Closed Feed Water Heaters

Ideal Regenerative Rankine Cycle

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.

A Closed Feedwater

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.

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.

Feedwater inlet and outlet pipes in front. On the side we see the heater drain pipes and throttling valves.

Schematic of a Power Plant Running an Ideal Regenerative Rankine Cycle with One Closed Feedwater Heater

T-S Diagram of an Ideal Regenerative Rankine Cycle with One Closed Feedwater Heater

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.

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.

Seamless vs. Welded and Cold worked

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.

Solution(1) Determine the fraction of steam extracted from the turbinesThe 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/kgState 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/kgState 3: saturated water P3 = 1 MPa (given) h3 = 762.81 kJ/kg v3 = 0.00113 m3/kgState 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/kgState 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/kgState 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/kgState 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