Kinetic Energy Formula. The Kinetic energy is the energy that an object has due to its motion. Ek, is the energy of a mass, m, in motion, v2. Kinetic Energy Formula Questions. 1) The mass of a baby elephant is 113 kg, and it walks at a constant velocity of 0.5 m/sec. Molecules have very little mass, but gases contain many, many molecules, and because they all have kinetic energy, the total kinetic energy can pile up pretty fast. Using physics, can you find how much total kinetic energy there is in a certain amount of gas?
If the pressure causes the acceleration, how can the kinetic energy and 'flow energy' be the same thing? Acceleration converts one form of energy into another (like dropping a rock converts potential energy into kinetic energy). Also, consider the non-accelerating, flow field: the kinetic energy is constant. The 'flow energy' may or may not be constant depending on if we're considering losses, but it can be anything and isn't tied to the kinetic energy (they are not exchanging energy).Consider the real-world example of a fan blowing air through a duct. The kinetic energy of an element is constant, but the 'flow energy' drops along the duct as energy is lost to friction against the duct walls.
Exactly what I was contemplating on! Flow Work (Flow Energy)Work is needed to push the fluid into or out of the boundaries of a control volume if mass flow is involved. This work is called the flow work (flow energy).
Flow work is necessary for maintaining a continuous flow through a control volume.Consider a fluid element of volume V, pressure P, and cross-sectional area A as shown left. The flow immediately upstream will force this fluid element to enter the control volume, and it can be regarded as an imaginary piston. The force applied on the fluid element by the imaginary piston is:F = PAThe work done due to pushing the entire fluid element across the boundary into the control volume isWflow = FL = PAL = PVFor unit mass,wflow = PvThe work done due to pushing the fluid element out of the control volume is the same as the work needed to push the fluid element into the control volume. I don't really like that description because what it glosses over is the fact that when the flow element moves to its next position in steady, lossless flow, that work expended does not get converted to kinetic energy. Indeed, it hasn't been converted to anything: it gets used again to move the flow element to the next position and the next position and the next position.
I don't like calling something 'work' when it hasn't changed/converted any energies.however, I can see how a physicist might prefer it because it is all-inclusive. But as an engineer, I try to ignore things that don't matter.In some ways, this is a bit like an argument between absolute and gauge pressure. If the tire gauge you use to pump up your car tires read in absolute pressure, it would be more inclusive - does that make it more correct? I don't care, but what I do care about is that makes it a pain in the a to use. You seem to understand exactly what I am trying to convey! It sure seems to me that the flow work term is included in elementary energy analyses in the form of the Bernoulli equation through the ##p## terms in the equation.
I feel like this was pretty well illustrated up there by where he showed the derivation of the Bernoulli equation directly from the work-energy theorem.On the other hand, if the confusion here is arising in view of the situation in a real pipe where you need a pressure gradient to maintain a constant flow speed with no acceleration, this is because of the effects of viscous dissipation, where the energy added due to the work done by the pressure gradient is being balanced by the energy dissipated by viscosity. This situation is not captured by Bernoulli's equation without modification because it assumes zero viscosity.
So, here's an example where they are called the same thing:I don't really like that description because what it glosses over is the fact that when the flow element moves to its next position in steady, lossless flow, that work expended does not get converted to kinetic energy. Indeed, it hasn't been converted to anything: it gets used again to move the flow element to the next position and the next position and the next position. I don't like calling something 'work' when it hasn't changed/converted any energies.however, I can see how a physicist might prefer it because it is all-inclusive. But as an engineer, I try to ignore things that don't matter. It sure seems to me that the flow work term is included in elementary energy analyses in the form of the Bernoulli equation through the ##p## terms in the equation.
I feel like this was pretty well illustrated up there by where he showed the derivation of the Bernoulli equation directly from the work-energy theorem.On the other hand, if the confusion here is arising in view of the situation in a real pipe where you need a pressure gradient to maintain a constant flow speed with no acceleration, this is because of the effects of viscous dissipation, where the energy added due to the work done by the pressure gradient is being balanced by the energy dissipated by viscosity. This situation is not captured by Bernoulli's equation without modification because it assumes zero viscosity.
![Kinetic Energy Formula Liquid Kinetic Energy Formula Liquid](http://www.chemistrytutorials.org/ct/images/matters/change_of_phase1.png)
Sir the confusion arised because the 'p' used in the 'pv' term in the Bernoulli equation should actually represent a pressure difference across a fluid element not absolute pressure at a point in fluid as it is basically expression for work done by the net pressure force on a fluid element.Some textbooks have termed it as 'flow energy' or 'pressure energy' which is technically incorrect as Sir Jano L. Pointed out and in practice we substitute this pressure difference across a fluid element by the thermodynamic pressure at a point in flow.This messed up things when I was trying to relate the 'pv' term as some kind of energy associated with the fluid as is commonly misinterpreted since its basically work. Sir the confusion arised because the 'p' used in the 'pv' term in the Bernoulli equation should actually represent a pressure difference across a fluid element not absolute pressure at a point in fluid as it is basically expression for work done by the net pressure force on a fluid element.Some textbooks have termed it as 'flow energy' or 'pressure energy' which is technically incorrect as Sir Jano L. Pointed out and in practice we substitute this pressure difference across a fluid element by the thermodynamic pressure at a point in flow.This messed up things when I was trying to relate the 'pv' term as some kind of energy associated with the fluid as is commonly misinterpreted since its basically work.
But it needn't be a pressure difference in that term. Bernoulli's equation essentially considers the flow in a control volume. Using an absolute pressure at two ends of it gives you all the pressure differential you need for the analysis. One of the drawbacks is therefore ignoring everything that happens between those two points you choose.If you want something that is valid more universally and in a differential sense, you would have to move to the energy equation typically used in concert with the Navier-Stokes equations.
Sir Boneh3ad even if we consider an energy balance for a control volume and try to derive the Bernoulli equation from the Steady Flow Energy Equation we have the term specific enthalpy h=u+pv on both sides the latter term on the right hand side of the equation being defined as the work done by pressure forces on a fluid parcel that is about to enter the C.V. By the fluid immediately upstream to push it within the C.V. Or the work done on a fluid parcel that is about the leave the C.V. So in light of the previous, it should be noted that ##p## is absolute pressure, not a pressure differential. In the illustrations, they show a large box because it is illustrative, but in reality, we are talking about the pressure of an infinitesimal fluid element moving past a point and the energy passing through that point as a result. That is the definition of enthalpy, after all.
It is the energy carried along by the flow due to internal energy (##u##) and the specific flow work (##pnu##) where ##p## is absolute pressure.
![Kinetic Kinetic](https://study.com/cimages/multimages/16/ee8cdff5-78b5-4ffd-a0b5-7c5453b3088c_kinetic_energy__of_a_gas.png)
Kinetic Energy ProblemsOn this page I put together a collection of kinetic energy problems to help you understand the concept of kinetic energy better. The required equations and background reading to solve these problems are given.Problem # 1Two solid uniform cylinders are connected with a belt wrapped around both cylinders, as shown in the figure below. The mass and radius of the smallest cylinder is m A and r A, respectively, and the mass and radius of the largest cylinder is m B and r B, respectively.
When the cylinders are rotating, what is the ratio of the kinetic energy of the largest cylinder to the kinetic energy of the smallest cylinder?Problem # 2In the previous problem, suppose that the distance between the center of both cylinders, A and B, is equal to d, and the linear density of the belt is ρ. What is the kinetic energy of the belt if the smallest cylinder is rotating at a speed of S revolutions per minute?Problem # 3In the figure below, a liquid is shown flowing through a horizontal pipe in steady-state fashion. The cross-sectional area of the pipe at location A is larger than at location B. Using the Bernoulli equation and the concept of mass conservation, explain why there is a difference in the kinetic energy of the liquid between locations A and B.Problem # 4In the figure below, wind is blowing through a wind turbine at constant velocity v 1 upstream of the wind turbine, and at constant velocity v 2 downstream of the wind turbine.
What is the maximum kinetic energy that the wind turbine can obtain, as a percentage of the kinetic energy of the wind upstream of the wind turbine?Problem # 5The figure below shows a linkage with a wheel on each end, and each wheel can move freely inside two slots that are perpendicular to each other. The mass of the linkage is m, its rotational inertia about the center of mass is I G, its length is L, and the angle that the linkage makes with the bottom slot is θ. If the bottom wheel is pulled to the left at a constant velocity v, what is the kinetic energy of the linkage as a function of θ? Ignore the mass of the wheels. Note that 0 V A, by mass conservation. Therefore, according to the Bernoulli equation, the kinetic energy, per unit mass of liquid, at location B is greater than at location A. This means that the fluid pressure at location A is greater than the fluid pressure at location B.Answer for Problem # 4The answer is 59.3%.
The solution to this problem comes from the well known Betz' Law for wind turbines, which can be easily referenced online. Take the time to look this up, since it's an interesting problem, originally solved by Albert Betz in 1919.Answer for Problem # 5To make things easier to visualize, first imagine an xy-coordinate system with origin located at the bottom right corner, where the line of travel of the center-point of both wheels meet, and positions to the left of, and above, this point are considered positive. Then, the x-position of the center of mass of the linkage is x = (1/2) Lcos θ, and the y-position of the center of mass is y = (1/2) Lsin θ. Now, take the derivative with respect to time to obtain velocity. So, dx/dt = v/2 = -(1/2) Lsin θ(d θ/dt) (equation 1), and dy/dt = (1/2) Lcos θ(d θ/dt) (equation 2). From equation 1, solve for d θ/dt = - v/( Lsin θ). Substitute this into equation 2 and we get, dy/dt = -(1/2) v/(tan θ).
The kinetic energy of the linkage is then KE = (1/2) I G(d θ/dt) 2 + (1/2) m((dx/dt) 2 + (dy/dt) 2).Answer for Problem # 6The total kinetic energy of the car is equal to the kinetic energy of the car body + driver + wheels. Therefore, KE = (1/2) M S 2 + 4×(1/2) I G w 2 + 4×(1/2) m S 2, where w is the angular velocity of each wheel, which is equal to w = S/ r.Answer for Problem # 7The kinetic energy of the propeller is equal to KE = (1/2) m v 2 + (1/2) I G w 2.report this ad.