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 2000 vessels, each 100 μm in diameter, what is the upward speed in mm/s of the sap in each vessel? The density of tree sap is 1040 kg/m³.

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Hey, everyone. Here's a problem that asks about flow rate in regards to capillary action. So capillary action causes water to rise in a tube due to adhesive forces between water and the walls of the tube. Suppose a capillary action pump moves water through a compound pipe with a net flow of 50 g per hour. So the first given the problem tells us is that the net flow rate is 50 g per hour. So because the units are grams per hour with grams in the numerator, that means that this is something called mass flow rate, which is usually written as M dot So the rate of change of the mass of the fluid. So the flow of the fluids mass and it's given as 50 g per hour. The problem goes on if the compound pipe has 3500 bores. So I'm gonna say N equals 3500 for the number of bores which holes basically and each bore has a diameter of 600 micrometers, then find the water speed in the bores. So I'm gonna say the water speed is V and that's what we're trying to find. The problem. Specifies that we're looking for that speed in units of millimeters per second. The problem also tells us to assume that the water has a density of 1010 kg per cubic meter and this is a multiple choice problem. So we have four options here and notice that all four options have units of millimeters per second, which makes sense because the problem asks us to find the answer with those units. So this is a problem about flow rate. So we want to find some connection between the speed, the water speed that we're trying to find. And the mass flow rate which we're given now one way we can do this is to connect mass flow rate to the volume flow rate. Since volume flow rate, we have a pretty basic formula for that using the continuity equation. We know from the continuity equation that the volume flow rate which is usually written as Q is equal to the speed of the flow times the area through which it travels. So va but we need to find some connection between Q and M dot And there is a formula for this which states that M dot The mass flow rate is equal to the density of the fluid times the volume flow rate. Now you may or may not have seen this formula before if you haven't seen this formula before and it confuses you. There actually is an intuitive reason for it because if you were to solve this formula for a row. Then you'll notice that row in terms of this mass flow rate formula is equal to M dot The mass flow rate divided by the volume flow rate, which you might notice is basically just another way of stating the typical density formula that density is equal to mass over volume. So that's where this formula intuitively comes from. But since we're given density but not given the volume flow rate, let's expand this out a little bit more. So as we mentioned earlier, the volume flow rate, Q is equal to V times A as per the continuity equation. So row times V times A and M dot is given density is given V is what we want to find. So A is the last thing we need to expand out. We're not given the surface area of any of the bores. But since we're told about a diameter of them, we we can assume that they are circular holes and recall that the formula for the area of a circle is equal to pi times the square of the circle's rating. Yes, but don't get excited, don't get excited and just put pi R squared in for a because there's something else we should consider. And it's the fact that there are many holes, there are multiple bores, there are specifically 3500 of them as given by the problem. So that means that the area through which the fluid is flowing is not just the area of one bore but the area of many bores. So instead of subbing pi R squared in for A, instead I am going to sub in for A N times pi R squared because the area through which the water flows is proportional to the number of those bores. And so that's what we have to consider. That's the last thing you have to consider with that, with this expression. But fortunately for us, now looking at this problem, uh pretty much everything here is known except for V what we're trying to find, we technically don't have R either, but we are given the diameter of the bores. So recall that radius is equal to diameter divided by two a diameter is equal to twice the radius. So if you want to find the radius of the bore, we just need to take 600 microns and divide it in two. So the radius then is just going to be equal to microns. So now all that's left for us to do is take this formula and solve it for V by dividing both sides of the equation by row N pi R squared, dividing that by both sides. And we find a formula stating that the water speed is equal to and the dot the mass flow rate divided by the density of the fluid times N the number of bores times pi times the square of the radius. And So now we've just got to take the things given to us in the problem and plug them into our variables in this equation though, note that our units are not very consistent in order for this equation to work, we need all of our units to be properly aligned. So in other words, like for example, the mass flow rate is given to us in grams per hour. But since we know that we're looking for an answer with seconds, we're gonna want to convert from hours to seconds. And since we have a a density with kilograms, we're gonna want to convert from grams to kilograms. So I'm going to write 50 g per hour. And then we're gonna use our trusty old chain link conversion to convert from grams into kilograms. So I'm gonna multiply by one kg over 1000 g and also multiply by one hour over 3600 seconds and divide the whole thing by 10, 10 kilograms per cubic meter times an end value of 3500 times pi and then lastly times the square of 300 microns, which again, we wanna make our units consistent. So instead of writing 300 microns, I'm gonna write 300 times 10 to the power of negative six m. Since that's what will happen if we convert from microns to meters and then don't forget to square it. And if we were to put this into our calculator, we'll find a water speed of about 1. times 10 to the power of negative five m per second. But note that the problem asks us to find this in millimeters per second. So we're gonna do one final conversion on this to convert from meters into millimeters. So a final unit conversion goes millimeters are in one m. And so if you put that your calculator, then finally, we find 1.39 times 10 to the power of negative two millimeters per second. And that is our final answer. And if we were to look above at the problem, we'll notice that we do indeed have that as an option. It is option A. So the answer to this problem is part a 1. times 10 to the power of negative two millimeters per second. And that is it for this problem. I hope this video helped you out if you'd like to learn more. Uh Please view some of our other tutoring videos and you'll find other practice problems that will hopefully get you a a better, more confidence with these types of problems. And that's it. That's how you would solve this problem. Bye bye.

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Key Concepts

Transpiration

Fluid Dynamics

Cross-sectional Area