991761

=n991761i=1

[u(x)]2 (26)


n991761i=1

[u(x)]2

= nu2

(27)


u =

991770 (vhm

primeprime

h)2sum

n (28)


(vhmprimeprime

h)2991761

= 75

1048667m2

s2

1048669 (29)


u = u(1) = u(n) =

991770 752

n =

75radic n

983131m

s

983133 (30)





11



denser tubes is




12



mprimeprime

h = mtot

S tot=

φhL

NLπdu(32)


m1 = S 1S tot


NLπdumtot =

N minus 3n

N mtot (33)


˙V 1 = m1vh =

N

minus3n

N mtotvh (34)


u1 =V 1A1

=N minus3nN

mtotvh1 8Lndu

(35)











10

991761 =

n991761i=1

1048667 u(x)

vh(x)mprimeprime

h(x)

10486692

(25)


(vhmprimeprime

h)2991761

=n991761i=1

[u(x)]2 (26)


n991761i=1

[u(x)]2

= nu2

(27)


u =

991770 (vhm

primeprime

h)2sum

n (28)


(vhmprimeprime

h)2991761

= 75

1048667m2

s2

1048669 (29)


u = u(1) = u(n) =

991770 752

n =

75radic n

983131m

s

983133 (30)





11



denser tubes is




12



mprimeprime

h = mtot

S tot=

φhL

NLπdu(32)


m1 = S 1S tot


NLπdumtot =

N minus 3n

N mtot (33)


˙V 1 = m1vh =

N

minus3n

N mtotvh (34)


u1 =V 1A1

=N minus3nN

mtotvh1 8Lndu

(35)











11



denser tubes is




12



mprimeprime

h = mtot

S tot=

φhL

NLπdu(32)


m1 = S 1S tot


NLπdumtot =

N minus 3n

N mtot (33)


˙V 1 = m1vh =

N

minus3n

N mtotvh (34)


u1 =V 1A1

=N minus3nN

mtotvh1 8Lndu

(35)











12



mprimeprime

h = mtot

S tot=

φhL

NLπdu(32)


m1 = S 1S tot


NLπdumtot =

N minus 3n

N mtot (33)


˙V 1 = m1vh =

N

minus3n

N mtotvh (34)


u1 =V 1A1

=N minus3nN

mtotvh1 8Lndu

(35)









Documents

Ene-39 4024 Shell and Tube Condensers 2