A U-tube open at both ends is partially filled with water (Figure (a)). Oil (rho

Question

A U-tube open at both ends is partially filled with water (Figure (a)). Oil (rho = 750 kg/m^3) is then poured into the right arm and forms a column l= 4.99 cm high (Figure (b)). Determine the difference h in the heights of the two liquid surfaces. (The density of water is 1.00 times 10^3 kg/m^3.) 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 (Figure (c)). Determine the speed of the air being blown across the left arm. Assume the density of air is 1.29 kg/m^3.

Explanation / Answer

In case B, the pressures at location 1 and 2 are equal P1=P2 and therefore

P0+wg(Lh)=P0+ogL

w(Lh)=oL

1000*(4.99-h)=750*h

h= 2.85m

In case C, the fast traveling air above the left side of the U-tube will create smaller pressure. The right side of the U-tube will still have atmosphere pressure (P0) and will push the liquid up to the left side. Again, the pressures at location 1 and 2 are always equal: P1=P2

. So we have

P+wgL=P0+ogL.

Therefore

P = P0(wo)gL

To find the speed needed for creating the smaller pressure, we can set up the Bernoulli along the dashed line shown in the figure below, which gives

P+1/2airv2=P0

Here the vertical position (y) is identical on both sides of the equation so it is canceled. The speed of air on the right hand side is zero, so only have the P0 term.

Then we have

1/2airv2= P0 – P

1/2airv2 = (w o)gL

1/2*1290*v^2 = (1000-750)*9.8*4.99

v= 4.35m/s

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