Distributed Resistances and Fan Models Chapter 4

Training Manual May 15, 2001 Inventory # Objectives Model the effects of flow field features without modeling the feature geometry. Establish the concept of distributed resistances –Define the term “distributed resistance” –List examples of distributed resistance applications State how distributed resistances are applied. –Implementation –Directional dependence.

Training Manual May 15, 2001 Inventory # Lesson 1 Objectives (continued) Modeling form factor losses. Modeling friction losses. Modeling permeability losses. Distributed resistance modeling guidelines. Describe the fan model. Example 1 Example 2

Training Manual May 15, 2001 Inventory # The Distributed Resistance Concept The purpose of a distributed resistance is to simulate the pressure drop of a geometrical feature of the problem domain without modeling the details of the feature’s geometry. In creating the fluid flow model, distributed resistances can facilitate the modeling process.

Training Manual May 15, 2001 Inventory # The Distributed Resistance Concept (continued) Examples of physical phenomena which you might want to model with distributed resistances fall into four categories: 1. Friction losses Flow within the tube bundle of a heat exchanger Flow through a duct of irregular cross-section - modeled as a pipe.

Training Manual May 15, 2001 Inventory # The Distributed Resistance Concept (continued) 2. Form factor losses Flow through bends or tees in pipes Flow through valves 3. Permeability Ground water flows Flow through a filter 4. Fans (or pumps) Fan Pump

Training Manual May 15, 2001 Inventory # Form factor losses Friction losses Permeability losses How Distributed Resistances are Applied Distributed resistances are treated as Source terms (Ri) in the momentum equations. They are applied via Real Constants as element quantities.

Training Manual May 15, 2001 Inventory # Directional Dependence There are three types of directional dependencies in distributed resistance relationships: Type 1 - Isotropic – The resistance factors are the same for all directions.

Training Manual May 15, 2001 Inventory # Type 2 - Forcing flow in one direction –Set resistance factors in the x, y or z direction –Resistance factors in the non-flow directions are automatically set very high. Type 3 - Directionally dependent resistance –Each direction has its own set of parameters Directional Dependence (continued)

Training Manual May 15, 2001 Inventory # Form Factor Losses K is the form loss factor per unit length. K values typically obtained from a handbook such as Idelchik Shows the effect of pressure drop of a flow cross-section change such as an obstruction Typically applied to a very localized region

Training Manual May 15, 2001 Inventory # L is the length in the finite element model over which the resistance is applied. Note that K FLOTRAN = K handbook / 2L Note that g c =1 in FLOTRAN, imposed by using consistent set of units. Form Factor Losses (continued) While K is dimensionless in Idelchik, in FLOTRAN the units are 1/L (per unit length since it is applied as an element quantity.

Training Manual May 15, 2001 Inventory # Friction Losses Applies to a flow passage The friction factor is unit-less. Not a wall roughness friction factor, FLOTRAN assumes a smooth wall The FLOTRAN friction factor is equal to half the Darcy friction factor

Training Manual May 15, 2001 Inventory # From the Literature, the Darcy Equation for Head Loss: Moody chart relationship for Laminar Flow Substituting and rearranging, Friction Losses (continued)

Training Manual May 15, 2001 Inventory # In FLOTRAN, the friction pressure losses are expressed as: where a and b are user supplied constants. For Laminar flow:a=32b=1 For turbulent flow:a=0.158b=0.25 Note that values of the constants are half the Darcy factors (and twice the Fanning factors). Friction Losses (continued)

Training Manual May 15, 2001 Inventory # Q = rate of flow k = Darcy’s coefficient of permeability L = length of the sample A = sample cross-sectional area h = hydraulic head Permeability Generally used to model flow through porous media Units of C are 1/L 2 Darcy’s Law

Training Manual May 15, 2001 Inventory # K = permeability, an empirical constant (length 2 )  fluid = the unit weight of the fluid (mass/length 3 )  = fluid viscosity (mass*time/length 2 ) Permeability (continued) k, Darcy’s coefficient of permeability depends on both the properties of the fluid and the porous material. k has units of length/time The porous material permeability characteristics are characterized by K where

Training Manual May 15, 2001 Inventory # Permeability (continued) The FLOTRAN permeability value is the inverse of the porous material permeability.

Training Manual May 15, 2001 Inventory # Application of Distributed Resistances in ANSYS/FLOTRAN In the preprocessor (PREP7) menu select Real Constants Choose Add. For this example we have already defined element type 1 to be FLUID141. OK

Training Manual May 15, 2001 Inventory # Application of Distributed Resistances in ANSYS/FLOTRAN (continued) Select the distributed resistance or fan model of your choice. Enter values as appropriate. Note: Real constant set number must be > 1 since FLOTRAN ignores real constant set 1.

Training Manual May 15, 2001 Inventory # Distributed Resistance Modeling Guidelines Any combination of distributed resistance loss type is possible with any combination of directional dependence type. –Isotropic Friction loss + Isotropic permeability –Permeability in the x-direction + Isotropic Form factor loss Distributed resistances can be used in any coordinate system and/or with rotating flows.

Training Manual May 15, 2001 Inventory # Distributed Resistance Modeling Guidelines (continued) A localized distributed resistance produces a large gradient in the flow resistance direction. This indicates a need for mesh refinement !! Artificial viscosity can be used to smooth velocity fields in the neighborhood of severe resistance changes. Mesh refinement is even more effective than artificial viscosity in achieving an accurate, converged solution for distributed resistances. The mass imbalances that may occur with distributed resistances when using a coarse mesh are localized.

Training Manual May 15, 2001 Inventory # Distributed Resistance Modeling Guidelines (continued) If the distributed resistance itself causes significant turbulence, then the turbulence model should be disabled. The distributed resistance generally is calculated to completely model the pressure effect (Δp) of the geometry represented. To disable the turbulence model, first select all the nodes in the distributed resistance region.

Training Manual May 15, 2001 Inventory # D, ALL, ENKE, 0 D, ALL, ENDS, 1 Distributed Resistance Modeling Guidelines (continued) Next, set turbulent kinetic energy (ENKE) to zero and turbulent energy dissipation (ENDS) to one (again,in the distributed resistance region) using the boundary condition commands, i.e.,

Training Manual May 15, 2001 Inventory # Fan Model A fan (or pump) can be modeled as a momentum source. The fan (pump) model does NOT describe the details of the flow near the fan (pump). The fan (pump) model does approximate the effects of the fan (pump) on the problem domain.

Training Manual May 15, 2001 Inventory # Fan Model (continued) Equation for the fan model is in terms of a pressure gradient in the direction of the flow. Need to know the “length” of the fan Head-capacity curves are often in terms of volumetric flow. Use the actual fan (pump) area to calculate the velocity at the outlet.

Training Manual May 15, 2001 Inventory # P = pressure s = distance through fan V = velocity through fan Fan Model (continued) Fan model constants are input as Real Constants. Use consistent set of units !

Training Manual May 15, 2001 Inventory # Fan Model (continued) Type 4 fan model acts along a single specified coordinate direction. Type 5 fan model may act in an arbitrary direction, oriented at an angle to the global coordinate system.

Training Manual May 15, 2001 Inventory # Example 1 A K-Factor can be used to produce a pressure drop at the end of a channel V inlet Evaluate mesh sensitivity and artificial viscosity effects

Training Manual May 15, 2001 Inventory # Example 1 (continued) The parameters for the channel without distributed resistance are: –ΔP ~ 0.05 psi –Re ~ 4 x 10 6 –Vaverage = in/sec –Effective viscosity ~ to Add a pressure drop of ΔP = 1.50 psi

Training Manual May 15, 2001 Inventory # Example 1 (continued) Calculate K = Expected  P total = 1.55 psi

Training Manual May 15, 2001 Inventory # Example 1- Mesh A (continued)

Training Manual May 15, 2001 Inventory # Example 1 - Mesh B (continued)

Training Manual May 15, 2001 Inventory # Example 1 - Mesh C (continued)

Training Manual May 15, 2001 Inventory # Example 1 (continued) Comparison of results –Accuracy of the pressure drop –Mass balance

Training Manual May 15, 2001 Inventory # Example 1 (continued) Expected pressure drop = 1.55 A.V. is Artificial Viscosity

Training Manual May 15, 2001 Inventory # Example 1 (continued) Inlet mass flow rate =

Training Manual May 15, 2001 Inventory # Example 1 (continued) Conclusions –Modest mesh refinement pays big dividends. –Artificial viscosity refines the solution

Training Manual May 15, 2001 Inventory # Example 2 We want to model flow through a pipeline which includes a heat exchanger. The friction factor can be used to simulate the flow through 24 parallel small pipes (the heat exchanger tubes) with one large pipe. Each of the small pipes has a radius of 0.5 Fluid properties : –  –  – Re = 200 (for small pipe) – Pipe length is 5

Training Manual May 15, 2001 Inventory # Example 2 (continued) Determine the pipe geometry and distributed resistance which well enable matching the mass flow with a specified pressure drop of 20 through a length of 5. To match the mass flow, the equivalent pipe must have the following area:

Training Manual May 15, 2001 Inventory # Example 2 (continued) From the Reynolds number, The known pressure drop (20) is used to calculate the friction factor. The equivalent pipe length is also chosen to be 5.

Training Manual May 15, 2001 Inventory # Example 2 (continued) For laminar flow, For the small pipes: Next we find the pressure drop in the larger pipe and make up the rest with a distributed resistance.

Training Manual May 15, 2001 Inventory # Example 2 (continued) For the larger pipe, the Reynolds number is calculated: The large pipe pressure drop is:

Training Manual May 15, 2001 Inventory # Example 2 (continued) We need to make up the pressure loss with a distributed resistance. By analogy with the equation for the small pipe, the following parameters for distributed resistance would make up the remaining pressure drop:

Training Manual May 15, 2001 Inventory # Example 2 (continued) The resulting mass flow per radian should be

Training Manual May 15, 2001 Inventory # Example 2 (continued) Conclusions: –Optimum results are obtained with slightly lower values of the resistance coefficient. –Results are accurate to within about 10 percent –Distributed resistance is an inexact science. Don’t expect exact results.