Saturday, August 15, 2009

Estimating Heat Loss from Buried Pipe

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Plant in country experience subzero ambient temperature, one of the common natural phenomenon is increased heat loss from piping to ambient lead to low temperature fluid. Low temperature fluid will results fluid characteristic change i.e. low reactivity, increased viscosity, promote freezing, crystallization, etc and subsequently lead to several problem such as low productivity, high pressure drop, blockage, scaling, fouling, etc. It is important to design plant system to minimize heat loss to ambient.

Water system at subzero may freeze and lead to blockage and potentially cause pipe damage. One of the effective and non-costly method is utilize low conductivity of soil to reduce heat to ambient by burying water pipe. This may involve a typical heat loss calculation.



Above image is a typical buried pipe in soil.
Heat loss,
Q = S x Km x (T1 - T2)

with Shape factor S
S = 2 x PI x L / Ln [ 4 z / D ]

where
Q = heat loss from pipe, W
S = Shape factor, m
Km = Average soil conductivity, W/m-K
T1 = Pipe skin temperature, degC (or K)
T2 = Soil surface temperature, degC (or K)
PI = 3.141592654
L = Buried pipe length, m
z = Distance between the ground surface and the center of the buried pipe, m
D = Outside diameter, m

Above equations are applicable when L is large compare to D and z is larger than 1.5D.

Example
A 30m long 100mm diameter hot water pipe of district heating system is buried in the soil 500mm below the ground surface (from pipe center line). Ground surface exposing wind chilling is 10 degC whilst pipe pipe skin temperature is expected to be 80 degC. Determine heat loss from pipe if soil average conductivity is about 0.9 W/m-K.

Km = 0.9 W/m-K
T1 = 80 degC
T2 = 10 degC
PI = 3.141592654
L = 30 m
z = o.5 m
D = 0.1 m

L is much larger than D and z = 0.5 larger than 1.5(0.1) = 0.15
Above equations may be used.

Shape factor,
S = 2 x PI x L / Ln [ 4 z / D ]
S = 2 x PI x 30 / Ln [ 4 x 0.5 / o.1 ]
S = 62.9 m

Heat loss,

Q = S x Km x (T1 - T2)
Q = 62.9 x 0.9 x (80 - 10)
Q = 3963 W

Ref : Section 3-7, Heat Transfer A Practical Approach by Yunus A. Cengel, 2nd Ed.


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