A nuclear power plant draws 3.0 x 10⁶ L/min of cooling water from the ocean. If the water is drawn in through two parallel, 3.0-m-diameter pipes, what is the water speed in each pipe?

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Hi everyone in this practice problem. We will have to calculate the velocity of the water moving in the pipe. We will have a water tank containing a large volume of water with three pipes each with a radius of 2.54 m draining the water from the tank at a combined rate of 8.0 times 10 to the power of seven gallons per hour. When assume that the pipes carry equal volumes throughout the tree. And we're being asked to calculate the velocity or how fast does the water move in the pipes? The options given are a 5.53 m per second. B 49.8 m per second. C 1.38 m per second and D 12. m per second. So we are going to assume that the three pipes are going to be identical. Therefore, the speed of water in each pipe is going to be the same. What we know is the combined rate which is going to be the Q which is going to be given by 8.0 times 10 to the power of seven gallons per hour. So I am going to first convert the gallons per hour into meter cube per seconds to make it easier for us to actually work with. So I'm gonna multiply this with one m cube for every 264.1 gallons and multiplied that again with one hour for every 303, seconds. That will give us the queue of 84. m cube per seconds or the combined rate of of 84.12 m cube per seconds. Fully meric fluid can be calculated by multiplying the velocity of the water with the cross sectional area where P is the velocity and A is the cross sectional area of the pipe. Since we have three pipe, then the Q will be equals to three multiplied by V multiplied by A and in this case, that will be equals to 84.12 m cube per seconds. So in this case, what we want to do is to then rearrange this so that we will be able to find the velocity of the water moving into the pipes. So velocity will then be equals to 84.12 m cube per second divided by three A A or the cross sectional area will be pi R squared. So I'm gonna substitute that N so V will equals to 84. m Q per seconds divided by three multiplied by pi multiply that again by R squared and R squared is given to be 2.54 m squared. So I'm gonna input that here 2.54 m squared with the A being pi R squared just like. So, so calculating this, we will then be able to find the velocity or the speed of the water within the pipes, which will then come up to a value of 1. m per second. So the velocity of the water moving into the pipes or moving inside of the pipes is going to be 1.38 m per second, which will correspond to option C in our answer choices. So option C is going to be the answer to this particular practice problem and 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, but that will be it for this one. Thank you.

A nuclear power plant draws 3.0 x 10⁶ L/min of cooling water from the ocean. If the water is drawn in through two parallel, 3.0-m-diameter pipes, what is the water speed in each pipe?

Verified Solution

Key Concepts

Flow Rate

Cross-Sectional Area

Continuity Equation