Bernoulli's Equation and its Applications

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. In a strong storm, the wind blows sideways, across the top of a house. And yet roofs get lifted straight up and thrown off. How can a sideways wind push a roof upwards? Strange, right? One equation explains it: Bernoulli's equation. Questions on pipes, tanks and jets of water come up often in JEE, so let's begin.

0:29Bernoulli's equation: energy along a streamline

So, look at the figure: water flowing through a pipe that narrows and rises. At section one, capital P one is the pressure, capital A one the area, and v one, the coloured arrow, the speed. Small h one is the height above the dashed reference line, and section two has the same labels with subscript two. Now tell me: what stays the same as the water flows? The energy per unit volume, yes! P, plus half rho v squared, plus rho g h, equals a constant. Rho, the Greek letter, is the density of the fluid, and g is the acceleration due to gravity. The figure's last line writes it between the two sections, beside continuity: A one v one equals A two v two. Okay? But careful: the flow must be steady, with the velocity at each point fixed in time, and both points must lie on the same streamline. Right? The fluid must also be ideal: incompressible, so its density stays fixed, and non-viscous, so no energy is lost to friction inside it. Clear? Now the three terms. P is pressure energy per unit volume. Half rho v squared is kinetic energy per unit volume: half m v squared, with rho, the mass of one unit volume, in place of m. And rho g h is potential energy per unit volume. So this is just conservation of energy. Got it?

1:59Faster flow means lower pressure

Now the most useful case: horizontal flow. h is the same everywhere, so rho g h is the same on both sides and drops out. Then P plus half rho v squared is constant. Okay? In the narrow part of a pipe, the water is squeezed, and it moves faster. So is the pressure there higher, or lower? Many students say higher, because the water is squeezed. No, no, careful! The speed goes up, so half rho v squared goes up, and the sum is fixed. So, as the double arrow on screen says: v larger, P smaller. Faster flow, lower pressure. With me? And now, our puzzle. In a storm, fast air rushes over the roof, while the air inside the house is still. Lower pressure above, so the higher pressure inside pushes the roof up! Mystery solved. Clear? Now let us put numbers on it, from one example on this page: wind at forty metres per second, written m s to the power minus one, and air of rho one point two kilograms per cubic metre, written k g m to the power minus three. The pressure above drops by half rho v squared, nine hundred and sixty pascals, P a for short. Okay? On a roof of one hundred square metres, that is about ten to the power five newtons, upwards! The wavy equals sign means approximately, and capital N is the newton. Got it? The same idea gives lift to an aeroplane wing, the aerofoil: air moves faster over its top. Atomisers work this way too, and a spinning cricket ball swings because air moves faster past one side: the Magnus effect. Right?

3:47Torricelli: water out of a hole in a tank

Alright, now a wide tank with a small hole in its side, at depth small h below the open surface. Okay? Use Bernoulli between the surface and the hole. How fast does the water come out? See, both are open to the air, so the pressures cancel, and in a wide tank the surface barely moves. So rho g h equals half rho v squared, and v equals root two g h, the square root of two g h. This is Torricelli's law, the speed of efflux: just like a stone falling freely through h. Clear? Now stand the tank on the ground, with capital H the liquid height. The hole is then capital H minus h above the ground, and the jet falls through that height. Right? Just like a ball thrown horizontally, the time to fall, t, is the root of two times capital H minus h, over g. Range x equals v times t, and look, g cancels! x equals two times the root of h times capital H minus h. Got it? Now think. The deepest hole gives the fastest jet. So should the hole be at the very bottom for the longest jet? No, no, careful! At the very bottom, the jet has no height to fall, so the range is zero. We want h times capital H minus h to be largest. Their sum is fixed at capital H, so the product is largest when they are equal: small h equals capital H over two, and the range is then capital H itself. Okay?

5:22Example 1: pressure in the narrow part of a horizontal pipe

Now, example one. Water flows through a horizontal pipe, from a section of area ten square centimetres, with speed two metres per second and pressure two times ten to the power five pascals, to a section of area five square centimetres. Find the pressure there. Pause the video, and try it first. Done? Okay, let's check. Continuity first: A one v one equals A two v two, so ten times two equals five times v two. Clear? So v two is four metres per second. The areas appear as a ratio, so no need to convert square centimetres. Right? Before calculating, predict: will P two, in the narrow section, be more than P one, or less? Less! Faster flow, lower pressure. Our pipe is horizontal, unlike the figure's, so the rho g h terms cancel: P one minus P two equals half rho, times v two squared minus v one squared. With me? For water, rho is one thousand kilograms per cubic metre. Half times one thousand, times four squared minus two squared, that is sixteen minus four: six thousand pascals. Okay? So P two equals two times ten to the power five, minus six thousand: one point nine four times ten to the power five pascals. Got more than P one? Then check your sign! Clear?

6:52Example 2: range of a tank jet and the best hole depth

Now a harder one, example two. Here, a tank on the ground holds water twenty metres deep, and a hole is made five metres below the surface. How far from the tank does the jet land? And where should the hole be for the greatest range? Pause the video, and try both parts. Done? Okay, let's check. Careful with the letters: in our formula, is small h five metres, or fifteen? Five metres, yes! Small h is the depth below the surface, not the height above the ground. Capital H is twenty, so the hole is fifteen metres above the ground. Right? So x equals two times the root of five times fifteen: two root seventy five, which is ten root three, about seventeen point three metres. Got it? Second part: not at the bottom, remember! The hole should be at half the depth: capital H over two, ten metres below the surface. With me? And the range is then capital H itself: x max, x with the subscript max, equals twenty metres, up from about seventeen point three. Clear?

8:04Quick revision: Bernoulli, efflux speed and maximum range

So, quick revision. Bernoulli: P, plus half rho v squared, plus rho g h, is constant, for steady ideal flow along one streamline. Faster flow, lower pressure: wings lift, roofs fly off, spinning balls swing. From a tank, the efflux speed is root two g h, and the range is largest, capital H, for a hole at half the depth. Now try the practice problems on this page. All the best!