ECONOMIZER DESIGN
Inlet gas temperature 255C
Outlet gas temperature ?
Gas mass flow rate 49.86 kg/sec
Specific heat 1.020505 kJ/Kg C
Inlet water temperature 40C
Outlet water temperature 196.284 -26.284 = 170C
Enthalpy inlet hf40C 167.53 kJ/Kg
Enthalpy outlet hf170C 719.08 kJ/Kg
Mass flow of water 5.75 kg/sec
The outlet temperature of water is taken short of the saturation temperature at operating
pressure of 13 bars, in order to prevent steaming.
Now applying energy balance;
Energy gain by water = Energy lost by the gases
Energy gain by feed water =m (h out h in)
E gain = 5.75(719.08 167.53)
E gain = 3171.41 kW
Therefore energy gain by water is equal energy loss by the gases
q Gain, HO =q lost, gas
3171.41 KW=m C p T
3171.41 =49.86 1.020505 (Tin Tout)
(Tin Tout) gas =62.32C
Tout =255 -62.32
T out , gas =192.67C
TEMPERATURE FOR TRANSPORT PROPERTY
Average temperature of gas
255+192.67
T g, a v g= =223.83C
2
Average temperature of water
170+40
T w, a v g= =105C
2
Average film temperature of gas
Properties of fluid such as density viscosity prandtl number etc are calculated at the film
temperature (temperature of the fluid at solid fluid interface). Since temperature is
changing continuously along the flow direction, therefore finding the mean of mean
temperature of water and flue gases.
( , + , )
Tf = =164.42C
2
Log-mean temperature-difference (LMTD)
Since the purpose of economizer is to preheat the fluid, therefore counter flow
configuration is used. Otherwise disturbance would be created through steaming,
especially when the flow is downward.
Log mean temperature difference of parallel flow is less than that of counter flow. We select
parallel flow configuration.
( )
T LMTD = Ta
In
Tb
Ta = Tg, out Tw, in
T b = Tg , in T w , out
T gas, in 255C
T gas, out 192.67C
T water, in 40C
T water, out 170C
[(192.6740) (255170)]
T LMTD =[ 192.6740 ]
ln( )
255170
T LMTD =115.55C
ASSUMING A SUTAIBLE TUBE SIZE
Assuming the diameter of tubes and thickness in accordance with mass flow rate.
Outer diameter
OD =38 mm=0.038m
And thickness
t=4mm=0.004m
Now internal diameter would be;
OD -2(t) = ID
Di =ID = 38-(4x2)
Di =ID =30mm=0.03m
Area for one tube
A = (Di) 2=/4(0.03)2
4
A = 7.068610-4m2
Number of tubes
Nw=
AV
For density taking average temperature of water in economizer
170+40
T avg, w = = 105C
2
1 1
f 105C = = = 957 kg/m3
0.001045
m = 5.76kg/sec
A = 7.068610-4m3
V=1.5m/sec (Assumed)
5.75
N w=
(9577.0686104 1.5)
N w=5.66 6
6 tubes are quite low; it will increase the length of tubes so;
V=0.5m/sec (Assumed)
5.75
N w=
(9577.06861040.5)
=17.000117 tubes
Thus 17 tubes are quite reasonable for lesser frontal area. This would curtail the length of
the tubes.
TUBES SPECS FOR ECONOMIZER
Outer diameter 38mm
Inner diameter 30mm
Thickness 4mm
Number of tubes 17
Area 7.068610-4m2
FIXING THE GAS MASS VELOCITY
Gas mass velocity is basically the mass flux (amount of exhaust gas passing through per
unit area). The range for the gas flux typically ranges from 6 to 16 kg/m2sec depending
upon the type of gas which ascends from dirtiest to clean.
m=AV
=V=G
Setting up =G=8kg/m2sec for cement industry exhaust gas;
[Reason: acoustic cleaning system]
m
G= A ff
49.86
Aff = 8
Aff=6.232m2
Now length of economizer tubes can be calculated.
PITCH OF TUBES
Length of the tube can be evaluated by the formula
A ff= L x (ST-do) N w
Setting up length up to 4 meters
Since length of the tubes is fixed therefore only variable is the transverse pich in the
equation.
A ff
ST= LN w + do
Then;
6.232
S T= +0.038
(417)
ST=0.13m
WIDTH OF TUBES
Since number of tubes are known and we have find the pitch. Now setting the width for
bank of tubes.
Width=S TN w
=0.1317
=2.204m
ESTIMATING THE OVERALL HEAT TRANSFER CO EFFICIENT
Overall heat transfer coefficient is useful for finding the surface area for heat transfer. Since
convective heat transfer coefficient of gas side has much lower value than water side,
therefore gas side coefficient is the governing parameter. Even though they dont have
much impact, but for higher accuracy we have considered it.
Also we have to consider the fouling factor on gas side and corrosion factor on water side.
Therefore the equation for overall coefficient becomes
1 do 1 do do
= + + ln( )+ffid+ffod
hidi ho 2KL di
For water side
Inner diameter=30mm=1.181 inch
(Average temperature)HO=105C=221F
Mass flow rate=5.75kg/sec =20.7tonns/hr
=45635.68 lb/hr.
At average temperature of water 105C properties are
f=957kg/m3
f=26010-6 Ns/m2
Prf =1.61
Kf=68310-3 W/m.k
Using correlation for inside heat transfer coefficient
NU=0.023(Re)0.8(Pr)0.4
NU =0.023(VD/)0.8 (Pr)0.4
9570.50.03 0.8
NU=0.023 ( ) (1.61)0.4
260106
NU= 173 = = 173
1730.683
hi= h=3.94 KW/m2K
0.03
For water side NU=173
hi=(3.94)KW/m2K
For gas side
Properties of gases evaluated at the film temperature
255+192.67
Tg , avg = 255
2
C p=0.2219495 Kcal/kg
C p=0.932188 KJ/kg
=0.653566 kg/m.hr
=1.8143210-4 KJ/m sec
K=0.285340 kcal/m.hr
K=3.3289610-4 KJ/m sec
PRANDTL NO.
Pr= = 0.5081
=0.5 KG/m3
= =3.628610-4 m2/sec
REYNOLD NO.
Gd o
Re= =80.038/1.8143210-4
Re=1676
NUSSELT NO.
For outside heat transfer co efficient using correlation
Nu = 0.33FH FN Re0.6 Pr0.33
Nu =0.3311(1676)0.6(0.5081)0.33
Nu =22.701
hodo
= 22.701
(22.7013.3289104 )
ho =
0.038
ho = 198.865 W/m2
(h water side 20h gas side)
Using tube material to be mild steel thermal conductivity=55W/m.k
Now;
1 do 1 do do
= + + ln( )+ffid+ffod
hidi ho 2KL di
1 0.038 1 0.038 0.038
=( ) +( )+( )In( )+0.002+0.000179
U 0.033940 198.865 2554 0.03
1
= 3.21510-4 + 5.0310-3 + 0.001 + 0.000179 + 6.4984
U
U= 152.97W/m2K
FLOW AREA REQUIRED
3171.41103
A= =
UTLMTD (152.97115.55)
A = 179.42m2
NUMBER OF TUBES HIGH REQUIRED
1
NH=
179.42 1
NH=
17 0.0384
NH=24.3 25
