Iso-12241.pdf 2c1b1h

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<0,40m. Equation (18 a) is also used for ducts with rectangular cross-section, having a width and height of similar magnitude.

4.1.2.4 External surface resistance The reciprocal of the outer surface coefficient h, is the external surface resistance. For plane walls the surface resistance ß s e , in m2.K/W, is given by

ß ,

=- I

rn2.K/W

hse

For pipe insulation the linear thermal surface resistance Rie is given by: ßie = h se

1

m. K/W .De

For hollow spheres the thermal surface resistance Rh e is given by

14 COPYRIGHT International Organization for Standardization Licensed by Information Handling Services

STD.ISO 122LiI-ENGi 1 9 9 8 0

Lid51403 0739LilcI L 9 T

IS0 12241:1998(E)

IS0

4.1.3

Thermal transmittance

Thermal transmittance U is given by U=---

'

W/(m2. K)

ei -e,

For plane walls the thermal transmittance U can be calculated 1 1 1 - = -+ R + -= ßSi+ ß + ß , U hi hS€?

m2-K/W

For pipe insulation the linear thermal transmittance U, can be calculated 1 U1

- --

1

hi . 71:. Di

+ß,+

1

h,,

*

71:

. De = ßli + ßI + Rie

m2.K/W

For hollow spheres the thermal transmittance Usph is given by:

The surface resistance of flowing media in pipes ßsi(in the cases predominantly considered here) is small and can be neglected. For the external surface coefficient h,,, equations (19) and (20) apply. For ducts one also has to use the internal surface coefficient. The reciprocal of thermal transmittance U is the total thermal resistance ß T for plane wails and respectively the total linear thermal resistance ßTI for pipe insulation and ß T sph for hollow spheres insulations. The thermal transmittance of a duct with rectangular cross sections can be obtained by eq. (25) by replacing ß by ßd(eq. 14).

4.1.4

Temperatures of the layer boundaries

The general equation for the heat loss in a multi-layer wall may be written in the following general form:

(28)

and RT

= ßsi+

ßt

+ ß 2 + ... + ß, + ßse m2WW

where Ri, ß2... are the thermal resistances of the individual layers and ß,,, &are the thermal surface resistances of the interior and exterior surtace.


4

4

Figure 8: The temperature distribution for a multi-layer plane wall in relation to the thermal surface resistance and the thermal resistances of layers The ratio between the resistance of each layer or the surface resistance with respect to the total resistance will give a measure of the temperature change across the particular layer or surface in K.

esi -e, =-.(ei R1

-ea)

K

e1-e2=-.R2 (e¡ - e a )

K

RT

RT

e,, -ea =-.(ei RS,

-ea)

K RT ßT is defined for plane walls according to equation (25), for cylindrical pipes according to eq. (26), and for spherical insulations by equation (27).

4.2

Surface temperature

The surface temperature can b e calculated by using eq. (30)


0

IS0 12241:1998(E)

IS0

For operational reasons it is often stipulated in practice that a certain surface temperature or temperature of the surface higher than that of the ambience should be maintained. The surface temperature is no measure for the quality of the thermal insulation. This depends not only on the heat transmission but also on operating conditions which cannot be readily determined or warranted by the manufacturer. These include among other things: ambient temperature, movement of the air, state of the insulation surface, effect of adjacent radiating bodies, meteorological conditions etc. Further, it will be necessary to make assumptions for the operating parameters. With all these parameters it is possible to estimate the required insulation thickness using equation (30) or diagram 1 (see reference [lo]). It must be pointed out, however, that these assumptions will correspond to the subsequent operating conditions only in very rare cases. Since an accurate registration of all relevant parameters will be impossible, the calculation of the surface temperature is inexact and the surface temperature cannot be warranted. The same restrictions apply to the warranty of the temperature difference between surface and air, also called excess temperature. Although it includes the effect of the ambient temperature on the surface temperature it assumes that the heat transfer by convection and radiation can be covered by a total heat transfer coefficient whose magnitude must also be known (see 4.1.2). However, this condition is generally not fulfilled because the air temperature in the immediate vicinity of the surface, which determines the convective heat transfer, mostly departs essentially from the temperature of other surfaces with which the insulation surface is in radiative exchange. Diagram 1: Determination of insulating layer thickness for a pipe at a given heat flux density or for a set surface temperature (see next page)



STD.ISO 12241-ENGL L778 m 4851903 073743q a35 m 0

I S 0 12241:1998(E)

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The equation for the thickness parameter C’ is derived from equations (24) and (26) by elementary transformations. Equation (a) permits calculation of the necessary insulation thickness for a given linear density of heat flow rate, whereas equation (b) allows calculation of the required insulation thickness for a given temperature difference between the pipe surface (with insulation) and the ambient temperature.

In both cases h,, must be assumed or calculated (see example 8.7).

4.3

Prevention of surface condensation

The surface condensation does not only depend on the parameters affecting the surface temperature but also on the relative humidity of the surrounding air which very often cannot be stated accurately by the customer. It is all the more difficult to state the higher the relative humidity is, in which case fluctuations of the humidity or of the surface temperatures make themselves felt strongly. Unless other data are available assumptions have to be made as in diagram 3 (see clause 9) to calculate the necessary insulation thickness to prevent dew formation on pipes. Using equation (30) the necessary insulation thickness to prevent dew formation can be obtained by iterative techniques. The allowed temperature difference (in OC) between surface and ambient air for different relative humidities at the onset of dew formation is given in table 3.

5

Calculation of the temperature change in pipes, vessels and containers

5.1

Longitudinal temperature change in a pipe

To obtain the accurate value of the longitudinal temperature change in a pipe with a flowing medium, ¡.e. liquid or gas, the following equation applies:

\orm- o,) = loim- o,/. ePn”

K

where

Sf,

4, 0, c, fi

I

is the final temperature of the medium, in O C ; is the initial temperature of the medium, in O C ; is the ambient temperature, in OC; is the specific heat capacity at constant pressure of the flowing medium, in kJ/(kg.K); is the mass flow rate of the flowing medium, in kg/h; is the length of the pipe, in m; is the linear thermal transmittance, in W/(m.K).

Equations (31) and (32) can also be used for ducts with rectangular cross section. Then U, has to be replaced by Ud (es. 25). Since, in practice, the allowed temperature change is often small, for approximate calculation the following equation applies:

dû ql

is the longitudinal temperature change, in K; is the linear density of heat flow rate, in W/m.


For the calculation of 91 equation (24) can be used or equations (5) and (7) if the external surface coefficient can be neglected. Equation (33) will yield results of satisfying accuracy only for relatively short pipes and a relatively small temperature change [A8 I0,06. (& - %)l.

5.2

Temperature change and cooling times in pipes, vessels and containers

The allowed cooling time t, for a given temperature change is calculated by

where

9 A

t, rn

is the density of heat flow rate, in W/m2; is the surface area of the container or vessel, in m2; is the cooling time, in h, producing the temperature drop; is the mass of contents, in kg; is the specific heat capacity, in kJ/(kg.K), of the medium.

For a spherical container 9.A is replaced by the heat flow rate Qsph from equation (11). The accurate calculation of the time-dependent temperature change is performed according to section 5.1, using equation (31) and replacing I by t and a by a’. The approximate time-dependent temperature drop can be calculated by equation (36): with a’=

U * A .3,6 m.

-

NOTE In calculating the cooling time it is assumed that no heat is absorbed by the media during cooling. The obtained cooling time is the fastest, which means there is a safety factor built in by modelling (design calculation). For small containers the heat capacity of the container itself is taken into and in equation (34) an analogous term as in equation (37) is added.


6

Calculation of cooling and freezing times of stationary liquids

6.1

Calculation of the cooling time for a given thickness of insulation to prevent the freezing of water in a pipe

It is impossible to prevent the freezing of a liquid in a pipe, although insulated, over an arbitrary long period of time.

As soon as the liquid (normally water) in the pipe is stationary the process of cooling starts. The linear density of heat flow 91 of a stationary liquid is determined by the energy stored in the liquid ,*m, and in the pipe material p-mpas well as by the freezing enthalpy required to transform water to ice. If mp.p<< mw., then mP+c,,may be neglected. The time until freezing starts is calculated using the following equations :

where

and

I 6, Of,

0,

c, mw m P

is the length of the pipe, in m; is the initial medium temperature, in OC; is the final medium temperature, in OC; is the ambient temperature in OC; is the specific heat capacity, in kJ/(kg.K); is the mass of water, in kg; is the mass of the pipe, in kg.

In practice, for the calculation of 9, the exterior thermal surface resistance should be neglected for insulated pipes.

If a comparison is made between uninsulated and insulated pipes the influence of the surface coefficient of the uninsulated pipe must be taken into consideration. The density of heat flow rate of the uninsulated pipe is given by:

As an approximation the cooling time is given by:


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STDeISO 1224L-ENGL 1998

LIA51703 0737437 59'4

I S 0 12241:1998(E)

The time until freezing starts is calculated by using the procedure above with point of the liquid.

0

e,,

tso

equal to the freezing

Diagram 2 shows examples for the cooling time before freezing starts for a range of pipe diameters and ambient temperatures, for water initially at 5 OC.

6.2

Calculation of the freezing time of water in a pipe

The freezing time is dependent on the heat flow and the diameter of the pipe. It is given by:

with

*and f Qp

hi r Pice

is the percentage of water that is frozen; is the interior pipe diameter, in m; is the specific enthalpy = latent heat of ice formation = 334 kJ/kg; is the density of ice at O"; Pice = 920 kg/m3.

The percentage fof water that is frozen shall be chosen according to a requirement, ¡.e 25% ( f = 25). The allowable cooling time may be taken as well from diagram 2. Due to the reduction of the cross-section of slides, taps and fittings cooling and freezing times are reduced as well. It is advised to decrease the cooling and freezing times twp and 4, given in 6.1 and 6.2 by 25%. The allowed cooling times may also be taken from diagram 2.

7

Thermal bridges

Pipe mountings, s and armatures may be thermal bridges which cannot be calculated by normal means. They cause additional heat losses, which can be taken into consideration in different ways. For pipes, components in the insulating layer like spacers and s are taken into by an additional term AA to the reference thermal conductivity l. of the insulation material (see clause 9):

The effect of valves, slide valves and flanges may be taken into according to table 4 by adding a fictitious pipe length Al to the given length I:

Like in pipes, the real temperature drop in containers is much affected by thermal bridges. A substantial increase in thickness of the insulation in containers is necessary.


EID-IS0 L Z Z L i l - E N G L 1778 @

IS0

8

-

'48517U3 0739'418 YA0 W

IS0 12241:1998(E)

Underground pipelines

Pipelines are laid in the ground with or without thermal insulation either in channels or directly in the soil.

8.1

Calculation of heat loss (single line)

At present, the following two methods are used for pipe-laying without channels:

The heat flux per metre of an underground pipe is calculated from

is the medium temperature, in OC; e,, is the surface temperature of the soil, in O C ; RE is the thermal resistance for a pipe laid in homogeneous soil, in m-W ; LE is the thermal conductivity of the ambient soil, in W/(m.K); HE is the distance between the centre of the pipe and the sutface, in m.

4

The thermal resistance for the ground (figure 9) is calculated in accordance with equation (46) RE =

1

2.TC.A-E

2.HE

. arcosh -

m.K/W

Di

I

-_ Figure 9

@O ~~~

1

- Underground pipe without insulation 23

COPYRIGHT International Organization for Standardization Licensed by Information Handling Services

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I S 0 12241:1998(E)

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whereby equation (46) is simplified for HEID,> 2 to

For underground pipes with insulating layers in accordance with figure 10, the thermal resistance is calculated in accordance with equation

i

i

I

j

l

I

I

I

Figure 1O - Underground pipe comprising several concentric layers, e.g. consisting of insulating material and sheathing (e.g. jacket pipe), embedded in a bottoming (e.g. sand) with a square cross-section The square cross-section of the outer layer with side length a is thereby taken into consideration with an equivalent diameter O,= 1,073. a


(49)

@

IS0

I S 0 12241:1998(E)

Internal diameter Di is identical to Do ( where j = 1). The thermal resistance of the ground ßEresults for this case at

Calculation methods are available for the determination of the heat flow rate and temperature field in the ground for several adjacent pipes, ¡.e double lines or laid systems, see references [I21 to [14] in Annex C. In the case of commonly used jacket pipes which are laid adjacent to each other, if Al

I

-

-

'STDDISO LZilLIL-ENGL L778 W 4853703 0737q23 T 7 5 I S 0 12241 :1998(E)

9

0 IS0

Tables and diagrams

Table 3: The allowed temperature difference in "C between surface and ambient air for different relative humidities at the onset of dew formation Ambient air temperature "C

-20 -15 -10 -5 O

2 4

6 8 10

12 14 rl.6 18 20 22 24 26

28 30 35 40 85 50

Relative air humidities. Yo

40 30 35 45 50

- - - -12,3 12,9 13,4 13,9 14,3 14,7 15,l 15,6 16,O 16,5 16,9 17,4 17,8 18,l 18,4 18,6 18,9 19,2 19,5 20,2 20,9 21,6 22,3

10,4 10,8 11,3 11,7 12,2 12,6

9,1 9,6

9,9 10,3 10,7 ll,o i3,O 11,4 13,4 13,8 14,2 14,6 15,l

15,5 15,7 15,9 16,l 16,4 16,6 16,9 17,l 17,7 18,4 19,o 19,7

8,O 8,3

799 7,3

47 9,o 9,3 9,7 10,l

7,6 7,9 61 8,5 8,9

11,8 10,4 9 2 12,2 10.8 9,6 12.6 11,2 10,o 13,O 11,6 10,l 13,4 11,7 10,3 ?3,6 11,9 10,4 1 3 3 12,l 10,6 14,O 12,3 10,7 14,2 12,5 10,9 14,4 12,6 11,l 14,7 12,8 11,2 14,9 13,O 11,4 15,l 13,2 11,6 15,7 13,7 12,o 16,l 14,2 12,4 16,7 14,7 12,8 17,3 15,2 13,3

- - - --

Example: At an ambient temperature of 20" C and 70 % relative humidity the allowed surface temperature is 20 "C - 5,6 O C = 14,4 "C. a) V a k s and slide valves To for the presence of valves and slide valves in a piping system, add additional length in metres from table 4 to the real length of the pipeline before calculating the heat loss. These values for the valve and its own flanges, but not for the flanges where the valve mounts in the piping system [see b)]. Values in table 4 assume typical industrial insulation thicknesses for the temperatures given, and thermal conductivities of A = 0,08W/(m.K) at 100 "C mean temperature, and A = 0 , l O W/(m.K) at 400 "C mean temperature.


i

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STD-IS0 12241-ENGL 1 9 9 8 0

4851903 0739422 901

IS0 12241:1998(E)

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Table 4: Additional heat losses due to components in a pipeline

1

Pipe diameter Medium

Pipe inside

Qin m

I

1

0,lO

1

1

0,50

1

' 1

@ in"C

100

295 15 690 495

100

3,o

22 8,O 6,O

400

25 10,o 795

9

490

I

I

I

I

non-insulated valve 213 insulated valve 3/4 insulated valve

400 16 630 5,o

6 3,0

non-insulated valve 2/3 insulated valve 3/4 insulated valve

I

Pipe outside

I

19 7,o 64

32 11,o

83

b) Pair of flanges To for the heat losses from a pair of flanges in a piping system (including the flange pair when a valves is mounted ): Non-insulated flanges:

From the table above, use one third of the length given for a valve of the same diameter. Add this to the real length of the piping before calculating the heat losses.

Insulated with flange boxes: To the real length of the piping, add one meter for each flange with flange box, before calculating the heat losses. Insulated flanges:

No adjustment required; calculate heat losses based on real length.

c) Pipe suspensions Add to calculated heat loss (without previous compensation for other components). in interior spaces: in the open air without wind: in the open air with wind:

15 ?Ao of the heat loss 20 Yo of the heat loss 25 % of the heat loss

d) s for sheet-metal pipelines jackets Additions to thermal conductivity:

for steel s 0,Ol O W/(m.K) for ceramic s 0,003 W/(m.K)


STD.ISO 12241-ENGL 1778

4851703 0737423 848

IS0 12241:1998(E)

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Diagram 2: Determination of cooling times from 5 O C to O O

IS0

C

The maximum allowed cooling time of water in pipes of different diameter and with different insulation thicknesses at different ambient temperatures to avoid freezing of the water in the pipe. The initial water temperature e= 5 "C,the wind speed v = 5 m/s, 1 = 0,04 W/(m.K), h,=20 W/(m2K)

O

100

200

300

400

500 O

Pipe diameter in mm

20.

40

60

80

100

120

140

160

Time in hours

Example: For a given pipe diameter of 300 mm with an insulation thickness of 60 mm and an ambient air temperature of - 1O "C, the maximum allowed cooling time is 40 h.


STD.IS0 12291-ENGL 1778 0

9851703 0737929 784

Is0 12241:1998(E)

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The required insulation thickness to prevent dew formation

Diagram 3: Ex1

Relatire air humidify 80 %

diam

mm

+151+10]+5 10

1.5

I

I

Medium temperalure in T 1.10 1-15 1-20 1-25 1-301-35 1-40 1-45 1-50 1-55 1-60 1-65 1-70 1.751+151+10[+5 10

Relatire air humidily 85 O/.

exlem

.. . Pix

1-5

Medium iemperalute in "C diam 1-10 1-15 1-20 1-25 1-30 1-351-40 1-45 1-50 1-55 1-60 1-65 1-70 1-75 mm

The required insulation thickness in mm for refrigerant pipes of different diameters and different temperatures at different relative humidities of the ambient air. is A = 0,04W/(m.K), The thermal conductivity of the insulation at 8 = iO The thermal conductivity of the insulation at 8 = -100 "C is A = 0,033 W/(m.K), Ambient air temperature is 20 "C, hse=6 W/(m2-K) O


~

~

I

STD.IS0 12241-ENGL 1 7 7 8 M 'i851703 0737425 b1O I S 0 12241:1998(E)

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IS0

Annex A (informative) Comments on thermal conductivity

There is a distinction between the laboratory and the declared thermal conductivity as well as the design value of the thermal conductivity of an insulation.

A.l Laboratory thermal conductivity

&,I,

An important description of the quality of a thermal insulation (see I S 0 8302) is the laboratory thermal conductivity, measured with a guarded hot plate in accordance with I S 0 8302 or the heat flow meter in accordance with IS0 8301 on plane samples. It is dependent on the kind of thermal insulation, its composition, structure and density and on the temperature. The laboratory thermal conductivity is measured with unused, dry samples in the given temperature regime in steps of 20 K, 50 K or 100 K (see IS0 9251). The laboratory thermal conductivity for plane products is called A&, and is given as a function of the mean test temperature. The laboratory thermal conductivity of dry thermal insulations of hollow cylindrical specimens like pipe sections (see IS0 9229) in different diameters and thicknesses is determined with the pipe testing apparatus according to IS0 8497. This value includes besides the temperature difference parameters, which are due to the test conditions, for example the influence of longitudinal or transverse ts and of single or multiple insulation layers, ¡.e. effects of workmanship. It is given in the relevant temperature regime as a function of the mean temperature. This value is called &b,R

A.2 Declared thermal conductivity

heC

The declared thermal conductivity stated by the manufacturer must take production-related fluctuations into consideration. The declared value for plane products is based on the laboratory thermal conductivity ;ilab,p and the declared value of pipe sections on &,,R. A method to derive the declared thermal conductivity from the laboratory thermal conductivity is given in I S 0 13787. Another approach for the derivation of the declared and designed values is given in reference [8].

A.3 Designvalue

h

The design value of an insulation is warranted by the contractor who does the actual application. The evaluation of the design value is done either on the basis of the declared thermal conductivity or on the basis of the laboratory thermal conductivity. In addition the thermal conductivity has to be increased by allowance factors taking into the influemes of the actual temperature difference of the installed material, of workmanship, changes in density or structural changes (see BS 5422). If, in the case of pipe insulation, the declared thermal conductivity is used as basis, the value might already include these parameters. However, one has to prove that they are of sufficient magnitude. If a vertical thermal insulation is permeable to air and if there exists an air layer within the insulation, then ,an appropriate allowance has to be considered.


STD*ISO 122LiL-ENGL 1778 0

IS0

'+A51703 0737426 557 IS0 12241:1998(E)

The influence of other thermal bridges, which are due to the installation of the insulation material such as spacers, carrying and constructions has to be included with allowance factors according to clause 7 and table 4 of this standard. The resultant design value is only a warrantable quantity if the allowance factors due to thermal bridges are known with sufficient accuracy. For other kinds of constructions the allowance factors have to be determined either experimentally or by calculation. NOTE - More advanced calculation techniques for thermal bridges are given in reference [14].

The laboratory thermal conductivity of a specimen, taken from an installed insulation may be only checked if no material or structural changes occurred during mounting.


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STD.ISO 1 2 2 4 1 - E N G L 1998

4851903 0739427 493

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Annex B (informative) Examples

B.l Calculation of the necessary insulation thicknesses for a double layered wall of a firebox For this example the following boundary conditions are given: Inside temperature: Outside temperature: Height of the wall: Maximum density of heat flow rate: Wind velocity

4 = 850 OC ûe= 20OC H=4m q = 500 W/m2 v = 3 m/s

The insulation should consist of the following materials: 1" layer: alumino silicate fibre 2"d layer: :mineral wool with a galvanized sheet iron lining Temperature at the boundary layers:

e;= 600 OC

Neglecting the inner surface resistance, the insulation thickness of the 1" layer can be found from eq. 2 in 4.1.1 .with a thermal conductivity A,= 0,20 W/(m-K) at O,,= 725 O C .

To calculate the necessary thickness of the second insulation layer, the external surface coefficient hse has to be calculated from eq. 15 using eq. 16 and eq. 18f and estimating an external surface temperature O,, = 60 OC: With C,= 1,47

X

W/(m2.K4),which is equivalent to E = 0,26 and a temperature factor a, of

a,. = 4 ~ 3 1 3 , 1 5=~1 , 2 3 ~ 1 80 K3 h, is given by h, = 1,23 x 1OE x 1,47 x 1O-8 = 1,81 W/(m2.K) and the convective term h,, according to eq. 18f is

64

h,, = 5,763-

= 103 W /(m2-K)

giving a total external surface coefficient:

hse= 12,31 W/(m2.K)


4853703 0737Li28 32T m IS0 12241:1998(E)

STD.ISO 12243-ENGL 3798 0

IS0

The practical thermal conductivity of the mineral wool at O,, = 330 "C is A2 = 0,110 W/(m.K), to which an additional term for steel s has to be added AA = 0,Ol O W/(m-K) (see clause 7 ), giving ; Ie f f =

0,120 W/(m.K)

Using again eq.1 for the calculation of the insulation thickness of the Sndlayer, which this time reads:

The necessary insulation d2 is given by a simple mathematical transformation:

(

d2 = 0,120.

600 - 20 = 0,130 m 500 12,31

L)

Control of density of heat flow rate and temperature distribution using eq. 24 in 4.1.3:

'

850 - 20 = 499 W/m2 = 0.10 0.130 1 --+-+0,20 0,120 12,31

The calculation of the temperature distribution according to eq. 30 leads to a t temperature at the boundary layer e,= 600,5 "C and an external surface temperature e, = 59,9 OC, which is in a good agreement with the above assumption. B.2

Heat flow rate and surface temperature of an insulated pipe

For an insulated hot air supply pipe with a dusty sheet iron lining, the density of heat flow rate and the external surface temperature are calculated. Boundary conditions: Medium temperature (air): Ambient air temperature: Pipe diameter: Insulation thickness: Practical thermal conductivity of the insulation at 8=165 OC: Radiation coefficient of the sheet iron lining of the insulation: Outer pipe diameter De= D,+ 2 d

Q =3OO"C 8, = 20 "C 4 = 0,324 m d = 0,200 m

A

=0,062 W/(m.K)

C, = 2,49 W/(m2.K) De = 0,724 m

The given thermal conductivity A of the insulation material has to be increased according to clause 7 by AA = 0,Ol W/(m-K)

&, = 0,062 + 0,Ol = 0,072 W/(m.K)


I

STD.ISO 1 2 2 9 1 - E N G L 1778

= 9851703

~

h,=

5,30

+ 0,05-(30 - 20) = 5,8

hse

_

_

@

IS0

0737q27 2bb

I S 0 12241:1998(E)

Using eq. 19 from 4.1.2.3, the external surface coefficient surface temperature of ese= 30 OC.

~~

can be calculated with an estimated

W/(m2.K)

The inner surface resistance is neglected. The linear density of heat flow rate is then calculated according to eq. 24 , inserting eqs. 5 and 26.

gl =

n . (300 - 20) = 151,l W/m 1 0,724 1 .In--0,324 + 5,8 .0,724 2 '0,072

The external surface temperature is then found from eq. 30

e,,

0,24 53

= 20+-.(300-20)

= 31,6oc

which isiin ,good agreement with the above assumption of 30 OC, B.3 Tasiaperature drop in a pipe Calculafion of the longitudinal temperatures drop of a hot steam pipe. Boundary conditions: Medium temperature (hot steam): Medium mass flow: Specific heat capacity: Ambient air temperature: Pipe diameter: Pipe lemgth: Insulation thickness: Thermdl conductivity of the insulation (design value) at $v=120 OC: Outer pipe diameter De=D,+2d:

S,,= m = Ç, = 4 =

4

=

I d

=

A

=

=

0, =

250°C 45000 kg/h 2,233 kJ/(kg*K) -10°C 0,40m 2500m 0,12 m 0,061 W/(m.K) 0,64 m

The inmr and outer surface resistances are neglected in this example. This gives a linear density of heat flow rate by using eq. 5 and 6 of qi = 212 W/m which gives according to eq. 33 a longitudinal temperature drop of approximately: AO =

212 . 2 500.3,6 = 9,0 oc 45 000.2,233

The accurate temperature drop is calculated using eq. 31 and 32: O,,

+

+

= -1 O 1250 101x exp - (2,9 x 1

Therefore the accurate temperature drop is 250

-


x 2 500) = 2318 OC

231,8 = 18,2 OC

~

S T D - I S 0 L22LiL-ENGL 1998 0

=

Li851903 O737Li30 T88 I S 0 12241 :1998(E)

IS0

8.4

Temperature drop in a container

Calculation of the temperature drop of a spherical hot water supply container in 15 hours. Boundary conditions:

= =

Medium temperature (hot water): Specific heat capacity: Ambient air temperature: Sphere diameter: Corresponding mass of water: Insulation thickness: Thermal conductivity of the insulation (design value) at O,, = 30 OC: Outer sphere diameter 0, = Di+2d:

%Il

e,

=

Di = m = d =

A = De =

80 "C 4,18 kJ/(kg*K) -15 "C 2,50 m 818 1 kg 0,15 m 0,05 W/(m.K) 2,8 m

The inner and outer surface resistances are neglected in this example. This gives a heat flow rate by using eqs. 9 and 10 of: (80 + 15)

SPsph

= 696 W

@sph =

which gives according to eq. 36 a temperature drop of approximately:

A8 =

696 . 15.3,6 = /I OC 818 1.4,18

The accurate temperature drop is calculated using eqs. 34 and 36 with Of, and being the temperatures at the start and the end of the cooling period. Of,,, =-15+(80+15).exp-

( ~

: :8

Therefore the accurate temperature drop is 80

e,,,in eqs. 34 and 36

15)

-

78,9 = 1, I "C


S T D - I S 0 122LiL-ENGL 2 9 9 8 W Li851703 0739Y3L 914 IS0 12241:1998(E)

8.5

0

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Cooling and freezing times in a pipe

Determination of the cooling time down to zero "C and the time for the partial freezing of the water (25%

of the volume). Boundary conditions: Interior pipe diameter: Interior insulation diameter: Water temperature at the start of cooling : Ambient temperature: Insulation thickness: Thermal conductivity: Heat capacity of water: Latent heat of freezing: Specific heat capacity of water: Density of ice:

= 0,090m

= 0,1079m =

+1ok

=

-1Ok

= 0,100m = 0,04 W/(m-K) = 26,7 kJ/K = 334 kJ/kg = 4,2kJ/(kg*K) = 920 kg/m3

The heat Tiow is calculated, neglecting the surface coefficient h,,, using eq. 38 in 6.1 : x . [lo-(-1 O)] -

qwp-

1 2 x 0,04

0,3079 0,1079

= 4,79 W/m

.In----

The corresponding cooling time down to the freezing point, neglecting the heat capacity of the pipe is given by eq. 37: 20 20.26,7 . In twp =

l o =21,5h

4,79.3,6.1

or, using eq 40 -

twp

26,7.1O

- 4,79.3,6.1

= 155 h

The heat flow rate and the freezing time of 25% of the volume of the pipe is given by eqs. 41 and 42 : 7t.10

4fr = -. 1 In 0,307 9 0,08 0,107 9

= 2,40 W/m

'I'

and

25 920.n . (0,09)* .334 = 56,6 h 2,4 .3,6 .4

tf, = -. 1O0


S T D - I S 0 L 2 2 4 L - E N G L 1998 W 4853903 0739Li32 8 5 0 0

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I S 0 12241 :1998(E)

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6.6 Underground pipeline

Determination of the heat flow rate of an insulated underground metal pipeline, protected by an outer polyethylene pipe: Boundary conditions: External diameter of metal pipe: Depth below surface: Insulation thickness: Thermal conductivity at 55 O C : PE pipe internal diameter: Thickness: Soil: Temperature: Thermal conductivity: Medium temperature:

De HE d

= = =

k

=

Q

= =

dPE

0,219 1 m 1,0 m 0,061 m 0,028 W/( m-K) 0,341 m 0,007 m 3 "c 1,75 W/(m.K) 100 "c

According to eqs. 45, 47 and 48 the heat flow rate is:

TC.(^ O0 - 3) 4-1 In2.1,75 0,355

qi= ___. 1

+

1 O,341 .In2.0,028 0,2191

= 35,5 W/m

The temperature difference between the PE pipe surface and the surrounding soil is calculated according to 4.1.4: AO = -.35,5 _ _1_ . In- 4 . 1 -7,8K 7~ 2.1,75 0,355

Thus the PE pipe surface esetemperature is:

e,,

=3,0+7,8=10,8oc


B.7

Required insulation thickness to prevent surface condensation

Determination of the necessary insulation thickness to prevent dew formation on the surface of the thermal insulation of a refrigerant pipe. Boundary conditions: = e, = Q = ( P =

Medium temperature: Ambient air temperature: Pipe diameter without insulation: Relative humidity of the ambient air:

-20 "C 20 "C 0,273 m 90 %

@ln

Insulation material with galvanized sheet metal cladding Table 3, clause 9 gives an allowed maximum temperature difference of 1,7 K, hence

e,

= i8,3

o c

The thmmal conductivity of the insulation at

e,,=

(-20

+ 18,3)/2 = -0,85

OC:

;t = 0,039 W/(m.K)

The external surface coefficient can be estimated from eq. 19: h,, = 5,3 + 0,05-1,7 = 5 3 9 W/(m'-K) which leads to the parameter C from diagram 1, eq. (b)

C' = 2.0,039 (-20 5,39

'[

- 20)

18,3-20

1'

- 1 = 0,326

The necessary insulation thickness may be found directly from diagram 1 The value for the thickness seems to be slightly higher than 120 mm. The following value is chosen: d = 125 mm Control calculation: Equation 6 gives the linear thermal resistance of the insulation, Ri = 2,65 (m.K)NV. The linear thermal surface resistance, using h,, = 5,39 W/(m2.K), is calculated by equation 22, RI, = 0,113 (m-K)NV. Then the surface temperature can be calculated using equation 30:

,e,

= 20

0,113 + 0,113 + Z,65

*

(-20

- 20) = 18,37

C

which is higher than the allowed minimum temperature of the pipe surface and hence prevents dew formation.


STD-IS0 1 2 2 r l l - E N G L 1998 0

‘4851903 0739434 b 2 3 I S 0 12241:1998(E)

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Annex C (inf or mative) Bibliography [l] I S 0 8301 :1991, Thermal insulation - Determination of stea-dy state thermal resistance and related properties - Heat flow meter apparatus. [2] IS0 8302:1991, Thermal insulation - Determination of steady state thermal resistance and related properties - Guarded hot plate apparatus. [3] I S 0 8497:1995, Thermal insulation - Determination of sfeady state thermal transmission properties of thermal insulation for circular pipes. [4] I S 0 9229:1991, Thermal insulation - Thermal insulation materials and products - Vocabulary. [5] I S 0 9251:1987, Thermal insulation - Heat transfer conditions and properties of materials Vocabulary.

[6] I S 0 9288:i 989, Thermal insulation - Heat transfer by radiation - Physical quantities and definitions. 171 I S 0 13787:-”, Thermal insulation of building equipment and industrial installations - Procedure for the determination of declared and design thermal conductivity. [8] ASTM C680-89, Practice for determination of heat gain or loss and surface temperatures of insulated pipe and equipment systems by the use of a computer program. [9] Standard international units of the international union for pure and applied physics (IUPAP); Document SUN 75-5. [l O] VDI 2055:1994, Thermal insulation for Heated and Refrigerated Industrial and Domestic Installations. [l 11 BS 5422:1990, Method for specifying thermal insulating materials in pipes, ductwork and equipment (in the temperature range -40°C to +7OO0C). [12] Vidal, J.: Determination of heat losses in underground conduits. Brussels: Editions SIC, 1961. [13] Brauer,H.: Calculation of heat losses from underground pipes. €nergie,15 (1963) No. 9, pp. 354-365. 1141 Zeitler, M.: Calculation method for determining heat loss from various underground pipe systems. fernwarme International (1 980) N0.3,pp. 170-179.

1) To be published.


IS0 12241 :1998(E)

ICs 91 .I 20.1O; 91.140.01 Descriptors: thermal insulation, buildings, pipelines, fluid pipelines, pipes (tubes), equipment, heat transfer, heat losses. rules of calculation. Price based on 39 pages


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