I like to complete my discussion of similitry by giving some examples. So, firstly the notation, let's suppose we have an airship like this, and we're interested in modeling, for example, the drag force on it. So the shape of it looks like this. And the model of course looks exactly the same only smaller, if it's geometrically similar, which it must be. And normally, we denote the forscale or the size situation as the prototype and the bottle as the model. And we use subscript p for prototype and m for model. So for example the length of the prototype here could be lp 60 meters some width wp is five meters. The corresponding length in the model let's say lm is six meters and the width in the model is unknown. We want to calculate that. So the ratio of these two we'll denote by r = p / m. In other words, the ratio of prototype to model. And note that some people invert that, but I will take r as being the ratio of prototype to model. So for example, the length of the prototype divided by the length in the model is equal to lr, the length ratio, and this we refer to as a scale ratio or the model scale. So for full geometrical similitude, or similarity, will require that lp/lm is equal to wp/wm is equal to the ratio of any corresponding length scales is equal to the length scale ratio, or the scale ratio. So in this case we have LP / LM = LR is 60/6 is 10. So we would refer to this as a model scale of 1 to 10 or one-tenth scale model. So therefore, the diameter of the model, Wm Is the prototype diameter divided by the length scale ratio is five over ten is equal to 0.5 meters. So an example, a one to twelve scale model is to be built to study flow over a spillway. The flow in the prototype is 200 cubic meters per second. The flow rate in the model should be which of these? Now firstly, we note that generally speaking, a volume flow rate is a velocity times an area. In other words, the ratio of a volume flow rate in the prototype, to that in the model, Qp over Qm is equal to area on the prototype times velocity in the prototype, divided by area model times V model. But in turn, the areas are proportional to length squared, so Qp/Qm = Lp squared Vp Over lm squared. Vm, or rearranging, I get this or this. Qr is equal to lr squared times Vr. In other words, the volumetric flow rate ratio is equal to the len square ratio squared, multiplied by the velocity ratio. And that equation is generally true for any dynamically similar model. However, the actual ratio depends on what kind of a model it is. In this case, we have flow over spillway which is dominated by gravity so this is a froude number model. In otherwise, for similitude we require Froude number prototype to be equal to Froude number in the model. Substituting in the definitions of Froude number we have this. And gravity is of course constant so that cancels from both sides of the equation and rearranging we get that the ratios of velocity vp over vm is equal to the square root of lp over lm. Or the velocity ratio is the length scale ratio raised to the one-half power, and that relationship is generally true for a Froude number model. If I combine those two equations together, I find that the volume ratio, QR is the length scale ratio raised to the five halves power. But this relationship and this relationship remember applies only to a food member model. So now I can apply this to this problem and I have flow rate in the model is Qp divided by LR to the five-halves. QP is 200 cubic meters per second. The scale ratio is 12 to the five-halves is equal to 0.4 cubic meters per second, and the closest answer is A. Next, I have a number of questions related to this situation. The spillway for a dam is 20 meters wide and is designed to carry 125 cubic meters per second. A 1:15 scale model is to be constructed. Firstly, the width of the model should be most nearly which of these alternatives? So the solution is the length scale ratio is LP over LM which is also equal to the width, WP over WM and we're given that this is 15. It's a one to 15 scale model. Therefore, the width of the model is the width of the prototype divided by 15. Which is equal to 1.33 meters, or answer C. The next question, the flow rate in the model should be most nearly which of these? So, again, we have a general relationship, Qr = Lr^2 Vr, which is always true, But in this case specifically for a Froude number model from the previous slide we have Qr is Lr to the five halves power. So therefore, Q model is Qp over lr to the five-halves is equal to that, and the answer is 0.143 cubic meters per second. So the closest answer is B, and that is the flow rate we have to run the model act to get similitude. Finally, the time for the model corresponding to 24 hours in the prototype is most nearly which of these. So how much time in the model corresponds to a time of 24 hours in the prototype? Well, to answer this, we note that generally speaking, time is proportional to some distance divided by velocity. Or conversely, velocity is proportional to a distance divided by time. So therefore, the ratio of time in the prototype to time in the model is length in the prototype divided by V in the prototype, divided by LM over VM; or rearranging, TR is equal to the length-scale ratio, divided by the velocity ratio. And that equation will be generally true for any dynamically similar model. However, now we have to invoke our Froude number similitude for this case. And previously we showed the the velocity ratio for a Froude number model is the square root of the length scale ratio. So combining those two equations we find that the time ratio in a Froude number model is the square root of the length-scale ratio. And again, these equations apply only to a Froude number model. So, now we can substitute in, so the time in the model is Tp divided by the time ratio which is Tp over the square root of the length-scale ratio 24 over square root of 15 is 6.2 hours, and the answer is C, is the corresponding time. The next example has water flowing through a large check valve of 2 feet diameter at a rate of 30 cubic feet per second. If the inlet diameter in the model is 3 inches, and assuming the same properties, the required flow rate in the model is most nearly which of these? So in this case, gravity forces are negligible, but viscous forces are important. So this is a Reynolds number model. Okay, so we have a general flow equation again that applies Qr = Lr squared Vr. That always applies but in this case, we have a Reynold's number model. In other words, the Reynold's number in the prototype is equal to the Reynold's number in the model. Or Vp, lp of nu p is Vm lm over nu m. Were given that we have the same water flowing in the model and prototype so the properties are equal. So nu p and nu m are the same. Cancels out, and rearranging we have the velocity ration, vp over vm, is equal to lm over lp, or the velocity ratio is the inverse of the length scale ratio. But again, this equation now, only applies to a reynold's number model. Combining those two together, we find that the volume flow ratio, QR, is equal to the length scale ratio, but again, only for a Reynolds number model. So, now, we can substitute in the numbers, the flow rate in the bottle is the flow rate in the prototype, divided by the length scale ratio, which is 30. And the link scale ratio the diameter is two feet which is 24 inches, or 3 inches in the model, so the link scale ratio is 24 over three and the answer is 3.75 cubic feet per second. So, the closest answer is D. The last example is a torpedo which is to be tested in the wind tunnel. The model is the same size as the prototype and the kinematic viscosity of air is 1.46 times 10 to the minus 5. The kinematic viscosity of the water is 1.2 time ten to the minus six. If the velocity of the torpedo in the water is 8 meters per second, the air velocity in the wind tunnel is must be nearly which of these? So, in this case, again, viscus forces are important here. Gravity is not important, so this is a Reynolds number model- Reynolds number in the prototype is equal to Reynolds number in the model, which is this equation. Or, rearranging, in this case we're given that it's a full-scale model, the sizes are the same. So lp and lm are the same, so they cancel so VM is equal to VP. Nu P over Nu M where Nu is the kinematic viscosity. In that case, we have eight times kinematic viscosity of the prototype is 1.12 times ten to the minus five is water. The kinematic viscosity of the air 1.12 times 10 to the minus six. And the answer is 104 meters per second. So the closest answer is c, 105 meters per second. So notice that to achieve similitude here, the model must be run much faster than the proto type. And another very important thing that this brings out is that the model is being tested in air whereas the prototype is in water. So an important thing to keep in mind is that similitude places no requirements on the types of fluids which are being used. The only requirement is that the Reynolds number in the prototype Be equal to Reynolds numbers in the model and it places no particular restrictions on what the fluids are. We can easily test a torpedo which is to be operated in the water in a wind tunnel. That completes our discussion of similitude and dimensional analysis.