Name Student ID# ------------ ----------CHE 324F -Final exam December 16, 2017

1) (total 50 points) Consider the refrigeration cycle shown in the process flowchart below: ~-----------~4f-------------~

COMP01 COl>001

EVAP01

This refrigeration cycle is designed to produce the necessary refrigerant at -50C which is employed to condense the ammonia in the Haber process we designed this year. For your calculations, assume refrigerant Rl34a (1,l,1,2-Tetrafluoroethane, MW= 102 g/mol) is used, having the following properties: Tc= 374K, Pc= 40.6 bar, ro=0.34, log(Psat, bar)= 4.30-930/(T0c+242), liquid density= l.3g/cm3

Cpiiq= 151 J/mol-K, Cpgas = 41.6 J/mol-K, Lllivap (k:J/mol) @-28°C= 22.2 k:J/mol The overall cryogenic process consists of 3 refrigeration cycles in cascade. To simplify, we only concentrate in the last cycle (shown above) where stream 4 is vapour at T4= -49°C (P4= 0.30 bar), Stream 2 is saturated liquid at-28 °C (P2= 0.92 bar). You can assume that the pressure drop across the condenser and the evaporator is negligible. You can also assume that the valve operates in an isenthalpic manner (h3 = h2) and that T3= -50°C. The compressor has an isentropic efficiency of75%. For this process, calculate:

a) (5 pts) The real discharge temperature of the compressor (Tl, Kelvin) Tl= ___ K

b) (5 pts) Considering the conditions of stream 2 as reference (saturated liquid at -28°C) h2=0 k:J/mol, calculate the values ofh3 and h4 (k:J/mol). h3 = ___ k:J/mol h4= _____ k:J/mol

c) (5 pts) The molar flowrate (molls) ofR134a required to achieve the duty of the evaporator (set by the Haber process) of2MW. n3 = _____ molls

d) (5pts) Calculate the required power for the compressor (kW) Power = _____ kW

e) (5pts) Determine the pipe diameter for stream 2. Diameter= _____ inches

f) (7 pts) Determine the thickness and schedule for the pipe of stream 2 constructed of grade A carbon steel seamless pipe. Thickness= _____ mm Schedule: ___ _

g) (9 pts) Assume that the temperature of the Haber gasses cool down (in the tube side of EV APO 1) from 40°C to -20°C in a linear fashion. The refrigerant Rl34a is introduced in the shell side of EV APOI and enters at -50°C and leaves at -49°C. Under these conditions, size the tube-shell heat exchanger (EV APOI), indicating the number of shells needed, assume F=0.8, U= 500 W/m2-K and neglect any fouling.

h) (9 pts) Use the bare factor module method to determine the installed cost of the heat exchanger in 2018, assuming a CEPCI= 610 for 2018. You should plan to make this exchanger out of carbon steel as we do not expect any mayor corrosion issues (no water in the system). Remember that the pressure in the shell side (Rl34A) is 0.3 bar, and the pressure in the tube side (Haber gases) is about 180 bar. Cost: ____ USD (2018)

Steam In

l 1) . (20 marks) The following figure contains a PFID obtained from www.instrumentationtoolbox.com.

· The PFID corresponds to a heat exchanger used to heat a cold feedstock with steam, where the steam_is introduced through the shell of the exchanger.

----~ w ~ '\iii}

@----

Based on the diagram, explain:. a) How is the temperature of the heated @

feedstock controlled (in your description , use the appropriate instrument, relays : numbers, what type of signals are used, ~ Fcv-

101

_etc.) ~~--1---'-

b) How does FIClOl control the flow of r cl) feedstock? What type of pump could I •·101

17

1

. respond to that form of control? Explain. roldFuedSlod:in : 101

c) Explain what alarms exist in the system (;)1

-,

0

c

1

______ ! and what role could they play as a layer i Of protection? Condonsato

~ ~

I

---©•RC ___ j 102 :

I

w ~

s~oc.'<r.u:

d) Explain what process parameters are measured in the steam line. What is the purpose of measuring them? None of the steam line parameters are controlled, how could you justify that for this drawing. Are any of these signals going to a DCS?

2) (20 points) Complete the PFID for the heat exchanger EV APO 1 and the liquid ammonia separator SEPOl shown in the dotted area below. The evaporation ofR134a occurs in the shell of EV APOl producing the condensation in the tubes of the ammonia in the Haber gases. You cannot.control the incoming flow ofHab_er gases. If you need to control the flow ofR134a, consider that Comp 01 is a variable speed drive compressor. Remember to include any necessary safety device needed as a layer of protection. If you decide to include alarms, explain what should the operator do in each case. Include an explanation of the control strategy, particularly how would the system respond to changes in flow or composition of ammonia in the Haber gases. *Comment how would you achieve a SIL level 4 in your liquid level in SepO 1 if your pump POI only has a SIL level 2. *If the pressure of the P18 line is 180 bar, and the vapour pressure of ammonia at -20C is about 1 bar, can I place the pump 30 cm below lower level of SEPO 1?

. . . . .. . . . .

CondOl

P-13

. ...........•...............................•..•........•............•...•.....

P-15 Uncondensed Haber gases

Evap01

P-16

Sep01 P-17 Liquid NH3

P-18

Haber gases P01 . .

~--··························································································

: . ,,,... ,.. ~ ~ , . .

4 (10 pts) The following information was obtained from the SDS-.ofrefrigerant R134a (Dupont­generated document):

SECTION 2. HAZARDS IDENTIFICATION

Emergency Overview Rapid e_vaporation of the liquid may cause frostbite.

Potential Health Effects Skin

1, 1 , 1,2-T etrafluoroethane

Eyes 1, 1 , 1 ,2-T etrafluoroethane

Inhalation 1, 1, 1,2-T etrafluoroethane

Carcinogenicity

Contact with liquid or refrigerated gas can cause cold burns and frostbite. May cause skin irritation. May cause: Discomfort, itching,- redness, or swelling.

Contact with liquid or refrigerated gas can cause cold burns and frostbite. May cause eye irritation. · May cause: tearing, Redness, Discomfort.

Misuse or intentional inhalation abuse may cause death without warning symptoms, due to cardiac effects. Other symptoms potentially related to misuse or inhalation abuse are: Anaesthetic effects, Light-headedness, dizziness, confusion, incoordination, drowsiness, or unconsciousness, irregular heartbeat with a strange sensation in the chest, heart thumping, apprehension, feeling of fainting, dizziness or weakness. Vapours are heavier than air and can cause suffocation by reducing oxygen available for breathing.

None of the components present irnnls material at concentrations equal to or greater than 0.1 % are listed by IARC, NTP ~ or OSHA, as a carcinogen.

For the refrigeration cycle of question 1 identify 10 potential hazards ( considering the hazards identified in the SDS but without repeating them), including at least 1 in each of the categories: chemical energy, P-V energy, thermal energy, kinetic energy, potential energy, and electrical energy.

Hazard 1: Energy:

Hazard 2: Energy:

Hazard 3: Energy:

Hazatd4: Energy:

Hazard 5: Energy:

Hazard 6: Energy_:

Hazard 7: Energy:

Hazard 8: Energy:

Hazard 9: Energy:

Hazard 10: Energy:

; ... ~ ' ,

~.

'·

;/

Formulas Mass balances with and without reaction. Differential mass balance=> accumulation is dm/dt (it also works on moles, considering reaction).

ddmti·= "\""' mi,j - "\""' mi,k +ii(rateof gene~ationor;onswnption) L j (streams in) L k (streams out) · 0 ri = n*V ; ri = rate of reac;tion (~ol/L-s) ; simple potential rate law for a reactant: ri= -k*cin

Residence time = V/0 V =Volume of reactor/ volumetric flow rate.

%Conversion= (moles consumed/moles fed)*100%

Selec:five yield of A to produce B == moles of A consumed to produce Bl total moles of A consumed

Reaction equilibrium constant for. gases Keq = ll (Yit (P/Pref) vi Pref= 1 atm, Yi is the molar fraction of "i"

Example reaction: aA+ bB ~ cC , Keq = {(yc)c/[(YAY1(y8 )b]}* (P/Pref)Cc-a-b)

Phase equilibrium:

Antoine equation Psaturation = 1 O"(A-B/(T +C)}. Be careful with the units, for each set of A, B, and C values. Hemy's law (dilute non ideal solutions) xl *H = yl *P, H = Remy's constant.

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Raoult's law (ideal gas, ideal solutions) xl *Pl sat=:= yl *P

Relative saturation (humidity)= 5r= YiP/Pisat %R.H.= 100%*(yw*P)IP0 w

Molal saturation= Sm= moles of vapour/moles of vapour-free gas= y/(1-yi)

Abso\ute saturation(humidity) = sa=yM/[(l-yi)Mc1ry]

Bubble point (xi= Zi, LYi =1), Dew point (Yi= Zi, LXi =1)

General flash calculation equation: ~ki-l)z/(1-(v/f)+ki(v/f)) = 0 where ki=y/xi

Energy balance with reaction and in steady state.

(Ln/hj)out -(Lni*hi)in + M<.E +Af>E =Q +W => remember to use standard enthalpies of formation (h0 fi) when calculating the enthalpy of each component in each stream.

Sensible heat (change of enthalpy without phase change) => Ah = CPaverage *AT= fcp*dT

Latent heat ( change of enthalpy with phase change) => Ahvaporization , Ahfusion

Mlvap = R*Tc*[7*(1-Tr)0.354 +11 *oo*(l-Tr)°-456

]

ro: accentric factor, for water use 0.344

Differential energy balance (neglecting changes in kinetic and potential energy), unsteady state:

dU/dt = (L0 nj*hj)in -(L0 ni*hi)out + Q +W

U= internal energy; H =U+PV; U=m*u, du= CvdT (no phase change)

Cv = Cp for liquids and solids, Cp=Cv+R for ideal gases.

Rate of heat transfer Q0 = UH* A *(T surroundings - Tsystem); where UH is the overall heat transfer coefficient

Universal gas constant (R): R = 8.314 J/mol-K = 0.082 atm L/mol-K = 1.986 cal/mol-K

Pressure units: 1 atm = 1.013 bar= 101.3kPa = 101325 Pa= 760mmHg = 14.7 psi= 10.3 m water

Energy units: 1J = 0.239 cal= 9.486E-4 Btu; Power units: lKW = lKJ/s = 1.341HP= 0.9486 Btu/s

Mass units: 1kg = 1000g = 0.001 metric ton= 2.20462 lb= 35.27392 oz

Volume: lm3 = lO00L = 264.17 gal= 35.3145 ft3 =1056.68 qt

Temperature: T, Kelvin= T,Celsius +273.15 ; T, Rankine= T, Fahrenheit +459.67 T(°F) = 1.8*T(0 C) +32

Hydraulics.

Bernoulli equation ( ~-+ vz + hz) = ( ~ + vz + hz) where hz is the elevation and V is the average velocity pg . 2g 1 pg 2g 2

• -··· . 1t·D2 ..... • ..... Flow and velocity in a pipe V = volumetric flow rate = 4 V where V is the average velocity in the pipe .

.Volumetric flowrate of gas V= n · Z · R ·TIP where n is the molar flowrate and Z the compressibility factor.

Pressure drop (or head loss) due to friction (LID method): ilPfriction = p · g · hzoss = p · f · (~) · (:2) Pressure drop ( or h~ad loss) due to friction (k1 method): /JPfriction = P · g · hzoss = P · g · L Kzoss · (:)

Haaland' s approximation for Darcy's frictioD. factor f = (-1.8 log [ :; + ( l1)"

11

])-,

Reynolds number Re = DV p/µ

Pressure drop (friction) in a packed bed, Ergun equation LlP = (l~E) (( ~'! ) + 1. 75) .!:_ pV/ . . € !!.....!.....1! Dp . ~◄ .

Where e: = fluid volume :fraction; Dp: particle diameter; Vs=superfi.cial velocity= V/cross se~tion;tl area Valves

Cv of for liquid flow Vgal/min = Cv (:~-Pz):st Ltqu

Cv vs Ktoss equivalence for liquids Cv = 29.9D2 /.JKLoss

Cv of for gas flow Vstd L/min = 6950 · Cv · P1 bar · (1 -2

<Pi-Pz)) • 3P1

SGgas = specific gravity, for gasses= density of the gas/density of air~ MW gas/(28.96 g/mol) SGtiquid = specific gravity, for liquid= density of liquid/density of water Vessel and pipe thickness {hoop stress) Maximum allowable stress (S) = tensile strength ( or ultimate strength)/Design safety factor

ANSI B3 I .3 equation for minimum pipe thickness (t) t = tc + tth + Ccs=~PY)) · (10~~~0z)

Where tc => corrosion allowance (0 no corrosion, 0.0625" typical, 0.125" maximum), tth =>'thread depth (0.11" for pipes larger than 2" diameter), P => maximum allowable pressure, D => outside diameter, E => longitudinal well joint factor (I seamless pipes, 0.95 electric fusion, 0.85 electric resistance weld, 0.6 furnace butt weld). Toi => % tolerance, 12.5% for pipes smaller than 20", and I 0% for larger diameters. Y => welding factor (0.4 for ferrous materials below 900°F, and 0.7 for higher temperatures). ASME Boiler and Pressure Vessel Code (BPV C) vessel thickness ... same as ANSI B3 I .3, but no 1th.

Compressors and pumps

Discharge temperature from an isentropic (ideal) compressor with an ideal gas= !2ideal = T1(Pi/P1)"((y-l)/ y)

y = Cp/Cv for ideal gas, Cp = Cv +R Ideal power of the compressor (flow system)= Wideal = :n:Cp· (T2idea1 -T1) where n is the molar flowrate

Real work of the compressor (flow system)= Wreal = Wideal ITJisentropic = n·Cp·(T2rea1 -T1) Isentropic efficiency TJisentropic => Centrifugal~ 70-80%, Reciprocating 72-88%, Rotatory~ 65-75% Pumps, W = V · dPITJpump

N SH _ Po h Ptiap V2 h P Available - - + - - - - - losses pg pg 2g

Driver efficiency=> Electrical motors~ 90%, Turbines 40% to 70%, gas-powered ~30%

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P.

... f•-

..

! ·~.

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Heat exchangers

. 8Tm = Tl-t2 8 Tour T2-tl

: Q = U* A *F* 8 Tim =>A= Q/(U*F* ~ Tim)

T1 -T2 R=--­

t2 - t1

When Nshells are satisfied, F=0.8

Udirty = 1 1

-·Rf: fouling factor. --+Rt

· Uclean ·

Cost estimation - Bare module factor

Cp at time t, ·= Cpo@ *Fbm

Cp @t2, = Cp @tl *CEPCI@t2/CEPCI@tl

For heat exchangers and other equipment,

FBM =Bl+ B2*Fp*FM

For shell and tube exchangers, B1=1.63, B2=1.66

--. '

r1!

! ! ! ! I

i IV

l i I•

! Best if P<0.7 j

, If not, use multi stages i

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t1( 1 ... . j

:... 0.9 a ] 0.8 g

HP.:ait P.icr.h:ainnP.rl r:f /kW\

-~ 0.7

8 0.6 l--f--1--1-!--H--i ·1-+-~++-+-Y--1-;-..;,,,..+->i--!\--i+-1

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

(c) Single-pass cross-now wilh both nuids w1111ixcd

0.4 0.5 0.6 0.7 0.8 0.9 1.0

(,/) Single-poss cross-now with one nuid mixed and the other w1111ixed

! !

Metals-~

Ferrous Cast Irons High Cort>on S1eels Medium Carbon Steels Lo'N Caroon Steels L~, Alloy St;,els ~.tlinlc'SS Steels

Nor..ferrous Aklrrrilm Alloys Copper AIIO'}'s Le3dAJ,:;ys Magnesium Aloys Nickel Alloys Tio.nium AJlc;·s ZincAkM<

Ceramics Glasses Borosa:ate Glass ('}

Gb~s Cer.amic {'} • 'SiiC3 Glass (') •" Soda-Lir?E Glass (')

Porous 2lick (') Ccncre,te, typical {'} Stone (')

Tedinical A.unina (') Aluminun N'1ride {') Boron Carbide ('} Silicon(') Silicon Carbide ( •) Silicon tl;tride {'} Tunast:n Carbide ('l

Comiposites Metal Maminun/Silioor. Carbide

Pol:,mer CFR? GFRP

Natural i!,;a:nboo Cori! . L-=..the, Wood, lypical (Longitudinal) Wood.,;,;..,..,, fTrar.sversal

)ata courtesy of Granta Design Ltd)

UPPER FIGURES INDICATE WALL THICKNESS IN IN.

PIPE 0.0. 5 10 20 SIZE in IN.

¼ .405 .035 .049 .. 1383 .1863

¼ .540 .049 .065 .2570 .32!)7

1/, .675 .049 .065 . .3276 .4235

½ .840 .065 .083 .5383 .6710

¾ 1.050 .065 .083 .6838 .8572

1 1.315 .065 .109 .8678 1.404

1¼ 1.660 .065 .109 1.107 1.806

1½ 1.900 .065 .109 1.274 2.085

2 2.375 .065 .109 1.604 2.638

2½ 2.875 .083 .120 2.475 3.531

3 3.500 .083 .120 3.029 4.332

3" /7 4.0 .083 .120 3.472 4.973

4 4.50 .083 .120 3.915 5.613

4½ 5.0

- - --- ··- ·-.

Civ (MPa) Cits (MP.:i) , ..

215 . 780 350. • IODO 4/)j . 1155 550 . 1o40 3•:15 . 900 410 . 12'10 2~ . 395 345 . 550 400 1100 460 . lZID 170 1000 480 - 2240 30 . 500 58 . 5:,1 30 . 500 100 . 5E,O

.s . 14 12 . 20 70 . 400 t.55 . 475 70 . 1100 ::-45 . 1:..10

200 . 1245 2.00 . 1625 ~ 450 135 . 520

2b4 . 3e4 22 . 32 750 - 2129 62 - i7i

1100 11,00 45 . 155 360 420 31 . 35

50 . 140 7 . ,4 32 . 60 2 . a 34 . 248 5 - 17

690 5500 35D 665 1970 . 270,0 187 . 270 2583 . 5687 35D . 560 3200 . 34~0 160 . I.SD 1000 5250 370 . 550

524 . 5500 69D - .SOD 3347 6833 37D 55D

2ao . 32li 2~D - 365 55-'l . 1050 550 . 1050 110 . 192 138 . 241

35 . ¥, 38 . 45 0.3 . 1.5 0.5 . 2.5

5 . 10· 20 . 26 30 70 6D . 100 2 . 6 4 - 9

Ci,, (MP.i} Cits (MP.:i)

.Polymers 1

S:;i;tomg 6uty1Ruro:r 2 - .. 3 5 - 10 EVA 12 18 l5 2•J lsoprer.: (~} 20 . 25 2'J - 25 Na1ur.il Rwber (NR) 21'.) 30 22 . 32 N~prenetCR) 3.4 24 3.4 - 24 Polyuretlale Elastomm ( el PU} 25 51 25 51 Silicone Elastom:rs 2.4 5,5 :?.4 5.5

Tne~las~c ABS 18.5 51 21.0 . 05.2 C,..llulos~ Polyrrlas (CA) 23 ,f:i, 25 . 50 looom.<(I) 8.3 15.9 1i2 3-12 Nylons (PA) QiJ . 94.8 {A) 165 Pol~roonate (PC) 59 70 c-0 . i2.4 PE K 65 05 70 . 103 Polyethylene (PE) i7.9 29, :l:0.7 . «.a PET 56.5 62.3 48.3 - 72.4 Acrylic (PMMA) 53.8 . 72.4 48.3 . iS.6 Acetll(POM) 48.5 i2.4 c-:l aa Polypro~ (PP) 20.7 37.2 2,.0 . <!:i.4 Poly5¥er,e (PS) 28.7 56.2 35.9 5.ti.5 Pol'.(Urelrol\e Thermoplas1ics {1pPI)) 40 . 53.8 3i - 62 PVC 35.4 . :i2.1 40.7 . es., T &flon (PTi'E} 15 25 2'I . 30

TBe1moset Epoxies 36 71.i 45 . a-~.a Phenoi,,s 27.6 49.7 34.:. . 62,1 Pol""'ster 33 40 41.4 89.6

Polymer f03ms Fls?xibi= Pol)'l116 Foam (VLO) 0.01 . 0,12 0.24 . 0.85 A:xibla Polyrr,er Foam (LO) 0.02 0.3 0.24 . 2.35 ~xi~ Polyn>B" Foam (MD) 0.05 0.7 0.43 . 2.95 Ri!;!id Polym:r foam (LO) 0.3 1.7 0.45 . 2.25 Rigid Polymi;r Foam (MD) 0.4 3,5 0.55 5-.1 R\Qid P,.,;-=r .foam (HD) o.s 12 1.2 . 12.4

1 For full 11.1n1.es. and acronyms of polymers - see Section V.

(*) 1\1B: For ceramics, yield stress is replaced by compres.sh'e strength, which is more relevant in ceramic design. Note that ceramics are. of the order of 10 times stronger in compression than in tension.

TABLE 9.7...;_ANSI PIPE SCHEDULES (courtesy of Natural Gas Processors Suppliers Assn.)

LOWER FIGURES INDICATE WEIGHT PER FOOT IN LBM

30 40 STD. GO 80 XH 100 120 140 160 XXH

.068 .068 .095 .095

.2447 .2447 .3145 .3145

.088 .088 .119

.4248 .4249 .5351

.091 ,091 .126 .126

.5676 .5676 .7388 .7388

.109 .109 .147 .147 .187 .294

.8510 .f.1510 1.088 1.0f.lf.l 1.304 1.714 .113 .113 .154 .154 .218 .308

1.131 1.131 1.474 1.474 1.937 2.441 .133 .133 .179 .179 .250 .358

1.679 1.679 2.172 2.172 2.844 3.659 .140 .140 .191 .191 .250 .382

2.273 2.273 2.997 2.997 3.765 5.214 .145 .145 .200 .200 .281 .400

2.718 2.718 3.631 3.631 4.859 6.408 .154 .154 .218 .218 .343 .436

3.653 3.653 5.022 5.022 7.444 9.029 .203 .203 .276 .276 .375 .552

5.793 5,793 7.661 7.661 10 01 13.70

.216 .216 .300 .300 .437 .600 7.576 7.576 10.25 10.25 14.32 18.58

.226 .226 .318 .318 .636 9.109 9.109 12.51 12.51 22.85

.237 .237 .281 .337 .337 .437 .531 .674 10.79 10.79 12.66 14.98 14.98 19.01 22.51 27.54

.247 .355 .710 12.53 17.61 32.53 --- --- --- --- --- --- --··

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\ -

/SME SA515 • Grade 55 Carbon Steel ·

/ -

~ ~SME SA285 • Grade A Carbon Steel

f -' - ASTM 1060 - Temper O ~-

I t 01rmlk \ i I I I I \ ' ASTM 5052 .- Temper 0 \ _ Aluminum \ .. \

-.........._ 0 ... -- -· . . - ··- .. - -- - ----- . ··- -- - .. -

• I . . I . -400 -300 -200 -100 0 100 200 300 400 500 600 700 800 900 1000

Temperature, °C

. 1400 -----....---------.------....-----------,

_,

600 ASME SA-240 • Grade 304-----li-----!lr--"t-'t-~---------t Stain~ess Steel'

o-------------------------200 -100 0 100 200 300 400 500 600 700 800 900 1000

Temperature, °C

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2. TO $1~.?ur, RO\JTl~C Cf ?R()C(SS rt.ow U!IE. SWF. PIECES CF EO~~PIJENT ~AY 1-P?EAA l't IWi(£ fK;.1'1 ON[ PL.AC( CU n-:£ FLOW 0IAta;.U. E.OUlfUJ,ifH SOOOl'UCAIED will 0C INOIC~ltO BY OAS>IEO LIN(S.

l. ~•Sllllf~[lff IIJ(NTI!1CATl()l;S •S IU.VS'IT!AltD •!<£ e•.srn C,I l'flE INS~l.''V(NT SOCIETY er AMERlCA SlANOARO S-5.J..

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~- CIVZNSICl'I f'RCLt er.mm er LC OAt!..OOU 10 TANGENT u:~ ~ 80-IT0'-1 Of HORJZONTAL \'ESstl.. INDICAiCS 1'CIUI.IJ. U:VEL

SCALE: ~ONE

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EXAMPLE PROJECT

P&ID AND PFD STANDARD LEGEND ANO-SYMBOLOGY

VALVES, INSTRUMENTS AND PIPING SYMBOLS

S~•s:one & Wsbs!Br, Inc.

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........, ... . .

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10 100 1,000 l0,000

Heat Transfer Area, A (m2) .... 0) Q. ... 10,00(1-.----.---.-. . -.. ----.-.. -.--.. -.----.-.-.~------

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CEPCI =397 (Se l 2001)

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H~at Transfen\tea. A'(rt.2)

· . . Kettle Raboiler : ..•

CEPCI = 397 l . (Sept 2001) t

U-Tube

100 1,000 10,000

H~at Transfer Area, A (m2)

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