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