The air-standard Brayton Cycle is the ideal cycle for a simple gas turbine and has been proposed as a direct cycle ... Reversible-adiabatic expansion ...

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Heat Exchanger 2

3

Comp.

Turbine

. Wnet

Heat Exchanger 1

4 . QL

Figure 1: Air Standard Brayton Cycle The processes involved in the Brayton Cycle are: 1) Constant pressure heat addition from the high temperature source 2) Reversible-adiabatic expansion in the turbine 3) Constant pressure heat rejection to the low temperature sink 4) Reversible-adiabatic compression in the compressor Figure 2 illustrates these processes on T-S and P-υ diagrams. We can analyze the performance of the Brayton Cycle by noting that for a single-phase gas,

η =1−

C pL (T4 − T1 ) QL = 1− C pH (T3 − T2 ) QH

(1)

where we have taken into account the fact that mass flow rate is constant. If we can further assume that the specific heat is constant, then

η =1−

(T4 − T1 ) T (T / T − 1) =1− 1 4 1 (T3 − T2 ) T2 (T3 / T2 − 1)

(2)

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3 3

2

Constant Pressure Τ

P

4

Constant Entropy

Constant Entropy

2 Constant Pressure

1

1

4 υ

S

Figure 2: T-S and P-υ diagrams for the Brayton Cycle Note, that since P2 = P3 and P1 = P4 , then

P2 P3 = P1 P4

(3)

In addition, recall the thermodynamic relationships

P2 T = 2 P1 T1 where k ≡

Cp Cv

k ( k −1)

=

P3 T = 3 P4 T4

k ( k −1)

(4)

and we have used the equality of pressure ratios given in Equation 3. From Equation 4

T2 T3 = T1 T4

T3 T4 = T2 T1

(5)

which may be substituted into Equation 2 to give for the efficiency

η = 1−

T1 T2

(6)

or in terms of pressure

η =1−

1 ( P2 / P1 ) ( k −1)

k

(7)

The efficiency is then a direct function of the isentropic pressure ratio. One important feature of the Brayton Cycle is the large amount of compressor work required relative to the turbine work, with the compressor requiring as much as 40% to 80% of the turbine output. This contrast dramatically with the Rankine cycle where only 1% or 2% of the turbine output was required to operate the pumps. The difference in pump work and compressor work is due to the difference in the specific volume between gases and liquids in the expression

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Wcomp m

=

e i

υdP

(8)

and can have a devastating effect on the performance of an actual cycle where turbine and compressor efficiencies must be included. The development of high efficiency gas turbines and compressors is therefore a high priority for these cycles to be effective. Modifications to the Brayton Cycle

The performance of the Brayton Cycle can be improved, by noting that in Figure 2 the temperature of the gas leaving the turbine is higher than the gas temperature leaving the compressor. We can therefore transfer heat from the turbine exhaust to the high pressure gas leaving the compressor, prior to entering the high temperature heat exchanger. This is accomplished through the use of a regenerator, which is similar in concept to the feed heaters employed in the Regenerative Cycle. The Brayton Cycle with regenerators is illustrated below.

Comp. 2

x

3

Heat Exch.

1

. Wnet

Turbine

. QH

4

Regenerator

Heat Exch.

y

. QL

Figure 3: Brayton Cycle with Regenerators These modifications are also indicated on the T-S and P-υ diagrams below.

3

Τ

x 2

.x

2

Constant Pressure 4

P

3 Constant Entropy

Constant Entropy y Constant Pressure

1

.y

1

S

4

υ

Figure 4: T-S and P-υ Diagram with Regenerator

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The cycle efficiency is most conveniently computed using work

η=

wturb + wcomp

(9)

qH

where the heat transfer rate is now

q H = C p (T3 − T x )

(10)

wturb = h3 − h4 = C p (T3 − T4 )

(11)

and the turbine work can be expressed as

For an ideal regenerator, the temperature leaving the regenerator is equal to the turbine exhaust temperature, i.e.

Tx = T4

q H = wturb

giving for the cycle efficiency

η=

wturb + wcomp qH

=

wturb + wcomp wturb

=1+

wcomp wturb

(12)

The compressor work is given by

− wcomp = h2 − h1 = C p (T2 − T1 )

(13)

such that

η =1−

C p (T2 − T1 ) C p (T3 − T4 )

= 1−

(T2 − T1 ) T (T / T − 1) =1− 1 2 1 (T3 − T4 ) T3 (1 − T4 / T3 )

(14)

for constant specific heat. Again, expressing the temperature ratios in terms of pressures,

η =1−

T1 (T2 / T1 − 1) T [( P / P ) ( k −1) k − 1] =1− 1 2 1 T3 (1 − T4 / T3 ) T3[1 − ( P4 / P3 ) ( k −1) k ]

(15)

where as before

P2 P3 = P1 P4 such that

η =1−

T1[( P2 / P1 ) ( k −1) k − 1] T P =1− 1 2 ( k −1) k T T3[1 − ( P1 / P2 ) ] 3 P1

( k −1) k

[1 − ( P1 / P2 ) ( k −1) k ] [1 − ( P1 / P2 ) ( k −1) k ]

(16)

or upon simplification

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T P η =1− 1 2 T3 P1

( k −1) k

(17)

The efficiency is now a function of both the isentropic pressure ratio and the ratio of the minimum to maximum temperatures in the cycle. Example: An HTGR is to be operated on the Brayton Cycle. The coolant is helium with a compressor inlet temperature of 91 F and 380 psia. The coolant exits the compressor at 1025 psia. The reactor exit temperature is 1560 F. Determine the cycle efficiency. Solution:

For the simple Brayton Cycle,

η = 1−

1 ( P2 / P1 ) ( k −1)

k

where P2 = 1025 psia and P1 = 380 psia. For helium, k = 1.667 such that the cycle efficiency is

η =1−

1 = 32.77 % (1025 / 380) (1.667 −1) 1.667

If we now compare to the efficiency using a regenerator, then

T P η =1− 1 2 T3 P1

( k −1) k

(91 + 460) 1025 = 1− (1560 + 460) 380

(1.667 −1) 1.667

= 59.43 %

It would appear based upon these examples, that reactor concepts based on these cycles offer the potential of significantly higher thermal efficiencies than those available in current Light Water Reactor designs.

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