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Tuesday, 27 January 2015

DESALINATION AND THE SOLAR SOLUTION

Here we will discuss this post with one example 

In this concept, salty water will be extracted from a bore with the help of a pump through a filter i.e. suction filter. There is one more pump which is basically utilized for increasing the pressure of the water before it enters the reverse osmosis filter. 

The reverse osmosis filter will extracts some drinkable water from the high pressure flow of the saline water. However it must be noted that most of the salty water is discharged as untreated water and it will be used for general purpose. 

Before the discharge water leaves the system it passes through a device which forms part of pump B designed to recover as much energy from the high pressure flow as possible. The potable water flowing out of the reverse osmosis filter is then stored in a stainless steel tank for later distribution. 

The other important point is this that from where we are securing energy for driving the pump and for this requirement we are using here solar panel and battery system.

Technical analysis

In technical analysis we have to understand the complete specifications of storage steel tank such as
The tank volume (V)
The dimensions of the tank radius (R) and depth (DT)
The minimum head (HMIN)
The total length (L) of steel required for the four supports
The area (A) of sheet metal required to fabricate the tank
The mass of the empty steel tank (M)

Requirement that we should have

R1. The system must produce potable water at a minimum rate of 4000 L per day 
R2. The system must be able to store at least 5 days of water supply, in case of break downs.
R3. The system must provide a minimum gauge pressure at the outlet of 100kPa

We have some information as mentioned below

Total height should be less than the 4 times of diameter of tank 
The steel supports are to be concreted in to the ground to a minimum depth of 2 M
The steel supports material is available in 6 M length
Tanks are made of stainless steel
Other information is given in following table

Quantity
variable
value or equation 
unit
Density of potable water 
ρw (rho w)
1000
kg/m3
Density of stainless steel
ρS (rho s)
8000
kg/m3
Thickness of stainless steel plate
t
1.5
mm
Cost per square metre of stainless steel
CS
100
$/m2


Volume of tank

As we have information that system should have ability to produce potable water at a minimum rate of 4000 L per day. 

If we consider this capacity of system only then it seems that minimum 4000 L/day will be consumed. We have also informed that system must have ability to store at least 5 days of water supply, in case of break downs. 

Hence, capacity of storage tank must be approximate 200, 00 L. As this is design problem, we have to think in real use and hence we will take some factor of safety of 25 %. 

So, volume of tank must be 125% of 200,00L
Volume of storage tank = 250, 00 L
Volume of storage tank =25 cubic meter

Height and Radius

In order to determine the required specification, we must assume that what will be the height to diameter ratio for water tank that we are going to design. 

Let the diameter to height ratio is
D/H= 1/3
Or H/D= 3
Where, height = H    and diameter = D
Volume of tank = πR2H
πR2H= πD2H/4= πD2 x 3 D/4

Volume of tank = 3.142 D3  
25 = 3.142 D3 
D= 1.99 meter
Height of tank = 5.98 meter
Diameter = 1.99 meter

As we have information, that structure for support material will be purchased in 6 m length and it must be assure that 2 m length of supporting material must be inside ground for better supporting
We will purchase 6 m length of supporting materials and we will cut this in two equal portion of 3 m length each. 

In each 3 m length, 2 m will be inside the ground and hence
Min = 1 m 

As there are four support of having length of 3 m each, hence we have to use two length of supporting material with total length of steel of 12 m.

Total sheet metal required in m2 for fabricating the above specified tank

A = 2πRH + 2πR2
R= 0.995 m
H = 5.98 m
A= 43.61 m2

Thickness of metal sheet that we have selected is 1.5 mm
Mass of empty tank = volume of metal x density
Mass of empty tank = A x t x ρ = 43.61 x1.5/1000 x 8000
Mass of empty tank =523.32 K

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