Client: XXXProject No: XXXProject Title: XXXDocument: Colour KeySheet Ref: Manual InputRevision: ResultsLast Updated: 03/01/2011 Do not use

Revision detail: Assumptions and important notesSources and titles

Reference Method Used:

Main Data InputPhysical Properties Units Air density at related temperature and pressure

Liquid Air Room Pressure P 101.325 kPaMol. Wt of air M 29 kg/kmol

Liquid in the tank Etylenediamine (EDA) Gas const R 8.31 kJ/kmol K

Density,ρ 897 1.34 Vapour/air Temp t -10Specific Heat,Cp 2.8 1.005 kJ/kg K T 263.15 K

2847 1005 J/kg K Air Density, PM/RT 1.34Viscosity,µ 1.8 - cP or m.Pa.s

0.0018 0.0000198 kg/m.s Thermal ConductiviEtylenediamine (EDA)

Thermal conductivity,k 0.257 0.0257 W/m.K

Co-efficient of volumetric expansion, ß 0.000108 0.00343 1/K Thermal Conductivit k = 0.224 W/m.K

Molecular Mass of liquid,M 60.1 - kg/kmol

11.14 -

Units

Wet wall 5000 Source: Chemical Engineering Design by Coulson and Richardson, Volume 6, Page 640

Thermal Conductivities/thickness Units

45 W/m K Source: Metal wall thickness 3.68 mm

0.00368 m

0.038 W/m K Source: Insulation thickness 25 mm

0.025 m

Surface Emissivity Units

Wall, ε 0.9 Assumed - less than 1

Gravitational constant, g 9.81

Pipe dimensions Units

Inside pipe diameter Di,p 0.041 mOutsidepipe diameter Do,p 0.048 mMean pipe diameter Dm,p 0.045 mOutisde diameter insulation Do,i 0.098 mLog mean diameter insulatioDlm,i 0.070 m

e, absolute roughness 0.00005 m

Temperature Units

21 Temperature just after loading Summary Units

Problem Description:

Important values and calculations

kg/m3 oC

ρair = kg/m3

k = 3.56 x 10 -5 x Cp ( ρ4/M)1/3 ------------> from Coulson & Richardson. Vol 6, Page 321

Melting Point, oC oC

Assumed fouling coefficient, hF

W/m2 K

Metal walls (Carbon Steel, max 0.5% Carbon),kM Engg Toolbox : Thermal Conductivity of some common Materials

Insulation (Armaflex), kI Engg Toolbox : Thermal Conductivity of some common Materials

m/s2

Engg Toolbox: Surface roughness several materials

Liquid in pipe, TLoC

-10 velocity 1.00 m/swind factor 6.2 -ambient -10 ˚C

Summary of temperatures used in calcs Units Heat loss/unit length 9.8 W/m

294.2 K

263.2 K

First Guess 278.7 K

After iteration 293.7 K 293.5

First Guess 278.7 K

After iteration 265.0 K 265.6

Summary of flow conditions in pipe Units

Velocity 1.0 m/s

Reynolds Number 20401.7666666667 Colebrook equation for friction factorA 6.07406 A=-2.0*LOG[(e/(D*3.7))+(12/Re)]B 5.93527 B=-2.0*LOG[(e/(D*3.7))+(2.51*A/Re)]C 5.94915 C=-2.0*LOG[(e/(D*3.7))+(2.51*B/Re)]f 0.02827 f=[A-(B-A)^2/(C-2B+A)]^-2

Calculation

2.63E+08

2.63E+08

1.55E+08

1.55E+08

= Cp x µ /k

19.95

0.77

Outside air, TAoC

Liquid in pipe, TL

Outside air, TA

Tw=(TL + TA )/ 2

Tw Tw=TL-(Utot/hi)(TL-TA)

Tws=(TL + TA )/ 2

Tws Tws=(Utot/(hRo + h*wo))(TL-TA)+TA

NRe = (ρ x v x Di) / µ

Calculation of Grashof Number (N GR)

Grashof Number, NGr = L3 x ρ2 x g x ß x ΔT /µ2

NGr for the liquid phase ( ρ2 x g x ß x /µ2 ) ( ρ2 x g x ß x /µ2 ) L3 x ΔT x L3 x ΔT

NGr for outside air ( ρ2 x g x ß x /µ2 ) ( ρ2 x g x ß x /µ2 ) L3 x ΔT x L3 x ΔT

Calculation of Prandtl Number (N Pr)

Prandtl Number,NPr

NPr for the liquid phase

NPr for outside air

Calculation of Rayleigh Number (N Ra)

Rayleigh Number,NRa = NGr x NPr

Coefficient of liquid at pipe wall at no flow conditions, hwi Coefficient of liquid at pipe wall at flowing conditions, hwi

Iteration:

Put the right values manually into respective yellow cells untill difference between the two values approache zero

L=Di 0.04 m

0.45 KL=Di 0.04 m

2.63E+08 19.95

8.12E+03 20402f 0.02827

1.62E+05

Reference: Incropera Page 515

11.24 235.87

o.k o.k

o.k

o.k

Nusselt Equation (Perry 5-13) Nusselt Equation (Perry 5-13)

70.55 1480.08

L=Do,i 0.10 m

1.85 K

1.55E+08

2.72E+05

2.11E+05

9.58

o.k

Nusselt Equation (Perry 5-13)

2.50

------------- Equation 21

------------- Equation 22

12228.26 ------------- USING Equation 21

1.52 ------------- USING Equation 22

ΔT = TL – Twl

NGr x L3 x ΔT NPr

NGr NRe

NRa,l

For horizontal cylinders, Nusselt Number, NNu For horizontal cylinders, Nusselt Number, NNu

NNu ={0.60 + (0.387 x (NRa)1/6)/[1+(0.559/NPr)9/16]8/27}2 Ra ≤ 1012 NNu =(f/8)(NRe-1000)(NPr)/[1+12,7(f/8)1/2(NPr2/3-1)]

NNu NNu

Where Ra ≤ 1012 Where NPr ≤ 2000

Where NRe ≤ 5e6

Where NRe ≥ 3000

Coefficient of liquid at wall, hi = NNu x k / Di

Coefficient of liquid at wall, hi = NNu x k / Di

Coefficient of liquid at wall, hi W/m2 KCoefficient of liquid

at wall, hi W/m2 K

Outside coefficient of air at pipe wall/insulation, h'wo

ΔT = Tws- TA

NGr x L3 x ΔT

NGr

Nra,A

For horizontal cylinders, Nusselt Number, NNu

NNu ={0.60 + (0.387 x (NRa)1/6)/[1+(0.559/NPr)9/16]8/27}2 Ra ≤ 1012

NNu

Where Ra ≤ 1012

Coefficient of outside air at wall,hAwV,cyl = NNu x k /Do

Coefficient of outside air at wall,hAwV,cyl W/m2 K

Conduction coefficient for metal wall and insulation, hM and hI

hM = kM /tM

hI = kI /tI

hM W/m2 K

hI W/m2 K

------------- Equation 24

2.341 ------------- USING Equation 24

Summary

70.55

1480.08

2.50 Do NOT use this value

15.53

12228.2608695652 6.2

1.52

5000

2.341

1.40

0 m/hr

3.16 m K/W

Total heat loss per unit length

Q/L 9.8 W/m

Radiation coefficient for pipewall to air (hRO)

hR = 0.1713 ε [((Tws + 460)/100)4 - ((TA + 460)/100)4]/( Tws - TA)

hR,A W/m2 K

Coefficient ( W/m2 K)

Coefficient of liquid at pipe wall at no flow (free convection), hwi

Coefficient of liquid at pipe wall at flow (forced convection), hwi,f

Outside coefficient of air at pipe wall, h'wo

Coefficient of outside air at cylindrical wall considering wind enhancement factor for the assumed wind velocity, h*wo

Obtained by multiplying above value by wind enhancement factor

Conduction coefficient for metal wall hM

Conduction coefficient for insulation hI

Fouling coefficient, hFi

Radiation coefficient pipewall (hRO)

Overall coefficient,Utot

Overall Heat Transfer Coefficient per unit length, U tot,l

Overall coefficient, Utot,l per unit length at wind velocity of

1/Utot,l = 1/(hwi x πDi) + tm/(km x πDm,p) + ti/(ki x πDlm,i) + 1/((h*wo + hrd ) x πDo,i) + 1/(hfi x πDi)

1/Utot,l

Q/L= (TL-TA)/Utot,l

Reference: Incropera Page 515

No Flow Flow

Windforce Q[˚C] [-] [W/m]

5 0 4.4 4.55 3 4.7 4.95 5 4.8 5.15 6 4.9 5.10 0 5.7 60 3 6.2 6.50 5 6.3 6.60 6 6.4 6.7

-10 0 8.5 8.9-10 3 9 9.6-10 5 9.2 9.8-10 6 9.5 9.9

Ambient temperature

Heat input

[W/m]

101010101010101010101010

-12 -10 -8 -6 -4 -2 0 2 4 60

2

4

6

8

10

12

Heatbalance EDA feedlineFlow conditions

Heat loss @ quiescent airPolynomial (Heat loss @ quiescent air)Heat loss @ Beaufort 3Polynomial (Heat loss @ Beaufort 3)Heat loss @ Beaufort 5Polynomial (Heat loss @ Beaufort 5)Heat loss @ Beaufort 6

Ambient temperature [˚C]

heat

loss

/inpu

t [kW

]

-12 -10 -8 -6 -4 -2 0 2 4 60

2

4

6

8

10

12

Heatbalance EDA feedlineFlow conditions

Heat loss @ quiescent airPolynomial (Heat loss @ quiescent air)Heat loss @ Beaufort 3Polynomial (Heat loss @ Beaufort 3)Heat loss @ Beaufort 5Polynomial (Heat loss @ Beaufort 5)Heat loss @ Beaufort 6

Ambient temperature [˚C]

heat

loss

/inpu

t [kW

]