2012 Lecture BKC Insulations & Refractory

8/4/2019 2012 Lecture BKC Insulations & Refractory

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RAMAKRISHNA

INSULATIONS &

REFRACTORIESPurpose and Applications

Types and Some TerminologiesPerformance Evaluation

Efficient System OperationAssessment of Systems

Energy Saving OpportunitiesCases and Numerical Problems

Use of SoftwareReferences


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-: INSULATIONS :-

PURPOSE AND APPLICATIONS A thermal insulator is a poor conductor of heat and has a

low thermal conductivity. Insulation is used in buildingsand in manufacturing processes to prevent heat loss or heat

gain. The multi-purpose benefits include: Reduces over-all energy consumption

Offers better process control by maintaining processtemperature

Prevents corrosion by keeping the exposed surfaceof a refrigerated system above dew point

Provides fire protection to equipment

Absorbs vibration


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TYPES .

Classified into three groups according to the temperature rangesfor which they are used:

Low Temperature Insulations (up to 90oC); Eg. Cork, Wood,85% magnesia, Mineral Fibers, Polyurethane and

expanded Polystyrene (i.e. thermocol), etc. applied inrefrigerators, cold and hot water systems, storage tanks,etc.

Medium Temperature Insulations (90oC to 325oC) ; Eg. 85%Magnesia, Asbestos, Calcium Silicate and Mineral Fibers

etc. applied in low temperature, heating and steam raisingequipment, steam lines, flue ducts etc.

High Temperature Insulations (325oC and above) ; Eg.Asbestos, Calcium Silicate, Mineral Fibre, Mica andVermiculite based insulation, Fireclay or Silica basedinsulation and Ceramic Fibre applied in super heated


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TYPES

Insulation materials can also be classified intotwo:

Organic insulations are based onhydrocarbon polymers, which can beexpanded to obtain high void structures. Eg:Thermocol (Expanded Polystyrene) and Poly

Urethane Form(PUF).

Inorganic insulations are based onSiliceous/Aluminous/Calcium materials in

fibrous, granular or powder forms. Eg. Mineralwool Calcium silicate etc.


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AND SOME

TERMINOLOGIES Thermal conductivity of a material is the heat loss per

unit area per unit insulation thickness per unittemperature difference. The unit of measurement is

W-m2/mC or W-m/C. The thermal conductivity ofmaterials increases with temperature. So thermalconductivity is always specified at the meantemperature (mean of hot and cold surfacetemperatures) of the insulation material.

Critical Insulation Thickness: Insulation thicknesscorresponding maximum heat transfermostly consideredin electrical wiring.

Economic Thickness of Insulation:Insulation thicknesscorresponding to the lowest total cost of insulation

plus cost of fuel as a result of heat loss through


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SOME USAGE

Calcium Silicate: Used in industrial process plant

piping where high service temperature and compressive

strength are needed. Temperature ranges varies from 40C to 950 C.

Glass mineral wool: These are available in flexible

forms, rigid slabs and preformed pipe work sections.

Good for thermal and acoustic insulation for heating

and chilling system pipelines. Temperature range of

application is10 to 500 C

Thermocol: These are mainly used as cold insulation

for piping and cold storage construction.


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SOME USAGE

Expanded nitrile rubber: This is a flexible material that formsa closed cell integral vapour barrier. Originally developed forcondensation control in refrigeration pipe work and chilled

water lines; now-a-days also used for ducting insulation for airconditioning. Rock mineral wool: This is available in a range of forms from

light weight rolled products to heavy rigid slabs includingpreformed pipe sections. In addition to good thermal insulationproperties, it can also provide acoustic insulation and is fire

retardant. Use of Moulded Insulation:Lagging materials can be obtained

in bulk, in the form of moulded sections; semi - cylindrical forpipes, slabs for vessels, flanges, valves etc. The main advantageof the moulded sections is the ease of application and

replacement when undertaking repairs for damaged lagging.


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temperature

NUMERICAL EXAMPLE TO


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NUMERICAL EXAMPLE TOEXPLAIN ECONOMIC

THICKNESS OF INSULATIONConsider an 8 bar steam pipeline of 6 dia

having 50-meter length. We will evaluate thecost of energy losses when we use 1, 2and 3 insulation to find out the mosteconomic thickness. A step-by-stepprocedure is given below.

1. Establish the bare pipe surface temperature,by measurement.

2. Note the dimensions such as diameter,length & surface area of the pipe sectionunder consideration.


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NUMERICAL EXAMPLE

3. Assume an average ambient temperature. Here,we have taken 30oC.

4. Since we are doing the calculations for

commercially available insulation thickness,some trial and error calculations will be requiredfor deciding the surface temperature after puttinginsulation. To begin with assume a value between55 & 65 C, which is a safe, touch temperature.

5. Select an insulation material, with known thermalconductivity values in the mean insulationtemperature range. Here the mean temperature is111 C. and the value of k = 0.044 W/m2oC for

mineral wool.


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NUMERICAL EXAMPLE

6. Calculate surface heat transfer coefficients of bare andinsulated surfaces, using equations discussedpreviously. Calculate the thermal resistance andthickness of insulation.

7. Select r2 such that the equivalent thickness of insulationof pipe equals to the insulation thickness estimated instep 6. From this value, calculate the radial thickness ofpipe insulation = r2-r1.

8. Adjust the desired surface temperature values so that thethickness of insulation is close to the standard value of1 ( 25.4 mm).

9. Estimate the surface area of the pipe with differentinsulation thickness and calculate the total heat loss from

the surfaces using heat transfer coefficient, temperaturedifference between pipe surface and ambient.


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NUMERICAL EXAMPLE

10. Estimate the cost of energy losses in the 3scenarios. Calculate the Net Present Value ofthe future energy costs during an insulation life

of typically 5 years.

11. Find out the total cost of putting insulation onthe pipe ( material + labor cost)

12. Calculate the total cost of energy costs and

insulation for 3 situations.

13. Insulation thickness corresponding to thelowest total cost will be the economic thicknessof insulation


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REFRACTORIES

The general requirements of a refractory material can be

summed up as :

Ability to withstand high temperatures.

Ability to withstand sudden changes of temperatures.

Ability to withstand action of molten metal slag, glass,

hot gases, etc.

Ability to withstand load at service conditions.

Ability to withstand load and abrasive forces.

Low coefficient of thermal expansion.

Should be able to conserve heat.

Should not contaminate the material with which it

comes into contact.


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SOME TERMINOLOGIES

Melting Point, Size, Bulk Density, Porosity, ColdCrushing Strength, Creep, Volume stability,expansion, and shrinkage at high temperatures,Reversible thermal expansion, Reversible thermalexpansion, Thermal Conductivity, etc.

Fireclay Refractories, High alumina refractories,Silica brick, Magnesite, Chromite refractories,

Zirconia refractories, Oxide refractories(Alumina), Molybdenum based refractories,Monolithics, Ceramic Fibres, High EmissivityCoatings, Castables and Concretes


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AND CLASSIFICATION


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Questions and Answers

In determining the optimal economic insulation thickness for a

steam pipeline, thickness which of the following factorsneed not be considered a) annual hours of operation b)

calorific value c) pipe material d) cost of fuel.

Magnesite, chrome-magnesite, dolomite are examples of ------

--- type of refractory a) acid b) basic c) neutral d)none of the above

Which is not a property of Ceramic fibre insulation

a) low thermal conductivity b) light weight c) high heat

storage d) thermal shock resistantThe insulating material made by blending and melting of

alumina and silica at a temperature of 1800-2000oC is

known as a) insulating brick b) high alumina brick c) fire

brick d) ceramic fibre


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A steam pipeline of 100mm diameter is not insulated for 100 metre

length, supplying steam at 10 kg/cm 2.Find out the fuel savings if it is

properly insulated with 50mm insulating material.Assume 8000 hours

of operation per year. Given:

Boiler efficiency :80%

Cost of fuel oil :Rs20,000/tonne.

Gross Calorific value of fuel :10300k.cal/kg

Surface temperature without insulation :1700C

Surface temperature after insulation :500C

Ambient temperature :300C

Solution: Existing Heat Loss:

Surface heat loss S= 10+(TsTa)/20 x(Ts-Ta)

Where Ts =Hot surface temperature0

CTa =Ambient temperature0C

S =Surface heat loss in k.Cal/hrm2

Substituting values S= 10+(17030)/20 x(170-30)

=2380 k.Cal/ hr-m2


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Modified System :

After insulating with 50mm insulating material the surfacetemperature has reduced to 500C

Substituting values S= 10+(50

30)/20 x(50-30)=220 k.Cal/ hrm2

Calculation of Fuel savings:

Pipe dimension =100 metre length and 100mm diameter

Surface area existing =3.14x 0.1x 100 =31.4 m2

Surface area after insulation =3.14x0.2x100 =62.8 m2

Total heat loss in existing system=2380x31.4 =74732 k.Cal/hr

Total heat loss in modified system=220x62.8 =13816 k.Cal/hr

Reduction in heat loss =74732-13816 =60916 k.Cal/hr

No of operating hours per annum =8000

Savings in heat per year =60916x8000 =487328000 k.Cal/year

Annual savings in fuel oil=487328000/(10300x0.8)x1000

= 59.142 Tons

Monetary savings per annum =59.142x20000

=Rs11.83 lakhs


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REFERENCES

Book2, Study Materials for CEM and

CEA Examinations, BEE, New Delhi

www.energymanagertraining.com

Website of Wikipedia, the free

encyclopedia


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ACKNOWLEDGEMENTS:

REFERRED BOOKS,WEBSITES & IISWBM

Questions?

THANK YOU

2012 Lecture BKC Insulations & Refractory

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