Year 1 physics: Fluid Dynamics with Tree Sap replacing Water

Click For Summary

Homework Help Overview

The discussion centers around a fluid dynamics problem involving the upward flow of tree sap in response to water loss through transpiration. The original poster presents a scenario where tree sap replaces water lost at a specified rate, and they seek to determine the upward speed of sap in the vessels of the tree trunk.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to convert the water loss rate into a volumetric flow rate and calculates the volume of sap needed per vessel. Some participants question whether the replacement of water is by weight or volume, noting the relevance of sap density in this context. Others suggest considering the relationship between flow speed and vessel diameter.

Discussion Status

The discussion is ongoing, with participants providing comments and questions that encourage deeper thinking about the problem. The original poster acknowledges the need to reconsider their approach based on feedback received.

Contextual Notes

There is uncertainty regarding the interpretation of the problem, particularly concerning whether the sap replacement should be calculated based on weight or volume. The original poster expresses confusion about how to proceed from their current calculations.

ConquestAce
Messages
4
Reaction score
0

Homework Statement


A tree loses water to the air by the process of transpiration at the rate of 110 g/h. This water is replaced by the upward flow of sap through vessels in the trunk.

If the trunk contains 1900 vessels, each 100 μm in diameter, what is the upward speed of the sap in each vessel? The density of tree sap is 1040 kg/m3.

Known :
110g/hr water loss
d = 100μm
D = 1040kg/m^3

Homework Equations


berulinni's equation and contintuty equation :
p + 0.5*rho*v^2+rho*g*y = c
(v_1)(A_1) = (v_2)(A_2)

The Attempt at a Solution


Initially I converted the water loss to kg/s which was (110g/hr)*(1hr/3600seconds) and since 1 kg of water = 1L of water the water loss to be replaced by sap rate was 3.055*10^-5 m^3/s

Then calculated the volume of the sap to be 3.054x10-8 m^3 using Vsap=Vwater

Divided by vessels to get 1.60737*10^-11 m^3

And from now on, I only have uncertainties:
and a lot of possibilies but not sure how I can proceed.
 
Physics news on Phys.org
First of all, it is not clear from the question whether the water is replaced by weight or by volume. The fact that you are given the density of the sap suggests replacement by weight, although it will not matter much since the density is very similar to that of water.

Second, given that you have computed how much sap must flow through one vessel, how fast must it flow depending on the vessel diameter?
 
I am still stuck on this question if anyone could help out.
 
You have already gotten relevant comments, including a direct question. If you cannot answer this question, please state your thoughts on the matter and why this is not clear to you. If you are looking for someone to solve the problem for you you have come to the wrong place.
 
You're right, I didn't want to think about it. I will reattempt the question properly with your post in mind.
 

Similar threads

Undergrad Calculating Water Pool Depth: Factors to Consider in Fluid Dynamics
  • · Replies 13 ·
Replies
13
Views
2K