# Treeometry – An Exercise In BID Analysis When you and physics disagree, physics wins. Biologically inspired design (BID) can provide tremendous innovation in product design and development, but not every instance of bio-inspiration is destined for success. Even when the biology appears to align closely with a design challenge, it may not yield an effective solution. It all comes down to the math. Without at least a basic quantitative analysis to ensure feasibility, bio-inspired design is just bio-inspired dreaming – worse yet, frivolous development may lead to bio-inspired debt.

Consider this example: trees can grow very tall. In fact, giant redwoods can tower over 100 m. Still, they manage to transport water from their roots to their leaves without any massive turbine-based pump. They rely on a combination of capillary pressure and evapotranspiration to do the work for them. Genius! Why don’t we use this to get water to the rooms of hotels and high-rises? We’ve just saved a bunch of energy! But is it really that simple?

First, let’s consider the underlying principle of capillary pressure, specifically Laplace pressure inside of a tube. The Laplace pressure arises at curved fluid interfaces as a result of the energy required to form such a surface. Consider a spherical droplet of water in air. Let the surface tension at the interface be γ (roughly 0.0728 at 20ºC). The total surface energy is 4πr2γ, and therefore the pressure differential necessary to balance that is: PL is the Laplace pressure, V is the droplet volume, and r is the radius of curvature of the surface. The pressure is, by convention, positive when the water surface is convex relative to the air (droplet) and negative when the surface is concave relative to air (bubble), as indicated in Figure 1. Figure 1. Depiction of Laplace pressure orientation.

In a narrow cylindrical tube, the liquid-air interface will form a portion of a spherical surface to minimize the interfacial surface area for any given contact angle; this assumes a scale small enough ignore deformation due to gravity. The contact angle denotes the angle of the liquid-air boundary at the point of contact with the solid – in this case the wall of the tube. This angle results from a balancing of forces due to the three competing surface energies: liquid-solid, liquid-air, solid-air. Given a contact angle θ, the radius of curvature, rcurv, is: It is important to remember that the pressure differential is the initial pressure minus the final pressure; hence flow occurs from high pressure to low pressure, and a negative Laplace pressure encourages liquid to flow toward the liquid-air interface. When the Laplace pressure is negative, and one end of our cylindrical tube is placed in a liquid source (assuming infinite supply), the liquid will get pulled through to the other end of the tube. The effect, however, is limited by gravity in the vertical direction. The pressure contribution due to gravity is ρgh, where ρ is the liquid density (~1000 kg/m3 for water), g is gravitational acceleration, and h is the height above the source. The resulting differential pressure, ΔP, is: Now, let’s look at flow rate. For our operating conditions, we can assume laminar flow, and the guiding metric here is the Hagen-Poiseuille equation. This equation relates laminar volumetric flow rate with the pressure differential over the length, L, for a cylindrical tube of a given radius, rtube, and a fluid with some dynamic viscosity, μ: For a derivation, check out this Wikipedia article.

In the ideal case, we are removing water from the end of the tube quickly enough to achieve the above flow rate. If not, the water pressure will balance out at the end of the tube and the flow will stop. To simplify, let us assume the tube length and height are effectively equivalent (completely vertical). The opposing radial terms (r3tube and -r4tube ) indicate that the flow rate can be maximized for any given height.

To maximize flow with respect to radius, set the relevant derivative equal to zero: Now comes the fun part: what are some reasonable parameters to throw at this equation?

A reasonable advancing contact angle (θ during forward flow) might be 30º, and the we can take the height of the building to be 100 m. Plugging in the other values at 20ºC, we get: That’s a tremendously small flow rate. I mean, if you were sipping your 16-oz. soda through a straw at that rate, it would take you 45 billion years to finish – the universe isn’t even that old! Still, for a complete picture, we need to consider the cross-sectional area of this flow: This is also tiny: there are viruses with larger cross-sections. The quotient of the two values yields the average fluid velocity: While this value is less small, the velocity still amounts to the proverbial molasses in winter. In order to support even one person in the U.S. with a typical water consumption of 100 gallons/day (about 4.38 · 10-6 m3/s), the total fluid cross-section would have to be: This is over 4000 square feet – a nicely sized home – and does not even account for the thickness of the tube walls. Trees must do better than this. What are we missing?

The previous calculations presumed a set of parallel tubes with constant radius. I know I never provided an alternative, but did you stop to consider that there might be? In fact, had we considered tree biology more carefully, we might have expected our lackluster result. The channels in trees actually branch and get narrower as they stretch from roots to canopy. This is a learning moment: a superficial understanding of a biological system can lead to significant design oversights.

The optimization math gets pretty ugly here, so rather than attempting to find an analytic solution, let’s take a more pragmatic approach. Let’s assume that we only care about the flow rate near the top of the tubes; for everything else we only need to ensure that filling would occur. This means that the pressure at earlier points need only counter gravity. Let l be the position along the tube: Let’s still ignore wall thickness for simplicity. While we could probably have the flow area vary, it might be nice to maintain a more or less constant footprint from floor to floor in our high-rise. So let’s make the flow cross-section constant. Let A be the flow area and ntube be the number of tubes at a given radius: Flow rate and pressure can be analogized with current and voltage in an electric circuit, respectively. There, current is equal to the voltage divided by the resistance. Here, we take Ω to be the resistance to fluid flow: Note that the final resistance is over the entire cross-section of the number of tubes; hence, it appears in the equation (ntube).

We want to vary the radius over the length, so we must consider the differential resistance, and integrate over the entire length. We’ll apply the maximum radius determined earlier: This is the minimal flow resistance we can achieve with a constant cross-section, while permitting fluid transport along the entire length of the tube system.

Now, we need to maximize flow rate for the system already filled with water. We want to tack on a small section that provides the capillary pressure we need to maximize flow rate. However, we have to ensure that the tube radius changes slowly enough not to undermine our flow assumptions. To that end, let’s take the additional length, lΔ, to be at least 100 times the difference between the radius at the beginning of lΔ, rL, and the radius at the end of lΔ, rL+lΔ: Let’s further assume that the radius changes linearly over lΔ: Let’s define the extra resistance, ΩΔ, as a fraction, f, of the rest of the resistance, ΩL: Note that, for f = 0, there is no flow, and as f goes to infinity, flow increases towards a constant value. This is the only term that remains: This implies that flow is maximized when the end tube radius, rL+lΔ, is 0. This is unphysical, but tells us that we want to make the final radius as small as possible, while still allowing the fluid to flow according to the equations we are applying. Water molecules have a diameter of about 0.275 nm. Let’s say we want the tube to be at least 100 water molecules thick; this means a minimum radius of 50 water diameters or about 13.8 nm.

What does this mean for f? For a primary length, L, of 100 m: That is definitely small; more than twice that fraction of people get struck by lightning each year in the U.S. Now, the cross-section needed to get 100 gallons/day (or 4.38 · 10-6 m3/s) is: For a 100 m redwood with a 4 m radius, we get: This is actually on the order of what one finds online for the water uptake of redwoods. Trees have constraints that we don’t have to worry about, so one might expect our idealized results to far surpass their uptake. However, redwoods are not beholden to idealized mathematical constructions. The xylem may achieve effective pore diameters below our chosen target; the Hagen-Poiseuille equation may not even apply under the conditions near the xylem terminus, possibly providing for a higher flow rate. Also, the fluid in the xylem can be doped with salts, polysaccharides, and proteins to achieve higher surface tensions (although this will likely increase viscosity, as well). Finally, redwoods can use other tricks to generate flow, like osmotic gradients — but we’ll leave that discussion for another day.

Unfortunately, matching the redwood isn’t good enough for us. We would need at least one redwood-sized volume for every 6 people, which amounts to a small forest for even a single 32-floor high-rise.

So would this system ever be useful? Let’s consider the results for a shorter building. If we reduce the height to 25 m (about 8 floors): This flow rate is almost 100 times faster than the previous one. This might actually be viable. 8 floors with 16 households per floor and an average of 2 people per household yields 256 people. This would require 1.12 · 10-6 m3/s of water, which amounts to an area of 31.6 m2. This is roughly an 18 ft. by 18 ft. square rising through every floor. While not a huge footprint, it’s by no means negligible — and manufacturing constraints make the 13.8 nm pore target a bit unrealistic, anyway. If we instead limit pore size to 1 micron and reduce our aspirations to a 3 story (10 meter) house: For a family of 4: This is a small room’s worth of area on each floor.

Takeaway: the mechanism could be employed, but it imposes rather stringent constraints and provides little practical value. For the most reasonable application investigated (a 3-story home) there is no notable advantage over the water pressure already provided by the municipality. Moreover, we have yet to determine how to get water out of the structure fast enough to maximize flow. Trees can actually offer more insight there, but we’d have employ a different set of principles from those currently being discussed.