solar powered water purification system

DESIGN OF SOLAR POWERED WATER PURIFICATION SYSTEM

CHAPTER ONE

1. INTRODUCTION 1.1. Background of the Project Purifying water is a crucial process for drinking safety and requires not only removal of ino rganic material but also bacterial treatment. To address this problem, there are numerous ongoing efforts to provide fresh water for impoverished communities. In Mekelle this project is important drilling wells to tap into the groundwater. However, groundwater is limited and the wells must be dug deeper as the resource becomes depleted over time. A desalination system removes the salt from underground water, transforming it into potable water through one of many purification processes. Distilling, or boiling the dirty or salty water to produ ce clean vapor that is then then condensed back to water, is perhaps the simplest technique. The process requires boiling salt water at temperatures over 100 °C, a temperature at which bacteria is killed. The major objective of this Solar-Powered Water Purification System is to produce clean and drinkable water using a simplified system that powered solely by the sun. To become a marketable, worldwide product, the system must be portable and durable enough to transport through all terrains and conditions over long periods of time. In addition, it should be inexpensive and userfriendly. 1.2.

Problem statement

We know that our country Ethiopia is known as the “water tower of Africa’’. However some par ts ts especially in Tigray regions of the country, the water is very salty and cannot drink easily. Due to this problem, people who live in these areas are suffering with drinking salty water problems. Indeed we have seen this challenge when we were in mekelle campus during our educational trip this year. Even though our country Ethiopia has a sufficient solar radiation, but it is not utilized into a useful form of energy. Due to these and other facts, our study is a useful and critical one to address the above problem by “designing of solar powered water purification system for drinking purpose” for the students of Mekelle University to provide safe and pure drinking water.

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Objectives of the project

1.3.

1.3.1. General objective: 

To design and analyze the th e Solar-Powered Water Purification System.

1.3.2. Specific objectives: 

To utilize our solar resources into a useful form of energ y.



To improve our living standard by using solar PV technology.



To design solar PV system.



To design parabolic trough for water heating.



To design water storage tanks.



To design water piping systems.



To simulate solar water purification system components.


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Objectives of the project

1.3.

1.3.1. General objective: 

To design and analyze the th e Solar-Powered Water Purification System.

1.3.2. Specific objectives: 

To utilize our solar resources into a useful form of energ y.



To improve our living standard by using solar PV technology.



To design solar PV system.



To design parabolic trough for water heating.



To design water storage tanks.



To design water piping systems.



To simulate solar water purification system components.


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CHAPTER TWO 2. LITRATURE REVIEW Solar powered water purification system utilizes the power of concentrated sunlight through its parabolic trough. The intense in tense heat created by b y this set-up immediately vaporizes the salty water, removing dissolved solids solids and the Salt. The vapor is then condensed; and the resulting water is deemed safe to drink. Direct Solar purification system of our project is able to produce 10000 liters (37900 gallons) of clean water per day using only solar power. (Text book of solar water pumping, By Jeff Kenna,)

2.1. Working principle of the System HTF, known as Duratherm-450, circulates through the troughs; picking up energy from the concentrated solar water evaporates in one area (leaving contaminants behind) and condenses as clean water in another area. This project is intended for coastal communities, communities, an abundant supply of brine water should be readily available. While not in use, the solar troughs should be covered or sheltered from direct sunlight. Protecting the troughs from the sun is important because, if left in direct sunlight while the system is not operating, the heat transfer fluid (HTF) along the length of the troughs will continue to pick up heat, potentially reaching temperatures that begin to damage system components. The insulation had the lowest temperature rating of 120 °C, and therefore was the maximum temperature the system could reach. At system start-up, salty water coming from the ground, originally stored in the saltwater storage tank, flows into the boiler, filling it just above the level of the heat exchanger (HEX). Two float switches located inside the boiler control the opening and closing of a solenoid valve (controlling the flow of salt water into the boiler). A combination of two float switches and a solenoid valve maintain the water level in the boiler, preventing the boiler from overflowing or running dry. The fluid then flows through the HEX inside the boiler, giving off the energy as heat to the salt water. As the salt water reaches the target temperature of 100°C, it boils and causes a vapor pump to switch on. The vapor pump pulls the newly formed water vapor from the boiler, directing it to IOT/UOG/MECHANICAL IOT/UOG/MECHANICAL ENNGINEERING ENNGINEERING

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a helical coil condenser located inside the salt-water storage tank. Heat will leave the condensing vapor to preheat the salt water in the salt-water storage tank. The condensed vapor liquefies and is collected in a clean water storage tank. The distillation process for the Solar-Powered Water Purification System happens in the boiler. The salt water vaporizes when the water within the boiler is boiled from the heat provided by the hot loop. The hot loop uses a HTF that collects heat from the solar parabolic troughs. That HTF runs through a HEX in the boiler and transfers the heat to the salt water. This causes the water to boil and separates water vapor from undrinkable und rinkable debris and contaminants. All the contaminants and other waste in the salt water are killed at atmospheric pressure because of the boiling process. After the vapor leaves the boiler, it is then condensed into purified drinking water. The flow of fluids throughout the system is made possible through the use of system pumps. A vapor pump is used to pull the water vapor from the boiler through the con denser. A HTF pump is used to circulate the HTF through the solar thermal collection loop (trough) and the salty water is pulled by the the submerged centrifugal centrifugal pump from from the ground to the the salt water tank. The three pumps are powered through the use of a PV panel. The power requirement for the pumps will be seen in chapter five.


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Fig.1. System assembly of the project

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2.1.1. Input and Outputs The system has two inputs: salt water and solar energy (in both thermal and electrical form). The source of the salt water was assumed to be 35,000 ppm, which is a typical salinity level for the underground water. It enters the system from the saltwater storage tank to be boiled. Solar energy is provided through the sun, and was assumed to be 850 W/m2. By utilizing the sun for thermal and electrical power, the system’s energy requirements are fulfilled and allow for a completely sustainable design. The system has two outputs: clean wat er, and thermal energy. The clean water is obtained through the condensation o f water vapor through the condenser. There is an abundance of thermal energy carried within the water vapor as it leaves the boiler. Therefore, by placing the condenser in the salt water storage tank, a preheating process can occur. The heat is transferred from the vapor to the salt water before it enters the boiler. Although some thermal energy is lost to the environment, this was minimized using insulation. 2.2.

Main Components of Solar Powered Water Purification System

The Solar-Powered Water Purification System functions to remove chemical and microbial impurities from saltwater through the use of solar energy. The system itself is comprised of main sub-components: 

Solar collector,



HEX,



Salt-water storage tank,



Clean water storage tank,



Boiler,



Condenser,



System pumps,



PV panel,



Motor,



Trough/PV panel frame,



Piping system,



Cables and switches, and



Troughs


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2.2.1. Boiler

The distillation process for the Solar-Powered Water Purification System happens in the boiler. The salt water vaporizes when the water within the boiler is boiled from the heat provided by the hot loop. The hot loop uses a HTF that collects heat from the solar parabolic troughs. That HTF runs through a HEX in the boiler and transfers the heat to the salt water. This causes the water to boil and separates water vapor from undrinkable debris and contaminants. All the contaminants and other waste in the salt water are killed at atmospheric pressure because of the boiling process. After the vapor leaves the boiler, it is then condensed into purified drinking water. 2.2.2. Solar Collectors The solar collectors (solar parabolic troughs) function as the system’s solar thermal energy harnessing source. The largest advantage of a solar parabolic trough over other methods is its ability to concentrate the energy from the sun to yield higher energy collection efficiency for a low cost. A major material of the solar parabolic trough is the reflective surface. 2.2.3. Drive System A drive system was employed to facilitate the movement of the two solar collectors and PV panel. This single axis rotation is necessary in order to accurately track the sun’s movement across the horizon throughout the day. A motor is employed to power the rotation of the solar collectors. The motor’s direction of movement was controlled by light sensor. 2.2.4. Heat Exchangers 

Heat Transfer Fluid (HTF)

The HTF absorbs heat as it flows through the troughs and transfers this heat to HEX located in the boiler. A single-phase flow was desired to avoid pressure changes that would quickly lead to leaks within the piping, as well as reduced heat transfer capabilities of the fluid. Salt water boils at approximately 100°C, so the HTF must be in a liquid phase above that temperature to transfer heat in the HEX. Three main fluids were considered: water, glycol, and an oil-based fluid. Water is considered an ideal HTF because of its low maintenance, cost, and environmental impact. However, it boils at 100°C and, therefore, a single-phase flow would not be maintained. Gl ycol is a water-sugar mixture that can reach higher boiling temperatures at certain concentrations and it


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boils 102-1880c. It also has a low environmental impact because it is biodegradable. However, it requires extensive maintenance, as sugar levels must be regularly monitored. Ultimately, an oil-based fluid called Duratherm-450 was selected. It has a boiling point of 232°C and requires low maintenance due to additives, including antioxidants, corrosion inhibitors, defoaming agents, seal and gasket extenders, suspension agents, and metal deactivators. Although it has a moderate environmental impact, and must be disposed of with other waste oils, it is not considered toxic and does not cause skin irritation when accidental contact occurs. 

Boiler Heat Exchanger (HEX)

A HEX was located within the boiler, and allowed for transfer of energy from the HTF to the salt water for boiling. To maximize the surface area for heat transfer, and guided by the boiler dimensions, the HEX was designed to be a helical shape. Therefore, the major design variables to select were the tube diameter, number of coils, and HTF flow rate. 2.2.5. Condenser The condenser was located within the salt-water storage tank and allowed heat transfer from the vapor to the salt water. In doing so, the vapor would change to distilled water while the salt water would be preheated. 2.2.6. Control System Logic The system controls the following elements: HTF Pump, Vapor Pump, and Solenoid Valve. The control algorithm works in the following manner: 1. The system starts by reading the ambient light level as an analog value. If the ambient light

reading is less than 500 decibel (db), all components are off. If the ambient light reading is greater than 500 db, the HTF pump is turned on 2. The boiler temperature is read by the thermocouple and the vapor pump remains off until the

boiler temperature is greater than 100°C. 3. If the state of both float switches are read as ‘low,’ the solenoid valve opens until the state of

both float switches are read as ‘high. 4. Process starts again.


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Start

Start

Ambient light

No.

>500?

Everything is off

Yes Turn on HTF pump Yes Read Bioler tempratu off

Float switch=1?

Valve open

Yes vapor pump

Temprature>100?

off

No.. Flow control valve closed

Vapor pump on Fig.2. System logic control (closed loop)


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CHAPTER THREE 3. METHEDDOLOGY The methodology that follows in our project is:  Examining the economic living standard of the students.  By asking the different organization about climate chan ges of their surroundings.

To analyze the raw data we will may follow the following steps; The following twelve steps can be used in the design process for a PV-powered water purification system. These steps will help you ensure that the system functions properly and that water is supplied for the operation in the amounts and at the locations required.

Step 1: Know the amount of water Requirement. Step 2: Determine the water Source. Step 3: System Layout.

The third step in the system development process is to determine the layout of the entire system, including the locations and elevations of the following components: • Water source. • Pump. • PV panels. • Storage tanks. • Points of use (i.e. water troughs). • Pipeline routes. Step 4: Determine the size of water Storage. Step 5: Determine Solar Insolation (peak Sun hours) and PV Panel Location. Step 6: Design Flow Rate for the Pump. Step 7: Total Dynamic Head (TDH) for the Pump. Step 8: Pump Selection and Associated Power Requirement. Step 9: PV Panel Selection and Array. Step10: System layout obtains solar PV information from various solar dealers both locally and

internationally. This information should include; type, cost, size, weight… etc IOT/UOG/MECHANICAL ENNGINEERING

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Step 11: PV Array Mounting and Foundation Requirements.

Finally we should provide a descriptive summary of the completed system to the University that includes the following information: • All system components and their specifications. • System operating characteristics, such as required voltages, amperages, w attages etc.


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CHAPTER FOUR 4. SLECTION OF MATERIALS FOR THE PROJECT Materials have their own inherent property. So selection of materials is a very important task in an Engineering design. As a result, in material selection the properties that are taken into consideration are durability, corrosion resistivity, yield and tensile strength, industrial applicability, electrical and thermal conductivity, availability and cost of the material. Due to the above reasons and considerations we have selected the materials for our project as follows: 

Material for salt water tank.

 Concrete of grade 20 

Material for heat exchanger and condenser.

 Stainless steel with AISI No 304 , tensile strength = 1600Mpa , yield strength = 276Mpa 

Material for pure water tanker.

 Carbon steel with AISI No 1045, yield strength = 63kpa , tensile strength = 107kpa 

Material for electrical wires.

 Copper wires.  Due to high conductivity of copper. 

Material for bolts.

 Carbon steel. 

Material for piping systems. 

Pipes of heat transfer fluid.

 Copper pipes due to high heat transfer property of copper. 

Pipes of underground water, heat exchanger and condenser.

 PVC grade 1 piping material with yield strength of 8.3kpa

4.1.

System Pumps 

Heat Transfer Fluid (HTF) Pump

The HTF pump drove the circulation of the HTF through the hot loop, and was selected based on various criteria, of which the main ones were: overcoming the static pressure of the loop while providing the initial target flow rate of ranging 0.1 – 0.6 kg/s, consuming a maximum of 230 - 280 W of DC power, and compatibility with the HTF. Based on the length of copper piping in the IOT/UOG/MECHANICAL ENNGINEERING

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system and number of bends, the static pressure of the system was calculated to be 20 -80 KPa. From the solar panel capacity of 300- 800 W of DC power this constraint was further justified because an AC/DC converter would decrease efficiency while increasing costs, and DC power would allow for easier adjustments to the flow rate during testing. Based on these design considerations, an N991-32 Series Gear Pump was selected. This pump has Viton seals, which are compatible with the HTF. This pump also has a temperature rating of 150°C, and was placed at the inlet to the solar collectors. Thus, the pump was exposed to the HTF at its coolest point, minimizing the possibility of overheating. 

Vapor Pump

The vapor pump drove the vapor out of the boiler and into the condenser. The main criteria for selection of this pump included its capacity, ability to run wet and dry, and a maximum power consumption of 50- 100 W. The goal of the system was to provide 9.81*10-6 m3/s of vapor, and therefore the pump was required to pull at least that amount of vapor. In addition, because control of the vapor pump may require it to run when no vapor is present, the pump was required to be able to run dry without failing. Finally, due to the solar panel capacity, the 300- 800 W power constraint was included. Based on these design co nsiderations, a Greylor RF-100 pump is selected. The pump had a temperature rating of 120°C, and was placed at the highest point between the top of the boiler and the inlet to the condenser. This allowed for optimal performance of the pump, as recommended by the manufacturer. ( Text book of system design of solar power water pumps, by Justin Bonnel, in 1987 ) 

Piping

Piping within the system was conducted using standard 150mm copper piping, and 50mm for PVC pipes and soldered together using a lead-free solder material. The design of the piping was conducted in a logical fashion, and many break points along the length of the piping were created to allow for quick removal and replacement of damaged or faulty piping/fittings. This modular design was specifically intended to expedite the maintenance process. A number of valves were also included which allow for the bypassing of the solenoid valve into the boiler for testing purposes or for manual operation of the in-flow to the boiler. A drain valve was implemented at the exit of the solar trough line to allow for the quick removal of heat transfer fluid when IOT/UOG/MECHANICAL ENNGINEERING

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maintenance and replacement is required. A number of nylon fittings were attached to the boiler to separate dissimilar metals from each other, reducing the risk of galvanic corrosion to these otherwise vulnerable areas. 

Electrical Subsystem

The purpose of the electrical and control system is to power and control the actions of the system components. This includes the HTF pump, vapor pump, solenoid valve. A microcontroller will be used to obtain ambient light readings, boiler water level, and boiler temperature.

These

measurements are used to optimize system performance by providing data that will be useful to evaluate system ramp up time and overall power consumption. 

Sensors

To sense the ambient light levels as an analog value, photo resistors are used as the top resistor in a voltage divider circuit. A photo resistor is a resistor that exhibits photoconductivity, where its resistance decreases with increasing light intensity. In this application, three cadmium sulfide photo resistors are connected in parallel as the top half of the voltage divider circuit, while a 12kΩ resistor is placed on the bottom half. The photo resistors are angled so that the ambient light levels can be read consistently throughout the day with varying sun position. The light sensor is placed in the same housing as the single axis. 

Float Switch

Two vertically mounted, high-temperature, float switches are used to determine the water level in the enclosed boiler. The float switches will be used to enable cyclic loading of underground water into the boiler to optimize the distillation process. Each float switch will be wired in series. The states of the float switches will be read by the microcontroller using a digital input pin to control the operation of the solenoid valve. 

Thermocouple

A thermocouple is used to measure the boiler temperature. Thermocouples use the thermoelectric effect to measure temperatures. Two dissimilar metals with known thermal expansion coefficients are soldered together at one end. Two wires extend from this point; and, together, they produce a voltage difference across them that is directly related to the temperature at the soldered end, known IOT/UOG/MECHANICAL ENNGINEERING

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as the See beck voltage. Thermocouples can respond to changes in temperature very rapidly, but they are also sensitive to electromagnetic noise. This high sensitivity can compromise the quality of the readings if it is not taken into account. Therefore, it is important that the thermocouple wires are isolated from the pumps, and motors throughou t the system. 

Power Switching

To enable full control over the electrical components in the system, power switching is necessary. Switching transistors are used to control the components.

Electromechanical switches are

controlled using NPN switching transistors that control the vapor pump, solenoid valve, and the HTF pump. 

Material for the Body of the trough

Aluminum was selected over steel because of its lightness, lower cost, ease of fabrication and energy effectiveness in use of material. Its light weight reduces the overall weight of the trough. 

Material for the Reflecting Surface

To reduce the overall weight of the solar water heater, a light glass mirror of 2mm thickness, of high surface quality and good specular reflectance was selected. A glass mirror was selected over polished aluminum surface because its reflectivity of 95% is b etter than that of aluminum (85%). Also, glass surface is easier to clean than aluminum surface. 

Material for the Absorber

Aluminum was selected over copper and steel because of its lower cost, light weight, ease of fabrication and energy effectiveness in use of material. Its light weight reduces the overall weight of the solar water heater. 

Material for the Absorber Surface Coating

Black paint was selected for the absorber coating. It is selected over other coatings b ecause of its higher absorptivity at angles other than normal incidence, adherence and durability when exposed to weathering, sunlight and high stagnation temperatures, cost effectiveness and protection to the absorber material. 

Heat Transfer Fluid


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Duratherm- 450 was selected as the heat transfer fluid for the solar heater because of its stability at high temperatures, low material maintenance and transport costs, safe to use, and is the most commonly fluid used for domestic heating applications. 

Material for the Vertical Support of the trough

A rectangular, hollow, steel bar was selected for the support of the trough. This is because of its strength, rigidity, resistance to deflection by commonly encountered winds, and its ability to withstand transverse and cross-sectional loads of the entire heating portion of the trough. 

Material for the Base of the trough

A combination of angle and flat, steel bars were chosen for the base which supports the whole solar water heater structure. Flat and angle bars w ere chosen to provide solid and rigid support for the rectangular, vertical axis steel bar which supports the parabolic trough.


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CHAPTER FIVE 5. ANALYSIS OF SYSTEM SPECIFICATION System specification and doing an engineering analysis is common in any design concept. Accordingly for our project we have used the following parameters and assumptions. 5.1. Design specifications for the system 

Amount of water to be pumped per day = 10m3 = 10000L



Amount water required for 3000 Students, each per day = 3L



Water density = 1000 kg/m3



Acceleration due to gravity, g = 9.8 m/s2



Solar PV module used = 300W



Operating factor = 0.8 (PV panel mostly does not operate at peak rated power).



Pump efficiency = 40%.



Mismatch factor = 0.85 (PV panel does not operate at peak rated power point).



Total vertical lift = 13m.



Friction losses = 5% of the total vertical lift.



Type of water source = well.



Static water level = 6m.



Dynamic water level = 8m.



Quality of water = impure and salty.



Water storage = Circular Tank



Salt water tank location = Fixed above the ground.



Distance from solar panel to the pump = 5m



Distance from tank to the pump = 5m



Type of pump selected = submersible pump.



Pump power = g*ρ*h*q



Height of salt water tank = 4m,



Depth of underground water = 7m


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5.2.

Design of trough

 Available Energy

For simplification we assumed the trough to have a horizontal axis, because it allows for a more conservative approximation; however we plan to angle the trough collector at an angle equal to the regions latitude, which will increase the average energ y gain per meter. There are many ways to measure the available radiation from the sun. Most commonly used is a pyranometer, which measures diffuse plus beam radiation from the sky and sun. These devices measure absorbed radiation by detecting the temperature difference between two concentric silvered rings, one coated in magnesium oxide and the other Parson’s black.  Theoretical background

Several parameters are used to describe solar concentrating collectors. Given below are brief descriptions of some of these parameters: The aperture area Aa, is the area of the collector that intercepts solar radiation. The Acceptance angle, is defined as the angle through which a source of light can be moved and

still converge at the receiver (Hsieh, 1986). A concentrator with small acceptance an gle is required to track the sun continuously while. a concentrator with large acceptance an gle needs only seasonal adjustment. The absorber area A abs, is the total area of the absorber surface that receives the concentrated

solar radiation. It is also the area from where useful en ergy can be extracted. The Concentration ratio C, is defined as the ratio of the aperture area to the absorber area. 5.3.

Design calculations

 Sizing of the Parabolic Trough

The heat demand load of the heater is such that it will heat about 15000 liters of water in a day, from ambient temperature to 100 OC. Thus at an average uniform rate of solar insolation, the heater will make 4 cycles of almost equal lengths in time to heat the quantity of water required. The absorber (trough) of the heater will be a cylinder of outside diameter Dabs, internal diameter dabs, height h, and thickness, tx = 4mm. The amount of solar radiation collected is highly dependent on the geometry of the trough, which acts as our aperture for solar collection. This can be observed with the concentration ratio, (C). IOT/UOG/MECHANICAL ENNGINEERING

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C = Aa/Ar……………………………………………………………………………..[1] The aperture area is directly proportional to the concentration ratio. This means that the higher the concentration ratio the higher the temperatures that can be reached. This is because the number of images, formed by the reflection of sunlight, seen by the receiver pipe will increase. However, the objective of this system is to heat water to vaporization, and not to produce high quality steam; therefore a medium concentration ratio is used. For the trough analysis first a basic shape was introduced as follows: The internal volume of the cylinder is the same as the volume of water, Vw, to be heated. Therefore Vw

  =  ∗h

For simple solution of the equation and optimum design of the absorber the height h is made to be the same as the diameter of the absorber. 15m3

= d/4 so,

dabs = 2.67m = habs

Dabs = dabs + 2t = 2.67 + 2 (0.002) = 2.674m The effective surface area of the absorber is given as: A abs

=   /4 + Dh

= 3.14*2.674 2 /4 +3.14*2.674 *2.67 = 28m2

Concentration ratio = Aa/Aabs To reduce the frequency of tracking the sun C is set at 10 (Magal,1993). Aa = C*28 = 280m2 The aperture diameter, Da is given by 



=280

Da = 18.88m The half-acceptance angle,

= sin-1

 /

= Sin-1

The optimum rim angle

is given by (Garg and Prakash, 2000):

√ 1/10 = 18.43

0

is given by


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The focal length, f , of the dish is obtained from (Stine and Harrigan, 1985): f = Da(1+cos71.570) /(4sin71.57) f = 6.52m The height, h, of the trough is given by: h = Da2 /16f =18.82/16*6.52 h = 3.38m The estimated useful energy for one cycle of the designed trough is given by:

Where, I b = beam radiation ῂ = thermal efficiency The efficiency range of most solar concentrators is 40% - 60% (Magal, 1993). Hence for Kaduna take 0.6 (Mohammed, 2009) I b = ID = 3000W/m2 qu .= 0.6*3000w/m2*280m2 = 504kw For four cycles, the total useful energy is Q’u = 4q’u = 4*504kw = 2016 kw The useful energy is also given by q’u = MwC pw(Tw-Ta) = heat gained by water Where; Mw = the rate of heating the water and C pw = the specific heat capacity at constant pressure of the water. is obtained from tables of properties of water (Rogers and Mayhew, 1981) as 4186J/kgK (at 25 oC). Ta = ambient temperature Tw = water boiling temperature



mrW = ῂ.ID. Da2/4C pw (Tw-Ta); where, mrW is mass flow rate of water. mrW = 0.6*3000*3.14*18.82 /4*4186(100-25) =1.605kg/s mrW = 5778.4kg/hr Mrw = density of water *V w /t; where v w = volume of water t = time taken to heat the water The density of water is evaluated at the temperature of 250C = 997kg/m3 IOT/UOG/MECHANICAL ENNGINEERING

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t = (997*15)/1.605 = 9317.7seconds = 155.3 minutes = 2.5hrs For four cycles the total time to heat the required amount of water it takes 4*2.5hr T = 10hrs This is very much time consumable, as a result we use two parabolic troughs for faster operations. The energy, Pabs, absorbed by the absorber is determined as follows. Optical efficiency of the absorber is given b y: ῃ o = Pabs /Aa Pabs = ῃ o / ῃ (ῃ AaID) ,taking optical efficiency = 0.7 P abs = ῃ o / ῃ *q’u = 0.7/0.6*qu = 1.167qu, so for qu = 504kw Pabs = 504*1.167 kw = 588 kw. Solar insolation is defined as average intensity (radiation per solid angle) or the measure of solar radiation received on a surface at some time. Average insolation on the Earth’s surface is approximated to be 250 W/m2 or 6 kWh/m2/ day. 1 kWh/m²/day = 1,000 W * 1 hour / ( 1 m² * 24 hours) = 41.67 W/m² With the data presented an average daily insolation of 522 W/m2 was used for the system design calculations.

5.4.

Trough Geometry


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Fig.3: Solar collector


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Ultimately the best trough geometry was chosen, as follows: o

Diameter = 0.4m of aperture

o

Focal length = 7m

o

Depth = 0.5 m

o

Rim angle = 75 degrees

o

Trough length = 1.3 m = 4.27 ft

o

Aperture, A = 280 m2

o

Receiver, A = 28 m2

o

Receiver outer diameter, Do = 0.15 m

o

Receiver inner diameter, Di = 0.144 m

o

Cover outer diameter, Do = 0.033 m

o

5.5.

Cover inner diameter, Di = 0.031 m

Selecting the type of heat transfer fluid (HTF)

The HTF absorbs heat as it flows through the troughs and transfers this heat to HEX located in the boiler. A single-phase flow was desired to avoid pressure changes that would quickly lead to leaks within the piping, as well as reduced heat transfer capabilities of the fluid. Salt water boils at approximately 100°C, so the HTF must be in a liquid phase above that temperature to transfer heat in the HEX. Water is considered an ideal HTF because of its low maintenance, cost, and environmental impact. However, it boils at 100°C and, therefore, a single-phase flow would not be maintained. Some of the HTFS used in industrial applications are listed below. 

Therminol VP1



Therminol- 59







Syltherm- 800 Therminol- 66 Xceltherm- 600



Therminol XP



Water, salt (60% NaNO3, 40% KNO3)



Dowtherm-450



Hitec X


Page 23


Ultimately we have selected an oil based fluid called Duratherm-450 for our project due to the following reasons.  It has a boiling point of 232°C and requires low maintenance due to additives, including

antioxidants, corrosion inhibitors, de-foaming agents, seal and gasket extenders, suspension agents, and metal deactivators.  It does not cause skin irritation when accidental co ntact occurs  Duratherm- 450 is specifically engineered for applications requiring process heating and

cooling efficiently between -1OC-232.20C.  A colorless, clear and bright fluid  It is economically and thermally stable  It is an efficient oxidative, long lasting and environmentally friendly heat transfer fluid.

i.e. nontoxic, non-hazardous and possesses no ill effect to worker safety and does not require special handing.  After its longer service life it can be disposed of with other waste oils.

It has superior resistance for slugging than other fluids.



Insolation and solar panel location

5.6.

The location for the system is Mekele Adihaki Business and Economics Campus. Insolation for this area is taken averagely as follows. Table.2. Insolation peak sun hours of mekelle per month

Location of Mekelle=13.3 0 N 39.90 E Month

Sep

Oct

Nov

Dec

Jan

Feb

Mar

Apr

May

Jun

Jul

Aug

Sun h

5

6

7

7

9

10

10

9

6

4

3

3

Source; Metrological statistics of Mekelle RE. Energy assessment (2004)

The number of peak sun hours is defined as the number of hours at an irradiance level of 1 kW/m2 required to produce the total solar energy density available in 1 day. Peak sun hours are determined from the monthly mean solar irradiance values and are an indicator of how many hours per day solar arrays will operate at peak power output. So, designed flow rate for the submerged pump Q = volume / average time. Average time (Sun hours) = 5+6+7+7+9+10+10+9+6+4+3+3 /12 = 6.58hrs IOT/UOG/MECHANICAL ENNGINEERING

Page 24


Now, Q = 10m3 /6.58hr = 1.52 m3/hr = 25.33L/minute = 4.2*10-6m3/s Taking the standard flow rate form table, Q = 27L/min Total dynamic head, TDH = HS + HD +HP +HF Where, Hs = vertical lift, HD = Dynamic head, HF = friction head, HP = pipe head, HS = 8m+7m = 15m and HP = 5m Total pipe length = 15+5 = 20m


Page 25


5.7.

System lay out of the project HTF tanker

Heat

Pumps

transf er

Parabolic troughs

Pipes

Solar anel

Boile

Valve

Water

Condenser

vapor

Controller Salt water tank

Pure water tanker

Ground level

Underground water tanker

7m Submerg ed pump

Fig: 4: System layout of the project.

From standard table of PVC pipes, diameter of pipe should be between 3 /4’’/4’’ -2 1/4’’ or (19-57mm). (19-57mm). A pipe with larger diameter is highly expensive. So we have selected a pipe diameter of 2 ¼’’ IOT/UOG/MECHANICAL IOT/UOG/MECHANICAL ENNGINEERING ENNGINEERING

Page 26


Then we take 50mm diameter of standard value for the system. 

We assume the fitting of each pipe is at 900 bend.



PVC Grade 1 Type V- piping material



Yield strength = 8.35 ksi



Surface roughness, k = 0.1 mm



Pressure rating = 50N /mm2



Maximum allowable pressure, Pmax = 2σ 2σst /do ………………………B ………………………Barlow arlow formula. formula.



Friction loss = 3.9x20/100 = 0.78, Where 3.9 is friction factor of a pipe (from table)

In order to account the losses of bends, valves and other minor losses doubling the friction losses is recommended. So, friction loss = 2x0.78 = 1.56 Finally, TDH = 20 + 1.56 = 21.56m, take TDH = 22m

5.8. Pump selection for underground water According to the above analysis we have selected by comparing the following submersible pump types as follows. Table 3. Centrifugal pump specifications

Centrifugal pumps

Helical rotor pumps

Head = 0 to 80m

Head 50 to 150m

Q = 6 to 20m3 per day

Q >20m3 per day

For the selection of solar pump we use the following standard flow rate versus ve rsus power charts taken from metrological data provided by NASA of LORENTZ solar pump systems, (Appendix A: graph). From the above analysis and graphs a solar pump of 400W is required for salty water pumping purpose.

5.9.

Designing the heat transfer fluid pump

The design is based on the length of piping and number of bends. The total head loss of the pump is calculated by using the following formula. IOT/UOG/MECHANICAL IOT/UOG/MECHANICAL ENNGINEERING ENNGINEERING

Page 27


HL = (f L/D+∑K)V2/2g………………………………………………………(1) Where, f = friction factor. K = loss factor (resistance coefficient) D = diameter of piping material. V = desirable velocity in piping systems = 2.5m/s

Fig.5: PVC pipe

To make the analysis lets use the following standard tables: Table 4. Pressure loss through fittings.

Pipe Outside Diameter

16

25

32

50

63

75

90

110

900 elbow

0.34

0.5

0.65

1.0

1.26

1.5

1.88

2.58

450 elbow

0.16

0.24

0.32

0.52

0.63

0.75

0.95

1.33

900 bend

0.1

0.16

0.22

0.34

0.44

0.56

0.75

1.00

Tee in line flow

0.123

0.19

0.23

0.36

0.45

0.56

0.69

0.95

Tee in line to branch

0.77

1.17

1.47

2.21

2.98

3.68

6.00

0.22

0.31

0.37

0.51

0.80

1.11

1.34

(mm)

flow

Reducer

1.58

Table 5: Pipe friction data for commercial pipes


Page 28


5.10.

Nominal Size (mm)

Fraction factor (f)

15

0.027

20

0.025

25

0.023

32

0.022

40

0.021

50

0.019

65-80

0.018

100

0.017

125

0.016

150

0.015

200-250

0.014

300-400

0.013

450-600

0.012

Design Parameters heat transfer fluid pump:

 Length of trough = 1.3m

pipe for D-450 = 1.3m + 0.5m = 1.8m (we have two troughs)  Length of Cu –  Horizontal length of pipe from heat exchanger = 0.2m  Vertical height of the pipe from bend to the pump = 0.6m  Vertical height from pump to the next pipe = 0.3m  Horizontal length from bend of pipe = 0.5m  Vertical length of pipe from bend to copper pipe = 0.4m  Horizontal length of pipe from bend to heat exchanger = 0.4m  Number of bends = 7

From the above parameters let’s calculate the head loss of the pumping system by using the abov e equation (1) HL= (f L/D +∑K) V2 /2g Where: - HL = head loss f = friction factor IOT/UOG/MECHANICAL ENNGINEERING

Page 29


K = loss factor (resistance coefficient) D = diameter of piping material V = desirable velocity in piping systems = 2.5m/s From the above standard tables; f = 0.019 for 90o bend K = 0.34 for 90o bend D = 50mm G = 9.81m/s2 Total length of pipe = 0.4m + 0.2m +1.8*2m +0.5m + 0.4*2m + 0.9m+1.8m = 8.2m The head loss, HL = [0.019*8.2/0.05+(7*0.34)]*2.52/2*9.81 = 1.75m Then total dynamic head, TDH = 8.2m + 1.75m = 9.95m = 10m Q = A*V; taking the fluid velocity = 2m/s Q = 3.14*0.052/4*2m/s = 3.9*10-3 m3/s So, the pump power requirement, P = ρghQ P = 9.81 m/s2*717.95kg/m3*10m*3.9*10-3m3/s P = 274. 68W 5.11.

Designing Vapor Pump

 Design Parameters  Vertical length of pipe from heat exchanger to the pump = 0.5m  Vertical length of pipe from pump to bend = 0.5m  Horizontal length of pipe from bend to the next pipe = 1.5m  Vertical length of pipe from bend fitting to condenser = 1m  Number of bends of pipe material = 2

From the above parameters, total length of pipe = 0.5m = 0.5m+1.5m+1m = 3.5m So, the loss HL = {0.34*3.5/0.05+(2*0.019)}*2.52/2*9.81 HL = 7.5m Then total dynamic head, TDH = 3.5+7.5 = 11m Flow rate Q = AV Taking the vapor velocity = 0.5m/s (from previous analysis) Q = 3.14 *0.052/4 *0.046 IOT/UOG/MECHANICAL ENNGINEERING

Page 30


Q = 9.8*10-4m3/s Pump power requirement, P = ρghQ P = 9.81*900*11*9.8*10-4 P = 95 W Since all the pumps are driven by solar power. So, by adding the above power requirements; Ptot = (274.68+95+400) W Ptot = 769.68W; take P = 770 Wh After calculating the power requirement of each pump we must to calculate the time of operation for each pump as follows. Since we know that Q = AV, Where, A = area and V = the flow velocity A = 3.14 *0.052 /4 A = 1.96*10-3m2 . Now, the time of operation

 

Q = AV = A

then rearranging this T =

 

Where, S = total dynamic head of the pump T = time of operation Q = flow rate of each fluid

∗ .∗ So, time of submerged pump T = .∗^− = 2.85hr ∗ .∗ Time of vapor pump, T = .∗  = 1.9 hr ∗ .∗ Time of heat transfer fluid pump T, = .∗^− 10 = 1.7 hrs

Fig.6 centrifugal pump with motor


Page 31


5.12. DESIGN OF SOLAR PV SYSTEM FOR THE PUMPS The first step to sizing a solar electric system is to determine the average daily energ y consumption. The average daily energy consumption should be as accurate as possible, and ways to conserve power should be considered as well because the t otal energy consumption will determine the size of the system. Table 6: Solar energy demand for the pumps

Pump type

Power (W )

Hour of operation Daily total load demand per day (hr)

(W hr)

Submerged pump

400

2.85

1140wh

Vapor pump

95

1.9

180.5wh

274.68

1.7

467wh

Heat

transfer

fluid

pump Total = 1787.5wh

Since solar panels are not 100% efficient as a result we h ave to estimate system losses. To account this system loss pv panels should be over sized by at least 25% of the total daily demand of energy. So actual daily demand of energy that a solar pv system can produce = 1787.5wh*1.25 = 2234.375wh For our analysis of PV system take the following parameters  Location of site (Mekelle) = 13.3 N 39.29 E ,elevation = 2212m  PV module of mono-crystalline silicon = 300w p  Operating factor of PV modules 0.6 to 0.9  Battery efficiency = depth o discharge = 80%  Efficiency of panel = 90%  Inverter efficiency (distribution loss) = 98%  Nominal battery voltage = 48V  Battery capacity = 150 Ah  Peak hours of the location = 8 hr (sun light available in a day)

During actual operation times we cannot get the exact amount of energy from a PV module. As a result let’s take the operating factor = 0.8 IOT/UOG/MECHANICAL ENNGINEERING

Page 32


So actual output power of a PV module = 300w*0.8 = 240W

    cu u u w f u∗k su us .ℎ = 0.966 take number of PV modules = 1 = ∗ℎ

The number of PV modules =

N modules 

Battery Sizing

The size of the battery bank required will depend on the storage capacity required, the maximum discharge rate, the maximum charge rate, and the minimum temperature at which the batteries will be used. Battery load =

daily energy demand battery voltage

=

Battery ampere hour requirement =

.ℎ  

= 46.55wh

 ∗   ℎ  ℎ∗ 

Taking days of autonomy for the battery at least 3 days Battery Ah

=

.∗ = 178.125 Ah .∗.

Number of batteries needed =

.ℎ  ℎ  =   () ℎ

= 1.1875

Take number of battery = 2 As a result we select the following battery specifications o

Battery type LP150

o

Voltage = 48v

o

Length = 505mm

o

Width = 221mm

o

Height = 255mm



Charge Controller Sizing

The controller size was determined as follows Current rating =

5.13.

  w ∗ffcc f   s

=

.ℎ∗. ℎ

= 83.79 Amperes.

Array Configurations

As Mekele city is situated at 13.30 N / 39.290 E


Page 33


So we will be setting up the panels at 30 0 N – E angle so that our fixed arrays can give us maximum possible output. 

Approximation of the Required Space

The PV module we have selected has a dimension of 1580 x 808mm.

Fig 8: Solar array configuration

On an average in Mekelle university (Adihaki campus) the Sun ray falls in 600 angle, and we are setting up our PV Modules at 300 angle with the ground. Where, the northern part of the module will be at height. Here, in the above figure the ‘E’ arm is PV module and ‘A’ arm is considered to be the Sun ray. From here we get E = 808mm D = 808Sin30 = 404mm D = ASin60 A=

  

= 466.5mm

B = A*Cos60 = 466.5 x Cos60 = 233.25mm C = E*Cos30 = 808mm x Cos30 = 699.75mm


Page 34


So the total Space required for the Panel in Y-axis is (699.75 +233.25) mm = 933mm and as the Length (X-axis) of the Panel is 1580mm, it will need 1580mm x 933mm Space for each panel to set up, which will be 1474140mm2. So the total space required for 2 modules to setup is around 2948280 mm2 = 2.95m2.

5.14. Design of underground water tank  Assumptions o

Tanker capacity = 20m3

o

Depth of tanker well = 7m

o

Wall thickness = 200mm

o

Concrete class = compressive strength of 20 N/mm2

Tensional strength of concrete is about 10% o f compressive strength, which in this case is 2 N/mm2  Analysis

Water density = 1000 kg/m3 Acceleration due to gravity, g = 9.8N/s2 Maximum water pressure = h ρ g;

where, h = height,

ρ = Density of water and g = Acceleration due to gravity Pmax = 7 x 1000x9.8 = 68.6kN/m2 From the above specifications the diameter D, of the underground tank is calculated as follows.

  V = A*h = *h = 20m3  D2 = 20*4/(7*3.14) D2 = 1.9m (take 2m) Design load (pressure) = Actual load x factor of safety which is given as 1.6 Design load (pressure) is therefore 68.6x 1.6 = 1 09.76kN/m2 Since the tank is cylindrical, the predominant stress on the walls is tensional. Maximum tensional stress at the bottom of the tank is given by σt = pr/2t Where: σt = tensional stress, p = design pressure, IOT/UOG/MECHANICAL ENNGINEERING

Page 35


r = radius of tank and t = thickness of tank

σt = 109.76 x 1 / 2 x 0.2 = 274.4 k N/m2 = 0.2744 N/mm2 Characteristic tensional strength of concrete is 2 N/mm2. Design strength = Characteristic strength divided by material factor of safety. For concrete, is given as 1.5 Therefore, design strength = 2 / 1.5 = 1.33 N/ mm2 = 1.33 N/ mm2 is greater than the tensional stress and therefore adequate. Shear acting at the bottom wall of the tank is given by: τ = pr/ (π d) t / 2

where, τ is the shear stress

τ = 109. 76 x 1 / (π x 1x0 .2/2) = 349.55kN/m2 τ = 0.34955 N/mm2…..say 0.35 N/mm2 Factored Shear strength of concrete is given by 0.8 x square root of its compressive strength up to a maximum of 5 N/mm2. (This shear strength has already been factored using = 1.25. Since the shear strength = 0.8 x square root of 20 = 3.57 N/mm2. Note: The reinforcement is adequate. 5.15.

Design of incoming water tanker

A water tank is used to store water to facilitate the daily requirements of habitats. Circular tanks have minimum surface area when compared to other shapes for a particular capacity of storage required. Hence the quantity of material required for circular water tank is less than required for other shapes. But the form work for a circular tank is very complex and exp ensive when compared to other shapes. Square and Rectangular water tanks are generally used underground or on the ground. Circular tanks are preferred for elevated tank s. Storage tanks are built for storing water, liquid petroleum, petroleum products and similar liquids. Analysis and design of such tanks are independent of chemical nature of product. They are designed as crack free structures to eliminate any leakage. o

Material for the Tanker = Concrete.

o

Volume = 15000LS

o

Tanker capacity = 15.5m3

o

Height of tanker = 4m


Page 36


o

Wall thickness = 200mm

o

Concrete class = compressive strength of 20 N/mm2

o

Tensional strength of concrete is about 10% of compressive strength, which in this case is 2 N/mm2

 Analysis

Water density = 1000 kg/m3. Acceleration due to gravity, g = 9.8N/s2 Maximum water pressure = h ρ g; Where, h = height, ρ = Density of water and g = Acceleration due to gravity Pmax = 4 x 1000x9.8s = 39.2kN/m2 From the above specifications the diameter D, of the underground tank is calculated as follows.   V = A*h = *h = 15.5m3



D2 = 15.5*4/(4*3.14) D2 = 4.936m…take D = 2.5m Design load (pressure) = Actual load x factor of safety which is given as 1.6 Design load (pressure) is therefore 39.2 x 1.6 = 62.72 kN/m2 Since the tank is cylindrical, the predominant stress on the walls is tensional. Maximum tensional stress at the bottom of the tank is given b y; σt = pr/2t. Where, σt = tensional stress, p = design pressure, r = radius of tank and t = thickness of tank = σt = 62.72 x 1.8 / 2 x 0.2 = 112.9 kN/m2 = 0.1129 N/mm2. Characteristic tensional strength of concrete is 2 N/mm2 Design strength = Characteristic strength divided by material factor of safety. For concrete, is given as 1.5 Therefore, design strength = 2 / 1.5 = 1.33 N/ mm2 = 1.33 N/ mm2 is greater than the tensional stress and therefore adequate.


Page 37


Shear acting at the bottom wall of the tank is given by: τ = pd/ (π d) t / 2 where, τ is the shear stress τ = 62.72 x 1.8 / (π x 1.8x0 .2/2) = 199.68 kN/m2 τ = 0.19968 N/mm2…say 0.2 N/mm2. Factored Shear strength of concrete is given by 0.8 x square root of its compressive strength up to a maximum of 5 N/mm2. (This shear strength has already been factored using = 1.25. Since the shear strength = 0.8 x square root of 20 = 3.57 N/mm2). Note: The reinforcement is adequate. 5.16.

Design of helical heat exchanger for water boiling



Basic design procedure and theory



Design Procedure.

Here is a simple procedure for designing an HCHE: Determine the heat-transfer coefficients. To calculate the heat-transfer coefficients in the coil and the annulus, the following parameters must be kn own: 1. The length of coil, L, needed to make N turns:

 ()+

L =N

2. The volume occupied by the coil, Vc:

2

Vc =

3. The volume of the annulus, Va:

Va

= ( /4) (^2  ^2)

4. The volume available for the flow of fluid in the annulus, Vf

Vf = Va-Vc 5. The shell side-equivalent diameter of the coil tube De;

De = 4Vf /

 

The heat-transfer in the annulus, ho, can now be calculated by using following equation For Reynolds numbers, NRe, in the range of 50-10,000. hoDe/k = 0.6NRe0.5 N pr 0.31 The heat transfer coefficient of the fluid based on inside coil diameter hi is taken from R e Vs jh chart. IOT/UOG/MECHANICAL ENNGINEERING

Page 38


Then heat transfer coefficient inside the coiled tub e based on inside diameter, hic hic = hi [1+3.5(D/Dh) ]

Then the heat transfer coefficient inside the coil b ased on outside diameter, hio hio = hic (D/do) The overall heat transfer coefficient, U 1/U =1/ho+1/hio+t/kc+Rt+Ra Where- t = the thickness of the coil wall Kc = the thermal conductivity of the coil R a = sell side fouling factor R t = tube side fouling factor The general equation for heat transfer across a surface is: Q = UA ΔTm Where: Q = heat transferred per unit time, W,(heat load) U = the overall heat transfer coefficient, W/m2°C, A = heat-transfer area, m2, ΔTm = the mean temperature difference, the temperature driving force, °C. ΔTm = FtΔTlm Where: ΔTlm true temperature difference, the mean temperature difference for use in the design equation. Ft = the temperature correction factor. The correction factor is a function of the shell and tube fluid temperatures, and the number of tube and shell passes. It is normally correlated as a fun ction of two dimensionless temperature ratios: R = T1-T2 /(t2-t1) Where; R is equal to the shell-side fluid flow-rate times the fluid mean specific heat; divided by the tube-side fluid flow-rate times the tube-side fluid specific heat. S = t2- t1/(T1-T2) Where: S = a measure of the temperature efficiency of the exchanger. For a 1 shell: 2 tube pass exchanger, the correction factor is given by:


Page 39


And

Where, Tlm = log mean temperature difference, T1 = inlet shell-side fluid temperature, T2 = outlet shell-side fluid temperature, t1= inlet tube-side temperature, t2= outlet tube-side temperature. The following assumptions are made in the derivation of the temperature correction

Factor Ft , in addition to those made for the calculation of the log mean temperature difference: 1. Equal heat transfer areas in each pass. 2. A constant overall heat-transfer coefficient in each pass. 3. The temperature of the shell-side fluid in any pass is constant across any cross section. 4. There is no leakage of fluid between shell passes.

An economic exchanger design cannot normally be achieved if the correction factor F , falls below about 0.75. In these circumstances an alternative type of exchanger should be considered which gives a closer approach to true counter-current flow. The use of two or more shells in series, or multiple shell-side passes, will give a closer approach to true Counter-current flow, and should be considered where a temperature cross is likely to o ccur. Where both sensible and latent heat is transferred, it will be necessary to divide the temperature profile into sections and calculate the mean temperature difference for each section. 5.17.

GEOMETRY OF HELICAL COILS


Page 40


Fig.9 Helical coil

Fig. 10 Coil in boiler

The prime objective in the design of an exchanger is to determine the surface area required for the specified duty (rate of heat transfer) using the temperature differences available. The overall coefficient is the reciprocal of the overall resistance to heat transfer, which is the sum of several IOT/UOG/MECHANICAL ENNGINEERING

Page 41


individual resistances. For heat exchange across a typical heat exchanger tube the relationship between the overall coefficient and the individual coefficients, which are the reciprocals of the individual resistances, is given by:

Where: Uo = the overall coefficient based on the outside area of the tube, W/m2°C, ho = outside fluid film coefficient, W/m2°C, hi = inside fluid film coefficient, W/m2°C, hod = outside dirt coefficient (fouling factor), W/m2°C, hid = inside dirt coefficient, W/m2°C, kw = thermal conductivity of the tube wall material, W/m°C, dt = tube inside diameter, m, do = tube outside diameter, m. The magnitude of the individual coefficients will depend on the n ature of the heat transfer process (conduction, convection, condensation, boiling or radiation), on the physical properties of the fluids, on the fluid flow-rates, and on the physical arrangement of the heat-transfer surface. 5.18. 

Parameters of helical coils Dimensions

Tube diameters in the range | in. (16 mm) to 2 in. (50 mm) are used. The smaller diameters | to 1 in. (16 to 25 mm) are preferred for most duties, as they will give more compact, and therefore cheaper, exchangers. Larger tubes are easier to clean b y mechanical methods and would be selected for heavily fouling fluids. The tube thickness (gauge) is selected to withstand the internal pressure and give an adequate corrosion allowance. The British standard BS 3274 covers exchangers from 6 in. (150 mm) to 42 in. (1067 mm) diameter; and the TEMA standards, exchangers up to 60 in, (1520 mm). Up to about 24 in. (610 mm) shells are normally constructed from standard, close tolerance, pipe; above 24 in. (610 mm) they are rolled from plate. Table 8. Standard dimensions for steel tube.


Page 42


Outside diameter (mm)

Wall thickness (mm)

16

1.2

1.6

2.0

---

---

20

---

1.6

2.0

2.6

---

25

---

1.6

2.0

2.6

3.2

30

---

1.6

2.0

2.6

3.2

38

---

2.0

2.6

3.2

50

---

2.0

2.6

3.2

According to the above table and standards we specify the following dimensions. 

DO =1000mm



Thickness of boiler (heat exchanger cylinder) =4mm



B =340mm



C = 460mm



LC =1500mm



Dh = 400mm



do = 30mm





p = 0.5do = 15mm D = 25mm

Where; DO = Outer diameter of the cylindrical heat exchanger(boiler) LC = length of heat exchanger cylinder B = outside diameter of inner cylinder C = inside diameter of outer cylinder D = inside diameter of coil Dh = mean diameter of the helix do = outside diameter of coil P = pitch 

Specifications of Duratherm -450

o

Inlet temperature = 2320c

o

Required outlet temperature = 1000c


Page 43


o

Heat capacity, c p = 2.721KJ/kg .k

o

Thermal conductivity = 0.128w/m .k

o

Viscosity = 0.43centipoise = 0.00043Ns/m2 = 1.55kg /m hr,

o

Density = 717.95kg/m3

o

Vapor pressure = 22.27kpa

o

Mass flow rate = 2160 kg/hr

All the above values are taken from table of Appendix C at the inlet temperature of duratherm450. 

Inlet shell side fluid temperature = 250C of salty water



Out let shell side fluid temperature = 1000C of water vapor



In let tube side temperature = 2320C



Out let tube side temperature = 2320C-1000C =1320C

Calculate the shell side heat transfer coefficient h0

√(2(0.2)2  (0.0045)2⁄

Since the length of coil needed, L = N

=1.257N

Where, N = number of turns The volume available for fluid flow in the annulus, Vf, = Va-Vc

= /4 ∗ L -/4(22)

Vf = 2.504*10-3 N The sheel side equivalent diametre De = 4Vf/

d0L = 4*2,504*10 N/∗0.03∗1.257 -3

De = 0.0845m. The mass velocity of the fluid ;V m = M’/

[(/4)((C – B ) – (D 2

2

h2

2

– Dh12)) ]

where , Dh1 = inside diametre of helix = 0.43m Dh2 = out side diametre of helix = 0.37m M’= mas flow rate of the fluid (kg/ hr ) Vm = (2160) / [3.14/4((0.46^2 -0.34^2) – (0.43^2 – 0.37^2)) ] Vm = 57325 kg/m2hr Ren = De*Vm /µ = 0.0845*57325 / 1.55 = 3125 Then calculate the ho and hio values as follows For reynold’s number R e,in the range of 50 -10000 is ho = k*0.6*R e0.5*N pr 0.31 IOT/UOG/MECHANICAL ENNGINEERING

Page 44


where ,N pr = prandtl number, the Prandtl number can range from 0.3 for cooling to 0.4 for heating. Then take 0.4. ho = heat transfer coefficient outside the coil ho = 0.6 *0.128*31250.5*0.40.31 =38.24 J /hr m2 k hio = hic*(D /do) Where, hio = heat transfer coefficient inside the coil ba sed on outside diameter of the coil hic = heat transfer coefficient inside coiled tube based on inside diameter the fluid velocity is V = q/Af Where Af = cross section area of coil

 = D/4 = 3.14 *0.025 /4 = 4.909*10^-4 m 2

2

q = the volumetric flow rate of the fluid = M /density = 2160 /717.95 = 3.01m3/hr Then, V = 3.01/4.909*10^-4 = 6131.6m/hr = 2m/s The Reynolds number (tube side), Re = D *V *density /viscosity = 0.025*6131*717.95 /1.55 Re = 70995.9 From standard chart jH vs R e, for a Reynold’s num ber of the above value we get culburn factor for heat transfer jH, = 220 hi = heat transfer coefficient inside a straight tube based on inside diameter Then hi = jH (K /D)(N pr )1/3 = 220(0.128 /0.025 ) (0.4)1/3 = 150.2w /h m2 k Now, hic = hi [1+3.5(D /DH)] = 150.2 [1+3.5(0.025 /0.4) ] = 183 .1W/h m2 k Thus, hio = hic * (D /do) = 183.1 *(0.025 /0.03) = 152.6 w/h m2 k Calculate the overall heat transfer coefficient U, The coil wall thickness, t is = (d o – D ) /2 = ( 0.03-0.025) /2 = 0.0025m Then, 1/U = 1/ho+1 / hio+ t/ k c+ R t +R a Where, R t and R a are the fouling factors depend on the nature of the liquid. In this case R t and R a are 0.00082 h m2 0c /kcal = 0.0007m2k/w  The thermal conductivity of stain less steel is 14kcal /h m

0

c = 16.28w/m k

,then

substituting the values 1 /U = 1 /38.24 + 1/152.6 + 0.0025/16.28 + (0.0007) *2 = 0.0342W/h m2 k U = 29.33 W/ h m2 k And we have to determine the required area, A, from the previous formula; The log mean temperature difference , ΔTLM = (T1 – t2) – ( T2 – t1) / ln [ (T1 – t2) /(T2 – t1)] IOT/UOG/MECHANICAL ENNGINEERING

Page 45


Where, T1 = inlet shell side fluid temperature T2 = out let shell side fluid temperature t1 = in let tube side temperature t2 = out let tube side temperature, by substituting the numerical values ΔTLM = (25 -100) – (100 -232 )/ ln ((25-100)/100 - 232 )) = 100.880 C To account for perpendicular flow use the correction factor 0.99, so the corrected log mean temperature difference, Δtc = 0.99 *100.88 = 99.87120C = 372.87 K Since liquid Duratherm - 450 is flowing inside a helical coil, so the mean temperature difference, Δ Tm = 232 0 C -1000 C = 132 0 C = 405 k Now the Heat load per a unit area, Q = AU Δ Tm = (1) (29.33) (405) = 11.88Kw/hr Then the required area A, = Q/U* Δtc = 11.88 *1000 /29*372.87 = 1.0986m2

( )

Calculating the number of turns of coil required N = A /

Since L = 1.257N

N = 1.0986/(3.14*0.03*1.257) = 9.27, (take N = 10) Therefore, L = 9.27 *1.257 = 12m. Where, L = length of coil. The height of the cylinder needed to accommodate 10 turns of coils is, H = n p +do H = 10*0.15m +0.03m = 1.53m 5.19. 

Design of condenser Design Parameters of Condenser  Mass flow rate of water vapor = 3kg/sec = 10800kg/hr  Pressure of condenser = 5bar  Inlet saturated temperature to condenser, Tsat = 100oC  Condensation temperature, Tcond = 4oC  Average molecular mass of water vapor = 18kg/mol  Enthalpy of water vapor = 2668KJ/kg  Enthalpy of condensate = 16.78KJ/kg o  Available cooling water temperature = 4 C

 Condensation completed temperature, T = 10oC  Limited rise temperature, T = 6oC

From previous standard tables of heat exchanger d esign, let us take the following tube dimensions.  Material type = Stainless steel.


Page 46

DESIGN OF SOLAR POWERED WATER PURIFICATION SYSTEM  Outer diameter of cylinder (D) = 500mm.  Inner diameter of cylinder = 1.058* D mm  Thickness of cylinder , t = 3 to 5mm  Outer diameter of tube = 0.211*D mm  Inner diameter of tube = 0.145*D mm  Diameter of connecting tube = 0.2*D mm  Diameter of connecting PVC pipe = 0.3*D mm  Length of tube = 3000mm

According to the above design parameter of the condenser; Heat transfer from the vapor, Q, = m’* Δh = 10800/3600 (2668-16.78) = 7954kW Cooling water flow rate, m = Q/( ΔTCpH2O) = 7954 KW/(100oC-4oC)*4.2KJ/kgo m = 19.7kg/s From standard table of overall coefficients of condensers, U = 1000 to 1500 W/m2 oC, Taking, U = 1200W/m2 oC Calculate the range and saturation temperatures of con densers as follows:

− − − And, S = − R=

Where ,Tlm = log mean temperature difference, T1 = inlet shell-side fluid temperature, T2 = outlet shell-side fluid temperature, t1 = inlet tube-side temperature, t2 = outlet tube-side temperature. R = 100oC-10oC/(10oC-4oC) = 15 And; S = 10-4/(100-4) = 0.06 The log mean temperature difference, ΔTLm = (100-10)-(10-4) / ln((100-10 /10-4)) ΔTLm = 33.23oC And mean temperature difference, ΔTm = Ft*ΔTLm Where Ft = temperature correction factor; take Ft = 0.99 for vertical coil tubes. So, ΔTm = 0.99*33.23 = 32.9oC IOT/UOG/MECHANICAL ENNGINEERING

Page 47


Trial area, AT = Q/(U* ΔTm) = 7954*1000/(1200*32.9) = 201.5m2 Surface area of one tube, Asur = 100 * 10-3*3.14*3 = 0.942m2 So number tubes = 201.5/0.942 = 213.9; take 214 Pitch, P = 1.25*Dtubes = 1.25*100 = 125mm Tube bundle diameter , D b = 100 (214/0.158)1/2.263 = 2387.6mm Number of tubes in center of row, Nr = 2387.6/125 = 19 

Shell side heat transfer coefficient

Mean temperature of shell side and tube side is calculated as follows: Shell Side = (100+10) /2 = 55oC And; Tube Side = (10+4) /2 = 7oC Estimate tube wall temperature, Tw; assume condensing coefficient of 1500 W/m2°C, (55-Tw)*1500 = (55-7)*1200; Tw = 16.6oC Mean temperature of condensate = (55+16.6) /2 = 35.8oC From thermo table heat transfer coefficient hc at 35.8 0C = 1480 w/m2 0C closer enough to the assumed value. 

Tube-side coefficient

Tube Side Cross sectional area 3.14/4 (72.5*10-3)2*214/4 = 0.22m2 The number 4 accounts to the number of passes. Density of water at 7oC = 1000kg/m3 Velocity of tube = 19.7/1000*(1/0.22) = 0.09m/s By taking into consideration fouling factor from standard table of cooling water 3000 to 6000W/m2 o

C

Taking 5000W/m2oC for our condenser, the heat transfer coefficient of the inside tube; hi = [5000(1.35+0.02*7)0.090.8]/( 72.50.8) = 3.526kw /m2 0C Fouling factors: as neither fluid is heavily fouling, use 6000 W/m2°C for each side, kw = 50 W/m°C 

Overall coefficient

1/U =1/1480+1/6000+ [100*10-3ln(100/72.5) /2*50 ] +100/72.5*1/6000+100/72.5*1/3526 1/U =1168.843W/m2 oC This is highly closer to the estimated value, therefore the d esign is safe. IOT/UOG/MECHANICAL ENNGINEERING

Page 48

DESIGN OF SOLAR POWERED WATER PURIFICATION SYSTEM 

Pressure drop analysis Shell-side pressure drop

Shell inside diameter = 530mm Clearance = 70mm Then Shell inside diameter = 70+530 = 600mm Cross flow area, As = P-Do*Ds2/P =(125-100)(*6002*10-6) /125 = 0.072m2 Mass flow rate based on inlet conditions, m = 10800/3600*1/0.072 = 41.67kg/sm2 Equivalent diameter, De = 1.27(1252-0.785*1002)/100 = 98.74mm Vapor viscosity = 0.008*10-3 Ns/m2 Then Reynolds Number, R e = m*De*10-3/(0.008*10-3) = 41.67*98.74/(0.008) R e = 514311.97 From chart of Re Vs jf (friction factor); jf = 1.8*10 -3 Fluid velocity = mass flow rate/density of vapor = 41.67/900 = 0.504m/s Take pressure drop as 50 per cent of that calculated using the inlet flow From the graph, a factor k = 3 ΔP = 1/2 [k*jf*(600/De)(L/Do)*900(0.046)2] =1/2 [3*1.8*10-3(600/98.74)(3/0.6)*900(0.046)2/2] ΔP = 0.078N/m2 Negligible pressure drop, because the pressure of the condenser was 5bar. That means the pressure drop is very much less than 50% of condenser pressure, so the analysis is safe. 

Tube-side pressure drop

Viscosity of water = 0.6*10-3 Ns/m2 Then, Re = velocity of the tube*Density of vapor *Dit /u Re = 0.09*900*72.5*10-3/(0.6*10-3) Re = 9787.5 Then from chart friction factor, jf = 3.5*10^-3 ΔPtube = 4[3*3.5*10-3(3/72.5*10-3)+2.5]*(900*0.092/2) ΔPtube = 64.5894N/m2…it is an acceptable. 5.20. 

Design of pure water tanker General Design Consideration

 Design pressure


Page 49


A vessel must be designed to withstand the maximum pressure to which it is likel y to be subjected in operation.For vessels under internal pressure, the design pressure is normally taken as the pressure at which the relief device is set. This will normally be 5 to 10 per cent above the normal working pressure, to avoid spurious operation during minor process upsets. When deciding the design pressure, PD:

p D



po



po 

10 100

where, po is the operating pressure of pure water tanker and pipes

PD = 50N/mm2 +50*0.1 PD = 55 N/mm 2 

Design temperature

The strength of metals decreases with increasing temperature so the maximum allowable design stress will depend on the material temperature. The design temperature at which the design stress is evaluated should be taken as the maximum working temperature of the material, with due allowance for any uncertainty involved in predicting vessel wall temperatures. Then

0

T

O



15 C

From typical design stress table find the stress for low alloy steel:Design stress: -  D



Tensile strength: - 

t

240 N/mm



2

550 N/mm

2

For cylindrical shell thickness required to resist internal pressure can be determined from the formula. Now assume outer diameter of the shell is 1 m. Then check for which option that our pressure is safe. Assume thickness of shell = 3mm. For carbon and low alloy steel corrosion allowance is

2mm should

be used

Therefore, T = 3mm+2mm = 5mm Taking the length of the shell = 1m


Page 50


L/D = K for P > 3.34 Mpa; Where, K = cons….4 < K < 6 But for economic purpose select

K



4

Then, L = D To calculate the volume of the Tanker V=

π

(−) 

, But Di =Do-2t Di =990mm

V

 −)∗ ( = 

= 15.6m3. So the design of our tanker is safe, because

15.6m3 > 10m3 from previous assumption.

Fig: 11. Pure water tanker


Page 51


5.21.

Design of Head for pure water tanker

All water tanker must be closed at the end by heads. The ends of cylindrical tankers are closed by head various shapes. This are:o

Flat plates head

o

Hemispherical head

o

Ellipsoidal head etc

But our design head is ellipsoidal head. Most standard ellipsoidal are manufactured with a major and minor axis ratio of

2 : 1 the

following equation can be calculated required thickness.

Assume: o

Thickness, T = 3mm

o

Considering corrosion allowance = 2mm

So, thickness of head = 5mm To calculate ellipsoidal head volume; Taking height of head = 50mm

 ∗  Now volume of head, V =  3.14∗12∗ 0.05 = 0.026m V = 6 h

3

h



Calculate the stress on the water tanker using lame’s equation o

Tangential stress: - is the maximum tensile stress and it is known as

circumferential stress. o

Radial stress : - is the maximum compressive stress and it is negative.

The negative sign indicates that the radial stress is opposite to design stress equal. Since, if the stress is less than the maximum tensile strength of the material, then the design is safe. Now, to calculate the value of stress; IOT/UOG/MECHANICAL ENNGINEERING

Page 52


a) Tangential stress

 ∗()  = ()−()

  [1 ] =233.06N/mm

2

b) Radial stress

∗^ =  −^

  [1  ] = - 49.5N/mm 2

compressive

c. Longitudinal stress

 ∗  =  −

= 67.72N/mm2

5.22. Design of support for the Boiler  

Total weight of the pressure vessel (dead weight)

The major sources of dead weight loads are: 1. The vessel shell. 2. The vessel fittings: manways, nozzles. 3. Internal fittings: plates (plus the fluid on the plates); heating and cooling coils. 4. External fittings: ladders, platforms, piping. IOT/UOG/MECHANICAL ENNGINEERING

Page 53


5. Auxiliary equipment which is not self-supported; condensers, agitators. 6. Insulation. 7. The weight of liquid to fill the vessel.

For preliminary calculations the approximate weight of a cylindrical vessel with domed ends, and uniform wall thickness, can be estimated from the following equation of steel vessels: Ws = 240*Cv*Dm*(Hv+0.8Dm)t Where; Ws = total weight of the shell, excluding internal fittings, such as plates, Cv = a factor to account for the weight of nozzles, man ways, internal supports, etc; which can be taken as: Cv



1.08 for

Cv



1.15 For

vessels with only a few internal fittings, distillation columns, or similar vessels, with several man ways,

Hv = Height, or length, between tangent lines (the length of the c ylindrical section) t = Wall thickness, mm Dm = Mean diameter of vessel Dm = (Di+t*10-3)m Since, DO of heat exchanger = 1000mm (from previous analysis) H = 1.5m (from the previous analysis) Di = DO-2t = 992mm Dm = 0.992m+0.004m = 0.996m Thus, WC = 240*1.08*0.996m (1.5m+0.8*0.996m)*0.004m Wc = 2.372N To find the weight of fluid which score the maximum weight; The density of water = 1000kg/m3 WF = mf g = vf ρf g, Where v f



 f

the volume of fluid in the cylinder = 15m3 (from assumption)



the density of water .

Thus, WF = 15m3*1000kg/m3*9.81m/s2 WF = 147150N To find the total weight of the system will be: W = Wc +WF = 2.372+147150 IOT/UOG/MECHANICAL ENNGINEERING

Page 54


W =147152.372N We choose round bar as support that is skirt support is preferable to vertical position. The three skirt support is welded at 1200 with the cylindrical part of the cylinder W total

Therefore weight each support carries

Let

W each

P





W total 3

3

=

of load.

.K =49.05KN 

The length of weld part (x) is subject to pure shear and the bar weld at two parts.

Fig. 12: boiler and its supports analysis

p  2  0.707  S   allo  x

Where S  weld thickness allo 



allowable shear stress

p  weight of each load


Page 55




y

allo 



n  factor of safty

3n

Where

n  2.3  y

Take



 t

in the standard table = 550Mpa

 

allo



y

3 n



550 Mpa 3  2.3

Mpa

 138.06

S 10mm

Take

Now, X =

49.05∗1000N 2∗0.707∗10mm∗138.06Mpa

= 25mm

Adding in 10.2mm starting and stopping welding X = 10.2mm+25mm = 35.2mm 

Then find the diameter of support

The diameter of a support is determined by buckling consideration 2

P e 

 EI

N 2

Where pe  bulcking load eulerian load  E  elastic

mod ules

of carbon steel

N  length from the ground I  2nd moment of inertia Pe = n*p Pe = 2.3*49.05*1000N = 112815N Take N = 2mm


Page 56


2

I 

pe  N

2   E

But

I 

 D

4

64

E = 70Mpa D4 = 64Pe N2/3.143*E =

∗∗ .∗

= 10.74mm

Take D = 12mm

Boile

Pipe

Pure water tanker

Salt water tanker Support

Plate Support Fig. 13: Assembly of salt water tanker, condenser, boiler and pure water tanker.

5.23. Environmental Impact Water distillation requires a large quantity of energy to take place. This energy can be supplied from a variety of sources, such as electricity and fossil fuels. However, burning fossil fuels result in large quantities of carbon dioxide emissions to the atmosphere. To minimize the environmental impact of the water distillation system, solar energy, a renewable clean source of energy, was used. Solar parabolic troughs are used to collect heat from the sun by reflecting its rays onto a central focal length, and transferring its heat to a non-toxic HTF. PV panels are used to produce the electricity necessary for operation of the system’s electrical components. IOT/UOG/MECHANICAL ENNGINEERING

The total power Page 57


consumption of the system is 770 W when all of the components are in use. The vapor pump, submerged pump, and HTF pump consume 95W, 400W and 274.68W, respectively.

By

implementing a control system, these electrical components can be turned on and off as needed to minimize the power consumption and size of the PV panel. A 300-W mono crystalline PV panel is used to power the entire system, making it completely off-grid. In addition to having no carbon dioxide emissions, the Solar-Powered Water Purification System produces very little noise.

CHAPTER SIX 6.

COST ANALYSIS

6.1. Cost of PVC piping material for underground water

Diameter, D = 50mm Density of PVC = 1400kg/m3 Area, A = 3.14*D2/4 = 3.14*0.052/4 = 1.96*10-3m2 V = A*L = 1.96 *10-3 *20mm = 0.0392m3 Mass, m = density *Volume = 1400*0.0392 = 54.88kg Unit cost of PVC pipe = 10ETB/kg Now cost = mass*unit cost /kg Cost = 54.88 * 10 = 548.8 ETB 6.2.

Cost of PVC piping for condenser D pipe = 150 mm Area, A = 3.14*0.152/4 = 0.0176m2 Length of pipe L p 3.5m V = A*L = 0.0176*3.5 = 0.062M3 Mass , m = 1400*0.062 = 86.6kg Now cost = mass*unit cost /kg Cost = 86.6 *10 = 866ETB

6.3. Cost of copper piping for HTF


Page 58


Density of copper = 8960kg/m3 Length of pipe LP = 8.2m D = 0.05m Area*D2/4 = 1.96*10-3m2 V A*L = 1.96*10-3 * 8.2 = 0.016m3 m = Density *V = 143.4kg Now, cost = mass*unit cost /kg Unit cost = 20ETB/kg Cost = 20*143.4 = 2868 ETB 6.4. Cost of water cooled condenser

Material density = 7800kg/m3 (stain less steel) D = 1000mm Height, h = 700mm V = 3.14/4[Do – Di]h = 3.14/4[12-0.992]0.7 = 0.01094m3 m = density* V = 7800*0.01094 = 85. 3kg Unit cost of stainless steel = 30ETB/kg Now cost = mass*unit cost /kg Cost = 85.3*30 = 2559ETB 6.5. Cos of heat exchanger (boiler)

Material = Stain lees steel Density of Steel = 7800kg/m3 D = 1m H = 1.5m V = 3.14/4(1^2 -0.99^2)*1.5 = 0.0234m3 Mass = 7800kg/m3 *0.0234 = 182.52kg Now cost = mass*unit cost /kg Cost = 182.5*30 = 5475ETB


Page 59


6.6. Cost of pure water tanker

Material = carbon steel D = 1m H =1m Density of C steel = 7850 kg /m3 V=

.  (Do^2 – Di^2)*h

=

. 3  (1-0.992^2)*1 = 0.0125m

Mass = 7850*0.0125 = 98.25kg Cost = mass*unit cost /kg Unit cost = 15ETB/kg Cost = 98.25*15 = 1473.7ETB 6.7. Business plan

Business plan is document which spells out the goals and objective of a business and clearly outlines how and when the project will be achieved. 6.7.1 •

Brake even analysis Break-even analysis is a tool to determine the level of production/ sales at which the business will cover both fixed and variable costs

•

It indicates the minimum amount of revenue that a business must earn in order to cover the total cost incurred so that it does not incur any loss.

The break-even analysis is a useful tool in that it provides answers to the following questions: 1. Does it make sense at all to engage in a business? Is the expected profit stable and big

enough to allow for unforeseen risks, and for d rawings for you as owner? 2.

At what level of production will the business be able to cover all its costs? What is the minimum price required for the product to be viable at different levels of production?

3. What happens if financial assumptions of costs or prices are changing? What are the

best, worst and probable scenarios of the project?


Page 60


Table 9: General cost list of components of the project.

Component name list

Quantity

Unitcost (ETB)

Total cost

Boiler

1

5475

5475

Condenser

1

2559

2559

Parabolic trough

2

1750

3500

Absorbersurface coating

1

500

500

Salt water storage tank

1

10000

10000

Clean water storage tank

1

1474

1474

Submerged pump

1

2000

2000

Vapor pump

1

1500

5000

HTF pump

1

3000

3000

Copper Pipe

1

2868

2868

Wires

3

25

75

Sensors

3

2000

6000

1

2400

2400

1

1000

1000

50

200

Fixed cost

300w PV panel Thermocouple Trough

and

panel 4

support Drive motor

1

8000

8000

Trough and panel mount

2

30

60

Switches

2

20

40

Duratherm – 450 Fluid

2

940

1880

PVC pipes

2

548+866

1414


Page 61


Boiler support

3

Maintenance

60

180

1000

1000

Float switches

2

500

1000

Guard salary

1

700

700

Admin cost

1

2000

2000

3000

3000

250

500

Installation cost Flow control valve

2

Total fixed cost

65

825

ETB Variable cost

Labor

(digging,

10000

10000

Transportation

2000

2000

Delivery cost

500

500

masonry)

Total fixed cost

12 500 ETB

 Maximum Capacity= 3 650 000 Liters per year  Total Fixed Cost = 65 825 ETB  Sales Price = 1 ETB  Variable Cost = 0.8 ETB  Contribution =Sales Price - Variable Cost = 1-0.8 =0.2

Volume of Production At BEP =

  

= 329 125units

You need to sell 329 125 units to break even IOT/UOG/MECHANICAL ENNGINEERING

Page 62


Sales revenue = Production units X Sales price

= 329125 X 1ETB= 329 125 ETB

   ∗  ()

OR

= 329 125 ETB

 The result can also be calculated based on the equation that at break even sales revenue

and all costs (fixed and variable) are the same, i.e. Capacity Utilization =

       ∗ *100 =     

= 9.017 %



This is the most popular method of expressing break-even.



Margin of safety =100-9.017 = 90.98 % the higher the margin of safety the higher is our profit beyond the break even.



If you produce more, you make a profit. It is a quick indicator for steering your business. Revenue

Total sales

(ETB)

(ETB)

329 625

PROFIT

Total cost (ETB)

BE POINT 329 125

FIXED COST

LOSS

10000 L

20000 LITERS

329 125 Liters


658 250 Liters

Output sales (units)

Page 63


Fig.14 break even analysis of the project

Yearly cash flow statement of the project Table.10 Yearly cash flow of the project

Particulars

Pre operation period

Cash at the

Year (E.C) 2009

2010

58,175

152,350

2011

2012

2013

2014

beginning of the year Cash in flow

Equity

30,000

Borrowing

100,000

Cash sales Total cash in flow

182,500 130,000

240,675

Cash out flow

Investment capital

65,825

Pre operation cost

6000

Operational cost

0

78,325

Tax

0

10,000

Total cash out flow

71,825

88,325

Cash at the end o

58,175

152,350

(fixed+ variable)


Page 64


CHAPTER SEVEN 7. RESULT AND DISCUSSION A successful water treatment approach requires a total s ystem approach to: 

Prepare the water before it goes to the boiler



Maximize the potential of condensate



Provide internal and external protection of the components and



Avoid shut down problem of the system

Even though there are a number of water treatment techniques such as: Osmosis, softening, filtration, and disinfection, the solar powered water purification system for drinking purpose has greater purification ability longer life span and better solar resource utilization project. The system has two inputs: salt water and solar energ y (in both thermal and electrical form). The source of the salt water was assumed to be 35,000 ppm, which is a typical salinity level for the underground water. It enters the system from the saltwater storage tank to be boiled. Solar energy is provided through the sun, and was assumed to be 850 W/m2. By utilizing the sun for thermal and electrical power, the system’s energy requirements are fulfilled and allow for a completely sustainable design. The system has two outputs: clean water, and thermal energy. The clean water is obtained through the condensation o f water vapor through the condenser. There is an abundance of thermal energy carried within the water vapor as it leaves the boiler. Therefore, by placing the condenser in the salt water storage tank, a preheating process can occur. The heat is transferred from the vapor to the salt water before it enters the boiler. Although some thermal energy is lost to the environment, this was minimized using insulation.


Page 65


Table: 11. Result and discussion list of component dimensions.

Part name Underground

Diameter(mm)

Length(mm)

Thickness(mm)

Height(mm)

2000

200

7000

3600

200

4000

water tanker Salt water tanker PVC pipe

50

8200

5

Copper pipe

150

3500

4

Parabolic

400

1300

4

Battery

505

255

PV panel

1580

5

trough

Condenser

1000

1500

5

Boiler

1000

1500

4

Pure water

1000

1000

5

808

tanker HTF tanker

500


3

550

Page 66


CHAPTER EIGHT Conclusion and Recommendation  Conclusion

The Solar-Powered Water Purification System is designed with the parameters of being sustainable, portable, cost effective, easy to use, an d scalable. These parameters were achieved b y using solar troughs, which are cost effective, scalable, and can reach boiling temperatures for the distillation process, which is a sustainable and cost effective way to produce clean water. A control system was also used and powered by a PV panel to make the system easy to use.  Recommendation

Finally, when we compare the efficiency of solar powered water purification system with other systems for drinking purpose, it is better due to its cost saving ability, environmental friendly, longer life span, renewable resource utilization and for perfect water purification of drinking purpose of any areas having salty water problems our project is a vital solution and we suggest those areas to apply our project.


Page 67


PART DRAWINGS


Page 68



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Page 79


Part number 12 11 10 9 8 7 6 5

4 3 2 1

Part name

Quantity

Dimension

9 1 1 1

Material specification Steel Concrete Stainless steel Carbon steel

Elbow Foundation Boiler Pure water tanker Condenser Copper pipe Parabolic trough Panel & trough supports Solar panel Motor with pump PVC pipe Underground water tanker

1 1 2

Stainless steel Copper Aluminum

(D1000*H1500)mm D150mm

Remark

D6*,90O T 5mm (D1000*H1500)mm (D4000*H4000)mm

(D1300*L1500)mm 4

Carbon steel

D50mm

1 1

Silicon

(L505*W221*H255)mm

2 1

PVC concrete

(D50*L1100)mm (D3600*H4000*T200)mm

REFERENCES IOT/UOG/MECHANICAL ENNGINEERING

Page 80


1. Allen Barnett et al, “Milestone toward 50% efficient solar cell modules”, presented in 22rd European Photovoltaic Solar Energy Conference, Milan, Italy, 2007 2. Bruce Anderson , the solar home book-heating, cooling and designing with the sun. 3. Dunn, PD, Heffers Renewable Energy Sources, Conversion and Application, Peter Peregrinus Publishing Ltd, Cambridge, 1986. 4. Jemes M.Gere, strength of materials 5. Jan. F.Kreider and Frank kreith , solar heating and cooling second edition. 6. W. H. Bross, Advances in Solar Technology Volume 1, Pergamon Press publishing, 1987. 7. www. Heat transfer-fluid.com

APPENDIXES Appendix A: Flow rate Vs head chart for the selection of pump power. IOT/UOG/MECHANICAL ENNGINEERING

Page 81

DESIGN OF SOLAR POWERED WATER PURIFICATION SYSTEM (BERNTLORENTZ GmbH & Co K, 2015).

22m

Appendix B: Properties of duratherm-450 Table 1. Property Vs Temperature Chart


Page 82

DESIGN OF SOLAR POWERED WATER PURIFICATION SYSTEM Temperature

Density

Kinematic

Dynamic

Thermal

Heat

Vapor

(Celsius)

(kg/m3)

Viscosity

Viscosity

Conductivity

Capacity

Pressure

(centistokes)

(centipoises)

(W/m.k )

(kJ/kg.k )

(kpa )

-12

884.54

40.50

35.82

0.145

1.985

0.00

-7

880.85

28.70

25.26

0.145

2.001

0.00

-1

877.01

21.20

18.59

0.145

2.018

0.00

4

873.17

16.20

14.14

0.144

2.035

0.00

10

869.48

12.70

11.04

0.144

2.052

0.28

16

865.64

10.20

8.83

0.144

2.068

0.34

27

858.11

7.00

6.01

0.142

2.102

0.48

32

854.26

5.90

5.04

0.142

2.119

0.55

38

850.58

5.10

4.34

0.142

2.135

0.62

43

846.74

4.40

2.73

0.142

2.135

0.68

49

842.89

3.90

3.29

0.142

2.169

0.74

54

839.21

3.50

2.94

0.140

2.186

0.80

66

831.52

2.80

2.33

0.140

2.219

0.96

71

827.83

2.50

2.07

0.140

2.235

1.04

77

823.99

2.30

1.90

0.138

2.252

1.14

82

823.99

2.10

1.72

0.138

2.269

1.24

88

820.30

2.00

1.63

0.138

2.286

1.35

93

816.46

2.80

1.46

0.138

2.303

1.48

99

812.68

2.70

1.36

0.137

2.319

1.61

104

800.92

2.70

1.29

0.137

2.336

1.76

116

801.24

1.50

1.12

0.137

2.370

2.09

121

797.56

1.40

1.03

0.137

2.386

2.26

127

793.71

1.30

0.95

0.135

2.403

2.55

132

789.87

1.20

0.94

0.135

2.420

2.86

138

786.19

1.20

0.86

0.135

2.437

3.20

143

782.34

1.10

0.86

0.135

2.453

3.59

149

778.66

1.10

0.86

0.133

2.470

4.02

154

774.81

1.00

0.77

0.133

2.487

4.51

160

770.97

1.00

0.77

0.133

2.504

5.06

166

767.28

0.90

0.69

0.133

2.520

5.67

171

763.44

0.90

0.69

0.133

2.537

6.35

177

759.60

0.80

0.61

0.132

2.554

7.12


Page 83

DESIGN OF SOLAR POWERED WATER PURIFICATION SYSTEM 182

755.91

0.80

0.60

0.132

2.571

7.98

188

748.22

0.80

0.60

0.132

2.584

8.94

193

744.54

0.70

0.52

0.132

2.604

10.02

199

740.69

0.70

0.52

0.130

2.621

11.23

204

736.85

0.70

0.52

0.130

2.638

12.99

210

733.16

0.70

0.51

0.130

2.654

14.11

227

721.79

0.60

0.43

0.128

2.705

19.87

232

717.95

0.60

0.43

0.128

2.721

22.27

www.heat-transfer-fluid.com Appendix C: Table 7: Solar tech 0.37-55kW solar pump model.

Model

Pump

Pump

Water

Daily

Outlet

Adapting

Open

MPP

Spec.

Power

Head

Water

Dia.

Well dia.

Circuit

Voltage

flow SPA4370010

3PH22V50Hz

0.37kw

47-

1-10m3

32m SPA4370010-

3PH22V50Hz

0.37kw

2 SPA4370020

SPA4370020-

3PH22V50Hz

3PH22V50Hz

0.38kw

0.39kw

3PH22V50Hz

0.55kw

1-10m3 30mm

3PH22V50Hz

0.55kw

29-

10-

30mm

20m

20m

1’’ ¼

29-

10-

30mm

20m

20m

1’’ ¼

70-

1-10m3 30mm

2 SPA4550020

70-

0.55kw


100mm

100mm

100mm

100mm

1’’ ¼ 1-10m3 30mm

48m 3PH22V50Hz

100mm

1’’ ¼

48m SPA4550010-

30mm 1’’ ¼

32m

2 SPA4550010

47-

voltage

100mm

1’’ ¼

40-

10-

30mm

28m

20m3

1’’ ¼

100mm

350 -

280 -

450VDC

350VDC

180 -

150 -

450VDC

350VDC

350 -

280 -

450VDC

350VDC

180 -

150 -

450VDC

350VDC

350 -

280 -

450VDC

350VDC

180 -

150 -

450VDC

350VDC

350 -

280 -

450VDC

350VDC

Page 84



Page 85

solar powered water purification system

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