3.5 Analyze regenerative feedwater heating | University of Alberta MEC E 340: Applied Thermodynamics

University of Alberta MEC E 340: Applied Thermodynamics

Vapour Power Cycles

Heater balances, extraction flow, and cycle performance

The working fluid is water in a steam power cycle. State 1 is the turbine inlet, state 2 is the extraction point, and state 3 is the final turbine exhaust. Feedwater states are identified by their positions around the heater. Regenerative feedwater heating diverts some expanding steam to warm water returning toward the boiler. Analyze the heater as a steady-flow control volume, then include its flow split in the cycle work calculation. The examples supply all enthalpies and work values as problem data; they are not claimed steam-table values. Unless stated otherwise, assume negligible kinetic and potential energy changes and negligible heat loss from the heater. Turbine work is output, pump work is input, and boiler heat is input.

What you will learn

Explain how extracted turbine steam warms feedwater in a steam cycle.
Distinguish open and closed feedwater heaters using their mass and energy balances.
Calculate an extraction flow from supplied enthalpies.
Account for extraction and component flow rates when calculating cycle work and efficiency.

1. Why regenerate, and what to compare

In a steam cycle without regeneration, liquid leaving the condenser is pumped toward boiler pressure and then heated in the boiler. With regeneration, some steam is extracted during turbine expansion and transfers energy to the returning feedwater. For a given boiler-outlet state, warmer feedwater generally means less heat must be supplied in the boiler.

The extracted steam does not continue through the remaining turbine stages. Regeneration therefore changes both boiler heat input and turbine work. A feedwater temperature rise by itself does not establish whether cycle performance improves; calculate work and heat on the same mass basis. The comparison also depends on the supplied cycle states and component work, so do not assume a numerical efficiency improvement without calculating it.

Heater energy transfer is not boiler heat input. For cycle efficiency, use heat supplied in the boiler, and subtract pump work from turbine output to obtain net work. Whether the heater is open or closed determines the appropriate heater balance.

Thermal efficiency is net cycle work divided by boiler heat input, both on the same mass basis:

\eta_{th}=\frac{w_{net}}{q_{in}}

.

Extracted steam warms feedwater but no longer expands through later turbine stages.
Compare cycles using net work and boiler heat input on one consistent basis.
An open heater mixes streams; a closed heater transfers energy between separate streams.

2. Define the basis and balance the heater

For a heater connected to a turbine extraction point, a useful basis is one kilogram entering the turbine. If the fraction extracted is

y

, then

y

kilograms enter the heater from the turbine and

1-y

kilograms continue through the later turbine section. State this basis before writing any balance so that each enthalpy flow is multiplied by the correct mass.

In an open heater, feedwater and extracted steam enter separately and leave as one mixed stream. Let their inlet enthalpies be

h_f

and

h_b

, and the mixed outlet enthalpy be

h_o

. On the one-kilogram turbine-inlet basis, feedwater flow is

1-y

, extracted-steam flow is

y

, and outlet flow is one kilogram. With no heater heat loss or work, the energy balance gives the extraction fraction.

A closed heater keeps the feedwater and extracted-steam streams separate. On a basis of one kilogram of feedwater, let its enthalpy rise from h_{f,in} to

h_{f,out}

. If

y

kilograms of bleed steam enter at

h_b

and leave as drain at

h_d

, the feedwater energy gain equals the bleed-stream energy loss. Use the stated drain condition or supplied drain enthalpy; do not assume an unprovided value.

These simplified balances assume steady operation and negligible kinetic and potential energy changes. If a problem specifies heat loss, work, or a different flow basis, retain those terms and use that basis instead of applying the simplified relation unchanged.

y=\frac{h_o-h_f}{h_b-h_f}

A mass basis makes the stream flow rates explicit.
An open heater requires a mixing balance; a closed heater has separate stream balances.
Use supplied enthalpies or values from an identified property source at the stated conditions.

3. Carry the extraction through the cycle

For a turbine with one extraction point, one kilogram enters the first turbine section, and only

1-y

kilograms enter the later section. If the states are turbine inlet 1, extraction point 2, and final exhaust 3, the supplied cycle arrangement determines whether the extracted stream does any further work. In Example 3, the extraction stream leaves the turbine at state 2 and has no further turbine path, so turbine work per kilogram entering the cycle is the first-section work plus the flow-weighted work of the second section.

Pump work also depends on flow. A pump that handles only condenser flow contributes work in proportion to that flow; a pump that handles the full feedwater flow contributes work on the full basis. Read the cycle arrangement and use the stated or calculated flow through each pump. Do not count a pump's specific work as though it applied to a different mass flow.

Find net work by subtracting pump input from turbine output. Divide by boiler heat input on the same mass basis to obtain thermal efficiency. Check that the extraction fraction is between zero and one, that the continuing flow is nonnegative, and that net work and efficiency have physically consistent signs.

w_t=(h_1-h_2)+(1-y)(h_2-h_3)

Weight each turbine section's work by the mass flowing through it.
Include any further work from extracted steam only if the cycle arrangement provides a turbine path for it.
Scale pump work by the flow through that pump.
Use net work, not turbine work alone, in the efficiency calculation.

4. Property data and a reliable solution sequence

First identify the working fluid, control volume or cycle, numbered states, pressure and phase information supplied, mass basis, and assumptions. Draw or read a schematic only to understand the flow arrangement; a schematic is not a source of enthalpy or an exact state location. A heater with multiple inlets may be clearest as a written flow description rather than a simplified one-inlet device drawing.

Next write the mass and energy balances for the heater. Obtain enthalpies from the problem or a named property source, matching each value to the correct state and stated conditions. Carry units, normally kilograms and kilojoules per kilogram for these examples, so energy per unit mass is in kilojoules per kilogram.

Finally, use the extraction flow in the turbine and pump calculations, calculate net work and efficiency, and check the balances and flow fractions. If a required property or drain condition is missing, do not invent it; state what additional information would be needed. The first-law balance checks energy accounting; the supplied state information and sensible flow directions help check whether the result is physically consistent.

A diagram clarifies connections but does not replace property data.
State the assumptions before simplifying the steady-flow balance.
Check flow rates, energy units, and performance results at the end.

Worked example

Example 1: Extraction fraction for an open heater

On a one-kilogram turbine-inlet basis, feedwater enters an adiabatic open heater at

h_f=210\ \mathrm{kJ/kg}

. Extracted steam enters at

h_b=2800\ \mathrm{kJ/kg}

, and the mixed outlet enthalpy is

h_o=640\ \mathrm{kJ/kg}

. Find the extraction fraction

y

. Treat the enthalpies as supplied problem data.

Set the flow basis
One kilogram of main-cycle flow reaches the heater connection. The extracted stream contributes $y$ kilograms, and feedwater contributes $1-y$ kilograms. Together they leave as one kilogram of mixed outlet flow. Assume steady operation, no heat loss or work, and negligible kinetic and potential energy changes.
Apply the heater balance
The steady energy balance says incoming enthalpy flow equals outgoing enthalpy flow under these assumptions. Substitute the supplied enthalpies; the mixed outlet flow is one kilogram on this basis.
$(1-y)(210)+y(2800)=640$
Solve and check
Solving gives a positive fraction less than one. The outlet enthalpy also lies between the two inlet enthalpies, as expected for adiabatic mixing.
$y=\frac{640-210}{2800-210}=0.166023\approx0.1660$

Answer: The extraction fraction is

y\approx0.1660

, or about 16.6% of the turbine-inlet flow.

Check: The continuing flow is

1-y\approx0.8340

kg per kilogram entering the turbine. The outlet enthalpy is between the inlet enthalpies, and the balance is satisfied to rounding.

Worked example

Example 2: Bleed fraction for a closed heater

A closed feedwater heater is steady and adiabatic to its surroundings. On a basis of one kilogram of feedwater, its enthalpy rises from

600

to

800\ \mathrm{kJ/kg}

. Extracted steam enters at

h_b=2800\ \mathrm{kJ/kg}

and leaves as drain at

h_d=500\ \mathrm{kJ/kg}

. These enthalpies are supplied data. Find the bleed flow

y

in kilograms per kilogram of feedwater.

Set the stream basis
One kilogram of feedwater passes through the heater, while $y$ kilograms of extracted steam enter and leave as drain. The two streams remain separate. Assume negligible kinetic and potential energy changes and no work or heat transfer to the surroundings.
Balance energy
The feedwater gains $800-600$ kilojoules per kilogram of feedwater. The bleed stream loses $2800-500$ kilojoules per kilogram of bleed steam, so its contribution must be multiplied by $y$ .
$800-600=y(2800-500)$
Solve and check
Divide the feedwater enthalpy rise by the enthalpy drop per kilogram of bleed steam. The resulting positive, small flow supplies the stated feedwater energy gain.
$y=\frac{200}{2300}=0.0870$

Answer: The bleed flow is

0.0870

kg per kilogram of feedwater on the stated basis.

Check: The feedwater gains

200\ \mathrm{kJ}

per kilogram. The bleed stream loses

2300\ \mathrm{kJ/kg}

, and

0.0870\times2300\approx200\ \mathrm{kJ}

per kilogram of feedwater.

Worked example

Example 3: Extraction in turbine work and efficiency

A steam cycle has one turbine extraction point. State 1 is turbine inlet, state 2 is the extraction point, and state 3 is final turbine exhaust. On a one-kilogram turbine-inlet basis, supplied enthalpies are

h_1=3400

,

h_2=2800

, and

h_3=2200\ \mathrm{kJ/kg}

. Use

y=0.166

. Supplied pump works are

5\ \mathrm{kJ/kg}

of condenser flow and

10\ \mathrm{kJ/kg}

of full feedwater flow. Supplied boiler heat input is

2750\ \mathrm{kJ/kg}

of turbine-inlet flow. Find turbine work, net work, and thermal efficiency. All values are problem data.

Regenerative steam cycle with one extraction point

Water is the working fluid in a closed steam power cycle. Schematic P–v plot, not to scale; state locations and paths are qualitative, and no property values are implied. One kilogram enters the turbine at state 1; fraction $y$ is extracted

Calculate turbine work
One kilogram flows through the first turbine section. The remaining $1-y=0.834$ kilograms flow through the second section. Add the enthalpy drop in the first section to the flow-weighted enthalpy drop in the second.
$w_t=(3400-2800)+(0.834)(2800-2200)=1100.4\ \mathrm{kJ/kg}$
Calculate pump work
The condenser-flow pump handles $0.834$ kilograms on this basis, while the full-feedwater pump handles one kilogram. Weight each supplied pump-work value by its actual flow.
$w_p=(0.834)(5)+(1)(10)=14.17\ \mathrm{kJ/kg}$
Find net work and efficiency
Subtract pump input from turbine output, then divide net work by the supplied boiler heat input on the same basis. The result is positive and below one.
$w_{net}=1100.4-14.17=1086.23\ \mathrm{kJ/kg},\quad \eta_{th}=\frac{1086.23}{2750}=0.3950$

Answer: Turbine work is

1100.4\ \mathrm{kJ/kg}

, net work is

1086.23\ \mathrm{kJ/kg}

, and thermal efficiency is approximately 39.5%.

Check: Net work is less than turbine work because pumps require input. The efficiency is positive and below one. The supplied boiler heat input is used directly; no additional property values are assumed.

Common mistakes and how to avoid them

Using one kilogram through every turbine section after steam extraction.

Correction: On a one-kilogram turbine-inlet basis, only

1-y

kilograms pass through sections downstream of the extraction point.

Applying an open-heater mixing balance to a closed heater.

Correction: For an open heater, streams mix and require a mass balance. For a closed heater, keep the streams separate and balance their energy changes.

Assuming a drain enthalpy or another property not given in the problem.

Correction: Use supplied data or an identified property source. If a needed state condition or property is missing, state what information is required.

Calculating thermal efficiency from turbine work alone.

Correction: Subtract pump work to obtain net work, then divide by boiler heat input using the same mass basis.

Lesson summary

Regenerative feedwater heating uses extracted turbine steam to warm cycle feedwater.
Open heaters mix streams; closed heaters transfer energy while keeping streams separate.
Use the appropriate steady mass and energy balances to find the extraction or bleed flow.
Carry the extraction fraction into downstream turbine work and pump-flow calculations.
Check property sources, units, flow fractions, and performance results.

Check your understanding

Question 1

On a one-kilogram main-flow basis, what flow continues past an extraction point if

y=0.12

?

$0.12\ \mathrm{kg}$
$0.88\ \mathrm{kg}$
$1.12\ \mathrm{kg}$
$1.00\ \mathrm{kg}$

Show answer and explanation

0.88\ \mathrm{kg}

The continuing flow is

1-y=0.88

kg per kilogram entering the cycle.

Question 2

Which description matches an ideal adiabatic open feedwater heater?

Incoming streams mix, and the outlet enthalpy flow equals the sum of inlet enthalpy flows.
The streams remain separate, so the heater has no mass balance.
The heater produces shaft work equal to its outlet enthalpy flow.
The bleed flow must always equal the feedwater flow.

Show answer and explanation

Incoming streams mix, and the outlet enthalpy flow equals the sum of inlet enthalpy flows.

An open heater mixes its incoming streams. Its steady mass and energy balances describe the mixing and determine the outlet or extraction fraction when the necessary data are given.

Question 3

For an adiabatic closed heater, what does the simplified energy balance equate?

The feedwater enthalpy gain to the bleed-stream enthalpy loss, accounting for each stream's mass flow.
The feedwater and bleed-stream outlet enthalpies.
The feedwater enthalpy gain to turbine work.
The feedwater enthalpy gain to zero, because the streams do not mix.

Show answer and explanation

The feedwater enthalpy gain to the bleed-stream enthalpy loss, accounting for each stream's mass flow.

The streams remain separate, but energy transfers between them. Their enthalpy changes and mass flows must balance when heat loss and work are negligible.

Key terms

Regenerative feedwater heating: Warming cycle feedwater using steam extracted during turbine expansion.
Extraction fraction: The fraction of main turbine flow diverted at an extraction point, denoted by $y$ on the stated basis.
Open feedwater heater: A heater in which extracted steam and feedwater mix directly.
Closed feedwater heater: A heater in which extracted steam and feedwater exchange energy while remaining separate streams.
Thermal efficiency: Net cycle work output divided by heat supplied in the boiler.

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3.5 · Analyze regenerative feedwater heating

University of Alberta MEC E 340: Applied Thermodynamics

Heater balances, extraction flow, and cycle performance

What you will learn

1. Why regenerate, and what to compare

2. Define the basis and balance the heater

3. Carry the extraction through the cycle

4. Property data and a reliable solution sequence

Example 1: Extraction fraction for an open heater

Example 2: Bleed fraction for a closed heater

Example 3: Extraction in turbine work and efficiency

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through MEC E 340

About this lesson and its review