#6 The optimal concentration of nectar-I

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There are nectar-feeding species in bees, butterflies, moths, ants, hummingbirds, honeyeaters, sunbirds, and bats. The more concentrated the nectar is, the more energy per unit volume it contains. However, the nectar would also become more viscous and therefore compromise the maximum flow rate. Does this phenomenon place any limitation on nectar-feeding species? How would animals utilizing different nectar-feeding mechanisms be affected differently? That’s what we are going to discuss in these serial episodes and also the research problem that Kim, Gilet and Bush were interested in.

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The authors of the suggested reading consider 3 different nectar-feeding mechanisms. They are

(1) Active suction utilized by butterflies, moths, and some bees and ants. These animals produce negative pressure in their oral cavity to suck in the nectar.

(2) Capillary suction employed by hummingbirds, honeyeaters, and sunbirds. These birds intake nectar by capillary flow in the hollow tube formed by their tongue.

(3) Viscous dipping exploited by bats, some bees and ants. They place their tongue into the nectar and rapidly retract it back to their mouth.

In these series, we will talk about active suction in episode I, capillary suction in episode II, viscous dipping in episode III, and have extra explanation about LLD theory in episode IV.

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Let’s consider the physics mechanism of active suction. Assume the mouthpiece of butterfly a hollow tube with inner radius a, tube length L. Assume nectar density ρ, viscosity μ. Assume mean flow velocity in the tube u. Assume the negative pressure produced by butterflies ΔP. According to the Newton’s second law, we could write down the following for the nectar in the tube:

Inertial force = Pressure difference X area + Gravitational force + Viscous force

or
$(\pi a^2L\rho) \textup{d}u/\textup{d}t= \Delta P(\pi a^2) - (\pi a^2L\rho)g – 8\pi\mu uL$

The viscous force might be a little bit unfamiliar. Please refer to the end of this episode. There are 4 terms in the above equation and it looks terrible, so we are going to remove some terms by physics reasoning.

How much negative pressure could be produced by butterflies and moths? Biophysicists believe the work producible by unit volume of muscles is a constant and therefore the force producible by muscle is proportional to its cross-section area. (This is deducible from scaling laws. Please refer to our previous articles: Scaling law of locomotion-I：https://goo.gl/2olyQH & Scaling law of locomotion-II：https://goo.gl/61l7vv) Consequently, if it is possible to define a characteristic length L for a creature, the force producible by it is F ~ L^2. The pressure it could achieve is therefore
$P \sim F/L^2 \sim L^0$

That is to say all creatures could produce approximately the same negative suction pressure and that was validate experimentally. The maximum negative pressure produced by human and mosquitos are both around 10kPa.

What about the gravitational force? The length of their mouthpiece is about 1cm. Consider the ratio of gravitational force and pressure force:

$\frac{\pi a^2 L\rho g}{\Delta P \pi a^2}=\frac{\rho gL}{\Delta P}\sim 0.01$

It is small enough compared with the pressure force, so we discard it from our equation.

And finally let’s consider the ratio between the inertial force and viscous force (This is actually the well-known Reynolds number introduced by Stokes in 1851):
$\frac{(\pi a^2L\rho)\textup{d} u/ \textup{d}t}{8\pi\mu uL} = \frac{(\rho a^2)\textup{d}u/\textup{d}t }{8\mu u }= \textup{Re}$
To estimate the term du/dt, assume the suction frequency of butterflies f, which is experimentally observed to be 10 Hz. The acceleration of nectar in the mouthpiece could then be estimated as
$\textup{d} u/ \textup{d}t \sim u/\Delta t\sim uf$

So the Reynold number could now be formulated as
$\textup{Re} = \frac{\rho a^2uf}{8\mu u }\sim\frac{\rho a^2 f}{\mu}$
The inner radius of the mouthpiece of butterflies is about 100μm. Substitute the real scale of these variables back and we will get Re ~ 0.1. The inertial force is rather small so we discard it, too.

Finally the equation becomes
$0 = +\Delta P(\pi a^2) - 8\pi\mu uL$
That is just Poiseuille flow.

Ultimately we would like to know the relations between the flow rate and viscosity, for they are most easily observed experimentally. A naïve way to deduce the relations would be done by assuming the suction pressure ΔP a constant. Then
$Q = (\pi a^2)u = \Delta P(\pi a^4)/8\mu L \propto \mu^{-1}$
Nevertheless, the experimentally observed power law is
$Q \propto \mu^{-0.5}$
What shall we do? Another possible hypothesis proposed by Pivnick is that the power during active suction is a constant. That is to say
$\dot{W}= Q\Delta P =\Delta P^2(\pi a^4)/8\mu L = \textup{Constant}$
Because
$\Delta P =8\mu LQ/(\pi a^4)$
We now arrive at
$\dot{W} = Q^2(8\mu L)/(\pi a^4) = \textup{Constant}$
So
$Q \propto \mu^{-0.5}$
And that is the correct power law observed experimentally.

/*Supplementary: Poiseuille flow */

In 17^th century, Newton quantitatively discuss the viscosity of fluid first. He believed that there would be viscous force whenever there is a velocity gradient between fluid layers. And the viscous stress τ (the viscous force per unit area) was “assumed” to be

$\tau = \mu\frac{\textup{d}u}{\textup{d}y}$
The fluid following this assumption is said to be Newtonian. Those violate this assumption is termed non-Newtonian instead.

That’s then consider a hollow tube streaming Newtonian fluid. Assume the flow is driven by the pressure difference ΔP, the tube length L, inner radius R, fluid density ρ, viscosity μ. Consider the “fluid column” with radial distance less than r. At steady state, the pressure force will balance out the viscous force. We could then write down that:
$-\Delta P(\pi r^2) = \tau (2\pi rL) =\mu \frac{\textup{d}u(r)}{\textup{d}r}(2\pi rL)$
or
$\frac{\textup{d}u(r)}{\textup{d}r}= \frac{-\Delta P r}{2 \mu L}$
Integrate it with the no-slip boundary condition (flow velocity = 0 at r = R) and we will get
$u(r) = \Delta P(R^2 - r^2)/(4\mu L)$
The velocity profile is quadratic and it has its maximum value at the central of the tube.

After knowing the velocity profile, it is easy to get the volumetric flow rate Q: $Q=\int_{r=0}^{R}u(r)(2\pi r)\textup{d}r = \int_{r=0}^{R}\frac{\Delta P(R^2 - r^2)}{4\mu L}(2\pi r)\textup{d}r =\frac{\pi\Delta PR^4}{8\mu L}$
And this is the well-known Poiseuille’s law for incompressible fluid.

Often we would love to express the flow rate in terms of mean velocity u by assuming
$Q=\pi R^2 u$

Substitute it into Poiseuille law and we will get

$u =\frac{\Delta P R^2}{8\mu L}$
Because the above equation comes from the balance between viscous force and pressure force, the viscous force could be estimated as
$F_{\textup{viscous}} = -\Delta P(\pi R^2) = -(8u\mu L/R^2)(\pi R^2) = -8\pi u\mu L$

As what we stated previously.

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*Suggested reading:

Kim, W., Gilet, T. and Bush, J. W. M. (2011). Optimal concentrations in nectar feeding. PNAS 108(40): 16618-16621.

‪#‎biophysics ‪#‎3minBiophysics ‪#‎生物物理 ‪#‎三分鐘生物物理
‪#‎fluid ‪#‎mechanics ‪#‎Poiseuille

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