Topic 3

Question 1

Bernoullis equation

Basic principle

Answer 1

During flow of an ideal fluid
  - The mechanical energy is not converted into other forms
    of energy (eg. heat)
The work done on a fluid equals the change in mechanical energy
Av = constant -->  v constant

Question 2

Bernoullis equation

Kinetic energy

Answer 2

The Kinetic energy of the fluid in the section -> A*∆X remains the same, but the potensial energy changes as the fluid rises

Question 3

Bernoullis equation

The netforce on the fluid

Answer 3

The netforce on the fluid in the tube due to the surrounding
Fluid is (Pa-Pb)A
If the fluid in a section moves a short distance ∆X , then the work done is:
w = (Pa-Pb)A * ∆X = (Pa-Pb)∆V

Question 4

Bernoullis equation

Potensial energy

Answer 4

The work done on the fluid must equal to the increase ∆Epot = (δ∆V)gyB - (δ∆V)gyA

From w = ∆Epot we get:
PA-PB = δg (yB - yA) or PA+ δgyA =PB+δgyB v is constant

P= Preassure pluss
The potential energy pr unit volume = δgy of the fluid is the same everywhere in the follow tube if the velocity remains constant.

Question 5

Bernoullis equation

If the cross sectional area of the following tube changes

Answer 5

Then the fluid velocity V also changes and therefor the kinetic energy per unit volume will also change.
( 1/2 δv^2 )

The work done on a fluid must then be set equal to the change in the potential plus kinetic energy of the fluid

Question 6

Bernoullis equation

The equation

Answer 6

PA + δgyA + 1/2 δvA^2 = PB + δgyB + 1/2 δvB^2 or

P+ δgy + 1/2 δv^2 = Constant

Question 7

Bernoullis equation

Answer 7

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