Moran M.J., Shapiro H.N. Fundamentals of Engineering Thermodynamics

Подождите немного. Документ загружается.

regenerator from such an ideal regenerator. The regenerator effectiveness is defined

as the ratio of the actual enthalpy increase of the air flowing through the compressor

side of the regenerator to the maximum theoretical enthalpy increase. That is,

reg

2 h

(9.27)

As heat transfer approaches reversibility, h

approaches h

and h

reg

tends to unity

(100%).

In practice, regenerator effectiveness values typically range from 60 to 80%, and

thus the temperature T

of the air exiting on the compressor side of the regenerator

is normally well below the turbine exhaust temperature. To increase the effectiveness

above this range would require greater heat transfer area, resulting in equipment

costs that might cancel any advantage due to fuel savings. Moreover, the greater heat

transfer area that would be required for a larger effectiveness can result in a signifi-

cant frictional pressure drop for flow through the regenerator, thereby affecting over-

all performance. The decision to add a regenerator is influenced by considerations

such as these, and the final decision is primarily an economic one.

In Example 9.7, we analyze an air-standard Brayton cycle with regeneration and

explore the effect on thermal efficiency as the regenerator effectiveness varies.

regenerator effectiveness

9.7 Regenerative Gas Turbines 523

Evaluating Thermal Efficiency of a Brayton Cycle with Regeneration

c c c c EXAMPLE 9.7 c

A regenerator is incorporated in the cycle of Example 9.4. (a) Determine the thermal efficiency for a regenerator

effectiveness of 80%. (b) Plot the thermal efficiency versus regenerator effectiveness ranging from 0 to 80%.

SOLUTION

Known:

A regenerative gas turbine operates with air as the working fluid. The compressor inlet state, turbine

inlet temperature, and compressor pressure ratio are known.

Find: For a regenerator effectiveness of 80%, determine the thermal efficiency. Also plot the thermal efficiency

versus the regenerator effectiveness ranging from 0 to 80%.

Schematic and Given Data:

Fig. E9.7a

Combustor

234

TurbineCompressor

reg

80%

cycle

Regenerator

= 1400 K

= 300 K

= 100 kPa

= 300 K

p = 1000 kPa

p = 100 kPa

= 1400 K

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524 Chapter 9 Gas Power Systems

Engineering Model:

Each component is analyzed as a control volume at steady state. The control volumes are shown on the

accompanying sketch by dashed lines.

2. The compressor and turbine processes are isentropic.

3. There are no pressure drops for flow through the heat exchangers.

4. The regenerator effectiveness is 80% in part (a).

5. Kinetic and potential energy effects are negligible.

6. The working fluid is air modeled as an ideal gas.

Analysis:

(a)

The specific enthalpy values at the numbered states on the T–s diagram are the same as those in Example

9.4: h

5 300.19 kJ/kg, h

5 579.9 kJ/kg, h

5 1515.4 kJ/kg, h

5 808.5 kJ/kg.

To find the specific enthalpy h

, the regenerator effectiveness is used as follows: By definition

reg

2 h

Solving for h

5 h

reg

2 h

21 h

5 10.821808.5 2 579.921 579.9 5 762.8 kJ

With the specific enthalpy values determined above, the thermal efficiency is

➊ h 5

22 1W

2 h

22 1h

2 h

11515.4 2 808.522 1579.9 2 300.1

11515.4 2 762.82

➋ 5 0.568 156.8%2

(b) The IT code for the solution follows, where h

reg

is denoted as etareg, h is eta, W

comp

is Wcomp, and so on.

// Fix the states

T1 = 300//K

p1 = 100//kPa

h1 = h_T(“Air”, T1)

s1 = s_TP(“Air”, T1, p1)

p2 = 1000//kPa

s2 = s_TP(“Air”, T2, p2)

s2 = s1

h2 = h_T(“Air”, T2)

T3 = 1400//K

p3 = p2

h3 = h_T(“Air”, T3)

s3 = s_TP(“Air”, T3, p3)

p4 = p1

s4 = s_TP(“Air”, T4, p4)

s4 = s3

h4 = h_T(“Air”, T4)

etareg = 0.8

hx = etareg*(h4 – h2) + h2

// Thermal efficiency

Wcomp = h2 – h1

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Wturb = h3 – h4

Qin = h3 – hx

eta = (Wturb – Wcomp) / Qin

Using the Explore button, sweep etareg from 0 to 0.8 in steps of 0.01. Then, using the Graph button, obtain the

following plot:

What would be the thermal efficiency if the regenerator effec-

tiveness were 100%? Ans. 60.4%.

➌ From the computer data, we see that the cycle thermal efficiency increases

from 0.456, which agrees closely with the result of Example 9.4 (no regen-

erator), to 0.567 for a regenerator effectiveness of 80%, which agrees closely

with the result of part (a). This trend is also seen in the accompanying graph.

Regenerator effectiveness is seen to have a significant effect on cycle ther-

mal efficiency.

➊ The values for work per unit of mass flow of the compressor and turbine

are unchanged by the addition of the regenerator. Thus, the back work ratio

and net work output are not affected by this modification.

➋ Comparing the present thermal efficiency value with the one determined

in Example 9.4, it should be evident that the thermal efficiency can be

increased significantly by means of regeneration.

➌ The regenerator allows improved fuel utilization to be achieved by transfer-

ring a portion of the exergy in the hot turbine exhaust gas to the cooler air

flowing on the other side of the regenerator.

Ability to…

❑

sketch the schematic of the

regenerative gas turbine and

the T–s diagram for the

corresponding air-standard

cycle.

❑

evaluate temperatures and

pressures at each principal

state and retrieve necessary

property data.

❑

calculate the thermal

efficiency.

✓

Skills Developed

9.8 Regenerative Gas Turbines with

Reheat and Intercooling

Two modifications of the basic gas turbine that increase the net work developed are

multistage expansion with reheat and multistage compression with intercooling. When

used in conjunction with regeneration, these modifications can result in substantial

increases in thermal efficiency. The concepts of reheat and intercooling are introduced

in this section.

9.8 Regenerative Gas Turbines with Reheat and Intercooling 525

Fig. E9.7b

0.1 0.3 0.5 0.70 0.2 0.4 0.6 0.8

Regenerator effectiveness

Thermal efficiency

0.1

0.2

0.3

0.4

0.5

0.6

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526 Chapter 9 Gas Power Systems

9.8.1

Gas Turbines with Reheat

For metallurgical reasons, the temperature of the gaseous combustion products enter-

ing the turbine must be limited. This temperature can be controlled by providing air

in excess of the amount required to burn the fuel in the combustor (see Chap. 13).

As a consequence, the gases exiting the combustor contain sufficient air to support

the combustion of additional fuel. Some gas turbine power plants take advantage of

the excess air by means of a multistage turbine with a reheat combustor between the

stages. With this arrangement the net work per unit of mass flow can be increased.

Let us consider reheat from the vantage point of an air-standard analysis.

The basic features of a two-stage gas turbine with reheat are brought out by con-

sidering an ideal air-standard Brayton cycle modified as shown in Fig. 9.16. After

expansion from state 3 to state a in the first turbine, the gas is reheated at constant

pressure from state a to state b. The expansion is then completed in the second tur-

bine from state b to state 4. The ideal Brayton cycle without reheat, 1–2–3–49–1, is

shown on the same T–s diagram for comparison. Because lines of constant pressure

on a T–s diagram diverge slightly with increasing entropy, the total work of the two-

stage turbine is greater than that of a single expansion from state 3 to state 49. Thus,

the net work for the reheat cycle is greater than that of the cycle without reheat.

Despite the increase in net work with reheat, the cycle thermal efficiency would not

necessarily increase because a greater total heat addition would be required. How-

ever, the temperature at the exit of the turbine is higher with reheat than without

reheat, so the potential for regeneration is enhanced.

When reheat and regeneration are used together, the thermal efficiency can

increase significantly. The following example provides an illustration.

reheat

c c c c EXAMPLE 9.8 c

Determining Thermal Efficiency of a Brayton Cycle

with Reheat and Regeneration

Consider a modification of the cycle of Example 9.4 involving reheat and regeneration. Air enters the compres-

sor at 100 kPa, 300 K and is compressed to 1000 kPa. The temperature at the inlet to the first turbine stage is

1400 K. The expansion takes place isentropically in two stages, with reheat to 1400 K between the stages at a

constant pressure of 300 kPa. A regenerator having an effectiveness of 100% is also incorporated in the cycle.

Determine the thermal efficiency.

SOLUTION

Known:

An ideal air-standard gas turbine cycle operates with reheat and regeneration. Temperatures and pres-

sures at principal states are specified.

Fig. 9.16

Ideal gas turbine with reheat.

Reheat

combustor

Turbine

stage 2

cycle

Combustor

Turbine

stage 1

Compressor

4′

a b

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9.8 Regenerative Gas Turbines with Reheat and Intercooling 527

Find: Determine the thermal efficiency.

Schematic and Given Data:

What percentage of the total heat addition occurs in the reheat

process? Ans. 52%.

Analysis: We begin by determining the specific enthalpies at each principal state of the cycle. States 1, 2, and 3

are the same as in Example 9.4: h

5 300.19 kJ/kg, h

5 579.9 kJ/kg, h

5 1515.4 kJ/kg. The temperature at state

b is the same as at state 3, so h

5 h

Since the first turbine process is isentropic, the enthalpy at state a can be determined using p

data from Table

A-22 and the relationship

5 p

5 1450.52

300

1000

5 135.15

Interpolating in Table A-22, we get h

5 1095.9 kJ/kg.

The second turbine process is also isentropic, so the enthalpy at state 4 can be determined similarly. Thus

5 p

5 1450.52

100

300

5 150.17

Interpolating in Table A-22, we obtain h

5 1127.6 kJ/kg. Since the regenerator

effectiveness is 100%, h

5 h

5 1127.6 kJ/kg.

The thermal efficiency calculation must take into account the compressor work,

the work of each turbine, and the total heat added. Thus, on a unit mass basis

h 5

2 h

21 1h

2 h

22 1h

2 h

21 1h

2 h

11515.4 2 1095.921 11515.4 2 1127.622 1579.9 2 300.19

11515.4 2 1127.621 11515.4 2 1095.92

➊ 5 0.654 165.4%2

➊ Comparing the present value with the thermal efficiency determined in part

(a) of Example 9.4, we can conclude that the use of reheat coupled with

regeneration can result in a substantial increase in thermal efficiency.

Ability to…

❑

sketch the schematic of the

regenerative gas turbine with

reheat and the T–s diagram

for the corresponding

air-standard cycle.

❑

evaluate temperatures and

pressures at each principal

state and retrieve necessary

property data.

❑

calculate the thermal

efficiency.

✓

Skills Developed

Engineering Model:

Each component of the power plant is analyzed as a control

volume at steady state.

2. The compressor and turbine processes are isentropic.

3. There are no pressure drops for flow through the heat exchangers.

4. The regenerator effectiveness is 100%.

5. Kinetic and potential energy effects are negligible.

6. The working fluid is air modeled as an ideal gas.

Fig. E9.8

Regenerator

exit

= 300 K

p = 1000 kPa

p = 100 kPa

p = 300 kPa

= 1400 K

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528 Chapter 9 Gas Power Systems

9.8.2

Compression with Intercooling

The net work output of a gas turbine also can be increased by reducing

the compressor work input. This can be accomplished by means of mul-

tistage compression with intercooling. The present discussion provides an

introduction to this subject.

Let us first consider the work input to compressors at steady state,

assuming that irreversibilities are absent and changes in kinetic and

potential energy from inlet to exit are negligible. The p–y diagram of Fig.

9.17 shows two alternative compression paths from a specified state 1 to

a specified final pressure p

. Path 1–29 is for an adiabatic compression.

Path 1–2 corresponds to a compression with heat transfer from the work-

ing fluid to the surroundings. The area to the left of each curve equals

the magnitude of the work per unit mass of the respective process (see

Sec. 6.13.2). The smaller area to the left of Process 1–2 indicates that the

work of this process is less than for the adiabatic compression from 1 to

29. This suggests that cooling a gas during compression is advantageous

in terms of the work-input requirement.

Although cooling a gas as it is compressed would reduce the work, a

heat transfer rate high enough to effect a significant reduction in work is

difficult to achieve in practice. A practical alternative is to separate the work and

heat interactions into separate processes by letting compression take place in stages

with heat exchangers, called intercoolers, cooling the gas between stages. Figure 9.18

illustrates a two-stage compressor with an intercooler. The accompanying p–y and

T–s diagrams show the states for internally reversible processes:

c Process 1–c is an isentropic compression from state 1 to state c where the pressure

is p

c Process c–d is constant-pressure cooling from temperature T

to T

c Process d–2 is an isentropic compression to state 2.

intercooler

Fig. 9.17 Internally reversible compression

processes between two fixed pressures.

Adiabatic compression

′

Compression with cooling

int

rev

∫

vdp

___

Fig. 9.18 Two-stage

compression with intercooling.

22′

2′

s = c

T = c

Compressor

stage 1

Compressor

stage 2

Intercooler

out

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The work input per unit of mass flow is represented on the p–y diagram by shaded

area 1–c–d–2–a–b–1. Without intercooling the gas would be compressed isentropically

in a single stage from state 1 to state 29 and the work would be represented by

enclosed area 1–29–a–b–1. The crosshatched area on the p–y diagram represents the

reduction in work that would be achieved with intercooling.

Some large compressors have several stages of compression with intercooling

between stages. The determination of the number of stages and the conditions at

which to operate the various intercoolers is a problem in optimization. The use of

multistage compression with intercooling in a gas turbine power plant increases the

net work developed by reducing the compression work. By itself, though, compression

with intercooling would not necessarily increase the thermal efficiency of a gas turbine

because the temperature of the air entering the combustor would be reduced (com-

pare temperatures at states 29 and 2 on the T–s diagram of Fig. 9.18). A lower tem-

perature at the combustor inlet would require additional heat transfer to achieve the

desired turbine inlet temperature. The lower temperature at the compressor exit

enhances the potential for regeneration, however, so when intercooling is used in

conjunction with regeneration, an appreciable increase in thermal efficiency can result.

In the next example, we analyze a two-stage compressor with intercooling between

the stages. Results are compared with those for a single stage of compression.

9.8 Regenerative Gas Turbines with Reheat and Intercooling 529

Evaluating a Two-Stage Compressor with Intercooling

c c c c EXAMPLE 9.9 c

Air is compressed from 100 kPa, 300 K to 1000 kPa in a two-stage compressor with intercooling between stages.

The intercooler pressure is 300 kPa. The air is cooled back to 300 K in the intercooler before entering the sec-

ond compressor stage. Each compressor stage is isentropic. For steady-state operation and negligible changes in

kinetic and potential energy from inlet to exit, determine (a) the temperature at the exit of the second compres-

sor stage and (b) the total compressor work input per unit of mass flow. (c) Repeat for a single stage of com-

pression from the given inlet state to the final pressure.

SOLUTION

Known:

Air is compressed at steady state in a two-stage compressor with intercooling between stages. Operating

pressures and temperatures are given.

Find: Determine the temperature at the exit of the second compressor stage and the total work input per unit

of mass flow. Repeat for a single stage of compression.

Schematic and Given Data:

Fig. E9.9

= 300 kPa

Compressor

stage 1

Compressor

stage 2

Intercooler

out

= 100 kPa

= 300 K

= 1000 kPa

= 300 K

= 100 kPa

= 300 K

= 1000 kPa

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530 Chapter 9 Gas Power Systems

Engineering Model:

The compressor stages and intercooler are analyzed as control volumes at steady state. The control volumes

are shown on the accompanying sketch by dashed lines.

2. The compression processes are isentropic.

3. There is no pressure drop for flow through the intercooler.

4. Kinetic and potential energy effects are negligible.

5. The air is modeled as an ideal gas.

Analysis:

(a)

The temperature at the exit of the second compressor stage, T

, can be found using the following relationship

for the isentropic process d–2

5 p

With p

at T

5 300 K from Table A-22, p

5 1000 kPa, and p

5 300 kPa,

5 11.3862

1000

300

5 4.62

Interpolating in Table A-22, we get T

5 422 K and h

5 423.8 kJ/kg.

(b) The total compressor work input per unit of mass is the sum of the work inputs for the two stages. That is

5 1h

2 h

21 1h

2 h

From Table A-22 at T

5 300 K, h

5 300.19 kJ/kg. Since T

5 T

, h

5 300.19 kJ/kg. To find h

, use p

data

from Table A-22 together with p

5 100 kPa and p

5 300 kPa to write

5 p

5 11.3862

300

100

5 4.158

Interpolating in Table A-22, we obtain h

5 411.3 kJ/kg. Hence, the total compressor work per unit of mass is

5 1411.3 2 300.1921 1423.8 2 300.1925 234.7 kJ

3 located on the accompanying p–y diagram. The temperature at this state can

be determined using

5 p

5 11.3862

1000

100

5 13.86

Interpolating in Table A-22, we get T

5 574 K and h

5 579.9 kJ/kg.

The work input for a single stage of compression is then

5 h

2 h

5 579.9 2 300.19 5 279.7 kJ

This calculation confirms that a smaller work input is required with two-stage com-

pression and intercooling than with a single stage of compression. With intercooling,

however, a much lower gas temperature is achieved at the compressor exit.

In this case, what is the percentage reduction in compressor

work with two-stage compression and intercooling compared to a single

stage of compression? Ans. 16.1%.

Ability to…

❑

sketch the schematic of a

two-stage compressor with

intercooling between the

stages and the corresponding

T–s diagram.

❑

evaluate temperatures and

pressures at each principal

state and retrieve necessary

property data.

❑

apply energy and entropy

balances.

✓

Skills Developed

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Referring again to Fig. 9.18, the size of the crosshatched area on the p–y diagram

representing the reduction in work with intercooling depends on both the tempera-

ture T

at the exit of the intercooler and the intercooler pressure p

. By properly

selecting T

and p

, the total work input to the compressor can be minimized. For

example, if the pressure p

is specified, the work input would decrease (crosshatched

area would increase) as the temperature T

approaches T

, the temperature at the

inlet to the compressor. For air entering the compressor from the surroundings, T

would be the limiting temperature that could be achieved at state d through heat

transfer with the surroundings only. Also, for a specified value of the temperature T

the pressure p

can be selected so that the total work input is a minimum (cross-

hatched area is a maximum).

Example 9.10 provides an illustration of the determination of the intercooler pres-

sure for minimum total work using a cold air-standard analysis.

9.8 Regenerative Gas Turbines with Reheat and Intercooling 531

Determining Intercooler Pressure for Minimum Compressor Work

c c c c EXAMPLE 9.10 c

For fixed inlet state and exit pressure, use a cold-air standard analysis to show that minimum total work input

for a two-stage compressor is required when the pressure ratio is the same across each stage. Assume steady-state

operation and the following idealizations: Each compression process is isentropic, there is no pressure drop through

the intercooler, temperature at the inlet to each compressor stage is the same, and kinetic and potential energy

effects can be ignored.

SOLUTION

Known:

A two-stage compressor with intercooling operates at steady state under specified conditions.

Find: Show that the minimum total work input is required when the pressure ratio is the same across each stage.

Schematic and Given Data:

Engineering Model:

The compressor stages and intercooler are analyzed as control vol-

umes at steady state.

2. The compression processes are isentropic.

3. There is no pressure drop for flow through the intercooler.

4. The temperature at the inlet to both compressor stages is the same.

5. Kinetic and potential energy effects are negligible.

6. The working fluid is air modeled as an ideal gas.

7. The specific heat c

and thus the specific heat ratio k are constant.

Analysis: The total compressor work input per unit of mass flow is

5 1h

2 h

21 1h

2 h

Since c

is constant

5 c

2 T

21 c

2 T

Fig. E9.10

variable

Compressor inlet state specified

s = c

T =

specified

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532 Chapter 9 Gas Power Systems

With T

5 T

(assumption 4), this becomes on rearrangement

5 c

2 2b

Since the compression processes are isentropic and the specific heat ratio k is constant, the pressure and

temperature ratios across the compressor stages are related, respectively, by

5 a

1k212



and





5 a

1k212

In the second of these equations, T

5 T

by assumption 4.

Collecting results

5 c

1k212

1 a

1k212

2 2 d

Hence, for specified values of T

, p

, and c

, the value of the total compressor work input varies with the

intercooler pressure only. To determine the pressure p

that minimizes the total work, form the derivative

01W

1k212

1 a

1k212

2 2 df

5 c

k 2 1

bca

b1 a

5 c

k 2 1

1k212

2 a

1k212

When the partial derivative is set to zero, the desired relationship is obtained

➊

By checking the sign of the second derivative, it can be verified that the total

compressor work is a minimum.

➊ This relationship is for a two-stage compressor. Appropriate relations can be

obtained similarly for multistage compressors.

Air enters an ideal two-stage compressor with intercooling at

100 kPa. The overall compressor pressure ratio is 12. What pressure, in

bar, would minimize the total work input required? Ans. 3.464 bar.

Ability to…

❑

complete the detailed

derivation of a thermody-

namic expression.

❑

use calculus to minimize a

function.

✓

Skills Developed

9.8.3

Reheat and Intercooling

Reheat between turbine stages and intercooling between compressor stages provide

two important advantages: The net work output is increased, and the potential for

regeneration is enhanced. Accordingly, when reheat and intercooling are used

together with regeneration, a substantial improvement in performance can be real-

ized. One arrangement incorporating reheat, intercooling, and regeneration is shown

in Fig. 9.19. This gas turbine has two stages of compression and two turbine stages. The

accompanying T–s diagram is drawn to indicate irreversibilities in the compressor

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