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Showing posts with label Thermo Fluid - Bernoulli Theorem. Show all posts
Showing posts with label Thermo Fluid - Bernoulli Theorem. Show all posts

Gas Laws

Calculations using Boyle's Law

Concepts

Boyle's Law states that the product of the pressure and volume for a gas is a constant for a fixed amount of gas at a fixed temperature. Written in mathematical terms, this law is

P V = constant

A common use of this law is to predict how a change in pressure will alter the volume of the gas or vice versa. Such problems can be regarded as a two state problem, the initial state (represented by subscript i) and the final state (represented by subscript f). If a sample of gas initially at pressure Pi and volume Vi is subjected to a change that does not change the amount of gas or the temperature, the final pressure Pf and volume Vf are related to the initial values by the equation

Pi Vi = Pf Vf



Exercise

Objective

  • Use Boyle's Law to predict the final pressure for a gas that has undergone an expansion.

In both parts of this exercise, you will work with two glass bulbs connected by a tube containing a stopcock. Initially, the bulb on the right is completely evacuated (that is, it contains no gas) while the bulb on the left contains a sample of gas at pressure Pi. A manometer is provided to measure the pressure in the left bulb.

When the stopcock is opened, the gas in the left bulb will expand to occupy both bulbs and the pressure in the left bulb will change accordingly.

Information on the volumes of the two bulbs will be given to you, and you may read the initial pressure from the manometer. Use Boyle's Law to calculate the final pressure. Then open the stopcock and see if your prediction is correct.



Part 1

Each time you press the "New Conditions" button, the experiment will be reset with new volumes for the two bulbs and a new initial pressure. After calculating the final pressure, press the "Open Stopcock" button to allow the gas to expand and check your prediction.

Practice the calculations until you are able to solve the problem consistently.


Volume of Left Bulb
L

Volume of Right Bulb
L



Part 2

In performing the Boyle's Law calculations in Part 1, you will no doubt have observed that it is not necessary to know the initial and final volumes. It is only necessary to know the volume ratio: Vi/Vf.

This part of the exercise is the same as Part 1 except you are not given the volumes of the two bulbs. Instead you are given the ratio of the two volumes: Vright/Vleft.

Repeat the exercise using the volume ratio in the calculations. Bear in mind that Vright/Vleft is not the same as Vi/Vf .


Vright/Vleft =

Bernoulli's Principle

This is an important principle involving the movement of a fluid through a pressure difference. Suppose a fluid is moving in a horizontal direction and encounters a pressure difference. This pressure difference will result in a net force, which by Newton's 2nd law will cause an acceleration of the fluid. The fundamental relation,



$\textstyle \parbox{4.5in}{\vspace*{5pt} work done = change in kinetic energy \vspace*{5pt}}$

in this situation can be written as



$\textstyle \parbox{4.5in}{\vspace*{5pt} - (change in pressure) x area x distance = change in kinetic energy, \vspace*{5pt}}$

which furthermore can be expressed as



$\textstyle \parbox{4.5in}{\vspace*{5pt} change in pressure + change in ( kinetic energy / volume ) = 0. \vspace*{5pt}}$

In other words,



\fbox{\parbox{4.5in}{\vspace*{7pt} Pressure + ( kinetic energy / volume ) = constant \vspace*{7pt}}}

which is known as Bernoulli's principle. This is very similar to the statement we encountered before for a freely falling object, where the gravitational potential energy plus the kinetic energy was constant (i. e., was conserved).

Bernoulli's principle thus says that a rise (fall) in pressure in a flowing fluid must always be accompanied by a decrease (increase) in the speed, and conversely, if an increase (decrease) in , the speed of the fluid results in a decrease (increase) in the pressure. This is at the heart of a number of everyday phenomena. As a very trivial example, Bernouilli's principle is responsible for the fact that a shower curtain gets ``sucked inwards'' when the water is first turned on. What happens is that the increased water/air velocity inside the curtain (relative to the still air on the other side) causes a pressure drop. The pressure difference between the outside and inside causes a net force on the shower curtain which sucks it inward. A more useful example is provided by the functioning of a perfume bottle: squeezing the bulb over the fluid creates a low pressure area due to the higher speed of the air, which subsequently draws the fluid up. This is illustrated in the following figure.


Figure 7.2: Action of a spray atomizer
\begin{figure} \begin{center} \leavevmode \epsfysize=5.5 cm \epsfbox{figs/heat-6.eps} \end{center} \end{figure}

Bernouilli's principle also tells us why windows tend to explode, rather than implode in hurricanes: the very high speed of the air just outside the window causes the pressure just outside to be much less than the pressure inside, where the air is still. The difference in force pushes the windows outward, and hence explode. If you know that a hurricane is coming it is therefore better to open as many windows as possible, to equalize the pressure inside and out.

Another example of Bernoulli's principle at work is in the lift of aircraft wings and the motion of ``curve balls'' in baseball. In both cases the design is such as to create a speed differential of the flowing air past the object on the top and the bottom - for aircraft wings this comes from the movement of the flaps, and for the baseball it is the presence of ridges. Such a speed differential leads to a pressure difference between the top and bottom of the object, resulting in a net force being exerted, either upwards or downwards. This is illustrated in the following figure.


Figure 7.3: Lift of an aircraft wing
\begin{figure} \begin{center} \leavevmode \epsfysize=5.5 cm \epsfbox{figs/heat-7.eps} \end{center} \end{figure}

Bernoulli Theorem Design Equations and Formulas Calculator
Fluid Mechanics and Hydraulics

Bernoulli Equation


Solving For Head Loss

Head Loss


Inputs:
static head or elevation (Z1)
static head or elevation (Z2)
pressure (P1)
pressure (P2)
velocity (V1)
velocity (V2)
density (p)
acceleration of gravity (g)


Conversions:
Z1 = 0
= 0 foot
Z2 = 0
= 0 foot
P1 = 0
= 0 pound/foot^2
P2 = 0
= 0 pound/foot^2
V1 = 0
= 0 foot/second
V2 = 0
= 0 foot/second
p = 0
= 0 pound/foot^3
g = 0
= 0 foot/second^2


Solution:
head loss (h) = HAS NOT BEEN CALCULATED


Other Units:


Change Equation

Note Bernoulli Equation Assumes:
1. flow is streamline
2. steady state flow
3. inviscid fluid
4. incompressible fluid

Select an equation to solve for a different unknown
head loss Solve for head loss
static head or elevation at point 1 Solve for static head or elevation at point 1
pressure at point 1 Solve for pressure at point 1
velocity at point 1 Solve for velocity at point 1

Where
h = head loss
Z = static head or elevation
P = Pressure
V = fluid velocity
p = fluid density
g = acceleration of gravity
Q = flow rate


Reference - Books:


1) P. Aarne Vesilind, J. Jeffrey Peirce and Ruth F. Weiner. 1994. Environmental Engineering. Butterworth Heinemann. 3rd ed.

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