Tank Heat loss calc - Liquid 1 Client:XXXProblem Description:
Six Storage tanks [rectangle in shape] are filled up with
Isopropanaol (liquid 1) at the initial temperature of 45 oC and it
is required to maintain the liquid temperature at 30oCProject
No:XXXAll the tanks are placed adjacent to each other with minor
gap in between. Tanks are stored in a closed building maintained at
room temperature let's say 10oC and a wind speed of around
2miles/hr.Project Title:XXXHow long will it take to drop down
temperature from 45 to 30oC with/without insulation OR after what
time tanks are required to heat up with/without insulation ? Data
is given as below.Document:Storage Tank Heat Loss CalculationColour
KeySheet Ref:Tank Heat Loss Calcs - Liquid 1Manual
InputRevision:1ResultsLast Updated:31/12/2010Do not useRevision
detail:1) Correction cell reference C94 for Grashof number vap.
PhaseAssumptions and important notes2) Note added regarding
calculation of equivalent diameter cell C142Sources and titles3)
Recalculation introduced to iterate to more accurate coefficients
(see cells C316 and below)4) C replaced by K for correct
calculation in SI units5) Cooling time formula modified to
compensate for non-linearityReference Method Used:Predict Storage
Tank Heat Transfer Precisely - By J.Kumana and S.KothariImportant
values and calculationsImportant Notes:1) Uniform temperature
inside the tank2) Provision is made to select back wall area -
However, to calculate the maximum heat loss back wall area should
be considered-- (see the cell number C330 and C335)3) Provision is
made to select insulation thickness --- (See the cell number
C221)Main Data InputPhysical PropertiesUnitsAir density at room
temperature and pressureLiquidAirVaporRoom PressureP101.325kPaMol.
Wt of airM29kg/kmolLiquid in the tankISOPROPANOLGas
constR8.31kJ/kmol KDensity,7901.251.25kg/m3Room Tempt10oCSpecific
Heat,Cp3.01.0051.005kJ/kg KT283K300010051005J/kg KAir Density,
PM/RTair =1.25kg/m3Viscosity,2.42--cP or
m.Pa.s0.002420.00001980.0000198kg/m.sThermal Conductivity
ofISOPROPANOLThermal conductivity,k0.1990.02570.0257W/m.Kk = 3.56 x
10 -5 x Cp ( 4/M)1/3 ------------> from Coulson &
Richardson. Vol 6, Page 321Co-efficient of volumetric expansion,
0.000750.003430.003431/KThermal Conductivityk =0.199W/m.KMolecular
Mass of liquid,M60--kg/kmolBoiling Point, oC82--oC355.15KAssumed
fouling coefficient, hFUnitsDry wall7000W/m2 KSource:Chemical
Engineering Design by Coulson and Richardson, Volume 6, Page 640Wet
wall5000W/m2 KSource:Chemical Engineering Design by Coulson and
Richardson, Volume 6, Page 640Roof7000W/m2 KSource:Chemical
Engineering Design by Coulson and Richardson, Volume 6, Page
640Bottom4000W/m2 KSource:Chemical Engineering Design by Coulson
and Richardson, Volume 6, Page 640Thermal ConductivitiesUnitsMetal
walls (Carbon Steel, max 0.5% Carbon),kM54W/m KSource:Engg Toolbox
: Thermal Conductivity of some common MaterialsInsulation (Glass
wool), kI0.04W/m KSource:Engg Toolbox : Thermal Conductivity of
some common MaterialsGround (Earth), kG1.5W/m KSource:Engg Toolbox
: Thermal Conductivity of some common MaterialsSurface
EmissivityUnitsWall and roof, 0.9Assumed - less than
1TemperatureUnitsVapour in tank, TV33oCAssumed - just below the
liquid tempLiquid in tank. TL35oCMinimum temp requirement by
processOutside air, TA10oCAssumed - as tanks are inside the
buildingGround, TG12oCAssumed - just above ambient
temperatureVapour in tank, TV306.15KLiquid in tank.
TL308.15KOutside air, TA283.15KGround, TG285.15KGravitional
constant, g9.81m/s2CalculationCalculation for Grashof Number
(NGr)Grashof Number, NGr= L3 x 2 x g x x T /2NGr for the liquid
phase( 2 x g x x /2 )7.84E+08( 2 x g x x /2 ) L3 x T7.84E+08x L3 x
TNGr for the vapour phase( 2 x g x x /2 )1.34E+08( 2 x g x x /2 )
L3 x T1.34E+08x L3 x TCalculation for Prandtl Number (NPr)Prandtl
Number,NPr= Cp x /kNPr for the liquid phase36.44NPr for the vapour
phase0.77Coefficient of vapour at wall, hvwNote: as an initial
approximation, assume that the wall temperature is the average of
the vapour and outside air tempTw= (TV + TA )/ 2First
GuessTw295.9KAfter iteration see belowtotal height of the tank,
L2.55m% of liquid full (in terms of height)95%L =2.55mProportional
height in contact with liquid, Lw2.42mT = Tv -
Tw10.25KLwProportional height in contact with vapour, L -
Lw0.13mNGr2.85E+06For vertical plates and cylinders, Nusselt
Number, NNu3.1 mNNu= 0.138 x (NGr)0.36 x ((NPr)0.175 -
0.55)-------------Equation 15NNu11.812.3 mCoefficient of vapour at
wall, hvw= NNu x k /( L - Lw)Nusselt Equation (Perry
5-13)Coefficient of vapour at wall, hvw2.38W/m2 KCoefficient of
liquid at wall, hLwNote: Here, neither NPr nor (NGr NPr) falls
within the range of application of the below equations. Therefore,
again apply equation Equation 15 using average temp TwNNu= 0.138 x
(NGr)0.36 x ((NPr)0.175 - 0.55)------------- Equation 15Where,
0.1100 and 104 < (NGr NPr) < 109T = TL - Tw0.65KNote: In
article, Equation 16 is used but as NPr and (NGr NPr) do not fall
in the range, we can't apply Equation 16 directly. Therefore, used
Equation 15NGr7.25E+09NNu648.79------------- USING Equation
15Coefficient of liquid at wall, hLw53.36W/m2 K------------- USING
Eq (a)Coefficient of vapour at roof, hVrfor the surfaces facing
down, NNu= 0.27 x (NGr NPr)0.25-------------Equation 20Where, 2 x
107 < (NGr NPr) < 3 x 1010for the surfaces facing down, hVr=
(0.27 x k/D) (NGr NPr)0.25------------- Eq (b)NGr NPr =5.55E+10~ 2
x 107 < (NGr NPr) < 3 x 1010295.9KFirst guessNote: We will
apply equation (b) assuming roof of diameter and Tw =286.6KAfter
iteration see belowNote: Applied Equation 20 as (NGr NPr) is very
close to the above rangeNGr =1.34E+08x L3 x TWhere,T= Tv -
TwKT19.55KL= DmComment Extra Large: Not sure if the Characteristic
Length/hydraulic diameter shoud be taken instead. See:
http://www.engineeringtoolbox.com/hydraulic-equivalent-diameter-d_458.htmlL3.01mCalculated
equivalent diameter from the roof/bottom area of tankEquivalent
Diameter for Roof/Bottoml2.3mNGr
=7.17E+10w3.1mArea7.13m2hVr1.12W/m2 K------------- USING Eq
(b)Equivalent Diameter3.01mCoefficient of liquid at the tank
bottom, hLbfor the surfaces facing up, NNu= 0.14 x (NGr
NPr)0.33-------------- Equation 19Where, 2 x 107 < (NGr NPr)
< 3 x 1010for the surfaces facing up, hLb= (0.14 x k/D) (NGr
NPr)0.33------------- Eq ( c)NGr NPr =4.30E+11 2 x 107 < (NGr
NPr) < 3 x 1010Tw= (TL + TG) /2Note: Applied Equation 19, though
(NGr NPr) is out of range???Tw296.65KFirst GuessNote: We will apply
equation (c) assuming tank bottom diameter and Tw =307.6KAfter
iteration see belowNGr =7.84E+08x L3 x TWhere,T= TL - TwKT0.55KL=
DmL3.01mNGr1.18E+10hLb63.90W/m2 K------------- USING Eq
(c)Coefficient of outside air at roof, h'Arfor the surfaces facing
up, NNu= 0.14 x (NGr NPr)0.33-------------- Equation 19for the
surfaces facing up, h'Ar= (0.14 x k/D) (NGr NPr)0.33-------------
Eq (d)Note: Assume Tws = Tw since the roof is uninsulated and get
the coefficient for still air from equation (d)295.9KFirst
guessTws286.6KAfter iteration see belowNGr1.34E+08x L3 x TWhere,T=
Tws - TAKT3.45KNGr1.26E+10h'Ar2.37W/m2 K------------- USING Eq
(d)Coefficient of outside air at wall, h'AwNote: Assume that the
temperature drop across the film is one-fourth of the drop from the
inside fluid to the outside air (averaged for the wet and dry
walls) and use Equation 15 and (e) to find the co-efficientNNu=
0.138 x (NGr)0.36 x ((NPr)0.175 - 0.55)------------- Equation 15h=
0.138 x (NGr)0.36 x ((NPr)0.175 - 0.55) x k/L------------- Eq
(e)applicable for the vertical plates taller than 3ftT= ((TL + TV )
/ 2 ) - TA) / 4T6KNGr1.34E+08x L3 x
Twhere,L2.55mNGr1.33E+10NNu247.51------------- USING Equation
15h'Aw2.49W/m2 K------------- USING Eq (e)Conduction coefficients
for ground, metal wall and insulation ( hG, hM, hI)hM= kM
/tM------------- Equation 21hI= kI /tI------------- Equation 22hG=
8 kG/*D------------- Equation 23Where,tM6mmthickness of
metal0.0060mtI25mmthickness of insulationPredict Storage Tank Heat
Transfer Precisely - By J.Kumana and S.KothariEngg Toolbox :
Thermal Conductivity of some common MaterialsEngg Toolbox : Thermal
Conductivity of some common MaterialsEngg Toolbox : Thermal
Conductivity of some common Materials
