Power in Flows: Conveyor Belts, Wind and Water

Transcript

0:00Introduction

Hello everyone! Let's start in a factory. Picture a conveyor belt running at a steady speed, with sand pouring onto it from above. To stop the belt from slowing down, a motor has to supply extra power. Now here is the strange part. Only half of that extra power ends up as kinetic energy of the sand. Exactly half, whatever the speed of the belt. So where does the other half go? Hold on to that question. Today we find the power when mass keeps flowing: conveyor belts, wind, and water. Questions like these do come up in JEE, so let's go.

0:38Conveyor belt: the extra force

So, let's set it up. The belt moves at a constant speed v. The sand falls vertically, so it lands with no speed along the belt, okay? And mu, the Greek letter on screen, is the rate at which the sand lands, in k g s to the power minus one: kilograms per second. Clear? Now think. The belt's speed is not changing. So why does the motor need any extra force at all? Because new sand keeps landing! Yes, exactly! Each second, mu kilograms of sand must be taken from rest up to the belt's speed v. With me? So, each second, the belt gives the sand a momentum of mu times v. And force is the rate of change of momentum, remember? So the extra force is F equals mu v. Okay? And the power? Power is force times velocity. So P equals F v, that is, mu v times v: mu v squared. Write this down. Got it?

1:39Where does the motor's power go?

Now, back to our puzzle. The motor supplies mu v squared every second. So tell me. Each second, mu kilograms reach the speed v. How much kinetic energy is that? Half m v squared, with mu in place of m: half mu v squared. Correct! K E on screen is short for kinetic energy. But wait. That is only half of what the motor gives, right? So where does the other half go? Into heat! When the sand lands, it is not moving along the belt, but the belt is. So the sand slips on the belt for a short while. Friction acts during that slipping and turns the other half, half mu v squared, into heat. Half to kinetic energy, half to heat. Puzzle solved! And careful in the exam: if you write the motor's power as half mu v squared, you have forgotten the heat. That is the trap. Clear?

2:35Wind: the power goes as v³

Now, wind. Air of density rho moves at speed v through an area A, the area swept by the turbine's blades, okay? Rho is the Greek letter that looks like a small p. In one second the air moves a distance v, so a column of volume A times v passes through. Its mass is rho A v. That is the mass flowing every second. With me? Each second, this air carries kinetic energy: half, times rho A v, times v squared. So the wind carries kinetic energy at the rate half rho A v cubed. See? One v comes from the mass, and two from v squared. Right? But a turbine cannot take all of this energy. It extracts only a fraction of it. So, for a given turbine, its power is proportional to v cubed. The sign that looks like an infinity sign, open on one side, means proportional to. Okay? Now tell me. If the wind speed doubles, what happens to the power? Four times, because kinetic energy goes as v squared? No, no, careful! The mass flowing each second also doubles. So the power goes up by two cubed: eight times! Remember this one. Got it?

3:51Falling water and water jets

Next, water falling down a dam to a turbine. The m with a dot on top, m dot, is the mass rate: kilograms of water per second. Each second, m dot kilograms fall through a height h and lose potential energy m dot g h, where g is the acceleration due to gravity. So the power is P equals eta, m dot, g, h. That eta in front is the Greek letter like an n with a long tail. Okay? And what is eta? It is the efficiency: the fraction of that energy turned into electrical power. Clear? Now, one more. Water shoots out of a hose nozzle as a jet. Which formula gives the power of the jet? Think of the wind. Yes, exactly the same as the wind! The jet has speed v and comes out of a nozzle of area A, right? Same reasoning, so its power is half rho A v cubed, with rho now the density of water. Same idea, new place. Got it?

4:54Example 1: power output of a wind turbine

Alright, let's try one. Wind at ten metres per second passes through the fifty square metre swept area of a turbine. The air density is one point two kilograms per cubic metre. Find the wind power, and the electrical output if the turbine converts forty percent. Pause the video and try it. Done? Okay, let's check. Wind power is half rho A v cubed. So, half times one point two, times fifty, times ten cubed. Right? Half of one point two is zero point six. Times fifty, thirty. Times ten cubed, which is one thousand: thirty thousand watts. Capital W is the watt, and k W is the kilowatt, so that is thirty kilowatts. And look at that ten cubed on screen: the speed is cubed, not squared. Square it by mistake and your answer is ten times too small. Got it? Now, the turbine converts forty percent of this. Forty percent of thirty kilowatts: twelve kilowatts of electrical output. That's it!

5:59Example 2: sand on a conveyor belt, force and power

Now one with three parts, and it brings back our puzzle. Sand falls vertically at four kilograms per second onto a conveyor belt moving at one point five metres per second. Find the extra force and the extra power needed to keep the belt's speed constant, and how the power is shared. Pause the video and try it. Done? Okay, let's check. Here mu is four kilograms per second, and v is one point five metres per second. F equals mu v: four times one point five, six newtons. Capital N is the newton. Clear? Now the power. Many students write half mu v squared here. Should we? No, no, careful! That is only the kinetic energy part. The motor's power is F times v: six times one point five, nine watts. Okay? Now, how is it shared? Kinetic energy gained per second: half times four times one point five squared. One point five squared is two point two five; times two, four point five. So four point five joules every second, that is, four point five watts. Right? And the rest, nine minus four point five, which is four point five watts, becomes heat as the sand slips on the belt. Exactly half and half, just like our puzzle at the start! See how neat that is?

7:29Quick revision: conveyor belt, wind and falling water

So, quick revision. Conveyor belt: the extra force is mu v, the extra power is mu v squared, and it splits into half kinetic energy, half heat. Wind or a water jet: half rho A v cubed, so the power goes as v cubed. Falling water: P equals eta m dot g h. Now go and try the practice problems on this page. You will enjoy them. All the best!