I want to continue our discussion of continuity in energy equations now by doing some examples of the energy equation or the Bernoulli equation. And in this example, we have a horizontal pipeline with a sudden enlargement and the energy in the hydraulic grade lines are shown, and the distances between them. If the fluid is incompressible, the ratio of the larger to the smaller diameter is most nearly which of these alternatives? So, to solve this, we recognize, first of all that the distance between the hydraulic and energy grade lines is the local velocity head, V squared over 2g. In other words, this distance here, from here to here, is the local velocity head, V squared over 2g in the pipe. Now, in this example, they show the energy grade line as being constant. In other words, horizontal, so that we're neglecting any energy losses. In fact, that is not realistic, and in a sudden expansion like this, in reality we would have a head loss, or an energy loss that we would have to account for, but we're going to neglect it here. So V squared over 2g is the distance between them or if the distance between these two lines is H then the local velocity is the square root of 2gh. We're given the distances, so therefore the up stream velocity here, V1 is 2g times 5, which is 9.9 meters per second. Similarly, the downstream velocity here at 2, is square root of 2 times 9.81 times 2, which is 6.26 meters per second. Next we invoke continuity for an incompressible fluid. The volumetric flow rate is area times velocity or a1 v1, which is also equal to a2 v2. So, substituting for the diameters and rearranging this expression for the diameter ratio, which is what we're looking for, we have d2 over d1 is square root of v1 over v2. Plugging in the numbers, we have 1.26, so the closet answer is D. In this example, we have air which is flowing steadily from a pressured tank and we're given that the pressure in the tank is 3 kilopascals, and the diameters are shown, and it exits, finally, to air, or to atmosphere, at the right hand side. We're given the nozzle diameters, and the specific weight of air, and the question is, the air flow rate is which of these alternatives? So to solve this, we'll apply Bernoulli equation between the tank at 1 and the open nozzle at 3. So there's that Bernoulli equation, and then we can start to simplify it. If the cross-section area of the tank is very large, then V1 is negligible, and that goes out. If the pipeline is horizontal, we can neglect any elevation changes, and finally at the open atmosphere here where the jet is discharging into the atmosphere, the pressure there is atmospheric pressure, which is 0 gauge. So P3 also goes out. So, we're left with P1 over gamma is V3 squared over 2g, or V3, the exit, or nozzle velocity is square root of 2g P1, divided by specific weight. Plugging in the given numbers, we have V3 is equal to 70 meters per second. Which represents a conversion from the pressure head inside the tank to velocity head V3 squared over 2g of the jet exiting the nozzle. Next, we invoke continuity, so the volume metric flow rate of an incompressible fluid is velocity times area or V3 pied by 4 D3 squared. Plugging in the numbers, we find that Q, the volumetric flow rate, is 0.005 cubic meters per second, and the best answer is therefore D. The next problem is a continuation of this problem. Except that now, the question is, what is the pressure in the pipe here, P2? Is it most nearly, which of these four alternatives? So we follow the same procedure again. We apply the Bernoulli equation, only now I'm going to apply the Bernoulli equation between station 1 inside the tank here to Station 2, some point along the pipe. And the equation is as given. For the same reasons as before, V1 is negligible, z1 and z2 go out. But now we have to leave those two terms, p2 and V2 squared, in there. So rearranging, we have P1 is equal to that, or P2, which is what we're looking for, is equal to P1 minus scamina V2 squid over 2g. And to solve the problem, obviously we need to know what the velocity in the pipe is. Now, if we hadn't done the previous problem, we would have had to do this in two steps. First, applying Bernoulli equation from station 1 and 3 to find the exit velocity of the volume flow rate. And then apply continuity, which is the next step, to compute the velocity in the pipe. So from continuity, we know that the volume flow rate at station 2 is equal to the volumetric flow rate at station 3, or V2, the velocity in the pipe is V3 over A3 over A2. Substituting in for the areas, and substituting the numbers, we find that the velocity in the pipe, V2, is 7.78 m/s. And now, we're in a position to go back to this equation and solve for p2. So p1 is 3 Pa, which is 3 times 10 cubed. Substituting in the numbers, we find that the pressure, P2, is 2,960 pascals, or 2.96 kilopascals, and the closest answer is C, 3 kilopascals. Next problem, we have kerosene with a specific gravity of 0.85. Flowing through this pipe and constriction as shown, the gage here reads the pressure difference between the upstream station one and the throat at two. We're given the upstream velocity and the specific weight and the question is, the gage reading, in other words the pressure difference between one and two is which of these alternatives? So, again, we apply our Bernoulli equation between upstream and the throat as shown, where gamma k there is a specific weight of kerosene and, usual, for a horizontal pipe, Z1 and Z2 cancel out or, rearranging for the pressure difference we have p1 minus p2 is gamma k times v1 squared, etc. So, next we have to invoke continuity to find the velocities to solve that equation. And assuming the kerosene is incompressible as usual, we have V1 A1 equals V2 A2, so V2 is given by this expression. We know the diameters substituting in, so the velocity at the throat is 17.8 meters per second. The specific weight of kerosene, by definition, is specific gravity multiplied by the specific weight of water. The specific gravity is 0.85, so the specific weight of the kerosene is 8,330 Newtons per cubic meter. And now we can substitute back into this expression for the pressure difference. So P1 minus P2 is given by that which is equal to a 117 times 10 cubed pascals or a 117 kPA. And so, the closest answer is B, 120 kilopascals. And this completes this discussion of continuity and energy equations.
