20°C water flows at 1.5 m/s through a 10-m-long, 1.0-mm-diameter horizontal tube and then exits into the air. What is the gauge pressure in kPa at the point where the water enters the tube?

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Hey, everyone. Let's go through this problem here. Water at a temperature of 30 degrees Celsius is transported through a horizontal pipe that is 50 m long and has a diameter of five centimeters flowing at a speed of two m per second. The water then exits the pipe into the air and falls into an artificial lake. Calculate the pressure difference between the water's entry pressure and the ambient atmospheric pressure. The viscosity of water at 30 degrees Celsius is 300.7972 milli pascal seconds. And we have four multiple choice options. Option A 2.2 multiplied by 10 to the power of one pascals. Option B 1.3 multiplied by 10 to the power of two pascals. Option C 1.0 multiplied by 10 to the power of three pascals. And option D 4.1 multiplied by 10 to the power of four pascals. For this problem. We're looking for a pressure difference and we want to relate it to multiple variables including a viscosity. And the only equation we have at our disposal that involves viscosity is the qua equation. To recall that the quasi equation states that the volume flow rate of a fluid Q is equal to pi multiplied by the radius of the flow rate of the power of four, multiplied by the pressure difference divided by eight, multiplied by the viscosity ada multiplied by the length of the tube. L. So first off based on the equation based on the variables given to us in the problem, the length of the tube L is 50 m. The radius isn't given but the diameter is the diameter is given as five centimeters. And recall that radius is just half of a diameter. So that means 5.0 centimeters divided by two is our radius. So that's just 2.5 centimeters. And the viscosity A is given to us in the problem as 0.7972. And then it's given as mili pascals many mini pascal seconds. So to get this into our standard units, I'm gonna multiply this by 10 to the power of negative three. So that it's just in units of pascal seconds. Now, everything is known to us except the pressure difference, which is fine because that's what we want to find. And the volume flow rate, which is not given to us in the problem. Fortunately, for us. However, we have another formula for the volume flow rate. Recall that the volume flow rate is also equal to the cross sectional area through which the flow travels multiplied by the flow speed. Also recall that the surface area of a circle is equal to pi multiplied by the square of its radius. So instead of writing this as a V, we can write it as pi multiplied by R squared V. Everything in this equation is known. So we can set the two cues equal to each other. So pi R squared V then is equal to pi are to the power of four delta P divided by 88 L. Now let's get to work, simplifying this. The pi is cancel out since they're on both sides of the equation. And two of the R's will cancel out because we have an R squared on one side and an R raise the power of four on the other side. So that means this is equal to V equals R raise the power of two multiplied by delta P divided by eight A L. The problem is asking us to solve for the pressure difference. So we want to solve this equation for delta P doing a simple algebraic rewrite. We find that delta P is equal to eight multiplied by a multiplied by L all divided by. And then there's also a V in the numerator A be all divided by are squared. We can get this formula pretty simple just by taking the original simplified equation that we found a minute ago and just multiplying both sides of the equation by eight A to L and then dividing both sides of the equation by R squared. So now to solve this, all we have to do is take our new equation for delta P and substitute in all the variables given to us in the problem. So that's eight and multiplied by a which we established earlier is just 0.7972, multiplied by 10 to the power of negative three pascal seconds multiplied by 50 m or L multiplied by two m per second or V. And all of this divided by the square of the radius, as we mentioned earlier, the radius of the flow is 2. centimeters. So converting that into meters, that's 2.5 multiplied by 10 to the power of negative two m squared. And if we put this into a calculator, then we find a pressure difference of about 0.42 pascals. If we were to rewrite this in scientific notation, then it's about 1.0 times 10 to the power of three pascals. And so that means that this is our final answer to the problem. And if we look at our multiple choice options up above, we can see that that option, that answer does exist as an option as option C option C is 1.0 multiplied by 10 to the power of three pascals. So that is the answer to this problem. I hope this video helped you out. If you'd like more practice, please check out some of our other videos which will hopefully, hopefully give you more experience with these kinds of problems, but that's all for now. I hope you have a lovely day. Bye bye.

20°C water flows at 1.5 m/s through a 10-m-long, 1.0-mm-diameter horizontal tube and then exits into the air. What is the gauge pressure in kPa at the point where the water enters the tube?

Verified Solution

Key Concepts

Fluid Dynamics

Bernoulli's Principle

Gauge Pressure