Managing Energy Efficiency
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Managing Energy Efficiency
Question 1: PC VS LAPTOP
Total expected yearly energy usage (kWh), energy cost ($):
PC Energy Consumption (kWh)
Maximum:
Annual hours: (350 X 4) hours = 1400 hours per year
Wattage = 195.04
Maximum energy consumption for PC per day = Wattage X hours/100
= (195.04 X 4) ÷ 1000 = 0.7802 kWh
Annual maximum consumption = 0.7802 X 1400 = 1092.28 kWh
Low-power sleep mode:
Annual Hours: (350 X 7 hours) = 2450 hours per year
Wattage: 1.75
Energy consumption per day: (1.75 X 7) ÷1000 = 0.01225
Annual consumption: 0.01225 X 2450 = 30.0125
Off mode:
Annual Hours: (350 X 13 hours) = 4550
Wattage: 1.03
Daily Energy Consumption: (1.03 X 13) ÷ 1000 = 0.01339 kWh
Annual energy consumption: 0.01339 X 4550 = 60.9245 kWh
Total Energy consumption:
Total Daily consumption: 0.7802 + 0.01225 + 0.01339 = 0.80584 kWh
Total Annual consumption: 1092.28 + 30.0125 + 60.9245 = 1183.217 kWh
Total for 275 PCs = 325384.675 kWh
Laptop Energy Consumption (kWh)
Maximum:
Annual Hours: (350 X 4) hours = 1400 hours per year
Wattage: 53
Daily Energy Consumption: (1.1 X 4) ÷ 1000 = 0.0044 kWh
Annual consumption: 0.0044 X 1400 = 6.16 kWh
Low-Power sleep mode:
Annual hours: (350 X 7 hours) = 2450 hours per year
Wattage: 1.1
Daily Energy consumption: (1.1 X 7) ÷ 1000 = 0.0077 kWh
Annual consumption: (2450 X 0.0077) = 18.865 kWh
Off mode:
Annual hours: (350 X 13 hours) = 4550
Wattage: 1.1
Daily Energy consumption: (1.1 x 13) ÷1000 = 0.0143 kWh
Annual consumption: 0.0143 x 4550 = 65.065 kWh
Total daily laptop energy consumption: 0.0044 + 0.0077 + 0.0143 = 0.0264 kWh
Total Annual: 6.16 + 18.865 + 65.065 = 90.09 kWh
Total for 275 laptops: 24774.75 kWh
Energy Costs of the PC ($)
Peak hours Energy Costs Calculation:
Stand by 2pm-7pm = 5 hours
Annual stand by and peak hours = (5 X 250 weekdays) = 1250
Wattage: 1.75
Daily consumption: (5 x 1.75) ÷ 1000 = 0.00875 kWh
Total annual consumption: 0.00875 x 1250 = 10.9375 kWh
Off 7pm-8pm = 1 hour
Annual off and peak hours = (1 x 250) = 250
Wattage: 1.03
Daily consumption: (1 x 1.03) ÷ 1000 = 0.00103 kWh
Total annual kWh: 0.00103 x 250 = 0.2575 kWh
Total annual peak energy costs = (0.2575 + 10.9375) x $ 0.25 = $ 2.79875
Shoulder hours Energy Costs
Off 7am-8am = 1 hour
Annual shoulder and off hours: (250x 1) = 250
Wattage: 1.03
Daily Consumption: 0.00103 kWh
Total annual: 0.2575 kWh
Stand by 8am-10am = 2 hours
Annual stand by and shoulder hours: (2 x 250) = 500
Wattage: 1.75
Daily consumption: (1.75 x 2) ÷ 1000 = 0.0035 kWh
Total annual consumption = 0.0035 x 500 = 1.75 kWh
Maximum 10am-2pm = 4 hours
Annual shoulder and maximum hours: (4 x 250) = 1000
Wattage: 195.04
Daily Consumption: (195.04 x 4) ÷ 1000 = 0.78016 kWh
Total annual consumption: 0.78016 x 1000 = 780.16 kWh
Weekends
Maximum 10am-10pm = 12 hours
Annual maximum and weekend hours: (12 x 100) = 1200
Wattage: 195.04
Daily consumption: (195.04 x 12) ÷ 1000 = 2.34048 kWh
Annual consumption: 2.34048 x 1200 = 2808.576 kWh
Off 7am-8am = 1 hour
Annual hours: (1 x 100) = 100
Wattage: 1.03
Daily consumption: (1.03 x 1) ÷ 1000 = 0.00103 kWh
Annual Consumption: 0.00103 x 100 = 0.103 kWh
Standby 8am-10am = 2hours
Annual hours: (2 x 100) = 200
Wattage: 1.75
Daily consumption: (1.75 x 2) ÷ 1000 = 0.0035 kWh
Annual consumption: 0.0035 x 200 = 0.7 kWh
Total shoulder hours cost: (0.7 + 0.103 + 2808.576 + 780.16 + 1.75 + 0.2575) x 0.19 = $ 682.4
Off-Peak hours Energy Costs
Weekdays
Off 8pm-7am = 11 hours
Annual hours = (11 x 250) = 2750
Wattage: 1.03
Daily consumption: (1.03 x 11) ÷ 1000 = 0.01133 kWh
Annual consumption: 0.01133 x 2750 = 31.1575kWh
Weekends 10pm-7am 9hrs
Annual hours: (9 x 100) = 900
Wattage: 1.03
Daily consumption: (1.03 x 9) ÷ 1000 = 0.00927 kWh
Annual consumption: 0.00927 x 900 = 8.343 kWh
Total off peak costs: (31.1575 + 8.343) x 0.11 = $ 4.345055
Total annual energy costs for one PC = (4.345 + 682.4 + 2.799) = $ 689.544
Energy costs for 275 PCs = $ 189624.6
Energy Costs of the Laptop ($)
Peak hours Energy Costs Calculation:
Stand by 2pm-7pm = 5 hours
Annual stand by and peak hours = (5 X 250 weekdays) = 1250
Wattage: 1.1
Daily consumption: (5 x 1.1) ÷ 1000 = 0.0055 kWh
Total annual consumption: 0.0055 x 1250 = 6.875 kWh
Off 7pm-8pm = 1 hour
Annual off and peak hours = (1 x 250) = 250
Wattage: 1.1
Daily consumption: (1 x 1.1) ÷ 1000 = 0.0011 kWh
Total annual kWh: 0.0011 x 250 = 0.275 kWh
Total annual peak energy costs = (0.275 + 6.875) x $ 0.25 = $ 1.788
Shoulder hours Energy Costs
Off 7am-8am = 1 hour
Annual shoulder and off hours: (250x 1) = 250
Wattage: 1.1
Daily Consumption: 0.0011 kWh
Total annual: 0.275 kWh
Stand by 8am-10am = 2 hours
Annual stand by and shoulder hours: (2 x 250) = 500
Wattage: 1.1
Daily consumption: (1.1 x 2) ÷ 1000 = 0.0022 kWh
Total annual consumption = 0.0022 x 500 = 11 kWh
Maximum 10am-2pm = 4 hours
Annual shoulder and maximum hours: (4 x 250) = 1000
Wattage: 53
Daily Consumption: (53 x 4) ÷ 1000 = 0.212 kWh
Total annual consumption: 0.212 x 1000 = 212 kWh
Weekends
Maximum 10am-10pm = 12 hours
Annual maximum and weekend hours: (12 x 100) = 1200
Wattage: 53
Daily consumption: (53 x 12) ÷ 1000 = 0.636 kWh
Annual consumption: 0.636 x 1200 = 763.2 kWh
Off 7am-8am = 1 hour
Annual hours: (1 x 100) = 100
Wattage: 1.1
Daily consumption: (1.1 x 1) ÷ 1000 = 0.0011 kWh
Annual Consumption: 0.0011 x 100 = 0.11 kWh
Standby 8am-10am = 2hours
Annual hours: (2 x 100) = 200
Wattage: 1.1
Daily consumption: (1.1 x 2) ÷ 1000 = 0.0022 kWh
Annual consumption: 0.0022 x 200 = 0.44 kWh
Total costs: (212 + 11 + 0.275 + 763.2 + 0.11 + 0.44) x 0.19 = $ 187.53475
Off-Peak hours Energy Costs
Weekdays
Off 8pm-7am = 11 hours
Annual hours = (11 x 250) = 2750
Wattage: 1.1
Daily consumption: (1.1 x 11) ÷ 1000 = 0.0121 kWh
Annual consumption: 0.0121 x 2750 = 33.275 kWh
Weekends 10pm-7am 9hrs
Annual hours: (9 x 100) = 900
Wattage: 1.1
Daily consumption: (1.1 x 9) ÷ 1000 = 0.0099 kWh
Annual consumption: 0.0099 x 900 = 8.91 kWh
Total off peak costs: (33.275 + 8.91) x 0.11 = $ 4.64035
Total annual energy costs for one laptop = (4.64 + 187.535 + 1.788) = $ 193.963
Energy costs for 275 laptops = $ 53339.825
- NPV for Option 1:
Annual Energy costs
Laptop: 53339.825
PC: 189624.6
NPV of Option 1:
R = 189624.6
n = 5
i = 7
PV = 189624.6 x (1- (1.07%) ^ (-5))/ 7% = $ 831, 923.1796
NPV= 831,923.179 – 412500 = 419423.179
NPV for Laptops:
NPV = 53339.825 x (1- (1.07%) ^ (-5))/ 7% = $ 234,013. 081
234013.081 – 440000 = (- 205986.919)
C
The best financial option of the two is the laptop option because it registers a higher return in future than the laptop option. The negative NPV shows a higher future return on investment for the laptops than Option 1.
The best option energy wise is the laptop because the amount of energy consumed by the laptops is lower than that consumed by option 1 and thus less costly.
E
A rise in electricity costs by 10% mean that the cost paid by the institution for electricity will also rise by 10%:
Laptop: 53339.825
PC: 189624.6
The above original costs will rise by 10% i.e.:
Laptops energy costs: 58673.805
PCs energy costs: 208, 587.06
PV for the laptops: 58673.805 x (1- (1.07%) ^ (-5))/ 7% = 257414.378
NPV: 257414.378 – 412500 = -155085.622
PV for the PCs: 208587.06 x (1- (1.07%) ^ (-5))/ 7% = 915115.497
NPV: 915115.497 – 440000 = 475115.497
F
If electricity costs rise by 10% in the next five years, the laptops will register a positive NPV while the PCs will register a negative NPV. This means that with an increase in energy costs the PCs will be causing losses to the company. A ten percent increase in energy costs will lead to negative returns from the PC option while the laptop option will register positive returns due to its energy saving quality. It is therefore prudent and cost saving to choose the laptop option over the PC option.
QUESTION 2
Pressure drops:
Duct 1 3 Pa/m x 0.36 = 1.08
Duct 2 1 Pa/m x 5.625 = 0.5625
Duct 3 0.25 Pa/m x 1 = 0.25
Calculating velocity:
Volume of ducts:
1: 0.006 x 0.006 = 0.000036 m2
2: 0.0075 x 0.0075 = 0.00005625 m2
3: 0.01 x 0.01 = 0.0001m2
Duct velocity:
vm = qm / Am
vm = air velocity
qm = air flow
Am = area of duct
A
Duct | Average velocity (m/s) | Pressure for frictional losses | Pressure to supply air | Total pressure | Suitability in kW |
1 | 39.33 | 1.08 | 1092500 | 1,011574 | 5.9 |
2 | 8.39 | 0.5625 | 148977 | 264849 | 3.4 |
3 | 1.18 | 0.25 | 11800 | 47200 | 2.7 |
B
Fan and duct combination | (Electricity costs) Weekday ($) | Weekend day ($) | Week ($) | A year ($) | Ten years ($) |
0.75/ duct 1 | 8.5 | 10.089 | 62.7 | 3009.6 | 30096 |
0.75/duct 2 | 5.6 | 4.845 | 37.7 | 1583 | 15830 |
0.75/duct 3 | 5.2 | 3.8475 | 33.6 | 1411.2 | 14112 |
1.1/duct 1 | 16.7442 | 12.331 | 108.162 | 4542.8 | 45428 |
1.1/duct 2 | 9.6492 | 7.106 | 62.5 | 2623 | 26232 |
1.1/duct 3 | 7.7 | 5.643 | 49.8 | 2091 | 20910 |
C
QUESTION 3: COGENERATION
A.
Peak = 2pm-8pm
Shoulder = 7am -2pm and 8pm-10pm
Weekends 7am-10pm
Off peak = 10pm- 7am weekdays and weekends
Peak costs for the aquatic pool:
1961.47 x 0.23 = 451.14
Shoulder 1635 x 0.17 = 277.95
1426 x 0.17 = 242.42
1631 x 0.17 = 277.27
Off peak 738 x 0.09 = 66.42
Total daily energy costs for the aquatic plant = $ 1315.2
B.
Graph of energy consumption against time.
C.
Cogeneration unit maintenance costs:
The electricity maintenance costs for the unit are charged at 0.03 kWh. These charges can be spread across the charging system to get the overall cost.
Peak costs= 0.03kWh x 6hrs x $ 0.23 = $ 0.0414
Shoulder; 0.03kWh x 9hrs x $ 0.17 = $ 0.0459
Off peak: 0.03 x 10 x 0.09 = 0.027
Weekends 0.03 x 14 x 0.17 = 0.0714
0.03 x 10 x 0.09 = 0.027
Daily maintenance costs (Weekdays):
0.0414 + 0.0459 + 0.027 = $ 0.9
Total Daily electricity maintenance (Weekends):
0.027 + 0.0714 = $ 0.0984
Cost per week:
(0.9 x 5) + (0.0984 x 2) = 4.5 + 0.1968 = $ 4.6968
Cost per year:
4.6968 x 52 = $ 244.2336
D.
Total gas consumption for the year:
The gas consumption for the year is a sum of the total monthly gas consumption levels. Therefore the gas consumption for the year = 60157 Gj
The gas charge per GJ is $ 11. Hence, cost of gas per year= 60157 x 11.00 = $ 661727
E.
Cogeneration unit 1:
1200 – (606+106+591) = (-103)
Cogeneration unit 2:
1600 – (790+134+796) = (-120)
Cogeneration unit 3:
2000- (978+178-1012) = (-168)
F.
Each of the cogeneration units listed demands a high amount of electricity to run. The cogeneration units should run during times of the day when the demand for electricity and heat by the plant is higher.
Cogeneration unit 3: This unit produces more electricity but also uses more it should therefore run between 8pm and 10pm. During this time, the energy requirements of the plant are moderate and can sustain the unit and there is enough demand for heat to warrant its operation. Cogeneration unit 2 should run when there is a higher electric consumption and demand by the plant. In this case, the times between 12pm and 4pm would be ideal. Cogeneration unit 1 has a lower production of electricity and is suitable for the time between 11pm and 4am.
G.
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