1 – Using 20c/kWh calculate:
a – Ten 100W bulbs used 6 hours per day.
Daily consumption = 100W x 6h = 600Wh
Monthly consumption = 600Wh x 30 days = 18kWh
Monthly Cost = 18kWh x 20c/kWh = $3.60
b – Colour TV set for 5hrs per day.
Colour TV uses approximately 100W. Therefore,
100W x 5h = 500Wh /day
Per month = 500Wh x 30 = 15kWh
15kWh x 20 cents = $3.00
c – Night light of 1W left on continously.
1W x 24hrs x 30 days = 720Wh
720Wh = 0.72kWh
0.72 x 0.2 = 14.4 cents
2 – Benefits of Energy efficient lighting
Energy efficient lighting is a significant use of energy in the industrial sector. Industrial use of lighting may account for 5-10% of energy costs, as for commercial space lighting can be up to 40% of total energy consumption. It is therefore paramount to have an efficient and effect lighting system.
The installation of energy efficient lighting has multiple benefits a beyond the immediate saving provided by using a lower wattage light. The lower consumption of power by newer lighting systems reducing the air conditioning load required to cool the space. This cost can be significant. The use of improved lighting and new technology means that morale and mood can be affected from total light, and colour temperature.
The facility Manager want to replace 34 W F40T12 lamps with 10 W LED lamps in 200 4-lamp fixtures which are operated for 3128 hours /year in an air conditioned facility. Assume F40T12 34-watt costs $2 each and lasts 20,000 h, while 10 W LED costs $12 each and lasts about 30000 h. Electric energy cost is $ 0.3 per kWh. The demand charge is $6 per KVA/month. Air-conditioner runs one third of the lamp annual operating time. The EER (Energy Efficiency ratio) of the air conditioner is 3. EER in this context is the ratio of cooling output (kW) to the electric power input to the air conditioner (kW).
a ) the annual cooling energy saving on the air-conditioner load (in kWh)
b) the overall annual electricity energy saving, which can be achieved (in kWh)
c) the payback period of this energy saving opportunity investment (10 marks)
All the light energy of both lamps will be converted into heat energy and instantly added to the cooling load.
Power factor of the all equipment is one
AC runs for 12 months for purpose of cooling.
The lights operate for 12 months but do not operate continuously for 24hrs/Day.
The overhead costs, such labour cost for replacement is not included. Assume any replacement cost per each fixture.
The BF (Ballast factor) for both Florescent and LED bulbs equal 1.25.
200, 4 lamps fixtures, 800 Lamps total used 3128 h/pa,
Existing Lamp – 34W, 20,000h, $2
LED Lamp – 10W, 30,000h, $12
Electricity – $0.3/kWh, $6/kVA/Month
AC time = 1/3 Lighting tIme = 1042.6 h/pa
a) Find the annual cooling saving.
Ballast Factors are identical and therefore ignored in this case
First find the net energy saving.
Current System Consumption per year = 800 x 0.034 x 3128 = 85,081kWh/pa
New System Consumption per year = 800 x 0.01 x 3128 = 25,024kWh
The AC system uses 1kWh for every 3kWh of cooling provided, and runs one third of the time of the lighting
Therefore to the saving in annual cooling is (85081 – 25024) / 3 / 3 = 6673 kWh/pa
At 30 cents a kWh this represents $2001.90 of saving
b) Find the overall energy saving.
This is found by summing the lighting saving as well as the air conditioning saving.
Total Annual energy saving = (85081 – 25024) + 6673 = 60057 + 6673 = 66730 kWh
At 30 cents a kWh this represents a saving of $20,019
c) Find the payback period
Cost of implementation = $12 * 800 = 9,600
Energy Savings = demand reduction costs and energy saving costs
Demand reduction costs = 800 * (0.034 – 10) * 6 = $115.2 per month, is $1382.40 per year
Consumption savings = 113,441 x $0.3 = $34,032.30 p.a.
Total savings = $35,414.70
The simple payback period is = 9,600 / 35414.70 = 0.27107 years or 3.25 months.
4. What is coeeneration and how can energy recovery be used to improve efficiency.
Traditional energy generation is inefficient due to the high amount of thermal losses suffered to the system by heat rejection (think radiator on a car, cooling lakes for power plants). Cogeneration and trigeneration captures that ‘waste’ heat and uses it for heating water or another fluid, to provide heating for hot water systems or air conditioning. In a trigeneraiton system an absorption chiller is also used (which acts like a fridge in reverse) that allows coolth to be produced as well as heat and electricity. The absorption chiller produces cold water that can be used for air conditioning.
Due to the usual waste heat generated from turbines, cogeneration is able to increase the efficiently of a system significantly. A usual 40% efficient system can be improved to 80 or 90+% efficiency by use of capturing waste heat. Energy recovery allows for (typically) thermal energy to be captured and used in any system. The captured energy does not need to be for hot water, it may provide thermal heat to improve heat pumps or work in a chiller cycle, providing air conditioning for a building.
Other methods of energy recovery include the use of thermal insulation, which protects a surface from emitting or receiving heat energy, or material matrix energy capture systems whereby a ‘core’ is heating and cooled with a fluid such as air or water.
5. What are heat pumps? What makes them an efficient method of air conditioning? Should they be used in humid climates such as Singapore?
A heat pump works a bit like a fridge in reverse except instead of capturing coolth and rejecting heat, the reverse is used. By using a refrigerant and a compressor, a closed refrigeration system can be used whereby a gas is compressed and heated up. The heated gas goes through a heat exchanger to pass the heat to the medium such as hot water. Once passing through this the gas is allowed to expand, cooling it further, to a much colder temperature than the ambient air. The refrigerant is then able to ‘capture’ heat from the ambient air to warm up, allowing the system to recharge and be able to go through the cycle again.
By use this way the system can be used to heat or cool something with additional ‘free’ energy being provided by ambient temperature. This can allow heat pumps to have very high efficiencies. For example a hot water system can get 4kWh of energy for every 1kW of electrical power.
Heat pumps are particularly useful in humid climates for two reasons. One is that it is typically warm, therefore providing higher efficiency for the pumps, as well a heat pump cycle can act as a dehumidifier, providing additional ‘coolth’ if required.