box dT test increase Pa

Air pressure increase in response to warming

1 Introduction

The objective of this illustration is show the verification of the coupled thermal / air flow module in

SEEP/W (AIR/W) and TEMP/W. TEMP/W will compute air temperatures in a closed box. The warming air should expand, and the density and pressure in the box should increase. The processes are coupled and solve simultaneously. SEEP/W solves for air flow, pressure and density while TEMP/W computes convective heat transfer due to flowing air.

2 Feature highlights

GeoStudio feature highlights include:

• Coupled thermal driven density dependent air flow • Comparison with known solution using Pv = mRT

3 Geometry and boundary conditions

This is a simple but important example to verify the thermally coupled density dependent air flow

formulation, because the computed results can be compared with exact solutions. In this example, a cubic meter of “air” in an enclosed box is to be heated by 10 degrees Celsius over a 100 second time span. If we assume that air behaves as an ideal gas, then it will follow the ideal gas law Pv=mRT, where P is the air pressure, R is the universal gas constant, T is the temperature and m/v is mass over volume or density. There are four analyses in the example file. Two of these establish the initial thermal and pressure (air and water) conditions, and two are the coupled thermal–pressure transient analyses.

The boundary conditions for the initial temperature file are a temperature of zero Celsius applied to all nodes in the box, as shown below on the left. The initial water pressure is set to a pressure head of

-1000m and applied to all nodes, as shown in the middle image. The -1000m is set to ensure that, based on the water content function for the material, the water content is fixed at a very low value, which in turn maximizes the air content. The right image below shows an atmospheric Pa = 0 condition applied to the top of the model for initial air pressures.

Figure 3-1 Initial condition boundary conditions for TEMP/W, SEEP/W and AIR/W analyses AIR/W Example File: box dT test increase Pa.doc (pdf) (gsz)

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The boundary conditions for the transient model seepage equation are the same as shown above for the hydraulic case. That is, Pw = -1000m water pressure at all locations. The thermal and air pressure

conditions are different, however. The air temperature in the transient analysis is a function over time, as shown below, and it is applied to all nodes in the model. The surface air pressure boundary condition is removed and no air pressure BC is applied in the model. This is done to simulate a closed container with an initial air pressure of zero at the top of the box. No air can escape the box during the heating process. If no air can escape, then as the air warms, the air pressure must increase according to the ideal gas law.

Applied Air Temperature108Temperature (°C)2002040Time (sec)6080100 Figure 3-2 Applied air temperature boundary condition function

In this example, TEMP/W is coupled to SEEP/W with the air flow option engaged. This means that TEMP/W will be the master process and AIR/W will be a slave process. A master process controls the start of the model as well as the time stepping in the transient process. TEMP/W will be launched to start the AIR/W solution because, in this case, the air temperature at the start of the process determines the air density in the box at the start. TEMP/W will compute the temperatures and pass this data to AIR/W, which will in turn solve the air flow equation for air pressure and velocity. The velocity will be passed back to TEMP/W, which will use this information to determine the convective heat flow in the moving air.

4 Material properties

Since this example includes solving the thermal, seepage and air flow equations at the same time, it is necessary to set up material properties for all three equations. There are different options for material models, depending on the complexity of the analysis. In this case, with full thermal coupling of

air/water/heat, it is better to use the Saturated / Unsaturated SEEP/W with AIR/W model. This requires that a water content function, hydraulic conductivity function and air conductivity function be defined. The three functions are shown below.

Recall that the hydraulic boundary condition set to -1000m. If you consider the water content function below, and look up the water content at a pressure of -10,000 kPa, you will see it is almost zero. If the

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water content is fixed at almost zero, then the area above the line is the air content, which is now almost 100%.

sand1.0e-041.0e-051.0e-06)c1.0e-07e/sm1.0e-08( ty1.0e-09viticu1.0e-10dno1.0e-11C-X1.0e-121.0e-131.0e-141.0e-150.010.11101001000Matric Suction (kPa) Sand1.0)0.8m³/³m( tnet0.6noC rteaW0.4 l.oV0.20.00.010.1110100100010000Matric Suction (kPa) AIR/W Example File: box dT test increase Pa.doc (pdf) (gsz) Page 3 of 6

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sand1.0e+001.0e-011.0e-02Air X-Conductivity (m/sec)1.0e-031.0e-041.0e-051.0e-061.0e-071.0e-081.0e-091.0e-101.0e-111.0e-121.0e-130.00.20.40.60.81.0Degree of Saturation The thermal functions are not important, because all nodes have a defined temperature boundary condition, so the solution to the thermal equation is actually known and therefore not dependent on material properties. Therefore, the Simplified thermal model can be used as shown below.

Figure 4-1 Simplified thermal model properties

5 Comparison of results

The objective in this example was to confirm that the ideal gas law is valid for thermal coupled density dependent air flow. Recall that the initial air pressure analysis had a Pa = 0 boundary condition applied to the model. This means the air pressure is zero at the top of the box. While the density of air is very small when compared to water, it must still increase with depth according to Pa = Rho_air x depth. For a 1m high box, the air pressure should increase to 0.012 kg/m3 over a 1m depth. The image below confirms this is the correct starting condition.

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Figure 5-1 Initial air pressure profile

According to the ideal gas law, if the temperature increases by 10 degrees, then the air pressure should increase by 3.72 kPa. The following image confirms this is correct.

Figure 5-2 Increase in air pressure due to warming

Finally, at the end of warming, the air pressure is still dependent on the elevation. You can see in the image below that while the overall pressure increases by 3.72 kpa, the profile with depth is still maintained.

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Figure 5-3 Pa profile at the end of simulation

Finally, the air density was supposed to increase if the air is warmed in a closed box. It does….

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