CondensateFeedwaterSystem Part1.ppt

The Condensate-
Feedwater System
Part 1
Assoc.Prof.Sommai Priprem, PhD.
Department of Mechanical Engineering
Khon Kaen University

รศ.ดร.สมหมาย ปรีเปรม 2
References:
1. El-Wakil, M.M., Powerplant Technology, International ed., McGraw-Hill
Book Co., 1984.
2. Kam W. Li.,and Paul Priddy A.,Power Plant System Design, John Wiley
& Sons, Inc., 1985.
3. Drbal Lawrence F.(editor),et.al., Power Plant Engineering: by Black
Veatch, Chapman & Hall Book, 1996.
4. Hicks Tyler G.(editor), Power Generation Calculations Reference Guide,
McGraw-Hill Book Co., 1985.
Related Websites:
http://www.egat.co.th/english/index.htm
http://www.egat.co.th/english/powerplants/namphong.htm
http://en.wikipedia.org/wiki/Power_plant
http://www3.toshiba.co.jp/power/english/thermal/index.htm
http://www.cheresources.com/ctowerszz.shtml

Outline of this chapter:
• Overview of condensate & feedwater
• Direct contact condenser
• Surface condenser
• Surface condenser calculations
• Closed feedwater heater
• Open feedwater heater
• Boiler makeup and treatment

Turbine
wt
Condenser Qc
1
2
Steam Power Plant with Feedwater Heater
Cooling
Water
3
Feedwater
Heater
Pump
4
5
6
Feedwater
Pump
7
Turbine
wt
Boiler
QB
Condenser Qc
1
2
Cooling
Water
3
7
8
Feedwater
Heater
Pump
4
6 5
Open type feedwater heater Closed type feedwater heater
Boiler
QB

FWH 2 FWH 1
Turbine
wt
Condenser
Cooling
Water
Pump
FWH 4 FWH 3
Pump
Boiler
QB
Steam Power Plant with 4-Feedwater Heaters
Vent
Feedwater

T
s
P1
2
3
Condenser:
2-importants functions
1. to condense exhaust steam  recover
the high-quality feedwater (reduce cost)
2. to create low back pressure (vacuum)
of the turbine.  more turbine work 
higher efficiency
Condenser
2
Cooling
Water
P2a
P2b
2
3
2-type of condenser
1. Direct Contact
2. Surface Condenser

Feedwater Heater
Purpose: to improve cycle efficiency by heating
the condensate and feedwater, by bled steam
from stesm turbine, before returning to the
steamgenerator.
2-types of feedwater heater
1. surface or shell-and-tube type.
2. open or direct-contact or deareating type

6-2 Direct-Contact Condenser
Non-condensable
to SJAE
Dry Cooling Tower
To plant feedwater system
Condenser
Turbine exhaust
2
Pump
3
4
5
Steam is directly mixed with
spray cooling water causes
the steam to condense and
becomes “Condensate” 5
3
3
2
3
5
2
2
5
3
5
2
3
5
2
h
h
h
h
m
m
m
m
m
:
balance
Mass
h
m
h
m
h
m
:
balance
Energy
o
o
o
o
o
o
o
o








Example 6.1 Find the ratio of circulating water to steam
flows if the condenser pressure is 1 psia and the
coolig tower cools the water to 60 F. Assume turbine
exhaust at 90 percent quality.
5
3
3
2
3
5
2
2
5
3
5
2
3
5
2
h
h
h
h
m
m
m
m
m
:
balance
Mass
h
m
h
m
h
m
:
balance
Energy
o
o
o
o
o
o
o
o







Solution h2 = hf + xhfg
h2 = 69.73 +(0.9 x 1036.1)
= 69.73 Btu/lbm
h3 = hf @P3 = 69.73 Btu/lbm
h5 ~ hf @ 60oF = 28.06 Btu/lbm
mo
5/mo
2 = 22.38 answer

Barometric and Jet Condenser
pipe
tail
in
drop
Pressure
ΔP
Pressure
Condenser
P
pressure
c
atmospheri
P
)
s
*
m(N
*
1kg
or
)
s
*
ft(lb
*
lbm
32.2
factor,
conversion
g
m/s
or
ft/s
acc.,
nal
gravitatio
g
m
or
ft
tailpipe,
of
height
h
kg/m
or
/ft
lb
mixture,
of
density
ρ
P
P
P
g
g
h
f
cond
atm
2
2
f
c
2
2
3
3
3
m
f
cond
atm
c









 

Barometric Condenser
Jet Condenser

Water Box
Discharge
water outlet
Water Box
Condenser
Shell
Air-Vapor outlet
Tubes
Cold Water inlet
Condensate outlet
Steam inlet
Surface Steam Condenser
• Most common used in powerplants.
• It is shell-and-tube type heat exchanger.
• Size up to 1 million ft2 (93,000 m2) heat transfer area.

Discharge
water outlet
Water Box
Condenser
Shell
Air-Vapor outlet
Cold Water inlet
Condensate outlet
Steam inlet
2-Pass Steam Condenser

• Number of pass: Single pass, Two pass
• Single and Multipressure Condenser.
• Tube Sizes (appendix K)
– OD, Gage (BWG), Thickness, ID, A/ft
• Tube Materials (table 6.2)
– stainless steel, copper, Alluminum-Brass, Aluminum-Bronz
– type 304 staimless steel is most popular used.
6-3 Surface Condenser

Table 6-1 Condenser Dimension Sheet: Westinghouse electric Corporation.

• Noncondensable gases must be removed.
• Noncondensable ie, air :
– Leak into the system at setion that operates below Patm,
such as condenser section.
– Decomposition of water to H and O by thermal or chemical
reaction.
• Problems of having noncondensable gases:
– raise total pressure  higher condenser pressure  less
turbine output  lower plant efficiency.
– They “Blanket” the heat transfer area, ie, condenser.
– Cause chemical activities, ie. O2  corrosion.
Deaeration

• Process to remove noncondensable gases is called
‘Deaeration’
• Most Fossil powerplants have a deaerating feedwater
heater.
• Condenser must have good deaeration.
• Good deaeration in condenser required time,
turbulence, and good venting equipment.
• Noncondensables are cooled to reduce volume before
‘Pump out’.
• Venting equipment: Reciprocating compressor (Dry
vacuum pump); Steam-Jet air Ejector (SJAE)
Deaeration

First-stage ejector
Second-stage ejector
Intercondenser
Aftercondenser
Noncondensible in
steam
Drain
Drain
Venet
Condensate out
Condensate in
A two-stage steam-jet air ejector (SJAE)

• Heat transfer Surface Area
– Heat Load
– Tube material and size
– Fluid properties
– Flow velocity
• Pressure drop
– Tube size
– flow velocity
6-4 Surface-Condenser Calculations

Steam
Schematic diagram of a condenser tube.
ri ro
Water
Tw
q
Ts
(2)
)
T
T
UA(
q
(1)
A
h
kL
)
r
/
r
ln(
A
h
T
T
q
w
s
o
o
o
i
i
i
w
s






1
2
1

L

(5)
h
U
therefore,
:
h
k
and
h
(4)
h
kL
)
r
/
r
ln(
A
A
h
A
U
(3)
A
h
A
kL
)
r
/
r
ln(
A
h
U
(2)
and
(1)
eqn
Combine
i
i
i
o
o
o
i
o
i
i
o
o
o
o
i
o
i
i
i
i









1
2
1
2
1
1



Steam
L
T
ITD
TR
TTD
LMTD
Ts
T1
T2
Temperature variation in fluids
ITD = Internal Temperature Difference
TTD = Terminal Temperature Difference
TR = Temperature rise
LMTD = Log Mean Temperature Difference

• Method proposed by the Heat Exchange Institute Standards for
Steam Surface Condenser:
Q = UAΔTm (6-5)
where: Q = heat load on condenser,
U = overall condenser heat-transfer coefficient, based
on outside tube area,
A = total out side tube surface area,
ΔTm = log mean temperature difference in the
condenser:
ΔTm = (ΔTi – ΔTo)/ln(ΔTi /ΔTo) (6-6)
U = C1C2C3C4 √V (6-7)
Surface-Condenser Calculations

V cooling water velocity
C1 factor depend upon tube OD
C2 Correction factor for cooling
water inlet temp.
C3 Correction factor for tube
material and gage.
C4 Cleanliness factor

• Heat balance: mo
w Cp(T2 – T1) = Q (6-8)
• Pressure drops
ΔPtotal = ΔPWater Box + ΔPtubes
ΔPtotal = ρh (g/gc) (6-9)
Circulating-water flow and pressure drop

Fig 6-10 Pressure drop in a condenser
water boxes, express as head in feet, A-
one pass, B-two pass.
Fig 6-11 Pressure drop in a condenser tubes, express as
head in feet per foot length of tubes.

Calculation Procedure
1. Select number of pass (1 or 2)
2. Selected Tube; Material, Size, Thickness.  k, h (Table K)
3. Selected Water velocity in tube
4. The value U can be determine ( Table 6-2)
5. Calculate amount of heat transfer rate, Q required (total load
in condenser)
6. Heat transfer surface area, A, can be calculated (eqn 6-8)
7. Total length of tube, L, can be calculated.
8. With given length per tube  no. of tubes is calculated
9. determine head loss in water boxes (fig 6-10)
10. determine head loss in tubes (fig 6-11)
11. Total head loss in condenser can be determined.(eqn. 6-9)

Example 6-2 Design a condenser that would
handle 3x106 lbm/h of 90% quality steam at 1
psia, as well as 360,000 lbm/h of 112 oF drain
water from the low-pressure feedwater heater,
and 1,875 lbm/h of 440 oF drains from the jet
air ejector. Fresh water is available at 70 oF.

Solution
Heat transfer calculations: select:
1. A two pass condenser
2. Type 304 stainless steel tubing
3. Tube: 50 ft in length, 7/8” OD, 18 BWG
4. TTD = 6 oF
5. Inlet water velocity = 7 ft/s
Heat Load Q = Turbine exhaust + Low-pressure drain + SJAE drain
= 2.802 x 109 Btu/h
ΔTi = Tsat – Twi = 101.74 – 70 = 31.74 oF
ΔT0 = TTD = 6 oF
ΔTm = (ΔTi – ΔTo)/ln(ΔTi /ΔTo) = 15.45 oF
U = 263 x 1.00 x 0.58 x 0.85 x √7 = 343.0 Btu/(h ft2 oF)
A = Q/U ΔTm = 528,675 ft2

for 7/8” OD tube A/L = 0.2291 ft2/ft (App.K)
therefore:
total length of tube = 528,675 ft2 / 0.2291 ft2/ft
= 2,307,615 ft
and, number of tube = 2,307,615 ft / 50 ft/tube
= 46,150 tubes
Water calculation:
T2 – T1 = (ΔTi – ΔTo) = 31.74 – 6 = 25.74 oF
for Cp = 0.99 Btu/(lbm oF)
mo
w = Q / Cp(T2 – T1)
= 2.802 x 109 Btu/h /[0.99 Btu/(lbm oF) x 25.74 oF]
= 1.0996 x 108 lbm/h

Volume flow rate, = mov
= (1.0996 x 108 lbm/h)(0.01605ft3/lbm)/(60 min/h)
= 29,414 ft3/min;
= 29,414 x 7.481 = 220,043 gal/min
Pressure drop in water box ( fig. 6-10) = 2.7 ft
Pressure drop in tubes ( fig. 6-11) = 0.32 ft/ft
Tube:allow for 1.2” thick tube sheet  actual length of each tube
= 50 + 2x(1.2/12) = 50.2 ft
total pressure drop = 0.32 ft/ft x 50.2 ft/tube x 46,150 tubes
= 741,353.6 ft
total head loss = 2.7 ft + 741,353.6 ft = 741,356.3 ft
total pressure drop ,ΔPtotal = ρh (g/gc)
= (62.3 lbm/ft3)(741,356.3 ft)(32.2/32.2)
= 46,186,497.49 psi

Volume flow rate, = mov
= (1.0996 x 108 lbm/h)(0.01605ft3/lbm)/(60 min/h)
= 29,414 ft3/min;
= 29,414 x 7.481 = 220,043 gal/min
Pressure drop in water box ( fig. 6-10) = 2.7 ft
Pressure drop in tubes ( fig. 6-11) = 0.32 ft/ft
Tube:allow for 1.2” thick tube sheet  actual length of each tube = 50 +
2x(1.2/12) = 50.2 ft
total pressure drop = 0.32 ft/ft x 50.2 ft/tube x 2 = 32.13 ft
total head loss = 2.7 ft + 32.13 ft = 34.83 ft
total pressure drop ,ΔPtotal = ρh (g/gc)
= (62.3 lbm/ft3)(34.83 ft)(32.2/32.2)= 2,169.91 lbf/ft2
Wo = mov ΔP
= (1.0996 x 108 lbm/h)(0.01605ft3/lbm)(2,169.91 psf)
= 3.8295 x 109 ft-lbf/h = 3.8295 x 109 /(60 x 33,000) = 1,915 HP

CondensateFeedwaterSystem Part1.ppt

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