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P1b(ii) Keeping Homes warm

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Presentation on theme: "P1b(ii) Keeping Homes warm"β€” Presentation transcript:

1 P1b(ii) Keeping Homes warm
P1b(ii) Keeping Homes warm You will learn about: How energy is lost from uninsulated homes About payback time About energy efficiency About Sankey diagrams

2 Saving Energy www.PhysicsGCSE.co.uk
The roof is insulated with fibreglass. Fibreglass contains air pockets which prevent heat escaping via conduction. As less energy escapes and more energy stays in the house it means it is energy efficient. This log cabin has a snowy roof. It tells us that it must have really good loft insulation. If it didn’t then the snow would have melted. The fireplace in this house is NOT efficient. Most of the heat conducts through the walls out of the room. Only the heat facing forward can get into the room. The fireplace in this house IS efficient. The heat radiates into the room in all directions. Less heat is wasted so is more efficient. For every 100J of energy released as heat only 25J are transferred to the room. For every 100J of energy released as heat 90J are transferred to the room. efficiency = 𝑒𝑠𝑒𝑓𝑒𝑙 π‘’π‘›π‘’π‘Ÿπ‘”π‘¦ π‘œπ‘’π‘‘π‘π‘’π‘‘ π‘‘π‘œπ‘‘π‘Žπ‘™ π‘’π‘›π‘’π‘Ÿπ‘”π‘¦ 𝑖𝑛𝑝𝑒𝑑 π‘₯ 100 efficiency = 25 𝐽 100𝐽 π‘₯ 100=25% 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑑 efficiency = 90 𝐽 100𝐽 π‘₯ 100=90% 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑑

3 Sankey Diagrams The light energy arrow is 9 times smaller than the heat energy arrow. This is because 10J is 9 times smaller than 90J. This means the Sankey diagram is to scale. Sankey diagrams are used to show energy transformations. An arrow to the right shows the USEFUL energy output. Remember: there may be more than one arrow so you would need to add up all of the values Inside the arrow shows the TOTAL INPUT energy An arrow pointing downwards shows the WASTE energy output. Remember: there may be more than one arrow so you would need to add up all of the values The values outside = the values inside. 100J = J J

4 Payback Time www.PhysicsGCSE.co.uk
There are many ways a home loses heat: Roof Walls Doors Floors Windows The house is the source of the energy The atmosphere is the sink. It can absorb lots of energy without changing its temperature by very much. The Payback Time equation will tell you which insulation method is the most cost-effective. π‘ƒπ‘Žπ‘¦π‘π‘Žπ‘π‘˜ π‘‡π‘–π‘šπ‘’ π‘¦π‘’π‘Žπ‘Ÿπ‘  = π‘π‘œπ‘ π‘‘ π‘œπ‘“ π‘–π‘›π‘ π‘’π‘™π‘Žπ‘‘π‘–π‘œπ‘› π‘Žπ‘›π‘›π‘’π‘Žπ‘™ π‘ π‘Žπ‘£π‘–π‘›π‘” Insulation Typical cost in Β£ Typical annual fuel saving in Β£ Cavity wall 250 125 Double glazing 5000 100 Loft 150 Double glazing π‘ƒπ‘Žπ‘¦π‘π‘Žπ‘π‘˜ π‘‡π‘–π‘šπ‘’ π‘¦π‘’π‘Žπ‘Ÿπ‘  = Β£5000 Β£100 =50 π‘¦π‘’π‘Žπ‘Ÿπ‘  This means that it will take 50 years to earn back the initial start up cost of Β£5000. After 50 years all savings are profit. Cavity wall insulation only takes 2 years to pay back the initial cost

5 Quick Questions A coal fire transfers 5000 J of energy stored in the coal into 3000 J of useful heat. Calculate the efficiency of the fire. Using the table on slide 4, calculate the payback time for loft insulation. Draw a Sankey diagram for a lightbulb that has 100 J of electrical energy that transfers into 75 J of useful light energy. A fire is 25% efficient. It burns 20kg of coal each day. The energy output into the room is 46.3 MJ. Calculate the energy in 1 kg of coal

6 Quick Questions A coal fire transfers 5000 J of energy stored in the coal into 3000 J of useful heat. Calculate the efficiency of the fire. Using the table on slide 4, calculate the payback time for loft insulation. Draw a Sankey diagram for a lightbulb that has 100 J of electrical energy that transfers into 75 J of useful light energy. A fire is 25% efficient. It burns 20kg of coal each day. The energy output into the room is 46.3 MJ. Calculate the energy in 1 kg of coal 20kg coal releases 46.3 MJ so 1kg coal releases 46.3/20 = MJ Turn 25% to 100% need to multiply by 4 So MJ x 4 = 9.26 MJ efficiency = 𝑒𝑠𝑒𝑓𝑒𝑙 π‘’π‘›π‘’π‘Ÿπ‘”π‘¦ π‘œπ‘’π‘‘π‘π‘’π‘‘ π‘‘π‘œπ‘‘π‘Žπ‘™ π‘’π‘›π‘’π‘Ÿπ‘”π‘¦ 𝑖𝑛𝑝𝑒𝑑 π‘₯ 100 efficiency = 𝐽 5000 𝐽 π‘₯ 100=60% π‘ƒπ‘Žπ‘¦π‘π‘Žπ‘π‘˜ π‘‡π‘–π‘šπ‘’ π‘¦π‘’π‘Žπ‘Ÿπ‘  = π‘π‘œπ‘ π‘‘ π‘œπ‘“ π‘–π‘›π‘ π‘’π‘™π‘Žπ‘‘π‘–π‘œπ‘› π‘Žπ‘›π‘›π‘’π‘Žπ‘™ π‘ π‘Žπ‘£π‘–π‘›π‘” π‘ƒπ‘Žπ‘¦π‘π‘Žπ‘π‘˜ π‘‡π‘–π‘šπ‘’ π‘¦π‘’π‘Žπ‘Ÿπ‘  = =1.67 π‘¦π‘’π‘Žπ‘Ÿπ‘ 


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