(b) A pressure gauge reads 50 kPa as water flows at 10.0 m/s through a 16.8-cm-diameter horizontal pipe. What is the reading of a pressure gauge after the pipe has expanded to 20.0 cm in diameter?

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Hello, everyone. Let's go through this problem. Oil of density 850 kg per cubic meter circulates through a horizontal hose system inside an olive oil filling machine. First, the oil flows at a speed of eight m per second through a hose with a radius of one centimeter. An oil manometer indicates a pressure of seven multiplied by 10 to the power of four pascals. A horizontal hose elbow connects the one centimeter radius hose to a smaller hose with a radius of 10.75 centimeters, calculate the pressure inside the smaller hose. And we're given four options to choose from option A 1.1 multiplied by 10 to the power of four pascals. Option B 9.3 multiplied by 10 to the power of four pascals. Option C 2.1 multiplied by 10 to the power of five pascals and option D 7.8 multiplied by 10 to the power of five pascals. So, recapping what the problem is describing, we have some hose of some radius which then connects to a s a smaller hose, the smaller radius. So I'm representing that in this diagram by using two different sizes of the sections of hose and I'll label the first section one and the second section two. And as for the variables, we're given the problem, we're told that V one is equal to eight m per second. We're told that our one is equal to one centimeter. We're told that R two is 0. centimeters. And we're told that P one is equal to seven, multiplied by 10 to the power of four pascals because we're trying to find a relationship between the different variables of the fluid that exist in the two different sections of the tube. What we can do is use the Bernoulli equation which states that the pre that along a streamline of fluid, the pressure plus one half the density multiplied by the square of the flow speed plus the density multiplied by the gravitational accelerate acceleration multiplied by the vertical position is constant anywhere along a streamline in the fluid. So if we use the left hand side of this expression, the Bernoulli equation for the different sections of the tube and set them equal to one another, then we can find a relationship between the different variables. So let's start. So P sub one for the variables in the larger thicker hose, P sub one plus one half multiplied by the density of the fluid. The sub one squared plus row G in the vertical position, Y sub one equals P sub two plus one half row V sub two squared plus row G Y sub two. Now, one way we can simplify this is by realizing that it's entirely up to us to define what the positions are of the streamline we're looking at. So if we look at the, at the hose as being completely horizontal with no significant change in elevation, then we can define our positions for points one and points two to be at the exact same height along the stream line. Which means that we can assume that Y sub one is equal to Y sub two. This means that the potential energy terms can cancel out. So we'll now have a simpler equation to work with P sub one plus one half row V one squared equals P sub two plus one half row V two squared. The problem is asking for us to calculate the pressure inside the smaller hose. So we're essentially looking for P sub two. So let's solve the beli equation we found for P sub two. So first, I'm going to subtract from both sides of the equation one half row V sub two squared. And then we can simplify this by factoring the one half row out of the two V terms. So P sub two is equal to P sub one plus one half row multiplied by in parentheses, B sub one squared minus V sub two squared. Now we need to find a relationship between V sub one and V sub two before we can finally solve this. So another way we can get a relationship between the two speeds is to use the continuity equation which states that the cross sectional area of the flow multiplied by the speed of the flow is constant everywhere throughout the flow. So in other words, a one, V one is equal to a two V two. So let's expand this out a bit using what we know to find a relationship between V sub one and V sub two. And first off recall that the formula for the cross sectional area of a circle is equal to pi multiplied by the square of its radius. So let's apply that to the continuity equation. So instead of a sub one, I'm going to write pi R sub one squared multiplied by V one equals pi R sub two squared multiplied by V sub two. Now the P can cancel out and are just left with R one squared. V one equals R two squared V two. Now let's algebraically rewrite this to solve for one of the V S. It doesn't really matter which I'm just going to solve for V sub two by dividing both sides of the equation by R sub two squared. So we then find that V sub two is equal to V sub one multiplied by R sub one squared divided by R sub two squared. So now that we've found a formula for V sub two in terms of V sub one we can plug that in to the V sub two that we had in our Bernoulli equation. So P sub two is equal to P sub one plus one half row. And then in parentheses V one squared minus and then the square of the formula we found for V sub two which is V sub one multiplied by R one squared over R two squared. And that whole thing in parentheses is also squared. And now to solve this down further, let's distribute the exponents. So P one plus one half row multiplied by, in parentheses. V sub one squared minus V sub one squared multiplied by R sub one, raise the power of four divided by R sub two, raise the power of four to simplify this further. Let's factor out the V one squared. So P one plus one half row multiplied by V sub one squared times or multiplied by in parentheses, one minus R sub one divided by R sub two, all raise the power of four. And now all that's left for us to do is to plug in the values that we were given in the problem. So for P sub one, we know that that has been given to us as seven multiplied by 10 to the power of four m, seven multiplied by 10 to the power of four uh pascals, not meters whoops plus one half multiplied by the density which was given to us as 850 kg. Per cubic meter, 100 and 50 kilograms per cubic meter multiplied by the square of V sub one which is eight m per second, multiplied by one minus, then raise the power of four of the ratios of R one to R two. So that's one centimeter divided by 0.75 centimeters. And we don't need to do any unit conversions with that because the centimeters will cancel out. So even if we wanted to convert those radii from centimeters to meters, it would be ultimately pointless. But if we were to then put all that into a calculator and will then find a final pressure of about 1.1 multiplied by 10 to the power of four pascals. And that then is our answer to the problem that is the pressure in the second section of the pipe. And if we look at our options, we can see that option. A says that 1.1 multiplied by 10 to the power of four pascals. So that is the answer to this problem. I hope this video helped you out. If you're looking for more practice, please check out some of our other videos which will hopefully give you more experience with these types of problems. But that's all for now. I hope you all have a lovely day. Bye bye.

(b) A pressure gauge reads 50 kPa as water flows at 10.0 m/s through a 16.8-cm-diameter horizontal pipe. What is the reading of a pressure gauge after the pipe has expanded to 20.0 cm in diameter?

Verified Solution

Key Concepts

Bernoulli's Principle

Continuity Equation

Pressure Measurement in Fluids