Solution exercise 6

Cold side (tap water) hot side (condensate)

cm& = 30000 kg/h = 8.333 kg/s hm& = 50000 kg/h = 13.889 kg/s Tc,in = 17 C Th,in = 67 C Tc,out = 40 C cp,h = 4184 J/kgK vc 1.5 m/s ph = 1.2 bar

Geometry

DS = 0.38735 m Rfo = 0.000176 mK/W do = 0.019 m Rfi = 0.000176 mK/W di = 0.016 m = 60 W/mK Lmax = 5 m B = 0.25 m Parallel layout with PT = 0.024 m Nt = 124/2 = 62 Np = 2

Solution Average temperature cold side: Tc = 5.282

4017=

+C

Tap water (28.5C) (in the tubes)

cp,c = 4182.72 J/kg K c = 612.296.10-3 W/mK c = 995.80 kg/m c = 842.338.10-6 kg/ms Prc = 5.7738

( ) 801656,,

== incoutccpcTTcmQ &&

W

( ) 2.53,,,

== outhouthinhhph TTTcmQ && C

Average temperature hot side: Th = 1.602672.53

=

+C

Condensate (1.2 bar)

cp,h = 4184 J/kgK h = 652.977.10-3 W/mK h = 982.72 kg/m h = 478.742.10-6 kg/ms Prh = 3.078

First estimate of the average wall temperature: Tw = 3.4425.281.60

=

+C

Viscosity at the wall: w = 625.37.10-6 kg/ms

Calculation of the convection coeficient at the inside Hydraulic diameter: inner diameter di

6713.0

42

=

=

ic

t

c

c

d

Nm

vpi

&

m/s < 1.5 m/s OK!

12698Re ==c

cicdi

dv

> 2300

Flow in a tube: Turbulent: Gnielinski

f = ( ) 0073683.028.3)ln(Re58.1 2 = di

Nu = ( )

82.9111Pr

27.121

Pr1000Re2 3

2

32

=

+

+

Ld

f

fi

(when neglecting correction for di/L)

Large temperature difference between the fluids: correction

Nucorr = 363188.9411.0

=

==

i

c

ii

w

c hdhNu

W/mK

Calculation of the convection coefficient for the shell side (Kern)

Equivalent diameter: 019599.04

42

2

=

=

o

o

T

e d

dP

Dpi

pi

m

C = PT do = 0.005 m

Area the fluid flows through: 02017.0==T

s

s PBCD

A m

Mass flux 6.688==s

hs A

mG&

kg/ms

28190Re ==h

se

s

GD

4705PrRe36.014.0

3155.0

=

=

u

w

hhs

h

eu hDh

W/mK

uuu

fui

u

i

fi

ii AhAR

ldd

AR

AhUA1

2

ln11

++

++= pi

with l = Ltube . Np (for 1 conduction path)

Ai = pi.di.l

Au = pi.du.l

Hence u

fuui

u

i

u

fii

u

i hRd

dd

dd

Rdd

hU1

2

ln11

++

++= pipi

U = 1050.65 W/mK

Q = U.A.F.LMTD

Logaritmic mean temperature difference: ( ) ( ) 38.314067172.53ln

4067172.53,

=

= tsLMT C

F = 0.95 (1 shell pass and 2 tube passes, P= 0.46, R = 0.6) (fig 2.4)

m 59.25=

=

LMTFUQA

A = 46.3= tubetubetpo LLNNdpi m, rounded 3.5 m.

The amount of baffles can now be calculated: Nb = Ltube/B 1 = 13 baffles

Pressure loss

pi?

fcorr = 0068396.025.0

=

w

cf

44812

142

=

+= pcc

i

tubecorri N

v

dLfp

Pa

po? ( )( ) 253884.0Reln19.0576.0 == sef

( ) 175932

114.0

2

=

+=

w

heh

sbso

D

DNGfp

Pa