 A long, thin, flat plate is parallel to a 10 meter per second stream of water at 20 degrees Celsius. At what distance from the leading edge will the boundary layer thickness be 2 centimeters? So we recognize that we have developed equations for the boundary layer thickness for laminar and turbulent flow. We're going to be using a delta of 2 centimeters. We can figure out the Reynolds number at our position, and we're solving for x. The real question is, do we use the laminar equation or the turbulent equation? We don't have enough information to determine if the flow is going to be laminar or turbulent at that position, so we're going to have to make a guess and see if we are correct. Let's start by guessing that we have laminar flow. So for laminar flow, our equation is going to be delta over x is equal to 5 divided by Reynolds number to the 1 half power. Delta is known, x is the thing that we're solving for, and the Reynolds number at that position is going to be the Reynolds number where x is our characteristic length. So we're using velocity times length divided by the kinematic viscosity. Alternatively, we could use the dynamic viscosity in the denominator at which point we would have the density in the numerator, but since we have water at 20 degrees Celsius, it's easiest for us to just use the kinematic viscosity. The kinematic viscosity for water at 20 degrees Celsius is going to come from table A1, where I can see at 20 degrees Celsius that our kinematic viscosity of water is 1.005 times 10 to the negative 6th meter squared per second. 1.005 e to the negative 6 meter squared per second. That came from table A1. Length here is actually x, because we are using Reynolds number where x is the characteristic length. And for that velocity, we are using what's called the free stream velocity, which is the velocity of the flow around the boundary layer. That's 10 meters per second. Remember that inside the boundary layer, we have a lower velocity that increases to the free stream velocity, so it's important to keep track of what the velocity is referring to. So I'll make that substitution symbolically at which point we have delta over x is equal to 5 divided by velocity to the one-half power times x to the one-half power divided by the kinematic viscosity to the one-half power. Now I can begin to solve for x. Remember that if I'm taking x divided by x to the one-half power, that's equivalent to taking x to the 1 times x to the negative one-half power and x to the A times x to the B is going to be x to the A plus B. Therefore x to the first times x to the negative one-half is equivalent to x to the 1 minus one-half, which is just x to the one-half power. So I have delta is equal to 5 times the kinematic viscosity to the one-half power divided by the free stream velocity to the one-half power times x to the one-half power. Now x is what we're looking for. So we're going to take delta times velocity to the one-half power divided by 5 times the kinematic viscosity to the one-half power. We are going to square that entire term in order to yield just x. Since we're looking for x in centimeters, I guess I don't ask for a specific unit. Since we're looking for x in a length dimension, we want the inside of these square brackets to be in the length dimension to the one-half power. That way when I square it, I get just a length dimension quantity. So I have 2 centimeters divided by 5 times 10 meters per second. And then I'm going to take everything to the one-half power. So 10 to the one-half times meters to the one-half times seconds to the one-half. And then I'm dividing by the kinematic viscosity, which for water at 20 degrees Celsius was 1.005 e to the negative 6th meter squared per second. So that's going to be 1.005 e to the negative 6 meters squared to the one-half power, which is just going to be meters to these seconds to the one-half power. Seconds to the one-half power is going to cancel seconds to the one-half power. Meters to the one-half power and meters and centimeters are going to eventually cancel. Meters to the one-half power divided by meters is going to yield meters to the negative one-half power. And then since I have centimeters in the numerator, that's going to eventually yield just a length dimension to the one-half power, which is what I want. So I will convert centimeters to meters and then meters cancels meters, leaving me with meters to the one-half power. Presumably this length is going to be rather small, so I will eventually probably express this in centimeters. But for now, I'm getting meters to the one-half power inside of the parentheses, which when squared is going to yield meters as a result. So if I pop up my calculator, I'm going to take 2 times 10 to the one-half power divided by 5 times 1.005 e to the negative 6 to the one-half power times 100. And then I'm taking that entire quantity and squaring it and I get 159.204 and that's going to be in meters. So if I had a laminar flow development, it would take until I reached 159.2 meters of length before the boundary layer reached 2 centimeters. So is that the answer to our question? We have to check first to see if we were actually correct in our laminar assumption. So if we figure out the Reynolds number at this position and we get a Reynolds number that is less than 5 times 10 to the 5th, then we were correct that it's laminar the whole time, at which point this is our answer. If we find out that somewhere before 159.2 meters it transitioned to turbulent flow, then we're going to have to adapt our analysis to the turbulent correlation for the boundary layer thickness. So we have 10 meters per second divided by 1.005 e to the negative 6 meters squared per second multiplied by 159.204 meters. Seconds cancel seconds, meters and meters cancel meters squared, leaving me with a dimensionless parameter. So I'm going to take 159.204 times 10 divided by 1.005 e to the negative 6 and I get 1.584 times 10 to the 9th. That number is greater than 5 e5, therefore turbulent, therefore somewhere before a position of 159.2 meters along this flat plate, the flow transitioned from laminar to turbulent, therefore I cannot use the laminar equation for the boundary layer thickness. So this is not actually my answer for the question. I'm going to have to repeat my analysis using turbulent flow. So for turbulent flow, I have delta over x is equal to 0.16 divided by Reynolds number with respect to x to the 1 seventh power. So like I did earlier, I'm going to make the substitution for the Reynolds number symbolically so that I can solve for x. That's going to be velocity to the 1 seventh power times x to the 1 seventh power divided by the kinematic viscosity to the 1 seventh power. And I'm going to have delta times velocity to the 1 seventh power divided by 0.16 times kinematic viscosity to the 1 seventh power is equal to x over x to the 1 seventh. So just like earlier, this is going to be x to the first power multiplied by x to the negative 1 seventh power, which is equivalent to x to the 1 minus 1 seventh power, which is x to the 6 seventh power, which I'm sure sounds great on the microphone. Lots of sibilant sounds. x to the 6 seventh power. Anyway. Then to solve for x, I'm going to have to take everything on the right to the power of 7 sixths. That's 7 over 6. So delta times velocity to the 1 seventh power divided by 0.16 times the kinematic viscosity to the 1 seventh power all raised to the 7 sixths power. And I need to take everything inside of that kinematic viscosity and raise it to the 1 seventh power. So that's going to be 1.005 times 10 to the negative 6th to the 1 seventh power seconds to the 1 seventh power. And then for meter squared, I'm going to have meter squared times the 1 seventh power because x to the a raised to the b is equivalent to x to the a times b. So I'm taking meters to the power of 2 times 1 seventh, which is just two sevenths. And then I'm taking that entire quantity. And I'm raising it to the power of 7 sixths. And in order to make the units cancel, I'm going to have to convert centimeters to meters. 100 centimeters is one meter. And then I'm left with meters to the 1 seventh power times meters to the first power times meters to the negative 2 over 7th power, which is equivalent to meters to the 8 sevenths power minus 2 sevenths power, which is meters to the 6 sevenths power, which when taken to the 7th sixth power is going to yield just meters. So calculator, if you would please we have 2 times 10 to the power of 1 over 7. Neither of those numbers were correct calculator. And then we are dividing by 0.16. Yep, multiplied by 1.005 times 10 to the negative 6th raised to the 1 seventh power. And we're raising that entire quantity to the power of 7 over 6. And I used a subtraction instead of a minus sign, which honestly I should have known by now. So if it transitions to turbulent flow along the way, we are going to have forgotten our centimeters to meters conversion. Much more better. Okay, if it transitions from laminar to turbulent along the way, we will end up with a position of 1.296 meters, which is the point at which the boundary layer thickness becomes 2 centimeters. And then again, just for good practice, we can double check that it is indeed turbulent at that position by taking 10 meters per second multiplied by 1.2963 divided by 1.005 times 10 to the negative 6 meters squared per second. And we get a Reynolds number of 1.289E7, which is greater than 5E5, therefore turbulent. So for flow along this flat plate, the boundary layer will reach 2 centimeters at a position of 1.296 meters.