ข้อสอบสามัญเครื่องกล Plant 3/2549

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2

1 50 (X) ( 1 )1. Fission

. 2-3 . 200 MeV

. 3 Fission fragments .

2. Moderator, Control rod Coolant . Moderator Fission

Fission . Control rod

Fission. Coolant .

3. Fission Heat exchanger/Steam generator Secondary loop Turbine . Pressurized water reactor (PWR).. Boiling water reactor (BWR).. Pressurized heavy water reactor (PHWR)..

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4. Pressurized water reactor (PWR) Boiling water reactor (BWR)

. PWR Turbine . BWR 2 Coolant loops Secondary loop Turbine. Primary loop

PWR BWR.

5. Moderator Pressurized heavy water reactor (PHWR).. Low pressurized water (H2O).. High pressurized water (H

2O).

. Heavy water (D2O).

. Liquid sodium.6. PWR

. Coolant Primary loop

. Turbine . .

7. BWR. Coolant Primary loop . Turbine . .

8. . Greenhouse gas . 18

. Capacity factor 90% ( USA) base-load.

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9. Radioactive decay _____

. .

. .

10. (t1/2) ()

()

. () . -244 ( 80.8 ) . -233 ( 20.9 ) .

11. . . . .

12. Combined cycle

. SOX. NO

X

. Particulates. CO

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13. Fossil

. SOX. NOX. Particulates. 3

14. . .

. . 3

15. SOX

. . . .

16. SOX. Greenhouse effect. . . 3

17. NOX. NO N2O. NO2 NO3. NO NO2. 3

18. NOX Fossil. NO. NO2

. N

2

. N2O3

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19. NOX

. . 2 . 3 . 4

20. NOX. N O2. N O2

. N Flame chilling. N

21. Load-duration curve . . .

. 22. Load factor

. . . .

23. kWh. . . Operating cost. . .

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24. (Variable cost)

. . . .

25. . . Load-duration curve

. .

26. 2 Demand Demand . (Fixed costs) Operate. (Variable costs).

. 27. 15-20%

. . Fixed cost . .

28. Peak Off-peak. Off-peak. Off-peak. Supply Demand . Peak

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29. Base load

. Fixed cost . . . Heat rate

30. 4 .

. Combined cycle. .

31. Cogeneration . .

. . 2 Power Cycles Combined Cycle

32. Cogeneration Topping Cycle . Power Cycle

Process. Process

. Power Cycle

Process. 2 Power Cycles (Binary Power Cycles)

33. Cogeneration . Combined Cycle. Gas Turbine Cogeneration

. Steam Turbine Cogeneration

. Internal Combustion Engine Cogeneration

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34. Cogeneration

. Gas Turbine Cogeneration. Steam Turbine Cogeneration. Combined Cycle. Internal Combustion Engine Cogeneration

35. Back Pressure Steam Turbine Turbine . .

. .

36. Steam Turbine Cogeneration . Steam Turbine Cogeneration. Gas Turbine Cogeneration. Combined Cycle Cogeneration. Internal Combustion Engine Cogeneration

37. Cogeneration 10 . Steam Turbine Cogeneration. Gas Turbine Cogeneration. Combined Cycle Cogeneration. Internal Combustion Engine Cogeneration

38. Cogeneration . . . .

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39. Steam Turbine Steam Turbine Cogeneration

. Back-pressure Turbine. Extraction Turbine. Pass-out Turbine.

40. Cogeneration Part Load. Gas Turbine Cogeneration. Steam Turbine Cogeneration

. Combined Cycle Cogeneration. Internal Combustion Engine Cogeneration

41. 1 MW Cogeneration . Steam Turbine Cogeneration. Gas Turbine Cogeneration. Internal Combustion Engine Cogeneration

. Combined Cycle Cogeneration42. Cogeneration 1 MW

4 MW . 40%. 45%. 50%. 60%

43. Cogeneration 1 MW 1 MW Heat toPower Ratio . 1. 2. 4. 6

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44. Cogeneration Steam Turbine 2

. Steam Turbine Cogeneration. Gas Turbine Cogeneration. Combined Cycle Cogeneration. Internal Combustion Engine Cogeneration

45. Cogeneration . Annual Operating Cost . Annual Operating Cost

. Cogeneration . Cogeneration, Annual Operating Cost

46. Gas Turbine Cogeneration Combined Cycle Cogeneration . By-pass Stack. Steam Turbine . 2 . Condenser

47. Cogeneration . Internal Combustion Engine Cogeneration. Gas Turbine Cogeneration. Combined Cycle Cogeneration. Steam Turbine Cogeneration

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48. QHh

E e 2 .

.

.

.

49. Cogeneration . Steam Turbine Cogeneration. Gas Turbine Cogeneration. Combined Cycle Cogeneration. Internal Combustion Engine Cogeneration

50. Internal Combustion Engine Cogeneration

. . . .

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2 2 ( 25 )

1. 1 Air Standard Cycle 1 bar 33 o C Compressor 8 1,100 o C Cp = 1.005 kJ/kg-K k = 1.4

. Isentropic Compressor Gas Turbine.1 Net Work Cycle kg .2 Heat Input Cycle kg .3 Cycle Efficiency

T2T

2= T

1{P

2/P

1}

(k-1)/k= (33+273) (8)

(1.4-1)/1.4= 554.3 K

Qin = 2Q3 = cp (T3 T2) 1.005 x ((1100+273)- 554.3) = 822.79 kJ/kg .2 T4T

4= T

3/{P

2/P

1}

(k-1)/k= 1373 / (8)

(1.4-1)/1.4= 757.95 K

Qout = 4Q1 = cp (T1 T4) = 1.005 x (303 757.95) = - 454.21 kJ/kg

Net work, W

W = Qin + Qout = 822.79 - 454.21 = 368.57 kJ/kg .1

Cycle Efficiency = W/Qin = 365.55 / 822.79 = 0.4479 = 44.8 % .3

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. Isentropic Compressor Gas Turbine 90 %

.1 Net Work Cycle kg .2 Heat Input Cycle kg .3 Cycle Efficiency

Isentropic Compressor work, WCI = cp (T

1 T

2) = 1.005 x (33+273 - 554.3 ) = - 249.54 kJ/kg

Compressor work, WC = cp (T1 T2) = WCI/Eff = - 249.54/0.9 = -277.27 kJ/kg

T2

= 277.27 / 1.005 306 = 581.89 K

Isentropic Turbine work, WTI = cp (T3 T4) = 1.005 x (1100+273 - 757.95 ) = 618.11 kJ/kg

Turbine work, WT = cp (T3 T

4) = WTI X Eff = 618.11 x 0.9 = 556.31 kJ/kg

Net work, W = WT + WC = 556.31 - 277.27 = 279.03 kJ/kg .1

Qin = 2Q3 = cp (T3 T2) 1.005 x ((1100+273) 581.89) = 795.06 kJ/kg .2

Cycle Efficiency = W/Qin = 279.03 / 795.06 = 0.4479 = 35.06 % .3

2. (Gas Turbine Cogeneration Plant) 1 Air Standard Cycle 1bar 33 o C Compressor 8 1,100 o C 10 MW Heat Recovery Steam Generator, HRSG (Saturated Steam) 10 bar 120 o C HRSG (Saturated Water) 10 bar Enthalpy = 763kJ/kg (Saturated Steam) 10 bar Enthalpy = 2,778 kJ/kg

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ton/hr

T4T

4= T

3/{P

2/P

1}

(k-1)/k= 1373 / (8)

(1.4-1)/1.4= 757.95 K

T2T2 = T1{P2/P1}

(k-1)/k= (33+273) (8)

(1.4-1)/1.4= 554.3 K

Qin = 2Q3 = cp (T3 T2) 1.005 x ((1100+273)- 554.3) = 822.79 kJ/kg

T4T

4= T

3/{P

2/P

1}

(k-1)/k= 1373 / (8)

(1.4-1)/1.4= 757.95 K

Qout =

4Q

1= c

p(T

1 T

4) = 1.005 x (303 757.95) = - 454.21 kJ/kg

Net work, W

W = Qin + Qout = 822.79 - 454.21 = 368.57 kJ/kg

Mass ()Mg = Power/Net work = 10,000/368.57 = 27.13 kg/s

Heat from hot gas, Qg = Mg X cp X (Tin Tout)

Qg = 27.13 x 1.005 x (757.95 - (120 + 273)) = 9,951.4 kW

Steam mass. Ms = Qg/(h2 h1) = 9,951.4/(2778 763) = 4.94 kg/s = 4.94 x 3.6 = 17.78 ton/h