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

 

1. 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:

v_m = q_m / A_m

v_m = air velocity

q_m = air flow

A_m = 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.