The 30-cm-long left coronary artery is 4.6 mm in diameter. Blood pressure drops by 3.0 mm of mercury over this distance. What are the (a) average blood speed and (b) volume flow rate in L/min through this artery?

Video transcript

Hi, everyone in this practice problem, we are being asked to first determine the average fluid velocity. And second, the following flow rate, we will have a fiscus fluid flowing to a 25 centimeter long pipe with a radius of 2.6 centimeter. And the pressure of the fluid drops to 250 pascals along its length. The coefficient of physical custody of the fluid is 2.5 times 10 to the power of negative three pascal seconds. And we're being asked to determine first, the average fluid velocity and second, the following fluoride in liters per minute going through the pipe. The options given are a 1st 33.8 m per 2nd and 2nd, 4.31 times 10 to the power of three liters per minute. B 1st 43.8 m per 2nd and 2nd, 3.31 times 10 to the power of three liters per minute. C 1st 34.8 m per 2nd and 2nd, 4.41 times 10 to the power of three liters per minute. And lastly d 1st 23.8 m per 2nd and 2nd 3.51 times 10 to the power of three liters per minute. So in this practice problem, we are going to employ um the equation where we'll have the quasi equation for fiscus flow to be Q to be equals to pi multiplied by R to the power of four multiplied by delta P of that divided by eight mu multiplied by L just like. So and at the same time, we know that Q or the volume flow rate is going to be calculated by multiplying the area which is pi multiplied by R squared multiply that by the velocity average or the average velocity which is va V G here. So we, we can um equate this two equation right here and combine them together to get an equation for the average velocity or average fluid velocity to give an answer for the first part. So equating this two equation, we will then get pi multiplied by R to the power of four multiplied by delta B, all of that divided by eight mu multiplied by L equals to pi multiplied by R squared multiplied by V average. And we can cross out both PS and the R to the power of two. So the left side will just have R to the power of two. And now we will have the equation for V average to then be equals to delta P multiplied by R to the power of two off that divided by eight M L just like. So, so we are given all of this information in the problem statement. So we will want to actually substitute all that into this equation right here delta P is given to be 250 pascals which is the pressure drop through the length. So 250 pascals multiply that by R squared. So the R here is given by the radius which is 2.6 centimeter. So um converting that in two m, we will have 2.6 times 10 to the power of negative two m. And we want that to be squared off, that is going to be divided by eight multiplied by mu. And mu is going to be the viscosity of the fluid given by 2.5 times 10 to the power of negative three pascal seconds, 2.5. And lastly, the L which is going to be the length is going to be given by centimeter and converting that two m, we will have 25 times 10 to the power of negative two m just like. So, so putting this into our calculator, we will then get for the first part, the average fluid velocity or V average to then be 33.8 m per second just like. So, so now we can actually utilize uh the V average to then calculate our Q or our volume flow rate in liters per minute to then get the answer to the second part. So for the second part, we will utilize the equation of Q equals to pi multiplied by R squared and multiplied by V average. And this will then give us our Q. We can actually also also utilize the first equation or the quasi equation that we have here to determine what Q is. But since the other equation is easier, we can just utilize that because we already have the V average here. So then we will substitute everything we will have pi multiplied by R squared and R squared is 2.6 times 10 to the power of negative two m squared multiply that by the average which is going to be 33.8 m per second, and that will give us the Q to then equals to 0.7178 meter cu per seconds. However, we're asked to calculate the Q or the falling flu rate in liters per minute going through the pipe. So we are uh we're going to convert that by multiplying this wit 1st 1000 liter for every one m cube and also multiplying it again with 60 seconds for every one minute. So multiplying the Q with those two conversion, we will then get our answer to. Then B equals to 4306. liters per minute, which will be uh rounded up to be 4.31 times 10 to the power of three liters per minute. So the answer for the first part, which is the average flow velocity is 33.8 m per second. And the answer for the second part, which is the volume flow rate is going to be 4.31 times 10 to the power of three liters per minute. And this will then give us the answer to be equal to option A in our answer choices. So answer A is going to be the answer to this particular practice problem. That will be it for this video. If you guys still have any sort of confusion, please make sure to check out our other lesson videos on similar topics and that'll be it for this one. Thank you.