UNICAMPFULLY DEVELOPED & PERIODIC FLOWS

Source of Information on Phoenics Lecture: fully developed flows

Includes:

1. Introduction, 2. Single Slab fully developed flows, 3. Activation terms, 4. Periodic flows and 5. References.

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FULLY DEVELOPED & PERIODIC FLOWSAdvantages of fully developed regime

• The case of developed flow in a duct or passage of constant cross-section will be referred to as single-slab fully-developed flow, since such flows can be simulated by performing calculations on a single-slab of computational cells.

• This is useful when it is desired to compute a fully-developed pipe-, plane-walled-channel- or Couette-flow situations, without, as was formerly necessary, simulating the flow in a long duct, either elliptically or parabolically.

• In other words, the Friction factor and Nusselt numbers can be determined without having to solve the entrance region problem.

• The problem of hydrodynamically and thermally developed flow inside constant cross-section ducts is of interest in numerous practical applications. Examples are flows in pipes, canals, sewers, and ventilation systems.

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FULLY DEVELOPED & PERIODIC FLOWSHydrodynamic Definition

• For constant property flow in a duct of constant cross section, the velocity distribution becomes independent of the streamwise coordinate at sufficiently large distances from the inlet.

• Such unchanging velocity distribution is said to fully developed.

Le ≅ 0,06(d)Re - Laminar

Le ≅ 4,40(d)Re(1/6) - Turbulent

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FULLY DEVELOPED & PERIODIC FLOWSThermal Definition

• For the temperature field, the fully developed regime is not as easily characterized as that for the velocity.

• In fact, the temperature does not become independent of the streamwise coordinate.

• The usual definition of thermally developed regime is thathe SHAPES of the temperature distributions at sucessivestreamwise locations are the same, so that the sucessivedistributions can be brought together by a suitable scale.

• In such thermally developed flow, the heat transfer coefficient is independent of the streamwise coordinate.

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FULLY DEVELOPED & PERIODIC FLOWSThermal Definition

• The dimensionless form of the temperature is given as function of the wall and the bulk temperatures:

• In such thermally developed flow, the heat transfer coefficient is independent of the streamwise coordinate.

bw

wTTTT

−

−=θ

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FULLY DEVELOPED & PERIODIC FLOWSEquations set

• The fluid flow conservation equations (constant properties) are greatly simplified for the fully developed regime.

• Without the streamwise convection terms the mass, momentum and energy equations reduces to:

( ) ( )

( ) ( )

( ) ( )

( ) ( )

∂

∂+

∂

∂+

∂∂

=∂

ρ∂+

∂

ρ∂→

∂

∂+

∂

∂µ+

∂−=

∂ρ∂

+∂ρ∂

→

∂

∂+

∂

∂µ+

∂−=

∂ρ∂

+∂ρ∂

→

∂

∂+

∂

∂µ+−=

∂ρ∂

+∂ρ∂

→

∂∂

+∂∂

=→

2

2

2

2z

"pp

2

2

2

2

2

2

2

2

2

2

2

2

yT

xTk

zq-

yTvC

xTuC

HEnergy

yv

xv

dyP

yvv

xvu Y Momentum

yu

xu

dxP

yvu

xuu X Momentum

yw

xw

dzdP

yvw

xuw ZMomentum

yv

xu0 Mass

&

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FULLY DEVELOPED & PERIODIC FLOWSEquations set

• The mass equation does not have dw/dz because the flow is fully developed!

• The Z momentum, the pressure gradient has to be specified (external source), i.e., it is not deduced from the equations set.

• The convection terms retain terms relative to the secondary flow components only.

yv

xu0

∂∂

+∂∂

=

( ) ( )

∂

∂+

∂

∂µ+−=

∂ρ∂

+∂ρ∂

2

2

2

2

yw

xw

dzdP

yvw

xuw

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FULLY DEVELOPED & PERIODIC FLOWSAdvantages of fully developed regime

• The u and v momentum equations represent the secondary flow.

• Notice they are uncoupled from the Z momentum.• The secondary flow is evaluated with the transverse

pressure gradient simultaneously.• The calculations are performed only in one Z slab!

( ) ( )

( ) ( )

∂

∂+

∂

∂µ+

∂−=

∂ρ∂

+∂ρ∂

∂

∂+

∂

∂µ+

∂−=

∂ρ∂

+∂ρ∂

2

2

2

2

2

2

2

2

yv

xv

dyP

yvv

xvu

yu

xu

dxP

yvu

xuu

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FULLY DEVELOPED & PERIODIC FLOWSEquations set

• Likewise Z momentum, the energy equation retains:• the energy convective transport terms due to the secondary

flow,

• The energy diffusive transport terms along the x and y directions,

• And the axial energy flux has to be specified (external source),i.e., it is not deduced from the equations set.

( ) ( )

∂

∂+

∂

∂+

∂∂

=∂

ρ∂+

∂

ρ∂2

2

2

2"zpp

yT

xTk

zq-

yvTC

xuTC &

2

2pz

zTkwTCq

∂

∂+ρ−=

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FULLY DEVELOPED & PERIODIC FLOWSEquations set

• The fluid temperature is:

• For uniform wall heat flux :

• For constant wall temp :

• Notice, for uniform wall heat flux the bulk and wall temperatureare related to the wall heat flux, but for constant temperature the axial gradient of both temperatures are equal but there is no explicit relationship with the wall heat flux.

( ) wwb TTTT +−θ=

p

"zbw

CmPq

dzdT

dzdT

⋅⋅

=≡&

&

dzdT

dzdT bw ≡

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FULLY DEVELOPED & PERIODIC FLOWSEquations set

• The axial energy gradient, dqz/dz, is expressed as a volumetric source S as a function of the wall and bulk temperatures, which for

•

• uniform wall heat flux is:

• Where P is the pipe perimeter and qz is the known wall heat flux (W/m2)

• constant wall temp is: ∀θρ−=dz

dTCWS bp

( ) ∀

⋅⋅

ρ−≡∀ρ−≡∀ρ−=p

"z

pb

pw

p CmPqWC

dzdTWC

dzdTWCS

&

&

( )[ ]∀+−θρ−≡∀ρ−= wwbpp TTTdzdWC

dzdTWCS

UNICAMPSingle-slab fully-developed flows

• The analysis of such problems is now possible in PHOENICS by:

1. performing the calculation on a single IZ slab of computational cells;

2. deactivating the axial convection terms for all solved-for variables;

3. and deactivating the built-in pressure-gradient term in the W1 momentum equation.

• If the duct wall is circular or an infinite plane, the flow is one-dimensional and then the pressure P1 and cross-stream velocity (U1 or V1) need not be solved. Consequently, all convection terms can be de-activated. For any other duct cross-section, the flow in the slab will be two-dimensional and so the pressure P1 must be solved and only the axial-convection terms can be deactivated.

UNICAMPGiven Pressure Gradient

• Since the user must render the W1 pressure-gradient term

inactive, it is then usually necessary to prescribe a mean

axial pressure gradient in order to create finite axial

velocities, so that the mass flow rate is an outcome of the

PHOENICS calculation.

UNICAMPGiven the bulk or wall temperature gradient

• When it is the mass flow rate that is known and the user requires the pressure gradient, this procedure requires a number of separate calculations to be made where the pressure gradient is prescribed and then adjusted from calculation to calculation until the computed mass flow rate matches the specified value.

• Therefore, a more economical alternative has been provided whereby PHOENICS automatically calculates the pressure gradient from overall continuity for a given mass flow rate. The pressure adjustment procedure is based on that proposed by Patankar and Spalding [1972] for parabolic flows, and the PHOENICS implementation is described by Madhav [1992].

UNICAMPGiven energy rate (J/sec)

• Under certain heating and cooling conditions, such as constant heat flux or uniform temperature at the duct walls, a fully developed temperature profile can exist in the sense that the dimensionless temperature profile remains invariant with distance along the duct.

•There are a large number of possible thermal boundary conditions for a rectangular duct, since any of the four walls may be adiabatic, of constant heat flux or constant temperature, or thermal conduction in the surrounding wall may be significant. The only cases presently supported by PHOENICS are surface heat flux constant and surface temperature constant in the peripheral direction.

UNICAMPFlow Activation Settings

• Use VR to set fluid properties, domain , objects and boundary conditions, etc. But do not use VR (3.5.1) to activate model ‘fully developed’.

• Make sure that your 2D (xy) grid has NZ=1• Go to Q1 GROUP 7 and write:

1. SOLVE (P1,U1,V1,W1) if your domain is 2D (XY)2. FDSOLV(FLOW, mass flow) to prescribe the mass flow in

kg/sec going thru Z slab OR choose step 3 instead3. FDSOLV(DPDZ, press grad) to prescribe the pressure

gradient in Pa/m thru Z slab• Go to Q1 GROUP 19 and write:

1. PARSOL=F2. FDFSOL=T

UNICAMPHeat Activation Settings

• Several tests were performed with PHOENICS v 3.5.1 but they failed to predicted the Nusselt number for a laminar flow with constant temperature in a square duct.

• The Nu expected value is of 2.98 while the Nu PHOENICS value is of 3.5. Certainly there is some problem with the built in routine which will be fixed in the future.

• Another bug is that the routine works only for enthalpy equation. If one try temperature (Tem1) equation the results are ‘more’ non-sense.

• Tests for h1 equation and constant wall heat flux also were not successful.

• Building the thermal fully develop routine using In-Form is a good exercise, try it !

UNICAMPAvailable Phoenics Library Case Numbers

quasi-one-dimensional (ie fully-developed 2D)Couette flow, Reynolds number = 10 951Couette flow, various Reynolds no. and other options 9581D channel; fully-developed flow; k-eps model 6051D Laminar Pipe Flow with Heat Transfer. 5001D Laminar Pipe Flow with Mass Transfer. 5011D Turbulent Pipe Flow with Heat Transfer. 5021D Laminar Couette Flow with Heat Transfer. 5051D Laminar MHD Channel Flow. 5061D Laminar MHD Couette Flow. 507

quasi-two-dimensional (ie fully-developed 3D)Fully-developed flow in duct of square cross-section 8962D Laminar Duct Flow with Heat Transfer. 5032D Turbulent Duct Flow with Heat Transfer. 504

UNICAMPWORKSHOP 1

• We are going to explore the fully developed fluid flow and heat transfer in a square channel.

1. Using VR set up the (XY) square with 1mx1m with NX=NY=21

2. The fluid is air (0)3. Activate the solution for velocity

(laminar regime) and pressure4. Set sweeps = 50, and relaxation

settings for W1 = 1E+09 5. At last use four plate objects to

define the channel walls and create a pressure relief cell.

UNICAMPWORKSHOP 1

• It is given the axial pressure gradient, dp/dz is 4.425E-04 Pa/m. The problem is to find the W1 velocity field corresponding to this dp/dz

• GROUP 71. SOLVE(P1 ,U1 ,V1 ,W1)2. FDSOLV (DPDZ, 4.425E-04)• GROUP 191. FDFSOL=T; PARSOL=F• Check solution, at channel center, w1

= 1.725 m/s• P1 equation was not used, therefore

its residual is meaningless. In fact one could set up: SOLVE(w1) only.

• If you got trouble in obtaining this solution download wksh_FULLY-1

UNICAMPWORKSHOP 2

• It is the same as WKSH 1 but now is given the mass flow rate (10g/sec air) and you are asked to find the pressure gradient.

• GROUP 71. SOLVE(P1 ,U1 ,V1 ,W1)2. FDSOLV (MASS,1)• GROUP 191. FDFSOL=T; PARSOL=F• Change w1 to RELAX(W1 ,FALSDT,

1.000000E+03)• Check solution, at channel center, w1

= 1.725 m/s. • You must have got the same solution

of wksh-fully1. Check the result file for mass flow rate and pressure gradient.

• If you got trouble in obtaining this solution download wksh_FULLY-2

UNICAMPWORKSHOP 3

• We start from wksh fully 1 with dp/dz=4.425E-04 Pa/m but now the SOUTH wall is not stationary but moving U1 = 1 m/sec.

• Consider the flow turbulent and use the LVEL model to determine new velocity field.

• Find the proper relaxation coef such that 500 sweeps are enough!

• Noticed you just solved a 3D velocity field using just one-slab!

• Try to explain why the w1 field is not symmetrical • If you got trouble in

obtaining this solution download wksh_FULLY-3

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END OF THE WORKSHOP ON FULLY DEVELOPED FLOW

UNICAMPPERIODIC FLOWS

• A significant number of technically important duct configurations do not admit fully developed solutions of the type presented. • Among such ducts are those in which the flow cross section is not constant but varies in a periodic manner along the flow direction.

• Ducts with periodic area change are frequently found in plate-fin heat exchangers

FLOW

PERIODIC SECTION

UNICAMPThe Periodic Regime

• Consider the flow in the (XY) plane with the main direction parallel to the x axis.• If the periodic section has length L along the x direction then the velocity profile at the channel inlet and outlet are the same, i.e.

u(x,y) = u(0,y)=u(L,y) and v(x,y) = v(0,y)=v(L,y)

• The equation above characterizes the peridic fully developed regime for velocity field, replacing the conventional condition dw/dz = 0.• The variation of u and v in a single module which encompasses the period length L is capable of representing the changes of u and v through out the entire flow field.

UNICAMPThe Periodic Regime

• The pressure difference along the module outlet and inlet is constant,

•And its profile along the channel is represented below

( ) ( ) ( ) ( )y,L2xpy,Lxpy,Lxpy,xp +−+=+−

p

x

UNICAMPThe Periodic Regime

The outlet to inlet pressure drop is responsible for the global mass flow in the positive x-direction, defining:

Where β is a constant. It is then reasonable to split the pressure field into two components: the βx related to the global mass flow and P(x,y) related to detailed local motions inside the module

( ) ( )( ) β=+− Ly,Lxpy,xp

( ) ( )y,xPxy,xp +β−=

p

x

-βx

UNICAMPThe Periodic Regime

•The conservation equations for mass and momentum for constant property flow are:

•The term βx represents the mean pressure term. It has to be specified externally while the P(x,y) is evaluated internally.•The terms involving d2/dx2 were retained in recognition of the fact that large local streamwise gradients may occur in periodically fully developed flows.

y2

2

2

2

x2

2

2

2

gy

vx

vyP

yvv

xvu

tv

gy

ux

uxPx

yuv

xuu

tu

0yv

xu

ρ+

∂∂

+∂∂

µ+∂∂

−=

∂∂

+∂∂

+∂∂

ρ

ρ+

∂∂

+∂∂

µ+∂∂

−β=

∂∂

+∂∂

+∂∂

ρ

=∂∂

+∂∂

UNICAMPCharacteristics of the Periodic Flow Built-in Solver

• Periodic fully-developed flows can be simulated in PHOENICS by using the built-in x-cyclic (XCYCLE=T) option provided that: 1. the calculation does not employ cylindrical polar coordinates ( for

then there is no z-cyclic option ); 2. the lateral or transverse flow is not cyclic as well; or the flow does

not have complex thermal boundary conditions.• In any of these events, the user is advised to adopt the

approach of repeatedly transferring the downstream exit values of all variables, except pressure, to the inlet plane at IZ=1;

• Whichever approach is adopted, the solution domain is restricted to a single geometrical module which repeats itself in identical fashion in the actual geometry.

UNICAMPPHOENICS Settings

•• Define the domain, models, fluid properties, numerics and grid using VR

• You may also define plates and other types of boundary conditions BUT DO NOT DEFINE INLET AND OUTLET, leave them blank.

• In group 6 type: XCYCLE=T

•In group 13 specify the mean flow pressure gradient: ∆P/L (β)

• PATCH(PRESDROP,VOLUME,1,NX,1,NY,1,NZ,1,LSTEP)

•COVAL(PRESDROP,U1,FIXFLU,∆P/L)

• do not forget to create a pressure relief

UNICAMPWORKSHOP 4

• We are going to explore the periodic fully developed fluid flow in a rectangular channel(1x1/2) with periodically spaced thin plates.

1. Using VR set up the (XY) rectangle with 1mx0.5m with NX=NY=512. The fluid is air (0)3. Activate the solution for velocity (turbulent regime use LEVEL) and pressure4. Set sweeps = 5000, you may try finding good relaxation settings5. At last use four plate objects to define the channel walls, a thin plate as a

dividing wall at the mid plane and create a pressure relief cell.

• We are going to explore the periodic fully developed fluid flow in a rectangular channel(1x1/2) with periodically spaced thin plates.

0.2 m

0.2m

0.2m

1 m

0.5 m

0.3 m

UNICAMPWORKSHOP 4

• It is given the mean axial pressure gradient, β = ∆p/dz = 0.01 Pa/m. The problem is to find the W1 velocity field corresponding to this dp/dz

• GROUP 61. XCYCLE=T

• GROUP 13• PATCH (PRESDROP,VOLUME,1,51,1,51,1,1,1,1)• COVAL (PRESDROP,U1 , FIXFLU , 1.000000E-02)

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• If you got trouble in obtaining this solution download wksh_PERIOD-1

UNICAMPWORKSHOP 4

• Comments:1. Use AUTOPLOT and compare the inlet and outlet pressure fields (i.e.)

for x1 and x 51. At the domain openings they are almost identical as they should. Notice that PHOENICS solves for P(x,y), the actual pressure field is recovered introducing βx component.

2. Use AUTOPLOT and compare the inlet and outlet V1 velocity (i.e.) for x1 and x51. They are almost identical as imposed by the periodicboundary condition.

3. Finally, use AUTOPLOT and compare the inlet and outlet U1 velocity (i.e.) for x1 and x51. You may notice that they are not coincident, as they should. But in fact you are not comparing U1 at the same slabs. The C.V. for U1 is staggered. Therefore U1 values at west face of the 51th C.V. should be the U1 inlet values for the east face of the 1st C.V., but these values are not stored. The solved values are for the west face of the 1st C.V., that is the reason why the velocities are not coincidental.

UNICAMP

END