Continuity Theory - Fluids

[Fila Delphia]

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Hello all, hope you can help me out with this question.

The conditions presented were that there is a barge flowing in a river from an upper dock to a lower dock, 100 km apart.

The river has steady laminar flow, a gentle grade making the river not turbulent **all qualities satisfying Bernoulli's Principle.

There are no tributaries that enter the river and no water is removed.

Thus, we can infer that all parts of the section of the river are CONTINUOUS and have the same constant flow rate Q1 = Q2 (Av1 = Av2)

The question
Near the upper dock, the barge moves 50% faster than at the lower dock. What inference could one make between the cross-sectional area of the river at the upper, AU, and lower, AL, docks?

the Answer was AU = 2AL/3

I chose AU = AL/2 thinking that if Q = Av, and if the upper dock the barge moves 50% faster, than the area decreases so AU's area is half of AL's. Why is this thinking wrong?

The explanation was just confusing to me, and I hope someone can explain it in a more direct way:

All parts of this section of the river have in the same (constant) flow rate since there are no extra sources or outlets. The flow speed being 50% faster means that vU = 3vL / 2. According to the Continuity Equation, this means that the areas will have the reciprocal relationship. Just thinking one is twice the other could lead to choices A or C. If you get the algebra reversed, you may get choice D.

Thanks your help

 

[ashtonjam]

I think the mistake here was that you assumed the barge moving 50% faster meant moving twice as fast. If the barge did move twice as fast, then your assumption would be right (the area of the upper is twice as small). Instead, you have to plug in 3/2 instead of 2 to get the right answer.

Upper flow rate = Lower flow rate
Q1 = Q2
A1v1 = A2v2

The velocity in the upper dock is 3/2 that of the lower. Therefore, v1 = 3/2 v2
A1 (3/2 v2) = A2v2
3/2 A1 = A2
A1 = 2/3 A2