OpenStax College Physics, Chapter 12, Problem 54 (Problems & Exercises)

This is College Physics Answers with Shaun Dychko. We're going to calculate the Reynold’s number for water flowing through a firehose and then also through the nozzle of the firehose and we're going to begin by writing down all the things that we're told that the diameter of the hose is 6.4 centimeters. But we want to divide that by two since the radius is what going to be used in our Reynold’s number formula. And so we convert it into meters by multiplying by times 10 to the minus two and divide that by two. We get 3.2 times ten to the minus two meters is the radius of the hose. The volume flow rate through the hose and the nozzle. And they have to have the same volume flow rate according to the equation of continuity, because the water is not compressible. So the rate at which the volume goes through the hose has to be the same as the rate at which the volume goes through the nozzle So there's no subscript here in other words, there's no ‘H’ or ‘N’ needed because it's the same volume flow rate in both portions. We're going to convert that into a cubic meters per second by multiplying by one cubic meter for every 1000 liters. And there's 40 liters per second becomes 40 times ten to the minus three cubic meters per second. The pressure at the beginning of the hose is this and the hose goes up this height of 10 meters. But it turns out that these two pieces of information are not necessary. They are extraneous, but we don't really know that at this point. So let's just write down all the stuff we know. And the radius of the nozzle is half three centimeters which is 1.5 times 10 to the minus two meters. So Reynold’s number is two times the density of the fluid times its speed through the tubet times the radius of the tube divided by the viscosity of the fluid. And so the Reynold’s number in the hose is two times water density times the speed of the water through the hose times the hose radius divided by viscosity. And the volume flow rate is the speed of the fluid through the hose times the cross-sectional area of the hose. And we can solve this for V H because we want to make a substitution for it in our Reynold’s number formula. And so we divide both sides by area of the hose and we have V H is Q over A H. And the cross-sectional area of this cylindrical hose is a circle. And so we have pi times hose radius squared is the area and all of this gets substituted in for V H in our Reynold’s number formula. And we do that here and the R H squared in this denominator here one of them cancels with this R H here and we're left with R H to the power of one. And we can write that down here multiplied by this pi times the viscosity and that's two times rho Q over all that. So, the Reynold’s number in the hose is two times density of water 1000 kilograms per cubic meter times the volume flow rate, 40 times ten to the minus three cubic meters per second divided by pi times the radius of the hose, 3.2 times ten to the minus two meters multiplied by the viscosity of water which we have to look up in table 12.1 and we'll assume that water is at 20 degrees Celsius say, so that's 1.002 milla pascals seconds. And the ‘milla’ means multiply by 10 to the minus three Pascal seconds. This works out to 7.94 times 10 to the five and that is more than 3000 and 3000 is this threshold beyond which the flow is definitely turbulent. And so there we go, turbulence is happening in the hose. Now for the Reynold’s number in the nozzle, we're gonna do a slight trick to save us from having to write down so many big numbers. And we have this, the logic is the same for getting the formula for the it's the same formula for the Reynold’s number in the nozzle as it was for the hose. But we use the radius of the nozzle instead of the radius of the hose. Everything else is the same. It's the same volume flow rate, same viscosity of water, same density of water. And now instead of plugging in all these different things, which we could very well do, and that would be fine. I'm choosing instead to multiply R N by the number one, because it's the number one, you can do that always you are allowed to multiply by the number one. That doesn't change any quantity. But we're writing this one in a strategic way as radius of the hose divided by radius of the hose and doing so, it causes us or allows us to write this. Right here. And this is the Reynold’s number in the hose, and we're left with this portion here, we're dividing all of that by R N over R H which is a bit confusing to divide a fraction by a fraction. So instead I’m going to multiply by the reciprocal of all this, which is R H over R N. So the radius of the hose divided by the radius of the nozzle gets multiplied by the Rennold’s number in the hose. So that's 3.2 centimeters divided by 1.5 centimeters and because we're dividing the same units, we don't need to convert this into meters. And so that saves us a little bit more writing, times by the answer from part up here. Reynold’s number for the hose and we get 1.69 times 10 to the six is the Reynold’s number in the nozzle, which also is more than 3000. And so it is also turbulent flow in the nozzle. And we expected turbulent flow in the nozzle because you can hear turbulence in the nozzle at the end of a garden hose. That sound of the water rushing out of a constricted nozzle is caused by the turbulence.