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