 A hydrofoil half a meter long and two meters wide is placed in a seawater flow at 12 meters per second at 20 degrees Celsius. I want us to figure out the boundary layer thickness at the end of the plate and then figure out what the drag force would be with all three variations of turbulent flow that we could consider. That is, laminar to turbulent transitional flow across the plate, fully turbulent across the entire plate if it's smooth, and turbulent rough flow for a roughness of 0.001 meters. So our hydrofoil, which is like an airfoil except for water, is being modeled as a flat plate. Immersed in seawater where the velocity is 12 meters per second and the temperature is 20 degrees Celsius. I want us to figure out what the boundary layer thickness is at the end of the plate. It is half a meter long and two meters wide. So we will start with a Reynolds number calculation at the trailing edge. Use the value of the Reynolds number to figure out if we use the laminar or turbulent equation for the boundary layer thickness. And for that calculation we're going to need to look up the properties of seawater. For that we will jump into table A3 where I see that seawater with 30 salinity is going to give me a density of 1,025 kilograms per cubic meter. Then I also want a dynamic viscosity and that is 1.07 times 10 to the negative third kilograms per meter second. So a Reynolds number at the trailing edge is going to be velocity times our total length times density of our seawater all divided by our dynamic viscosity. We have to do density divided by dynamic viscosity because we do not have a kinematic viscosity for seawater. Our velocity was 12 meters per second. Our length was one half meter. Our density was 1,025. Our dynamic viscosity is 1.07 times 10 to the negative third. That was in kilograms per meter second. So kilogram cancels kilogram seconds cancel seconds meters and meters and meters cancel cubic meters. Giving me a unitless proportion. So calculator if you would please that's 12 times 0.5 times 1,025 divided by 1.07 times 10 to the negative third. Calculator says 5.75e6 which is definitely turbulent. Which was kind of implied by the fact that I had asked for the drag force for a variety of turbulent flow characteristics. So first up I have the boundary layer thickness and for that we're going to jump into our turbulent equation for boundary layer thickness which is delta over x is equal to 0.16 divided by the Reynolds number with respect to x to the 1 seventh power. x here is at length so that's delta is equal to length times 0.16 divided by the Reynolds number with respect to length to the 1 seventh power because again the x position that we're evaluating is at the trailing edge so the x is L that's 0.16 divided by the Reynolds number to the 1 seventh power. And again because 0.16 divided by the Reynolds number to the 1 seventh power is a unitless proportion whatever units I plug in for L will be when I get out for delta. If I plug in meters I will get out meters which is probably way too big but we'll cross average when we get to it. 0.5 times 0.16 divided by our Reynolds number to the 1 seventh power and we get 0.00866 meters or 8.66 millimeters then for the coefficient of drag for laminar to turbulent transitional flow I'm going to use this equation so 0.031 divided by the Reynolds number to the 1 seventh power minus 1440 divided by the Reynolds number that is again 0.031 divided by the Reynolds number to the 1 seventh power minus 1440 divided by the Reynolds number so I'm going to take 0.031 divided by the Reynolds number to the 1 seventh power minus 1440 divided by the Reynolds number and we get 0.003105 just double check that I used the correct numbers 1440 0.031 we did hooray next I can calculate the drag force so arbitrary arrangement of variables we have one half times let's go with the density first this time times the coefficient of drag times the velocity square times the area my density is going to be the density of seawater which we looked up coefficient of drag we just calculated velocity is going to be 12 meters per second and the area of effect is going to be two times 0.5 times two meters it's two times because we have drag on both sides of our hydrofoil so that's one half times 1025 kilograms per cubic meter times 0.003105 times 12 squared meters squared per second squared times two times 0.5 meters times two meters let's assume we want an answer in newtons and we can adapt to kilo newtons if we need to a newton is a kilogram meter per second squared so kilograms cancel kilograms meters cancels one of the meters second squared cancels second squared square meters and meters cancels cubic meters leaving me with newtons so 0.5 times 1025 times 0.003105 times 12 squared times two times 0.5 times two we get a drag force of 458.3 newtons for part c i want us to determine what the drag force would be if the coefficient of drag were generated using the assumption that it was smooth turbulent flow across the entire plate as opposed to laminar at the beginning of the plate and turbulent at the end of the plate so the only difference between b and c is that coefficient of drag is calculated using this equation so that equation is 0.031 divided by the reynolds number to the one seventh power that's 0.031 divided by that number to the power of one over seven and we get 0.003355 then we're going to take that and plug it into our drag force calculation and we get 495.2 newtons so i repeat the process one more time except this time i have turbulent flow with a roughness of 0.0001 meters 0.0001 meters that was four zeros right no three zeros before the one and after the decimal so for rough turbulent flow we are discarding the laminar region entirely and we are using this equation for the coefficient of drag so cd is going to be 1.89 plus 1.62 times log base 10 of the length of the plate over epsilon quantity raised to the negative two and a half power i have a unitless proportion inside of the log base 10 term which means that everything is going to be unitless which means that my coefficient of drag will be unitless so calculator we're gonna take 1.89 plus 1.62 times log base 10 i will go looking for of the length of our plate which was 0.5 meters divided by our roughness value which is 0.0001 meters and then we are taking that entire quantity to the power of negative 2.5 i have to use a negative sign instead of a subtraction sign or my calculator will get mad we get a coefficient of drag of 0.005733 an alternative way to come up with that number would have been to use figure 7.6 on figure 7.6 i could have found the blue line corresponding to my proportion of length over roughness that would have been 0.5 divided by 0.0001 which is 5000 the line for 5000 is right here my runnels number is 5.75 times 10 to the sixth so i'd be using a value oh probably right around here so that would intersect my blue line and give me a coefficient of drag maybe 0.0055 somewhere between 0.0055 and 0.006 so that's an alternative method but the calculation is easier than the colbrook equation so there's not much value in direct lookups from that chart the chart is more useful for generalizations or by seeing how trends occur if our runnels number increases or decreases past a certain point so one more time we're going to do our drag force calculation with our new coefficient of drag so we're using 0.005733 and that gives us a drag force of 846 newtons so the presence of just a little bit of roughness the bumps that are about a tenth of a millimeter tall on average increases our drag force from almost 500 newtons to 850 newtons that almost doubles just by having a surface that has bumps that are an average of a tenth of a millimeter tall so it's in our best interest to try to smooth the surfaces that are dragging against water especially in applications where our velocity is relatively high