2. Introduction
• A feedwater heater is used in a conventional power plant to
preheat boiler feed water. The source of heat is steam bled
from the turbines, and the objective is to improve the
thermodynamic efficiency of the cycle.
4. Description of C.F.W.H
• Closed feed-water heaters are typically shell and tube type heat
exchanger where the feed-water passes throughout the tubes and is
heated by turbine extraction steam. These do not require separate
pumps before and after the heater to boost the feed-water to the
pressure of the extracted steam as with an open heater.
5. Advantages of C.F.W.H
• Reduces the irreversibility involved in steam generation and hence
increase efficiency.
• It helps to avoid thermal shock to the boiler metal when the feed-water
is introduced back into the steam cycle.
• Streams generally need to be at the same pressure to be reversibly
mixed.
• After stream 4 transfers heat to the boiler feed in the feedwater
heater, it can either be pumped up to the boiler pressure and added
to the boiler feed as shown here, or it can be allowed to irreversibly
mix with the condenser feed.
6. Feedwater inlet and
outlet pipes in
front. On the side
we see the heater
drain pipes and
throttling valves.
7.
8. Schematic of a Power
Plant Running an Ideal
Regenerative Rankine
Cycle with One Closed
Feedwater Heater
9. T-S Diagram of an Ideal Regenerative Rankine
Cycle with One Closed Feedwater Heater
10. CFWH vs OFWH
• Compared with open feedwater heaters, closed feedwater heaters
are more complex, and thus more expensive. Since the two streams
do not mix in the heater, closed feedwater heaters do not require a
separate pump for each heater. Most power plants use a combination
of open and closed feedwater heaters.
11. Materials used in MFG (CFWH)
• Processing and testing advancements on the welded and cold worked
tubing developed over the last 65 years offer many technical and
commercial advantages over the seamless product. Although
seamless carbon and alloy steel feedwater heater tubing is still used,
the vast majority of stainless steel feedwater heater tubing is in the
welded, cold-worked, and annealed condition. Even though the
seamless stainless tubing enjoys an ASME Code advantage of 15%
higher stress level allowing a thinner wall, little, if any, is used in
global feedwater heaters. The welded and cold-worked tube
manufacturers have developed standard proprietary manufacturing
processes and testing focused toward feedwater heater applications
that most seamless producers have not followed.
16. Sample Problem
A steam power plant operates on an ideal reheat-regenerative Rankine
cycle with one open feedwater heater, one closed feedwater heater,
and one reheater. The fractions of stream extracted from turbines and
the thermal efficiency of the cycle are to be determined.
Assumptions:
All the components in the cycle operate at steady state.
Kinetic and potential energy changes are negligible.
17.
18. Solution
(1) Determine the fraction of steam extracted from the turbines
The enthalpies at various states and the pump work per unit mass
of fluid flowing through them can be determined by using the
water tables.
State1: saturated water
P1 = 10 kPa ( given)
h1 = 191.83 kJ/kg
v1 = 0.00101 m3/kg
State 2: compressed water
P2 = 1 MPa (given)
wpump,in = v1 (P2 - P1)
= 0.00101(1,000 - 10) = 1.0 kJ/kg
h2 = h1 + wpump,in = 192.83 kJ/kg
State 3: saturated water
P3 = 1 MPa (given)
h3 = 762.81 kJ/kg
v3 = 0.00113 m3/kg
State 4: compressed water
P4 = 16 MPa (given)
wpump,in = v3 (P4 - P3)
= 0.00113(16,000 - 1,000) = 16.9 kJ/kg
h4 = h3 + wpump,in = 762.81+ 16.9 = 779.71 kJ/kg
State 5: saturated water
P5 = 16 MPa (given)
h5 = 1,650.1 kJ/kg
State 6: superheated vapor
P6 = 16 MPa (given)
T6 = 600oC (given)
h6 = 3,569.8 kJ/kg
s6= 6.6988 kJ/(kg-K)
State7: superheated vapor
P7 = 5 MPa (given)
s7 =s6= 6.6988 kJ/(kg-K)
h7 = 3,222.4 kJ/kg
State 8: superheated vapor
P8 = 5 MPa (given)
T8 = 600oC (given)
h8 = 3,665.6 kJ/kg
s8= 7.2731 kJ/(kg-K)
State 9: superheated vapor
P9 = 1 MPa (given)
s9 =s8= 7.2731 kJ/(kg-K)
h9 = 3,138.1 kJ/kg
State 10: saturated mixture
P10 = 10 kPa (given)
s10 =s8= 7.2731 kJ/(kg-K)
x8 = (s8 - sf@10 kPa)/sfg@10 kPa = 88.3%
h10 = hf@10 kPa+ x8hfg@10 kPa = 2,304.76 kJ/kg