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# Shell and Helical Tube Heat Exchanger CFD Simulation, ANSYS Fluent Training

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In this project, the heat transfer between two hot and cold working fluid inside a shell and helical tube heat exchanger has been investigated.

This ANSYS Fluent project includes Mesh file and a Training Movie.

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## Introduction

The helical heat exchanger consists of several coils (tubes or helical tubes) with spring-like curves that are placed inside a cylindrical chamber called a shell. To increase the level of heat transfer, several spirals are usually used instead of one. But many helical heat exchangers use only one helix. The working principle of a helical heat exchanger is similar to that of a conventional shell and tube heat exchanger. The only difference between the two heat exchangers is the optimal use of space in the helical heat exchangers.

## Project description

In this project, the heat transfer between two hot and cold working fluid inside a helical heat exchanger has been investigated. The hot fluid enters the computational domain with a mass flow rate of 0.05 kg/s a temperature of 313K. the cold fluid enters the helical tubes with a mass flow rate of 0.0333 kg/s and a temperature of 289 K. the energy equation is activated to calculate the temperature changes inside the computational domain. RNG k-epsilon model with standard wall functions is exploited to solve turbulent flow equations.

## Shell and Helical Tube Heat Exchanger Geometry and mesh

The geometry of this project is designed in ANSYS design modeler and meshed in ANSYS meshing. The mesh type used for this geometry is unstructured and the element number is 1796590.  ## Shell and Helical Tube Heat Exchanger CFD simulation settings

The key assumptions considered in this project are:

• Simulation is done using pressure-based solver.
• The present simulation and its results are considered to be steady and do not change as a function time.
• The effect of gravity has been taken into account and is equal to -9.81 in Y direction.

The applied settings are summarized in the following table.

 Models Viscous model k-epsilon k-epsilon model RNG near wall treatment standard wall function Energy on Boundary conditions Inlet Mass flow inlet Cold water 0.033333 kg/s Cold water temperature 289 K Hot water 0.05 kg/s Cold water temperature 313 K Outlets outflow Walls Stationary wall Inner wall, baffle coupled Outer wall Heat flux 0 W/m2 Solution Methods Pressure-velocity coupling coupled Spatial discretization Pressure Second order Momentum second order upwind turbulent kinetic energy first order upwind turbulent dissipation rate first order upwind Initialization Initialization method Standard gauge pressure 0 Pa Velocity (x,y,z) (1.22479,0,0) m/s Turbulent kinetic energy 0.005625416 m2/s2 Turbulent dissipation rate 0.2661231 m2/s3 Temperature 289 K

## Results

Contours of pressure, velocity, temperature, etc. are obtained and presented. Now if we calculate the cold water outlet average temperature, the software will give us the value of 308.04081 K. Hence if we use the formula,  and implement the values of cold water inlet and outlet temperature which are 289 and 308.04081 K respectively, we can calculate the heat transfer rate which will be equal to almost 2651.65 W. provided comparing this calculated value with the value which fluent report gives us (2998.196 W), we can observe 11% percent error in our calculated parameter.

Mesh file is available in this product. By the way, the Training File presents how to solve the problem and extract all desired results.

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