ข้อสอบสามัญเครื่องกล power plant 1/2547

8/14/2019 Power Plant 1/2547

1/14

2 (Power Plant / Gas Turbine) 1/2547

6 5 20 1. 5 MW

compressor 97 k Pa, 30 0C compressor 2 k Pa compressor 5.5 : 1 isentropic compressor 84 % (transmission efficiency) 98 % turbine 900 0C (C12 H26) 43,100 kJ/kg

97 % 3 % isentropic turbine 88 % turbine 100 k Pa

specific fuel consumption thermal efficiency

(CPa) = 1.005 kJ/kg K , (CPg) = 1.147 kJ/kg K , isentropic index (a) = 1.4 , (g) = 1.33 gas constant ( R ) = 0.287 kJ/kg K

(20 )

1

8/14/2019 Power Plant 1/2547

2/14


c = 84 %

m = 98 %T03 = 900

oC = 1,173 KHV = 43,100 kJ/kg

b = 97 %

compressor

adiabatic process stagnationh0a = h01 T0a = T01

T0a ~ Ta = T01 = 30 + 273 = 303 K

P1 ~ P 01 P01 = 97 2 = 95 k Pa

P02 = 5.5 P01 = 5.5 x 95 = 522.5 kPa P03 = P02 0.03 P02 = 0.97 x 522.5

= 506.8 k Pa

2

Wnet = 5 MW

Pa = 97 kPaTa = 30

oC

Pi = P0a - P01 = 2 k Pa

01

02

= 5.5

Pb = P03 - P02 = 3 % of P02t = 88 %

P04

= 100 k Pa

a = 1.4

g = 1.33R = 0.287 kJ/kg K Cpa = 1.005 kJ/kg KCpg = 1.147 kJ/kg K

8/14/2019 Power Plant 1/2547

3/14


compressor

01

20

T

T

= a1a

01

02

P

P

= 4.1

14.1

)5.5(

= (5.5)0.2857 = 1.6275

=20T 1.6275 x 303 = 493.1 K

c = 0.84 =0102

0120

TT

TT

T02 - T01 = 0.84303493.1

= 226.4 K

T02 = 529.4 K

Turbine

40

03

T

T

= g

g

1

)(04

03

= 1.33

11.33

)100

506.8(

= (5.068)0.248

= 1.4955

'04T =1.49551,173

= 784.3 K

t

= 0.88 =

4003

0403

TT

TT

T03- T04 = 342 K

T04 = 1,173 342 = 830.97 K

W = 5 MW = m a )(m

cW

tW

=

)T(T

0.98

1.005-)T1.147(Tm

01020403a

=

(226.4)

0.98

1.005-1.147(342)m a

3

8/14/2019 Power Plant 1/2547

4/14


= 160.1 am

am = 31.2 kg / s

bfHVm = m a Cpg ( T03 T02)

m f = 0.97x43,100)529.41,173(1.147x31.2

= 0.763 kg / s

f = af

m

m

= 31.20.763

= 0.024 kgf / kg

Specific fuel consumption (sfc) =Wf

= 600,3X160.1

0.024

= 0.55 kgf / kWh

Thermal efficieney () =HVfW

=43,100x0.024

160.1

= 15.5 %


4

8/14/2019 Power Plant 1/2547

5/14

2. Single stage gas turbine nozzle rotor (designed conditions) ( ) nozzle 700 stagnation pressure temperature 311 k Pa 850 0C static pressure 100 kPa total (stagnation) to static efficiency 87 % mean blade speed =500 m/s

stage 1. specific work done 2. Mach Number nozzle3. 4. total to total efficiency5. degree of reaction stage

(CP) isentropic ( ) 1.148 kJ/kgK 1.33

(20 )

Ca = 900 km/h = 250 m/s

F = A)P(P]CCf)[(1m aeaja ++am = = 100 kg/s

f = = 0.018 kgf /kgCj =

Ca = = 250 m/sPe = (nozzle)Pa = = 69 kPaA =

5

8/14/2019 Power Plant 1/2547

6/14


T04 = 970 K

T04 = T05 = T5 +gp

25

2C

C1

Check choking

2.49

69

172

P

P

a

04 ==

+

=

1g

g

)1

1

(1-1

1

cP

P

g

g

j

04

= 4

)]2.330.33(

0.951[1

1

Pc j

c

04

P

P= 1.914

choked


6

8/14/2019 Power Plant 1/2547

7/14

T5 = Tc =

+12

g

T04 = x 970 = 832.6 K

P5 = Pc = 1.914P04

= 1.914172 = 89.86 k Pa 1

25C = (970 832.6) x 2 x 1.147 = 315.2 x 10 3

C5 = Cj = 561.4 m/s

5 =5

5

TR

P=

832.6x0.287

89.86 = 0.375 kg/m3

5m = 5 A5 C5 = 0.375 A5 x 561.4 = am ( 1 + f )

= 100 x 1.018 = 101.8 kg/s A5 = 0.4836 m

2

F = 100 [ (1.018) 561.4 250 ] + (89.86 69) 0.4836

. = 42.24 kN

. (propulsive efficiency) , p

p =aC

C1

2

j+ =250

561.41

2

+ = 61.62 %

. (energy conversion efficiency) , e

e =(HV)m

/2)C(Cam

f

2a

2j

= 322

10x43,930x1.8

/2](250)[(561.4)100

. (overall efficiency) , o

o = 0.6162 x 0.1597 = 9.84 %


7

22.33

8/14/2019 Power Plant 1/2547

8/14

8/14/2019 Power Plant 1/2547

9/14

8/14/2019 Power Plant 1/2547

10/14

8/14/2019 Power Plant 1/2547

11/14


( 3- 4)

11

8/14/2019 Power Plant 1/2547

12/14

8/14/2019 Power Plant 1/2547

13/14

13

8/14/2019 Power Plant 1/2547

14/14


6. ( Combined Cycle Power Plant ) (Gass Turbine ) ( Steam Turbine) Heat Recovery Steam Generator, HRSG q q q Percentage

(20 )


14

ข้อสอบสามัญเครื่องกล power plant 1/2547

Documents