# Intro - Torricelli's Equation Sometimes, tanks burst - no engineer is perfect! However, as the backbone for said engineers it is our job to shuffle all the beautiful theories we have found so that they can turn it into a bad program, which they'll be reliant on as they refuse to be mathematicians. When a tank leaks, the water tends to come out like a jet, similar to [[Newton's Laws and SUVAT#Projectile Motion diagram REQUIRED|projectile motion.]] ![[Pasted image 20231009143255.png]] Yep, the cylinder's our tank, spitting out some sort of fluid, whatever it is. The question is, how do we find the velocity of this fluid? Well, Evangelista Torricelli had an idea! He'd purposely burst a bunch of water tanks just to model the speed of flow empirically, which by today's standards is a physics sin. Can't argue with results though, he gave us the equation: $v_{f} = \sqrt{ 2a \Delta h }$ **Where:** - $a$ is the acceleration of the fluid, given by the gravitational field strength of the object's surroundings - $\Delta h$ is the height from the rupture to the ground. Remember that's it's very, very important for you to understand where this comes from. I'll be adding a small proof after we're done with Bernoulli's Equation! The example question below shows us another method to apply Torricelli's Equation. The tank is only one way of doing this. >[!Example]- Problem Taster - Module 14-5 Problem 46 >Suppose that you release a small ball from rest at a depth of $0.600 m$ below the surface in a pool of water. If the density of the ball is $0.300$ that of water and if the drag force on the ball from the water is negligible, *how high above the water surface will the ball shoot as it emerges from the water?* > >(Neglect any transfer of energy to the splashing and waves produced by the emerging ball.) > >**Solution:** >Brainstorm: Let's think about this through [[Fluid Principles (Mechanics)#Archimedes' Principle|archimedes' principle]] first - let's try to assemble an equation that dictates how the ball moves! >Since the ball is going to be moving (and accelerating) upwards, we can apply Newton's Second Law of Fluids to create the equation: >$\rho_{w}Vg - \rho_{b}Vg = \rho_{b}Va$ >Now let's isolate $a$, the acceleration: >$a = \frac{\rho_{w}g}{\rho_{b}} - g$ >$a = g\left( \frac{\rho_{w}}{\rho_{b}} - 1 \right)$ >We know the density of the ball as a function of the density of water, so we can just substitute values (where $g = 9.81$) to get a value for our acceleration: >$a = 9.81\left( \frac{1}{0.3} - 1 \right) = 22.9 ms^{-2}$ >To find the speed, all we need now is to plug in the acceleration with Torricelli's equation. We know the height the ball travels within the fluid is $0.6m$, so: >$v = \sqrt{ 2(22.9) \times 0.6 } = 5.24 ms^{-2}$ >Back to [[Newton's Laws and SUVAT#Projectile Motion diagram REQUIRED|projectile motion]] for a bit - the ball is travelling vertically up, so we can just plug our values into the equation: >$h = \frac{v^2}{2g} = \frac{5.24^2}{2\times 9.81} = 1.40m$ # Bernoulli's Equation When going over the maths' textbook, a couple of you may have noticed "Bernoulli's Equation" being in an example. While maths often tries to overlook its real-life applications, Bernoulli's Equation is a pretty useful thing to learn, forming the backbone of third-year university fluid dynamics courses. There's a certain caveat to fluid flow - when you move a fluid into an asymmetric pipe that flows upwards, there's going to be an increase in the [[Gravitational Potential Energy|gravitational potential energy]] of the fluid. Yet there's no pump to be seen, and the fluid's still flowing out the other end. Why is that? Let's take the quintessential example for Bernoulli's Equation - the upwards pipe! ![[Pasted image 20230906191835.png]] Here, a fluid (for simplicity's sake we'll call it water for now) flows upwards, with a larger mouth to exit from at the end of the pipe. The pressures $p_1$ and $p_2$ are the pressures due to the fluid flow at the start and end respectively. There are two variables that change in this diagram - the volume of water flowing through one section of the pipe as well as the pressures. How can we reconcile this? Well, we can start by saying that energy and momentum in a system have to be conserved - momentum we can't really find as the mass of the water is also changing too. Let's instead settle on a conservation of energy relationship - Bernoulli's Equation! $p_{1} + \frac{1}{2}\rho v^2 + g\rho y_{1} = p_{2} + \frac{1}{2}\rho v^2 + g\rho y_{2}$ We'll go over more about why it's a conservation of energy relationship in the derivation below: >[!Abstract]- Proving Bernoulli's Equation - The Doable >Back to conservation of energy for this one! Note that trial and error notwithstanding you'll probably have to search for your starting ground on the internet (like I did). >Let's start off with defining what we need to know - that fluid flow in our case relies on conservation of energy. No change there! >However, when our pipe has a difference in height $h$, there's a slight problem - we need to do work on the fluid so that it can continue to flow. As the pipe is closed, [[The Equation of Continuity (Mechanics)|conservation of mass]] also applies, meaning: >$A_{1}v_{1} = A_{2}v_{2}$ >Let's put this into context with another diagram! > >![[Pasted image 20231124215757.png]] > >To begin the actual proof, we're going to have to look at the *work done* needed to overcome this change in the fluid flow. Remember - [[Power, Work Done and Energy (Mechanics, %%marked for rewrite + NEEDS LOADS OF DIAGRAMS, ASK EXTERNAL HELP!%%)|differentiating the work done gives us the force of the fluid flow.]] >The force is given by a pressure gradient at each end of the pipe, namely: >$dW = \int F(x) \, dx = p_{1}A_{1} dx - p_{2}A_{2}dx$ >T[](Power,%20Work%20Done%20and%20Energy.md)>$dE_{K} = \frac{1}{2}\rho dV(v_{2}^2-v_{1}^2)$ >$dU = mg(y_{1}+h) - mgy_{1}$ >Work Done is the total energy required of a system. This means we have another expression for the work done - the kinetic and potential energies combined! >$dW = \frac{1}{2}\rho dV(v_{2}^2-v_{1}^2) + mgy_{1} + mg(y_{1}+h)$ >Make an equivalence and isolate each variable number to one side. Just algebra from here on! >$(p_{1} - p_{2})dV = \frac{1}{2}\rho dV(v_{2}^2-v_{1}^2) - mgy_{1} + mg(y_{1}+h)$ >$p_{1} + mgy_{1} + \frac{1}{2}\rho v_{1}^2 = p_{2} + mg(y_{1}+h) + \frac{1}{2}\rho v_{2}^2$ >We've arrived at our final line of reasoning - Bernoulli's Equation itself. Give yourself a pat on the back! > >Pro Tip - Make sure you're working in terms of fluids' units! It'll be difficult (and not a good idea) otherwise. >[!Danger]- Proving Bernoulli's Equation - The Difficult >Apply conservation of energy onto the equation. Remember that [[Power, Work Done and Energy (Mechanics, %%marked for rewrite + NEEDS LOADS OF DIAGRAMS, ASK EXTERNAL HELP!%%)#Work Done|the work done]] is the change in the kinetic energy of the system, as follows: >$\Delta K = \frac{1}{2}mv_{2}^2 - \frac{1}{2}mv_{1}^2$ >$\D[](Power,%20Work%20Done%20and%20Energy.md#Work%20Done)mponent, there is both a horizontal and vertical work done. The work done against gravity, $W_g$, is dependant on the weight of the fluid flowing into a certain area at every moment $\Delta mg$ and the vertical displacement $y_2 - y_1$: >$W_{g} = -\Delta mg(y_{2}-y_{1})$ >$W_{g} = -g\rho \Delta V(y_{2} - y_{1})$ >Note that this deals with the *work done by gravity* - and as the fluid travels upwards - opposite the gravitational force, the work done value is negative. Now it's time for us to find the horizontal work done! >$W_{H} = F \Delta x$ >The force can be written as pressure times the area. This gives us: >$W_{H} = (pA)(\Delta x) = p \Delta V$ >We can split this into two values - one for the work done done by the flow of the fluid, another for the work done done by the system. The system will also have to apply a work done to displace fluid in front, leaving us with a total work done equation: >$W_{p} = -p_{2}\Delta V + p_{1}\Delta V = (p_{1}-p_{2})\Delta V$ >This gives us an expression for the total work done by the system: >$W = W_{g} + W_{p} = -g\rho \Delta V(y_{2} - y_{1}) + (p_{1}-p_{2})\Delta V$ >Thanks to the [[Power, Work Done and Energy (Mechanics, %%marked for rewrite + NEEDS LOADS OF DIAGRAMS, ASK EXTERNAL HELP!%%)#Work Done|work-kinetic energy theorem]] the total work done of the system (given by the expression above) will be equal to the original expression we derived for the work done, giving us an expression: >$-g\rho \Delt[](Power,%20Work%20Done%20and%20Energy.md#Work%20Done)with a "1" on one side and "2" on another. Let's start by dividing the expression by the change in fluid flow volume... >$-g\rho (y_{2} - y_{1}) + (p_{1}-p_{2}) = \frac{1}{2}\rho (v_{2}^2 - v_{1}^2)$ >Move the 'twos' to the right and the 'ones' to the left while bringing out the $g$... >$p_{1} + g\rho y_{1} + \frac{1}{2}\rho v_{1}^2 = p_{2} + g\rho y_{2} \frac{1}{2}\rho v^2$ >I WILL ADD DIAGRAMS TO THIS!! >[!Danger]- Torricelli's Equation - Bernoulli's Equation V2. >So you've made it past the 'gauntlet of deadly terror' - the equation that determines if your pipe's going to clog! > >![[Pasted image 20231116213657.png]] > >Here's one of the large vats water is held in in certain industries - such as the bottled water industry! As the water flows out of the container, there is a downwards velocity for water flowing down the pipe $v_1$, leading to a velocity $v_2$ as the water flows out of the hole. > >Our job is to find the outward velocity $v_2$. Time to use Bernoulli's Equation! First off, notice that the base of the container is going to be a lot larger than the leak. This means we can approximate $v_1$ as 0, as $v_2$ is going to be a lot larger than it regardless. Therefore: > >$P_{1} + \rho gh = P_{2} + \frac{1}{2}\rho v_{2}^2$ >$h$ as represented by the diagram - the change in the height between the top of the container and the leak. $\rho gh$ represents the fluid's potential energy, which is zero as it leaks out of the container (no desired resting state). > >Since the atmospheric pressure of the Earth is virtually homogenous over small distances, we can approximate $P1$ = $P2$. Also, this means that there's virtually no hydrostatic pressure on the surface. This brings our equation down to: >$\rho gh = \frac{1}{2}\rho v_{2}^2 $ >Now all we need to do is isolate the $v_2$! >$2gh = v_{2}^2$ >$v_{2} = \sqrt{ 2gh }$ >To generalise, the $g$ term just represents a situation where a liquid is an ideal fluid. If there are other forces present, we can substitute it for a more general $a$ term - because why not? >$v_{2} = \sqrt{ 2ah }$ >Nice! # What's Next? You're done! From here on, we're going to be diving into some more 'difficult' articles of fluid mechanics - first hydrostatics, then hydrodynamics. Who's ready for some turbulence? For the next instalment in fluids, go here: [[Hydrostatics (Mechanics)]]