Continuing our discussion of flow in pipes now I want to do some examples of computation of head loss in laminar flows, turbulent flows, and how we account for minor losses in systems. So first this example, oil is flowing in a horizontal pipe in a laminar flow. The flow rate is given some specific gravity and viscosity and if we have a head loss of 12 centimeters over a pipeline for 3 meters, the diameter of the pipe is most nearly, which of these alternatives. So, in this case, we're given that the, the flow is laminar, so automatically we know this is Poiseuille flow for which this equation applies, the relationship between the pressure drop and the flow rate, et cetera in the pipe. Now, in this case, we're given that the energy grade line here is sloping downwards and it drops by a height of 12 centimeters. How do we relate that to the pressure drop in the flow? Well, remember that the hydraulic grade line is below the energy grade line by a height equal to the local velocity head V squared over 2g. And the distance from the pipe's center line to the hydraulic grade line is the local pressure head p over gamma, which I'll call p1 over gamma here. Now, in this case, the diameter of the pipe is constant. Therefore, the cross sectional area is constant. Therefore, the velocity is constant. Therefore, V squared over 2g is constant. Therefore, the hydraulic grade line is parallel to the energy grade line. And the drop in the energy grade line is therefore equal to the drop in the hydraulic grade line. But the drop in the hydraulic grade line here, if the pressure head is p2 over gamma, the drop in the hydraulic grade line is the drop in the pressure head, p1 over gamma minus p2 over gamma. Which is equal to delta p over gamma, where delta p is the pressure drop in the pipe. So, in this case, you can see that delta p over gamma is equal to the drop in the pressure head or the energy grade line rather, which is 12 centimeters. So therefore, we have this equation here. The pressure head loss is the drop in the energy grade line is delta p over gamma is equal to 128 mu LQ over gamma pi D to the 4th. In this case, we're asked to calculate the diameter. So rearranging that equation to make D the subject, we get that. And now, we can substitute in the numbers 128 mu is point, viscosity is 0.38, etc. And the answer is the diameter is 0.10 meters or 10 centimeters, so the answer is B, 10 centimeters. In this example, we have water being pumped at a rate of 0.083 cubic meters per second. So, a 20 centimeter diameter pipeline and the relative roughness is given as 0.01. The water viscosity is given and the question is the friction head loss over a pipe length of 45 meters is most nearly, which of these alternatives. So firstly, we'd like to know whether the flow is laminar or turbulent. And to do that, we compute the Reynolds number of the flow, rho VD over mu. And first, we need to get the velocity, because in this case, we're given the volume flow rate and the area. So the velocity is Q over A is equal to Q over pi by 4D squared, which is 2.64 meters per second. And then the Reynolds number is equal to rho, which I'm assuming is 998 kilograms per cubic meter times the velocity 2.64 times the diameter divided by the viscosity, which is given and the Reynolds number is 5.92 times 10 to the 5th. So the flow is very obviously and very definitely turbulent. The relative roughness we're given, the relative roughness is epsilon over D is 0.01. So now the question is what is the friction factor, which is a function of epsilon over D for these two values of Reynolds number and relative roughness? So for this, we turn to the Moody Chart here. So the Reynolds number is approximately there, approximately 6 times 10 to the 5th. The relative roughness 0.01 is right there, looking at the intersection of those two curves. And reading off on the left-hand side, we see that the friction factor for this situation is 0.038. Next, we turn to the Darcy-Weisbach equation. Hf equals f L over D, V squared over 2g. And now we can evaluate that, f we've just found is 0.038. Pipe length is 45 and diameter is 0.2, etc. Which gives me a head loss of 3.04 meters, so the answer is A. Now next, I want to look at Minor Losses, which are other losses in the system separate from friction head losses and they would be due to valves, elbows, acute, other accoutrements in the city, in the setting. This is the term hm for minor losses and typically, we express a, a minor loss by an equation such as this one here. Hm is KL V squared over 2g, where KL is a so-called loss coefficient. V is a local reference velocity, so V squared over 2g is a local velocity head. Here is the corresponding section in the reference manual, where they express the mine losses they call it hf for fitting, but a more common term is a minor loss and the coefficient they express here as C, the loss coefficient. And the value of the loss coefficient generally speaking is not something you can predict. You have to look it up in tables or manufacturer's handbooks, etc. For example, different types of valves. Here are some examples of valves globe valves, angle valves, gate valves. Of course, the loss coefficient depends on whether the on how open the valve is. Or other examples for entrance losses from flow from a, a large reservoir or a large pipeline into a smaller one are given here. So, a re-entrant entrance like this has a very large coefficient, whereas a nicely flared one here has a very small energy loss or a very low loss coefficient. This is most easily illustrated by means of an example. So this example here, we have water flowing by gravity between two reservoirs. The left-hand one is higher than the other. The elevation differences between the two surfaces is 10 meters. The pipe diameter is 50 centimeters, length is 500 meters, friction factor is given. And we have a valve here with a loss coefficient of two. Question is neglecting of the minor losses, the velocity in the pipeline is most nearly, which of these alternatives? So we start off by applying Bernoulli equation between the two water surfaces. So the water surface of the upper level reservoir are labeled one and the lower level reservoir are labelled two, so the Bernoulli equation, the full Bernoulli equation with all losses and additions between those two surfaces is given here. Now we can simplify that right away. We don't have a pump in the system, so hp goes away. Where we don't have a turbine, so ht goes away. In the large reservoir here at one, this free surface is at atmospheric pressure, which is zero, so p1 over gamma goes out. Similarly, for the reservoir two, the pressure is atmospheric pressure, so that term goes out. The velocity in the large reservoir here we can neglect, so V1 goes out. Similarly, the velocity in the lower reservoir here is is very small, so V2 goes out. So, all we have left then is Z1 minus Z2 is equal to H, where H is the height difference here between the two levels is equal to hf, the head loss due to friction plus hm, the minor losses. And if you want to draw that in terms of the energy grade lines. The energy grade line for a large reservoir is coincident with the free surface, then the energy grade line slopes downwards due to friction in the pipeline here. Then we have a sudden loss across the valve, a sudden drop and then essentially constant and we end up at the free surface of the lower reservoir. So this drop here, the drop in the energy grade line due to friction is hf and the minor loss, the drop across the valve is hm. So the energy grade line here is simply a graphical representation of the Bernoulli equation. And you can see from this then that h this totally height is simply equal to hf plus hm, which is exactly the equation we obtained from the Bernoulli equation. Now continuing our equations for the losses for friction loss is the Darcy-Weisbach equation and the minor loss is KL V squared over 2g. So that is in turn is equal to that expression. And in this case, we're given the head and we want to calculate the velocity. So rearranging that, I get this equation and now we're ready to substitute the values in. We have 2gh height difference is, is given there divided by friction factor and the answer is 2.11 meters per second. So the final answer is B and that concludes our examples on calculating head loss in pipe systems
