Capture and Utilization of Capture and Utilization of Carbon DioxideCarbon Dioxide

Ethanol ProductionEthanol ProductionPresented By:Dana Al-Maiyas.

Supervised By:Prof.Mohamad A.Fahim.Eng.Yousif Ismael.

Table of ContentsTable of Contents1.Introduction of heat exchanger.2.Introduction of Shell & Tube heat

exchanger.3.Introduction of Plate heat

exchanger.4. Heater E-100 (Ethanol Plant ).5. Cooler E-101 (Ethanol Plant ).6. Heater E-100 (Car ).7.Heater E-101 (Car ).

8.Cooler E-102 (Car ).9.Heater E-103 (Car ).

Introduction

A heat exchanger is a device built for efficient heat transfer from one medium to another, whether the media are separated by a solid wall so that they never mix, or the media are in direct contact.

Types of Heat Exchanger.

* shell and tube heat exchanger

This type of heat exchanger consists of a shell (a large pressure vessel) with a bundle of tubes inside it. One fluid runs through the tubes, and another fluid flows over the tubes (through the shell) to transfer heat between the two fluids.

*Plate heat exchanger

-Plates are attractive when material costs are high.

-Plate Heat Exchanger are easier to maintain.- Plate Heat Exchanger are more suitable for highly viscous materials.

-The temperature correction factor ,Ft, will normally be higher with plate heat exchangers .

-Large heat transfer area afforded by the plates

Assumptions

1-Assume a counter current flow heat exchanger because it provides more effective heat transfer.

2-The value of the overall heat transfer coefficient was assumed in the shell and tube heat exchanger design.

3-Assume the outer, the inner diameter and the length of the tube on both the Shell and tube heat exchanger design.

Design Procedures

)A (Shell & Tube heat exchanger

1)Calculate Heat Load

Heat Load= m*cp*(T2-T1) Where: m= mass flow rate ( Kg/h) cp= Heat capacity ( KJ/kg°C) T2= inlet temperature of shell side°C T1= outlet temperature of shell side°C

2) Calculate Log mean Temperature difference

∆Tlm= ((T1-t2) – (T2-t1))/ln((T1-t2)/ (T2-t1)) where: ∆Tlm: Log mean temperature difference°C. T1 : Inlet shell side fluid temperature°C T2 : Outlet shell side fluid temperature °C t1 : Inlet tube side temperature°C t2: Outlet tube side temperature°C

3) Calculate the two dimensionless temperature ratios:

R= (T1-T2)/(t2-t1)S = (t2-t1)/(T1-t1)

Where:T1 : Inlet shell side fluid temperature T2 : Outlet shell side fluid temperature t1 : Inlet tube side temperature t2: Outlet tube side temperature

4) From Figure determine Ft by using R and S ratios Where : Ft : temperature correction factor .

5) Calculate True temperature difference ∆Tm= Ft *∆Tlm

6) Choose U from table depending on the type of flows in shell and tube sides. Where: U: Over all heat transfer coefficient .

7) Calculate provisional area A=(m/cp*(t2-t1))

Where: m: mass flow rate of the tube side flow (kg/s). cp: mass heat capacity ( KJ/kg°C). t1 : Inlet tube side temperature. t2: Outlet tube side temperature.

8) Assume inlet tube diameter, outlet tube diameter, and tube length.

9) Calculate area of one tube A = 3.14*L*do*10-3 Where: L: Tube length ( m) do: outlet diameter (mm)10) Define number of tubes Nt=( Provisional Area / Area of one tube )11) Calculate bundle diameter Db=Do(Nt/K1)(1/n1)

Where: Db: bundle diameter (mm). Do: tube outer diameter (mm). Nt: number of tubes. K1 and n1 are constants using triangle pitch of 1.25 .

12 (From Figure determine bundle diametrical clearance.

13) Calculate shell diameter Ds = Db+Dc Where: Db : bundle diameter( mm). Dc : clearance diameter( mm).

14) For Tube side coefficient calculate: - Mean temperature =((t1+t2)/2) - Tube cross sectional area A= (3.14/4)*di^2

- Tubes per pass = (Nt/2)

- Total Flow area = (Tubes per pass * A) Where: A : Tube cross sectional area , m2 - mass velocity = (m/ Total Flow area) Where: m : mass Flow rate Tube side , kg/s

- Linear velocity = (mass velocity/ ρ) Where: ρ : Tube Flow density , kg/m3

hi = ((4200*(1.35+0.02*t)*ut^0.8)/di^0.2)

Where: hi : Tube inside coefficient , W/m2°C. t : Mean temperature ( °C). ut : Linear velocity ( m/s). di : Tube inside diameter ( mm).Or; hi= ((kf/di)*jh*Re*(Pr)^0.33)

15) For Shell side coefficient calculate:

- baffle spacing = (Ds/5) - Tube pitch = 1.25 * do - As = ((pt – do)*DslB/pt)

Where: As : Cross Flow area , m2 pt : tub pitch lB : baffle spacing , m

Gs = (Ws / As) Where: Ws : Fluid flow rate on the shell side , kg/s - Equivalent diameter = (1.1/do)*(pt2-0.917do2) - Mean Shell side temperature = (T1+T2/2) - Reynolds number ( Re) = Gs de / µ - Prandtl number (Pr) = Cp µ / κ - hs = (kf*jh*Re*Pr(1/3)/de) and from figure @ Re we find jh .

16) Calculate over all coefficient

1/Uo= (1/ho)+(1/hod)+(doln(do/di)/(2kw))+(do/di)(1/hid)+(do/di)(1/hi)

Where:

Uo: the overall coefficient , W/m2°C ho: outside fluid film coefficient, W/m2 °C ,from table hi: inside fluid film coefficient.,from table hod : outside dirt coefficient (fouling factor) hid: inside dirt coefficient, W/m2°C kw: thermal conductivity of the tube wall material di : tube inside diameter, m do : tube out side diameter, m

17) Calculate pressure drop for:

- Tube side:

∆Pt = Np(8*jf*(L/di)+2.5)*(ρut^2/2)

Where: ∆Pt : tube side pressure drop , pa Np : number of tube side passes ut : tube side velocity (m/s) L : length of one tube ( m) Jf : tube side friction factor from figure @Re

-Shell side :

∆P = 8*jf*(Ds/de)*(L/lB)*(ρ*us^2/2)

Where: ∆P : shell side pressure drop (pa) Ds : shell diameter (m) lB : baffling spacing ( m) Jf : tube side friction factor from figure @Re

18) Thickness:

(t) = ((Pri)/(SEj-0.6P)) + Cc

Where : P: maximum internal pressure, kPa ri: inside radius of shell, m Ej: efficiency of joints as a fraction S: maximum allowable stress(for carbon steel), kPa Cc: allowance for corrosion, m

ResultsEquipmentHeater E-100

TypeShell and tube heat exchanger

Material of constructionCarbon Steel

Qtotal (kW) 2224

U(W/m² °C)146.4613845

Inlet temperture, Shell side °C500

Oultlet, Shell side °C100

Inlet temperature, Tube side °C24.12

Outlet temperature, Tube side °C305

Number of tubes562

Shell diameter, m0.725373277

LMTD, °C126.2069742

Heat exchanger area, m²103.8333078

InsulationGlass wool

Cost$ ,$49,800

EquipmentHeater E-101

TypeShell and tube heat exchanger

Material of constructionCarbon Steel

Qtotal (kW)4909

U(W/m² °C)79.999986

Inlet temperture, Shell side °C752.3

Oultlet, Shell side °C45

Inlet temperature, Tube side °C25

Outlet temperature, Tube side °C300

Number of tubes163

Shell diameter, m16.999544

LMTD, °C138.6193

Heat exchanger area, m²408.20342

InsulationGlass wool

Cost$ ,$92,400

)B (Plate heat exchanger

For the second plant (the car ) ,I used plate heat exchangr. The molar flow was too small.and I could not use shell & tube heat exchanger in the car .

By assuming ; d= 0.001 ft =0.000304 m L=0.1 m

C

T

TTT

T olm 37919244.23

)5025(

)10017.78(ln

)5025()10017.78(

ln1

2

12

)Number of transfer Unit ( NTU= (T1-T2)/∆Tlm

Typically , the NTU will range from 0.5 to 4, and for most applications will lie between 2 to 3 .

1:1 pass arrangement

Area required = ((Q*1000)/( U* Ft*∆Tlm))

Number of plates =(( Area required)/(3.14*L*d))

Results

EquipmentHeater E-100

TypePlate heat exchanger

Material of constructionMetal

Qtotal (kW)9.37E-02

U(W/m² °C)2000

Inlet temperture, process °C25

Oultlet, °C78.17

Inlet temperature, steam °C100

Outlet temperature, steam °C50

Number of transfer Unit2.274244508

LMTD, °C23.37919244

Number of plates 22

Heat exchanger area, m²0.002104957

InsulationGlass wool

EquipmentHeater E-101

TypePlate heat exchanger

Material of constructionMetal

Qtotal (kW)2.88E-01

U(W/m² °C)2000

Inlet temperture, process °C25.01

Oultlet, °C700

Inlet temperature, steam °C800

Outlet temperature, steam °C200

Number of transfer Unit5.036624733

LMTD, °C134.0163375

Number of plates13

Heat exchanger area, m²0.002104957

InsulationGlass wool

EquipmentHeater E-103

TypePlate heat exchanger

Material of constructionMetal

Qtotal (kW)1.88E-02

U(W/m² °C)2000

Inlet temperture, process °C25

Oultlet, °C96.8

Inlet temperature, steam °C150

Outlet temperature, steam °C70

Number of transfer Unit1.465734889

LMTD, °C48.98566618

Number of plates2

Heat exchanger area, m²0.000198958

InsulationGlass wool

EquipmentCooler E-102

TypePlate heat exchanger

Material of constructionMetal

Qtotal (kW)3.20E-01

U(W/m² °C)2000

Inlet temperture, process °C917.39

Oultlet, °C305

Inlet temperature, water °C20

Outlet temperature, water °C70

Number of transfer Unit1.186550523

LMTD, °C516.1095025

Number of plates3.342291517

Heat exchanger area, m²0.000319042

InsulationGlass wool