Storage Tank Heat Loss Calcs - Rev.1 31.12.2010 (1)

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XXX XXX XXX Storage Tank Heat Loss Calculation Tank Heat Loss Calcs - Liquid 1 1 12/31/2010

Problem 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 30oC All 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 10 oC and a wind speed of around 2miles/hr. How long will it take to drop down temperature from 45 to 30 oC with/without insulation OR after what time tanks are required to heat up with/without insulation ? Data is given as below. Colour Key Manual Input Results Do not use

1) Correction cell reference C94 for Grashof number vap. Phase Assumptions and important notes 2) Note added regarding calculation of equivalent diameter cell C142 Sources and titles 3) Recalculation introduced to iterate to more accurate coefficients (see cells C316 and below) 4) ˚C replaced by K for correct calculation in SI units 5) Cooling time formula modified to compensate for non-linearity Important values Reference Method Used: Predict Storage Tank Heat Transfer Precisely - By J.Kumana and S.Kothari and calculations Revision detail:

Important Notes: 1) Uniform temperature inside the tank 2) 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 Input Physical Properties

Units Liquid

Liquid in the tank Density,ρ Specific Heat,Cp Viscosity,µ Thermal conductivity,k Co-efficient of volumetric expansion, ß Molecular Mass of liquid,M Boiling Point, oC

Air

ISOPROPANOL 790 3.0 3000 2.42 0.00242 0.199 0.00075 60 82 355.15

7000 5000 7000 4000

Insulation (Glass wool), k I Ground (Earth), kG

Temperature Vapour in tank, TV Liquid in tank. TL Outside air, TA Ground, TG

Air Density, PM/RT

W/m2 K W/m2 K W/m2 K W/m2 K

54 W/m K

Source:

Engg Toolbox : Thermal Conductivity of some common Materials

0.04 W/m K

Source:


1.5 W/m K

Source:


Units

Units 0.9

Assumed - less than 1 Units

o 33 C o 35 C o

10 C o 12 C

Vapour in tank, TV

306.15 K

Liquid in tank. TL

308.15 K

Outside air, TA

283.15 K

Assumed - just below the liquid temp Minimum temp requirement by process Assumed - as tanks are inside the building Assumed - just above ambient temperature

P M R t T ρair =

101.325 29 8.31 10 283 1.25

kPa kg/kmol kJ/kmol K o C K kg/m3

Thermal Conductivity of ISOPROPANOL k = 3.56 x 10 -5 x Cp ( ρ4/M)1/3 ------------> from Coulson & Richardson. Vol 6, Page 321 Thermal Conductivity k= 0.199 W/m.K

Chemical Engineering Design by Coulson and Richardson, Volume 6, Page 640 Chemical Engineering Design by Coulson and Richardson, Volume 6, Page 640 Chemical Engineering Design by Coulson and Richardson, Volume 6, Page 640 Chemical Engineering Design by Coulson and Richardson, Volume 6, Page 640

Surface Emissivity Wall and roof, ε

0.0000198 0.0257 0.00343 -

3 1.25 kg/m 1.005 kJ/kg K 1005 J/kg K cP or m.Pa.s 0.0000198 kg/m.s 0.0257 W/m.K 0.00343 1/K kg/kmol o C K

Room Pressure Mol. Wt of air Gas const Room Temp

Source: Source: Source: Source:

Thermal Conductivities Metal walls (Carbon Steel, max 0.5% Carbon),k M

1.25 1.005 1005

Air density at room temperature and pressure

Units

Assumed fouling coefficient, hF Dry wall Wet wall Roof Bottom

Vapor

Ground, TG

285.15 K 2 9.81 m/s

Gravitional constant, g

Calculation Calculation for Grashof Number (NGr) 3 2 2 Grashof Number, NGr = L x ρ x g x ß x ΔT /µ NGr for the liquid phase ( ρ2 x g x ß x /µ2 ) ( ρ2 x g x ß x /µ2 ) L3 x ΔT

7.84E+08 3 7.84E+08 x L x ΔT

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

1.34E+08 3 1.34E+08 x L x ΔT

Calculation for Prandtl Number (NPr) Prandtl Number,NPr = Cp x µ /k NPr for the liquid phase

36.44

NPr for the vapour phase

0.77

Coefficient of vapour at wall, hvw Note: as an initial approximation, assume that the wall temperature is the average of the vapour and outside air temp Tw = (TV + TA )/ 2 First Guess Tw 295.9 K After iteration see below

total height of the tank, L % of liquid full (in terms of height) Proportional height in contact with liquid, Lw

2.42 m

ΔT = Tv - Tw Proportional height in contact with vapour, L - Lw

10.25 K 0.13 m

NGr

L =2.55m

2.55 m 95%

Lw

m

2.85E+06

3.

1

For vertical plates and cylinders, Nusselt Number, NNu NNu = 0.138 x (NGr)0.36 x ((NPr)0.175 - 0.55) NNu

2.3 m

Coefficient of vapour at wall, h vw = NNu x k /( L - Lw) Coefficient of vapour at wall, h vw

-------------Equation 15

11.81

2.38

Nusselt Equation (Perry 5-13) W/m2 K

Coefficient of liquid at wall, hLw Note: 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 Tw NNu = 0.138 x (NGr)0.36 x ((NPr)0.175 - 0.55)

------------- Equation 15

Where, 0.1 109

hLW =NNu x k/Lw

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

NPr

36.44

NGr NPr

2.64E+11

applicable for the vertical plates taller than 3ft

Tw = (TL + TA) / 2 Tw ΔT = TL - Tw

NGr

After iteration see below

----------------- Equation 16

Where, NPr >100 and 104 < (NGr NPr) < 109 Note: In article, Equation 16 is used but as N Pr and (NGr NPr) do not fall in the range, we can't apply Equation 16 directly. Therefore, used Equation 15

7.25E+09

NNu Coefficient of liquid at wall, h Lw

NNu = 0.495 x (NGr NPr)0.25

First Guess o 307.5 C 0.65 K

648.79

------------- USING Equation 15 W/m2 K

53.36

------------- USING Eq (a)

Coefficient of vapour at roof, hVr for the surfaces facing down, NNu = 0.27 x (NGr NPr)0.25

-------------Equation 20

for the surfaces facing down, h Vr = (0.27 x k/D) (NGr NPr)0.25

------------- Eq (b) 295.9 K 286.6 K

Note: We will apply equation (b) assuming roof of diameter and Tw =

First guess After iteration see below

Where, 2 x 107 < (NGr NPr) < 3 x 1010 7 10 5.55E+10 ~ 2 x 10 < (NGr NPr) < 3 x 10

NGr NPr =

Note: Applied Equation 20 as (NGr NPr) is very close to the above range

3 1.34E+08 x L x ΔT

NGr = Where, ΔT ΔT L L

= Tv - Tw K 19.55 K =D m 3.01 m

NGr = hVr

Comment Extra Large: Not sure if the Characteristic Length/hydraulic diameter shoud be taken instead. See: http://www.engineeringtoolbox.com/hydraulic-equivalent-diameter-d_458.html Calculated equivalent diameter from the roof/bottom area of tank Equivalent Diameter for Roof/Bottom l 2.3 m w 3.1 m

7.17E+10 W/m2 K

1.12

------------- USING Eq (b)

Area

2 7.13 m

Equivalent Diameter

3.01 m

Coefficient of liquid at the tank bottom, hLb for the surfaces facing up, NNu = 0.14 x (NGr NPr)0.33

-------------- Equation 19

for the surfaces facing up, h Lb = (0.14 x k/D) (NGr NPr)0.33

------------- Eq ( c)

Tw Tw

= (TL + TG) /2 296.65 K

Note: We will apply equation (c) assuming tank bottom diameter and Tw =

First Guess 307.6 K


Where, ΔT

= TL - Tw K 0.55 K

ΔT L L NGr hLb

=D m m

3.01

1.18E+10 63.90

NGr NPr =

7 10 4.30E+11 ≠ 2 x 10 < (NGr NPr) < 3 x 10

Note: Applied Equation 19, though (NGr NPr) is out of range???

3 7.84E+08 x L x ΔT

NGr =

Where, 2 x 107 < (NGr NPr) < 3 x 1010

W/m2 K

------------- USING Eq (c)

Coefficient of outside air at roof, h'Ar for the surfaces facing up, NNu = 0.14 x (NGr NPr)0.33

-------------- Equation 19

for 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) 295.9 K

First guess

Tws

286.6 K


3 1.34E+08 x L x ΔT

NGr Where, ΔT = Tws - TA ΔT

K 3.45 K

NGr h'Ar

Coefficient of outside air at wall, h'Aw

1.26E+10 2.37

W/m2 K

------------- USING Eq (d)

Note: 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-efficient NNu = 0.138 x (NGr)0.36 x ((NPr)0.175 - 0.55) h = 0.138 x (NGr)0.36 x ((NPr)0.175 - 0.55) x k/L

------------- Equation 15 ------------- Eq (e)

applicable for the vertical plates taller than 3ft

ΔT = ((TL + TV ) / 2 ) - TA) / 4 ΔT 6 K NGr

3 1.34E+08 x L x ΔT

L NGr

2.55 m 1.33E+10

where,

NNu h'Aw

247.51

------------- USING Equation 15 W/m2 K

2.49

------------- USING Eq (e)

Conduction coefficients for ground, metal wall and insulation ( hG, hM, hI) hM = kM /tM

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

hI = kI /tI

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

hG = 8 kG/π*D Where, tM

------------- Equation 23

6 0.0060 25 0.025

tI

mm m mm m

thickness of metal thickness of insulation

<-------------

ENTER INSULATION THICKNESS INSULATED

hM

W/m2 K

9000.00

2

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

hI

1.60

W/m K


hG

1.27

W/m2 K


Radiation coefficients for dry and wet sidewalls, and roof (hRd, hRw, hRr) 4

4

hR = 0.1713 ε [((Tws + 460)/100) - ((TA + 460)/100) ]/( Tws - TA) For the INSULATED side walls, assume Tws = TA + 0.25 (Tbulk - TA) Where, Tbulk is the temperature of liquid or vapour inside the tank if the surface is insulated For the UNINSULATED side walls, assume Tw = TA + 0.5 (Tbulk - TA) Where, Tbulk is the temperature of liquid or vapour inside the tank if the surface is uninsulated For the uninsulated roof, [assumed, roof is always uninsulated] Tw = TA + 0.5 (Tv - TA) Vapour in tank, TV

306.15 K

Liquid in tank. TL

308.15 K

Outside air, TA

283.15 K

Therefore, Tws for insulated DRYSIDE wall = TA + 0.25 (Tv - TA) Tws 286.5 K

First Guess After iteration see below

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

Therefore, Tws for insulated WETSIDE wall = TA + 0.25 (TL - TA) Tws 289.6 K


Therefore, Tws for uninsulated roof = TA + 0.5 (Tv - TA) Tws 286.6 K


Let us find hR for all above using Equation 24 hRd

2.548

W/m2 K


hRw

2.564

W/m2 K


2.549

W/m2 K


hRr

Summary Coefficient ( W/m2 K)

Dry wall

Wet wall

Roof

Bottom

2.38

-

-

-

Coefficient of liquid at wall, h Lw

-

53.36

-

-

Coefficient of vapour at roof, hVr

-

-

1.12

-

Coefficient of liquid at the tank bottom, hLb

-

-

-

63.90

Conduction coefficients for ground hG

-

-

-

1.27

Coefficient of outside air at roof, h'Ar [for still air]

-

-

2.37

-

Do NOT use this value

Coefficient of vapour at wall, h vw

Coefficient of outside air at roof considering wind enhancement factor for the assumed wind velocity, h*Ar

-

-

3.08

-

Obtained by multiplying above value by wind enhancement factor

Coefficient of outside air at wall, h'Aw [for still air]

2.49

2.49

-

-

Do NOT use this value

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

3.24

3.24

-

-

Obtained by multiplying above value by wind enhancement factor

Conduction coefficients for metal wall hM

9000

9000

9000

9000

Conduction coefficients for insulation hI Fouling coefficient, h F

1.6

1.6

-

-

7000

5000

7000

4000

(hRd, hRw, hRr)

2.548

2.564

2.549

-

Overall coefficient,Ud, Uw,Ur,Ub

0.821

1.225

0.932

1.243

2

m/hr

Radiation coefficients for dry and wet sidewalls, and roof

Overall Heat Transfer Coefficient, U Overall dry - sidewall coefficient, Ud at wind velocity of

1/Ud = 1/hvw + tM/kM + tI/kI + 1/(hAw + hRd ) + 1/ hFd 2 1/Ud 1.218 m K/W 2 0.821 W/m K

Ud Overall wet - sidewall coefficient, U w at wind velocity of

2

m/hr

1/Uw = 1/hLw + tM/kM + tI/kI + 1/(hAw + hRw ) + 1/ hFw 2 1/Uw 0.816 m K/W 2 1.225 W/m K

Uw Overall roof coefficient, Ur at wind velocity of

2

m/hr

1/Ur = 1/hVr + tM/kM + 1/(hAr + hRr ) + 1/ hFr 2 1/Ur 1.073 m K/W Ur

2 0.932 W/m K

Overall bottom coefficient, Ub 1/Ub = 1/hLb + tM/kM + 1/hG + 1/ hFb 2 1/Ub 0.805 m K/W Ub

Recalculation of Tws and Tw for next iteration

2 1.243 W/m K

which is obtained from the Graph/ 1.3 Figure 2 for the wind velocity of

2 m/hr

which is obtained from the Graph 1.3 /Figure 2 for the wind velocity of

2 m/hr

Tws,V = (Ud/(hRd + hAw))(Tv-TA)+TA)

286.4 K

Put this value manually in cell C253 untill difference approaches zero

Tw,V=Tv-(Ud/hVw)(Tv-TA)

298.2 K


Wet wall

Tws,W = (UW /(hRW + hAW ))(TL-TA)+TA)

288.4 K


Tw,L=TL-(Uw/hLw)(TL-TA)

307.6 K


Roof

Tws,R = (UR/(hVR + hAW ))(TV-TA)+TA)

287.0 K


Tw,R=TV-(UR/hVR)(TV-TA)

287.0 K

Put this value manually in cell D135 untill difference approaches zero

Tw,B=TL-(UB/hLB)(TL-TG)

307.7 K

Put this value manually in cell D157 untill difference approaches zero

Dry wall

Bottom

Surface area, A

2.55 m

Length, l Width, w Thickness, tM

2.3 m 3.1 m 0.006 m

Lw

95%

2.4225 m 0.1275 m

3.

Wet height, Lw Dry height, Ld

1

m

Assumption: % of tank full with liquid

=2.55m

Total height, L

L

Dimensions of one tank

2.3 m TOTAL dry side wall area, Ad = 2*((L - Lw)*l) + 2* ((L - Lw)*w) DO YOU WANT TO CONSIDER BACK WALL AREA? YES m2 Ad 1.38

<------------MAKE A SELECTION HERE

TOTAL wet side wall area, Aw = 2*(Lw * l) + 2* (Lw * w) DO YOU WANT TO CONSIDER BACK WALL AREA? YES m2 Aw 26.16

<------------MAKE A SELECTION HERE

Roof area, Ar = l * w Ar

7.13

m2

Bottom area, Ab = l * w Ab

7.13

m2

Overall Temperature Difference, ΔT Vapour in tank, TV

306.15 K

Liquid in tank. TL

308.15 K

Outside air, TA

283.15 K

Ground, TG

285.15 K

ΔT for dry side = TV - TA ΔT for wet side = TL - TA

23 K

ΔT for roof = TV - TA ΔT for bottom = TL - TG

23 K

25 K 23 K

SUMMARY Surface

U (W/m2 K )

Area (m2)

ΔT (K)

Dry wall Wet wall Roof Bottom

0.821 1.225 0.932 1.243

1.38 26.16 7.13 7.13

23 25 23 23

Individual Heat Loss, q (W) 26.00 801.31 152.87 203.76

Total

4.22

Heat loss from one tank, Q Q

41.8

1183.94

1183.94 W 1183.94 J/sec

Number of tanks

1

TOTAL heat loss from all the tanks, Q

1,184 J/sec

ASSUME NUMBER OF TANKS (Revision note: Note that heat loss is independant of number of tanks) TOTAL HEAT LOSS WITH INSULATION

HEAT LOSS AFFORDABLE BY LIQUID o 45 C 318.15 K 308.15 K

Temperature inside the tanks Temperature inside the tanks Temperature required to be maintained in the tank Specific heat capacity of liquid inside the tank

21.69 0.90

3000 J/kgK

Mass of liquid in one tank Total mass of the liquid in all tanks

13.65

T 13.65 T 13,645 kg

Mass content of liquid in the tank

Total volume of the tank ΔT Overall Heat Loss affordable by tank liquid

Time taken to drop the temperature up to 35 deg C Coulson & Richardson Vol.1 Page 430

10 K 409,356 kJ 409,356,473 J

ln((TA-Tstart)/((TA-Tend))= Utot x Atot/(m x Cp) x t

L= l= w= Volume = Total mass of liquid when tank is 100% full 100% liquid height equivalent to 95% liquid height equivalent to

=Lxlxw 2.55 2.3 3.1 3 18.18 m 14,363 14.36 14.36 13.65

kg T T T

78,069.19 sec 21.69 hrs 0.90 days

TOTAL TIME TAKEN TO DROP DOWN TANK TEMPERATURE AT DESIRED LEVEL WITH INSULATION

draulic-equivalent-diameter-d_458.html

and ΔT of

o 25 C

and ΔT of

o 25 C

Storage Tank Heat Loss Calcs - Rev.1 31.12.2010 (1)

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