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2011 Part 1 Answer Key

1 C 21 B 41 D 61 C

2 B 22 A 42 C 62 E

3 C 23 C 43 E 63 D

4 A 24 C 44 C 64 B

5 A 25 B 45 D 65 C

6 C 26 A 46 A 66 C

7 B 27 B 47 C 67 A

8 D 28 C 48 C 68 C

9 A 29 E 49 E 69 C

10 D 30 E 50 A 70 D

11 D 31 E 51 A 71 C

12 C 32 D 52 C 72 A

13 A 33 A 53 C 73 B

14 B 34 C 54 B 74 B

15 E 35 B 55 C 75 C

16 C 36 D 56 C 76 E

17 B 37 B 57 A 77 B

18 E 38 C 58 B 78 B

19 A 39 A 59 A 79 A

20 B 40 E 60 E 80 E

2011 JETS TEAMS Competition

Part 1 Answer Key

with Explanatory Solutions

2011 JETS TEAMS Competition www.jets.org 2

Part 1 Table of Contents

Note: The solutions provided apply to both the 9/10 and 11/12 levels. Part 1 Answer Key……………………………………………………………………………….……………………3

Scenario #1

Questions 1-10…………………………………………………………………………………………………….4-6 Scenario #2

Questions 11-20…………………………………………………………………………………………….…..7-8 Scenario #3

Questions 21-30…………………………………………………………………………………………..……9-11 Scenario #4

Questions 31-40………………………………………………………………………………………………12-15

Scenario #5 Questions 41-50……………………………………………………………………………………….…….16-19

Scenario #6

Questions 51-60………………………………………………………………………………………….….20-21

Scenario #7 Questions 61-70………………………………………………………………………………………….….22-25

Scenario #8 Questions 71-80…………………………………………………………………………………………..…26-28


2011 Part 1 Answer Key

1 C 21 B 41 D 61 C

2 B 22 A 42 C 62 E

3 C 23 C 43 E 63 D

4 A 24 C 44 C 64 B

5 A 25 B 45 D 65 C

6 C 26 A 46 A 66 C

7 B 27 B 47 C 67 A

8 D 28 C 48 C 68 C

9 A 29 E 49 E 69 C

10 D 30 E 50 A 70 D

11 D 31 E 51 A 71 C

12 C 32 D 52 C 72 A

13 A 33 A 53 C 73 B

14 B 34 C 54 B 74 B

15 E 35 B 55 C 75 C

16 C 36 D 56 C 76 E

17 B 37 B 57 A 77 B

18 E 38 C 58 B 78 B

19 A 39 A 59 A 79 A

20 B 40 E 60 E 80 E


Scenario #1 Deep Water Oil Rig Safety

1. Answer C

Failure Rate = = 0.00333

(0.0033)(100%) = 0.33%

2. Answer B

Reliability = 0.003333

Required Reliability = = 0.000000001

Adapters Needed = (0.003333)4 = 1.234 x 10‐10

Four Annular Preventers in stack are needed to meet our requirement.

3. Answer C

(22.8 min/day) (1 day/24 hr) (1 hr/60 min) = 0.01583

(0.01583) (100%) = 1.58% = Failure Rate

4. Answer A

Required Reliability = 10‐9


Five Ram Preventers in stack are needed to meet our requirement.


5. Answer A


Control Systems Needed = (0.01042)5 = 1.226x10‐10

Therefore, five control systems are needed.

6. Answer B



We need four monitoring systems to meet reliability standards. Using majority voting

(meaning more than half) we need at least 7 systems to have over half of them give a

positive or negative failure reading.

7. Answer B

Surface area of deck: (357 ft) 256 ft) = 91,392 ft2

To have one gallon per minute for ten minutes over the entire surface area, we need:

= (10 min) [1 gal/(min)(ft2)] (91,392 ft2) = 913,920 gallons

8. Answer D

Volume = (π) (R2) (H) = (3.14) (30 ft)2 (H)

Volume = 1,000,000 gal

1 ft3 = 7.48 gal of water

H = = 47.28 ft


9. Answer A

Surface Area = (π)(R2) = (3.14) (15 in)2 = 706.9 in2

Total pressure across the surface: (709.9 in2) (13,000 lb/in2) = 9.2 x 106 lbs

The amount of mud needed: = 1,021,000 gallons of mud.

Total Volume: = 136,499 ft3

10. Answer D

Volume of Sphere = (4/3) (π) (R3)

= (4/3) (3.14) (30 ft)3

= 113,097 ft3

Total Amount of Mud = (113,097 ft3)

= 845,968 gallons

Total Weight of Mud = (845,968 gallons)

= 7,613,712 lbs of Mud

Pounds per square inch (psi) =

= 10,770 lbs/in2

= 10,770 psi


Scenario #2 Wind Power

11. Answer D

Tip Speed of Blade = (8.4 mph) (4.5) = 37.8 mph

12. Answer C

Tip Speed of Blade = (3m) = 7.85 m/s

(7.85 m/s) = 17.56 mph

2.1

2.1 ≠ 4.5 The answer is no, you are not getting maximum power.

13. Answer A

(1 kW) (8,760 hr) ($0.10/kWh) = $876

14. Answer B

(9 years) (5,600 MWh/yr) = 50,400 MWh

(50,400 MWh) ($0.04/kWh) = $2,016,000

$2,016,000 ‐ $1,300,000 = $716,000

15. Answer E

Area = (π) (30m)2 = 2,826 m2

Power = (0.5) (1.225 kg/m3) (2,826 m2) (15 m/s)3 = 5,841,871 W = 5.845 MW


16. Answer C

Produced Electricity = (0.33) (5.845 MW) = 1.93 MW

17. Answer B

(15 m/s) (2.237 mph/mps) = 33.55 mph

Graph Estimate:

0.18

0.303 = 30%

18. Answer E

(291 Turbines) (1.5 MW) + (130 Turbines) (2.3 MW) =

436.5 MW + 299 MW = 735.5 MW

(735.5 MW) 220,650 homes

19. Answer A

= 114.75 MW/million people

20. Answer B

20

4

(20) (4) = 80

(20 rmp) (80) = 1,600 rpm


Scenario #3

Natural Gas New Discoveries

21. Answer B

A = (0.5) (530 Km) (480 Km) = 1.27 x 1011 m2

V2 = (1.4 x 1013 m3) = 0.01047 x 1013 m3 = 1.047 x 1011 m3

Th = = 0.82

22. Answer A

The annual production from a typical well is:

(100,000 m3/day) (365 days/yr) = 3.65 x 107 m3/yr

Dividing that into 10% of the annual usage yields the number of wells required:

(0.1)(645 x 109 m3/yr) / (3.65 x 107 m3/yr) = 1,767 wells

23. Answer B

V of cylinder = (π) (R2) (h)

(π) (π) (150m) = (π) (150 m)

= 10.59 m3

24. Answer C

ρ = 7,850 Kg/m3

V = (π) (1,500 m) = (π) [5.625 x 10‐3 – 4.225 x 10‐3] m2 (1,500 m)

= 6.594 m3

M = (V) (ρ) = (6.594 m3) (7,850 kg/m3) = 51,762 kg


25. Answer B

15 MPa = (1,200 kg/m3) (9.81 m/s2) (H)

H = = 1,274.21 m

26. Answer A

Cross Sectional Area = (π) = (π) = 0.09569 m2

Q = (0.8 m/s) (0.096 m2) = 0.076 m3/s

27. Answer B

= 450 movements

L = (450) (0.5 m/move) = 225m

L = (0.25) (2π) (H)

H = = 143m

This is the additional vertical length for a 90 degree turn.

28. Answer C

P =

= = 11.65 MPa


29. Answer E

Number of seconds in 30 hours = (30) (60) (60) = 108,000s

Number of Sand Trailers = = 162 Trailers

30. Answer E

Volume of particle = (4/3) (π) (R)3

= (4/3) (π) (D/2)3

= (1/6) (π) D3

= (1/6) (3.14) (1.2 x 10‐3m)3

= 0.904 x 10‐9m3

Number of sand particles:

N = = 93,922 sand particles.


Scenario #4

Passive Building Design

31. Answer E

North: (15) (0.4) (4,500 ft2) = 27,000 Btu/hr

South: (22) (0.4) (4,500 ft2) = 39,600 Btu/hr

East: (27) (0.4) (450 ft2) = 4,860 Btu/hr

West: (30) (0.4) (450 ft2) = 5,400 Btu/hr

Total Heat Gain = 76,860 Btu/hr

Where:

(30%) (1,000 ft) (15 ft) = 4,500 ft2

(30%) (100 ft) (15 ft) = 450 ft2

32. Answer D

North: (15) (0.4) (450 ft2) = 2,700 Btu/hr

South: (22) (0.4) (450 ft2) = 3,960 Btu/hr

East: (27) (0.4) (4,500 ft2) = 48,600 Btu/hr

West: (30) (0.4) (4,500 ft2) = 54,000 Btu/hr

Total Heat Gain = 109,260 Btu/hr


33. Answer A

Original Heat Gain:

Old East Windows: (4,500 ft2) (0.40) (27) = 48,600 Btu/hr

Old East Wall: (10,500) (0.05) (27) = 14,175 Btu/hr

Total: 62,775 Btu/hr

Heat Gain after replacing window with wall:

New East Window: (3,000 ft2) (0.40) (27) = 32,400 Btu/hr

New East Wall: (12,000 ft2) (0.05) (27) = 16,200 Btu/hr

Total: 48,600 Btu/hr

Savings: 62,775 Btu/hr – 48,600 Btu/hr = 14,175 Btu/hr

34. Answer C

Poor Construction: 1 ACH

Good Construction: 0.25 ACH

Savings of 0.75 ACH

For 800 hours and 225,000 ft of Air.

(1.2 Btu) (225,000 ft3) (800 hr) (0.75 ACH) = 162,000,000 = 162 MBtu

35. Answer B

162 MBtu = 1,620 Therms

1,620 Therms x $1.00/Therm = $1,620


36. Answer D

(19 kWh/ft2) (4.56 x 106 ft2) = 86.64 x 106 kWh

= 247.54 x 106 kWh Need to be generated.

(247.54 x 106 kWh) = 845.358 x 109 Btu

Amount of coal:

(845.358 x 109 Btu) = 78 x 106 lb of Coal

One Car’s Capacity =

(80 tons) (2,000 lb/ton) = 160,000 lb

Given 50 cars/train:

9.75 trains ~ 10 trains

37. Answer B

Total Electricity needs per year: 86.64 x 106 kWh

Energy Needs for Illumination: (25%) (86.64 x 106 kWh) = 21.66 x 106 kWh

= 28.67%

So we can reduce by 25% of that:

(28.67%) (25%) = 7.16%

Overall Reduction:

(7.16%) (9.75 trains) = 0.698 trains

(0.698 trains) (50 cars/train) = 35 cars.


38. Answer B

The electrical demand is = 9,735 homes

39. Answer A

A large facility can take the waste heat generated from a steam boiler and run it through a

waste heat recovery device and a steam turbine to generate electricity on site.

40. Answer E

Savings:

= 421.3 kWh/yr

(3.8 x 106 households) (421 kWh per year) = 1,600 x 106 kWh/yr

= 179,775 homes


Scenario #5

Solar Cell Devices and Optoelectronics

41. Answer D

Figure shows that the peak Solar Energy per second per unit area is around 475 nm

1 nm = 10‐9 m

Emax = = 2.61 eV or ZnSe

42. Answer C

(55 m2) (23%) = 2,530 W

43. Answer E

Part‐A, for 11/12 Level only:

VOC = ln

= ln = 0.4834V

Part‐B, for both 9/10 and 11/12 Levels:

FF = = 0.82


44. Answer C

(10%)

= 111.5 kW/year

(111.5 kW/year) = 5,576 m2

(5,576 m2) = $557,500

(977 x 103 kWh) ($0.12/kWh) = $117,240/yr

= 4.75 ~ 5 years

45. Answer D

Power needed by loads: (5) (120V) (0.02A) = 12W

Power needed to change the battery: (4 hours)

= 50W

Total Power Needs = 62W

Efficiency of Solar Cell:

= 15.5%


46. Answer A

Population Monthly Energy Power

10% 1.8 MWh 2,419W

20% 0.8 MWh 1,075W

30% 1.6 MWh 2,150W

40% 1.2 MWh 1,612W

Assumes 31 days per month, 24 hours per day

Average Power = (2,419W) (0.10) + (2,150) (0.30) + (1,612) (0.4) + (1,075) (0.2) =

1,746.7W

On the average we need: 1,746.7 W

We receive: (34%) =

= 250 households

= 25.68 m2

(25.68 m2) (250 households) = 6,420 m2 = 80m x 80m

47. Answer C

Green: 495 – 570 nm

At 495nm we need 2.5eV

At 570nm we need 2.17eV

GaP is at 2.26eV which corresponds to:

λ = = 549 nm which is in the range for green.


48. Answer C

= (2π) (0.2 cd) (1 – cos 35o)

= (6.28) (0.2) (0.18) = 0.226 Lumens = 89 LEDs

49. Answer E

P = (V) (I)

= (1.7V) (20mA)

= 34 mW

PTOTAL = (0.034W) (89 LEDS) = 3.026W

50. Answer A

1 stop light = 3,026mW

8 stop lights = 24,208mW

(0.20)

(24.2W) = 0.605 m2


Scenario #6

Oil Algae Production Engineering

51. Answer A

Acres of Corn = = 9.3 x 109 acres

52. Answer C

Acres of oil algae = = 9.3 x 106 acres

53. Answer C

Total MWh = (10 MW) = 87,600 MWh/year

54. Answer B

A screen = (60cm) (120cm) = 7,200 cm2

A total = (1.25 x 106 m2) = 12.5 x 109 cm2

Number of Screens = = 1,736,111

55. Answer C

Total = (4.2 x 109 MWh) (0.482) = 2.0244 x 109 MWh

Number of Test Systems = = 23,110

56. Answer C

Total Number of Needed Scrubbing Systems:

3,500


57. Answer A

Total Land Needed =

(3,500 scrubbing systems) (2,000 acres/system) = 7 x 106 acres

58. Answer B

Carbon Dioxide per Acre = = 9,855 tons/acre

Total Number of Acres = = 182,648 acres

59. Answer A

Average Farm size = = 283 acres per farm

60. Answer E

Butanol per Acre =

= 3,300,000 gallons/acre

Total Butanol = (3,300,000 gallons/acre) (400,000 acres)

= 1,320 billion gallons


Scenario #7

Energy Efficiency at Home

61. Answer C

The most beneficial action would be to insulate your attic.

62. Answer E

(5 hrs/day) [(4) (60W) + (2) (13W)] = 1,330 Wh/day

(2 hrs/day) [(6) (60W)] = 720 Wh/day

(4 hrs/day) [(3) (60W) + (2) (13W)] = 824 Wh/day

(2 hrs/day) [(5) (60W)] = 600 Wh/day

(3 hrs/day) [(3) (60W)] = 540 Wh/day

Total = 4,014 Wh/day = 4kWh/day

(365 days/year) (4 kWh/day) = 1,465 kWh/y

Errata Note, Question 62 Edit: How much electric energy is used per year in a U.S. household

with the currently placed lighting?

63. Answer D

As is:

Energy Cost = (1,465 kWh/y) (5 yr) ($0.10/kWh) = $732.5

Bulb Replacement Cost = (10 bulbs will need replacement in 5 yrs) ($0.50) Or $5

Total Cost‐1 = $732.5 + $5 = $737.5

CFL ONLY

(13W) [(6) (5) + (6) (2) + (5) (4) + (5) (2) + (3) (3)] = (13W) (81 hr) = 1.053 kWh/day

Energy Cost =

(1.053 kWh/day) (365 days/yr) (5 yr) ($0.10/kWh) = $192.17

Bulb Cost = (21 bulbs) ($3.00/bulb) = $63


Total Cost‐2 = $192.17 + $63 = $255.17

Total Cost Savings = [Total Cost‐1] – [Total Cost‐2] = $737.5 ‐ $255.17 = $482.33

Errata Note, Question 63 Edit: What is the total cost savings over the next 5 years associated

with switching from the current lighting in the house to all CFLs?

64. Answer B

Already calculated = $737.50 = Total‐Cost‐1

Lab Replacement Cost = ($35) (25 bulbs) = $875

(7W) [(6) (5) + (6) (2) + (5) (4) + (5) (2) + (3) (3)] = 0.567 kWh/day

Energy Cost = (0.567 kWh/day) (365 days/yr) (5 yr) ($0.10/kWh) = $103.48

Total Cost‐3 = $875 + $103.48 = $978.48

Cost Savings: Total Cost‐1 –Total Cost‐3 = $737.50 ‐ $978.48 = ‐$240.48

65. Answer C

T2 = 120oF = 48.9 oC

T1 = 40oF = 4.4 oC

T2 – T1 = (48.9 – 4.4) oC = 44.5 oC

Q = (m) (C) (T2 – T1)

m = (100 gallons/day) = 378 L/day = (378,000 mL/day) (1g/mL) = 378 kg/day

C = 4.186 J/g oC

Heat Energy Needed = (378 kg/day) (4.186 J/g oC) (44.5 oC)

= 70.4 MJ/day

Per Year = (365) (70.4) = 25,700 MJ/d

= 2.57 x 104 MJ/d

Electricity Needed =

= 2.86 x 104 MJ/d


66. Answer C

Electricity Needed as Calculated before: 2.86 x 104 MJ/y

Coal Needed: = 1.14 x 105 MJ/y = 11.4 x 104 MJ/y

100 MJ Natural Gas yields 71 MJ energy at home.

Gas heater efficiency = 70%

Heat Energy needed to enter the water heater: = 3.67 x 104 MJ/y

Heat Energy needed to come from the ground: = 5.17 x 104 MJ/y

Savings: = 0.546 = 54.6%

67. Answer A

= = 6.33 Per‐inch R‐value

Polyurethane Rigid Panel with R‐6.8.

68. Answer C

Current Heat Loss = = 0.083

Future Heat Loss = = 0.02

Savings = % = 76%


69. Answer C

Energy as is = (15 hr) (100 W) + (20 hr) (200 W) + (168 hr – 15 hr) (4 W)

+ (168 hr – 20 hr) (10 W) =

= 1,500 Wh + 4,000 Wh + 612 Wh + 1,480 Wh

= 7,592 Wh = 7.592 kWh

Energy with Power Strip = 1,500 Wh + 4,000 Wh = 5.5 kWh

% Savings = (100%)

= (100%) = 27.5%

70. Answer D

PC 200 W

Laptop 25 W

= = 0.875 = 87.5%


Scenario #8

Sustainable Transportation Fuels

71. Answer C

(# Acres) (Yield) (Ethanol Yield) =

(86,520,000 acres) (151 bu/acre) (2.74 gal prod/bu corn) = 35,796,784,000 = 35.8 BGY

= 41.9%

72. Answer A

76,330 Btu/gal – 58,609 Btu/gal = 17,721 Btu/gal

73. Answer B

1 gal of Ethanol = 76,330 Btu/gal

1 gal of Gasoline = 124,000 Btu/gal

(124,000 Btu/gal) = 4,769 Btu/mile

(85%) (76,330 Btu/gal) + (15%) (124,000 Btu/gal) =

64,880 Btu/gal + 18,600 Btu/gal = 83,480 Btu/gal

(83,480 Btu/gal) = 17.5 MPG

74. Answer B

Cathode: O2 + 2H+ + 4e‐ 2 H2O

Anode: 2H2 4H+ + 4e‐


75. Answer C

= 428 cells in series

76. Answer E

Part‐A for 11/12 Level only:

C8 H18 + O2 8 CO2 + 9 H2O

1 mole of Gasoline C8 H18

Yields 8 moles of CO2

CO2 = 44 gr/mol

C8 H18 = 114 gr/mol

114gr of C8 H18 yields:

8 x 44gr = 352gr of CO2

Part‐B for both 9/10 and 11/12 Levels:

Using density of gasoline to convert grams into gallons.

Density of Gasoline = 0.8 g/mL

1 gal = 3,787 mL

1 gal of C8 H18 equals (0.8 g/mL) (3,787 mL) = 3,029.6 gr

114 gr of C8 H18 equals = 0.0376 gal

and produces 352 gr of CO2.

= 9,354.5 gr of CO2 / gallon of gasoline.


77. Answer B

Carbon dioxide is used by plants during photosynthesis

78. Answer B

2.5 x 108 passenger vehicles

12,000 miles per vehicle per year

[All Combustion Vehicles]

(250 x 106 vehicles) (12,000 miles per vehicle per year) = 1.36 x 1015 gr/yr

[60% Electric Vehicles]

(1.36) (40%) = 0.544

(0.544 x 1015 gr/yr) + (150 x 106) (12,000) (301 gr/mile) =

1.0858 x 1015 gr/yr

1.36 x 1015 gr/yr – 1.0858 x 1015 gr/yr = 0.274 x 1015 = 274 x 1012 gr CO2 saved each year

79. Answer A

Internal combustion automobiles with petroleum gasoline, but with much higher fuel

efficiency would be a good recommendation to reduce our Nation’s consumption of fossil

fuels.

80. Answer E

All recommendations would help contribute to the transformative goals.

2011 JETS TEAMS Competition

Part 2 Solutions

©2011 JETS TEAMS Competition www.jets.org 2

Part 2 Table of Contents

Note: The solutions provided apply to both the 9/10 and 11/12 levels. Scoring Criteria………………………………………………………………………………………………………...3 Tasks #1 and 2…..…………………………………………………………………………………………………….4 Tasks #3 and 4…..…………………………………………………………………………………………………….5 Tasks #5 and 6…..……………………………………………………………………………………………….6 - 7 Tasks #7 and 8…..……………………………………………………………………………………………….7 - 8


Part 2 Scoring Criteria

Part 2 of the TEAMS competition consists of eight open-ended tasks. All Part 2 answers are considered final as scored. Answers to Part 2 Tasks are reviewed and scored using three main criteria:

1. Written structure/organizational style 2. Relativity to the topic(s) presented in the Task 3. Inclusion of critical elements as established by the author

Criteria 1: Written Structure/Organizational Style

Answers present the information clearly and concisely without excessive embellishment.

Writing is legible and able to be clearly read. Ideas are distinguished into sections/paragraphs. Each paragraph starts with a topic sentence.

Criteria 2: Relativity to the Topic(s) Presented

Answers include only facts supported by scientific and/or mathematical principles. Purpose of answer is to convince the reader of the proposed solution/position. Solutions appear well thought out and logical. Answers containing political viewpoints, racially insensitive ideas, or other

inappropriate comments are not accepted.

Criteria 3: Inclusion of Critical Elements as Established by the Author

Answers should contain as many of the critical elements as possible. New ideas and alternate solutions may expand beyond the critical elements yet are

logical and conceivable. Benefits of proposal are listed as compared to other pre-existing methodologies. Solutions include various impacts such as global, economic, environmental, ethical

and societal concerns.


Task 1

Where could energy be lost while we supply a building with electricity that has been generated in a coal-fired power Plant? How could this wasted energy be used for other useful purposes? For example: Coal-fired power plants use giant boilers to generate steam in order to run steam turbines. These boilers, however, just like residential furnaces, cannot turn 100% of the heat from burning the coal into useful steam. We call this boiler inefficiency.

Answers:

Boiler inefficiencies. Steam losses. Coal transportation cost. Coal waste products. Turbine losses/generator losses. Transmission line losses (can be large). Waste heat can be used in a combined heat and power (CHP) system. Low pressure steam can be used in nearby factory process (if available). Losses in Electrical Transformers. Plant equipment running at minimum speed but higher than actual load (nighttime

hours in summer).

Task 2

Why is an effort to improve energy efficiency usually more cost-effective than adding renewable energy to a building? What are some examples of low-cost energy efficiency improvement strategies for K-12 schools? For example: Compare the proposal of adding solar panels to power all the lights of a building for the entire day versus turning off unnecessary lights.

Answers:

Reduced resource demand since a building is already constructed. Less labor-intensive (less energy spent in construction). Reduced impact on environment (no new land is used for a new building). Reduce energy use without using energy to manufacture new equipment (smarter

building operation). Renewable technologies often require rare earth elements (energy

intensive/expensive mining). Better insulation leads to reduced heating/cooling loads. Higher-quality glass. Reduced amount of glass (less solar heat gain, also reduces cost). Occupancy sensors for lighting. Daylight sensors for lighting. Caulk and seal around windows and doors. Apply solar reduction treatment to windows. Unoccupied Temperature Setback. Less stringent corridor/lobby set-points. Planting Deciduous Trees for Summer Shading.


Task 3 What benefits does using the earth’s stored energy (geothermal) have over burning coal or natural gas to supply energy to a building? What are some possible environmental risks related to the use of geothermal energy? For example: Discuss the impact of geothermal fields on ground water tables and ground temperatures.

Answers/Benefits:

Earth acts to heat or cool water during the winter/summer, replacing or reducing boilers or cooling towers (reduced CO2, fuel consumption, reduced electricity use).

Use earth energy without burning hydrocarbons Reduced greenhouse gas emissions with energy production Ground temperatures are relatively constant (~55F) Ground can act as a “battery”: Energy added from cooling processes can later be

extracted for heating purposes After infrastructure is in place, no ongoing “utility cost” for the energy used

Answers/Risks:

If an open geothermal system is used, potential for contamination of water table. Drilling through aquifers can threaten water supply. New holes in the earth can disrupt natural flow of underground water. New hole provides a direct path for contaminants from the surface. May be unintended consequences for removing energy from the earth. Drilling may distrupt other natural resources captured underground (natural gas or

methane deposits). Damaged equipment may be left underground and not retrievable.

Task 4

What are some possible uses for solar energy at a typical high school? Consider all the related areas including the site, the landscaping, and the inside of the building. What are some of the technologies that are currently used in a typical high school that is being built today? For example: Parking lot lights that use solar panels to charge batteries.

Answers:

Main school signs (PV). Site lighting (PV). PV to grid PV to batteries Domestic Hot Water Heating Snow Melt Pool Heating Day-lighting

Winter Heating Ventilation Air Preheat Create buoyancy for Natural

Ventilation Grow food Grow other plants Grow renewable fuels


Task 5

In what ways is an electric car better than a gasoline car? In what ways are they worse? Take into consideration that the energy that the electric car needs to charge its batteries could come from burning coal, using nuclear power or wind (renewable) power. For example: Batteries used in electric cars, are heavily resource-dependant and need additional resources and energy to produce, that are not prevalent in a gasoline car. However, gasoline is a limited resource that also needs other resources and energy to process, refine and transport.

Answers:

Reduced fossil fuel dependence (+) Reduced emissions (+) Better efficiencies (+) Can have much less environmental impact with renewable utility (+) Charge (use energy) during off-peak periods – utility plant benefits(+) Source of distributed power(+) Power power source(+) Sell power during peak demand periods (make money if demand-based utility rates)

(+) Additional materials and process required to construct (batteries) (-) Requires charging from a coal-dependant utility, limiting emission reduction

(gasoline cars are independent of grid) (-) Short battery life expectancy (-) High battery replace cost (-) Little infrastructure in place (in US) to support electric cars (-)

Task 6

Why are mechanical systems (like pumps) that use water as the primary method to transport energy preferred over air systems (like fans)? Water has a much higher (four times higher) specific heat than air. Explain why water can transfer the same amount of energy in a much smaller quantity. How can this impact a building beyond energy implications? For example: The picture below illustrates this point: the one-inch diameter pipe can transport the same amount of energy in water as an 18 inch x18 inch air duct.


Answers:

Water can transport more energy per cubic foot than air. Piping requires much less space than ductwork to transport the same amount of

energy. Water has higher heat capacity than air. Energy in water source is easier to store than air. Equipment can typically transfer energy to/from water more efficiently than air. Water is typically easier to filter than air. Smaller space (mechanical room) typically require to house water based equipment

versus air. Water can more easily interact with the ground energy source (geothermal) than air

systems.

Task 7

What are the advantages and disadvantages of using electric heat to heat a building instead of heating with natural-gas? For example: Electric heating is 100% efficient in terms of converting the incoming electricity into heat, however the electricity originated in a power plant and we must consider what losses happen there.

Answers/Advantages:

Electric heat can have a reduced environmental impact if on a renewable utility than NG.

Electric heat systems typically have lower first cost. Electric heat systems typically have smaller footprint. Used during heating season when electricity demands and utility costs are often at

the lowest of the year. Electricity is often more readily available than natural gas at some remote sites.

Answers/Disadvantages: Electric heat is less than 35% efficient when considering the initial source. Electric heat’s source power (coal) pollutes more than NG. NG burns fossil fuels directly, but can often be 85-90%+ efficient. Typically higher Greenhouse emissions due to source of heat. Operating costs to heat with electrically are typically much higher than natural gas.


Task 8

How can architectural features on a building (like walls, windows etc.) impact the energy use of the building? For example: Windows can reduce lighting-energy when combined with daylight sensors that turn down or shut off the lights. However, sunlight will heat up the space which will increase the cooling required for the building. That will force the air-conditioning equipment to get larger to handle the additional cooling requirements. Answers:

Overhangs will shade the building, reducing cooling load in summer, but can let sunlight in during winter (reducing heating load).

Larger/heavier walls can insulate the building better. Building mass can shift the peak cooling loads to off-peak periods. Window orientation (N/W/E/S) can impact heating/cooling loads due to solar heat

gain. Window-to-wall ratio. Operable windows – natural ventilation. Placement of windows affects day-lighting and heat gain to the individual spaces. Skylight size and location affects heat gain and day-lighting properties. Perimeter versus volume ratio. Having a large interior space can have positive or

negative effects depending on the use of the space. Floor to floor height directly impacts the volume of space that requires conditioning. Color of the exterior surfaces influences heat transfer. Location relative to other buildings can affect the energy demand (summer/winter

shading from other structures). Large interior lobby spaces along a southern exposure can store energy. Location of adjacent spaces in a building can have a positive or negative. impact on

the conditioning of surrounding spaces. (i.e. placing a pool room nearby a space with low humidity requirements).

Documents

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