NATIONAL UNIVERSITY OF HO CHI MINH CITYHO CHI MINH UNIVERSITY OF TECHNOLOGY FACULTY OF CHEMICAL ENGINEERING DEPARTMENT OF UNIT OPERATIONS PROCESS AND EQUIPMENT DESIGN PROJECT DESIGN DIST
OVERVIEW
Ethanol, or ethyl alcohol, is a flammable and colorless organic compound commonly found in alcoholic beverages This transparent liquid has a distinct aroma, is lighter than water, and is highly volatile Additionally, ethanol is soluble in water, ether, and chloroform, making it a versatile substance in various applications.
When ethanol is produced as a mixed drink, it is pure cereal.
Ethanol can be used as an alcohol fuel (usually mixed with gasoline) Ethanol is also used in antifreeze products because of its low freezing point.
Refined ethanol and 95% ethanol are effective solvents, second only to water in common usage, and are widely utilized in products such as perfumes, paints, and medical alcohol Additionally, various ethanol-water mixtures can serve as solvents, with a 70% ethanol solution being particularly popular for its disinfectant properties.
Reaction at 300 o C, 70-80 atm with catalyst: wolframic acid or phosphoric acid:
Ethanol, utilized in alcoholic beverages and as a fuel source, is primarily produced through fermentation, where specific yeast species convert sugar in anaerobic conditions This process results in the production of ethanol and carbon dioxide, which can be summarized by general chemical reactions.
Water, a transparent, tasteless, and odorless substance with the chemical formula H2O, consists of one oxygen atom and two hydrogen atoms bonded covalently It exists in various forms, including precipitation as rain and aerosols like fog Clouds are created from suspended water and ice droplets, while finely divided crystalline ice can fall as snow.
Water is vital to the global economy, with around 70% of freshwater utilized in agriculture It serves as a primary food source through fishing in both salt and freshwater bodies Additionally, a significant portion of long-distance trade, including commodities like oil and natural gas, relies on waterways for transportation Water, ice, and steam are essential for heating and cooling in industries and homes As a versatile solvent, water is crucial in various industrial processes, cooking, and cleaning Furthermore, water, ice, and snow play a key role in numerous sports and recreational activities, including swimming, boating, fishing, diving, and skiing.
Table 1 : Liquid (x) - vapor (y) and boiling point of ethanol - water mixture at 1 atm x(%mol) 0 5 10 20 30 490 50 60 70 80 90 100 y(%mol) 0 33,2 44,2 53,1 57,6 61,4 65,4 69,9 75,3 81,8 89,8 100
Vapour Liquid Equilibrium of EtOH & Water
Equilibrium curve Line y=x x%mol y% m ol
Figure 1 : VLE diagram of EtOH & Water
Distillation is a separation technique that isolates individual components from a liquid mixture or a liquid-gas mixture, relying on the varying volatilities of the substances involved.
In the distillation process, a new phase is generated through evaporation or condensation, unlike absorption or desorption, which involves creating phase contact between two phases.
Considering a simple mixture with 2 components, we have:
● The top product consists of a large amount of most volatile component and a very small amount of less volatile component
● The bottom product consists of a large amount of less volatile component and a very small amount of most volatile component
Distillation allows us to separate multiple components, typically resulting in more than two products In a binary system like ethanol-water, the distillate primarily consists of ethanol with a small amount of water, while the bottoms are predominantly water containing only trace amounts of ethanol.
72.1 Classification based on working pressure
Distillation can be categorized into low pressure, normal pressure, and high pressure methods This technique fundamentally relies on the boiling temperatures of various components; when these temperatures are excessively high, reducing the working pressure effectively lowers the boiling point, facilitating the separation process.
72.2 Classification based on working principles
72.3 Classification based on heat – supplying method at tower bottom
In manufacturing processes, various types of towers are utilized, all sharing a fundamental requirement: a large contact phase surface area This is essential for effective fluid dispersion between different phases.
Distillation towers play a crucial role in various industries, particularly in oil refining, where large towers are essential for effective separation processes The dimensions of these towers, including their diameter and height, are influenced by factors such as the liquid and gas phases involved, as well as the desired purity of the final product This article explores two prevalent types of distillation towers: tray towers and packed towers, highlighting their distinct applications and efficiencies in the distillation process.
A tray tower features a cylindrical vertical design that incorporates various tray structures within its interior, effectively dividing the tower into equal segments In this setup, the liquid and vapor phases interact on the trays, facilitating the separation process The specific composition of these trays plays a crucial role in optimizing the tower's performance.
Bubble cap trays column: on Trays having bubble cap having form of circle, S,…, The structure is surrounded by grooves to pass the gas through and the transmission pipeline
A sieve tray column features trays with holes ranging from 3 to 12 mm in diameter, facilitating efficient separation processes In contrast, a packed column is a cylindrical tower composed of multiple segments joined by flanges or welding, where packing materials are arranged either randomly or sequentially to enhance mass transfer and efficiency.
Table 2 : Comparison of advantages and disadvantages of tower types
Packed column Sieve Tray column Bubble cap column
Easily working with dirty liquid
Relatively high efficiency, quite stable operation.
Less metal consumption than Bubble cap tray column.
Less energy consumption, so there are fewer trays.
The hindrance is quite high Complex structure. the distribution of phases according to the irregular tower cross section.
Requirement for rigorous installation -> Plate is flat.
For towers with a large diameter (> 2.4m), rarely use Sieve trays because the liquid is not evenly distributed on the tray
Great hindrance. Does not work with dirty liquids.
The distillation process utilizes various types of towers, and the selection of an appropriate distillation tower depends on factors such as purpose, performance, space, and economic conditions For our ethanol-water mixture, we opt for a bubble cap tray due to its distinct advantages.
High efficiency, stable operation, less energy consumption, so there are fewer trays.
MASS BALANCE
Feed concentration: x F = 0.60 (ethanol mass fraction)
Overhead concentration: x D = 0.90 (ethanol mass fraction)
Bottom concentration: x W = 0.05 (ethanol mass fraction)
The molecular weight of the inlet flowrate:
(1) The mole balance for distillation column: F= D+ B
(2) The ethanol conservation for the column: F x F = D x D + B x B
We therefore have the following system of equation:
The minimum reflux ratio represents the point at which the operating system aligns with an infinite number of theoretical trays At this ratio, while fixed costs become infinite, the operating costs, including raw materials, water, and pumps, reach their lowest levels.
(IX.24_P.58)[2] Determine the working reflux ratio from empirical formula (IX.25b_P.159)[2], we get:
12 Working equation with theoretical tray number
12.1 Working equation of rectifying section y= R R+ 1 x + x D
12.2 Working equation of stripping section y= R+ L R+1 x− L−1
To determine the theoretical trays number, we use this VLE graph shown below:
Realistic tray number: N Real = N Theo η avg (IX.59_P.170)[2]
11 Determine the average efficiency η avg
(IX.61_P.171)[2] With: x: Methanol mole fraction in liquid phase. y * : Methanol mole fraction in vapor phase.
Average liquid phase composition in the column:
The relative volatility for the whole mixture:
Based on liquid-vapor equilibrium T-xy diagram of ethanol-water, we obtain:
The viscosity for the whole mixture:
● Rectifying section: log log μ R = x R,ave × log log μ R, EtOH +( 1− x R,ave ) × log log μ R, H 2 O ¿ 0.5744 × log log(0.4387)+(1− 0.5744) ×log ( 0.3589) log log μ R =− 0.3949
● Stripping section: log log μ S = x S, ave × log log μ S, EtOH +( 1− x S, ave ) × log log μ S, H 2 O ¿ 0.1950 × log log (0.4093 )+(1−0.1950) × log log ( 0.3424 ) log log μ S =−0.4504
Determination of the average efficiency at each section:
12 Determine the realistic trays number
⇒ {N R ,real 035 trays N S,real =3.508 trays ⇒ { N R ,real trays N S,real = 4 trays
ENERGY BALANCE
According to the below equation:
21 Heat supply to preheater by hot water Q D1 :
With: D 1: The amount of steam for preheater (kg/h) λ 1 : Specific energy of steam (J/kg) r 1 : Latent heat of vaporization (J/kg) θ 1 : Temperature of condensate ( ° C )
C 1 : Water specific heat capacity (J/kg ° C ) With saturated steam at 2 at
Temperature of hot water at t s 9.62 ° C (T.43_P.42)[3]
22 Heat amount brought in by feed flow Q f :
With: F : The amount of feed product (kg/h)
C f : Feed product’s specific heat capacity (J/kg ° C ) t f : Initial temperature of feed product ( ° C )
23 Heat amount brought out by feed product Q F :
With: F : The amount of feed product (kg/h)
C F : Feed product’s specific heat capacity (J/kg ° C ) t F : Final temperature of feed product ( ° C )
24 Heat amount brought out by condensed water Q ng 1 :
With: D 1: The amount of steam (kg/h) θ 2 : Temperature of condensate ( ° C )
C 2 : Condensate’s specific heat capacity (J/kg ° C )
25 Heat loss to surroundings (about 5% of total heat loss) Q xq 1 :
With: D 1: The amount of steam (kg/h) r 1 : Latent heat (J/kg)
26 The amount of steam for preheater D 1 :
2II Energy balance for reboiler
According to the below equation:
21 Heat amount brought in the column by feed mixture Q F :
C F : Initial feed’s specific heat capacity (J/kg ° C ) t F : Temperature of initial feed ( ° C ) With t F 05 °C
22 Heat supply from reboiler to distillation column Q D2 :
With: D 2: The amount of steam needed to boil bottom mixture (kg/h) λ 2 : Specific energy of steam (J/kg) r 2 : Latent heat of water (J/kg) θ 2 : Temperature of condensate ( ° C )
C 2 : Water specific heat capacity (J/kg ° C ) With saturated steam at 2 at
Boiling point of this steam at t s 9.62 ° C (T43_P.42)[3]
23 Heat amount brought in by reflux stream the top section Q R :
With: G R : The amount of overhead reflux (kg/h)
C R : Overhead reflux’s specific heat capacity (J/kg ° C ) t R : Temperature of overhead reflux ( ° C ) With t R =t D x.68 ° C
24 Heat amount brought out by steam at the top section Q Y :
Q Y = D(1+ R) λ D (IX.159_P.197)[2] λ D =λ 1 y D + λ 2 (1− y D ) (IX.160_P.197)[2] With: D : Overhead product’s capacity (kg/h) λ D : Specific energy (J/kg) R: Reflux ratio λ 1 : Specific energy of ethanol in top section (J/kg) λ 2 : Specific energy of water in top section (J/kg) With t D x.68 °C
⇒ {r E 9 (kJ /kg )r W #12( kJ / kg ) (T.45 _P.38)[3] λ 1 =r E +t D C E 9 × 1000+ 78.68 × 3204=1.1011× 10 6 ( J / kg) λ 2 =r W +t D C W #12 1000+ 78.68× 4199=2.6424 ×10 6 ( J / kg)
25 Heat amount brought out by bottom product Q B :
With: B : The amount of bottom product (kg/h)
C B : Bottom product’s specific heat capacity (J/kg ° C ) t B : Temperature of bottom product ( ° C ) With t B 13° C
26 Heat amount brought out by condensed water Q ng 2 :
With: D 2: The amount of steam (kg/h) θ 2 : Temperature of condensate ( ° C )
C 2 : Condensate’s specific heat capacity (J/kg ° C )
27 Heat loss to surroundings (about 5% of total heat loss) Q xq :
With: D 2: The amount of steam (kg/h) r 2 : Latent heat (J/kg) With saturated steam at 2 at
Boiling point of this steam at t s 9.62 ° C (T43_P.42)[3]
28 The amount of steam reboiler D 2 :
With saturated steam at 2 at
Boiling point of this steam at t s = 119.62 ° C (T43_P.42[3]
(IX.163_P.198)[2] Heat duty of reboiler:
2III Energy balance for bottom cooler
The water amount we use to cool down bottom product:
2IV Energy balance for top condenser
In this system where we use complete condenser:
{r E 9( kJ / kg) r W #12(kJ / kg) (T.45 _P.38)[3] Applied formula: r =r E × x D +r W × ( 1−x D ) 9 × 0.9 +2312× (1−0.9)5.3 ( kJ / kg )
2V Energy balance for top cooler
In the system where we use complete condenser:
DISTILLATION COLUMN DESIGN
According to the below equation:
In a distillation column, the average vapor flow rate (V tb) and vapor speed (ω ytb) vary between the rectifying and stripping sections, leading to differences in the average vapor amount (g tb) and vapor density (ρ y) Consequently, this variation necessitates distinct diameters for the rectifying and stripping sections to optimize performance.
31.1 The average amount of vapor in distillation column g tb : g tb = g d + g 1
With: g đ : The outlet amount of vapor in the rectifying section (kg/h) g 1 : The inlet amount of vapor in the rectifying section (kg/h)
1.1.1 Determine the outlet amount of vapor in the rectifying section g đ : g đ =D ( R +1)46 ( 1.2981+ 1)= 4012(kg /h) (IX.92_P.181)[2]
1.1.2 Determine the inlet amount of vapor in the rectifying section g 1:
In the rectifying section, the amount of liquid in tray 1 is represented by G1 (kg/h), while the latent heat of the inlet mixture is denoted as r1 (J/kg) The relationship between these variables is defined by the equation g1 = G1 + D, where y1 is the vapor composition The latent heat of the inlet vapor, r1, can be calculated using the formula r1 = rE · y1 + (1 - y1) · rW, incorporating the latent heat of the outlet mixture, rđ (J/kg).
{r E 5( kJ / kg) r W #06( kJ / kg ) (T.45 _P.38)[3] r 1 = r E y 1 +( 1− y 1) r W = 845× y 1 +( 1− y 1) ×2306 ¿ 2306−1461 y 1 ( kJ / kg)
31.1.2.2 Determine the latent heat of the outlet vapor in the rectifying section r đ : y đ = y D= 0.8836 r đ = r EtOH × y đ +( 1 − y đ ) × r H 2O
⇒ {r E 9 (kJ /kg )r W #12( kJ / kg ) (T.45 _P.38)[3] r đ = r E × y đ +( 1− y đ ) × r W 9× 0.8836+( 1−0.8836) × 2312 ¿ 1019 (kJ /kg) With x 1 = x F =0.6∧ x D =0.9
31.2 The average speed of vapor mixture in the distillation column ω ytb :
To determine the average density of the vapor mixture (ρ ytb), the formula is given by ρ ytb = [y tb1 M EtOH + (1 − y tb1) M W] 273, where ρ xtb represents the average density of the liquid mixture in kg/m³, and ρ ytb denotes the average density of the vapor mixture in kg/m³ Additionally, the distance between two trays is specified as h = 0.3 m.
(IX.102 _P.183)[2] With: M EtOH : Molecular weight of Ethanol (kg/mol)
M H 2 O : Molecular weight of H2O (kg/mol) t tb : The average temperature of rectifying section ( ℃ ) y tb 1 : The average mole fraction of Ethanol in the rectifying section y tb 1 = y d 1 + y c1
2 =0.7022 With: y d 1: Mole fraction of Ethanol in vapor in feed tray y c 1 : Mole fraction of Ethanol in vapor in the rectifying section t tb = t F +t D
1.2.2 Determine average density of liquid mixture ρ xtb : ρ xtb = ( x ρ tb E + 1− ρ W x tb ) −1
(IX.104_P.183)[2] The average molar fraction: x tb = x F + x D
⇒ {ρ W 2(kg / m 3 ) ρ E s5( kg/m 3 ) (T.I.2_P.9)[1] ρ xtb = ( x ρ tb E + 1− ρ W x tb ) −1 = ( 0,7753 735 + 1− 972 0,7753 ) −1 w7.6 ( kg m 3 )
We choose h=0.3 m with the diameter range D= 0.6−1.2 m
32.1 The average amount of vapor in distillation column g' tb : g' tb = g' n + g' 1
To determine the outlet amount of vapor in the stripping section, denoted as g' n, use the formula g' n = g' 1 × 359 (kg/h), where g' 1 represents the inlet amount of vapor in the same section.
2.1.2 Determine the inlet amount of vapor in the rectifying section g ' 1:
In the stripping section, the amount of liquid in the first tray (G ' 1) is calculated as the sum of the inlet mixture's latent heat (r ' 1) and the product of the vapor's mass fraction (y B) and the flow rate (B) To determine the latent heat of the inlet vapor (r ' 1), it is essential to note that the vapor's mass fraction (y ' 1) is equal to the mass of the component (M E) multiplied by the vapor's mass fraction (y B).
⇒ {r ' E 1 (kJ /kg)r ' W "71( kJ /kg) (T.45 _P.38)[3] r ' 1 = r' EtOH y ' 1 +( 1− y ' 1) r ' H2O 1 0.3375 + (1 −0.3375 ) 2271 ¿ 1781.6( kJ / kg ) With x B =0.05
32.2 The average speed of vapor mixture in the distillation column ω' ytb :
To determine the average density of the vapor mixture (ρ ' ytb), use the formula: ρ ' ytb = [ y' tb1 M E + (1− y ' tb 1) M W ] × 273, where y' tb1 represents the mole fraction of the vapor component, M E is the molar mass of the vapor, and M W is the molar mass of the liquid Additionally, the average density of the liquid mixture (ρ ' xtb) is measured in kg/m³, and the distance between two trays is set at h = 0.3 m.
(IX.102 _P.183)[2] With: M EtOH : Molecular weight of Ethanol (kg/mol)
M : Molecular weight of H O (kg/mol) t ' tb : The average temperature of rectifying section ( ℃ ) y ' tb 1 : The average mole fraction of Ethanol in the rectifying section y ' tb 1 = y đ 1 + y w
2 = 0.3983 (kg /h) With: y 1: Mole fraction of Ethanol in vapor in feed tray y w : Mole fraction of Ethanol in vapor in the stripping section t ' tb = t B + t F
2.2.2 Determine average density of liquid mixture ρ ' xtb : ρ ' xtb = ( x ' ρ' tb E + 1−x ' ρ ' W tb ) −1
(IX.104_P.183)[2] The average molar fraction: x' tb = x F + x D
{ ρ' EtOH r7.3 (kg /m 3 ) ρ' Water = 966.3(kg / m 3 ) (T.I.2_P.9)[1] ρ ' xtb = ( ρ' x ' tb EtOH + 1− ρ ' Water x' tb ) −1 = ( 0.3824 727.3 + 1−0.3824 966.3 ) −1 = 858.43 ( kg m 3 )
We choose h=0.3 m with the diameter range D= 0.6−1.2 m
We choose the diameter for the distillation column D t =1.0 (m ), and the distance between two plates, choose h=0.3 (m)
3II Height of the tower H tower :
With: H : Height of the tower body (m)
N tt : Actual number of trays, N tt plates
H đ : The distance between two plate (m), H đ =0.3 (m) δ : The thickness of a plate (m), Choose δ =0.004 (m ) 0.8 ÷ 1: allowable distance between top and bottom
32 The bottom and the head of the tower
Use the bottom and the standard ellipsoidal head for the tower with welded body, placed vertically - internal pressure is greater than 7.104 N/m 2 (§.1_P.381)[2]
Calculate the bottom and the head completely the same.
Select parameters for the head (or bottom) according to the tower body diameter:
Straight Flange height of ellipsoidal head: SF =h flange % mm=0.025 m
The column height measures 6.63 meters, excluding the areas for the top and bottom reflux pipes, welding, and packed material for the column's flange When accounting for these additional components, we estimate that the total height of the tower will be approximately three times the distance between the trays, which is around 1 meter.
We calculate the weir length by computing the angle which covers our weir, α
We choose the segment so that its area equals to 7% of the total plate area.
The segment area is also known of as downcomer area (P.572)[8]
Areaof sector ˚ − Area of triangle= Areaof segment(1) ˚ weir length
3IV Bubble cap and downcomer calculation:
Choose the riser diameter d h = 100(mm)=0.1( m) (§.2_P.236)[2] The number of caps distributed on a plate n : n=0.1× D 2 d 2 h =0.1 × 1.0 2
(IX.212_P.236)[2] With: D : The inner diameter of the distillation column (m) d h : The riser’s diameter (m)
The height of the cap above the riser h 2: h 2 =0.25 d h =0.25 ×0.1=0.025 (m) (IX.213_P.236)[2] The bubble cap radius d ch : d ch = √ ❑ (IX.214_P.236)[2]
With: δ ch : The thickness of the caps (m), δ ch = 2÷ 3 (mm), choose δ ch =3 mm d ch = √ ❑ (IX.214_P.236)[2] ¿ 0.1457 =0.146( m)
The distance from plate’s surface to cap’s leg S:
S= 0 ÷0.025 (m), choose S= 0.0125 (m) (§.2_P.236)[2] The liquid level above the cap’s slot: h 1 = 15÷ 40(mm), choose h 1 = 30(mm) (§.2_P.236)[2]
The height of the cap’s slot b:
The distance between each slot c: c=0.003÷ 0.004(m), choose c=0.003 (m) (§.2_P.236)[2] The width of the cap’s slot a: a=0.002 ÷ 0.007(m ) (§.2_P.236)[2]
The height of cap’s slot b : b= ξω 2 y ρ y g ρ x (IX.215_P.236)[2]
With: ξ : Resistance coefficient, with ξ=1.5 ÷ 2 ρ y : vapor phase’s specific density (kg/m 3 ) ρ x : liquid phase’s specific density (kg/m 3 ) g : gravitational acceleration, g= 9.81 (m / s 2 )
Vapor’s velocity through the bubble caps ω y : ω y = 4 V y
With: V y : vapor’s flow rate (m 3 /h) d h : inner diameter of vapor pipe (m) n: number of caps on a plate
The number of slots in each bubble caps ⅈ :
With: c : The distance between each slot (m) b : The height of cap slot (m)
The minimum of cap’s step on the plate t min =d ch + 2 δ ch + l 2 (IX.220_P.237)[2]
With: l 2 : The minimum distance between each cap (m) l 2 5+0.25 d ch 5 +0.25 ×146I (mm)
We have in rectifying section:
{ ρ xtb w7.6 (kg / m 3 ) ρ ytb =1.3027( kg /m 3 ) g tb 642 ( kg/h)
(P236)[2] The height of cap’s slot: b= ξω 2 y ρ y g ρ x (IX.215_P.236)[2]
We have in stripping section:
{ ρ' xtb = 858.43(kg / m 3 ) ρ ' ytb =0.9853( kg/ m 3 ) g' tb (14.5 ( kg/ h)
(P236)[2] The height of cap’s slot: b= ξ ω 2 y ρ y g ρ x (IX.215_P.236)[2]
We have the relation of c, a, and the caps diameter:
⇨ a=2.88 3(mm), which satisfied the condition of a=0.002 ÷ 0.007(m ).
32 Weir and downcomer and riser
Assume the weir thickness is 3mm.
The height of liquid mixture above the weir h ow : h ow =2.84 E × ( Q L w L ) 2 ∕ 3
(E.5.3_P.110)[4] With: h ow : The height of liquid mixture above the weir (m)
Q L : Flow rate of liquid mixture (m 3 /h)
Liquid flowrate is determined by the formula:
With: M L : The average molecular weight of top product and feed (kg/kmol)
M D : The average molecular weight of top product (kg/kmol) ρ l : The specific weight of liquid in rectifying section (kg/m 3 )
2 4.315( kg/ kmol ) ρ l = ρ xtb w7.6 (kg / m 3 ) R=1.2981 D46 ( kg /h ) M F ).40 (kg / kmol) M D 9.87 (kg / kmol)
Liquid flowrate is determined by the formula:
With: M C : The average molecular weight of bottom product and feed (kg/kmol)
M D : The average molecular of top product (kg/kmol) ρ l : The specific weight of liquid in stripping section (kg/m 3 )
M F : The average molecular weight of feed (kg/kmol)
M L : The average molecular weight of bottom product and feed (kg/kmol)
2 #.99 (kg /kmol) ρ l =ρ ' xtb 8.43 ( kg/ m 3 ) R= 1.2981 D46 ( kg / h) F '00 (kg / h) M F ).40 ( kg/ kmol ) M D 9.87 ( kg / kmol) M w 58 (kg /kmol)
The average liquid flowrate in the distillation column
From X = 1.0581 and L D w = 0.8726, we search for the value of E:
We have the weir height h c : h c =(h¿¿1+b +S +h sr )− Δh¿ (IX.219_P.237)[2]
With: h 1: The liquid level above the cap’s slot, we chose h 1 = 30(mm) b : The height of the cap’s slot, b=0.035( m)
The selected distance from the plate's surface to the cap's leg is S = 0.0125 meters The height of the liquid mixture above the downcomer's weir is set at Δh = h_ow = 8.55 millimeters Additionally, the length from the cap's slot to the bottom of the cap is chosen to be h_sr = 5 millimeters.
We have the height of liquid on the plate h m : h m =h 1 + S+ h sr +b
With: h sr : The length from the cap’s slot to the bottom of cap, choose h sr =5( mm)
2.3.1 Gradient of liquid’s height on the plate Δ : Δ=C g Δ ' n
(E.5.5_P.111)[4] With: C g : Adjustable coefficient for vapor flowrate Δ ' : Gradient of liquiq’s height pass through a caps’ row n : The number of caps’ row, pick n = 3
With Adjustable coefficient for vapor flowrate C g = f (1.34 Q L
With: B m : the average width of caps installation (m)
With the average width of the area between two B m :
With: A : the average area between the two segments downcomer (m 2 )
I : the distance between two segments downcomers (m)
We have speed of vapor v : v= 4 V G π ( D t ) 2 =
We have the average molecular weight of gas phase in the column: ρ G = ρ ytb + ρ ' ytb
From X =5.8325 search for the value of C g from (P.5.10_P.111)[4]:
From these values, search from (5.11-5.14_P.112)[4], we get4 Δ ' :
4 ×3=1.8 (mm) Assume the riser height is 90 (mm): h riser = 90(mm)
3V Structure and resistances of bubble cap trays:
The resistance of the distillation column:
With: N tt : The realistic number of plates
∆ P đ : The total resistance on 1 plate ( N / m 2 )
∆ P s : Resistance due to surface tension ( N / m 2 )
∆ P t : Hydrostatic resistance caused by liquid on the plate ( N / m 2 )
(IX.137_P.192)[2] With: ω 0: vapor’s velocity through the bubble caps (m/s) ρ y : vapor phase’s specific density (kg/m 3 ) ξ : resistance coefficient, ξ= 4.5−5, choose ξ=4.5
Vapor speed in bubble caps slot ω 0: ω 0 = g tb
{ ρ ytb =1.3027 (kg/ m 3 ) g tb 642( kg/ h) n caps ix slots / capsa=0.00288 ( m) b=0.035 ( m ) ω 0 = g tb
Vapor speed in bubble caps slot ω 0: ω 0 = g' tb
{ ρ' xtb = 858.43(kg / m 3 ) ρ ' ytb =0.9853( kg/ m 3 ) g ' tb (14.5 (kg /h)n capsix slots /capsa=0.00288 (m )b=0.035 (m ) ω 0 = g ' tb
32 Resistance due to surface tension ∆ P s :
With: σ: The surface tension of the liquid on the tray (N/m 2 ) d td : Equivalent diameter of the bubble cap’s slot when fully open (m) When bubble caps’s slot fully open: d td = 4 × f x Π
With: f x : Surface area of bubble cap’s slot (m 2 ) Π : Equivalent perimeter of the bubble cap slot’s (m) d td = 4 f x Π = 4 a b
33 Hydrostatic resistance caused by liquid on the Tray ∆ P t :
∆ P t = ρ b × g× ( h b − h 2 r ) ;( N / m 2 ) (IX.139_P.194)[2] With: h r : The height of the bubble cap slot (m), h r =b=0.035 ( m) ρ b : Density of the foam (kg/m 3 ), With ρ b =(0.4 ÷0.6 ) ρ x h b : The height of foam layer on the plate (m) g : gravitational acceleration (m/s 2 ), With g= 9.81(m /s 2 ) h b = (h¿¿ c+ ∆− h x ) ( F− f ) ρ x + h x ρ b f +(h ch −h x ) f ρ b
With: h c : The height of weir raised above the plate (m), h c = 75.562u(mm)
∆ : The height of the liquid above the weir, ∆ =h ow =8.55 (mm) h x : The height of the liquid mixture (without foam) on the plate (m), h x =h m 5 ( mm)
F : The installation area of bubble caps (subtract 2 downcomers area) (m 2 ),
The formula for calculating the foam density (ρb) is given by F = A = 0.5969(m²) ρb, where ρb ranges from 0.4 to 0.6 times the average density of the liquid mixture (ρx) on the plate The total area of all bubble caps on the plate (f) is determined using the equation f = 0.785 dch².n, resulting in a value of f = 0.1673(m²) when n equals 10 and dch is 0.146 m The height of the bubble cap (hch) is specified as 8 mm.
{ ρ xtb w7.6 (kg / m 3 ) ρ ytb =1.3027 ( kg /m 3 )g tb 642( kg/ h) ρ b =( 0,4 ÷ 0,6) ρ xtb =( 311.04÷ 466.56 ) ( m kg 3 ) ,choose ρ b @0 ( m kg 3 )
{ ρ' xtb = 858.43(kg / m 3 ) ρ ' ytb =0.9853( kg/ m 3 ) g' tb (14.5 ( kg/ h) ρ b =( 0.4 ÷ 0.6) ρ x =(343÷ 515) ( m kg 3 ) , choose ρ b = 400 ( m kg 3 )
33.3 Total hydrostatic resistance of column:
∆ P=∆ P rectifying + ∆ P stripping =( N tt × ∆ P đ ) rectifying +( N tt × ∆ P đ ) stripping ¿ 15 × 182.3695+ 4 × 173.79431( N / m 2 )
34 Checking the operation of the distillation column
34.1 The openness of the bubble cap’s slot h S : h S =7.55 × ( ρ L ρ − G ρ G ) 1/ 3 × b 2 /3 × ( V S S y ) 2 /3 (E.5.2_P.108)[4] With: b : The height of the bubble cap slot (m), h r =b=0.035( m)
V y : Flow rate of vapor phase (m 3 /s)
S s : The total area of slot on a plate (m 2 ), S s =n.i a b
{ ρ xtb w7.6 (kg / m 3 ) ρ ytb =1.3027( kg /m 3 ) g tb 642 ( kg/h)
Percent of bubble caps’ openness h s b = 43.010
{ ρ' xtb = 858.43(kg / m 3 ) ρ ' ytb =0.9853( kg/ m 3 ) g' tb (14.5 ( kg/ h)
Percent of bubble caps’ openness h s b = 38.4520
The height of the liquid level with no foam for downcomer h d : h d =h w + h ow + ∆ + h t +h d ' (5.9_P.115)[4]
With: h w : The height of weir on plate (mm), h w =h c =0.075 (m) h ow : The height of liquid above the weir (mm), h ow = 8.55(mm)
The gradient of liquid height on the plate is represented as ∆, measuring 1.8 mm, while the pressure drop of the vapor phase through the plate is indicated by ht Additionally, hydraulic losses due to liquid flow from the downcomer to the plate are denoted as hd' To prevent flooding in the tower during operation, it is crucial that the height hd remains less than 1.
Hydraulic losses due to liquid flow from the downcomer to plate h d ' : h d ' =0.128 × ( 100 Q S L d ) 2 ( mm) (5.10_P.115)[4] With: S d : The area between the downcomer and plate (m 2 )
With the distance of two segment areas is I =0.7798 ( m), choose the length from weir to the below downcomer’ edge h i =0.125 (m)
⇨ The distance from downcomer below edge to plate is, h dp u−12.5b.5 (mm)
Total pressure drop of vapor phase through a plate h t : h t =h fv +h s +h ss +h ow + 1
The pressure drop caused by friction and airflow velocity changes through the caps in the absence of liquid is measured in millimeters (mm) The bubble cap's slot openness is also expressed in mm, with an average value of h savg at 0.5954 mm Additionally, the length from the liquid surface above the slot to the weir is represented as h ss, calculated using the formula h ss = h w − ∆∨ 75 − 82.5∨ 7.5 (mm).
The pressure drop due to friction and the change in airflow velocity through the caps when there is no liquid h fv : h fv '4 × K × ( ρ L ρ − G ρ G )( Q S G r ) 2 (mm) (5.8_P.115)[4]
With: S r : The total of riser area on one plate (m 2 )
We have K coefficient value search from table using the ratio of S s aj rj
With: S aj : The area between caps and riser (m 2 ), S aj = ( ⅆ ¿¿ ch 2 −ⅆ h 2 ) π
S rj : The area of the cap’s riser (m 2 ), S rj = ( ⅆ h 2 ) π
With S s aj rj =1.1316, search the table (T.5.16_P.115)[4] to get K coefficient:
{ ρ xtb w7.6 (kg / m 3 ) ρ ytb =1.3027( kg /m 3 ) g tb 642 ( kg/h) ρ' xtb 8.43 (kg / m 3 ) ρ ' ytb =0.9853 ( kg /m 3 ) g' tb (14.5 (kg / h)V y '96 (m 3 /h) V ' y = 2856.49 (m 3 /h) ρ G = ρ ytb + ρ ' ytb
⇨ The column will not flood under operation.
34.3 Check the liquid flow in the downcomer d tw :
The value to describe the flow into the downcomer which is uniform are not is illustrated as d tw : d tw =0.8 × √ ❑ (5.12_P.116)[4]
With: h o : The height free falling (m ), h o = H +h w + h d
H : The distance between each plate (m )
Assume the average breadth of the downcomer is 20% of the column diameter: d w =0.2× 1.0=0.2( m)
According to (P.116)[4], the value of d tw should not exceed 60% of the breadth of the downcomer: d tw =0.04394< 60 % × 0.2=0.12: satisfy the condition.
This proves liquid flow into the downcomer uniformly and will not have any impact on the column wall.
MECHANICAL CALCULATION
- Equipment working pressure: p mt =1 atm1325 N / m 2
The bottom of the distillation column is the most critically impacted area of the equipment, necessitating careful consideration of the static pressure exerted by the liquid mixture at this lowest section.
- We calculate the device structure under the maximum working temperature condition: t B 1 o C
- The mean density of the liquid phase of whole column ρ L = ρ xtb + ρ ' xtb
For an ethanol-water distillation column with a maximum working temperature of 95.1°C, to ensure the quality of the product, material stainless Steel X18H10T is selected
The cylindrical body of the column is designed using electric arc welding with two-sided welding techniques, ensuring it operates under normal pressure The various components of the column are securely connected through flanges, maintaining structural integrity.
In a case where the equipment’s walls are holeless and completely reinforced φ h =φ =0.95
The environment is a vapor - liquid mixture so the calculation pressure is the sum of vapor pressure p mt and hydrostatic pressure p l of liquid column:
⇨ The device is under internal pressure.
The thickness of cylindrical column body, working under internal pressure, is determined by the formula S:
With: D 1: Inner diameter (m) φ : Equipment wall’s vertical durable coefficient
C : additional coefficients (corrosion, abrasion and thickness tolerances) p : Inner pressure ( N / m 2 ) Choose:
Corrosion coefficient: time of use of equipment is 20 years, we assume corrosion rate of ethanol < 0.05 mm/year → C1 = 1 mm
When checking durability of device details, using the allowable stress is determined by the formula:
Additional coefficient due to corrosion, abrasion and thickness tolerances is determined by formula: C =C 1 +C 2 +C 3 + C 0 =0.0018 (m) (XIII.17_P.363)[2]
With: C 1: Additional coefficient due to corrosion, Choose C 1 =0.001(m)
C 2 : Additional coefficient due to abrasion (Material contains solid particles moving at high speed in the equipment), Choose C 2 =0( m)
C 3: Additional coefficient due to thickness tolerances, Choose C 3 =0.0008
0.1625536682 ×0.95 0 ≥25 (E.5-1_P.95-96)[5] Then with formula (XIII.17_P.363)[2] or (E.5-3_P.96)[5], the tower body thickness is:
⇨ Minimum thickness: Smin = 3.0 mm, choose S= 0.005 m=5 mm
We have to test whether S = 5mm satisfies the following condition:
The calculated test pressure is determined by the formula: p 0 = p th + p l (XIII.27_P.366)[2]
With: p l : Hydrostatic test pressure of water ( N /m 2 ) p th : Hydraulic test pressure ( N /m 2 ) p l = ρ Water g.H (XIII.10_P.360)[2] ¿ 818.015 ×9.81 × 7.63a229( N /m 2 ) p th = p l +1000001229 (N / m 2 ) (T.XIII.5_P.358)[2]
Conclusion: The thickness of cylindrical column body is S = 5.0 mm
4III Bottom and head thickness S ' :
Choose ellipsoidal head made by stainless steel X18H10T:
(XIII.47_P.385)[2] with: h b : Dished height of the bottom (N / m 2 ), h b =h t =0.25( m) k : Dimensionless coefficient, k =1
C : Additional coefficient, choose C =2 (mm) ( condition : S−C =5−1.8 =3.2 mm≤ 10mm) (S.1_P.386)[2]
{Steel sheet ' sthickness 4 −25 mmTensile strengthlimit [ σ k ]U0× 10 6 (N / m 2 )U0( N / mm 2 )
The bottom and head thickness are:
→ Choose: S ' =0.005 m for the easy fabrication.
After calculating the thickness, we need to check the wall stress according to the test pressure by the formula: σ = [ D t 2 + 2h b ( S' −C ) ] p 0
1.2 3.333 × 10 6 (N / m 2 ): satisfied condition Conclusion: The thickness of bottom and head is S’ = 5.0 mm
4IV Diameters of pipes - flanges connecting parts of equipment with pipes
Tubes are usually connected to the device by a removable or non-removable joint In this device, we use removable joints.
For removable joints, a pipe is made with a short tube and a flange or screw – thread to connect to the pipe
Flanged type commonly used with tube diameter d > 10 mm
Usually choose screw – thread with d ≤ 10 mm , sometimes we can choose d ≤ 32 mm
Pipe is made by stainless steel X18H10T
Flange is made by stainless steel X18H10T, structure is slip-on flanges.
Determine the pipe diameter when knowing the flow and speed: d= √ ❑
We have: Working pressure: P=0.1625536682( N m m 2 ) Search: The average speed of the fluid and gas (T.II.2_P.370)[1]
{t D x.68 o C x D =0.9 → {ρ EtOH s6.254 (kg /m 3 ) ρ Water = 972.726 (kg /m 3 ) ρ D = 1 x D ρ Ethanol + 1− x D ρ Water
Top vapor product pipe ρ ytb
The flange is an important part used to connect parts of equipment as well as to connect other parts to the body
When selecting a body joint flange, opt for a design featuring a bottom and head constructed from X18H10T steel The flange is a neckless type, which is integral to the equipment due to its methods of welding, casting, or forging This flange is primarily suited for applications involving low to medium pressure.
41 Flanges connected to equipment body
Choice of steel flanges, flange type 1 (T.XIII.27_P.421)[2]
Dt: internal diameter D: flange diameter Db: The length of two symmetric bolt. h: Height of flange db: bolt diameter
∆ h00 mm => Distance between flanges: 1800 m (T.IX.5_P.170)[2] Number of trays between 2 flanges: N P =6 trays
The number of actual flanges ¿ N N t
6 + 2=5.33=5(flanges ) There are 5 flanges used.
The tightness of a flange is primarily influenced by the packing material, which is softer than the flange itself When bolts are tightened, the packing layer deforms to conform to the flange's surface irregularities To maintain optimal tightness, we utilize a 3mm thick asbestos packing layer.
42 Flanges connecting equipment parts and pipes
Choose flanges made of black metal, type 1 (T.XIII.26_P.409-416)[2]
Internal diameter of the column (mm)
Search pipe fitting length (T.XIII.32_P.434)[2].
Table of length dimension values of pipe fittings
The mass of the whole tower is calculated by the formula: m=6 × m 1 flange + 20 ×m 1 tray + m body + m liquid +2 × m bottom /head
● Mass of 1 flange connected to equipment body:
Equipment made of stainless steel X18H10T: ρ X 18 H 10T y00 ( m kg 3 ) (T.XII.7_P.313)[2] m 1 flange = π
● Mass of liquid in the tower
Suppose the tower is filled with water
● Bottom and head mass: m bottom =m head G.9 kg (T.XIII.11_P.384)[2]
Choose supports that have parameters: (T.XIII.35_P.437)[2]
Allowable load on the supporting surface: q=0.56 × 10 6 ( N / m 2 )
AUXILIARY EQUIPMENT
Choose shell-tube heat exchanger type, tubes are made by stainless steel, size: 38 x 2 Horizontal equipment
External diameter: d ng 8 mm=0.038 m (T.VI.6_P.80)[2]
Internal diameter: d tr =d ng −2 δ t 8−2∗24 mm=0,034 m d mean = 0.038 + 0.034
Choose saturated steam as burner working at 2 atm:
Latent heat of vaporization: r "08000 J /kg (T.I.251_P.314)[1] Bottom stream temperature:
Choose reserve heat transfer, so:
∆ t LMTD = ∆ t max − ∆t min ln ln ( ∆ t ∆ t max min ) = 89.62 ln ( 89.62 38.57 −38.57 ) `.55 (K ) (V.8 _P.5)[2]
The heat transfer coefficient K is calculated by the formula as for flat walls:
With: α n : : heat supply coefficient of the steam burner ( m w 2 K ) α s : Heat supply coefficient of the feed outside tube ( m w 2 K )
∑ ❑ ❑ r t : Heat resistant through the tube walls and from the fouling factor ( m W 2 K )
52.1 Heat transfers through tube wall and fouling layer: q t = t w 1 −t w 2
With: t w 1 : Temperature of the wall exposed to steam (ºC) t w 2 : temperature of the wall exposed to the feed outside tube, ºC Thermal resistance through the pipe wall and fouling layer:
The thermal conductivity coefficient of steel X18H10T: λ t 3 W /m K
52.2 Heat supply coefficient of the feed outside the tube α s =7.77 × 10 −2 × ( ρ ρ h l − ×r ρ B h ) 0.033 × ( ρ σ l ) 0.333 × μ 0.45 λ ×c 0.75 0.117 × q ×T 0.70 s 0.37
{ λ EtOH =0.1651 (W /m K )λ Water =0.6536 (W / m K ) (T.I.130_P.135)[1] {σ EtOH =0.01936 ( N / m 2 ) σ Water = 0.06696 ( N / m 2 ) (T.24_P.25)[3] {c EtOH )11.825 ( J / kg.K ) c Water = 4186.644( J / kg K ) (T.1.154_P.172)[1] {r EtOH 2933( J / kg)r Water $26383( J / kg) (T.1.65_P.254)[1] Density of vapor phase of feed product: ρ h = P M HW
Density of liquid phase of feed: ρ l =ρ F 3.491( kg/ m 3 )
Bottom product solution dynamic viscosity:
{à EtOH =0.6402( cP) à Water = 0.5048(cP) log( à F ) =0.3699 ì log (0.6402 )+(1−0.3699) ì log ( 0.5048)
Thermal conductivity λ B = λ EtOH x F + λ Water ( 1 −x F ) −0.72 x F (1− x F )( λ Water − λ Ethanol ) (I.32_P.124)[1] ¿ 0.1651 × 0.6 + 0.6536 × ( 1−0.6 )−0.72 ×0.6 ( 1−0.6 ) (0.6536−0.1651 ) ¿ 0.2761(W / m K )
Specific heat coefficient c B =c EtOH x F +( 1− x F ) c Water ¿ 2911.825× 0.6 +(1−0.6 )× 4186.644 ¿ 3421.75 ( J / kg.K )
Latent heat of evaporation r B =r EtOH x F +( 1− x F ) r Water ¿ 882933 ×0.6 + ( 1−0.6 ) ×242638300313.116 (J / kg)
Based on previous work, we choose velocity of feed flow (outside the tube): v =0.25 ( m/ s)
Calculate Nusselt: Nu=0.021 × ε 1 × ℜ 0.8 × Pr 0.43 × ( Pr Pr t ) 0.25
With: ε 1 : The coefficient of influence of the heating factor: the ratio between the length
L and the diameter d of the pipe (T.3.1_P.110)[6] ⇒ choose ε 1 =1
52.3 Heat supply coefficient of steam burner in the tube α s =1.28 × √ 4 μ× r × λ ( t n −t 3 w × ρ 1 ) × d 2 ng ,W / m 2 K
Use iteration to determine t w 1 and t w 2 Choose t w 1 2.7 o C
Average temperature of condensate water: t mean = t n + t w 1
{water density : ρ N 5.88( kg/ m 3 )water viscosity : μ N =0.24198× 10 −3 ( Ns /m 2 )thermal conductivity : λ N = 0.68404(W / m K ) water ’ sheat of vaporization: r N "12707 (J /kg)
Consider if the heat loss is negligible then q t = q s 482(W /m 2 )
Choose number of tubes: n = 7 (tube), hexagon distribution.
= 2.5 0.034 t >50=¿ ε 1 =1 (satisfied condition) Number of tubes on the diagonal line: b = 3 (tube) (T.V 11_P.48)[2] t=1.2 ×d=1.2 × 0.038=0.0456( m )= 45.6( mm)
Internal diameter of the device:
Choose Kettle type, which is similar to 1-2 shell-tube horizontal heat exchanger.
The equipment is made by stainless steel, tubes size: 38 x 2
External diameter: d ng 8 mm=0.038 m (T.VI.6_P.80)[2]
Internal diameter: d tr =d ng −2 δ t 8−2∗24 mm=0,034 m d mean = 0,038 + 0,034
Choose saturated steam as burner working at 2 atm:
Latent heat of vaporization: r "08000 J /kg (T.I.251_P.314)[1] Bottom stream temperature:
Choose reserve heat transfer, so:
∆ t LMTD = ∆ t max − ∆ t min ln ln ( ∆ t ∆ t max min ) = 24.49 ln ( 24.49 19.62 −19.62 ) !.9651( K ) (V.8 _P.5)[2]
The heat transfer coefficient K is calculated by the formula as for flat walls:
(V.5 – P.3)[2]With: α n : : heat supply coefficient of the steam burner ( m w 2 K ) α s : Heat supply coefficient of the bottom product outside tube ( m w 2 K )
∑ ❑ ❑ r t : Heat resistant through the tube walls and from the fouling factor ( m W 2 K )
52.1 Heat transfers through tube wall and fouling layer: q t = t w 1 −t w 2
The temperature of the wall exposed to steam (t w 1) and the temperature of the wall exposed to the bottom product outside the tube (t w 2) are critical factors in assessing thermal resistance through the pipe wall and the fouling layer Understanding these temperature differentials is essential for optimizing thermal efficiency and ensuring effective heat transfer in industrial applications.
The thermal conductivity coefficient of steel X18H10T: λ t 3 W /m K
52.2 Heat supply coefficient of the bottom product outside the tube α B =7.77 ×10 −2 × ( ρ ρ h l − × r ρ B h ) 0.033 × ( ρ σ l ) 0.333 × μ 0.45 λ × c 0.75 0.117 × q × T 0.70 s 0.37
{ ρ EtOH q8.3133 (kg / m 3 ) ρ Water 9.7045 (kg / m 3 ) (T.I.2_P.9)[1]{ λ EtOH =0.1613 (W /m K )λ Water =0.6730 (W / m K ) (T.I.130_P.135)[1]{σ EtOH =0.01572 ( N /m 2 ) σ Water =0.05935 (N / m 2 ) (T.24_P.25)[3]{c 483.48 (J /kg K ) c B25.13( J / kg.K ) (T.1.154_P.172)[1]
{r EtOH 4.974 ( kcal/ kg)6317.1432(J / kg)r Water T1.435( kcal / kg)"66880.058( J / kg)
Density of vapor phase of bottom product: ρ h = P M HW
Density of liquid phase of bottom product: ρ l =ρ B 5.611( kg/ m 3 )
Bottom product solution dynamic viscosity:
{à EtOH =0.33927 (cP) à Water =0.29289(cP) log( à B ) =0.0201 ì log( 0.33927 )+(1−0.0201) ì log ( 0.29289)
Thermal conductivity λ B = λ EtOH x B + λ Water ( 1− x B ) −0.72 x B (1− x B )( λ Water −λ Ethanol ) (I.32_P.124)[1] ¿ 0.1613 × 0.05 + 0.6730 × ( 1−0.05 )−0.72 × 0.05 ( 1−0.05 ) (0.6730 −0.1613 ) ¿ 0.64543 (W / m K )
Specific heat coefficient c B =c EtOH x B +( 1− x B ) c Water ¿ 3483.48 ×0.05 +(1−0.05 ) × 4225.13 ¿ 4188.0475 (J / kg K)
Latent heat of evaporation r B =r EtOH x B +( 1− x B ) r Water ¿ 816317.1432× 0.05 +(1− 0.05 )× 2266880.058!94351.912 ( J /kg )
Based on previous work, we choose velocity of bottom product (outside the tube): v=0.25 ( m/ s)
Calculate Nusselt: Nu=0.021 × ε 1 × ℜ 0.8 × Pr 0.43 × ( Pr Pr t ) 0.25
With: ε 1 : The coefficient of influence of the heating factor: the ratio between the length
L and the diameter d of the pipe (T.3.1_P.110)[6] ⇒ choose ε 1 =1
52.3 Heat supply coefficient of steam burner in the tube α s =1.28 × √ 4 μ× r × λ ( t n −t 3 w × ρ 1 ) × d 2 ng ,W / m 2 K
Use iteration to determine t w 1 and t w 2 Choose t w 1 7.75 o C
Average temperature of condensate water: t mean = t n +t w 1
{water density : ρ N 3.986 (kg /m 3 ) water viscosity : μ N =0.2354 ×10 −3 ( Ns/ m 2 ) thermal conductivity : λ N =0.68467(W / m K ) water’ s heat of vaporization : r N "05835.5 ( J / kg)
Consider if the heat loss is negligible then q t = q s 3355.9(W /m 2 )
Choose number of tubes: n = 61 (tube), hexagon distribution
Number of tubes on the diagonal line: b = 9 (tubes)
Hence, the reboiler would consist a total of 9x2 rows.
The length of the tubes would be half of the total length:
Internal diameter of the device:
For optimal performance, select a shell-tube heat exchanger with stainless steel tubes measuring 38 x 2 mm The cooling water should have an inlet temperature of 25°C and an outlet temperature of 40°C Additionally, the steam exiting the tube should have a condensation temperature of tD x 68 (°C).
Average temperature of cooling water t mean = t 1 +t 2
Choose reserve heat transfer, so:
(V.8 _P.5)[2] Choose shell – tube heat exchanger, horizontal direction.
Heat transfer pipes are made of by X18H10T, size 38x2,
External diameter d ng 8 mm=0.038 m (T.VI.6_P.80)[2]
Internal diameter d tr = d ng −2 δ t 8 −2∗2 4 mm=0.034 m d mean = d ng +d tr
With: α n : : heat supply coefficient of the steam burner ( m w 2 K ) α W : Heat supply coefficient of the bottom product outside tube ( m w 2 K )
∑ ❑ ❑ r t : Heat resistant through the tube walls and from the fouling factor ( m W 2 K )
52.1 Heat transfers through tube wall and fouling layer q t = t w 1 −t w 2
● t w 1 : the temperature of the wall exposed to condensation vapor, ºC
● t w 2 : temperature of walls exposed to cold water, ºC
Thermal resistance through the pipe wall and fouling layer:
●The thermal conductivity coefficient of steel X18H10T: λ t 3 W /m K
52.2 Heat coefficient of water passes inside tubes
Select the velocity of water going through the tube: v n =1 m /s
Number of tubes: n = 4 G n ρ n π d tr 2 v n = 4 × 17.7056( kg/ s)
Choose n(tubes) (T.V.11_P.48)[2] Number of tubes on the diagonal line of hexagon: b=5 tubes
Actual velocity of water in the pipe: v n = 4 G n ρ n π d tr 2 × n = 4 ×17.7056 ( kg/ s)
Reynolds number: ℜ n = v n d tr ρ n μ n ¿ 1.0323 (m / s)× 0.034 (m) × 994.25 (kg / m 3 ) 0.765 × 10 −3 [ N s m 2 ] × [ kgm/ N s 2 ] E616> 10 4
Nusselt number: Nu n =0.021 ε 1 ℜ n 0,8 Pr n 0,43 ( Pr Pr w n ) 0,25
●ɛ1: The coefficient of influence of the heating factor: depends on ReN and the diameter d of the pipe Choose ɛ1 =1.
● Prw : Prandlt number of water at average temperature of wall.
The heat supply coefficient of water goes in the inner tube: α n = Nu n λ n d tr α N = Nu N λ N d tr = 342.1 × 0.625
52.3 Heat coefficient of overhead vapor condensed outside tubes
●Does not contain non-condensing air
●Vapor condensed on the outside of the pipe
From figure V.20 - page 30 - [2], 🡪 the coefficient depends on the tube layout and the number of tubes per vertical row: ɛtb = 0.75 (on aveage, each vertical row contains 3.8 tubes). α R=¿ɛ tb α 1 ¿
Heat load outside the tube: q condensed =q R =α R ( t D −t W 1) = α R ( 78.68−t W 1) = A ×(78.68− t W 1 ) 0.75 (IV.6)
From (IV.4), (IV.5), (IV.6), determine tw1, tw2.
Use iteration to determine t w 1 và t w 2 Choose t w 1 X o C
Average temperature of condensate: t ' mean = t D +t w 1
{liquid phase density : ρ R v4.33 ( kg/ m 3 )input viscosity : μ R =0.3996 × 10 −3 (Ns /m 2 )thermal conductivity : λ R =0.18264 (W / m K )latent heat of condensation : r R 17610( J /kg) α 1 =1.28 × √ 4 0.3996 1017610× × 10 −3 764.33 × (78.68 2 × −58) 0.18264 × 0.038 3 #59(W / m 2 K) α R =0.75 α 1 69.2 (W / m 2 K )
Consider if the heat loss is negligible then q t = q condensed 6587(W /m 2 ) t w 2 X−36587 × 4.675269727 × 10 −4 = 40.89 o C
Number of tubes on the diagonal line of hexagon: b=9 tubes
Tubes are arranged in an equilateral hexagon
0.034 2 >50 →ε 1 =1 (satisfied condition) Tubes are arranged in an equilateral hexagon t=1.2 ×d mean =1.2 × 0.036=0.0456 (m )E.6 ( mm)
Internal diameter of the device:
Choose tube-tube heat exchanger type, tubes are made by stainless steel, size: 38 x 2 Choose cooling water with inlet temperature: t1 = 25 o C, outlet temperature: t2= 40 o C. Flow out of the tube with temperature: t D x.68 (¿ o C )¿
Average temperature of cooling water t mean = t 1 +t 2
Choose reserve heat transfer, so:
(V.8 _P.5)[2] Choose shell – tube heat exchanger, horizontal direction.
Heat transfer pipes are made of by X18H10T, size 38x2,
External diameter d ng 8 mm=0.038 m (T.VI.6_P.80)[2]
Internal diameter d tr = d ng −2 δ t 8 −2∗2 4 mm=0.034 m d mean = d ng +d tr
With: α n : : heat supply coefficient of the steam burner ( m w 2 K ) α W : Heat supply coefficient of the bottom product outside tube ( m w 2 K )
∑ ❑ ❑ r t : Heat resistant through the tube walls and from the fouling factor ( m W 2 K )
52.1 Heat transfers through tube wall and fouling layer q t = t w 1 −t w 2
● t w 1 : the temperature of the wall exposed to condensation vapor, ºC
● t w 2 : temperature of walls exposed to cold water, ºC
Thermal resistance through the pipe wall and fouling layer:
●The thermal conductivity coefficient of steel X18H10T: λ t 3 W /m K
52.2 Heat coefficient of water passes inside tubes
Select the velocity of water going through the tube: v n =0.25 m/ s
Number of tubes: n= 4 G n ρ n π d tr 2 v n = 4 × 1.0361 ( kg/ s)
Number of tubes on the diagonal line of hexagon: b=3 tubes
Actual velocity of water in the pipe: v n = 4 G n ρ n π d tr 2 × n = 4 × 1.0361( kg/ s)
Reynolds number: ℜ n = v n d tr ρ n μ n ¿ 0.1640(m /s )× 0.034 ( m )× 994.25(kg / m 3 ) 0.765 × 10 −3 [ N s m 2 ] ×[ kgm N / s 2 ] r4750 →ε 1 =1 (satisfied condition) Tubes are arranged in an equilateral hexagon t=1.2 ×d mean =1.2 × 0.036=0.0456 (m )E.6 ( mm)
Internal diameter of the device:
Choose tube-tube heat exchanger type, tubes are made by stainless steel, size: 38 x 2 Choose cooling water with inlet temperature: t1 = 25 o C, outlet temperature: t2= 40 o C. Flow out of the tube with temperature: t D x.68 (¿ o C )¿
Average temperature of cooling water t mean = t 1 +t 2
Choose reserve heat transfer, so:
(V.8 _P.5)[2] Choose shell – tube heat exchanger, horizontal direction.
Heat transfer pipes are made of by X18H10T, size 38x2,
External diameter d ng 8 mm=0.038 m (T.VI.6_P.80)[2]
Internal diameter d tr = d ng −2 δ t 8 −2∗2 4 mm=0.034 m d mean = d ng +d tr
With: α n : : heat supply coefficient of the steam burner ( m w 2 K ) α W : Heat supply coefficient of the bottom product outside tube ( m w 2 K )
∑ ❑ ❑ r t : Heat resistant through the tube walls and from the fouling factor ( m W 2 K )
52.1 Heat transfers through tube wall and fouling layer q t = t w 1 −t w 2
● t w 1 : the temperature of the wall exposed to condensation vapor, ºC
● t w 2 : temperature of walls exposed to cold water, ºC
Thermal resistance through the pipe wall and fouling layer:
●The thermal conductivity coefficient of steel X18H10T: λ t 3 W /m K
52.2 Heat coefficient of water passes inside tubes
Select the velocity of water going through the tube: v n =0.25 m/ s
Number of tubes: n = 4 G n ρ n π d tr 2 v n = 4 ×1.1464 ( kg / s)
Number of tubes on the diagonal line of hexagon: b=3 tubes
Actual velocity of water in the pipe: v n = 4 G n ρ n π d tr 2 × n = 4 × 1.1464 (kg / s)
Reynolds number: ℜ n = v n d tr ρ n μ n ¿ 0.1815(m /s )× 0.034 ( m )× 994.25(kg / m 3 ) 0.765 × 10 −3 [ N s m 2 ] ×[ kgm N / s 2 ] 20