From the expression we found in the last note, we can see that the sap flow speed is an increasing function of leaf size \(l\). However, the gains in flow speed come at increasing cost since \(v_{sap}\) is a saturating function of \(l\). As we continue to increase leaf length \(l\), we get less and less of an increase in flow speed. In fact, in order to reach the maximum possible flow rate, we'd need to construct infinitely large leaves.

At the same time, building and maintaining leaves itself costs energy which must be taken into account when transforming sap flow speed into energy flow. The energy required to create and maintain larger leaves will, at some point, outweigh the benefits of a faster flow of energy in the form of sucrose.

We therefore expect to find a balance point \(l^*\) at which we maximize the flow of useful energy (energy from sucrose flow minus energy devoted to leaf upkeep) for a given height \(h\). This leaf size will provide the maximal flow rate of sap as a function of leaf length \(l\) and tree height \(h\).

Let's model the energy flow required for leaf maintenance as a linear function of the leaf size: \(\dot{E}_{leaf}(l) = \gamma_{leaf}l \).

Suppose that the tree can extract the energy \(E_{suc}\) from one mole of sucrose and that the concentration of sucrose in the tree sap is \(c_{suc}\) moles per liter. We can subtract the maintenance energy flux from the production energy flux to get the effective rate of energy production.

\[\begin{align} J_E(h,l) &= J_{production} - J_{maintenance} \\ &=E_{suc}c_{suc}v_{sap} - \gamma_{leaf}l \\ &= E_{suc}c_{suc}\frac{2 \Delta p L_p R^4\pi l}{R^3 + 16 h \eta l}- \gamma_{leaf}l \\ \end{align}\]

It is this quantity that captures the useful energy that the leaf supplies to the rest of the tree.

What is the maximum possible leaf size (in **cm**) in a tree that is 35 meters tall?

Hint: At what leaf size, \(l_{max}\), does the energy supplied by a leaf reach a maximum?

- \(\gamma_{leaf}=10^{-4}\)
**J/s/m**. - \(c_{suc}\) = 0.4
**M** - \(E_{suc}\) = 5800
**kJ/mole** - \(r=20\times 10^{-6}\)
**m** - \(\eta = 5\times 10^{-3}\)
**Pa s** - \(\Delta p= 10^6\)
**Pa**

Illustration by Maxicat Rhododendron

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