(II) A 6.0-cm-diameter horizontal pipe gradually narrows to 4....

Question

(II) A 6.0-cm-diameter horizontal pipe gradually narrows to 4.5 cm. When water flows through this pipe at a certain rate, the gauge pressure in these two sections is 36.5 kPa and 22.6 kPa, respectively. What is the volume rate of flow?

Accepted Answer

All for one. Let's take a look at this practice problem dealing with fluid flow. Now, in this problem, water flows through a horizontal hose that narrows from eight centimeters diameter to a four centimeter diameter. And the question wants to know was the volume flow rate. If the gauge pressures in these sections are 33 kg pascals and 25 kg pascals respectively. We're given four choices as our answers. O A is 1.3 multiplied by 10 to the negative 3 m cubed per second. OB is 2.3 multiplied by 10 to the negative three uh meters cube per second. OC is 5.2 multiplied by 10 to the negative 3 m cube per second. And choice D is 7.8 multiplied by 10 to the negative 3 m cubed per second. Now, we're interested in the volume flow rate. And so let's recall our formula for volume flow rate. So our volume flow rate, Q is equal to the cross sectional area multiplied by the speed of the fluid. So we have the cross sectional area a multiplied by the speed of the fluid B. So Q is equal to A multiplied by V. Now I'm given the diameter in two different sections of this pipe. So I ostensibly could calculate the area but I have no information about my speed. So in order to deal with the speed we need to use Bernoulli's equation. So recall Bernoulli's equation. So here we'll have P one plus row GH one plus one half row V, one squared is equal to P two plus row GH two plus one half row B two squared. So here P is the pressure rows, the density of the fluid, in its case, density of water, G is the acceleration due to gravity. H is the height of the fluid and V is the speed of fluid. So what this equation is basically saying is that the pressure plus this row GH one plus the one half row one squared has to be the same at both points. And this really comes from um conservation of energy. So it's basically saying that the energy at one point has to be equal to the energy of the other point. So I was told that we have a horizontal pipe that means H one and H two are the same. So those two terms will actually cancel out. And so I can simplify this a little bit. And I'll just have P one plus one half row B, one squared is equal to P two plus one half row B two squared. Now, I was given the two pressures and I have the density since I know that this is water. But I have again, no information about my speeds. So in order to figure out what the speeds are, I need to use the continuity equation. So we call for the continuity equation. That means that the area multiplied by the speed has to be the same at every single point. So here I'll have a one multiplied by V one is equal to a two multiplied by B two. Now, for the areas I have the cross sectional area of a circle since I was given diameters here. And so the area of a circle is just pi R squared. So for area one, I'll have pi multiplied by the quantity. So radius is just my diameter divided by two. So I'll have D one divided by two, that whole quantity squared multiplied by V one. I'll do the same thing for a two. So I'll have pi multiplying the quantity of D two divided by two in quantity squared multiplied by V two. Now I can solve this for either speed and then plug that back into the equation. So let's just solve this for V two. In doing so lots of things cancel. The pies will cancel, the one half squared will cancel. And what I'm left with is V two is equal to the quantity of D one divided by D two. All quantity squared multiplied by V one. So I can take that and plug that back into my Bernoulli's equation. And in doing so, I'll just have V one as my only unknown. So I'll have P one plus one half row. B one squared is equal to the left hand side of the equation remains the same, the right hand side, I'll have P two plus one half row. And in place of V two, I will have D one, the quantity D one divided by D two might he squared multiplied by B one. And then that whole quantity also gets squared. So let's just uh write that out and collect light terms on the same side of the equation. So I'm gonna collect all my B one terms on the left hand side of the equation. So I'll have one half row multiplied by the quantity of one minus. Since I've moved the um, term, the right to the left hand side, it has a minus sign and then I'll have D one divided by D two raised to the fourth power. I'll have the quantity of D one divided by D two raised to the fourth power. And that whole quantity of the one minus D one divided by D two is the fourth, that whole quantity is multiplied by B one squared. And so this is gonna equal to, equal to. And on the right hand side of the equation, I'll have P two minus P one. So I can now solve this for V one, we have V one, it's equal to the square root of two multiplied. And I'm actually gonna switch the signs here. So I'm gonna multiply both sides by negative one. And the reason for that is because I know that P one is greater than P two. So this will just allow me to keep positive quantities in the numerator and denominator. So I've P one minus P two divided by bro multiplied by the quantity of D one divided by D two quantity to the fourth power minus one. So I now have an expression for V one and I can plug that into my expression for my volume flow rate. So coming back to the volume flow rate, I'll have Q is equal to or the area we're just gonna choose um the first point. So we'll have pi multiplied by the radius squared. So this could be D one divided by two is my radius. That whole quantity is squared multiplied by V one. So I'm gonna have the square root of two multiplied by P one minus P two, all divided by row multiplying the quantity of D one divided by D two, that whole quantity raised to the fourth power minus one. So at this point, I can actually plug in values. So for D one, I will have the eight centimeters for D two, I'll have the four centimeters for P one, I'll have the 33 kg pascals. And for P two, I'll have the 25 kg pascals. So plugging all those values in, I'll have you is equal to pi multiplied by D one which is eight centimeters. But I need to convert that to meters. They're divided by 100. That'll give me 0.08 m divided by you. That whole quantity is squared multiplied by the square root of two multiplied by the quantity of the 33 kg pascal. So it's gonna be 33 multiplied by 10 to the three pascals minus 25 kg pascals. So 25 multiplied by 10 to the three pascals. And that gets divided by the quantity of the density is low. So the density of water is 1000 kilograms per meter cubed, that's a constant that we can look up multiplied by the quantity of D one which was eight centimeters divided by two, which was four centimeters race to the fourth power minus one. Now, I did not convert the centimeters and the denominator of that fraction. And that's because the units will cancel out. So if I were to convert those 2 m, um that conversion factor would also can cancel out. So this will just save me a little bit of time. So anytime you're divide in two quantities with the same unit that you would need to convert to si units, you can actually ignore that um since the units would cancel out. So everything on the right hand side is a number and so we can plug that into our calculator. And that turns out to be Q is equal to 5.2 multiplied by 10 to the negative three meters cubed for a second. So the place where um seconds actually comes in, if you don't see seconds at all, any of those units recall that pascals is um force per area. So that'd be newtons per meter squared. And Newton's is a kilogram meter per second squared. That's where the second comes in. And then everything else cancels out. And inside the square root sign, you're left with just meter squared per second squared. And when you take the square root of that, that gives you meters per second, which is the speed, which is what that square root actually represents. So now that we have our final answer of the 5.2 times 10 of negative three, we look up, we see that that corresponds to answer C. So just to recap, we were asked to find the volume flow rate. But in order to do that, we needed to combine our Benelli equation and our continuity equation in order to find the speed to use in our expression for the uh volume flow rate. But once we did that, we were able to calculate it to come up with our final answer. So I hope that this has been helpful and I'll see you in the next video.

(II) A 6.0-cm-diameter horizontal pipe gradually narrows to 4.5 cm. When water flows through this pipe at a certain rate, the gauge pressure in these two sections is 36.5 kPa and 22.6 kPa, respectively. What is the volume rate of flow?

Key Concepts

Continuity Equation

Bernoulli's Principle

Flow Rate

Watch next