A U-tube open at both ends is partially filled with water (Fig. P9.80a).?

A U-tube open at both ends is partially filled with water (Fig. P9.80a). Oil ( = 810 kg/m3) is then poured into the right arm and forms a column L = 4.00 cm high (Fig. P9.80b).



Figure P9.80 http://www.webassign.net/sf/p9_80.gif



(a) Determine the difference, h, in the heights of the two liquid surfaces. Assume that the density of air is 1.29 kg/m3.

cm

(b) The right arm is then shielded from any air motion while air is blown across the top of the left arm until the surfaces of the two liquids are at the same height (Fig. 9.80c). Determine the speed of the air being blown across the left arm.

m/s

A U-tube open at both ends is partially filled with water (Fig. P9.80a).?
1. The pressure is proportional to the height h of the column of liquid and its specific weight.

P=gph where

g - acceleration due to gravity

p- density

2. The pressure must be equal on ether side of the open tube



This all you need



Oh yes and the equation describing the state



P1= P2+P2

P1- pressure on the left (water only)

P2- pressure of water on the right

P3 -pressure of oil on the right



we have

gph1= gph2+gp'h3 where

p and p' are densities of water and oil respectively

or

ph1= gph2+gp'L



a) So we have to find

h=h1- (h2+L)

and since h1= ph2+p'L then

ph1 - ph2 = p'L

h1-h2=(p'/p)L now we have

h=(p'/p)L -L

h=L(p'/p) -1)

h=4(810/1000 -1)= - .79 cm (the negative sight suggests that h is higher than h1



b) Here you you employ Bernoulli's principle. The caped end on the right is at pressure of 1 atm the open end will experience lower pressure due to air movement. The speed is inversely proportional to pressure can be computed hence.

The original form of Bernoulli's equation is:



v^2/2 + gh + P/p=const.

where:

v = fluid velocity at a point on a streamline

g = acceleration due to gravity

h = height of the point on the streamline

P= pressure at the point on the streamline

p= density of the fluid at all points on the streamline



v^2/2 + gh1 + P/p=v^2/2 + gh2 + P/p

since the right end is capped.

v^2/2 + gh1 + P1/p= gh2 + P2/p

v=sqrt( g(h2 -h1) +( P2 - P1)/p)



h2-h1= L

P2-P1= mp'L

this is it!