Architecture 4.411 Building Technology Laboratory Spring 2004 Tutorial and design exercises for airflow studies using CONTAMW 2.0 Buoyancy-driven flows Please complete Table 1, using Table 2 as a template. It is not necessary to fill in all rows, but please provide a reasonable set of cases. Single-zone base case Open CONTAMW 2.0. Create a single-zone building by drawing a box or, alternatively, four walls Define an indoor zone by right clicking inside the box to reveal the icon pull-down menu, then selecting zone. Double click on the icon, name the zone and assign a volume, floor area, and indoor temperature. Note that the floor area and volume are automatically related by an assumed height of 3 meters. You can change this value by opening the Level pull-down menu and selecting Edit Level Data. Save your project before going too far. Right click somewhere on the perimeter – say, the upper wall – and select flow path. Double click to define the properties for the airflow path. First, select a flow element. n∆Click the New Element tab and try the power law for volumetric flow, Q = P C . Name this element (inlet, perhaps). The flow exponent is fine as is, at 0.5, but the flow coefficient needs to account for the area of the window. Let the flow coefficient be the window area in m2, multiplied by 0.77. The factor of 0.77 is Cd 2 , where the discharge ρ coefficient is 0.6 and the density is 1.2 kg/m3. Try 1 m2 for now, which means the flow coefficient should be 0.77. Then select the flow path tab and give the window a height relative to the floor. The default of 1.5 m is fine, for starters. Right click again on the perimeter to design a second flow path, on the lower wall. Repeat the process above to define its properties. Call the flow element “outlet.” Keep the default height of 1.5 m. Save your file. 1Select the weather pull-down menu and note that the ambient temperature is 20 oC. Open the simulation pull-down menu, run the simulation and note that nothing changes on the sketch pad. Click on either of the openings and note that the flow reported at the bottom of the CONTAMW window is zero. Open the simulation pull-down menu again, export the data as a text file, open the file, and note that there is no flow. Why? Think carefully: there are two reasons. Change the ambient temperature to 15 oC and try again. Still no airflow. Don’t bother with exporting data. When the sketch pad is unchanged, there will be no data. Now double click on the icon for the second flow path, click on the flow-path button, and change the relative height to 3 m. Again, run the simulation. Does the sketch pad change? It does. There are blue lines for flows and red lines for pressures. Note that the flows and pressures are the same for both openings. The direction of the lines differs: the set for the inlet points into the indoor zone and the set for the outlet points outward, denoting the flow direction. Export the data, open the data file, and note that the flows, in sm3/hr, are indeed equal and opposite in sign. The units are standard m3/hr, which account for the changes in temperature and therefore density. Standard volumetric flow is conserved, as is mass flow. Jot down the value for the flow and, even better, divide by the volume to obtain an air-change-per-hour value. Note also the indoor pressure relative to ambient and the pressure differences across the openings. While this tutorial will continue with changes to window height, you may at times want to add a new opening. In this case, you must pull down the View menu and turn off the sketchpad results to return to the normal sketchpad mode. Does height matter? Increase the height of the outlet to 4.5 m and re-run the simulation. Jot down the results. You have now doubled the height between the inlet and outlet windows. Does the airflow double? More than double? Less than double? Try a few more heights and tabulate the results. Plot or sketch the relation between airflow and height. Is a larger stack (chimney, stairwell, atrium) a good idea? Does temperature difference matter? Increase the temperature difference, by either warming the indoors or dropping the outdoor temperature. Try a few values of temperature difference. How does flow vary with temperature difference? Does it matter whether it is 25 oC in and 15 out, or 20 oC in and 10 out, both with a temperature difference of 10 oC? If it does matter, is the effect large or small? What does this say about solar chimneys or using heat gains in a double-façade cavity to draw air through occupied spaces? 2Does window size matter? Double the area of both the inlet and the outlet. What do you expect to happen to the airflow? What does happen? Now fix the total window area but make the two openings unequal. Start with a small inlet and a large outlet. Note the airflows and pressures. Flip things around and make a small inlet and a large outlet. Compare. 3Wind-driven flows Please complete Table 3, using Table 4 as an example. Single-zone base case Now let’s modify the same model to examine wind-driven flows. Set the height of the inlet and outlet windows to the same value. Ditto with the indoor and outdoor temperatures. Constant wind pressure Let’s deal with the wind in increasingly more realistic ways. First, for the inlet window define a constant wind pressure of 0.6 Pa, which corresponds to a 1 m/s wind. For the outlet, define the wind pressure as -0.6 Pa. Re-run the simulation and note, not surprisingly, that you obtain a flow. How does the value compare with those you obtained from buoyancy effects? If you prefer to estimate airflows due to a 2 m/s wind, please keep in mind that the wind pressure varies with the square of the wind speed and set the wind pressures to 2.4 Pa and -2.4 Pa. How does the airflow compare with what you calculated for 1 m/s? It doubles. Why, given what we did to the wind pressures? Because flow through an opening varies with the square root of the pressure drop across the opening. Here’s something else. Take the airflow for the wind pressure that corresponds to 1 m/s. Divide by 3600 to get m3/s. Recall that the window opening has a flow coefficient of 0.77, which corresponds to an opening area of 1 m2/s. Divide the airflow in m3/s by 1 m2 of area to obtain the speed of the air through the opening. How does it compare to the wind speed of 1 m/s? If you try the same procedure for