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| Section summary |
|---|
| 1. Heat losses
through uninsulated pipe |
| 2. Example of heat
loss calculation of uninsulated pipe : Step by Step guide |
| 3. Free Excel calculation tool for uninsulated pipe heat loss : pipe heat loss calculator Excel |
In factories, pipes are typically routed outside. This may not be a problem for most of products and circumstances, but in the case of winter conditions or if the product is sensitive to a loss of temperature, calculating the loss of heat through the insulated pipe is important to check if the temperature of the product is not dropping too much, in which case an insulation would be required.

The heat loss from an uninsulated pipe is basically made of 2 components : the heat loss through radiation and the heat loss through convection.
Total heat loss = radiation heat loss + convection heat loss
The pipe is emitting heat through radiation. The heat emitted can be calculated thanks to the following formula : Q/A = 0.1713 * ε * [(Ts/100)4 - (Ta/100)4]
With :
Q = heat loss (BTU/h)
A = heat transfer area (ft2)
Ts = surface temperature (R)
Ta = air temperature (R)
ε = emissivity of the pipe
The convection depends on the air velocity (wind speed), there are 2 cases for the calculation, either the wind speed is = 0, or the wind speed is > 0
The heat loss can be expressed with the following correlation :
Q/A = 0.27*ΔT1.25 / D0.25
With :
Q = heat loss (BTU/h)
A = heat transfer area (ft2)
ΔT = Ts - Ta
Ts = surface temperature (R)
Ta = air temperature (R)
D = pipe diameter (ft)
If wind is present, there is a tremendous influence on the heat transfer coefficient which can be estimated the following way : h = 0.11*c*G0.6/D0.4
With :
h = convection heat transfer coefficient (BTU/h.ft2.F)
c = specific heat (BTU/lb/F) = 0.24 for air
G = mass velocity of air (lb/h.ft2) = rho.v
rho = density (lb/ft3) = 0.075 for air
v = wind speed (ft/h)
D = pipe diameter (ft)
The total loss of heat is thus the sum of the heat loss through radiation and the heat loss through convection.
Total heat loss = radiation heat loss + convection heat loss
Considering a pipe a 66 mm external diameter and 10 m length that is transporting steam at 160c. What is the heat loss when the pipe is uninsulated, outside, in air at 20c and with wind at 10 km/h.
The pipe is in carbon steel, with an emissivity of 0.8 and a thermal conductivity of 142 W/m.K
It is assumed that the skin temperature of the pipe is equal to the temperature of the fluid, the resistance to heat transfer being on the outside of the pipe.
D = 66 mm = 0.2165 ft
L = 10 m = 32.8 ft
Ts = 160c = 320 F = 780 R
Ta = 20c = 68 F = 527.7 R
v = 10 km/h = 32808 ft/h
The equation given at 1.1 is used to calculate the heat loss due to radiation :
Q radiation = 0.1713*0.8*[(780/100)4 - (527.7/100)4] * 22.32 = 8930.9 BTU/s
With A = Pi*D*L = 22.32 ft2
Considering G = mass velocity of air (lb/h.ft2) = rho.v
G = rho.v = 0.075*32808 = 2498.8 lb/h.ft2
The heat transfer coefficient can be calculated from the correlation given in paragraph 1.3 above :
h = 0.11*c*G0.6/D0.4 = 0.11*0.24*2498.80.6 / 0.220.4 = 5.3 BTU/h.ft2.F
The heat loss due to convection can then be calculated via :
Qconvection = h*A*(Ts - Ta) = 5.3*22.32*(780-527.7) = 29934.8 BTU/h
Q total = 38865.7 BTU/h (11.4 kW)
Source
[Chopey] Handbook of Chemical Engineering calculations, Chopey et al, McGraw Hill, 2004