A soft drink (mostly water) flows in a pipe at a beverage plant with a mass flow rate that would fill 220 0.355-L cans per minute. At point 2 in the pipe, the gauge pressure is 152 kPa and the cross-sectional area is 8.00 cm^2. At point 1, 1.35 m above point 2, the cross-sectional area is 2.00 cm^2. Find the (b) volume flow rate. (c) flow speeds at points 1 and 2.

Question

Accepted Answer

Everyone in this problem. We have a water bottling plant that uses a pump that fills 900.5 liter bottles every five minutes. We have a gauge connected to the pipe originating from the pump at point A. That reads 135. Kill a pascal's in the pipeline. Cross section at that point is 20 centimeters squared. Point B. Is elevated two m from point A. And the pipe cross section is five centimeters squared. Were asked to determine the volume passing through a fixed point in the pipeline. Okay. In liters per second and the speed of the water. Okay, So the volume passing through and the speed of the water at points A. And B. Alright. So let's start with the first one. We're asked to determine the volume passing through a fixed point in the pipeline. Okay, well that's gonna be our volume flow rate. So volume flow rate, we're gonna call this Q. Okay, and this is just gonna be the volume per time. We want this in leaders per second. So thinking about the volume. Okay well the pump is filling Bottles. Okay? And each bottle is .5 L. So that's going to be 900 times 0. l Every five minutes. So per five minutes. That's alright. This many leaders in this amount of time. Now, we're told that we want this in Leaders for a second. Okay, So let's work with that. So in the numerator, if we multiply we're going to get 450 liters. Okay, in the denominator we still have five minutes and we want to convert two seconds. So let's multiply In every minute. OK. For one minute we have 60 seconds. Now we have a unit of minutes of the numerator and denominator. Those will divide out. And we're gonna be left with 450 liters per 60 seconds, which gives us a volume flow rate of 1.5 liters per second. Okay, so this is Q. R volume flow rate which is 1.5 liters per second. Alright, So if we look at our answer choices, okay, we see that Q. 1.5 in B. C. And D. Okay, So we can eliminate A. And we can eliminate E. Because those don't have the volume flow rate that we found. Okay, let's move on to the second part of this problem and try to figure out the speed of the water at points A. And B. All right, now, we're looking at the speed of the water, let's recall that our volume flow rate Q. Is equal to A. Okay, so the cross sectional area times the speed. Okay? So if we're trying to find the speed, well, that's gonna be equal to Q. Or flow rate that we just found divided by the area A. Which is gonna be 1.5 L per second divided by that cross sectional area. So now we just need to figure out what A. Is. Okay, at the point we're interested in. So we're interested in point A. We're interested in point B. And the problem tells us the cross section at those two points at point B. It is five centimeters per second And at a it's 20 cm/s. Alright, so at point A let's start there. We have that our speed V is going to be equal to 1.5 liters per second. Okay, That volume flow rate divided by the cross sectional area which were given in centimeters squared. Okay, so 1.5 liters per second divided by centimeters squared. Ok? We want to convert everything into meters so that it is in our standard unit. Okay, so on the right hand side, 1.5 liters per second. Okay, converting two m cubed. We can multiply we know that one m cubed is equivalent to 1000 liters. Okay, so we have to divide by 1000 and we get 0.15 m cubed per second. And similarly for our 20 centimeters squared we have centimeters squared times. We know that in one m squared we have centimeters squared. And so we get 0. meters squared there. Okay, and just be careful when you're converting centimeters squared, two m squared, it's not the same as converting centimeters to meters. Okay, there is that square on the unit. So we're dividing by 1000 instead of 100. Okay, so just watch that that's an easy place to make a mistake. Alright. So if we go back to our speed equation 0.0015 m cubed per second divided by 0.2 m squared. Okay when we divide these meters cubed per second, divided by meters squared, we're gonna be left with units of meters per second which is a unit for speed. We get zero points 75 m per second. Okay? Now if we want to go back two centimeters per second, which I think the answer choices are in. Let's just double check. Yes. The answer choices are in centimeters per second. So let's just go back two centimeters per second. Here, We're gonna multiply by 100. Okay. Which is going to give us cm/s. So we multiplied by 100 there for that conversion. Alright so we figured out the speed at point A. Let's do the same for point B. Okay. Alright. So let's just label this as V. A. So we don't get mixed up speed at A. Now for B. Same thing. Our flow rate, 1.5 liters per second divided by our cross sectional area at B. Which we're told is five centimeters squared. Okay now we've already done the conversion for 1.5 liters per second. Let's do it for five centimeters squared. Okay so we're gonna multiply by one m squared per 1000 m centimeters squared Like this. Okay. And we're gonna get 0.0005 m squared there. Alright. So we get back to our equation, we're looking at our speed. We have 0.15 m cubed per second. Okay. That volume flow rate divided by 0.5 m squared. Again our unit will be meters per second. We get a value of three m per second. Okay? And again if we want to convert two centimeters per second we multiply by 100. We're going to get 300 cm/s for the speed at Point B. Alright. So if we go back up and look at our answer traces, okay? We found that the volume passing through the fixed point Okay. Was 1.5 L /s. The speed of the water at point A. Was 75 centimeters per second and the speed of the water at point B was centimeters per second. And so we have answer choice D here. Thanks everyone for watching. I hope this video helped see you in the next one.

Verified Solution

Key Concepts

Mass Flow Rate

Continuity Equation

Bernoulli's Principle