biodiesel proudiction from sewage sludge 2222

AL-Qadisiyah university College of Engineering

Chemical Engineering Department

Biodiesel Production From Municipal

sewage sludge

set by : Neraan Abass Farah Ali

Supervised by : Dr.ALI A.JAZIE 2017-2018

: توقيع المشرف العلمي على المشروع

علي عبد الحسين جازع : الأستاذ الدكتور

: الإهداء

)ص(محمد المصطفى .. إلى من دنا فتدلى فكان قاب قوسين أو أدنى

) ع(مولانا علي المرتضى .... إلى فخر الورى

) ع( الزهراء مولاتنا فاطمة....إلى ابنة خديجة الكبرى

)ع(يدنا الحسن المجتبى ... إلى المسموم عدوانا وظلما

)ع(سيدنا أبي عبد الله الحسين ...إلى المذبوح من القفى

حبيبنا المهدي ... من يملا الأرض قسطا وعدلا بعدما ملئت ظلما وجورا )عج(المنتظر

نهدي لكم جهدنا هذا

ACKNOWLEDGMENTS

كتور علي عبد الحسين جازعالد

ألاالمثالية بدأت بابتسامتك وكبرت في تحيتك وعظمت في لسانك الذي لا يتحرك يا من رافقت خطواتنا ولم تبخل علينا بوقتك و . بالكلام الطيب والنصائح الجميلة

تقديرا ونقدم شكرنا . نصائحك التي كانت نبراسا أنار طريق النجاح لمشروعنا ترم المحلشخصك

الدكتور صالح عبد الجبار صالح

يا من زرعت التفاؤل في دربنا وقدمت لنا المساعدة والتسهيلات والأفكار

وتقديراوفاءا والمعلومات بارك الله جهودك وسدد بالخير والعطاء دربك لحضرتك واعترافا منا بالجميل نتقدم بجزيل الشكر

أساتذتنا الأفاضل

أعوام إلى في الحياة الجامعية من وقفة نعود الأخيرةخطو خطواتنا لابد لنا ونحن ن

لنا الكثير باذلين بذلك جهودا كبيرة قضيناها في رحاب الجامعة معكم يا من قدمتم ... من جديد الأمةفي بناء جيل الغد لتبعث

لكم لأنكم والتقدير والمحبة الشكر والامتنان آيات أسمى نمضي نقدم إنوقبل .. لنا طريق العلم والمعرفة و مهدتو .. رسالة في الحياة أقدس حملتم

لحظات عمرنا لكم كل الحب والاحترام أجملمن قضينا معكم زملائي وزميلاتي يا

أخراج كل من مدوا لنا يد العون والمساعدة في إلىتقدم بجزيل شكرنا نوأخيرا

وجه أكملهذا البحث على

Abstract

Sewage sludge generated in municipal wastewater treatment plants was used as a feedstock for biodiesel production via esterification/transesterification in a two-step process. In the first esterification step٬

greasy and secondary sludge were tested using acid and enzymatic catalysts. The results indicate that both catalysts performed the esterification of free fatty acids (FFA) simultaneously with the transesterification of triacylglycerols (TAG). This method of producing biodiesel is important and successful because it helps to get rid of accumulated municipal waste and produce millions of tons of renewable fuel in an easy and clean process.

TABLE OF CONTAINTS

Chapter one : Introduction Chapter two : partial part - Materials - Experimental set – up - Process

Chapter three : material balance

- Flow sheet - Material balance calculation

Chapter four : energy balance

- Introduction - Energy balance calculation

Chapter five : equipment design

- Distillation column design - Reactor design

Chapter six : cost estimation

Reference

List of table

Table 1-1 physical and chemical properties of municipal sewage sludge (MSS) Table 1-2 physical and chemical properties of methanol Table 1-3 physical and chemical properties of hexane Table 3-1 material balance of extractor Table 3-2 material balance of reactor Table 3-3 material balance of packed reactor Table 3-4 material balance of decanter Table 3-5 material balance of evaporator Table 3-6 material balance of wishing tower Table 3-7 material balance of neutralizer Table 3-8 material balance of distillation 1 Table 3-9 material balance of distillation 2 Table 6-1 total cost of equipment

Chapter one Introduction

1-1 Introduction

Biodiesel is a renewable biofuel that is usually obtained by reaction of a refined vegetable oil extracted in advance and simple aliphatic alcohol with an alkaline or acidic catalyst because it provides a similar energy density to nitrodiesel and is used in pure diesel engines or may be mixed with Diesel oil.

Sewage sludge is more effective when concentrations of fatty substances are more than 10% and are available and can be obtained from continuous yield or high food value Research around the world is currently focused on the biodiesel industry of converting cheap raw materials, especially renewable waste eg sewage sludge They are simple, economical, environmentally friendly and low cost

Wastewater usually contains an ample amount of free fatty acids and this is considered an obstacle to the production of fuel required when the alkaline catalyst is used as it will react with alkaline catalysts as a type of acid can be used to produce soap in the end so it will need consumption Most of the catalysts lead to a reduction in the process of production of bio-diesel. By comparison with the acid catalyst, the alkaline catalyst is the easiest Catalysts for the production of diesel began to take seriously and can be produced from cotton seed oil

Waste water (swage sludge ) get from treatment plant and contain a range of organic and inorganic compounds containing proteins ,fats and phenols .it also contains toxic and dangerous inorganic pollutants

Table 1. Some physical and chemical properties of municipal sewage sludge (MSS)

Characteristic

Bulk density (kg/L) 1.26–1.38 Particle density (kg/L) 2.40–2.56 Organic matter (g/kg) 418–592 Ash (g/kg) 345–440 Organic carbon (g/kg) 205–403 Oxygen (g/kg) 185–219 Nitrogen (g/kg) 45–49 Hydrogen (g/kg) 40–46 Mineral matter (g/kg) 112–153 pH 7.1–8.2 Higher heating value (MJ/k) 11.3-14.2

The production of biodiesel faced many challenges the first : the fats containing FFA s are usually extracted and then transesterified Impurities in the lipids from the sludge would have interfered with the catalytic process in the conventional production of biodeisel using generally hexane as a solvent Finally, the lipids are converted by catalytic transesterification into their corresponding FAME (biodiesel). Results indicated that among four sludge tested, primary sludge achieved the greatest lipids and biodiesel yields. The amount of extracted lipids for primary sludge was 25.3% compared to 21.9%, 10.1%, and 9.1% (dry weight basis) for blended, stabilized

history 2-1

transesterification of a vegetable oil was conducted as early as 1853 by scientists Biodiesel was developed in 1890 by the inventor Rudolf Diesel . the rise in fuel economy worldwide calls for the development of renewable biofuel The first experiments to produce biodiesel were on vegetable oil fuel by French government and Dr.diesel imagine that pure vegetable oils could be used for diesel engines for agriculture in remote areas of the world where petroleum not available at the time . During 1920 &1930 Belgium, France , Italy , the united kingdom , Portugal , Germany , Brazil , Argentina , Japan and chine were reported to have tested and useed vegetable oils as diesel fuels during this time . The high viscosity of vegetable oils cause some problems compared to petroleum diesel fuel which results in poor atomization of fuel in the fuel and often leads to deposist and coking of injectors combustion chamber and these problems attempts to over included heating of the vegetable oils blending it with petroleum diesel fuel or ethanol paralysis and creaking of the oil . Biodiesel was one of the most widely used alternative fuels because of its clean emission , ease of use and low cost to rival diesel oil .

The productive process involves two 3-1 :main phases

first step The process of conversion of the FFA (free fat acid )in the waste to the esters (biodiesel) the goal of this step is to reach the value of FFA less than 0.05 and thus can take the second step Second step Transesterefication conversion to biobiesel

sel es of biodi Benefit4-1

1- Reduce the emission of CO2 2- low smog formation rates 3- Elimination of sulfur emission 4- Low percentage of other pollutants ( carbon monoxide and soot ) 5- improve lubrication and lubrication properties 6- Clean the engine 7- biodiesel completely nontoxic 8- the flash point is high (160 C)

Production biodiesel method’s5-1

There are two method for the production the biodiesel: 1- homogeneous catalyzed method 2- Heterogeneous Catayzed method

the important of project 6-1

The important of project concentrated in being treated environmental problems for the accumulation of stream of municipal and not find a solution or utilized specially in Iraq At was produced a quantities in millions of items daily and go after expensive treatment to the river

material Properties of raw 7-1 1) methanol

CH3OH or CH4O Chemical formula Colorless liquid Appearance

0.792 g/ cm3 Density -97.6 c , -143.7 F , 175.6 K Melting point 64.7 C , 184.5 F , 337.8 K Boiling point

32.04 g/mol Molar mass 13.02 kpa (at 20 C) Vapor pressure

Table (1-2 ) properties of methanol

2) hexane

C6H14 Chemical formula Colorless liquid Appearance 0.6606 g/ cm3 Density

-96 to -94 C , -141F , 177K Melting point 68.5 C , 155.2 F , 341 K Boiling point

86.18g/mol Molar mass 17.60 kpa (at 20 C) Vapor pressure

Table (1-3 ) properties of hexane

3) sulfuric acid H2SO4

H2SO4 Chemical formula Colorless liquid Appearance

1.84 g/ cm3 Density 10 C , 50 F , 283 K Melting point

337 C , 639 F , 610 K Boiling point 98.079 g/mol Molar mass

0.001 mmHg (at 20 C) Vapor pressure

Table (1-4 ) properties of sulfuric acid

Chapter two Practical part

Materials 1-2

1- sludge 2- sulfuric acid 3- methanol 4- sodium hydroxide 5- hexane 6- calcium oxide

up– experimental set 2-2

1- filter 2- heat plate & condenser 3- rotary evaporater

4- center fuge 5-fourier transform infrared spectroscope (FTIR) 6- glassware (two nick flask, beaker , glass cylinder) 7- balance 8- thermometer

process 3-2 First step -Filtrating the raw material (municipal sewage sludge)

- weighted (10g) of sludge

- (1o g ) of sludge put it or melting in mixture of acid sulfuric , methanol and hexane

acid ( 100 ml of methanol + 2 ml of sulfuric +100 ml of hexane)

The ratios are 1: 16 : 0.037 Sludge : methanol : sulfuric acid

Experiments are performed at (60 c) for (8) hours Mixing rate (mixing speed ) 600 R/m

-Second step preparation of the catalyst *

- Including dissolve 10 g of NaOH in 200 ml methanol And mix them for 15 min

- Add 50 g of calcium oxide to the mixture -Mix the result for 1 hour

After that leave in furnace for drying it at 900 C for 2 hours

- Then grind the catalyst

Grind until it becomes a soft powder -

-Third step

- Add the catalyst to the first mixture (sludge , methanol , hexane , sulfuric acid) and mix for 8 hours in 60 c in mixing rate( 600 r/m)

- still impurities from the biodiesel by using center fuge

Center fuge

(Hettich )

These tubes contain a layer of impurities and pure biodiesel

- Remove the solvent (methanol ) from biodiesel efficiently by using rotary evaporator.

rotary evaporator (sturrt )

Pure biodiesel

- examines the product (biodiesel ) by Fourier transform infrared spectroscope (FTIR)

(FTIR)

Chapter three Material balance

3-1 Flow Sheet

Figure 3-1 flow sheet

Table 3-1 definition of abbreviation of flow sheet

Definition Abbreviation Extractor Ex -101 Reactor R -101,102 Decanter De -101 Dryer DR -101 Washing tower W -101 Neutralizer N -101 Distillation D -101,102 Tank T-101,102,103 Pump P -101,102,103,104,105

,106,107.108

3-2 Material Balance

Production rate =10000 ton/year

1000 kg 1 day year 10000 ton ton 24 hr 300 day year

= 1389 kg/hr

Yield = 95 %

Oil = production rate /yield = 1389 /0.95 = 1462 kg

From the practical

oil =10% waste waste = oil / 0.1 = 1462/0.1

= 14.62 kg/hr V waste = weight /density = 5 / 0.72 = 6.9444 L

Weight of methanol = 14.4 of waste weight = 14.4 * 14.62 = 210.528 kg Weight of hexane = 2.16 of waste weight = 2.16 * 14.62 = 315.792 kg

ExtractorB of .M

Input = output Waste+ methanol+ hexane = Oil +FFA+ methanol+ hexane

14.62 +210.528 +31.579 =1462 +(0.05 % oil ) +methanol +hexane

256.727 =1462 +0.0005(1462) +methanol +hexane Methanol +hexane = 1104.539 Hexane /methanol = 0.15 Hexane =0.15 * methanol Methanol +0.15 * methanol = 688.74 Methanol = 598.904 kg/hr Hexane =89.835 kg/hr

Output(kg) Input (kg) Substance

0 14.62 Waste 1462 0 Oil

598.904 210.528 Methanol 89.835 31.5792 Hexane 0.731 0 FFA

Table (3-1) material balance for extractor

Extractor Waste Methanol Hexane

Oil FFA

Methanol Hexane

ReactorB of .M Oil H2SO4 H2SO4 oil FFA hexane Hexane biodiesel 1 Methanol methanol

M.B of methanol Input = output =598.904 kg

M.B of hexane Input = output =89.835 kg

M.B of H2SO4

Input = 0.05 (Oil) = 0.05 * 1462 = 73.1 kg

Overall M.B

Oil +FFA +H2SO4 +methanol +hexane = H2SO4 +biodiesel 1 +Oil +methanol +hexane

1462 +0.731 +73.1 = H2SO4 +0.7 +1462 H2SO4 (out) =73.13 kg

Reactor

Output (kg) Input (kg) Substance 1462 1462 Oil

0 0.731 FFA 73.13 73.1 H2SO4 0.7 0 Biodiesel 1

598.904 598.904 Methanol 89.835 89.835 Hexane

Table (3-2) material balance for reactor

packed ReactorB of .M Oil biodiesel H2SO4 glycerol Biodiesel 1 methanol Methanol hexane Hexane H2SO4

M.B OF methanol input =output =598.904 kg/hr M.B OF hexan input =output =89.835 kg M.B OF H2SO4 input =output =73.13 kg Oil + biodiesel 1 = biodiesel + glycerol 1462 + 0.7 = biodiesel + glycerol 1462.07= biodiesel + glycerol n(oil)=wt/m.wt =1462/850 =1.72 kmol n(oil ) + n(glycerol ) = 1.72 wt(glycerol) = n* m.wt = 1.72 * 92.02 = 158.27 kg Biodiesel = 1304.43kg

Output (kg) Input (kg) Substance

0 1462 Oil 1304.43 0 Biodiesel

73.13 73.13 H2SO4 0 0.7 biodiesel 1

598.904 598.904 Methanol 158.27 0 Glycerol 89.835 89.835 Hexane

Table (3-3) material balance for packed reactor

Packed Reactor

)separator(Decanter B of .M Biodiesel Biodiesel Methanol Hexane methanol H2SO4 hexane Glycerol glycerol H2SO4

M.B of Biodiesel Input = output = 1304.43 kg/hr M.B of H2SO4 Input = output = 73.13 kg/hr From top : Methanol (out) = 0.005 * 598.904 = 2.94952 kg/hr Hexan (out) = 0.05 * 89.853 = 0.4492 kg/hr Glycerol (out) = 0.005 * 158.27 = 0.7914 kg/hr

From bottom : Methanol (out) = 598.904 - 2.949 = 586.964 kg/hr Haxan (out ) = 89.853 - 0.4492 = 89.403 kg/hr Glycerol (out) = 158.27 -0.7914 = 157.4786 kg/hr

Decanter

Output (kg) Input (kg) Substance

1304.43 1304.43 Biodiesel 73.13 73.13 H2SO4

598.904 598.904 Methanol 89.403 89.403 Glycerol

144.0704 144.0704 Hexane Table (3-4) material balance for decanter

washing tower B of .M

) (water

Biodiesel Methanol Hexane biodiesel Glycerol

T=60 C T= 40 C Water T= 35 C

Density (biodiesel ) = wt / v 894 = 1304.43/ v V = 1.46 V (water ) = 5 * V (biodiesel ) = 5 * 1.46 = 7.29

Washing Tower

Density (water ) = wt/v 1000= wt / 7. 29 Wt = 7295.47 kg

Biodiesel (out ) = biodiesel (in) + water = 1304.43 + 72.95 = 1377.38 kg/hr M.B of methanol input = output = 2.9495 kg/hr

M. B of Hexane Input = output = 0.449 kg/hr

M.B of Glycerol Input = output= 0.7914 kg

Output (kg) Input (kg) Substance 1304.43 1304.43 Biodiesel 7295.47 7295.47 Water

2.949 2.949 Methanol 0.7914 0.7914 Glycerol 0.449 0.449 Hexane

Table (3-5) material balance for wishing tower

)dryer (B of .M

Biodiesel biodiesel Water

Water

M.B of methanol Input = output = 2.949 kg/hr

M.B of Hexane Input = output = 0.449 kg/hr

M.B of Glycerol Input = output = 0.7914 kg

Water + biodiesel = water + biodiesel 1377.38 = 70.95 + biodiesel Biodiesel (out) = 1306.34 kg

Output (kg) Input (kg) Substance 1306.43 1377.43 Biodiesel

70.95 0 Water 2.9495 2.9495 Methanol 0.7914 0.7914 Glycerol 0.449 0.449 Hexane

Table (3-6) material balance for evaporator

Dryer

neutralizer B of .M

Methanol Methanol Hexane hexane Glycerol glycerol NaoH Na2SO4 H2SO4 H2O

2NaOH + H2SO4 Na2SO4 + 2H2O

N(H2SO4)= 73.13 / 98 = 0.746 kmol N(NaOH)=2 *0.46 =1.49 kmol Wt(NaOH) = 1.49 * 40 =59.6 kg N(Na2SO4)=1 * 0.746 =0.746 kmol Wt (Na2SO4)= 0.746 * 142 = 105.932 kg

N(H2O)= 2 * 0.746 = 1.492 kmol Wt (H2O)= 18 * 1.492 = 26.856


M.B of Hexane Input = output = 89.403 kg/hr

M.B of Glycerol Input = output = 157.4786 kg

PH

Output (kg) Input (kg) Substance 598.964 598.964 Methanol 89.403 89.403 Hexane

157.4786 157.4786 Glycerol 0 73.13 H2SO4

105.932 0 Na2SO4 0 59.6 NaOH

26.856 0 H2O Table (3-7) material balance for neutralizer

Distillation B of .M

Methanol Hexane methanol H2O Glycerol

Na2SO4

Methanol Hexane H2O Na2SO4 Glycerol

M.B of hexane Input=output = 89.403 kg/hr

M.B of Glycerol Input = output = 157.4786 kg/hr

M.B of Na2SO4

Input = output = 105.932 kg/hr M.B of H2O Input = output = 26.856 kg/hr

M.B of methanol

Methanol (input) = 598.964 kg/hr

Methanol (output) from top : Input * 0.99 = 581.094kg/hr

Dist.1

Methanol (output) from bottom : Input * 0.01= 5.86984 kg/hr


157.4786 157.4786 Glycerol 105.932 105.932 Na2SO4 26.856 26.856 H2O

Table (3-8) material balance for distillation 1

)2( Distillation B of .M

Methanol Hexane hexane H2O Glycerol

Na2SO4

Methanol Hexane H2O Na2SO4 Glycerol


M.B of Glycerol Input = output = 157.4786 kg/hr

M.B of Na2SO4

Input = output =105.932 kg/hr M.B of H2O Input = output = 26.856 kg/hr

M.B of Hexane

Hexane (out) from top : Input * 0.99 = 80.4627 kg/hr

Dist.1

Hexane (out) from bottom : Input * 0.01 = 0.89403 kg/hr


157.4786 157.4786 Glycerol 105.932 105.932 Na2SO4 26.856 26.856 H2O

Table (3-9) material balance for distillation 2

Chapter four

Energy balance

4.1 Introduction The calculations will be based on the first law of thermodynamic(The total quantity of energy is constant .When energy disappear in one form, it appears in other forms). Δ [(H + (1/2* U2) + Z*g) m] = Q - Ws (for open system) Where : Q = heat gained by system (positive) Ws = work done by system Assumptions : 1-Neglect kinetic and potential energy. ΔH = Q -Ws 2-For open system shaft work (Ws) =0 3-For open system with physical operation ΔH = Q 4- For open system with chemical reaction ΔH + ΔH reaction = Q ΔH = H out – H in H = m Cp mean ΔT Cp mean = Σ (Xi * Cpi) or Cp mean = Σ ( yi * Cpi ) ΔT = T – T reference , T reference = 298 K

4.2 Energy Balance Calculation 1- extractor

waste oil FFA Methanol hexane hexane methanol

Q = ΔH ΔH = m cp ΔT

H1=146.2*Cp * (25-25) = 0 H2 = 2105 * Cp * (25-25) = 0 H3 = 315.792 * Cp * (25 -25 ) = 0 ΔH1= 0

H4= 1462 *1.97 * (85 - 25 ) = 3090.3446 H5= 598.904* 0.086 * (85-25 ) = 4978.089 H6 = 89.853* 0.204 * (85 - 25 ) = 1099.5804 ΔH2= 176998.325

Q= ΔH2 - ΔH1 = 176998.325 - 0 = 176998.325 kJ

Reactor

Energy balance for equipment 2

Oil H2SO4 H2SO4 Oil FFA hexane Hexane biodiesel 1 Methanol methanol

ΔH2= 17698.325 H(H2SO4)= 373.1 * 0.147 * (85 - 25 ) =3305.919

ΔH3 =17698.325 + 3305.9191 = 21004.24

H1= 1462 * 1.97 * ( 65 - 25 ) = 115205.6 H2= 73.13 * 0.145 * (65 - 25 ) = 425.271 H3= 598.904* 0.083 ( 65 - 25 ) = 1988.36128 H4=89.835* 0.197 *(65 - 25 ) = 707.8998

ΔH4 = 118327.1318

Q= ΔH4 - ΔH3 = 118327.1318 - 182859.619 = 97322.8877 KJ

Reactor

Energy balance for equipment 3 Oil biodiesel H2SO4 glycerol Biodiesel 1 methanol

Methanol hexane Hexane H2SO4

∆H4 = 118327.1318

H1= 1304.43 * 0.123 * ( 60 - 25 ) = 5621.191 H2= 158.27 * 0.266 * (60 - 25 ) = 1475 .244 H3= 598.904 * 0.083 * ( 60 - 25 ) = 1739.816 H4= 89.835 * 0.1962 * (60 - 25 ) = 616.896 H5= 73.13 * 0.1447 * (60 - 25 ) = 370.538

∆H5 = 9823.685

Q = ∆H5 - ∆H4 =-108503.446 kj

Oil + 3methanol = 3 easter + glycerol ∆Hreaction = ∑Hf (product) - ∑Hf (reaction) Hf (methanol) = -238.42 KJ/mol Hf (glycerol) = - 669.6 KJ/mol Hf (0il ) = 9767.52 KJ/ mol

∆Hreaction=( 3* Hf(easter) + Hf (glycerol) - Hf(oil) +3* Hf (methanol) = 3*-22430.895+(-669.6) - 9767.52 + (-3*238.42) = -3854372745 KJ Q = ∆H +∆Hreaction = -108503.446 -3854372745 = -3854481248 kj

Packed Reactor

4Energy balance for equipment Biodiesel Biodiesel Methanol Hexane methanol H2SO4 hexane Glycerol glycerol H2SO4

T=60 ◦C T = 60 ◦C

∆H5 = 9823.685

∆H6 = ∆H5

Q = ∆H6 - ∆H5 = 0

Decanter


(T=25 C) water

Biodiesel Methanol Hexane biodiesel Glycerol

T=60 C T= 40 C Water T= 35 C

H(water)1 = 7295.47 * 4.18 * (25 - 25 ) = 0 H2= 1304.43 * 0.123 * (60 - 25 ) = 5620.13 H3= 2.949 * 0.0832 * ( 60 - 25 ) = 8.5874 H4=0.449 * 0.196 * (60 - 25 ) =3.08014 H5 = 0.7914 * 0.26631 * ( 60 - 25 ) = 7.376

∆H7= 5639.1735 kj/hr

H1= 7222.52 * 4.18 * (35 - 25 ) = 301901.386 H2= 1377 * 0.1318 * (40 - 25 ) = 2722.329 H3= 2.949* 0.0812 * (40 - 25 ) = 3.59182 H4=0.449 * 0.1906 * (40 - 25 ) = 1.283691 H5= 0.7914 * 0.2633 * (40 - 25) = 3.12576

∆H8= 304631.7103 kj/hr

Q = ∆H8 - ∆H7 = 298992.536

Washing Tower

Energy balance for equipment 6 (Dryer )

Biodiesel biodiesel Water

Water

Hwater = 70.96 * 4.18 * (100 - 25 ) = 229842.525 kj Hbiodiesel = 1306.34 * 0.1129 * (90 - 25) = 9590.62 kj Hmethanol = 2.949 * 0.08707 * (90 - 25 ) = 16.69001 kj Hhexan = 0.449 * 0.2062 * (90 - 25 ) = 7.64647 kj Hglycerol = 0.7203 * 0.2705 * (90 - 25 ) = 12.6691 kj

∆H9 = 239470.1506 kj

Q = ∆H9 - ∆H8 = -65161.559 kj

Dryer


Methanol Methanol Hexane hexane Glycerol glycerol NaoH Na2SO4 H2SO4 H2O

H (methanol) = 586.965 * 0.0832 * (60 - 25 ) = 1709.24204 KJ H (hexane ) = 89.403 * 0.196 * (60 - 25 ) = 613.3045 KJ H (NaOH) = 59.6 * 0.087 * (60 - 25 ) = 181.382 KJ H (glycerol ) = 157 .47 * 0.2663 * ( 60 -25) = 1467.699 KJ H (H2SO4 ) = 73.13 * 0.1447 * ( 60 -25 ) = 370.3668 KJ

∆H10 =4341.99442 kj

H (methanol) = 586.965 * 0.0832 * (60 - 25 ) = 1709.24204 KJ H (hexane ) = 89.403 * 0.196 * (60 - 25 ) = 613.3045 KJ H (Na2SO4) = 105.932 * 0.227 * (60 - 25 ) = 841.62974 KJ H (glycerol ) = 157 .47 * 0.2663 * ( 60 -25) = 1467.699 KJ H (H2O ) = 26.856* 4.18 * ( 60 -25 ) = 3929.0328 KJ ∆H11 = 8560.90808 KJ

Q = ∆H11 - ∆H10 = 4218.91366 kj

PH


Distillation column 1

QL Methanol QV QD 343 k Hexane methanol H2O Qf Glycerol

Na2SO4 338 k

Qw 343 k Methanol Hexane H2O Na2SO4 Glycerol

Cp mean = (Xmethanol * Cp ) + (X Na2SO4 * Cp ) + (X hexane * Cp ) + (X glycerol * Cp ) + ( Xwater * Cp )

Cp mean = ( 0.6962 * 2. 658 ) + ( 0.0741 * 1.596 ) + ( 0.1003 * 2.413 ) + (0.1101 * 1.263 ) + (0.0188 * 4.174 )

Cp mean = 2.4283 kj / kg .k

Q f = m cpmean ∆T

= 979.0056 * 2.4283 * (338 - 298 ) = 95092.771 KJ

QW = m cpmean ∆T Cpmean =(Xmethanol * Cp ) + (X Na2SO4 * Cp ) + (X hexane * Cp ) + (X glycerol * Cp ) + ( Xwater * Cp )

Dist.1

Cpmean = (0.0069 * 2.658 )+(0.0741 * 1.596 ) + (0.1003*2.413) +0.1101 * 1.263 ) + (0.0188 * 4.174) = 0.5961 Qw = 439.485 * 0.5961 * (343 - 298 ) = 11788.99167

At top

QD= m Cp ∆T = 581.0974 * 2.693 * ( 343 - 298 ) = 70419.8763 KJ

Qv = QD + QL + QC R= L/ D → L=R*D ( R=2) V= L+D V=2D +D → V = 3D V= 3 * 581.094 = 1743.282 Kg L = 2 D = 2 * 581.094 = 1162.188 Kg

λ methanol = 1085 kj/kg QV = m λ = 581.094 * 1085 =630486.99 QV= 630486.99 + 704119.8763 = 1334606.866 kj

QL = m Cp ∆T = 1162.188* 2.693 * ( 343 - 298 ) = 140839.7528 KJ

Qc = Qv - QD - QL

1123347.237 kj = QF +QR = QD + QW + QC

QR =1205556.105 kj

Energy balance for equipment 9 Distillation column 2

QL Methanol QV QD 343 k Hexane hexane H2O Qf Glycerol

Na2SO4 343 k

Qw 343 k Methanol Hexane H2O Na2SO4 Glycerol

Qf = m cp mean ∆T = 439.486 * 0.5961 * (343 - 298 ) = 11788.993 kj

QD = m cp ∆T = 89.403 * 2.413 *(343 - 298 ) = 9707.8247 kj

Qw = m cp mean ∆T

Cp mean = (Xmethanol * Cp ) + (X Na2SO4 * Cp ) + (X hexane * Cp ) + (X glycerol * Cp ) + ( Xwater * Cp )

Cp mean = (0.0004* 2.658 )+(0.0045 * 1.596 ) + (0.000047*2.413) +(0.0052* 1.263 ) + (0.00089* 4.174) = 0.0186

Qw = 301.11 * 0.0186 * (343 - 298 ) = 252.03452 kj

Dist.1

Qv = QD + QL + QC R= L/ D → L=R*D ( R=2) V= L+D V=2D +D → V = 3D V= 3 * 89.403 = 268.209 Kg L = 2 D = 2 * 89.403 = 178.806 Kg

λ hexane = 356 kj/kg QV = m λ = 268.209 *356 =95481.336 QV= 9707.8247 + 95481.336 = 105189.1607 kj

QL = m Cp ∆T =178.806 * 2.413* ( 343 - 298 ) = 19415.64951 KJ

Qc = Qv - QD - QL

76065.6864 kj = QF +QR = QD + QW + QC

QR = 74236.55263 kj

Chapter five Equipment

design

5-1 distillation column

Reflux ratio=��/� = 2

Ln=2*30.764 = 61.528 kmole/h Vn=Ln + D= 61.528 +30.764 =92.292 Kmole/hr

Figure(5-1)distillation column On the top Yn = Ln / Vn Xn+1 + D/Vn XD Yn = 0.667 Xn + 0.33 ……. (1) Lm= F+Ln Lm=32.27 + 61.528 =93.798 kmol/hr Vm=Lm - W Vm=93.798 - 1.506 = 92.292 kmol /hr

On the bottom : Ym = Lm / Vm Xm+1 - W/Vm Xw Ym= 1.016 Xm+1 - 0.0001

Number of theoretical stage = 7.99 say 8 Efficiency of column = 0.85 Number of actual stage =8 /0.85= 9.4 say 10 stages Assume tray spacing = 0.6 m Height of column = Number of actual stage * tray spacing Height of column = 10*0.6= 6 m

Physical properties :- Temperature at the top =70 Cº Temperature at the bottom =70 Cº Assume Pressure drop in column =10 kpa Pressure top =40 kpa Pressure down =50 + 10 = 50 kpa Components ρ L (kg/m3) Methanol 792 Hexane 655

Density of vapor =�� /��

R= 8.314 Kp.m3/Kmole. K At the top: ρv (methanol ) = 40 * 32.04 / 8.314 * (70 +273 ) = 0.456 ρv (hexane ) = 40 * 86.18 / 8.314 * (70+273) = 1.22 At bottom: ρv (methanol ) = 50 * 32.04 / 8.314 * (70 +273 ) = 0.5617 ρv (hexane ) = 50 * 86.18 / 8.314 * (70+273) = 1.511 ρL at the top = X methanol * ρL (methanol ) +X(hexane ) * ρL (hexane ) = 0.99*792 + 0.01*655 = 790.63 kg/m3

ρL at the bottom = X methanol * ρL (methanol ) +X(hexane ) * ρL (hexane ) = 0.01*792 + 0.99*655 = 656.37 kg/m3

ρL(average)= 790.63 + 656.37 / 2 = 723.5 kg/m3

ρv at the top = X methanol * ρv (methanol ) +X(hexane ) * ρv (hexane )

0.99 *0.456 + 0.01 * 1.22 = 0.457 kg/m3 = ρv at the bottom = X methanol * ρv (methanol ) +X(hexane ) * ρv (hexane ) = 0.01*0.5617 + 0.99 * 1.511 = 1.5015 kg/m3

ρv(average) = 0.457 + 1.5015 / 2 = 0.979 kg/m3

FLV=(Lm/Vm) *(ρV/ρL)0.5=(1.01) * (0.9790/723.5 )^0.5 = 0.03

Assume tray spacing = 0.6 m

From figure (11-27 ) vol.6

K= 0.0008

Correction for surface tensions surface tensions= 0.018 K1=(0.018/0.02)^0.2 *0.0008= 0.0783

uf= k1*(ρL -ρv /ρv)^0.5

= 0.0783*(723.5 - 0.979/ 0.979)^0.5 = 2.104 m/sec Design for 80% flooding at maximum flow rate

Uv=0.8 * 2.104 = 1.68

M.wt (average)= X*M.wt (top) + X*M.wt (bottom) /2 =( 0.99*32.04 +0.01*86.18)+(0.01*32.04+0.99*86.18)/2 =59.1055 kg/kmol

Maximum volumetric flowrate= Vn *M.wt (average)/ ρv * 3600

= 92.29 *59.105 /0.979 * 3600=1.547 m3/sec

Net Area required = Maximum volumetric flowrate/ Uv =1.547 / 1.68 = 0.921

As the first take down comer Area as (12%)

column cross Area = Net Area /(1- 0.12) = 1.0469 m2

column diameter = (A *4)/ π)^0.5 = 1.1548 m

liquid flow pattern

maximum volumetric liquid flow rate = Lm * M.wt(methanol)/ρL(avg) * 3600 = 93.798* 32.04 / 723.5 * 3600 = 0.001153 m3/sec

Nearest standard pipe size outside diameter 7 in(177.8 mm) inside diameter 199.4-200 mm

Thickness wall 3.76 mm

Provisional plate design Column diameter (Dc) =1.1548 m Column area (Ac) = 1.88 m2

Down comer area (Ad) = 0.12 *1.88 = 0.225 m2

Net area (An) = Ac – Ad= 1.88 - 0.225 = 1.65 m2

Active area (Aa) = Ac – 2Ad =1.88- 2*0.225= 1.43 m2

Hole area (Ah) take 10% (Aa) as first trial = 0.143 m2 Ad/Ac=(0.225/1.88)*100=11.9

From fig (11-31) vol.6 Lw/Dc=0.76 Weir length = 0.75* 1.1548 =0.866 m Take weir height = 50 mm Hole diameter = 5 mm Plate thickness = 5 mm

Check weeping Maximum liquid rate = (L* M.wt methanol )/ 3600 Maximum liquid rate = (61.528 * 32.04)/ 3600 =0.54kg/sec Minimum liquid rate at 70% turn – down = 0.7* 0.54 =0.38 kg/sec Maximum(how)=750*( Maximum liquid rate/ρL (avg)*Weir length)^ 2/3

= 750* (0.54/723.5 * 0.866)^2/3= 6.792 mm Minimum (how) =750*( Minimum liquid rate/(ρLaverage *Weir length )) 2/3= 750* (0.38/ 723.5 *0.866) ^2/3=5.373 mm At minimum rate hw+ how = 50 +5.373 =55.373 mm From figure (11.30) K2 = 30.2 ŭh(min) = (K2 – 0.90(25.4 – dh))/(PV average) 0.5

ŭh(min) = (30.2– 0.90(25.4 – 5))/(0.979) 0.5=11.96 m/sec Actual minimum vapor velocity = minimum vapor rate / Ah Actual minimum vapor velocity =(0.7 *1.88)/ 0.143= 9.2 m/sec So minimum operating rate will be well above weep point.

Plate pressure drop Dry plate drop Maximum vapour velocity through holes uˆh = 1.88/ 0.143 =13.14 m/sec

From figure (11.34) , for plate thickness/hole dia.= 1, and Ah / Ap = Ah / Aa = 0.1, C0 = 0.84 hd = 51 * (uˆh / C0) 2 *(PV average /PL average) hd = 51 * (13.14/0.83) 2*( 0.979/723.5)= 17.29 mm liquid Residual head hr = 12.5* 103 /PL average hr = 12.5* 103 / 723.5 = 17.27 mm liquid Total plate pressure drop ht = hd + (hw+ Maximum (how)) +hr ht=17.29 + (50 + 6.792 ) + 17.27 =91.352 mm liquid

Down comer liquid back-up Down comer pressure loss Take hap = hw – 10 = 40 mm. Area under apron, Aap = 0.09 * 40*10-3 = 0.0036 m2

As this is less than Ad = 0.225 m2

hdc = 166 *( Maximum liquid rate /(PL average * Aap))2

hdc = 166 *(0.54 /( 723.5* 0.0036))2= 7.135 mm Back-up in downcomer hb = (50 +6.792) +91.352+7.135 = 155.27 mm = 0.155 m 0.155 < 1/2 (plate spacing +weir height) So tray spacing is acceptable Check residence time tr = (Ad *hb *PL average) / Maximum liquid rate tr = (0.225 *0.1552 * 723.5) / 0.54 =46.8 sec > 3sec, satisfactory.

Check entrainment uv = Maximum volumetric flow rate / An uv =1.547/ 1.65 = 0.39 m/sec

per cent flooding = uv / uf =0.937 / 2.104 *100 = 44.5 say (45) FLV=0.050 , from figure (11.29) Ψ = 0.07, well below 0.1 Trial layout Use cartridge-type construction. Allow 50mm unperforated strip round plate edge : 50 mm wide calming zone

Diameter =1.1548 m Wire length = 0.866 m

Perforated area From figure (11.32) , at lw / Dc = 0.75 θc = 99° angle subtended at plate edge by unperforated strip=180-99=81° Mean length, unper forated edge strips= (1.1548– 50*10-3) *π *81/180 = 1.5599 m Area of un perforated edge strips = 50*10-3 *1.5599 = 0.0779 m2

Mean length of calming zone = (1.154– 50*10-3) sin(99/2) =0.839 m Area of calming zone = 2(0.839 *50*10-3) = 0.0839 m2 Total area for perforations, Ap=1.43– 0.0779–0.839 =0.5131 m2

Ah /Ap = 0.143/ 0.5131= 0.278 From figure (11.33) extra pollution lp / dh = 2.2

Number of holes Area of one hole = 1.964 * 10-5 m2

Number of holes = Ah / Area of one hole = 0.143/1.964 * 10-5 = 7281.05

Plate specification

Plate No 1 Plate I.D 1.154 m Hole size 5mm Active Holes 7281.05 Plate spacing 0.6 m Plate pressure drop 17.29 mmHg Plate material carbon steel Type of plate sieve plate

The mechanical design

Height between tangent line=6 m Diameter=1.154 m Skirt support height = 3 m sieve plates , mineral wool 75 mm thick ρmineral wool = 130 kg/m3 Material of construction , stain lees steel 18cr/8Ni unstabilised (304) Design stress 158 N/mm^2 at design temperature 70 C Operating pressure 0.1113 N/ mm^2 Take design pressure 10% above operating pressure Design pressure = 1.1*0.1113 =0.122 N/mm^2

Minimum thickness required for pressure loading

E= Pi * Di / 2 *f - Pi = 0.122 * 1154 / 2* 158 - 0.122 = 0.44 mm Add corrosion allowance of 2mm = 0.44 + 2 = 2.44 mm

As a first trial ,divide the column into five section ,with the thickness increasing by 2 mm per section ,try ( 2.44 , 4.44 ,6.44 ,8.44 ,10.44)

Dead weight of vessel Cv=1.15 vessel with plates Dm= 1.1548 + 0.00644= 1.16124 m H(hight) = 6 m t(wall thickness) = 6.44 mm Wv = 240 Cv Dm (Hv+0.8 Dm ) t = 240 * 1.15 * 1.16124 *( 6 + 0.8*1.16124)*6.44 = 16298.507 N = 16.298 KN

Weight of plates

Plate Area (Ac)= 1.88 m2 Wight of plates = 1.2 * 1.88 = 2.256

10 Plates = 10 * 2.256 = 22.56 KN Weight of insulation Mineral wool density = 130 kg/m3

Approximate volume of insulation =π * D * h * t

= 3.14 * 1.154 * 6 *75 * 10^-3 =1.52189 N Weight = 1.52189 *130 * 9.18 = 1940.87 N Double this to allow for fitting ECT = weight *2 /1000 = 3.88174 KN Total weight = weight (vessel ) + weight (plate) + w (insulation) =16.298 + 22.56 + 3.88174 = 42.740 KN Wind loading Take dynamic wind pressure as 1280 N/m3 Mean diameter including insulation = D + 2*(t+ insulation thickness) = 1.154 + 2*(6.44+75) * 10 ^-3 = 1.316 m Loading (per linear meter)f w = 1280 *1.316=1684.48 N/m Bending moment at bottom tangent line : Mx = Fw * x^2 /2 = 1684.48 * 36 /2 = 30320.64 Nm

Analysis of stresses At bottom tangent line Pressure stresses : GL= pi* Di / 4t = 0.122*1154/4*10.44 = 3.38 N/mm2 Gh = pi * Di / 2t = 0.122 * 1154/ 2* 10.44 = 6.742 N/mm2 Dead weight stress : Gw = total weight / π (Di +t)*t = 16324.95/ 3.14 (1154 +10.44)*10.44 =0.4276 Bending stresses: Do = 1154 + (2* 10.44) = 23518.5 mm Iv= π / 64 * (Do ^4 - Di ^4) = 3.14 / 64 *(23518.5^4 - 1154 ^4) = 1.501 * 10 ^16 Gb=(+,-) Mx / Iv (Di/2+t) =± (30340.91 * 10^3 / 1154 * 10^16)*(1.154/2)+10.44 = 1.2*10^-5

The resultant longitudinal stress is : Gz= GL + GW± Gb Gz= ( up wind ) = 3.384 - 0.4276 + 1.2*10^-5 = 3 Gz= (down wind ) = 3.384 - 0.4276 - 1.2*10^-5 = 2.9

The greatest difference between the principle stress (6.742 - 2.9 ) = 3.842

Well below the maximum allowable design stress

Check elastic stability (buckling)

Gc = 20000 * t/Do = 8.878 N/mm2

Vessel skirt support S = 90 Ө

Young modules = 200000 N/mm2 The maximum dead weight load on the skirt will occure when the vessel in full of methanol

Approximate = π /4 D^2 * h * 792* 9.81 = 48.76

Weight of vessel = 16.298507 Total weight = 16.298507 + 48.76 = 65.0585 Bending moment at base of skirt =0.22 *( 6+3/2) ^2 = 4.455

As the first trail take the skirt thickness as the same as that of the bottom section of vessel 10.44 mm Gbs= 4*Ms / π (Ds +ts ) * ts * Ds

4 *4.455 /3.14 (1154 +10.44 )*10.44 *1154 =0.405 = Gws (test ) = w/ π (Ds +ts ) * ts = 48760 / 3.14 *( 1154 +10.44 )*10.44=1.277 N/mm2

Gws (operating ) = 16298.507 / 3.14 * (1154+10.44)*10.44

= 0.426 Maximum Gs(compressive ) = Gbs + Gws = 0.405 +1.277 = 1.682

Maximum Gs (tensile ) == Gbs - Gws = 1.363- 0.426 = 0.937

Take the joint factor j = 0.85 Criteria for design Gs (tensile ) x > 0.85 * 158 * sin 90 0.937 x > 134.3

Gs(compressive ) x >0.125 * E * ts/Ds * sin 90 1.682 x > 0.125 * 200000 * 10.44 / 1154 * sin 90 1.682 x > 226.16

Both criteria are satisfied add 2 mm for corrosion gives a design thickness of 12.44

Base ring and anchor Approximate pitch circle dia = 1.154 +0.002 = 1.156 m

Circumference of bolt circle = 1.156 * π = 3.6298

Number of bolts required at minimum recommended Bolt spacing = 1156 * π / 600 = 6.049

Glosest multiple of 4 = 6 bolts

Take bolt design stress = 125 Ms = 4.455 Take w = operating value = 16.298 m

Ab = 1/Nb* f (( 4*Ms / Ds ) - 16298)=20.56

Blot root dia = (20.56*4/ π)^1/2 = 5.117 mm

backed reactor design 2-5

oil biodiesel H2SO4 glycerol Biodiesel 1 methanol Methanol hexan Hexane H2SO4

Reacted = FA * dxA Reacted = rA * dv FA *dxA = rA * dv

Catalysts Oil weight

1462 kg x

50 g 2 g

X = 58.48 kg = catalyst weight

τ = v/vₒ τ = residence time CA = f Aₒ /vₒ CA = concentration of component (A)

τ = Area under the curve * CAₒ rA = rate of reaction v = volume of reactor Xa =convertion of component (A)

FA = mole rate of component Volumetric Flow rate Vₒ

1- For first component (H2SO4) P = p*m.wt /R*T = 101.3 *(98)/8.314 *338 = 3.533 kg /m^3 V1 = m/P = 73.13 /3.533 = 20.699 2- for Oil P = p*m.wt /R*T =101.3*525/8.314*338 =9.0841

V2 = m/P = 1462/9.0841 = 160.94 m3/hr

3- For biodiesel P = p*m.wt /R*T = 101.3 *269.58 /8.314 *338 = 9.718 kg/m3

V3 = m/P = 0.7 /9.718 =0.072 m3/hr

4- For methanol

P = p*m.wt /R*T = 101.3 **32.04/8.314 *338 =1.155 kg/m3

V4 = m/P = 960.4686/1.155 = 831.59 m3 /hr

5- For hexan

P = p*m.wt /R*T = 101.3 * 86.18 /8.314 *338 = 3.107 kg /m3

V5 = 144.0704/3.107 = 46.375 m3 /hr

Vₒ = V1 +V2 +V3 +V4 +V5 = 1059.676 m3/hr Divided by 3600 Vₒ /3600 = Vₒnew = (0.2944)

:Catalyst volume Vc

From Perry (20-29) the superficial velocity uv = is between (0.15- 6) = 3.08 = 0.15+6/2 = 3.08 m/s

From Perry (10-17) Z/D =(1:4) D= diameter Z= packing height Vₒ = uv * A A= (3.14/4 )* D ^2 A = Vₒ/uv

= 0.2944/3.08 0.096 = (3.14/4 ) * D^2 D^2 = 0.1222 D = 0.421 m Z/D = 4 (max) Z = 4 *D Z =4*0.349 =1.3 Vc =( 3.14/4) *D^2 *Z Vc = 0.134 Vc = catalyst volume

:Total Reactor Volume

Vc /Vr *100 = 75 % Vr = Vc/ 0.75 = 0.134/0.75 = 0.178 m3

Vr = 3.14 /4 *D^2 * H H = Vr /Area = 0.178 /0.096 = 1.985 m Z = Packing height (m)

Z = 1.3 m D = 0.421 m H = 1.985 m say (2 m)

Mechanical Design

From harker process design page 215 Cylindrical shell thickness

e =( p *D /(2 ɽf - p)) + c e = shell thickness (mm) f = maximum allowable working stress (mN/m2) ɽ = weld efficincy factor c = corrosion allowable (mm)

from harker page 215 p = 54 atm = 0.873 mn/m2 D = 449 mm ɽ = max = 0.3 f = 70 mn/m2 for ɽ = maximum = 0.8 c-steel [vol 6 pg 809 table (13-2)] c =2 mm (vol 6 pg 810)

e = (0.873 * 349 /(2 *0.8*70*)-0. 873 )+ 2 = 9.408 mm 2- head thickness (ellipsoidal head ) From volume 6 page (816)

e = (pi*Di /2 f g -0.2 pi )+c

e = (0.873 *349/(2*70*0.8)-0.2 (0.873))+2 = 9.28 mm

: Bracket design saddles

At diameter 0.421 m

W= (30) From vol 6 pg 845 table (13-26 ) W = Fbs max load per bracket = w/4 Fbs = 7.5 = 7500 N Where tc = shell thick = 4.74 mm

Lc = Fbs / 60 * tc = 7500/ (60 *9.40) = 13.297 mm Lc = depth of bracket

Figures Used

Figure (11-27) flooding velocity sieve plate

Figure (11-31) relation between downcomer area and weir length

Chapter six Cost estimation

Total purchase cost of major equipment items(PCE)

EXTRACTOR -1 Vessel height = 15 m Diameter = 1 m From figure (6-5b) vol.6 Bare cost from figure =32,000 $ Material factor = 1.0 Pressure factor = 1.1 Cost = bare cost * material factor * pressure factor =32,000 * 1 *1.1 = 35200 $

Cost in 2018 = cost in 2014 * (index 2018/ index 2014 ) =35200 * 149.5 /111 = 47409 $

Fig. (5.4): Horizontal pressure vessels.

Reactor -2 From table (6-1) appendix B Suitable material is carbon steel C= 15000 $ S=capacity = 3m^3

n= 0.40 Ce= C S ^n Ce=15000(3) ^0.4 = 23277 cost in 2014

Cost in 2018 =cost in 2014 *(index 2018/ index 2014 ) Cost index in 2018 =149.5 Cost in 2018 = 23277 * 149.5/111 = 43472 $

packed reactor -3 From table ( 6-1) appendix B Suitable material is carbon steel C=2400 $ S=capacity = 10m^3

N= 0.6 Ce=2400(10) ^0.6 = 9554 $ cost in 2014 Cost index in 2018 = 149.5

Cost in 2018 = cost in 2014 * (index 2018/ index 2014 ) Cost in 2018 = 6554 * 149.5/111 = 12867 $

separating cost -4 From table ( 6-1) appendix B Suitable material is carbon steel C=2400 $ S=capacity = 10m^3

N= 0.6 Ce=2400(10) ^0.6 = 95543 cost in 2014 Cost index in 2018 =149.5

Cost in 2018 = cost in 2014 * (index 2018/ index 2014 )

Cost in 2018 = 9954 * 149.5/111 = 12867 $

washing tower -5 From table ( 6-1) appendix B Suitable material is carbon steel C=2400 $ S=capacity = 10m^3

N= 0.6 Ce=2400(10) ^0.6 = 95543 cost in 2014 Cost index in 2018 = 149.5

Cost in 2018 = cost in 2014 * (index 2018/ index 2014 )

Cost in 2018 = 9954 * 149.5/111 = 12867 $

evaporator -6 From table ( 6-1) appendix B Suitable material is carbon steel C=2000 $ S=capacity = 10m^3

N= 0.53 Ce=2000(10) ^0.53 = 67768 cost in 2014 Cost index in 2018 =149.5

Cost in 2018 = cost in 2014 * (index 2018/ index 2014 ) Cost in 2018 = 67768 * 149.5/111 = 91274 $

ryer D-7 From table ( 6-1) appendix B Suitable material is carbon steel C=35000 $ S=capacity = 10m^3

N= 0.53 Ce=35000 (10) ^0.53 = 72211 cost in 2014 Cost index in 2018 =149.5

Cost in 2018 = cost in 2014 * (index 2018/ index 2014 ) Cost in 2018 = 72211 * 149.5/111= 97257 $

equivalent cost- 8

From table ( 6-1) appendix B Suitable material is carbon steel C=24000 $ S=capacity = 10m^3

N= 0.6 Ce=2400 (10) ^0.6 = 9554 cost in 2014 Cost index in 2018 =149.5

Cost in 2018 = cost in 2014 * (index 2018/ index 2014 ) Cost in 2018 = 9554 * 149.5/111= 12867 $

distillation cost -9

Cost of vessel

H=6 m Dc = 1.154 m Pressure = 1.13 bar

From table ( 6-1) appendix B Suitable material is carbon steel C=30000 $ Pressure factor = 1 Materal factor = 1 Purchchased cost = bare cost * material factor * pressure factor = 30000*1*1=30000 in 2014 Cost index in 2018 =149.5 Cost in 2018 = 30000 * 149.5/111= 40405 $

Cost of plate

Dc = 1.154 m Type : sieve Material : carbon steel From table ( 6-3) appendix B C=320 $ Pressure factor = 1 Materal factor = 1 Purchchased cost = bare cost * material factor * pressure factor = 320*1*1=5760 $ in 2014

Cost index in 2018 =149.5 Cost in 2018 = 5760 * 149.5/111= 7757 $

Total cost of column = total cost of plate + cost of vessel = 40405 + 7757 = 48162 $

2 distillation 10 Cost of vessel

H=6 m Dc = 1.154 m Pressure = 1.13 bar From table ( 6-2) appendix A Suitable material is carbon steel C=30000 $ Pressure factor = 1 Materal factor = 1 Purchchased cost = bare cost * material factor * pressure factor = 3000*1*1=30000 in 2014 Cost index in 2018 = 149.5 Cost in 2018 = 30000 * 149.5/111= 40405$

Cost of plate

Cost of vessel Dc = 1.154 m Type : sieve Material : carbon steel Pressure = 1.13 bar From table ( 6-3) appendix A

C=320 $ Pressure factor = 1 Material factor = 1 Purchchased cost = bare cost * material factor * pressure factor = 320*1*1=320 $ in 2014 Total cost of plate = 10 * 320=3200 Cost index in 2018 =149.5 Cost in 2018 = 3200 * 149.5/111= 4309.9 $ Total cost of column = total cost of plate + cost of vessel = 4309.9 + 7757 = 12066.909 $

Cost Equipment 47409 Extractor 43472 Reactor 12867 Packed reactor 12867 Separator 12867 Washing tower 91277 Evaporator 97857 Dryer 12867 Equivalent 48162 Distillation 1

12066.9 Distillation 2 391109.9 Total cost of equipment

Table (6-1) total cost of equipment Estimation of fixed capital cost . from table (6-2) appendix B :

PCE = 391109 * 1.32 = 516263.88 $

f1 equipment erection = 0.4 f2 piping = 0.7 f3 instrumentation = 0.2 f4 electrical = 0.1 f5 buildings = 0.15 f6 utilities =0.5 f7 storage = 0.15 f8 site = 0.05 f9 ancillary building = 0.15

total physical plant cost PPC=516263.88 * (1+f1+f2+f3+f4+f5+f6+f7+f8+f9) PPC = 1239033.312 $

f10 design and engineering = 0.3 f11 contractors Fee = 0.05 f12 contingencies =0.10

fixed capital = 1239033.312* (1+0.3+0.05+0.1)=1796598.3 $ working capital . allow 5% of fixed capital to cover cost of the initial solvent charge = 1796598.3 *0.05= 89829.91 $ total investment required for project = 1796598.3 + 89829.91 =1886428.21 $

Annual operating costs. Reference table 6-3 appendix B operating time allowing for plant attainment = 300*0.95=285d/y ,285*24 = 6840 h/y

Variable costs

1- raw material solvent make up = 10 * 285 * 1 =2850 $ 2- miscellaneous material 10% of maintenance cost =89829.91*0.1 = 8982.991 $ 3- uitillite cost from table 6-5 appendix B Steam at 12 = 12 * 6840 * 200/1000 =16416 $ Cooling water at lpt = 1/100 * 6840 *7295 /1000= 498

Power at 2.2 p/mj =

(2.2/100)*300*285 =1881 $ 4- shipping and packaging and applicable Total variable cost = 14212.8 $

Fixed costs :

5= maintenance take as 5% of fixed capital =1796598.3 *0.05= 89829.91$

6- operating labour = No. of worker * annular salary = 25000 $

7- supervision = No. workers / 5 *annular salary of supervisor = 350/5 * 1000 $*12 = 840000 $

8- plant overheads .take as 50 % of operating labour=12500 $

9- laboratory . take as 30% of operating labour = 7500 $ 10- capital charges 10% of fixed capital (bank rate 8 %) = 1796598.3 *0.1=179659.8 $

11- insurance 1% of fixed capital = 1796598.3 *0.01=17969.5 $

12- local taxes = 0.02 * 1796598.3 =35931.9 $ 13- royalties = 0.05 * 1796598.3 = 89829.1 $ Total fixed capital = 1298217 Direct production cost = variable costs + fixed cost = 14212.8+1298217 =1312429 $

Annual cost operating =1312429 $ Basis = 1 day An annual production cost = annual operating cost / total production of biodiesel

= 1312429 /33336 kg/day =39.3 $

Profit = operating cost of liter *0.25 = 9 $

Price of one kilo of biodiesel = 10000/14212.8 = 0.7 $

Reference 1- https://ar.wikipedia.org/wiki

2- http://www.biodiesel.com/biodiesel/history/

2015.10.039.biortech.j/116-10//org.doi.dx://http -3

4- Revellome E.lternandez R.French w.ltomes w.Alley

E.J chemtechnol biotechnal 2010.614.20

5- Chemical engineering vol.6

6- Harker book

7- chemical properties hand book

8- Perrys chemical Engineers (2)

9- http://www.alrfeq.com/benifites of using the biofuel

10 - Physical-Properties-of-Pure-Methanol

11-http://www.london.ca/daspx/sewerand wastewater

/sewage treatment -index.htm

تم بحمد الله

biodiesel proudiction from sewage sludge 2222

Documents