Fluid Conservation: Worked Solutions

Try the questions on the lesson page first. Then compare both the answer and the reasoning. The worked steps here are original.

Question 1: In steady incompressible flow, a filled pipe diameter halves. At the same flow rate the speed…

In steady incompressible flow, a filled pipe diameter halves. At the same flow rate the speed…

1. Use circular area A = πd²/4. Why: The comparison specifies a diameter, while continuity uses area.
2. Halving d reduces area to A/4. Why: Area depends on the square of diameter.
3. With Av constant, the speed quadruples. Why: One-quarter the area requires four times the speed to keep the same volume flow rate.

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Question 2: Which term can be dropped when comparing two points at the same height with Bernoulli?

Which term can be dropped when comparing two points at the same height with Bernoulli?

1. Write Bernoulli with pressure, kinetic, and gravitational terms. Why: Energy can shift among these contributions along the ideal streamline.
2. Set y₁ = y₂ for the two compared locations. Why: Their gravitational terms ρgy are then equal.
3. Cancel the difference in gravitational terms only. Why: Pressure and kinetic terms can still differ and must remain unless further conditions are given.

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Question 3: A large open reservoir has a small hole depth h below the surface. Increasing h to 4h changes ideal exit speed by…

A large open reservoir has a small hole depth h below the surface. Increasing h to 4h changes ideal exit speed by…

1. Use v_exit = √(2gh). Why: Both the open surface and small outlet have atmospheric pressure and surface speed is negligible.
2. Replace h by 4h and take the ratio √(4h/h). Why: The square-root dependence sets the factor of change.
3. The exit speed doubles. Why: The depth is four times larger, but speed depends on its square root.

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Question 4: graph slope and a jet

A horizontal jet leaves a reservoir hole 0.20 m above a catch tray. A fitted graph of R² versus water depth h passes through the origin and has slope 0.72 m. Find the ideal expected slope and compare the measured range to the ideal range at h = 0.30 m.

1. The ideal projectile-plus-flow model gives R² = 4Hh, so expected slope is 4(0.20) = 0.80 m. Why: The jet speed comes from depth and its fall time comes from fixed outlet height.
2. At h = 0.30 m, calculate R_ideal = √(0.80×0.30) = 0.490 m and R_fit = √(0.72×0.30) = 0.465 m. Why: The graph plots range squared, so take square roots to recover actual ranges.
3. The slope is 10% below ideal; the range is 1 − √(0.72/0.80) = 5.1% below ideal. Why: A squared plotted quantity has a different fractional change from the quantity itself. Measurement uncertainty and possible losses are needed before naming a cause.

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