 An incompressible fluid is freely draining out of a large diameter 1 foot tall container through a 1 foot long, an incompressible fluid is freely draining out of a large diameter 1 foot tall container through a 1 foot long 0.004 foot diameter tube. The specific weight of the fluid is 58 pounds of force per cubic foot, and the fluid is draining out of the container at a rate of 0.15 cubic feet per hour. I want to determine the viscosity of this fluid in pounds of mass per foot second. This is a relatively common way to measure viscosity. You can take a bucket with a hose, record how long it takes for the bucket to drain, use the information from the beginning of the setup to calculate what the viscosity had to have been. In order to analyze the flow through the tube at the bottom, we're going to have to make an assumption about laminar or turbulent flow, and then check that assumption at the end. Granted, the word laminar appears in the example category, but let's ignore that for a moment. We can parse out some of the information in the problem statement. I have diameter 0.004 feet, referring to the diameter of the tube at the bottom. I was told the container is of a large diameter, which I'm going to use to assume that the velocity at the top of the fluid is approximately 0. I know the container is 1 foot tall, I know the hose is 1 foot long, meaning the total height difference between states 1 and 2 is going to be 2 feet. I have a volumetric flow rate, which I can use with the diameter of the tube to figure out an average velocity, and I know a specific weight, which I can use with standard gravity to calculate a density. So let's start with this number and this number. If my specific weight is density times gravity, and for this problem I know the specific weight is 58 pounds of force per cubic foot, and I assume standard gravity, I can take that specific weight divided by gravity to calculate a density of this fluid. Remember that a pound of force is the amount of force in one pound of mass multiplied by standard gravitational acceleration. So I'm going to say one pound of force is equal to one pound mass times 32.2 feet per second squared. At this point, pounds of force cancels pounds of force, second squared cancels second squared. Feet cancels feet. Leaving me with pound mass, that can't be right. Oh excuse me, I forgot the cubic feet. That's much more better. That leaves me with pounds of mass per cubic foot. So if I ask my calculator for some assistance, I can take 58 times 32.2 divided by 32.2 and I get 58. Thank you calculator, you were very helpful. That density is going to be one half of the required properties for the Reynolds number later on in the problem. Since I don't know a dynamic viscosity and I can't look one up because I don't know what the fluid is yet, I can't calculate Reynolds number yet. So I'm going to have to make an assumption about whether I have laminar or turbulent flow to calculate what the viscosity would have been, use that information to calculate a Reynolds number, use that information to see if it was indeed laminar or turbulent, and if I got it wrong repeat the process. The next step of the process is going to be to calculate the friction factor based on my conservation of energy equation. Next I can neglect terms that aren't relevant to this problem. I'm going to start by recognizing that state one and state two are both exposed to atmosphere. Therefore I'm going to assume that p1 and p2 are the same if we have an incompressible fluid with standard gravitational acceleration and approximately atmospheric pressure of both points. That means p1 over density times gravity will cancel on both sides of the equation. Furthermore, because of the quote large diameter on quote at the top, I'm assuming that that diameter is so large that the velocity is so small that I can treat it as being zero. I can't neglect the changes in potential energy because I do have two feet of height difference between state one and state two. I can neglect the turbine head and the pump head at which point I have z1 is equal to alpha two times v2 squared over two times gravity plus z2 plus the friction head. You'll remember from chapter three when we developed this equation that alpha is two if we have laminar flow and about one if we have turbulent flow. z1 is known, z2 is known, or more accurately the difference between z1 and z2 is known. So if I could calculate v2 I could determine a friction head. I could say z1 minus z2 which is two feet minus alpha two times v2 squared divided by two times gravity is equal to the friction head and then using my definition of the friction head for major losses I can write that as f times l over d times v squared divided by two times gravity. So which diameter, which velocity, and which length are these? Well for that I recognize that relative to the frictional losses along the big container walls the real frictional losses are in the tube. So as a result I'm going to use the length of the tube, the diameter of the tube, and the velocity of the tube in the right hand side of my equation over here. I'm using a length of one foot, a diameter of 0.004 feet, and then because the velocity has to be the same at the top of the tube as it is at the bottom because I have incompressible flow through the same diameter flowing steadily I'm going to plug in v2 here. So I'm using the length of the tube which is one foot one foot standard gravity and v2 squared. Then for the friction factor I'm going to plug in either the friction factor if I have turbulent flow or the friction factor if I have laminar flow. So for that I'm going to have to make an assumption. If I have laminar flow it's just 64 divided by the Reynolds number. If I have turbulent flow I have one over f to the one half power is equal to negative 2 times log bakes 10 of the relative roughness divided by 3.7 plus 2.51 divided by the Reynolds number times the friction factor to the one half power. I have to try one and if the Reynolds number doesn't match that means I got it wrong and I have to try the other one. One of these is easier to try first. It's the laminar one. So let's start by assuming that I have laminar flow. That way if it's wrong I can go back and do the turbulent flow if it requires it and if it's not wrong I didn't waste any effort doing the guess and check process required for turbulent flow. So since I'm assuming laminar flow that means friction factor is equal to 64 divided by the Reynolds number with respect to diameter. The Reynolds number with respect to diameter is the density times the velocity times diameter divided by the dynamic viscosity which means that I have 64 times dynamic viscosity divided by density times velocity times diameter. Now which diameter and which velocity are these? Well for the friction factor I'm assuming the frictional losses are along the tube primarily and as a result I'm going to use the diameter of the tube and the velocity of the tube. So when I plug this proportion into this equation I get z1 minus z2 minus alpha 2 which I know because I'm assuming laminar flow so I can just plug in 2 minus 2 times v2 squared divided by 2 times gravity is equal to 64 times the dynamic viscosity divided by density times v2 times d2 times L over d2 times v2 squared over 2 times gravity. I can simplify this a little bit by recognizing that v2 here is going to cancel the square in the v2 term on the right at which point I have z1 minus z2 minus 2 times v2 squared divided by 2 times gravity multiplied by density times d2 squared times 2 times gravity divided by 64 times the length of the tube times v2 is equal to my dynamic viscosity so z1 minus z2 I know it's two feet gravity I know I can figure out v2 by using the volumetric flow rate and the cross sectional area of the tube I know the density because I just calculated it I know the diameter state 2 0.004 speed I know gravitational acceleration I know the length of the tube it's one foot and v2 I can calculate in terms of the volumetric flow rate so everything is known or easily solvable generally speaking I'm in favor of plugging in as many aspects of the calculation symbolically as possible but in this case I can calculate a velocity at state 2 and plug in the number using my volumetric flow rate and the fact that that's equal to the average velocity times cross sectional area which for a circular tube is average velocity times pi over 4 times diameter squared that means velocity at state 2 is going to be 4 times the volumetric flow rate divided by pi times diameter squared so 4 times 0.15 cubic feet per hour divided by pi times 0.004 feet squared and then I recognize that I probably want feet per second for my calculation below which means that I'm going to use the conversion between hours and seconds I'll add to this one hour is 3600 seconds our cancels our square feet is going to cancel two of the three feet in the numerator leaving me with a velocity in feet per second so calculator I call on you once again that's going to be 4 times 0.15 divided by pi times the quantity 0.004 squared times 3600 so my velocity at state 2 is 3.31573 feet per second and armed with that velocity I now know everything I need to know to calculate a viscosity so I have two feet out front minus two times v2 squared minus two times 3.31573 squared feet squared per second squared divided by 2 times gravity divided by 2 times 32.2 feet per second squared my second squared is going to cancel second squared feet is going to cancel one of the feet in the square 2 is going to cancel 2 I have a quantity in feet inside of the parentheses and then I'm multiplying by density which I calculated above so multiplied by 58 pounds of mass per cubic foot multiplied by d2 squared 0.004 squared feet squared multiplied by 2 times gravity multiplied by 2 times 32.2 feet per second squared and then that is divided by 64 times the length of the tube so 64 times 1 foot times v2 which is 3.31573 here's that calculator by the way so I have feet inside of the parentheses and then outside of the parentheses I'm going to have square feet and feet canceling cubic feet I have second squared and seconds so one of the seconds in the square is cancelled then one of the feet in the denominator is going to be cancelled by the feet inside of the parentheses leaving me with pounds of mass per feet second the problem wanted an answer in pounds of mass per feet second so I'm good to go on unit conversions all I have to do now is compute a number so calculator again 2 minus minus calculator 3.31573 squared divided by 32.2 times 58 times 0.004 squared times 2 times 32.2 divided by 64 times 3.31573 yields 0.000467 0.000467 pounds of mass per foot second which I can also write as 4.67 e to the negative 4 and that gives me my viscosity the last thing to do is to check whether or not this viscosity would yield laminar flow through the tube if my Reynolds number is less than 2300 that I'm good to go if it's greater than 2300 that means I should have been calculating the viscosity using turbulent flow which means I have to go back to this step here when I plugged in a representation for f so my Reynolds number can be represented with respect to diameter as density times velocity times diameter divided by dynamic viscosity our density we know 58 pounds of mass per foot cubed that's foot cubed not foot question mark multiplied by our velocity at state 2 which was 3.31573 then I'm multiplying by my diameter at state 2 which is 0.004 and I'm dividing by my shiny new viscosity which is 4.67 e to the negative 4 pound mass per foot second so feet feet and feet cancel cubic feet seconds and seconds cancel pound mass cancels pound mass that gives me a dimensionless quantity which is what I want 58 times 3.31573 times 0.004 all divided by 4.67 times 10 to the negative 4th negative sine 4 I get 1647.21 which is less than 2300 which means that my assumption of laminar flow is correct therefore our viscosity for this fluid is 4.67 times 10 to the negative 4th pounds of mass per foot second