To determine energy consumption, it is essential to take into account the following parameters. The type and size of a house, the day and night occupancy, cooling and heating nature, weather, location, windows area and external wall area, occupants status and lifestyle, the effectiveness of lighting and appliances as well as the levels of loft insulation, floor, and walls.
To approximate the lighting/ heating consumption of a residence, it’s imperative to consider the temperature difference inside and outside the home. Thus, the heat losses of a particular house can be determined by the ventilation and fabric heat losses (John, 2014).
To calculate the fabric heat loss, the first step is to examine the total surface of uncovered external walls and then computing the u-values of every element. The heat lost through ventilation is dependent on air changes for each hour through windows and other mediums of infiltration that may be available as well as the cubic volume of every room.
Movement of heat by convection
Losses of energy by convection in a house can be stated as negligible in a home without wide-ranging calculations. For instance, in every piping system, air spaces exist between the insulation and the walls. The air gaps are estimated to be less than one-tenth of an inch. The spaces prevent the smooth flow of air and hence limiting heat loss by convection.
Movement of heat by radiation
Heat loss by radiation is a result of transmission of heat through waves by extremely energized molecules. Heat loss through this medium requires the hotter surface to be above the ambient temperature.
Relationship between Radiation, Convection, and Conduction
Practically, radiation and convection account for only 10% of the total energy lost in a system (Kokin, 2013). In light of the information from (Kokin, 2013), a general formula for calculating heat loss through the three mediums can be developed.
Movement of heat by conduction
Conduction denotes the transfer of heat through solids. In a house, the loss of heat by conduction is through the floor, ceiling, and walls. Conduction is the most considered mode of heat transfer. Heat tends to move from higher temperature solids to low-temperature ones. However, heat loss through conduction can be controlled through insulation.
The general equation of heat loss by conduction is,
Q= A/R*ΔT where Q refers to heat flow, A refers to Area, and R refers to the resistance commonly known as the R-factor and ΔT is the temperature difference (James, 2017).
Calculations
The house in question is a 3-bedroom terraced house made of two external walls and two walls with a separate the neighbours. The apartment consists of ground, first floor, and a loft.
Heat loss via walls
The total area of the walls is,
(10m ×5m) × 2 (walls) =100m2
(10m × 5 m) × 2 (walls) = 100m2
The flor area is 10m × 10m = 100m2
Thus, the total surface area is 300 with r value of 0.7 K/ (m2W)
Calculating total heat loss during winter
Night
300m2/0.7 K/ m2W× (6 × 12×30) days × (21-0 + 273.15) k= 272mj
Day
300m2/0.7K/M2W × (6×12×30) days × (21-7 + 273.15) k = 265.83 mj
Summer
300m2/0.7K/ (M2W × (6×12×30) days × (21-20 +273.5) k = 254.12 mj
300m2/0.7K/ (M2W × (6×12×30) days × (21-13 +273.5) k = 270.01 Mj
Therefore total heat loss (in a year) of the walls which are not insulated is 265.83 +254.12 +270.01 +272 =1061.96 mj
Heat loss through roof
The area of the roof is 10m by 10m which is a 100m2. The r-factor is 0.7 K/ (m2W)
Calculating heat loss during winter
Night- 100m2/0.7 K/ m2W× (6 × 12×30) days × (21-0 + 273.15) k= 90.76 mj
Day- 100m2/0.7K/M2W × (6×12×30) days × (21-7 + 273.15) k = 88.61 mj
Summer
Day- 100m2/0.7K/ (M2W × (6×12×30) days × (21-20 +273.5) k = 84.71 mj
Night- 100m2/0.7K/ (M2W × (6×12×30) days × (21-13 +273.5) k =90.00 Mj
Heat loss via windows and doors
Winter- Day-2m2 /3.5 K/(m2w) × (6×12×30) × (21-7 + 273.15) k + 1m2 /3.5 K/(m2w) × (6×12×30) × (21-7 + 273.15) k=1.01 mj
Night-2m2 /3.5 K/(m2w) × (6×12×30) × (21-0+ 273.15) k + 1m2 /3.5 K/(m2w) × (6×12×30) × (21-0 + 273.15) k=0.54mj
Summer
Day-[2m2×3 /3.5 K/(m2w) ×3× (6×7×30) × (21-20 + 273.15) k] + [1m2 /3.5 K/ (m2w) ×4 × (6×12×30) × (21-20 + 273.15) k]=0.53 mj
Night-[2m2×3 /3.5 K/ (m2w) ×3× (6×12×30) × (21-13 + 273.15) k]+ [1m2 /3.5 K/ (m2w) × (6×12×30) × (21-13 + 273.15) k] =0.53
Heat loss through uninsulated pipe
Winter
Day-(3.14×0.015×0.001) m2/500× (80-7+273.5) × (6×12×30) =0.18 mj
Night (3.14×0.015×0.001) m2/500× (80-0+273.5) × (6×12×30) =0.17 mj
Day-(3.14×0.015×0.001) m2/500× (80-20+273.5) × (6×12×30) =0.17 mj
Night-(3.14×0.015×0.001) m2/500× (80-13+273.5) × (6×12×30) = 0.18 mj
Heat loss through radiators
In this case the r-factor of the walls will be considered
The number of radiators is 5
Winter
Day-[1m2 /0.7 K/ (m2w) ×4 × (6×12×30) × (21-7 + 273.15) k]= 3.5 mj
Night-[1m2 /0.7 K/ (m2w) ×4 × (6×12×30) × (21-0 + 273.15) k]=3.63 mj
Summer
Day-[1m2 /0.7 K/ (m2w) ×4 × (6×12×30) × (21-20 + 273.15) k] =3.38 mj
Night-[1m2 /0.7/ (m2w) ×4 × (6×12×30) × (21-13 + 273.15) k]=3.60 mj
Heat loss through light bulb
Winter
Day-20×60 ×6×3×30×(21-7 + 273.15) k]= 185.43 mj
Night-20×60 ×6×2×30×(21-0 + 273.15) k]= 195 mj
Summer
Day-20×60 ×6×3×30×(21-20 + 273.15) k]= 178.12
Night-20×60 ×6×2×30×(21-13 + 273.15) k]=181.76
The total lost through is therefore the sum of all the heat and light lost which sums up to
2172.63 mega joules annually when the surfaces are not insulated.
Employing (Kokin, 2013) study that convection and radiation losses only 10% of the total heat, then the total heat loss is,
2172.63 × 100/90 = 2414.03 MJ
The budget available for insulation and change of lighting appliances to more efficient bulbs is £2500 (stambaek, 2014).
The cost of rock wool of 5.53 m2 is the £361.25 role with a total surface area of 200m2 rocks. After calculating the total cost involved in the insulation process, total energy consumption lowers by a whelming 23%.
Conclusion
There is an excellent potential for energy saving in apartments. An approach to energy conservation is to note the surfaces of the house that losses most energy. The determination of the energy in the 3-bedroom apartment was to illustrate the surfaces losing most energy. The walls, window, and electrical appliances contributed to the highest heat loss. Insulation of surfaces is the best approach to conserve energy as exemplified in the determination of energy loss in the calculation.
Bibliography
James, F., 2017. Heat TransferMediums. U.S Energy Information & Administration, 1(1), pp. 7-19.
John, A., 2014. Mechanisms of heat loss or transfer. E-Education Institute, 11(7), pp. 24-25.
Kokin, E., 2013. Modeling of Thermal Process in Apartment Houses. Estonian Academy of Science Engineering, 1(12), pp. 59-71.
stambaek, j., 2014. Heat Transfers. Biology Cabinet, 11(4), p. 15.
