Calculation on the energy

DAMON

2015-04-13 8:09 pm

Calculate the hourly heating load of a house in which all exterior surfaces have a thermal resistance of R = 2.0 m2-s-°C/J, the inside temperature is 25° C and the outside temperature is 10° C. The dimensions of the house are 9 m by 15 m by 3 m. The factor K for infiltration losses is 1 /hour or 1/3600 s.

K is the number of air changes per hour.

Use the following equations in your solution:

Rate of conductive heat loss,

Rate of heat loss due to infiltration, (in J/s)

Consider total heating load is equal to

Suppose you install new panels for your walls to improve thermal insulation. The new thermal resistance R for your walls is 3.0 m2-s-°C/J. How much are you going to save in heating cost per day? Consider the cost of heating in HK is HK$200 per 1 million kJ.
How much is your saving per year? Consider your average heating load remains same throughout the year.
The installation of new panels cost you HK$ 10000. Will you consider making such an investment?

回答 (1)

天同

2015-04-14 12:50 am

✔ 最佳答案

Total surface of 4 walls and roof = (144 + 135) m^2 = 279 m^2
Heat loss by conduction = (1/2) x 279 x (25 - 10) watts = 2093 watts
Heat loss by inflitration
= 1000 x 1.2 x (1/3600) x (9x15x3) x (25-10) watts = 2025 watts
Hence, total heat loss = (2093 + 2025) watts = 4118 watts

After installing new panels,
New heat loss by conduction = [(1/3) x 144 + (1/2) x 135](25-10) watts = 1733 watts
Total heat loss = (1733 + 2025) watts = 3758 watts

Reduction in heat loss = (4118 - 3758) watts = 360 watts
Reduction of heat loss per day = 360 x 24 x 3600 J = 3.11 x 10^7 J = 3.11 x 10^4 kJ
Saving of heating cost = 3.11 x 10^4 x 200/10^6 dollars = 6.22 dollars
Saving per year = 6.22 x 365 dollars = 2,270 dollars

No. The saving in heating cost in a year doesn't not make up the cost in installing the panels.

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