THE THERMAL BUDGET OF THE 1990-1992 WAHA'ULA LAVA TUBE
What controls the lengths of lava flows? Why are some lava flows on the Earth and other planets so much longer than Hawaiian flows? These are the kinds of questions that have motivated my research.
Lava rheology, effusion rates, total erupted volume, cooling, vent shape, and topography all play a role in shaping lava flows. However, it is not necessary to understand every bump and wiggle that a lava flow makes in order to understand why lava flows stop (or why some just don't seem to want to stop). A lava flow will quit moving when either (1) there is no more lava available or (2) the lava freezes. These are called the volume-limit and cooling-limit for lava flows. I have concentrated on the cooling-limit of lava flows and tried to determine the absolute longest flow that can be produced under a given set of conditions.
Figure 1. An active lava tube during the ongoing eruption.
The key to keeping lava warm and fluid is to insulate it from the cold outside world. This is best accomplished by lava tubes (Figure 1). In the current eruption, the lava cools only about 1 deg.C/km as it flows down the tube [Helz et al., 1991]. While this is remarkably efficient, it is hard to imagine lava flowing much further than 50 km with such a cooling rate. A drop of 50 deg.C should cause enough crystallization to stop a lava flow.
So, what processes control this 1 deg.C/km of cooling in Hawai'i? Can tubes be significantly more efficient? Under what conditions might a tube transport liquid lava for hundreds of kilometers?
In order to answer these latter questions, I started by quantifying the thermal budget of lava tubes. Heat is lost from a lava tube from conduction through its walls and roof, by air circulating in the cracked walls and roof, by radiation from, and wind out of, skylights, by boiling off rainfall, by the exsolution and escape of gases, and by thermal erosion of the tube floor. Heat is added into the tube by the flux of fresh hot lava, viscous dissipation, and the latent heat of crystallization. Thus the net cooling of the lava is just the difference between the heat lost and the heat input into the tube. The trick is to be able to calculate the heat in and the heat out.
Figure 2. Sketch map showing the Waha'ula tube within the flow field.
Numbers indicate tube sections defined in this study.
The heat lost by conduction is easy to calculate once enough assumptions are made. First, I assumed that the temperature along the tube system is in steady state. That is to say that temperatures change only with distance form the vent, not with time. This seems appropriate for a mature, well established lava tube. Further, I assumed that the lava tube is a cylinder of radius D with a roof of thickness ho. I also assume that the surface temperature above the tube is uniform and that the temperature of the inside of the tube is uniform. This ignores observed temperature variations on the order of 10 deg.C [e.g. Realmuto et al., 1992] but such variations should be negligible when compared to the ~1000 deg.C temperature contrast between the molten lava and the normal surface. The solution to the conduction problem of a hot cylinder in an infinite half-space is
qcond = 2k/cosh-1(2ho/D+1)
where qcond is the heat lost by conduction per unit length of tube, T is the temperature contrast between the lava and the surface and k is the thermal conductivity of the wall rocks [Incropera and DeWitt, 1990].
Experiments done on the cooling of hot wires embedded in permeable media [Cheng, 1985] can be used to estimate the heat lost by air convecting in the cracked walls of the tube. These experiments show that
qconv = 0.565 Ra1/2 T k
where Ra is the Rayleigh number defined by
Ra =air g K T D / ( )
where air is the density of the air, is the coefficient of thermal expansion of air, g is gravitational acceleration, K is permeability, is the viscosity of air and is the thermal diffusivity of the rocks.
The heat lost to boiling away rain can be calculated by
qrain = dw/dt Dw w Lw
where dw/dt is the rate of rainfall, Dw is the diameter across which the rain boils (larger than the diameter of the tube itself), w the density of water, and Lw the latent heat of vaporization of water.
I have also examined the heat lost out of skylights, from the exsolution from the lava of dissolved volatiles, and from thermal erosion and found these processes to be negligible for almost any reasonable conditions.
In principle, calculating viscous dissipation could be rather tricky. However, with the assumption of a steady state flow, the lava should not be accelerating and all the potential energy lost by flowing downhill must be converted into heat via viscous dissipation. Thus
qvisc = m g z/t = Q g z/x
where z/x is the slope and Q is the volumetric flux through the tube and is the density of the lava.
The net heat loss in the tube cools the lava. However, as the lava cools, it crystallizes, releasing latent heat. Latent heat can be taken into account as an additional term in the heat capacity of the lava:
Figure 3. Comparison of the calculated heat lost from the Waha'ula tube
through conduction, rain, and atmospheric convection in the wall rocks using
the nominal values in Table 1. Note that heat loss is smallest in section C
where the tube is going down the steep pali in a single strand. Also note that
in all sections, the three heat loss terms are about equal in magnitude.
qcool+latent = T/x Q r (Cp + L Xc/T)
where T/x is the cooling of the lava down flow, Cp is the heat capacity of the lava, L is the latent heat of crystallization, and Xc/T is the increase in volume fraction of crystals per degree of cooling. Now the entire thermal budget for a lava tube can be written as
qcool+lantent = qcond + qconv + qrain - qvisc.
Thus, in principle, it is possible to calculate T/x if the assumptions I have made hold and appropriate values can be found for all the input parameters.
The best way to test the model thermal budget is to compare it to an actual lava tube. I worked with the Waha'ula tube within the flow field because of the wealth of observations available from it. The Waha'ula tube extended 11 km from the lava lake to the coast near the Waha'ula Heiau (Figure 2). This tube was extensively monitored by the HVO staff via both geophysical measurements and direct sampling from skylights. The observations covered both the changes in time as well as along the length of the tube. The Waha'ula tube formed between 1986 and 1990, with the upper section of the tube system forming in the early part of the eruption. This stable upper section of the tube fed a series of lava tubes further downslope. The Waha'ula tube proper formed in 1990, began to decay in mid-1991, and finally passed away with the end of Episode 48 in early 1992.
The Waha'ula tube was overflown by the NASA C-130 aircraft in 1988, providing multi-spectral thermal infrared images of the entire flow field. The lava tube can be mapped out by tracing the 10-15 deg.C surface temperature anomaly directly over the tube [Realmuto et al., 1992]. These images show a surprisingly complicated picture of the tube system. For much of its length, the Waha'ula tube was actually a braided pair of tubes. Also, the path of the tube system was far from straight, implying that the lava flowed somewhat further than the 11 km straight line distance between and the coastal entries. The variations in the surface temperature anomalies are partially attributable to variations in the thickness of the roof on the lava tubes.
Given the complexity of the tube and the wealth of observations, it is possible to break the tube into 5 segments and select appropriate parameters for each segment. Table 1 lists my best estimates for the key input parameters. Using these values, I compute 11+/-4 deg.C of cooling from the pond to the ocean. This compares very favorably with the 8-10 deg.C of cooling actually observed in the tubes [Helz et al., 1991].
This simple model for the thermal budget of lava tubes seems to explain the cooling in the Waha'ula tube, but what does it tell us about lava tubes in general and long lava flows in particular? Figure 3 shows the relative efficiency of the processes extracting heat from the 5 segments of the lava tube. Other than the heat loss out of skylights (which is negligible), all the terms are roughly similar in magnitude. But is this only true for the case of the Waha'ula lava tube?
Heat loss by conduction is relatively insensitive to the input parameters. Heat loss is greatly enhanced only if the tube roof becomes less than a few tens of centimeters thick. Cooling by the rain, on the other hand, is completely governed by the local climate. My model would predict that lava tubes on the dry side of Hawai'i should be 10-20% more efficient than the Waha'ula tube. The heat loss by convecting air is controlled by the permeability of the wall rocks. This permeability is formed by cooling fractures and is lost as secondary minerals fill in the smaller cracks. Thus permeability can vary greatly even along a single lava tube. This variability in permeability is the single largest source of the uncertainty in my thermal model. If the permeability of the rocks surrounding a lava tube is significantly lower than my nominal estimate, the tube's efficiency could increase by 20-30%.
SEGMENT A B C D E # Branches 1 2 1 2 3 Length (km) 5 1.5 2 1.5 2 Diameter (m) 41 21 31 21 10.5 Roof Thickness (m) 10.5 10.5 1.50.5 21 31 Skylights (#/km) 0.60.2 0.50.2 0.50.5 00 00 Slope (%) 5.1 6.1 10.7 7.7 1.8 Rainfall (cm/year) 30050 25050 22550 22550 20050 100deg.C Isotherm 3010 205 3010 205 155 diameter (m) T/x (C/km) 0.92 1.2 0.44 0.93 1.2 Cumulative cooling (C) 4.6 6.4 7.3 8.7 11Thus favorable conditions might combine to make terrestrial lava tubes up to 50% more efficient than the Waha'ula tube. At first glance, this would still seem to preclude lava flows much over 100 km long. However, this is not the case. I have so far only discussed how to reduce the heat loss from the tube. We also need to examine how much lava this heat is being extracted from. Throughout most of its life, the Waha'ula tube had lava running only about knee-deep even though the tube was meters in diameter. This is exceedingly inefficient. In fact, the Waha'ula tube could have carried more than 10 times the volume flux of lava that it actually did.
If the volume flux of lava in the tube is increased, heat loss does not change, but that same heat is extracted from more lava. If the Waha'ula tube was completely full, the cooling of the lava inside the tube would have been less than 0.1 deg.C/km and a 500 km long tube would seem possible. Of course, it is not quite so simple. If the Waha'ula tube ran full, it would have had enhanced thermal erosion which would have enlarged the tube, possibly making it less thermally efficient. And it is not certain that a lava tube system similar to the Waha'ula tube would have formed if would have been putting out 30-40 m3/s of lava.
Still, the basic conclusion holds: tube-fed lava flows hundreds of kilometers long are possible given effusion rates only marginally greater than the current eruption.
Cheng, P. (1985) Natural Convection in a Porous Medium: External Flows, in Kakac, S., W. Aung, and R. Viskanta (editors) Natural Convection: Fundamentals and Applications. Springer-Verlag, Berlin, 1181 pp.
Helz, R. T., C. Heliker, M. Mangan, K. Hon, C. A. Neal, and L. Simmons (1991) Thermal History of the Current Kilauean East Rift Eruption. AGU Fall Meeting Abstracts, EOS, 72: 557.
Incropera, F. P., and D. P. DeWitt (1990) Fundamentals of Heat and Mass Transfer, Wiley, New York, 919 pp.
Realmuto, V. J., K. Hon, A. B. Kahle, E. A. Abbott, and D.C. Pieri (1992) Multispectral Thermal Infrared Mapping of the 1 October 1988 Kupaianaha flow field, Kilauea Volcano, Hawaii. Bull. Volc., 55:33-44.
reprinted from the Hawaii Center for Volcanology Newsletter, Volume 2, Number 1, December 1994
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