Sizing An Expansion Tank For A Heat Transfer System

Expansion tanks are important to the operation of heat transfer systems. They are designed to account for the volumetric expansion of heat transfer fluids as a result of temperature increase. Expansion tanks also remove moisture, non-condensables, degradation by-products and entrained air during startup and operation.

In this article, we present the sizing of an expansion tank using Relatherm HT-1, a high flash heat transfer fluid as an example.

Application Instance

Size an horizontal expansion tank for hot oil system using Relatherm HT-1 as Heat Transfer Fluid. The fluid is heated from 30 °C to 330 °C. System hot oil volume is 7.87 m³. Assume L/D to be used as 2.5. Thermal fluid density is 861.91 Kg/m³ at 30 °C and 752.17 Kg/m³ at 330 °C.

A minimum volume (10% ~ 20%) is considered in expansion tank at start-up in cold conditions. As heating is started volume expands and tank sizing should be such that expanded volume fills (70% ~ 80%) of tank volume.

If VSys = Volume of heat transfer system (piping and heat exchangers)

Vcold = Volume of the entire heat transfer (including piping, heat exchangers and cold fluid in expansion tank at start up)

Mcold = Total mass of thermal fluid in heat transfer system

ρcold = Density of thermal fluid in heat transfer system

Vhot = Volume of thermal fluid on expansion

Vexpansion = Volumetric expansion

Consider vessel volume to be V. Initial volume of hot oil at cold conditions is as following.

Vcold = 10% of V + VSys (in m³)

Mcold = ( 0.1 V + 7.87 ) ρcold (in Kg)

Volume on expansion is determined by dividing by density at hot conditions.

Vhot = ( 0.1 V + 7.87 ) ρcold/ ρhot (in m³)

Expansion in volume is obtained as

Vexpansion = ( 0.1 V + 7.87 )( ρcold/ ρhot – 1 )

Liquid level in expansion tank increases and fills up to 70% of the volume.

Vexpansion = 0.7 V – 0.1 V

Solving above equations for V provides capacity of expansion tank.

V = 1.96 m³

Volume for 2:1 elliptical horizontal tank is provided by.

V = πD²L/4 + πD³/12

With L/D = 2.5, solving above equation provides.

D = 1.35 m

L = 3.37 m