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Tank Example

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    TANK

    Sample Printout2014

    Lauterbach Verfahrenstechnik GmbH 1 2014

    Layout

    Input values: 1.234 or 1.234

    Calculated values: 1.234 or 1.234

    Critical values: 1.234 or 1.234

    Estimated values: 1.234 or 1.234

    Table of contents

    Table of contents ......................................................................... 1

    Calculation of heat loss of storage tanks ................................................. 2

    Properties tank medium*................................................................... 6

    Gas properties*........................................................................... 7

    Outside heat transfer coefficient, roof*.................................................. 8

    Inside heat transfer coefficient, bottom*................................................ 10

    Outside heat transfer coefficient, shell*................................................ 12

    Inside heat transfer coefficient, wet shell*............................................. 14

    Inside heat transfer coefficient, dry shell*............................................. 16

    Inside heat transfer coefficient, roof*.................................................. 18

    Heat transfer radiation inside, roof*.................................................... 20

    Physical properties of heating medium*................................................... 21

    Tube-side heat transfer*................................................................. 22

    Heat transfer coil around tubes*......................................................... 25

    Pressure drop in coil*................................................................... 27

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    TANK

    Sample Printout2014

    Lauterbach Verfahrenstechnik GmbH 2 2014

    Global conditions

    Product temperature tP 50 CInside pressure pi 1 bar

    Air temperature tL -20 C

    Wind speed Wv 10 m/s

    Ground temperature tB -20 C

    Geometry of storage tank

    Ground plan Round sketch

    Diameter DT 12 m

    Circumference 37.7 m

    Area 113.1 m

    Tank height HT 15 m

    Filling level HF 14 m

    Properties:

    Storage medium HFO 180

    Density 897.9 kg/m

    Specific heat capacity cp 1930 J/(kgK)

    Thermal conductivity 0.124 W/(mK)Dynamic viscosity 171 mPasKinematic viscosity 0.000190 m/sCoefficient of thermal expansion 0.000707 1/K

    Gas above storage medium

    Medium name air

    Density 1.078 kg/m

    Specific heat capacity cp 1008 J/(kgK)

    Thermal conductivity 0.02808 W/(mK)

    Dynamic viscosity 0.01964 mPasKinematic viscosity 0.000018 m/sCoefficient of thermal expansion 0.003101 1/K

    Wall thickness and thermal conductivity of the tank and the insulation

    Bottom Shell Roof

    Tank wall thickness 8 mm 8 mm 8 mm

    Thermal conductivity 52 W/(mK) 52 W/(mK) 52 W/(mK)

    Insulation thickness 250 mm 100 mm 100 mm

    Thermal conductivity 2.5 W/(mK) 0.04 W/(mK) 0.04 W/(mK)

    Heat losses

    Bottom Shell wet part Shell dry part Roof

    23.63kW

    14.32kW

    0.9186kW

    2.881kW

    Total heat loss Qtot 41.75 kW

    Calculation of heat loss of storage tanks

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    TANK

    Sample Printout2014

    Lauterbach Verfahrenstechnik GmbH 3 2014

    Bottom

    Thickness of the wall siB 8 mm

    Thermal conductivity of the wall iB 52 W/(mK)

    Thermal resistance of the wall iB 0.000154 mK/WThickness of the insulation saB 250 mm

    Thermal conductivity of the insulation aB 2.5 W/(mK)Thermal resistance of the insulation aB 0.1 mK/W

    Heat transfer coefficient outside aB 5 W/(mK)Heat transfer coefficient inside iB 28.67 W/(mK)

    Contact surface AB = 113.1 m

    Temperature inside tiB = 42.71 C

    Temperature (wall-insulation) tgB = 42.68 C

    Temperature outside taB = 21.79 C

    Heating performance QhB 0 kW

    Heat flow from inside QiB = 23.63 kW

    Heat flow to outside QaB = 23.63 kW

    Shell

    Thickness of the wall siM 8 mm

    Thermal conductivity of the wall iM 52 W/(mK)Thermal resistance of the wall iM 0.000154 mK/W

    Thickness of the insulation saM 100 mm

    Thermal conductivity of the insulation aM 0.04 W/(mK)Thermal resistance of the insulation aM 2.5 mK/W

    Heat transfer coefficient outside aM 41.67 W/(mK)

    Wet part

    Heat transfer coefficient inside iMb 17.68 W/(mK)

    Contact surface AMb = 527.8 m

    Temperature inside tiMb = 48.47 CTemperature (wall-insulation) tgMb = 48.46 C

    Temperature outside taMb = -19.35 C

    Heating performance QhMb 0 kW

    Heat flow from inside QiMb = 14.32 kW

    Heat flow to outside QaMb = 14.32 kW

    Dry part

    Heat transfer coefficient inside iMt 2.868 W/(mK)

    Contact surface AMt = 37.7 m

    Temperature inside tiMt = 41.5 C

    Temperature (wall-insulation) tgMt = 41.5 C

    Temperature outside taMt = -19.42 C

    Heating performance QhMt 0 kW

    Heat flow from inside QiMt = 0.9186 kW

    Heat flow to outside QaMt = 0.9186 kW

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    TANK

    Sample Printout2014

    Lauterbach Verfahrenstechnik GmbH 4 2014

    Roof

    Thickness of the wall siD 8 mm

    Thermal conductivity of the wall iD 52 W/(mK)

    Thermal resistance of the wall iD 0.000154 mK/WThickness of the insulation saD 100 mm

    Thermal conductivity of the insulation aD 0.04 W/(mK)Thermal resistance of the insulation aD 2.5 mK/W

    Emissivity of the roof D 0.8 -Emissivity of the product P 0.9 -Heat transfer coefficient outside aD 19.77 W/(mK)Heat transfer coefficient inside (total) iD 5.067 W/(mK)

    Contact surface AD = 113.1 m

    Temperature inside tiD = 44.97 C

    Temperature (wall-insulation) tgD = 44.97 C

    Temperature outside taD = -18.71 C

    Heating performance QhD 0 kW

    Heat flow from inside QiD = 2.881 kW

    Heat flow to outside QaD = 2.881 kW

    Balance

    Heating performance Qh_ges = 0 kW

    Heat flow from inside Qi_ges = 41.75 kW

    Heat flow to outside Qa_ges = 41.75 kW

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    TANK

    Sample Printout2014

    Lauterbach Verfahrenstechnik GmbH 5 2014

    Calculation of the heating coil

    Heating medium Transcal LT

    Mass flow m 7963 kg/hVolume flow V 10 m/h

    Pressure (abs.) P 3 bar

    Inlet temperature e 140 COutlet temperature a 131.9 CMean temperature m 136 C

    Tube outside diameter da 60.3 mm

    Tube wall thickness s 2.9 mm

    Tube inside diameter di 54.5 mm

    Thermal conductivity of tube material R 52 W/(mK)Fouling inside fi 0 mK/W

    Fouling outside fa 0 mK/W

    Properties of the heating medium

    Density 796.9 kg/mSpecific heat capacity cp 2342 J/(kgK)

    Thermal conductivity 0.125 W/(mK)Dynamic viscosity 1.183 mPas

    Kinematic viscosity 1.484 mm/s

    Properties of the storage medium at 87.85 CDensity 874.3 kg/mSpecific heat capacity cp 2062 J/(kgK)

    Thermal conductivity 0.1214 W/(mK)Dynamic viscosity 33.19 mPas

    Kinematic viscosity 0.000038 m/sCoefficient of thermal expansion 0.00072 1/K

    ResultTube-side velocity u 1.191 m/s

    Heat transfer coefficient inside i 1076 W/(mK)Heat transfer coefficient outside a 124.9 W/(mK)Overall heat transfer coefficient k 110 W/(mK)

    Pressure drop (straight tube without bends) P 5736 PaTube length L 23.33 m

    Area A 4.419 m

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    TANK

    Sample Printout2014

    Lauterbach Verfahrenstechnik GmbH 6 2014

    Properties of Heavy Fuel Oils (HFO)

    Selected oil: HFO 180

    Oil selection: 7

    State 1: State 2:

    Temperature 50 C 87.86 C

    Density 897.9 kg/m 874.1 kg/m

    Specific heat capacity cp 1930 J/(kgK) cp 2062 J/(kgK)

    Dynamic viscosity 171 mPas 33.19 mPasKinematic viscosity 0.000190 m/s 0.000038 m/sThermal conductivity 0.124 W/(mK) 0.1213 W/(mK)Coef. of thermal expansion 0.000707 1/K 0.000720 1/K

    Prandtl number Pr 2661 - Pr 564 -

    Thermal diffusivity a 7.155E-8 m/s a 6.731E-8 m/s

    Pr = /a = cp/a = /(cp)

    Properties tank medium*

    *Properties of heavy fuel oils

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    TANK

    Sample Printout2014

    Lauterbach Verfahrenstechnik GmbH 7 2014

    Properties of air

    State 1 State 2

    Temperature 1 50 C 2 CPressure p1 1 bar p2 Pa

    Density 1.078 kg/m kg/m

    Specific heat capacity cp 1008 J/(kgK) cp J/(kgK)

    Thermal conductivity 0.02808 W/(mK) W/(mK)Dynamic viscosity 0.01964 mPas mPasKinematic viscosity 0.000018 m/s m/sPrandtl number Pr 0.7046 - Pr -

    Thermal diffusivity a 0.000026 m/s a m/s

    Compressibility factor Z 0.9999 - Z -

    Specific enthalpy h 25180 J/kg h J/kgSpecific entropy s 84.9 J/(kgK) s J/(kgK)

    Thermal expansion 0.003101 1/K 1/KVelocity of sound w 360.5 m/s w m/s

    Molar mass M 28.96 g/mol

    Gas constant R 287.1 J/(kgK)

    Standard density S 1.293 kg/m

    Critical data

    Critical temperature Tc -140.6 C

    Critical pressure pc 3786000 Pa

    Critical density c 342.6 kg/m

    Validity range:

    -150 C 1000 C1 bar p 1000 bar

    Composition of the air:

    Mol-% Wt-%

    N2 : 78.12 75.570

    O2 : 20.96 23.161

    Ar: 0.92 1.269

    Normalization of Enthalpy and Entropy:

    h = 0 kJ/kg, s = 0 kJ/(kgK) at T = 298.15 K = 25C, p = 1.01325 bar

    for the pure components

    Gas properties*

    *Properties of air

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    TANK

    Sample Printout2014

    Lauterbach Verfahrenstechnik GmbH 8 2014

    Geometry:

    Heated plate length l 12000 mm

    Flow velocity w 10 m/s

    Properties:

    Mean pressure p 100000 Pa

    Mean temperature -20 C

    Density 1.078 kg/m

    Specific heat capacity Cp 1008 J/(kgK)

    Specific thermal conductivity 0.02808 W/(mK)

    Dynamic viscosity 0.01964 mPas

    Kinematic viscosity = 0.000018 m/s

    Prandtl number Pr = 0.7046 -

    Mean wall temperature W 44.97 C

    Prandtl number at wall temperature PrW 0.7046 -

    Phase (liquid = 1 / gas = 2) 2

    Exponent for liquids nF 0.25 -

    Exponent for gases nG 0 -

    Heat transfer:

    Reynolds number Re = 6587625 -

    Nusselt-number laminar Nulam = 1517 - (1)

    Nusselt-number turbulent Nuturb = 8312 - (2)

    Nusselt number average Nul,0 = 8449 - (5)Nusselt number with wall correction Nu = 8449 - (6)

    Heat transfer coefficient = 19.77 W/(mK)

    Outside heat transfer coefficient, roof*

    *Heat transfer in single-phase flow

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    TANK

    Sample Printout2014

    Lauterbach Verfahrenstechnik GmbH 9 2014

    Equations:

    l 19.77 12

    Nu = 8449 = 0.02808

    w l 10 12

    Re = 6587625 =

    0.000018

    Nu = Nu0 K 8449 = 8449 1 (6)

    2 2

    Nul,0 = Nulam + Nuturb (5)

    8449 = 1517 + 8312

    3

    Nulam = 0.664 Re Pr (1)

    3

    1517 = 0.664 6587625 0.7046

    0.037 Re.

    Pr

    Nuturb = (2)1 + 2.443 Re

    - . (Pr

    /- 1)

    0.037 6587625.

    0.7046

    8312 =1 + 2.443 6587625

    - . ( 0.7046

    /- 1 )

    Influence of the dependence of the properties on the temperature:

    Liquids:

    nF 0.25

    KF = (Pr / PrW ) 1 = ( 0.7046 / 0.7046 )

    Gases:

    nG 0

    KG = (T / TW ) 1 = ( 253.2 / 318.1 )

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    TANK

    Sample Printout2014

    Lauterbach Verfahrenstechnik GmbH 10 2014

    4. Horizontal plane surfaces

    Heat emission at upper side (lower surface cooled)

    Boundary conditions:

    Area of the body flown-around A 113.1 m

    Perimeter of the projection surface U 37699 mm

    Characteristic length l 3000 mm (11)

    Acceleration due to gravity g 9.81 m/s

    Temperature on the surface 0 42.71 CTemp. of fluid outside the boundary layer 50 CTemperature difference (0 - ) 7.287 K

    Properties:

    Mean temperature (0 + ) / 2 m 46.36 CDensity 897.9 kg/mSpecific heat capacity cp 1930 J/(kgK)

    Dynamic viscosity 171 mPasKinematic viscosity 0.000190 m/sThermal conductivity 0.124 W/(mK)Coefficient of thermal expansion 0.000707 1/K

    Chararcteristic values:

    Prandtl number Pr 2661 -

    Grashof number Gr 3.765E+7 - (3)Rayleigh number Ra 1.00E+11 - (4)

    Prandtl function f2(Pr) 0.9874 - (20)

    Nusselt number laminar Nu_l 121.1 - (18)

    Nusselt number turbulent Nu_t 693.7 - (19)

    Nusselt number Nu 693.7 -

    Heat transfer:

    Heat transfer coefficient a 28.67 W/(mK) (2)Exchange surface A 113.1 m

    Convective heat flux Q -23.63 kW

    Inside heat transfer coefficient, bottom*

    *Heat transfer by natural convection around immersed bodies

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    TANK

    Sample Printout2014

    Lauterbach Verfahrenstechnik GmbH 11 2014

    Equations:

    a = Nu / l (2)

    = 693.7 0.124 / 3 = 28.67 W/(mK)

    g l

    Gr = (3)

    9.81 3

    = 0.000707 7.287 = 3.765E+7 -

    0.000190

    Ra = Gr Pr = 3.765E+7 2661 = 1.00E+11 - (4)

    l = A / U = 113.1 / 37.7 = 3000 mm (11)

    1/5

    Nul = 0.766 Ra f2 (Pr) (18)

    1/5

    = 0.766 1.00E+11 0.9874 = 121.1 -

    1/3

    Nut = 0.15 Ra f2 (Pr) (19)

    1/3

    = 0.15 1.00E+11 0.9874 = 693.7 -

    11/20 -20/11

    f2 (Pr) = 1 + 0.322 / Pr (20)

    11/20 -20/11

    = 1 + 0.322 / 2661 = 0.9874 -

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    TANK

    Sample Printout2014

    Lauterbach Verfahrenstechnik GmbH 12 2014

    Heat loss in insulated pipelines (exposed)

    Parameters:

    Temperature medium inside i 50 CAir temperature a -20 CInside diameter of the pipe d1 12000 mm

    Inside heat transfer coefficient i 17.68 W/(mK)Wind velocity w 10 m/s

    Heat transfer:

    Wall thickness Thermal conductivity

    Tube s0 8 mm 0 52 W/(mK)Insulation 1 s1 100 mm 1 0.04 W/(mK)Insulation 2 s2 0 mm 2 1 W/(mK)

    Calculation:

    Outside diameter of the pipe d2 12016 mm

    Outside diameter of the insulation 1 d3 12216 mm

    Outside diameter of the insulation 2 d4 12216 mm

    Temperature difference i -a 70 CAuxiliary variable D 2.578 mK/W

    Outside heat transfer coefficient a 41.67 W/(mK)

    Heat loss per unit of length Q/l -1032 W/m

    Pipe length l mm

    Heat loss absolute Q kW

    Temperatures:

    Temperature medium inside i 50 CWall temperature inside Wi 48.45 CWall temperature outside Wa 48.45 CInsulation Iso -19.35 CSurface temperature O -19.35 CAir temperature a -20 C

    Outside heat transfer coefficient, shell*

    *Heat loss of walls and pipeworks

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    TANK

    Sample Printout2014

    Lauterbach Verfahrenstechnik GmbH 13 2014

    Equations:

    1 1 d2 1 d3 1 d4D = d4 + ln + ln + ln

    i d1 2 1 d1 2 2 d2 2 3 d3

    1 1 12.02

    = 12.22 + ln +

    17.68 12 2 52 12

    1 12.22 1 12.22

    + ln + ln = 2.578 mK/W

    2 0.04 12.02 2 1 12.22

    For static air (w=0) is valid:

    a = 8 + 0.04 O - a

    = 8 + 0.04 -19.35 - -20 = 8.026 W/(mK)

    For wind follows: 1/a = 0.024 = f( 10 ; 12.22 )

    d4 ( i - a )Q/l =

    D + 1 / a

    12.22 ( 50 - -20 )= = -1032 W/m

    2.578 + 1 / 41.67

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    TANK

    Sample Printout2014

    Lauterbach Verfahrenstechnik GmbH 14 2014

    2. Vertical areas (cylinders)

    Boundary conditions:

    Height of the cylinder h 14000 mm

    Diameter of the cylinder D 12000 mm

    Characteristic length l 14000 mm

    Acceleration due to gravity g 9.81 m/s

    Temperature on the surface 0 48.47 CTemp. of fluid outside the boundary layer 50 CTemperature difference (0 - ) 1.534 K

    Properties:

    Mean temperature (0 + ) / 2 m 49.23 CDensity 897.9 kg/m

    Specific heat capacity cp 1930 J/(kgK)

    Dynamic viscosity 171 mPas

    Kinematic viscosity 0.000190 m/sThermal conductivity 0.124 W/(mK)Coefficient of thermal expansion 0.000707 1/K

    Characteristic values:

    Prandtl number Pr 2661 -

    Grashof number Gr 8.056E+8 - (3)

    Rayleigh number Ra 2.14E+12 - (4)

    Prandtl function f1 (Pr) 0.986 - (13)

    Nusselt number for plate Nu_P 1995 - (12)

    Nusselt number Nu 1996 - (14)

    Heat transfer:

    Heat transfer coefficient 17.68 W/(mK) (2)Exchange surface A 527.8 m

    Convective heat flux Q -14.32 kW

    Inside heat transfer coefficient, wet shell*

    *Heat transfer by natural convection around immersed bodies

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    TANK

    Sample Printout2014

    Lauterbach Verfahrenstechnik GmbH 15 2014

    Equations:

    = Nu / l (2)

    = 1996 0.124 / 14 = 17.68 W/(mK)

    g l

    Gr = (3)

    9.81 14

    = 0.000707 1.534 = 8.056E+8 -

    0.000190

    Ra = Gr Pr = = 8.056E+8 2661 = 2.14E+12 - (4)

    1/6 2

    Nu_P = 0.825 + 0.387 Ra f1 (Pr) (12)

    1/6 2

    = 0.825 + 0.387 2.14E+12 0.986 = 1995 -

    9/16 -16/9

    f1 (Pr) = 1 + 0.492 / Pr (13)

    9/16 -16/9

    = 1 + 0.492 / 2661 = 0.986 -

    Nu = Nu_P + 0.87 h / D (14)

    = 1995 + 0.87 14 / 12 = 1996 -

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    TANK

    Sample Printout2014

    Lauterbach Verfahrenstechnik GmbH 16 2014

    2. Vertical areas (cylinders)

    Boundary conditions:

    Height of the cylinder h 1000 mm

    Diameter of the cylinder D 12000 mm

    Characteristic length l 1000 mm

    Acceleration due to gravity g 9.81 m/s

    Temperature on the surface 0 41.5 CTemp. of fluid outside the boundary layer 50 CTemperature difference (0 - ) 8.496 K

    Properties:

    Mean temperature (0 + ) / 2 m 45.75 CDensity 1.078 kg/m

    Specific heat capacity cp 1008 J/(kgK)

    Dynamic viscosity 0.01964 mPas

    Kinematic viscosity 0.000018 m/sThermal conductivity 0.02808 W/(mK)Coefficient of thermal expansion 0.003101 1/K

    Characteristic values:

    Prandtl number Pr 0.7046 -

    Grashof number Gr 7.789E+8 - (3)

    Rayleigh number Ra 5.488E+8 - (4)

    Prandtl function f1 (Pr) 0.3459 - (13)

    Nusselt number for plate Nu_P 102.1 - (12)

    Nusselt number Nu 102.1 - (14)

    Heat transfer:

    Heat transfer coefficient 2.868 W/(mK) (2)Exchange surface A 37.7 m

    Convective heat flux Q -0.9186 kW

    Inside heat transfer coefficient, dry shell*

    *Heat transfer by natural convection around immersed bodies

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    TANK

    Sample Printout2014

    Lauterbach Verfahrenstechnik GmbH 17 2014

    Equations:

    = Nu / l (2)

    = 102.1 0.02808 / 1 = 2.868 W/(mK)

    g l

    Gr = (3)

    9.81 1

    = 0.003101 8.496 = 7.789E+8 -

    0.000018

    Ra = Gr Pr = = 7.789E+8 0.7046 = 5.488E+8 - (4)

    1/6 2

    Nu_P = 0.825 + 0.387 Ra f1 (Pr) (12)

    1/6 2

    = 0.825 + 0.387 5.488E+8 0.3459 = 102.1 -

    9/16 -16/9

    f1 (Pr) = 1 + 0.492 / Pr (13)

    9/16 -16/9

    = 1 + 0.492 / 0.7046 = 0.3459 -

    Nu = Nu_P + 0.87 h / D (14)

    = 102.1 + 0.87 1 / 12 = 102.1 -

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    TANK

    Sample Printout2014

    Lauterbach Verfahrenstechnik GmbH 18 2014

    4. Horizontal plane surfaces

    Heat emission at lower surface (upper surface cooled)

    Boundary conditions:

    Area of the body flown-around A 113.1 m

    Perimeter of the projection surface U 37699 mm

    Characteristic length l 3000 mm (11)

    Acceleration due to gravity g 9.81 m/s

    Temperature on the surface 0 44.97 CTemp. of fluid outside the boundary layer 50 CTemperature difference (0 - ) 5.027 K

    Properties:

    Mean temperature (0 + ) / 2 m 47.49 CDensity 1.078 kg/mSpecific heat capacity cp 1008 J/(kgK)

    Dynamic viscosity 0.01964 mPasKinematic viscosity 0.000018 m/sThermal conductivity 0.02808 W/(mK)Coefficient of thermal expansion 0.003101 1/K

    Characteristic values:

    Prandtl number Pr 0.7046 -

    Grashof number Gr 1.24E+10 - (3)Rayleigh number Ra 8.768E+9 - (4)

    Prandtl function f1 (Pr) 0.3459 - (13)

    Nusselt number Nu 47.26 - (21)

    Heat transfer:

    Heat transfer coefficient a 0.4424 W/(mK) (2)Exchange surface A 113.1 m

    Convective heat flux Q -0.2515 kW

    Inside heat transfer coefficient, roof*

    *Heat transfer by natural convection around immersed bodies

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    TANK

    Sample Printout2014

    Lauterbach Verfahrenstechnik GmbH 19 2014

    Equations:

    a = Nu / l (2)

    = 47.26 0.02808 / 3 0.4424 W/(mK)

    g l

    Gr = (3)

    9.81 3

    = 0.003101 5.027 = 1.24E+10 -

    0.000018

    Ra = Gr Pr = 1.24E+10 0.7046 = 8.768E+9 - (4)

    l = A / U = 113.1 / 37.7 = 3000 mm (11)

    9/16 -16/9

    f1 (Pr) = 1 + 0.492 / Pr (13)

    9/16 -16/9

    = 1 + 0.492 / 0.7046 = 0.3459 -

    1/5

    Nu = 0.6 Ra f1 (Pr) (21)

    1/5

    = 0.6 8.768E+9 0.3459 = 47.26 -

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    TANK

    Sample Printout2014

    Lauterbach Verfahrenstechnik GmbH 20 2014

    Calculation of the heat flux by radiation

    Area Emissivity Temperature

    A1 113.1 m 1 0.9 - T1 50 C (= 50 C )A2 113.1 m 2 0.8 - T2 44.97 C (= 44.97 C )

    View factors:

    12 0.8466

    21 0.8466 21 = A1 /A2 12

    Calculation of the emitted energy:

    Ei = i Ti ; = 5.6704 10-

    W/(m K )

    E1 = 556.5 W/m E2 = 464.6 W/m

    Calculation of heat flux:

    . 1 2 A1 12Q12 = (T1 - T2 )

    1 - (1 - 1 )(1 - 2 ) 12 21

    .

    Q12 = 2.629 kW Q21 = 2.629 kW

    Equivalent heat transfer coefficient

    .Q

    =A (T1 - T2 )

    12 = 4.624 W/(mK)

    Heat transfer radiation inside, roof*

    *Surface radiation

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    TANK

    Sample Printout2014

    Lauterbach Verfahrenstechnik GmbH 21 2014

    Name of the oil Transcal LTMaterial structure naphthene base

    Manufacturer BP

    Former product/Comment

    Range of application

    State 1 State 2

    Temperature 136 C 126.3 C

    Density 796.9 kg/m 803.1 kg/mSpecific heat capacity cp 2342 J/(kgK) 2309 J/(kgK)

    Dynamic viscosity 1.183 mPas 1.358 mPas

    Kinematic viscosity 0.000001 m/s 0.000002 m/sThermal conductivity 0.125 W/(mK) 0.1255 W/(mK)Coef. of thermal expansion 0.000775 1/K 0.000781 1/KThermal diffusivity a 6.694E-8 m/s 6.766E-8 m/s

    Prandtl number Pr 22.18 - 24.97 -

    Specific enthalpy h 0 kJ/kg 0 kJ/kg

    Vapour pressure pD 604.7 Pa 427.1 Pa

    Pour point -54 C

    Initial boiling point 290 C

    Minimum operating temperature -20 C

    Maximum operating temperature 260 C

    Minimum temperature filling -20 C

    Minimum temperature startup 71 CMaximum film temperature 280 C

    Flash point 155 C

    Ignition temperature 240 C

    Neutralization number 0.01 mgKOH/g

    Coke residue 0.01 %

    Explosion limit Vol-%

    Molecular weight kg/kmol

    Temperature min max

    Density 900 kg/m 732 kg/m

    Specific heat capacity cp 1800 J/(kgK) 2770 J/(kgK)

    Dynamic viscosity 730 mPas 0.35 mPas

    Kinematic viscosity 0.0003 m/s 4.9E-07 m/sThermal conductivity 0.136 W/(mK) 0.118 W/(mK)Vapour pressure pD Pa 28000 Pa

    Physical properties of heating medium*

    *Properties of heat transfer oils

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    Lauterbach Verfahrenstechnik GmbH 22 2014

    Constant wall temperature

    Circular pipes

    Inlet temperature e 140 COutlet temperature a 131.9 CMean temperature m 136 CDensity of the fluid 796.9 kg/mSpecific heat capacity of the fluid cp 2342 J/(kgK)

    Thermal conductivity of the fluid 0.125 W/(mK)Dynamic viscosity of the fluid 1.183 mPas

    Kinematic viscosity of the fluid 0.000001 m/sPrandtl number Pr 22.18 -

    Prandtl number at wall temperature PrW 24.97 -

    Fluid: liquid or gas < 0 >

    0 = Circular pipes, 1 = Non-circular pipes < 0

    Pipe length l 23326 mmInside diameter of the pipe di 54.5 mm

    Cross sectional area of the pipe f 0.002333 m

    Perimeter of the pipe u 171.2 mm

    Hydraulic diameter dh = 54.5 mm

    Total mass flow Mg 7963 kg/h

    Total volume flow Vg 9.992 m/h

    Number of pipes with parallel flow Z 1 -

    Mass flow per pipe M = 7963 kg/h

    Velocity w = 1.19 m/s

    Reynolds number Re = 43683 -

    Balance: Q = Mg cp (a - e ) = -41.75 kW

    Results: Constant wall temperature

    Nu 469.5 0.125

    = = = 1076 W/(mK)dh 0.0545

    w dh 1.19 0.0545 796.9Re = = = 43683

    / 1000 1.183 / 1000

    Nu_m_ = (laminar non-disturbed flow Re < 2300) [6]Nu_m_ = (laminar entrance flow Re < 2300) [12]Nu_m_T = 475.7 (turbulent flow Re > 10000) [26]

    Nu_m = (transition zone 2300

    Re

    10000) [29]

    Factor K:

    0.11

    Liquids: K = (Pr/PrW ) = 0.987

    n

    Gases: K = (T/TW ) = for n = 0

    Nu = Nu_m(_,T) K = 469.5 [40,41]

    Heat transfer: Q = A log

    -41746 = 1076 3.994 -9.711

    Wall temperature W = 125.7 C

    Tube-side heat transfer*

    *Heat transfer in pipe flow

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    Calculations:

    Laminar flow: Re < 2300

    Laminar non-disturbed flow:

    HW = Re Pr dh/l = 2263

    1/3

    Nu_m__2 = 1.615 HW = [5]

    3 3 3 1/3

    Nu_m_ = 3.66 + 0.7 + Nu_m__2 - 0.7 =[6]

    3 3 3 1/3

    = 3.66 + 0.7 + - 0.7 =

    Laminar entrance flow:

    1/6

    2

    Nu_m__3 = HW =1 + 22 Pr

    1/6

    2

    = 2263 = [11]

    1 + 22 22.18

    3 3 3 3 1/3Nu_m_ = 3.66 + 0.7 + Nu_m__2 - 0.7 + Nu_m__3 =

    [12]

    3 3 3 3 1/3

    = 3.66 + 0.7 + - 0.7 + =

    Turbulent flow: Re > 10000

    -2 -2

    = [1.8 log Re - 1.5] = [ 1.8 log 43683 - 1.5 ] = 0.0213 [27]

    2/3

    /8 Re Pr dh

    Nu_m_T = 1 + =

    2/3 l

    1 + 12.7 /8 (Pr -1) [26]

    2/3

    0.0213 /8 43683 22.18 0.0545

    = 1 +

    2/3 23.331 + 12.7 0.0213 /8 ( 22.18 -1)

    Nu_m_T = 475.7

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    Transition zone: 2300 Re 100001/3

    Nu_m__2_2300 = 1.615 (2300 Pr dh/l) = 7.948 [32]

    1/6

    2

    Nu_m__3_2300 = 2300 Pr dh/l = 4.366 [33]1 + 22 Pr

    3 3 1/3

    Nu_m_L_2300 = 49.371 + Nu_m__2_2300 - 0.7 + Nu_m__3_2300

    Nu_m_L_2300 = 8.007 [31]

    2/3

    (0.0308/8) 10000 Pr dh

    Nu_m_T_10000 = 1 +

    2/3 l

    1 + 12.7 0.0308/8 (Pr -1)

    [37]

    Nu_m_T_10000 = 135

    Re - 2300 43683 - 2300

    = = = [30]10000 - 2300 10000 - 2300

    Nu_m = (1 - ) Nu_m_L_2300 + Nu_m_T_10000 = [29]

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    5. Horizontal curved areas (Cylinder)

    Cylinder:

    Outside diameter do 60.3 mm

    Wall thickness s 2.9 mm

    Inside diameter di 54.5 mm

    Thermal conductivity 52 W/(mK)Characteristic length l 94.72 mm

    Acceleration due to gravity g 9.81 m/s

    Temperatures and properties:

    Temperature on the surface 0 125.7 CTemp. of fluid outside the boundary layer 50 CTemperature difference (0 - ) 75.7 K

    Mean temperature (0 + ) / 2 m 87.85 CDensity 874.3 kg/mSpecific heat capacity cp 2062 J/(kgK)

    Dynamic viscosity 33.19 mPasKinematic viscosity 0.000038 m/sThermal conductivity 0.1214 W/(mK)Coefficient of thermal expansion 0.00072 1/K

    Characteristic values:

    Prandtl number Pr 563.8 -

    Grashof number Gr 315094 - (3)

    Rayleigh number Ra 1.777E+8 - (4)

    Prandtl function f3 (Pr) 0.9647 - (24)Nusselt number Nu 97.47 - (22)

    Heat transfer:

    Heat transfer coefficient (free convection) o 124.9 W/(mK) (2)Convective heat flux Q 41.75 kW

    Balance for the calculation of the surface temperature:

    Heating medium

    Specific heat capacity cp 2342 J/(kgK)

    Density 796.9 kg/m

    Mass flow m 7963 kg/h

    Volume flow V9.992

    m/h

    Inlet temperature in 140 COutlet temperature out 131.9 CMean temperature m 136 C

    Duty QH -41.75 kW

    Tube-side velocity u 1.19 m/s

    Heat transfer coefficient inside i 1076 W/(mK)Fouling inside fi 0 mK/W

    Fouling outside fo 0 mK/W

    Overall heat transfer coefficient k 110 W/(mK)

    Length of the cylinder L 23326 mm

    Area of the cylinder A 4.419 m

    Heat transfer coil around tubes*

    *Heat transfer by natural convection around immersed bodies

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    Equations:

    a = Nu / l (2)

    = 97.47 0.1214 / 0.09472 = 124.9 W/(mK)

    g l

    Gr = (3)

    9.81 0.09472

    = 0.00072 75.7 = 315094 -

    0.000038

    Ra = Gr Pr = 315094 563.8 = 1.777E+8 - (4)

    1/6 2

    Nu = 0.752 + 0.387 Ra f3 (Pr) (22)

    1/6 2

    = 0.752 + 0.387 1.777E+8 0.9647 = 97.47 -

    9/16 -16/9

    f3 (Pr) = 1 + 0.559 / Pr (24)

    9/16 -16/9= 1 + 0.559 / 563.8 = 0.9647 -

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    Straight pipes

    Parameters of the pipe:

    Length of pipe l 23326 mm

    Inside diameter of pipe di 54.5 mm

    Absolute roughness K 0.04 mm

    Number of pipes with parallel flow NR 1 -

    Properties:

    Density 796.9 kg/mDynamic viscosity 1.183 mPas

    Kinematic viscosity 0.000001 m/s

    Total mass flow mg 7963 kg/h

    Mass flow per pipe mR 7963 kg/hVolume flow per pipe VR 9.992 m/h

    Velocity of the fluid wi 1.19 m/s

    Result:

    Flow pattern turbulent (Re > 2320)

    Pressure drop p 5736 PaReynolds-Number Rei 43683 -

    Drag coefficient 0.02376 -

    Equations:

    wi di 1.19 796.9 0.0545Rei = = = 43683 -

    0.001 1.183 0.001

    Rei = 43683 > 2320 turbulent flow

    1 2.51 K / di= -2 lg +

    Rei 3.71

    1 2.51 0.00004 / 0.0545

    = -2 lg +

    0.0237643683

    0.02376 3.71

    l wi 23.33 796.9 1.19

    p = = 0.02376 di 2 0.0545 2

    p = 5736 Pa

    Pressure drop in coil*