# Sample calculations for the geometry of the double tube heat exchanger

We are searching data for your request:

Forums and discussions:
Manuals and reference books:
Data from registers:
Wait the end of the search in all databases.
Upon completion, a link will appear to access the found materials.

Tab. 1
Given:
Outer tube Inner tube daa= 11.5 mm dia= 9.5 mm dai= 7.5 mm dii= 5.5 mm
Tab. 2
Cold water circuit
Area of ​​the annulusEntire (wetted) perimeterEquivalent diameter
$A.q=((dia2)2−(dai2)2)⋅π$$U=(dai2)⋅2π+(dia2)⋅2π$$deq=4⋅A.qU$
$A.q=((9,5mm2)2−(7,5mm2)2)⋅π$$U=(7,5mm2)⋅2π+(9,5mm2)⋅2π$$deq=4⋅26,70mm253,40mm$
A.q= 26.70 mm2U = 53.40 mmdeq= 2 mm
Tab. 3
Hot water circuit
Area of ​​the circleEntire (wetted) perimeterEquivalent diameter
$A.q=(dii2)2⋅π$$U=(dii2)⋅2π$$deq=4⋅A.qU$
$A.q=(5,5mm2)2⋅π$$U=(5,5mm2)⋅2π$$deq=4⋅23,76mm217,28mm$
A.q= 23.76 mm2U = 17.28 mmdeq= 5.5 mm

Since the warm water flows through the inner pipe, the equivalent diameter in the experiment corresponds to the inner diameter of the inner pipe deq= dii= 5.5 mm. This calculation and that of the wetted area is not absolutely necessary.