Introduction

Cities and campuses often transfer utilities between buildings through buried tunnels. The energy forms involved may be steam piping, water, or electrical buses. Each of these forms may have parasitic energy loss, heating both air in the tunnel and subsequently the ground external to the tunnel. Calculation of these losses may be important to design. Clearly, losses can be minimized with insulation but will never be zero, and over the long run the losses can be significant. In some locales there are regulatory limits to temperature rise in the earth above. Additionally, there may be limits on the air temperature rise in the tunnel related to maintenance personnel access or equipment. Forced ventilation may be needed.

Textbooks1,2 provide solutions to each of the elements of this situation. A general model is described with a closed solution to the heat transfer with the surroundings and the temperature increase of ventilation flow, given the thermal properties of the materials and the dimensions of the design.  Figure 1 illustrates the problem to be solved.

Figure 1: Solution to Problem of Heat Transfer

This problem may involve a buried tunnel or a vault that may contain piping through which steam or a heated fluid is pumped. Alternatively, the vault could contain electrical conductors, which generate heat from resistive losses. This is the most general case. Piping or conductors may simply be buried without any encasement. In the case of a tunnel, there may be forced convection of air to provide cooling for materials or for maintenance. The air temperature will increase in the direction of flow. There may be no airflow, in which case heat is transmitted though the walls, into the surrounding earth, and hence to the environment. The internal temperatures will rise until there is equilibrium between the heat generated and the heat transferred. This solution encompasses such cases through specification of the appropriate overall U-value.

The Adiabatic Case

For reference, first consider the simple relation for the adiabatic case. In this case, there is the assumption of no heat transfer through the soil. Air flows though a closed duct or tunnel with heat being added at a uniform rate q per unit length. The expression for the temperature increase ΔT is the classic flow thermal energy relation:¹

ΔT = q*L/(F*Cp)

where:

q = rate of heat loss from pipe, watts/m
L = tunnel length, m
F = air mass flow rate, kg/s
Cp= specific heat of air, joules/kg

This equation is generally valid for any fluid and any set of consistent units.  For example, consider a tunnel with a length of 600 m   and steam piping with heat loss through the pipe’s walls and insulation of 1320 W/m.  Airflow rates at near atmospheric pressure typically are given in volumetric units, such as cubic meters per minute. Conversion to mass flow rate requires multiplication by the density and, if needed, conversion from minutes to seconds.  For an airflow of 500 m³/min at a density of 1.2 kg/m³, the air mass flow rate is about 10 kg/s.  The specific heat of air is about 1005 J/kg-K.   From the equation above, the adiabatic dry air temperature rise would be 79°C. If the inlet air temperature were 15°C, outlet air would be 94°C.

Heat Transfer through Tunnel Walls and Soil

When the ambient temperatures of the surrounding soil or at ground level are different from the air temperature inside the tunnel, energy will flow via that route at a rate dependent on the temperature difference and the thermal properties of the intervening materials. The overall resistivity (or its inverse, conductivity) can be calculated from the resistivities of the individual materials in the same manner as a resistive electrical circuit. The general computational process to obtain U is described in most textbooks.¹

Calculation of Resistivity through Tunnel Walls and Soil

The total resistivity is the sum of the resistivity of the component material in the heat path. This is shown in Figure 2.

FIGURE 2: Total Resistivity

The overall or total thermal resistance is therefore:

Rtotal = Rwall surf +  Rwall + Rsoil + Rsurf

Additional terms can be added if known. The overall conductivity U is then 1/R.

Correlations for most of the terms above are well known.1,2 Values for Rsoil are more difficult to estimate. The heat transfer mode is conduction, just as occurs through the tunnel walls. However, values for soil resistivity vary widely with soil composition and moisture
content.  Reference 2 is one source of values for different soils. This situation is referred to in the classical heat transfer literature as a “semi-infinite” boundary condition problem. One common practice for buried objects is to define a shape factor S related to the standard heat conduction relationship as follows:

Q = k*(A/L)*ΔT = S*k*ΔT

so that the shape factor S replaces the ambiguous term (A/L). Exact solutions are available for simply shaped objects such as a circular pipe (cylinder). Factors for many other shapes have been determined. These factors take into account the dimensions of the object as well as the depth of burial.

Closed Form Solution for Tunnel

As discussed above, an important task is to determine the overall value of the thermal resistivity from inside the tunnel to the surroundings.  However, once this is done and assuming it is a constant over the length of the tunnel, one can write a differential equation for the tunnel air temperature as a function of tunnel length and solve it formally. The derivation begins with a simple energy balance on a volume of air in a differential length of tunnel dL.  In words, Energy in + energy added = energy out + energy transferred through walls.

Substituting suitable expressions for each term and algebraically simplifying, the resulting differential equation is

dT/dx + (PU/mcp)T = (P U Tamb + dqgen/dx)mcp

where:

T = local tunnel air temperature
P = perimeter of tunnel
U = 1/R = overall thermal conductivity from pipe to surface
Tamb= ambient temperature
Tin = inlet air temperature at x = 0 (boundary condition)
m = mass flow rate of tunnel air
cp = specific heat of the tunnel air

The term (dqgen/dx) is the heat generation rate per unit length in watts per meter and is assumed to be constant in this derivation.

The solution³ to this equation is

T = (Tamb + (1/PU)(dqgen/dx)) + C e-(PUx/mcp)

where C = ((Tin – Tamb) – 1/PU))(dqgen/dx)

This equation is exact for the assumptions made, but of course the result is strongly dependent on the correctness of the property values that define U. In the limit, as R becomes very large (U → 0), the computed result will be the same as the adiabatic solution given previously. This expression is also valid if heat generation is zero or if the ambient temperature is higher than the tunnel air
temperature.

Example Calculations:

Assume district heating in a buried concrete tunnel. The parameters assumed are as follows:

	Heat loss from piping	1320 W/m
	Ventilation airflow	10 kg/s (500m³/s)
	Computer overall U	1.25 W/m²-K
	Perimeter	12 m
	Length	600 m

Table 1 summarizes these values.

Table 1 Cases and Comments

Case	Tamb(°C)	Tair in(°C)	Tair out(°C)	Comments
No soil heat transfer	n.a.	15	94	Adiabatic case
With soil heat transfer	15	15	67	Transfer through soil cools air
Smaller U (used half previous)		15	15	79°C For example, deeper burial
Higher Tamb no pipe heat	30	15	24	Transfer from ambient to tunnel
Higher Tamb with pipe heat	30	15	76	Ambient to tunnel plus pipe heat
If water pipe, prevent freeze	-20	10	-1	Providing 20W/m gives +1°C

References

1. Incropera, F.P., and DeWitt, D.P.  Fundamentals of Heat and Mass Transfer, 5th ed. New York: Wiley, 2002.

2. ASHRAE Handbook. Fundamentals, 2005 ed. Atlanta: ASHRAE.

3. Rainville, E.D. Elementary Differential Equations, 2d ed. New York, Macmillan, 1958.