 Consider a cheater attachment for the spout from the previous problem, where the outlet diameter is increased slowly by 5 degrees over 10 centimeters. Determine the flow rate of water out of the spout. So the question here is, as a result of this attachment, are we able to affect the flow rate? And better yet, can we actually increase it? And the answer is yes we can. That is the whole goal of this type of attachment. That's why it's called a cheater attachment. We are able to pull more water out of the reservoir than we would normally be allowed. The primary difference that is the reason that we are able to do this is because we are pushing back where atmospheric pressure is. As a result of the expansion before atmospheric pressure, we are allowing the velocity at state 2 now to be higher than it would have been otherwise. So we are able to push the water faster. And to explore that, we can set up a conservation of energy. So just like in the previous problem, I'm going to be assuming that I have atmospheric pressure on either end. Since I'm now analyzing from state 1 to state 3, I'm going to change all these 2s to 3s. So z1 minus z3 is still the 1.5 meters from the previous problem because I have a horizontal relationship between states 2 and 3. My velocity at state 3 is now going to be the thing that we calculate. With the wider diameter of the cheater pipe, we can figure out what that volumetric flow rate is or we can use that to figure out what the velocity is at state 2. I account for minor losses. Here I have two types of minor losses. Note that when I'm talking about the coefficient of loss for the entrance, I'm using v2. When I'm using the coefficient of the expansion process, I'm using v3. So when I plug those into my conservation of energy relationship, I end up with z1 minus z3. I'm still going to be assuming turbulent flow. So I will apply an alpha value of 1 and I have v3 squared over 2 times gravity plus k entrance times v2 squared over 2 times gravity plus the expansion coefficient times v3 squared over 2 times gravity. And just like before, I don't really care about the velocity. What I care about is the volumetric flow rate. And the volumetric flow rate can be written as pi over 4 times the diameter at either diameter because it has to be the same volumetric flow rate. The reason it has to be the same volumetric flow rate is because I have steady-state operation of incompressible flow. Steady-state implies that m dot 2 equals m dot 3. And when I write m dot 2 is equal to density times volumetric flow rate at state 2 and m dot 3 as density at state 3 times the volumetric flow rate at state 3. If the densities are the same, that means the volumetric flow rate has to be the same. So in place of v2, I can write 4 times the volumetric flow rate divided by pi times d2 squared. And in place of v3, I can write 4 times the volumetric flow rate divided by pi times d3 squared. And at that point, I now have z1 minus z3 is equal to 4 squared times the volumetric flow rate squared divided by pi times diameter at state 3 squared. Excuse me, to the fourth power times 2 times gravity plus the loss coefficient with the entrance times 4 squared times volumetric flow rate squared divided by pi times diameter at state 2 to the fourth power times 2 times gravity plus the loss coefficient associated with the expansion process times 4 squared times volumetric flow rate squared divided by pi squared times the diameter at state 2 again to the fourth power, excuse me, diameter at state 3 to the fourth power times 2 times gravity. Then I can factor out 4 squared volumetric flow rate squared pi squared and 2 times gravity squared of 1 over d3 squared plus the entrance loss coefficient divided by 2 squared d2 to the fourth power, excuse me plus the expansion loss coefficient divided by d3 to the fourth power. Then the volumetric flow rate would be pi squared times 2 times gravity divided by 4 squared times z1 minus z3 times 1 over d3 to the fourth power plus the entrance loss coefficient divided by d2 to the fourth power plus the expansion loss coefficient divided by d3 to the fourth power. And if all of those diameters are in centimeters then I will make a conversion at the end from centimeters to the fourth power to meters to the fourth power just to double-check that everything still cancels and then I'm going to take that entire quantity and put it into a square root so what I get out is the volumetric flow rate. I know gravity. I know pi. I know z1 minus z3. I know d2. So in order to be able to calculate the volumetric flow rate I have to look up the loss coefficients for the entrance the loss coefficient for the expansion process and I have to calculate what the diameter at state 3 will be. For the entrance we're going to use the same loss coefficient from the previous problem. That loss coefficient is just going to be a quarter of the way between 0 and 0.2 so we're going to use 0.05 again. For the expansion process we're going to have to look through section 6.9 to find where it allows for expansion processes and I see that the relevant figure is figure 6.23. So on figure 6.23 I need to locate the line that corresponds to my proportion of diameters and then look at the cone angle. So I was told 5 degrees of expansion on either side and remember that the cone angle here is actually 2 times theta which means that my cone angle itself is going to be 5 degrees and then for diameter at state 3 squared I'm going to have to figure out how much expansion there is. So D3 is going to equal D2 times, excuse me, plus 2 times this height which I'm calling H and I can write the tangent of 5 degrees is H over 10 centimeters which means H is equal to tangent of 5 degrees times 10 centimeters which means D3 is equal to D2 plus 2 times tangent of 5 degrees times 10 centimeters. So calculator I call on you once again. That is going to be fraught with the possibility of confusing degrees with radians. So 2.5 centimeters plus 2 times 10 centimeters times the tangent function of 5 degrees. So my diameter at state 3 is going to be 4.25 and then before I leave this I also have to calculate the proportion of diameters because I need that for figure 6.23. So that proportion of D2 over D3 is going to be 2.5 divided by 4.25 which calculator surely you will definitely get right the first time. You can do it. I believe in your calculator 2.5 divided by and I get 0.588. So 0.588 is going to be 88% of the way between this line and this line. So again, I'm going to call that like here ish and then I'm looking at a cone angle 2 times theta which means that my cone angle on this graph is actually going to be 10 degrees. So 10 degrees and 80% of the way. I'm going to call that like here which means that I'm looking at what do you think 0.17 0.175 maybe let's call that 0.175. I'll make the math a little bit neater. Now we have the diameter at state 3 is 4.25 centimeters. We can factor that out if we wanted to but let's just arbitrarily choose to not do that. Now I can plug everything in and then I should be getting meter squared per second squared. So this is on the wrong side. Not sure how that algebra was supposed to have worked out. Let's try that again. And we divide by 1 over 4.25 centimeters squared squared to the fourth power plus 0.05 divided by 2.5 to the fourth power times centimeters to the fourth power plus 0.175 divided by 4.25 to the fourth power centimeters to the fourth power we need that to be in 100 centimeters to the fourth power divided by 1 meter to the fourth power and then that entire thing to the square root. Okay, so I start off with 2 times 2 times calculator. 2 times pi squared times 9.81 times 1.5 divided by 4 squared 100 to the fourth power times 1 over 4.25 squared excuse me to the fourth power times 0.05 occurs to me that you guys can't see that well just imagine me typing it to and then back spacing and typing a four whole bunch of times 0.05 divided by 2.5 to the fourth power and then I supposed to be a plus sign 0.0 excuse me 175 divided by 4.25 to the fourth power and then about a thousand closing parentheses so before I take the square root I have 0.00037 let me double check that that all looks right or 25 to the fourth power 2.5 to the fourth power 4.5 to the fourth power entrance loss of 0.05 pension losses 0.175 we have the potential energy terms on the same side in the numerator which is good yet that looks right okay so then I'll wrap that entire thing in a square root at the end to get it into cubic meters per second so we're going to volumetric flow rate of 0.004 excuse me 006 098 cubic meters per second or about 0.0061 then because I have a bigger volumetric flow rate than the previous problem through the same diameter that means that I'm still must have a turbulent flow so my assumption is still reasonable interesting how the addition of that attachment multiplied our volumetric flow rate by what two and a half times I mean I have a calculator I can figure that out by 0.0026 yeah about 2.35 times we have a minor attachment to our Roman style reservoir and we are increasing our volumetric flow rate by 200% not too shabby hope the centurions don't find out that I'm stealing water from the emperor