Microsoft Word - Cengel and Boles TOC _2-03-05_.doc

When the cold-air-standard assumptions are utilized, it reduces to

(9–24)

A regenerator with a higher effectiveness obviously saves a greater

amount of fuel since it preheats the air to a higher temperature prior to com-

bustion. However, achieving a higher effectiveness requires the use of a

larger regenerator, which carries a higher price tag and causes a larger pres-

sure drop. Therefore, the use of a regenerator with a very high effectiveness

cannot be justified economically unless the savings from the fuel costs

exceed the additional expenses involved. The effectiveness of most regener-

ators used in practice is below 0.85.

Under the cold-air-standard assumptions, the thermal efficiency of an

ideal Brayton cycle with regeneration is

(9–25)

Therefore, the thermal efficiency of an ideal Brayton cycle with regenera-

tion depends on the ratio of the minimum to maximum temperatures as well

as the pressure ratio. The thermal efficiency is plotted in Fig. 9–40 for vari-

ous pressure ratios and minimum-to-maximum temperature ratios. This fig-

ure shows that regeneration is most effective at lower pressure ratios and

low minimum-to-maximum temperature ratios.

hth,regen
 1 a

T 1

T 3

b1rp 21 k^1 2>k

P

T 5 T 2

T 4 T 2

516 | Thermodynamics

510152025

0.7

0.6

0.5

0.4

0.3

0.2

0.1

η th,Brayton

Pressure ratio, rp

With regeneration

Without regeneration

T 1 /T 3 = 0.2

T 1 /T 3 = 0.25

T 1 /T 3 = 0.33

FIGURE 9–40

Thermal efficiency of the ideal
Brayton cycle with and without
regeneration.

3

5

s

T, K

4 a

1

qregen = qsaved

1300

300

2 a

qin

FIGURE 9–41

T-sdiagram of the regenerative
Brayton cycle described in
Example 9–7.

EXAMPLE 9–7 Actual Gas-Turbine Cycle with Regeneration

Determine the thermal efficiency of the gas-turbine described in Example

9–6 if a regenerator having an effectiveness of 80 percent is installed.

Solution The gas-turbine discussed in Example 9–6 is equipped with a

regenerator. For a specified effectiveness, the thermal efficiency is to be

determined.

Analysis The T-sdiagram of the cycle is shown in Fig. 9–41. We first deter-

mine the enthalpy of the air at the exit of the regenerator, using the defini-

tion of effectiveness:

Thus,

This represents a savings of 220.0 kJ/kg from the heat input requirements.

The addition of a regenerator (assumed to be frictionless) does not affect the

net work output. Thus,

hth

wnet

qin

210.41 kJ>kg

570.60 kJ>kg

0.369 or 36.9%

qin
h 3 h 5 
 1 1395.97825.37 2 kJ>kg
570.60 kJ>kg

0.80

1 h 5 605.39 2 kJ>kg

1 880.36605.39 2 kJ>kg

Sh 5 
825.37 kJ>kg

P

h 5 h 2 a

h 4 ah 2 a