Displacement Ventilation

The present example investigates the performance of a displacement ventilation system. Given measured values for inlet velocity, inlet temperature, and heat flux, this simulation yields field configurations of air temperature and velocity that are consistent with experimental measurements and analytic global models (Ref. 1).

Model Definition

In general, there are two classes of ventilation: mixing ventilation and displacement ventilation. In displacement ventilation, air enters a room at the floor level and displaces warmer air to achieve the desired temperature. Heating sources in the room can include running electronic devices, or inlet jets of warm air. A potential issue with the displacement ventilation approach is that significant temperature variation and strong stratification may arise.

The model geometry consists of a test chamber with the dimensions 2.5 m by 3.65 m by 3 m. A warm jet enters the chamber from an inlet located at the floor center. Fresh air, at constant temperature and relatively low velocity, is supplied through a wall inlet. Heat exits the chamber through an exhaust located in the center of the ceiling. The walls of the chambers are almost perfectly insulated.

Symmetry reduces the modeling domain to half of the chamber, as shown in Figure 1. The warm jet feeds 0.028 m3/s of air at a temperature of 45 °C into the room. The temperature of the fresh air is 21 °C and has a flow rate of 0.05 m3/s.

Figure 1: The modeling domain is reduced to half the chamber size due to symmetry.

Convection of heat can be either forced or free. Forced convection occurs if

(1)

where g is the gravity (m/s2), α(1/K) is the coefficient of thermal expansion, T (K) the temperature, U (m/s) the velocity, and L (m) refers to the characteristic length. Equation 1 states that the buoyancy force is small compared to the inertial force. In such a situation, the character of the flow field is described by the Reynolds number,

where ν (m2/s) is the kinematic viscosity. Natural convection occurs if Equation 1 is not fulfilled, in which case the flow field character is described by the Grashof number,

If the convective forces and buoyant forces are of the same order of magnitude, then

can be interpreted as the ratio between the inertial and viscous forces. That is, when the Grashof number is large, the flow becomes turbulent.

To investigate if Equation 1 holds, the air can approximated as an ideal gas in which case α = 1/T. Furthermore, ΔT ≈ 20 K, U ≈ 1 m/s, and L ≈ 2 m. This gives

Hence, it is the Grashof number that determines whether the flow is turbulent or laminar. Using the same approximations as above:

(2)

Equation 2 clearly indicates that the flow is turbulent.

Modeling Considerations

You model the flow using the k-ω model. The main reason for using the k-ω model over the k-ε model is that former is in general more reliable when it comes to predicting the spreading rate of jets (Ref. 2).

As can be seen in Figure 1, the inlets and the outlet have been extended with small domains. This is to avoid having velocity conditions perpendicular to the no-penetration conditions of the walls, which often turns out to be numerically unstable.

Results and Discussion

Figure 2 shows a streamline plot colored by the temperature. As expected, there is a stratification at z≈1 m with a complicated recirculation pattern above.

Figure 2: Streamlines colored by the temperature illustrating the velocity field.

A more quantitative picture is given in Figure 3 which shows isosurfaces of the temperature field. The stratification is even more clearly visible here. The result compares well with the experimental results in Ref. 1.

Figure 3: Isosurfaces of the temperature.

Figure 4 shows a comparison of the computed and measured temperature along the line (1.25,0,0) → (1.25,0,3), that is through the center of the jet. While the computational result captures the main trend with decreasing temperature with height, it still over predicts the experimental result with 2 °C at z=2.6 m. There are two possible reasons for this. The first is, as mentioned in Ref. 1, that the test chamber is not as well insulated as intended. The other possible explanation is that the buoyancy induced production in the k and ω equations (see for example Ref. 3) must be included in order to reproduce the experimental results more accurately.

Figure 4: Plume temperature.

References

1. D. Mazoni and P. Guitton, “Validation of Displacement Ventilation Simplified Models,” Proceedings of ‘Building Simulation '97’, the Fifth International IBSPA Conference, vol. I, pp. 233–239, International Building Performance Simulation Association (IBSPA), 1997.

2. D.C. Wilcox, Turbulence Model for CFD, 2nd ed., DCW Industries, La Canada, CA, 1998.

3. S. Tieszen, A. Ooi, P. Durbin, and M. Behnia, “Modeling of Natural Convection Heat Transfer,” Proceedings of the Summer Program 1998, Stanford: Center for Turbulence Research, 1998.

CFD_Module/Nonisothermal_Flow/displacement_ventilation

Modeling Instructions

From the menu, choose .

New

In the window, click

Model Wizard

In the window, click

In the tree, select ω.

Click .

Click

In the tree, select .

Click

Global Definitions

Parameters 1

In the window, under click .

In the window for , locate the section.

In the table, enter the following settings:


Name	Expression	Value	Description
H	3[m]	3 m	Room height
D	2.5[m]	2.5 m	Room depth
W	3.65[m]	3.65 m	Room width
Hd	0.5[m]	0.5 m	Diffuser inlet height
Ad	1.7[m^2]	1.7 m²	Diffuser inlet area
As	0.0324[m^2]	0.0324 m²	Source inlet area
Ao	0.04[m^2]	0.04 m²	Outlet area

Definitions

Variables 1

In the toolbar, click

and choose .

In the window for , locate the section.

In the table, enter the following settings:


Name	Expression	Unit	Description
Ms	0.028[m^3/s]	m³/s	Volume flow rate at source
Md	0.051[m^3/s]	m³/s	Volume flow rate at diffuser
Us	Ms/As	m/s	Source inlet velocity
Ud	Md/Ad	m/s	Diffuser inlet velocity
Tdiff	21[degC]	K	Diffuser air temperature
Tsource	45[degC]	K	Source air temperature
Tout	17[degC]	K	Outside temperature

Define a step function to be used when prescribing initial conditions for the temperature.

Step 1 (step1)

In the toolbar, click

and choose .

In the window for , locate the section.

In the text field, type 1.5.

Click to expand the section. In the text field, type 1.5.

From the list, choose .

Geometry 1

Block 1 (blk1)

In the toolbar, click

In the window for , locate the section.

In the text field, type D.

In the text field, type W.

In the text field, type 2*H.

Locate the section. In the text field, type -W/2.

In the text field, type -H.

Block 2 (blk2)

In the toolbar, click

In the window for , locate the section.

In the text field, type 0.05[m].

In the text field, type 2*Hd.

In the text field, type Ad/Hd.

Locate the section. In the text field, type -0.05[m].

In the text field, type -Ad/Hd/2.

In the text field, type -Hd.

Union 1 (uni1)

In the toolbar, click

and choose .

In the window for , locate the section.

Clear the check box.

Click in the window and then press Ctrl+A to select both objects.

Block 3 (blk3)

In the toolbar, click

In the window for , locate the section.

In the text field, type 4[m].

Locate the section. In the text field, type -1[m].

In the text field, type -2[m].

In the text field, type -4[m].

Difference 1 (dif1)

In the toolbar, click

and choose .

In the window for , locate the section.

Clear the check box.

Select the object only.

Find the subsection. Select the

toggle button.

Select the object only.

Block 4 (blk4)

In the toolbar, click

In the window for , locate the section.

In the text field, type 3[m].

In the text field, type 2[m].

In the text field, type 5[m].

Locate the section. In the text field, type -0.2[m].

In the text field, type -2[m].

In the text field, type -1[m].

Difference 2 (dif2)

In the toolbar, click

and choose .

In the window for , locate the section.

Clear the check box.

Select the object only.

Find the subsection. Select the

toggle button.

Select the object only.

Block 5 (blk5)

In the toolbar, click

In the window for , locate the section.

In the text field, type sqrt(As).

In the text field, type sqrt(As)/2.

In the text field, type H/2.

Locate the section. In the text field, type D/2-sqrt(As)/2.

In the text field, type -0.05[m].

Block 6 (blk6)

In the toolbar, click

In the window for , locate the section.

In the text field, type sqrt(Ao).

In the text field, type sqrt(Ao)/2.

In the text field, type 0.45[m].

Locate the section. In the text field, type D/2-sqrt(Ao)/2.

In the text field, type H-0.3[m].

Mesh Control Domains 1 (mcd1)

In the toolbar, click

and choose .

On the object , select Domains 2 and 5 only.

It might be easier to select the domains by using the window. To open this window, in the toolbar click and choose . (If you are running the cross-platform desktop, you find in the main menu.)