 A sharp, flat plate with a length of 50 centimeters and a width of 3 meters is parallel to a stream with a velocity of 2.5 meters per second. Determine the drag on one side of the plate and the boundary layer thickness for both air and water at 20 degrees Celsius and one atmosphere. So we are developing a boundary layer and that boundary layer is going to exert a drag force on the plate. We know the plate is 50 centimeters long and 3 meters wide, which is going to be in and out of the screen from this perspective. And I want to know the boundary layer thickness at the end and the drag force for both air and water, which means that we're doing two different analyses. And since we're doing two analyses, I will draw a big vertical line down the middle of my page and we will have air on one side and water on the other. So for air at 20 degrees Celsius, we're going to want density and kinematic viscosity at the very least. So for those properties, I will jump into table A2 and I can say that the density at 20 degrees Celsius is going to be 1.2 kilograms per cubic meter. And the kinematic viscosity for 20 degrees Celsius is going to be 1.5 times 10 to the negative fifth meters squared per second. And while I'm here, I might as well grab the properties for water. That would be table A1 at 20 degrees Celsius. The density of water is 998 kilograms per cubic meter. And the kinematic viscosity for water is 1.005 times 10 to the negative sixth meters squared per second. Then I'm calculating a boundary layer thickness and a drag force. For both of those analyses, I'm going to need to know if I have laminar or turbulent flow. So the first thing I will do is determine a Reynolds number with respect to length. My length here is 50 centimeters. So free stream velocity times the length divided by the kinematic viscosity of the fluid we are analyzing. The velocity here is two and a half meters per second. We are using a length of the plate of 50 centimeters. And then for air, my kinematic viscosity is going to be 1.5 times 10 to the negative fifth. And then to get the meter squared and the meters and centimeters to cancel, I need to convert 100 centimeters to meters. So my Reynolds number for the air flowing over the plate at a position at the end of the plate is going to be 2.5 times 50 divided by 1.5 times 10 to the negative fifth times 100. We get a Reynolds number of 8333, 83333, 0.3, which is 8.3 times 10 to the fourth. That's less than 5E5, which means I have laminar flow. The fact that I have laminar flow at the trailing edge of the plate implies that I have laminar flow across the entire plate, which means that I'm going to be using my developed equations for laminar flow. First up, I'll grab the boundary layer thickness, which is going to be delta is equal to x times 5 over Reynolds number to the 1.5 power. So at a position of 50 centimeters, that's going to be 50 centimeters times 5 divided by 83333.3 to the 1.5 power. And because 5 divided by the Reynolds number to the 1.5 power is a unit less proportion, whatever I plug in for units of x is what I'm going to get out for the units on the boundary layer thickness. So 50 times 5 divided by the Reynolds number to the 1.5 power gives me 0.866 centimeters. Then for our drag force, we're going to be using the flat plate theory. Remember that we have three equations. We have laminar flow across the entire plate, turbulent flow across the entire plate for smooth plates. And then we have flow that transitions from laminar to turbulent. We want the laminar form, which means that we're going to be using coefficient of drag is equal to drag force divided by 1.5 times density times the free stream velocity squared times the area. And that's equal to 1.328 divided by the Reynolds number with respect to length to the 1.5 power. So we're taking the second half of that equation to calculate coefficient of drag. That's 1.328 divided by 83333.3 to the 1.5 power. 333.3 to the 1.5 power. And we get, come on and calculate it, 1.328 divided by this number raised to the 0.5 power. We get 0.0046. That's a very low coefficient of drag, but that makes sense because it's a perfectly smooth flat plate in laminar flow. Then I'm going to use the first part of this equation to calculate the drag force. That's 1.5 times the coefficient of drag times the density times the free stream velocity squared times the area of effect. And by the way, every time I write this, I will probably write it in a different order because I just write down whatever variable comes to mind first and then kind of vomit letters until I get to the end. So if you see me write that as coefficient of drag times 1.5 times density times velocity squared times area, or coefficient of drag times area times 1.5 times density times velocity squared, or velocity times density times velocity times coefficient of drag times area over 2. All of those are representations of the same thing and it all comes from this definition. So coefficient of drag, we just determined 0.0046. The density is going to be the density of air, which we looked up. Free stream velocity we know, 2.5 m per second. What area do we use? Keywords in this sentence are drag on one side of the plate. So we are only caring about one side in regard to drag. Maybe the plate is sitting on top of a surface, in which case we don't care about any drag force on the bottom. So we are going to plug in 3 m times 50 cm as the area of one side, which is our area of effect. Then what unit would you guys like to express an answer in? How about newtons? Maybe we can start with newtons and then if we end up needing kilonewtons we can go there. But a newton is defined as a kilogram, meter per second squared, kilogram cancels kilograms, second squared cancels second squared. I have meters and cubic meters in the denominator and cubic meters and centimeters in the numerator, so I have to convert 100 cm to 1 m. And then all of the meters will cancel, leaving me with an answer in newtons. So 0.5 times 0.0046 times 1.2 times 2.5 squared times 50 times 3 divided by 100. And we have 0.025875. So for air blowing across the plate, we have very little drag force on the plate. Almost three hundredths of a newton. For water, we're going to start with the same relationship, the Reynolds number at the end of the plate, and we're going to deduce the flow characterization across the plate based on what we get for a Reynolds number at the trailing edge. So velocity times length divided by kinematic viscosity of water. So 2.5 times 50 divided by 100 times 1.005e to the negative 6 gives us a Reynolds number of 1.24e6. That's greater than 5e5, which means we have turbulent flow. So the fact that we have turbulent flow at the trailing edge of the plate does not mean that we're going to assume it's turbulent across the entire plate. It is more likely that we are going to start with laminar flow at the beginning of the plate and then transition to turbulent flow. So we have to account for both the laminar and turbulent parts of the flow in our drag force calculation. So again, we have laminar flow, turbulent flow, and transition from laminar to turbulent flow for our coefficient of drag for flat plates. Because our Reynolds number, which was 1.24 times 10 to the sixth is not greater than 8 times 10 to the seventh, we're going to use this equation for the coefficient of drag. So that coefficient of drag is 0.031 divided by Reynolds number to the 1 seventh power minus 1440 divided by the Reynolds number. As a fun fact, we could actually set up the turbulent analysis on the turbulent part of the plate and the laminar analysis on the laminar part of the plate, figure out how much length we had for the laminar region, subtract that from our turbulent region, and we would end up with this exact same relationship. But 0.0031 divided by our Reynolds number to the 1 seventh power minus 1440 divided by our Reynolds number, and we get a coefficient of drag of 0.003018. And from that coefficient of drag, we can calculate a drag force. Again, we can pick an arbitrary order of these variables. Area may be perhaps velocity squared. Density coefficient of drag times 50 centimeters times 3 meters times 2.5 squared meters squared per second squared times 998 kilograms per cubic meter times 0.003018. As a result of the higher drag coefficient and the larger density, we're going to have more drag force. And that quantity is going to be 0.5 times 50 times 3 times 2.5 squared times 998 times 0.003018. If we account for the conversion from centimeters to meters, that will give us newtons. So I'll add a divide by 100 at the very end. And I get 14.12 newtons. So we had a drag force on the laminar condition with air of 0.03 newtons, and with water, which makes it turbulent, we have 14.12 newtons. And it looks like I was in so much of a hurry to calculate the drag force that I jumped right over the boundary layer thickness calculation. So we'll jump back to that. For our turbulent flow, we're going to have delta is equal to x times 0.16 divided by the Reynolds number to the 1 seventh power. Again, we're calculating the boundary layer thickness at the trailing edge of the plate. And because 0.16 divided by a unitless quantity to the 1 seventh power is going to yield a unitless proportion, whatever units I plug in for x for units. So that's going to be 50 centimeters times 0.16 divided by our Reynolds number raised to the 1 seventh power. And because it's my calculator, it forgot the first digit. So we get 1.077 centimeters. That's our boundary layer thickness for the water. Scooch my drag force up. But it's like that never happened. There we go. For air, we had a boundary layer thickness at the trailing edge of 0.866 centimeters, primarily due to the fact that it was laminar. And then for water, we had a boundary layer thickness of 1.0775.