Following to article of "Can We Solve a Nonlinear Equation with
Many Variables? (Con)" posted on link: http://emfps.blogspot.co.uk/2016/10/can-we-solve-nonlinear-equation-with_19.html, the purpose of this article is to solve
a nonlinear equation with five variables in limited domain and range. I have
started it with a sample of thermal conduction in curtain wall system in
buildings in which this example pursues the energy saving and cost management
in shopfront glass in malls, shopping centers and mini-markets. This is a real
example because the people can directly use from the data and results of this
article to save energy.
Before starting of example, let me
tell you about a problem and a restriction when we are applying
this method to
solve the equations as follows:
1. As you understood, this method
gives us the same domain and range for all variables while we
need to different
domains and ranges to solve the equations in the fields of engineering. For
instance, domains for variables can be between 1 - 100 or 1 – 1000 but
sometimes we have to have real numbers for the variables including: x (0.0025,
0.48) or
y (-23.5, 0.78) or z (4, 563.75) and
so on. I can say to you that we are able to solve easily this problem by using
of some theorems in mathematics and change the domains and ranges of variables to
ideal and desired ones. I have shown you this changes in below example.
2. But, there is a restrictive
factor something like discrete functions instead of continuous functions so
that we have to follow this limitation forever. It means that we should
consider sub-intervals between domains and ranges for our variables. In fact,
the numbers which belong to variables between a domain and rage are discrete
not continuous. Of course, we can decrease and decrease sub-intervals step by
step. The best way is, to apply Arithmetic Progression for sub-intervals.
"Please be informed that we can
use many methods to reach our targets which are new data and results. Therefore,
the most important thing is, to extract new data and results by any method. In
fact, the analysis of these new data and results lead us to reach the gates of new
worlds. The methods are only tools to reach our targets."
A real example of energy saving:
Thermal conduction in shopfront glass
The law of thermal
conduction (Fourier's Law) gives us the great opportunity to calculate the rate of
energy transfer by heat for a slab of infinitesimal thickness (dx) and temperature
difference (dT) in which the rate of│dT/dx│is named the temperature gradient. Simple conductive heat transfer (in
watts) through a uniform body can be calculated from below equation:
P =
k. A.│∆T/ ∆x│ and ∆T =
Th – Tc
and Th >
Tc
Where:
- A is the area of the body (m2)
- ∆x is the body's thickness (m)
- Th is one face of the slab with high temperature (°C)
- Tc
is other face
of the slab with low temperature (°C)
- P is the rate of heat transfer (Watt)
The heat
transfer occurs only if there is a difference in temperature between two parts
of a slab.
For this example,
I used from manual of "CODE OF PRACTICE FOR USE OF GLASS IN
BUILDINGS" an Indian Standard which was adopted by the Bureau of Indian
Standards. You can find this paper posted on below link:
The page 27 of
above Standard shows us the boundaries for maximum area of designed normal
glass in compare with its thickness in shop fronts. (See Table 5.3)
According to Table 5.3, I chose below domains for A (area) and ∆x (thickness):
A = (6, 15) and ∆x = (15, 25)
And so, I consider domains for the temperature of outside and
inside as follows:
Th = (15, 32) and Tc = (-20, 4)
In this case, we will have a maximum of heat transfer equal to
41600 W and a minimum heat transfer equal to 2112 W. (kglass = 0.8 W/m°C)
P max = 41600 W, and Pmin = 2112 W
Therefore, the range for heat transfer is: P = (2112, 41600)
Now, I apply previous method and solve
above equation for five domains and range. The rate
of heat transfer (P) to the Number
of Results has been presented on below graph:
I only apply one sub-interval and get the total sum of results equal
to 195 in which we will have 30 results for P = 12800 W. If you spend more time
and apply many sub-intervals, it is possible, you will find much more results.
Analysis of findings and results
The most important part of this article is, to analyze the results
to reach our target which is the energy saving in the same conditions
accompanied by cost management.
According to the results, here is many outcomes which lead us to handle
the cost management and the energy saving. Let me present only several states
as follows:
1. For instance,
if we compare maximum transfer rate with another results, we will have
below analysis: (Please see below figure)
According to above table, if you decrease the temperature of your shop from 32 °C to 15°C and also increase the thickness of glass from 15 mm to 16.935 mm, then you will save about 40% of your energy consumption. What is your cost management?
Usually shop front glasses have around 8 years guarantee. If you open your shop 12 hour per day, total hours for energy consumption is: Total hours = 8*365*12 = 35040 hour
Then, you have around 40% (0.403846) energy saving which is equal to 16800 W. Therefore, you will be able to save the energy totally around 588772 KWh in the period of 8 years (16800*35040). The cost of electricity power is about 12 cent per KWh. In the result, you will save totally amount of USD70, 640 for 8 years and USD8, 830 per year. If we add future value of each annual saving with an average return rate (Cost of capital) equal to 11% per year, it means that you are saving amount of money around USD 104,718. What is your costs? You should purchase normal glass with thickness 17 mm instead of 15mm. Therefore, you should pay more USD 40 per square meter (Price of normal glass with 15mm thickness is around USD 130 and 17mm is around USD170). Consequently, your additional investment is USD 600. It means that you have still saved the amount of money around USD 104,000.
2. If we compare
some results together, sometimes we can see very exiting outcomes as follows:
According to above results, if we increase the thickness from 15mm
to 25mm, we are saving around 42% energy consumption and if we increase the
thickness from 15mm to 23 mm, we are saving around 35% energy consumption. For
calculation of cost management, you can apply the same method mentioned in Item
(1).
3. In the
reference with below results, if we are able to change the design for area of
7.74 m2 to 6 m2, we have already saved around 23% energy consumption.
As I told you, we can obtain 30 results for P = 12800 W. You can
find these results on below figure for your better analysis and design:
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