 A necked-down section in a pipe flow, called a venturi, develops a low-throat pressure that can aspirate fluid upward from a reservoir as shown in the figure below. Consider a venturi used to pull gasoline, with a specific gravity of 0.72, up from a fuel tank. The fuel in the tank is exposed to atmospheric pressure, and the air entering the venturi is traveling at 3 meters per second through a 5 centimeter diameter tube. The fuel needs to be pulled up a 10 centimeter section of hose measured from the fuel's surface. Wet-throat diameter would allow for the fuel to barely be pulled into the airstream. Furthermore, does this represent a maximum viable diameter or a minimum viable diameter? Let's begin with a few assumptions. First of all, I recognize that I have air flowing in, and then a converging cross-sectional area that then diverges, and I'm going to assume that states 1, 2, and 3 are all connected by a streamline here. So the air and then the air and fuel mixture is all flowing along a path that we are analyzing, and that path is parallel to the velocity of the air at every point. That streamline would essentially be in the center, but it could also be something like this as well. It doesn't really matter. We're just saying that states 1, 2, and 3 are all along the same streamline. Then I'm going to assume that I have uniform flow at states 1, 2, and 3, or you could also think of it like the velocity that we have is an average, and that the flow profile doesn't affect the operation of the device. Next, I'm going to assume that this height of fluid here is negligibly small, or rather, perhaps, that this distance to this distance is what's actually being described with the 10 centimeter length. Whatever the case, we are using the distance required for the fluid to be pulled up of 10 centimeters. Then I'm going to assume incompressible flow, and then steady state, and then that this flow is ideal, meaning there's, among other things, no losses due to friction. I recognize that this is a conservation of energy problem because we are talking about the enthalpy of the fluid being converted into kinetic energy as a result of the change in cross-sectional area. The increase in kinetic energy required to accomplish the same mass flow rate in a smaller cross-sectional area has to come from somewhere. It comes from the enthalpy of the fluid. As the enthalpy of the fluid decreases, the effective pressure of the fluid decreases, and the difference in pressure now between state 1 and the surface of the fuel is going to be what pushes the fuel up the hose. I will add to this that P1 is assumed to be about the same as atmospheric pressure. Then whatever the pressure decrease is between 1 and 2 is also going to represent the required pressure difference to get up the 10 centimeter height. This pressure difference required to push the fluid up the 10 centimeters is described with density, acceleration, and height, because we are talking about a pressure difference caused by a height of a column of fluid. And if the atmospheric pressure is higher than the pressure at state 2, it's going to push the fluid up that 10 centimeters. So here, that's going to be the density of gasoline times what we assume as standard gravity times 10 centimeters. I will add that to my assumptions as well. So the change in pressure across that 10 centimeter height is something that we can calculate. We have enough information to determine the density of the gas because we know the specific gravity of the gas and the specific gravity of the gas is going to be the density of the gas divided by the density of water at standard temperature and pressure, which we can look up from table A1 or table A3 as 998 kilograms per cubic meter. Then density of gas would just be specific gravity of gas times density of water. Then the delta P here is going to be specific gravity of gas times the density of water times gravity times 10 centimeters. And I know all four of those quantities. So I could compute a number, but I'm going to leave it symbolically for now. That pressure difference has to come from a decrease in enthalpy. So what we need to do is figure out how much kinetic energy increase is required to decrease the enthalpy enough to cause a pressure difference that is enough to push fuel up a 10 centimeter hose. So for that, we are going to use our conservation of energy and furthermore, because we recognize that we have such a simplified situation, we don't even need to use the full conservation of energy. We can jump to Bernoulli's equation, which remember is a simplification of the conservation of energy. The assumptions required to be able to apply Bernoulli's equation are steady flow without any friction along a streamline, that that flow is incompressible and that there are no opportunities for worker heat transfer. We meet all six of those requirements, which means that we can deploy this equation and describe those quantities at all three state points relative to some constant. And we don't actually care what the constant is. We just care that we can write p1 over density plus v1 squared over 2 plus gz1 as being equal to p2 over rho2 plus v2 squared over 2 plus gz2 equal to p3 over rho3 plus v3 squared over 2 plus gz3. Because I have incompressible flow, all three of these densities are the same. And it looks to me like this is approximately horizontal. So if we neglect changes in elevation, what we are neglecting is any changes in the potential energy. So all three of these terms disappear. Then I could say p1 over rho plus v1 squared over 2 is equal to p2 over rho plus v2 squared over 2 is equal to p3 over rho plus v3 squared over 2. And I don't actually care about state three. All I really care about is how much the pressure decreases from one to two and what change in velocity is required to do that. So I'm going to say p1 minus p2 over rho is equal to v2 squared minus v1 squared over 2. And this delta p is going to be the difference between p1 and p2. And remember, p1 was assumed to be about the same as atmospheric pressure, which means p1 minus p2 is the same as this delta p that is pushing the fluid up the hose. Therefore, I'm going to be saying the specific gravity of the gasoline times the density of water at standard temperature and pressure times gravity times 10 centimeters divided by density is equal to v2 squared minus v1 squared divided by 2. In that relationship, I will know everything except for v2, which I can solve for. I know v1 because it was given. I know the specific gravity of the gasoline because it was given. I know the density of water at standard temperature and pressure because we looked it up. I know gravity because we assumed it. I know 10 centimeters because I was told it was 10 centimeters. And I know this density because this density is the density of air. It is the density of the stream that we are describing from 1 to 2. I know this density is going to be at about standard temperature and pressure as well because we weren't given enough information in the problem to know anything else about the operation of this device. We don't know if the temperature of the incoming air is going to be particularly high or particularly low. And we assume that if we weren't told otherwise, it's about room temperature. The logic here is that if this is actually the fuel going into an engine, then this is going to be fresh air pulled from outside. And we use standard temperature for that anyway. So we're assuming the density of air is at standard temperature and pressure, which I will write under my assumptions. And that standard temperature and pressure is also the same as the assumption that we made earlier, that p1 was at atmospheric pressure. So half of that assumption was already in place. So in reality, this is only adding the standard temperature into the assumptions. So for the density of air, we will jump into our tables. And specifically what we want is going to be table A2 or A4. A2 is the density and viscosity of air at one atmosphere, which is what we have. So we can grab standard temperature, which is about 20 degrees Celsius, and use a density of 1.2 kilograms per cubic meter. Alternatively, if we were to jump to table A4, we have the properties of a variety of gases at standard temperature and pressure, of which one is dry air. And we could grab the specific weight of that substance here if we want it. But since what we need is density anyway, we can lug in 1.2 kilograms per cubic meter as per table A2. Then I know everything else, so I can solve for V2. And since what I want for this velocity is going to eventually be a quantity in meters per second, I should get the quantity inside of the parentheses to be in meters squared per second squared. That way it simplifies to meters per second when I take the square root. But before we get caught up in that calculation, let me just pause for a second to point out, why are we calculating velocity? I mean, that's not what the problem asked for. The problem asked for a diameter. So how do we relate diameter to velocity here? That's going to come from our conservation of mass. For this to be flowing steadily, we must have the same mass flow rate at state one and state two. And remember that we can describe mass flow rate as the density of a fluid times its average velocity times the cross-sectional area through which it is flowing. m dot one is going to equal m dot two, and the densities are the same because we have the same air at one and two, and we are assuming everything is an incompressible fluid, including air. So the densities will cancel and we will be able to write average velocity at state one times area one is equal to average velocity two times area two. And since we are assuming uniform flow here, which by the way, did I ever write that down? I didn't. So uniform flow. Then the average velocity at state one is the same as v one. And the average velocity at state two is the same as v two. So I'm able to say v one times a one is equal to v two times a two. That's from our conservation of mass. I have a circular cross-sectional area here, so I can write this as v one times pi over four times diameter one squared is equal to v two times pi over two, excuse me, pi over four times d two squared. And then I would be able to write diameter two as being the square root of d one squared times v one over v two. I will bring the d one out and just write this as d one times the square root of v one over v two. So the actual answer that I'm going to want for this problem is going to be taking this entire quantity down here, putting it into the denominator below v one, which we know, taking the entire square root of that quantity and then multiplying it by diameter at state one. Here, let me pull this down so that it can hopefully be legible or followable to someone who is just looking at this document. And then let's open up a new work page. Our two relevant equations right now are this. And of course, if you wanted to, you could calculate these separately. But I prefer to work as far as I can symbolically, as you guys know. So I'm actually going to keep this train going and make the substitution for v two so that we are eventually left with only one calculation to perform here. I should really just start over d one times v one over v two to the one half hour. And then d two would be this quantity in the denominator of v one, that entire quantity taken to the one half power all multiplied by d one. And since v one is in meters per second, if I get the denominator in meters per second, then the units will cancel and I will have the square root of a unit on this quantity, which is what I want. And then whatever units I put in for d one is what I will get out for d two. The answer asked for an answer in no particular unit. So I will answer in centimeters. And because we had a five centimeter diameter first day one, then I'm just going to plug in centimeters for D one as well. Well, let's get to calculating. Remember that I want meter squared per second squared when I take this first square root, so that I will have meters per second, canceling the meters per second and v one. Therefore I want this first quantity to be in meter squared per second squared. So I am going to be looking for meters squared per second squared. I recognize that I have cubic meters canceling cubic meters and kilograms canceling kilograms therefore to get into meters squared per second squared, I just have to convert centimeters into meters and there are 100 centimeters in one meter. At which point I will have meters squared per second squared plus a quantity in meters squared per second squared. That entire quantity will be taken to the one half power leaving me with meters per second which will cancel the meters per second in the numerator. And then I will have a unitless proportion multiplied by five centimeters. So with that, I can proceed to actually computing an answer. Five multiplied by a whole bunch of parentheses just for good measure. Three divided by another couple of parentheses, two times 0.72 times 998 times 9.81 times 10 divided by 1.2 times 100 and that quantity is added to three squared and then close parentheses and then one half power and then close parentheses and then one half power and let's see if that worked. No, did not have the correct number of parentheses here. No, still not right. Let's add more parentheses, shall we? You get a parentheses and you get a parentheses. Everybody is getting a parentheses. Five times 3 over 2 times 0.72 times 8 times 981 times 10 divided by 1.2 times 100 plus 3 squared. Entire quantity taken to the one half power. Nope, this should be here. Is that right? No. About now. Yeah, that's what I'm looking for. Okay, so the denominator is taken to the one half power, not the entire quantity that I take three divided by that quantity and raise that entire quantity to the one half power. Thank you Calculator for supporting pretty print. It makes this much easier to make sure that I have the correct parentheses in the correct place. So I end up with a diameter of state two of 1.4764. And the question that I had asked after that was, is this a minimum viable diameter or a maximum viable diameter? Well, think about that for a moment. We want a pressure drop that is going to be sufficient to pull gasoline up or rather be pushed up by atmospheric pressure, which is now higher than P2. If we had a larger diameter than this, then the increase in kinetic energy would not be sufficient enough to drop the enthalpy enough to yield a pressure difference that is enough to push the gasoline up. Therefore, this represents a maximum viable diameter because any smaller diameter would have the fluid moving faster through it. That would require a greater increase in kinetic energy, which would mean a greater drop in enthalpy, which would mean a bigger delta P, which means gasoline would still be coming in. It just wouldn't be barely coming in. It would be moderately coming in. So really, we want the diameter at state two to be less than or equal to 1.76 odd centimeters. And since we neglected friction, if you were actually constructing this, you would probably want to drop it a bit more than just 1.475, for example.