CN2125 Mini‐Project State Student Names and Matriculation Number Here 1.7 m 0.22064 5 mm 5 mm A B S We are embarking on a trip to Osaka, Japan in May. The temperature of Osaka at night is predicted to be around 15 0 C. Therefore we will need to wear slightly thicker clothing there. Despite the thick clothing, our body will continuously lose heat when we are outdoors. Hence, the objective of this mini project is to find out how long we can stay outdoors at night in Osaka before we start to experience mild hypothermia, which occurs at approximately 35 0 C. This models an UNSTEADY STATE 2-Dimensional heat transfer problem. - We have fixed the geometry of the human body to be a cylinder of 1.70m height and 0.22064 m in diameter - Prior to leaving for outdoors, we assume that initially, the human body has uniform distribution of heat throughout its core and extremitie. Hence, T A,o = T B,o = 37 o C. -Additionally, since the layer of clothing is very thin, we further establish that the temperature gradient across the clothing is negligible such that T s,o = 37 o C. - In order to determine the lower threshold time taken for the onset of mild hypothermia, we have to assume body heat generation to be zero since body heat generation only sets to prolong the onset time. Hence q generation = 0, to find this limiting case. Due to varying surface temperatures, we first approximate our film temperature, T using initial conditions. , ൌ ଷାଵହ ଶ ൌ 26 Ԩ ൌ . Properties of air at 299.15 K to be given as the following, k air (W m -1 K -1 ) Pr ࢼ2.61733175 × 10 -2 0.7082125 1.34774 ൈ 10 ݎܩൌ ఉ ఘ మ ఓ మ ܮଷ ∆ ൌ ሺ1.34774 ൈ 10 ሻሺ1.71 ଷ ሻሺ37 െ 15ሻ ൌ 1.48258 ൈ 10 ଵ ൌ .ଶଷସ ଵ.ଵ ൌ 0.13488 ଷହ బ.మఱ ൌ 0.1003029628 Since ଷହ బ.మఱ , a vertical cylinder can be evaluated using correlations for vertical plane walls Ra L = Gr L Pr = 1.0500 ൈ For vertical plane walls, ݑൌ ۉ ۈ ۇ0.825 0.387 ଵ ൬1 ሺ0.492/ݎሻ ଽ ଵ ൰ ଶ ی ۋ ۊଶ ൌ 256.10679 ൌ ݑ ܮൌ 256.10679 ൈ 2.61733 ൈ 10 ଶ 1.71 ൌ 3.9200 ଶ ܭଵ Example 1
Transcript
CN2125 Mini‐Project State Student Names and Matriculation Number Here
1.7 m
0.22064
5 mm 5 mm
A B S
We are embarking on a trip to Osaka, Japan in May. The temperature of Osaka at night is predicted to be around 150C. Therefore we will need to wear slightly thicker clothing there. Despite the thick clothing, our body will continuously lose heat when we are outdoors. Hence, the objective of this mini project is to find out how long we can stay outdoors at night in Osaka before we start to experience mild hypothermia, which occurs at approximately 35 0C.
This models an UNSTEADY STATE 2-Dimensional heat transfer problem.
- We have fixed the geometry of the human body to be a cylinder of 1.70m height and 0.22064 m in diameter - Prior to leaving for outdoors, we assume that initially, the human body has uniform distribution of heat throughout its core and extremitie. Hence, TA,o = TB,o = 37oC. -Additionally, since the layer of clothing is very thin, we further establish that the temperature gradient across the clothing is negligible such that Ts,o = 37oC. - In order to determine the lower threshold time taken for the onset of mild hypothermia, we have to assume body heat generation to be zero since body heat generation only sets to prolong the onset time.
Hence qgeneration = 0, to find this limiting case.
Due to varying surface temperatures, we first approximate our film temperature, T using initial conditions.
, 26 .
Properties of air at 299.15 K to be given as the following,
We further assume the human body to have the properties of water since water constitutes 70% of the human body. At 310K, properties of water: k = 0.6276, α=0.1515 X 10-6 m2/s
Bi =
. . .
. . . . .
Since 0.1<Bi<100, the Heislei charts are used (WWWR textbook Appendix F, Figure 7 & 8).
Therefore, the lower threshold time taken for the onset of mild hypothermia is approximately 2400s (40 minutes).
CN2125 Heat and Mass Transfer AY 2010/2011 Semester 2 Project Work — Instant Noodles State Student Names and Matriculation Number Here Introduction In this project, we will find out if the 3 minutes cooking time as suggested by the instant noodle manufacturer is valid. Typical net weight of noodles in 1 packet of instant noodles = 85g (information obtained from packaging of instant noodles) Diameter of noodles ≈ 1mm Initial temperature of uncooked noodles = 25˚C Calculating the convective heat transfer coefficient of water at 100˚C: Assumptions: (i) Temperature of boiling water is constant at 100˚C (ii) The composition of water remains unchanged, and hence the properties of water
remain unchanged (iii) Assume the 15cm by 15cm square base of the cooking pot is large such that it mimics
natural convection on a horizontal plate (iv) Base of the cooking pot is maintained at a constant temperature of 150˚C (v) Assume heating of water only occurs from the base of the pot and none from the sides.
∞
2150 100
2 125 At 125˚C, properties of liquid water:
193.1175 10 0.67775 / · 1.525
Values taken from Appendix I of WWWR textbook
∆ 193.1175 10 150 100 0.15 3.2588 10
· 3.2588 10 1.525 4.96975 10 For a hot surface facing up,
0.14 · 0.14 4.96975 10 514.722
⇒ · 514.722 0.67775
0.15 2325.687 / · Calculating the time taken to cook instant noodles in boiling water: Assumptions to simplify the calculation: (vi) Properties of noodles = properties of water at 100˚C since the water content in
noodles is high during cooking. (vii) Noodles are infinitely long cylinder (viii) Heat transfer into noodles is only by conduction (ix) Noodles do not expand during the course of cooking (x) Temperature of boiling water is constant at 100˚C (xi) Noodles are cooked once the temperature in the core reaches 90˚C (since it’s instant
noodles)
Example 2
At 100˚C, 0.682 / · 4211 / · 958.4 /
Values taken from Appendix I of WWWR textbook
.. . .
. 0.8525 The Temperature-Time Charts are used.
0.682
958.4 4211 1.689866 10 /
∞
∞
100 90100 25 0.1333
00.5 0
0.6822325.687 0.5/1000 0.58649
∴ from Fig.F5 of WWWR, 1.0
1.00.5100 1.689866 10 147.878 2.46
∴The time taken to cook 1 packet of instant noodles in boiling water is approximately 2.46 min, and this value corresponds to the recommended cooking time for instant noodles of 2 to 3 minutes as stated at on the packaging material. Error Analysis and Evaluation
(i) The property of noodles was assumed to be equivalent to that of water, which is not a very accurate assumption at the initial stage of cooking.
(ii) The noodles actually expand and absorb water in the process, thus its property changes and is also different from the assumed property of noodles.
(iii) The proximity of the strands of noodles is close enough such that it affects the heat transfer to noodles. This was not considered in our calculations.
(iv) The boiling water means that there are bubbles that rise from the bottom of the pan to the water surface. This affects the assumption that heat transfer to the noodles is by conduction and convection from the liquid water.
(v) For horizontal plate, the recommended Ra range for the equation of Nu is 2x10-7 to 3x10-10. But in this case, the Ra is out of this range. Hence extrapolation was used
Conclusion We found that the time needed to cook the noodle is 2.46 minutes. This value falls within the 3 minutes suggestion, however it should be noted that the assumptions we made will add uncertainty to our results. Hence our calculated value is only an estimation to verify if the recommended cooking time is valid. Ultimately, the best way to find out if the noodles is cooked to your preference (hard or soggy) or not is to try a noodle from the pot and decide if you want to turn off the fire or continue to let it boil. Afterall, the recommended cooking time is a range of 2 to 3 minutes, subjected to personal preference. References: WELTY J.R., WICKS C.E., WILSON R.E. & RORRER G.; Fundamentals of Momentum, Heat, and Mass Transfer 5th Edition, 2007. John Wiley & Sons
CN2125 Mini Project
State Student Names and Matriculation Number Here
R
Sugar δ
Lollipop Size and Consumption Time Introduction Lollipop has been a very popular candy snack among kids and juveniles for decades. In this mini project, the relation between lollipop size and time for lollipop consumption is explored and suggested size of lollipop is given to with current product design size. The mechanism for lollipop consumption is related to the diffusion of the components in saliva and continuous removal of it by swallowing motion. The pseudo steady state at the surface of the lollipop can be applied to calculate the disappearing rate of the candy sphere and Chilton-Colburn analogy is used to calculate the mass convection rate of components into saliva. Major ingredients of common lollipop are white sugar (sucrose), corn syrup (glucose) and other additives. Sucrose and glucose are used in our model. Data Sucrose (C12H22O11) Glucose (C6H12O6) Saliva (H2O) Diffusivity DS =4.3×10-6 cm2/s Diffusivity Dg =6.7×10-6 cm2/s Kinematic viscosity
Possible volume ratio of sugar and corn syrup is 1:1/3 to 1:½. Model calculation Saliva flows through the surface of lollipop (spherical geometry) and the average flow cross-section area is calculated as
δπδπ
RR
drrS
R
==∫
2
220 --- (1), where δ is the thickness of the layer on
the surface of lollipop. The concentration of components in the sugar film is assumed 100% of maximum solubility in the saliva (water).
Using Chilton-Colburn Analogy
32
2Sc
vkC
j cfD
∞
== (2)
πδυυυ
VRvvDD
22Re =×
== (3)
SV
=ν (4)
32
)(4υ
υ DiR
kci ×= (5)
where D
fCRe16
= since the saliva flow
rate is considered to be laminar Reynolds number where υ is the flow velocity of saliva on the surface of lollipop Mass convection factor where i= s, g. It has no relation with saliva velocity or flow rate.
Example 3
Pseudo steady state
dtdR
MRRN
i
ii
ρππ 22 44 = (6)
)( ∞−= iiscii CCkN (7) Calculation of average concentration in lollipop
gg
gs
s
saveT x
Mx
MMC ×+×==
ρρρ )( (8)
g
gg
s
ss
s
ss
s
MV
MV
MV
xρρ
ρ
+= sg xx −=1 (9)
Concentration of sucrose and glucose in sugar layer (from solubility) Sucrose solubility
molgcmgcm
molgg
mLwaterg
/18/1100
/3.342200
100200
33 ×=
(10)
0.095=+
=watersucrose
sucroses nn
ny ; 0.083=gy
Substitute all values into the equation
dtdR
MRRCkRCk avegscgsscs )(44)0(4)0( 222 ρπππ =−+−
Solve the equation, we have relation between time and R,
2)(4)(4
)/( 2
32
32
R
CD
CD
Mt
sgg
sss
ave ×
+
=
υυ
υυ
ρ
Ni is the flux of component i Cis is the surface concentration of component i and Ci∞ is the concentration of component i outside the film which is 0 in assumption. CT is the concentration of lollipop, xs and xg is the molar fraction of sucrose and glucose on the lollipop Vs and Vg is the volume fraction of sucrose and glucose in lollipop where ys is the maximum molar fraction of sucrose in sugar layer and yg is that of glucose Water molar concentration:
3/056.0 cmmolM
Cw
ww ==
ρ
swss yCC ×= and gwsg yCC ×=
Conclusion Set Vs:Vg=1:1/2, tabulate the t for R from 1cm to 3.5cm
The radius of 1.2-1.3 cm of lollipop will give a consumption time of around 30 minutes. The real product size of lollipop is about 2.7 cm diameters for Chupa Chups brand lollipop which meets the requirement of their customers to have a 20-minute consuming time (only licking and no crash of it). The best optimized time depends on people’s fulfillment and marginal pleasure changing with time which can be done by survey the targeting customers. The manufacturer can also increase diameter of the lollipop if they want the consumer to enjoy more time on individual lollipop.
Iron Man lies in the freezing tundra of Antarctica, battered and bruised from the recent battle.
“You might have won the battle, but you haven’t won the war Stane,” he groaned in a metallic modulation, while heaving a sigh of relief that he survived the blast of the AR-130 missile.
His internal computer system groans to a restart and indicates to him that while his suit is fully functional; his Arc Reactor battery has been damaged from the blast and requires immediate replacement. Iron Man painstakingly reaches down to his legs and opens the compartment containing various Arc Reactor battery replacements.
In the compartment storage unit, which stores the batteries safely at a temperature of 113K, he finds 3 usable replacements:
However, Iron Man is now faced with a dilemma. While these batteries are working, they require a core temperature of 345K in order to power up his metallic suit and to keep his life support unit going.
Iron Man’s backup power unit activates, indicating that he only has 6mins left to live. What he has to do now is to choose the battery with the geometry such that within 6mins of placing this battery inside the suit’s heating unit (h=100 W/m2K), it will be heated to the minimum operational core temperature of the battery of 345K. So which one does he choose? Will Iron Man live?
Example 4
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ron Man wilthat would bowever, the rs Iron Manthat Iron Ma
ate for him toboth of the le
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c Battery 65K
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average temp
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eries are assuure, To = 11
perature (29
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or 12.6 secengineering
attery. Mr Stprevent a re
umed to be o13K. The un
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g the equation
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e cubic shapeational tempbattery to reaconds to ma
g genius, wetark would aepeat of this
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Iron Man’s sx 104 kg/m3
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National University of Singapore Faculty of Engineering
Chemical & Biomolecular Engineering Department
CN2125 Heat and Mass Transfer Mini Project
Rib-eye Steak Cooking Time
State Student Names and Matriculation Number Here
Example 5
Jason, a guy who is keen to be a chef, decided to come out with a recipe of making rib-eye beef steak. Time for different degree of cooking was calculated by him after a thorough research on beef compositions, core temperature for different degree of cooking and other related information. Core temperature for different degree of cooking is summarized below:
Table 1. Degree of cooking and the corresponding core temperature To simplify the model, calculations are based on the following assumptions:
1. The shape of rib-eye steak is a cylinder. Usually dimension of the steak is 0.1m in diameter tabulated as D and 0.03m in thickness tabulated as H.
2. The steak will not shrink during cooking; 3. The cooking process is an one-dimensional heat transfer; 4. The convective heat transfer coefficient, h is obtained from literature as 15W/m K
(INCROPERA F.P. & DEWITT D.P., 2007). 5. The steak is flipped so frequently during cooking that there is no heat loss and
property change on the uncooked side when the other side is in contact with the pan surface.
6. During calculation, we assume symmetric and simultaneous heating from both sides. The actual time needed is twice the time calculated. The heat is transferred only in the axial dimension of the cylindrical steak.
7. Initial temperature for beef is set to be T0= 10 ; pan temperature, Tpan= 300 ; 8. The specific heat capacity of the steak above freezing, cp=2.81kJ/kg·K 9. The conductivity of the steak is calculated using model proposed by R.G.M. van
der Sman, from Agrotechnology and Food Sciences, Wageningen University based on its composition.
Table 2: Composition of a rib-eye steak and its conductivity The conductivity of the steak is given by (R.G.M. van der Sman, 2007):
k k1 / / 1 2Q δ
1 / 1 2Q δ
Where, , is volume fraction of the continuous phase. Assuming beep is of one continuous
phase, is 1 in this case. , are volume fraction of unfrozen solution, water, in this case, and
insoluble , fat and protein, in this case, respectively.
In meat emulsion, Q . The effective conductivity of the insoluble (fat and
Fat Protein Water Total Weight Percent 12% 26% 62% 100% Density(kg/m3) 0.9 1.35 1 1.057
Conductivity(W/mK) 0.222 0.410 0.633 0.669
And, δ 0.1725 (where k k )
Conductivity is calculated as 0.669. Cooking the steak is actually an unsteady conduction process of a cylinder. To
find out the time it will take to achieve a certain core temperature (ie. the temperature at the symmetry center), Heissler Charts can be used.
Take cooking raw steak, core temperature after cooking, T = 125 , as an example.
Calculation of the three dimensionless ratios:
Unaccomplished temperature change, Y = T TT T
= = 0.603;
Relative position, n = = .
= 0; (x1=H/2=0.015m)
Relative resistance, m = = ..
= 2.9733
Read from Appendix F in the textbook, the relative time, X = α = 2.15.
The thermal diffusivity of the steak, α = ρ = . = 2.25 10 m /s
Thus, t= X·α
= . ..
= 2150s = 35.83min
The cooking time for raw cooked steak, time = 2t = 71.6 min Similar calculation can apply on medium raw, medium, medium well and well
done cooked steak. The results are tabulated as below: Y n m X Time (min) Raw 0.603 0 2.973 2.15 71.6 Medium Raw 0.586 0 2.973 2.35 78.2 Medium 0.552 0 2.973 2.60 86.6 Medium Well 0.517 0 2.973 2.75 91.6 Well Done 0.483 0 2.973 2.90 96.6
Table 3. results for five degrees of cooking streak Thus in conclusion, to cook a normal size rib-eye beef steak, cooking time ranges
from 71.6 minutes to 96.6 minutes in order to achieve different degrees of cooking. Using the methods tabulated above, the cooking time for different sizes of rib-eye steak and different types of beef steaks can all be calculated. The theoretically calculated time is a good reference for new learners to control the degree of cooking for the beef steaks. Reference INCROPERA F.P. & DEWITT D.P. (2007). Fundamentals of Heat and Mass Transfer, 6th Edition. Publisher John Wiley & Son, New York. P284 R.G.M. van der Sman. (2007). Prediction of enthalpy and thermal conductivity of frozen meat and fish products from composition data. Elsevier, 13 (3), 16.
CN2125 Mini Proj
ProblemIn 2007,mosquitoNational mosquitomosquitowater (poThe granby NEAactive inthe insecas its efflast for 1we wanteffectivenModelinThe tranconstant respectivinitial Teinfested wTemephoAlso, it ithe mosq0.012ppmeffective SolutionFirst, it (specie BWilke-ChTemphosvolume oVC = 14.VP = 17.0
H
VA is fowater, μBsolution, obtained
ject
m Statement there were
oes are the rEnvironme
oes. One of oes, is to introtential breenular insecticA, have 1%ngredients. Ncticides to befectiveness a1 month (30t to determness for 1 m
ng the Diffusnsient diffus
temperaturvely. The graemephos conwater. Consos through tis assumed tquitoes oncem (Carvalhomosquito m
n: is necessary
B). Temphoshang correlas has a moleof Temphos .8cm3/mol 0cm3/mol
Hence, VA = (
ound to be 4B is 1.45cp.
T is 298.1:
State Stude
: e about 800root cause oent Agency the methodsroduce prescding ground
cides, as reco% TemephosNEA recomme added onc
against mosq0 days). In tmine the sizmonth. sion of Temsion of Temre and presanular insecncentration idering a limthe liquid bthat the Teme it reaches o 2004) of
mortality.
y to calculats is an orgaation is a suitecular formucan be evalu
VH = 3.7VS = 15.
(16)(14.8)+2
405.7cm3/mo The associ5K. Substit
..
ent Names an
0 dengue caof spreading
(NEA) advs NEA recomcribed amoud of mosquitoommended s as their mends that ce a month quitoes can this report, ze of the g
mephos in Wmephos in wssure of 29cticides, assuof 1ppm, ar
miting case woundary lay
mephos is imthe bulk meTemephos
te the diffusanic insectictable equatioula of C16Huated below:7cm3/mol .5cm3/mol
20(3.7)+(6)(
ol. Moleculaiation paramtuting the ab
ΦBMB
nd Matriculat
Page 1
ases and 20g dengue, thuvocates the mmends to e
unt of granuloes) is unavo
granular tha
Water Mediuwater mediu98.15K andumed to be re placed in where the re
yer surroundmmediately cedium, so Cin the cent
sivity of Tecide and a non to calcula
H20O6P2S3 . A:
VO = 7.Correct
7.4)+(2)(17.
ar weight ometer of watbove values
,
Fig
ion Number
0 dengue faus to prevenprevention
eradicate thelar insecticidoidable or di
at can main
um um occurs d 101325 Pspherical, h1L of larva
esistance of ding the grancarried away
CAS=0ppm. Itre of the g
emephos (spnon-electrolyate DAB. At normal b
4cm3/mol ion (2 benze
.0)+(3)(15.5
of water, MBter, ΦB is 2s into the W
.
gure 1: Structur
Here
Figure 2: S
atalities in Snt the sprea
of the bree breeding gde into placeifficult to rem
ntain the d
at Pa
has ae-the film manular is negy and consuIt is desiredgranular to
pecie A) in yte molecule
boiling point
ene rings) =
)-30 = 405.7
B is 18.0g/m.26. The tem
Wilke-Chang
/
x1
al Formula of Te
Spherical Granu
Singapore. Aad of dengueeeding of Agrounds of Aes where stagmove.
desired mor
ass transfer ogligible, so umed to eradd to have at achieve an
a water mee in water s
t, the molec
30.0cm3/mo
7cm3/mol
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1
emephos
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Aedes e, the Aedes Aedes gnant
rtality
of the kc=0.
dicate least 98%
edium o the
culare
ol
ity of of the n, we
Example 6
CN2125 Mini Project
State Student Names and Matriculation Number Here
Page 2
For the transient diffusion of Temephos in water medium, the Hessler’s Chart may be employed. The boundary conditions are: CA = CA0 =1 ppm at t=0 for 0≤ r≤R CA = CAS = 0 mg/mg at r=0 for t ≥ 0 CA = CAS = 0 mg/mg at r=R for t ≥0
Dimensionless ratio, Y CAS CACAS CA
. 0.012
Relative time, X DAB 2.64 x10 6cm2s 1 30days 24hr 60min 60s
Relative position, n 0
Relative resistance, m DAB 0, convective resistance is negligible
From Appendix F, Figure F.3, the corresponding value of X is 0.55.
X DAB 2.64 x10 6cm2s 1 30days 24hr 60min 60s 0.55
radius of granular x . Discussion of Results: It has been found that the minimum radius should be 3.53cm. However, most commercially available granular insecticides is less than 1cm diameter, thus they will not last for 1 month. This discrepancy can be due to that we have taken the mosquito mortality effectiveness to be 98%, while the commercially accepted effectiveness may be less than 98%. We have also assumed that all there is CAS = 0 and convective resistance is negligible, however in reality, there may be residual Temephos present in the bulk medium and there is significant surface resistance. Conclusion: Thus, in order to allow the insecticide to last for 1 month, it can be suggested that the granular insecticides be packed into spherical sachet of 3.53cm in radius, assuming that mass transfer only occurs at the exterior surface of the satchet. In fact, a similar approach has already been adopted by the Singapore Arm Forces (SAF) in the military camps, and it was reported that this helped to save SAF an estimated of $178000 annually (Mindef). Reference: Carvalho.M.S, C. D. (2004). Susceptibility of Aedes aegypti larvae to the insecticide temephos in the Federal District. Brazil: Rev Saude Publica
National Environment Agency. (2010). Campaign against dengue. Retrieved April 1, 2010, from National Environment Agency: http://www.dengue.gov.sg/
Welty.J. R., Wicks.C. E., Rorrer.G. L., Wilson.R. E. (2007). Fundamentals of momentum, heat, and mass transfer. United States of America: John Wiley & Sons.
Mindef (2008). Use of Sand Granular Sachet for Dengue Prevention. Retrieved April 1, 2010, from http://www.mindef.gov.sg/imindef/mindef_websites/topics/prideday/awards/Sand_Granular_sachet_for_dengue_prevention.html
1
NATIONAL UNIVERSITY OF SINGAPORE Department of Chemical & Biomolecular Engineering
DESIGN PROJECT STATEMENT: For Academic Year 2012-13
OVERVIEW Engineering design is an art that involves a creative and sound application of fundamental principles of science and engineering to produce a safe, practical, and cost-effective solution useful to the society. Holistic design experience is an essential part of any engineering education including Chemical Engineering. The final year design project aims to impart such a holistic experience in the Chemical Engineering curriculum, where the students get an opportunity to carry out the major steps involved in the design and evaluation of a new chemical manufacturing process or product. They develop and evaluate alternatives, perform rigorous simulation, size and optimize various processing units, analyse hazards and safety, develop schemes for control, estimate capital and operating costs, and finally assess project profitability. In other words, the design project serves as the capstone design course where students finally apply all their acquired knowledge and skills from previous years in an integrated fashion on a realistic, open-ended problem related to chemical manufacturing process or new product. To facilitate the above described final goal, you are receiving the following statement on the design project that you will eventually execute in your final year. The project is designed to serve as a central theme around which the learning outcomes of various modules can revolve. Lecturers of core modules will give you small, well-focused tasks such as tutorial problems, assignments, reviews and/or mini-projects related to the design problem given below. Students can also self-study at their own pace to learn more on the design problem. Such an integrated learning experience over several years highlights the importance and relation of core modules to the design project as well as motivates students. Further, we hope that the students will be better prepared for the design project and also learn how to solve open-ended problems by making critical design decisions with sound scientific justification and giving due consideration to cost and safety. DESIGN BRIEF Design a plant to produce drying oil (DO) and crude vinyl acetate from Acetylated Castor Oil (ACO). The process involves the use of Acetylated Castor Oil to produce Drying Oil, whose byproduct is Acetic Acid. This Acetic Acid is then reacted with ethylene and oxygen in the gas phase to produce Vinyl Acetate. Assume that the plant is to be located in Singapore and the operation time is 8000 h/yr. The raw material specifications are as follows. Acetylated Castor Oil: 298 K and 110 kPa. Ethylene: 99.9 mol% pure (0.1 mol% Ethane) 298 K, 1000 kPa. Oxygen: 100% pure, 298 K, 1000 kPa. The values given above are representative, and may vary in the design project that you will do in the final year (Semester 2 of the academic year 2012-13).
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PROCESS ACO at an elevated temperature produces DO and acetic acid via the following reactions. C15H31COOH(l) CH3COOH(g) + C14H28(l) C15H31COOH(l) C28H56(s) Gum (C28H56) is removed, DO is separated from acetic acid via distillation, and unreacted ACO is recycled (Turton et al.1). The production of VA involves the following exothermic reaction on a solid bed catalyst. C2H4 + CH3COOH + 1/2O2 CH3COOCHCH2 + H2O C2H4 + 3O2 2CO2 + 2H2O The reactions are irreversible and the reaction rates have an Arrhenius-type dependence on temperature. The reactor effluent is cooled and vapour and liquid streams are separated. The liquid stream from the separator becomes a part of the feed to the distillation column. The gas enters the bottom of an absorber, where the remaining vinyl acetate is recovered. Liquid acetic acid that has been cooled is fed into the top of the absorber to provide the final scrubbing. The liquid bottoms product from the absorber combines with the liquid from the separator as the feed stream to the distillation column (Luyben et al.2). REFERENCES
1. Turton R., Bailie R.C., Whiting W.B. and Shaeiwitz J.A., ‘Analysis, Synthesis, and Design of Chemical Processes’, Prentice Hall, 3rd Edition, 2009 (CL RBR TP155.7 Ana 2009).
2. Luyben, M.L., Tyreus, B.J. An industrial design/control study for the vinyl acetate monomer process. Computers & Chemical Engineering, 1998, 22, 867- 877.
3. ‘Encyclopedia of Chemical Processing’, editor: Sunggyu Lee. New York : Taylor & Francis, 2006. (TP9 Ency 2006).
4. ‘Ullmann's encyclopedia of industrial chemistry’, executive editors: Ullmann F, Gerhartz W, Yamamoto Y.S., Campbell F.T., Pfefferkorn R, et al. Weinheim, Federal Republic of Germany ; Deerfield Beach, FL, USA: VCH, 1985 (TP9 Ull).
5. ‘Encyclopedia of Chemical Processing and Design’, executive editor: John J. McKetta. New York : Marcel Dekker, Inc, c1976-<c1999> (TP9 Enc).
6. ‘Kirk-Othmer Encyclopedia of Chemical Technology’, executive editor: Kroschwitz JI New York : Wiley, 2004 (TP9 Kir 2004).
7. P A Schweitzer ‘Handbook of Separation Techniques for Chemical Engineers’, McGraw Hill, 1997.
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NATIONAL UNIVERSITY OF SINGAPORE Department of Chemical & Biomolecular Engineering
DESIGN PROJECT STATEMENT: For Academic Year 2013-14
OVERVIEW Engineering design is an art that involves a creative and sound application of fundamental principles of science and engineering to produce a safe, practical, and cost-effective solution useful to the society. Holistic design experience is an essential part of any engineering education including Chemical Engineering. The final year design project aims to impart such a holistic experience in the Chemical Engineering curriculum, where the students get an opportunity to carry out the major steps involved in the design and evaluation of a new chemical manufacturing process or product. They develop and evaluate alternatives, perform rigorous simulation, size and optimize various processing units, analyse hazards and safety, develop schemes for control, estimate capital and operating costs, and finally assess project profitability. In other words, the design project serves as the capstone design course where students finally apply all their acquired knowledge and skills from previous years in an integrated fashion on a realistic, open-ended problem related to chemical manufacturing process or new product. To facilitate the above described final goal, you are receiving the following statement on the design project that you will eventually execute in your final year. The project is designed to serve as a central theme around which the learning outcomes of various modules can revolve. Lecturers of core modules will give you small, well-focused tasks such as tutorial problems, assignments, reviews and/or mini-projects related to the design problem given below. Students can also self-study at their own pace to learn more on the design problem. Such an integrated learning experience over several years highlights the importance and relation of core modules to the design project as well as motivates students. Further, we hope that the students will be better prepared for the design project and also learn how to solve open-ended problems by making critical design decisions with sound scientific justification and giving due consideration to cost and safety. DESIGN BRIEF Design a plant to produce 99.9 mol% phenol starting from cumene, and explore its uses. The process involves the oxidation of cumene to form cumene hydroperoxide, and subsequent cleavage of hydroperoxide to phenol and acetone. Assume that the operation time is 8000 h/yr. The raw material specifications are as follows. Cumene: 298 K and 110 kPa Oxygen: 100% pure, 298 K, 1000 kPa The values given above are representative, and can vary in the design project which you do in the final year (i.e., in the Semester 2 of the academic year 2013-14).
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PROCESS The cumene-phenol process is based on the formation of cumene hydroperoxide and its cleavage to phenol and acetone. Two reaction steps form the basis of the production of phenol from cumene [4]:
1. Oxidation of cumene with oxygen to cumene hydroperoxide 2. Cleavage of cumene hydroperoxide in an acidic medium to phenol and acetone.
Phenol, acetone, unconverted cumene, and by-products are then separated in a series of distillation towers to recover high purity acetone and phenol, and recycle cumene. Phenol production processes based on the oxidation of cumene comprise the following sections [3]:
More details of phenol production can be found in Ullmann’s Encyclopedia [4]. REFERENCES
1. Turton R., Bailie R.C., Whiting W.B. and Shaeiwitz J.A., ‘Analysis, Synthesis, and Design of Chemical Processes’, Prentice Hall, 2nd Edition, 2003 (CL RBR TP155.7 Ana 2003).
2. ‘Encyclopedia of Chemical Processing’, editor: Sunggyu Lee. New York : Taylor & Francis, 2006. (TP9 Ency 2006).
3. ‘Encyclopedia of Chemical Processing and Design’, executive editor: John J. McKetta. New York : Marcel Dekker, Inc, c1976-<c1999> (TP9 Enc).
4. ‘Ullmann's encyclopedia of industrial chemistry’, executive editors: Ullmann F, Gerhartz W, Yamamoto Y.S., Campbell F.T., Pfefferkorn R, et al. Weinheim, Federal Republic of Germany ; Deerfield Beach, FL, USA: VCH, 1985 (TP9 Ull).
5. ‘Kirk-Othmer Encyclopedia of Chemical Technology’, executive editor: Kroschwitz JI New York : Wiley, 2004 (TP9 Kir 2004).
6. P A Schweitzer ‘Handbook of Separation Techniques for Chemical Engineers’, McGraw Hill, 1997.
