Gas Turbine Power Plants Gas Turbine Power Plants are lighter and more compact than vapor power plants. The favorable power-output-toweight ratio for gas turbines make them suitable for transportation.

Air-standard Brayton Cycle

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Q& CV W& CV − + (hin − hout ) 0= m& m&

For steady-state:

1Æ2

2Æ3 3Æ4 4Æ1

W&in Adiabatic compression Æ = (h2 − h1 ) m& Q& in Heat addition Æ = (h3 − h2 ) m& W& out Adiabatic expansion Æ = (h3 − h4 ) m& Q& out Heat removal Æ = (h4 − h1 ) m&

Cycle Thermal Efficiency:

η Brayton cycle

Q& out m& h −h = 1− = 1− 4 1 Q& in m& h3 − h2

Back work ratio:

W&in m& h −h = 2 1 bwr = W& out m& h3 − h4

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Ideal Air-standard Brayton Cycle (processes are reversible)

1Æ2 2Æ3 3Æ4 4Æ1

Isentropic compression Constant pressure heat addition Isentropic expansion Constant pressure heat removal

Qin

Qout

For the isentropic process 1Æ 2

P  Pr 2 = Pr 1  2   P1 

For the isentropic process 3 Æ 4

P  Pr 4 = Pr 3  4   P3 

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Ideal Cold Air-standard Brayton Cycle

For isentropic processes 1 Æ 2 and 3Æ 4 k −1  k

T2  P2 =   T1  P1 

Since

and k  k −1

T P2 P3 thus  2  = P1 P4  T1 

k −1  k

T4  P4 =   T3  P3  k  k −1

T =  3   T4 

→

T2 T3 = T1 T4

Thermal Efficiency

η Brayton = 1 − constk

h4 − h1 c (T − T ) T (T / T − 1) = 1− P 4 1 = 1− 1 4 1 h3 − h2 T2 (T3 / T2 − 1) c P (T3 − T2 )

T2 T3 T4 T3 recall = → = T1 T4 T1 T2

η Brayton = 1 − constk

T1 = 1− T2

1

(

k −1 P2 P1 k

)

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Efficiency increases with increased pressure ratio across the compressor

Back work ratio

W&in m& W& comp m& c P (T2 − T1 ) T2 − T1 = = = bwr = & & Wout m& Wturb m& c P (T3 − T4 ) T3 − T4

Typical BWR for the Brayton cycle is 40 - 80% compared to < 5% for the Rankine cycle. Recall, reversible compressor work is given by ∫12 vdP Since gas has a much larger specific volume than liquid much more power is required to compress the gas from P1 to P2 in the Brayton cycle compared to the Rankine cycle for which liquid is compressed. The turbine inlet temperature is limited by metallurgical factors, e.g., Tmax = 1700K

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Gas Turbine Irreversibilities

In the ideal Brayton cycle all 4 processes are assumed reversible, thus processes 2-3 and 4-1 are constant pressure and processes1-2 and 3-4 are isentropic. The constant pressure assumption does not normally incur any great errors but the compressor and turbine processes are far from isentropic

Ideal (reversible) processes: 1 - 2s and 3 - 4s Actual (irreversible) processes: 1 - 2 and 3 - 4

These irreversiblities are taken into account by:

ηturb

W&t   m&   = h3 − h4 =  h3 − h4 s W&t   m&   s

ηcomp

W& c   m&    s h2 s − h1 = = & h2 − h1 Wc   m&   

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Efficiency versus Power

Consider two Brayton cycles A and B with a similar turbine inlet temperatures T3

P  P  Since  2  >  2  Æ η A > η B  P1  A  P1  B

Since (enclosed area 1-2-3-4)B > (enclosed area 1-2-3-4)A  W& cycle   W& cycle  W& cycle, A m & A   >  Æ =  m&   m&  m& B W& cycle, B     B

A

In order for cycle A to produce the same amount of net power as cycle B, i.e., W& cycle, A = W& cycle, B , need m& A > m& B . Higher mass flow rate requires larger (heavier) equipment which is a concern in transportation applications 194

Increasing Cycle Power

The net cycle power is: W& cycle = W& t − W& c The cycle power can be increased by either increasing the turbine output power or decreasing the compressor input power. Gas Turbine with Reheat

The turbine work can be increased by using reheat, as was shown in the Rankine cycle

2

3

a

b

Compressor

1

4

The turbine is split into two stages and a second combustor is added where additional heat can be added

195

Recall:

T2 T3 so, isobars on T-s diagram diverge = T1 T4'

Q& in, 2

3

Q& in,1

T

b

Note: hb - h4 > ha - h4’

a

2

4 4’

1

s

The total turbine work output without reheat is: W& basic = [(h3 − ha ) + (ha − h4' )]m& The total turbine work output with reheat is: W& turbine = W& t ,1 + W&t , 2 = [(h3 − ha ) + (hb − h4 )]m& w / reheat

Since hb - h4 > ha - h4’

W& turbine

w / reheat

> W&basic

Since the compressor work h2 - h1 is unaffected by reheat W& cycle

w / reheat

> W& cycle

basic

The reheat cycle efficiency is not necessarily higher since additional heat Q& in, 2 is added between states a and b

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Compression with Intercooling

The compressor power can be reduced by compressing in stages with cooling between stages.

T2 T3 so, isobars on T-s diagram diverge = T1 T4'

Recall:

2’

2’

h2’ – hc > h2 – hd

d

197

The compressor power input without intercooling is: W& basic = [(h2' − hc ) + (hc − h1 )]m&

The total compressor power input with intercooling is: W& comp

w / reheat

= W& c ,1 + W& c , 2 = [(hc − h1 ) + (h2 − hd )]m&

Since h2’ – hc > h2 – hd Æ W& comp

w / reheat

< W&basic

Since the turbine work h3 – h4 is unaffected by intercooling W& cycle

w / reheat

> W& cycle

basic

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Different approach: The reversible work per unit mass for a steady flow device is ∫ vdP , so 2’

2 c 2'  W& c  = ∫ vdP = ∫ vdP + ∫ vdP   Without intercooling :  m&  basic 1 1 c = area b-1-c-2' -a

c 2 2  W& c  = ∫ vdP = ∫ vdP + ∫ vdP   With intercooling :  m&  w/ int 1 d 1 = area b-1-c-d-2-a

Since area(b-1-c-2’-a) > area(b-1-c-d-2-a)  W& c   W& c  >     m&  basic  m&  w / int 199

Aircraft Gas Turbines

Gas turbine engines are widely used to power aircraft because of their high power-to-weight ratio Turbojet engines used on most large commercial and military aircraft

Ideal air-standard jet propulsion cycle:

Diffuser

a

1

2

3

Nozzle

4

5

200

Normally compression through the diffuser (a-1), and expansion through the nozzle (4-5) are taken as isentropic

Q& in

Q& out

In the ideal jet propulsion engine the gas is not expanded to ambient pressure Pa. Instead the gas expands to an intermediate pressure P4 such that the power produced is just sufficient to drive the compressor, no net cycle power produced (W& cycle = 0 ), thus W& c W&t = m& m&

(h2 − h1 ) = (h3 − h4 ) After the turbine the gas expands to ambient pressure P5 which is the same as Pa. 201

Apply the steady-state conservation of energy equation to the Diffuser and Nozzle 2  Q& CV W& CV  Vin2   Vout   −  hout + 0= − +  hin +   m& m& 2   2  

Diffuser slows the flow to a zero velocity relative to the engine: Va2 V12 = ha + h1 + 2 2 Va2 h1 = ha + Diffuser (a Æ 1) 2 Va2 T1 = Ta + for constant k 2c P Nozzle accelerates the gas leaving the turbine (turbine exit velocity negligible compared to nozzle exit velocity):

Nozzle (4 Æ 5)

V52 V42 h4 + = h5 + 2 2 V5 = 2(h4 − h5 ) V5 = 2c P (T4 − T5 ) for constant k

202

The gas velocity leaving the nozzle is much higher than the velocity of the gas entering the diffuser, this change in momentum produces a propulsive force, or thrust Ft Ft = m& (V5 − Va )

Where V is flow velocity relative to engine For aircraft under cruise conditions the thrust just overcomes the drag force on the aircraft Æ fly at high altitude where the air is thinner and thus less drag To accelerate the aircraft increase thrust by increasing V5 In military aircraft afterburners are used to get very large thrust for short take-offs on aircraft carriers

An afterburner is simply a reheat device!

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Other Propulsion Systems

Turboprop

Turbofan

Subsonic ramjet

In turbofan bypass flow produces additional thrust for take-off. During cruise thrust comes from turbojet In a ramjet engine there is no compressor or turbine, compression is achieved gasdynamically. Ramjet engines produce no thrust when stationary thus must be coupled with a turbojet engine to get off the ground

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Supersonic Ramjet Engine

The flow is decelerated to subsonic velocity before the burner via a series of shock waves. Combustion occurs at constant pressure

Supersonic exhaust flow

Supersonic free stream flow choked flow

Turbojet-ramjet combination:

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Supersonic Combustion Ramjet (SCRAMJET) Engine

At very high Mach numbers the air temperature gets extremely hot after deceleration through the diffuser Va2 T1 = Ta + 2c P For Mach 6 flight speed, the air temperature just before the burner reaches about 1550K. At this temperature the air dissociates resulting in a drop in enthalpy At flight speeds greater than Mach 6 (hypersonic) better to burn fuel- in supersonic air stream

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US National Aero Space Plane (X-30)

Was to use 5 scramjet engines to achieve a Mach 12 flight speed To be used for travel to space and also as an airliner, a flight between any two points on earth would take less than 2 hours Canceled in 1993! Several countries have similar planes on the drawing board, Canada is not one of them!

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