Fluid Mechanics Bernoulli Equation Help I need help on number 43. Please show al

Question

Fluid Mechanics Bernoulli Equation Help

I need help on number 43. Please show all work ! Here are some hints to help....I just can't get this for some reason.

Hint: This one is another 2 equation, 2 unknown problem. You are asked for the discharge rate, but you don't have a velocity for the opening to work with to calculate that discharge rate. Come to think of it, you don't have a velocity for the other side either. So first come up with a relationship between Va and Vb. Next, the piezometers give you a height of a water column, which means we can find out the pressures at point A and B (in this instance, you aren't given a density for water. This doesn't actually matter, if you plug everything in correctly the density terms will cancel out one another). Using the Bernoulli equation, we now have another relationship between Va and Vb. Now with 2 equations, we can solve for Va and Vb, and thus the discharge rate.

Problem:

As water flows through the pipes, it rises within the piezometers at A and B to the heights hA = 1.5 ft and hB = 2 ft. Determine the volumetric flow. The volumetric flow of water through the transition is 3 ft3/s. Determine the height it rises in the piezometer at A if hB = 2 ft.

Explanation / Answer

Pb-Pa=0.5rho*va^2-0.5rho*vb^2=(0.5*12*0.0254)*1000*9.81

0.5*1000*(va^2-vb^2)=(0.5*12*0.0254)*1000*9.81

Also.

va*0.75^2=vb*1.25^2

Solving simultaneously,

va=1.853m/s

vb=0.667m/s

Volumetric flow=1.853*pi*(0.75*12*0.0254/2)^2=0.076m^3/s=76l/s

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