Question

state bernouli's theorem

Answer 1

HI

HERE IZ UR ANSWER

For the streamline flow of an ideal gas, the sum of K.E by unit volume, potential energy per unit volume and ressure energy per volume remains constant.

Refer to image for diagram,

By eqn of continuity, av = constant

area of face P is decreasing, so, velocity of flow of water increases, so Kinetic Energy per sec increases.

increase in Kinetic Energy = 1/2m (v2 sq. - v1 sq.)   m = mass, v2 = final velocity, v1 = initial velocity.

Also increase in potential energy due to increase in height = mg (h2-h1).

W = F.dx
   = P.A dx
   = PV

So, W = P1V1 - P2V2 , This is work done against pressure energy.
       
by conservation of mass, 
m1 = m2 (m = volume * density)
V1 * Row = V2 * Row (row se row cancel) row = density
V1 = V2

so, W = P1V - P2V

Work will compensate increase in K.E and P.E

Thru, Work Energy theorem,

P1V - P2V = mg(h2 - h1) + 1/2 m(v2 sq. - v1sq.)

P1V + mgh + 1/2mv1 sq. = P2V + mgh2 + 1/2mv2 sq.

dividing both sides by V.

P1 + row g h + 1/2 row v1 sq. = P2 + row g h2 + 1/2 row v2 sq.

P + 1/2 row v sq. + row g h = constant
P + 1/2 row v sq. = constant (when tube is horizontal)
So, V increases P decrease.

HOPE IT HELPS

Answer 2

Bernoulli's principle states that an increase in the speed of a fluid occurs simultaneously with a decrease in static pressure or a decrease in the fluid's potential energy.