Explosion protection

Explosion protection is used to protect all sorts of equipment or product against internal and external explosions & weather protection. 

Basic methods of protection, it's used mainly in Hazardous Area

1. Explosion Containment 

2.Segregation and 

3. Prevention

• Explosion containment: The only method that allows the explosion to occur but confines it to a well-defined area, thus avoiding the propagation to the surrounding atmosphere. Explosion-proof enclosures are based on this method.

• Segregation: A method that attempts to physically separate or isolate the electrical parts or hot surfaces from the explosive mixture. This method includes various techniques, such as pressurization, encapsulation, etc.

• Prevention: A method that limits the energy, both electrical and thermal, to safe levels under both normal operation and fault conditions. Intrinsic safety is the most representative technique of this method.

Most Used Protection Type

Flame proof enclosure Ex d

This is strong enough to withstand internal explosion. This design allows internal ignition sources, like sparks and (limited) hot spots.

Pressurized enclosures Ex p

This is keep dangerous atmosphere outside by overpressure.

Powder (quartz) filling Ex q

This is extinguish any ignition source in a quartz filled enclosure.

Increased safety Ex e

This is increased safety level for normally non sparking apparatus. This design does not allow any (unprotected) sparking components inside.

Intrinsic Safety Ex i

This is limitation of energy in the whole electrical circuit.

Non Incendive Ex n

Ex n consists of several sub types of protection.

Moulded encapsulation Ex m

This is moulding of the electrical circuit.

Optical radiation Ex op

This is safety by either limitation of radiation or protection of optical light.

Protection by enclosure Ex t (for dust only)

This is minimum IP degree for dust tight enclosure  (minimum IP rating of either IP5X or IP6X).

Equipment Protection Level

Where 

G for Gas

a for Very High Safety Level

b for High Safety Level

c for Normal Safety Level

D for Dust.

How to read the markings?

Ex db eb IIC T6 Gb

Ex : Explosion protected. 

db : Flameproof ’d’ with level of protection ’b’. 

eb : Increased safe ’e’ with level of protection ’b’. 

IIC : Group IIC 

T6 : Max. surface temperature classified T6 (85°C). 

Gb : This whole assembly has EPL Gb (zone 1).

Insulation Material Calculation for Pipe

 Insulation Material Calculation

Problem : How to Calculate Insulation Material & Cladding quantity for Pipe?

Solution: 

We have required following details:

1. Pipe Size

2. Pipe Length

3. Fittings quantities (like Valve, Flange, Elbow, Tee, Reducer, Cap etc.)

4. Insulation Thickness

Now we put these value in below formula:

RMT = Pipe Length + (Fittings quantities X Conventional Equivalent Lengths of fittings as per pipe size)

Note: Conventional Equivalent Lengths of fittings have selected from below Table.

Table

 Now you found insulation running meter (RMT) from above formula.

Convert RMT to Area from below formula:

Area = Pie X (Pipe outer diameter + 2 X Insulation thickness) X RMT

Note: All above value put in same measurement unit like (mm, cm, m).

Now we found cladding quantity:-

formula:

Cladding Quantity = Area X Cladding material weight X Safety Factor.

Note: Cladding material weight is same square measurement unit of area. for example you consider insulation area in m2 then cladding weight must in m2/kg.

Now we understand above theory in numerical concept.

Question: A 8" (OD-219.1mm) pipe having length is 100m including 2valves, 1flange, 2 elbow 45° & 1 elbow 90° & insulation thickness is 50mm. Cladding material is 24G plain aluminum sheet. Calculate area of insulation & quantity of cladding?

Answer: 

We Know about RMT formula, we put all values in Formula:

RMT = Pipe Length + (Fittings quantities X Conventional Equivalent Lengths of fittings as per pipe size)

RMT = 100 + (2X2.4 + 1X1.5 + 2X0.85 +1X1.4) 

(Pink color value consider from table)

= 100 + 9.4

RMT = 109.4 M

Now we calculate Area of insulation.

Area = Pie X (Pipe outer diameter + 2 X Insulation thickness) X RMT

Area = 3.142 X ((219.1 + 2 X 50)/1000) X 109.4

(we convert mm to m, hence we divide from 1000)

= 3.142 X 0.3191 X 109.4

=109.68 M2

Now we calculate Cladding Quantity.

Cladding Quantity = Area X Cladding material weight X Safety Factor.

Cladding Quantity = 109.68 X 1.52 X 1.1

(We consider 10% safety factor & weight is vary as per Cladding materials & gauge) 

Cladding Quantity =183.38kg.

Finally we get result that total insulation area is 109.68 M2 & Cladding Quantity is 183.38kg.

Reference for download IS 14164 Standard 

http://www.shristyenterprises.com/images/Standard/Standard%20-%20In%20Profile%20Page.pdf

Insulation Thickness Calculation for Pipe

 Sample Problem Statement

Calculate insulation thickness (minimum value) required for a pipe carrying temperature at 2000C. The pipe size is 10" and the maximum allowable temperature of outer wall of insulation is 600C. Thermal conductivity of the insulation material for the temperature range of the pipe can be taken as 0.046 W/m·C. The heat loss from steam per meter of pipe length has to be limited to 100 W/m.

Solution:-

For radial heat transfer by conduction across a cylindrical wall, the heat transfer rate is expressed by following equation,

For the given sample problem,

T1 = 600C
T2 = 2000C
r1 = 10" = 10 × 0.0254 m = 0.254 m
k = 0.046 W/m·C
N = length of the cylinder

Q = Heat loss

Q/N = Heat loss per unit length of pipe

Q/N = 100 W/m

Hence, inserting the given numbers in the radial heat transfer rate equation from above,

100 = 2π × 0.046 × (200-60) ÷ ln(r2/0.254)

ln(r2/0.254) = 2π × 0.046 × (200-60) / 100 = 0.4046

Hence, r2= r1 × e0.4046
r2= 0.254 × 1.4987 = 0.3806 m

Hence, insulation thickness = r2 - r1
thickness = 0.3806 - 0.254 = 0.1266 m or 126.6mm

Some margin should be taken on the insulation thickness because if the conductive heat transfer rate happens to be higher than the convective heat transfer rate outside the insulation wall, the outer insulation wall temperature will shoot up to higher values than 600C. Hence conductive heat transfer rate should be limited to lower values than estimates used in this sample problem. The purpose of this sample problem is to demonstrate radial heat conduction calculations and practical calculations of insulation thickness also require consideration of convective heat transfer on the outside of insulation wall.