Heat-Up Considerations

As per IEEE 515

Relationship between heat-up time and the heating device input for a pipe:-

t = H X Ln {(Qc – U X (Ti – Ta)) / (Qc – U X (Tf– Ta))} + {(P1 X Vc1 X hf) / (Qc – U X (Tsc– Ta))}

Where: -

t = Desired Heat-Up Time (s)

H = Thermal Time Constant (s)

Qc = Heating Cable Output (W/M)

U = Heat Loss per Unit Length of Pipe per Degree of Temperature (W/M·°C)

Ti = initial temperature of the pipe (°C)

Tf = final temperature of the fluid and the pipe (°C)

Ta = ambient temperature (°C)

P1 = density of product in the pipe (kg/m3)

Vc1 = internal volume of the pipe (m3/m)

hf = Latent Heat Of Fusion For The Product (J/Kg)

Tsc = Temperature at Which Phase Change Occurs (°C)

 

We calculate heat loss: -

 

U = 1 / ((1 / (Pie X D1 X hi)) + (Ln (D2 / D1) / (2 X Pie X k1)) + (Ln (D3 / D2) / (2 X Pie X k2)) + (1 / (Pie X D3 X hco)) + (1 / (Pie X D3 X ho)))

Where: -

U = Heat Loss per Unit Length of Pipe per Degree of Temperature (W/M·°C)

D1= Inside Diameter of the Insulation Layer (M)

D2 = Outside Diameter of the Inner Insulation Layer (M)

D3 = Outside Diameter of the Outer Insulation Layer (M)

k1 = Thermal Conductivity of the Inner Insulation Evaluated At Its Mean Temperature (W/M·°C)

k2 = Thermal Conductivity of the Outer Insulation Evaluated At Its Mean Temperature (W/M·°C)

hco = Inside Air Contact Coefficient of the Weather Barrier (W/M·°C)

hi = Inside Air Contact Coefficient From The Pipe To The Inside Insulation Surface (W/M·°C)

ho = Outside Air Contact Coefficient of the Weather Barrier To The Ambient Temperature (W/M2·°C)

 

We calculate Thermal Time Constant: -

H = ((P1 X Vc1 X Cp1) + (P2 X Vc2 X Cp2) + (0.5 x P3 X Vc3 X Cp3)) / U

Where:-

H = Thermal Time Constant (s)

P1 = density of product in the pipe (kg/m3)

Cp1 = specific heat of the product (J/kg·°C)

Vc1 = internal volume of the pipe (m3/m)

P2 = density of the pipe (kg/m3)

Cp2 = specific heat of the pipe (J/kg·°C)

Vc2 = pipe wall volume (m3//m)

P3 = density of the insulation (kg/m3)

Cp3 = specific heat of the insulation (J/kg·°C)

Vc3 = insulation wall volume (m3/m)

U = Heat Loss per Unit Length of Pipe per Degree of Temperature (W/M·°C)

 

If we consider some assumption:-

1.    Single layer insulation.

2.    (1 / (Pie X D2 X hco)) value is negligible due to mastic weather barrier &

3.    (1 / (Pie X D2 X ho))) Term may be omitted to give a conservative (high)       heat loss.

4.    (Pie X D1 X hi) Term is also negligible.

 

After that we find out result:-

U = 1 / (Ln (D2 / D1) / (2 X Pie X k))

Where: -

U = Heat Loss per Unit Length of Pipe per Degree of Temperature (W/M·°C)

D1= Inside Diameter of the Insulation Layer (M)

D2 = Outside Diameter of the Insulation Layer (M)

k= Thermal Conductivity of the Insulation Evaluated at Its Mean Temp. (W/M·°C)

 

Heat loss Calculation on Vessel

   Heat loss Calculation on Vessel

As per IEEE 515

Qins = ((Tp – Ta) / ((1 / hi) + (x / k) + (1 / hco) + (1 / ho))) X A

Where:-

Qins = Heat Loss for Region (W)

A = Vessel Surface Area of Insulated Region (m2)

Tp = Maintain Temperature (°C)

Ta = Minimum Ambient Temperature (°C)

x = Thermal Insulation Thickness (m)

k = Thermal Insulation Conductivity at Mean Temperature (W/m·°C)

hi = Inside Air Contact Coefficient from the Vessel to the inside Insulation Surface (W/m2·°C)

hco =  Inside Air Contact Coefficient From The Insulation Outer Surface To The Weather Barrier (W/m2·°C)

ho = Outside Air Film Coefficient From The Weather Barrier or Insulation Outer Surface To The Ambient Temperature (W/m2·°C)

 

We consider some assumption here: -

    (1 / hco) & (1 / hi) value is negligible due to mastic weather barrier.

   (1 / ho) Term may be omitted to give a conservative (high) heat loss.

Therefor

Qins = ((Tp – Ta) / (x / k)) X A

Qins = ((Tp – Ta) X k X A) / x

Now we consider safety factor, then:-

Qins = (((Tp – Ta) X k X A) / x) X SF

 

Where-

Qins = Heat Loss for Region (W)

A = Vessel Surface Area of Insulated Region (m2)

Tp = Maintain Temperature (°C)

Ta = Minimum Ambient Temperature (°C)

x = Thermal Insulation Thickness (m)

k = Thermal Insulation Conductivity at Mean Temperature (W/m·°C)

SF = Safety Factor

Unit of Qins:-

Qins = (°C X W/m·°C X m2) / m

Qins = W ( Watts)

 

Calculation of Vessel Area: -

A = (Pie X D X L) + ((Pie/2) X D X D)

Where-

A = Vessel Surface Area of Insulated Region (m2)

D = Diameter of Vessel (m)

L = Length or Height of Vessel (m)

 

Unit of A: -

A = m X m + m X m

A = m2

 

 

 

 

Example

Heat Loss for Region =?

Diameter of Vessel = 2.2m

Height of Vessel = 3.1m

Maintain Temperature = 55°C

Minimum Ambient Temperature = -2°C

Thermal Insulation Thickness = 50mm

Thermal Insulation Conductivity at Mean Temperature = 0.038W/m·°C

Safety Factor = 20%

 

Solution: -

Firstly we calculate Area of vessel: -

A = (3.14 X 2.2 X 3.1) + (3.14 X 2.2 X 2.2) / 2

A = 21.4148 + 7.5988

A = 29.01 m2

Now we convert insulation thickness mm to m

X = 50mm

X = 50/1000

X = 0.05m

Now we calculate heat loss of vessel:-

Qins = ((55 – (-2)) X 0.038 X 29.01 X (1+20 / 100)) / 0.05

Qins = (57 X 0.038 X 29.01 X 1.2) / 0.05

Qins = (57 X 0.038 X 29.01 X 1.2) / 0.05

Qins = 75.40 / 0.05

Qins = 1508.05 W

Heat loss Calculation on Pipe

Heat loss Calculation of Pipe Line

 

As per IEEE 515

It’s for Double Insulation on the pipe:-

Q = (Tp-Ta) / ((1 / (Pie X D1 X hi)) + (Ln (D2 / D1) / (2 X Pie X k1)) + (Ln (D3 / D2) / (2 X Pie X k2)) + (1 / (Pie X D3 X hco)) + (1 / (Pie X D3 X ho)))

Where

Q = Heat loss per unit length of pipe (W/m)

Tp = Desired maintenance temperature (°C)

Ta = Minimum design ambient temperature (°C)

D1 = Inside diameter of the inner insulation layer (m)

D2 = Outside diameter of the inner insulation layer (m) (inside diameter of the outer insulation layer when present)

D3 = Outside diameter of the outer insulation layer when present (m)

K1 = Thermal conductivity of the inner layer of insulation evaluated at its mean temperature (W/ m·°C)

K2 = Thermal conductivity of the outer layer of insulation, when present, evaluated at its mean temperature (W/m·°C)

Hi = Inside air contact coefficient from the pipe to the inner insulation surface when present, 

hco = Inside air contact coefficient from the outer insulation surface to the weather barrier when present (W/m2·°C)

ho = Outside air film coefficient from the weather barrier to the ambient air (W/m·°C). Typical values for this term range from 3 W/m2 °C to 284 W/m2 °C for low (below 50 °C) temperature applications.

It’s for single Insulation on the pipe:-

Q = (Tp-Ta) / (((Ln (D2 / D1) / (2 X Pie X k)) + (1 / (Pie X D2 X hco)) + (1 / (Pie X D2 X ho)))

(1 / (Pie X D2 X hco)) value is negligible due to mastic weather barrier &

(1 / (Pie X D2 X ho))) Term may be omitted to give a conservative (high) heat loss.

Final Formula is: -

Q = (Tp-Ta) / (((Ln (D2 / D1) / (2 X Pie X k))

Q = (2 X Pie X k) X (Tp-Ta) / (Ln (D2 / D1)

When we used safety factor then we use this formula: -

 Q = ((2 X Pie X k) X (Tp-Ta) / (Ln (D2 / D1)) X SF

Where

Q = Heat loss per unit length of pipe (W/m)

Tp = Desired maintenance temperature (°C)

Ta = Minimum design ambient temperature (°C)

D1 = Inside diameter of the inner insulation layer (m)

D2 = Outside diameter of the inner insulation layer (m)

D2 = D1 + 2t

t = Outer insulation layer on the pipe (m).

K = Thermal conductivity of insulation (W/ m·°C)

SF = Safety factor for heating system.

 

Example

Q =?

Tp = 85°C

Ta = 13.9°C

D1 = 0.0603m

D2 = 0.1603m

K = 0.0385W/ m·°C

SF = 10%.

Solution: -

Q = ((2 X Pie X k) X (Tp-Ta) / (Ln (D2 / D1)) X SF

Putting the value of each term:-

Q = ((2 X 3.14 X 0.0385 X (85-13.9) / Ln (0.1603/0.0603)) X 10%)

Q = (17.190558 / 0.978) X 10%

Q = 17.58 X 10%

Q = 19.33W/m

Companies for Electric Heat Tracing System

Companies for Electric Heat Tracing System

1.    Company Name:- Thermopads (India)

Headquarters: - Hyderabad, Telangana

Office in India: - Punjagutta, Hyderabad, Telangana

 

2.    Company Name:- Pentair (UK)

Headquarters: - London, UK

Office in India: - Verna Industrial Estate Verna, Goa 403722, India

 

3.    Company Name:- Heat Trace (UK)

Headquarters: - Frodsham, UK

Office in India: - Sector 63, Noida, UP

 

4.    Company Name:- Warmup (UK)

Headquarters: - London, UK

Office in India: - 

 

5.    Company Name:- Thermon (US)

Headquarters: - Austin, US

Office in India: - Mumbai, Maharashtra 

 

6.    Company Name:- Chromalox (US)

Headquarters: - Pittsburgh, Pennsylvania US

Office in India: - Janakpuri, New Delhi

 

7.    Company Name:- Emerson (US)

Headquarters: - Ferguson, Missouri, U.S.

Office in India: - Mumbai Maharashtra, Gurgaon Haryana, Mohali (SAS Nagar) Punjab & Pune 411 057, India

 

8.    Company Name:- BriskHeat (US)

Headquarters: - Columbus, US

Office in India: - 

 

9.    Company Name:- Parker-Hannifin (US)

Headquarters: - US

Office in India: - Navi Mumbai, India

 

10. Company Name:- Neptech (US)

Headquarters: - Highland, MI 48357, United States

Office in India: - Jafferkhanpet, Chennai, Tamil Nadu

 

11. Company Name:- QMAX (US)

Headquarters: - US

Office in India: - 

 

12. Company Name:- BARTEC (Germany)

Headquarters: - Bad Mergentheim, Germany

Office in India: - SECTOR-62, Noida, UP

 

13. Company Name:- Eltherm (Germany)

Headquarters: - Burbach . Deutschland

Office in India: - 

 

14. Company Name:- Danfoss (Denmark)

Headquarters: - Nordborg, Denmark

Office in India: - Tamil Nadu, India

 

15. Company Name:- Drexan (Canada)

Headquarters: - Western Canada

Office in India: - 

 

16. Company Name:- Technitrace (France)

Headquarters: - Toucy - France

Office in India: - 

 

17. Company Name:- Raychem (US)

Headquarters: - Menlo Park, California, United States

Office in India: - Mumbai, Maharashtra

 

18. Company Name: - Xicon International Ltd.

Headquarters: - Mumbai - 400072, India

Office in India: - Mumbai, Maharashtra

 

19. Company Name: - Vajraa Technical Services

Headquarters: - Pune - 411026, India

Office in India: - Pune, Maharashtra

 

20. Company Name: - Shaila Thermo Tek

Headquarters: - Secunderabad - 500003, India

Office in India: - Secunderabad, Maharashtra

 

21. Company Name: - SST Group

Headquarters: - Germany

Office in India: - Noida - 201301, India

 

22. Company Name: - Hytech Engineers

Headquarters: - Mumbai - 400072, India

Office in India: - Mumbai - 400072, India

 

23. Company Name: - Apurva Controls

Headquarters: - Pune - 411033, India

Office in India: - Pune - 411033, India

 

24. Company Name: - Heaton Engineering Pvt.Ltd. 

Headquarters: - Pune - 411033, India

Office in India: - Pune - 411033, India

Output of MI Heating Cable

 Output of MI Heating Cable

Here is how the output calculations for the MI heating cable are done:

 

Firstly we have some input:

V = Supply Voltage (V)

HL = Heat Loss (W/m)

TL = Total heating Cable Length (m)

 

Now we find Ideal resistance (Ri): -

Ri = (V X V) / (HL X TL X TL)

Unit of Ri: -

Ri = V2 / W / m X m X m

Ri = V2 / W X m

(V2 / W = Ohm)

Ri = Ohm/m

Now we will be select cable of less resistance from ideal resistance.

Now we will be calculating Output of heating cable (O):-

O = (V X V) / (Rc X TL X TL)

Where

Rc = Resistance of MI heating cable

Unit of Output: -

O = (V X V) / (ohm/m X m X m)

O = (V2) / (Ohm X m)

(V2 / Ohm = W)

O = W/m

 

Example: -

HL = 23.5 W/m

V = 230 V

TL = 26.3m

Output =?

Solution:-

Ri = (230 X 230) / (23.5 X 26.3 X 26.3)

Ri = 52900 / 16254.715

Ri = 3.2544 Ohm/m

Let now we assume we have some resistance like (5.6, 4.2, 2.3 & 1.6)

Then we select 2.3 ohm/m resistance cable therefore we calculated output: -

O = (230 X 230) / (2.3 X 26.3 X 26.3)

O = 52900 / 1590.887

O = 33.252 W/m