20°C water flows through a 2.0-m-long, 6.0-mm-diameter pipe. What is the maximum flow rate in L/min for which the flow is laminar?

Video transcript

Hi everyone. In this practice problem, we're being asked to calculate volume flow rate. We will have a 3.2 m long pipe with a radius of 12.7 millimeter attached to a tank. The tank itself is filled with water at 25 degrees Celsius and the water flows through the pipe at different rates depending on the water height above the pipe in the tank. We're being asked to calculate the greatest volume flow rate in meters cube per hour at which the flow will be laminar. The options given are a 0.128 m cube per hour. B 0.07 m cube per hour, C 0.256 m cube per hour. And lastly D 0.00101 m cube per hour. So in this particular practice problem, we will consider the water to be viscous. A flow is only laminar in the case where the Reynolds number is less than 2000. And Reynold's number itself is given by the equation of re to be equals to row multiplied by V multiplied by L divided by the viscosity mu. So the flow will only be laminar when the re number or the renaults number is less than 2000. So at 25 degrees Celsius, the viscosity of water is going to be known to be around 8.9 times 10 to the power of negative four pascal seconds. And the flow rate, which is what we're going to be interested to find is going to be represented by Q, which is going to be found by multiplying the area cross sectional area multiplied by the average velocity or the average area is going to be pi R squared. So in this case, Q will equals to pi R squared multiplied by the average just like this formula that we have right here. A is the cross sectional area and va VG is the average velocity. So in this case, the average velocity can be obtained from the Renault number formula formula here. So we want to rearrange the Renault number formula so that we can get the average to then equals to re NATS multiplied by the viscosity divided by row multiplied by L. So the density of water is known to be 1000 kg per meter cube DL is given in the problem statement, the mu is 8.9 times 10 to the power of negative four. And the renaults number will be obtained from the condition which is the la in their flow condition that we are interested to find. So in this case, the average maximum velocity, which is the velocity that we're interested to find for or the V average max will then equals to the maximum Reynolds number for the laminar flow multiply that by mu and divide that by row multiplied by L. So let's substitute all of this information in this uh part right here. So in this case, we will then get the average max to then equals to the average max equals to the renaults, the maximum renaults number which is going to be 2000 for a laminar flow condition which we have stated here. And then multiply that with the FICO which is going to be 8.9 times 10 to the power of negative four past 12 seconds to fight that with the row, which is going to be 1000 kg per meter cube. And lastly multiply that with the characteristic length which in this case, the only or the length that we are interested to find is going to be the diameter of the uh pipe itself. So we are given the radius for the pipe. So we have to multiply that by two in order to get the diameter. So two multiplied by 12.7 and we wanna convert that millimeter into meters. So multiply that again by 10 to the power of negative 3 m. So calculating this, we will then get the V average max or the maximum average velocity then equals to 0.0701 m per second. Awesome. So now we can use this uh maximum average velocity to get the greatest volume flow rate by employing the formula for Q. So we get Q to then equals two pi R squared multiplied by V average and looking for the maximum flow rate. So we have Q max to then equals to pi multiplied by R squared multiplied by V average max. So let's substitute all of our known information into this part into this formula. So Q max will then equals to pi multiplied by the radius which is going to be 12.7 millimeters. So convert that into meter multiply it by 10 to the power of negative 3 m squared multiply that by the V average max we just found which is 0.0701 m per second. And that will give us the Q max value in meter cu per second to then be 3.55 times 10 to the power of negative 5 m cube per seconds. What we are being asked for is the greatest fall in fluid in meter cube per hour. So we want to convert the meter cube per seconds into that. So we multiply this with 3600 seconds for every hour. So 3600 seconds divided by one hour. And that will give us the Q max value in meter cube per hour to then be 0.128 m cube per hour. So the greatest volume flow rate is 0.1 to 8 m cube per hour, which will actually be corresponding to option A and option A with the greatest volume rate to be 0.1 to 8 m cube per hour at which the flow will still be lighter will be the answer to this particular practice problem. So that will be it for this video. If you guys still have any sort of confusion, please make sure to check out our adolescent videos on similar topics, but that'll be it for this one. Thank you.