I've been stuck on this for an hour. (Probem is below)

The problem is with part b. I understand its the basic mass flow in - mass flow out with air. The equation being $$\rho_{air}A_4V_4 = \rho_{air}A_0V_0$$ where we are trying to solve for $V_4$ or the speed of air coming in through pipe 4.

We rearrange to $$V_1=\frac{\rho_{air}A_0V_0}{\rho_{air}A_4}$$

Everything should cancel to


The issue, what on earth is $V_0$???!

In part b, $\frac{dh}{dt}=0.1910\;\frac{ft}{s}$ and is the value I ASSUMED would be used for $V_0$ in this case but no, it gives a value of $27.5 \: \frac{ft}{s}$ which is not correct.

The value supposedly used for $V_0$ is 0.1484

What is this value? Where does it come from?

The only thing I can think of is that it has something to do with 'average velocity' as stated in the part b question.

Yet I'm drawing a complete blank. And have been for more time than I care to admit. Especially considering the solution is right in front of me.

Help is VERY appreciated at this point. enter image description here


1 Answer 1


In my opinion, this is a typo and your calculation is correct. In (a), the velocity at the boundary water-air is denoted $dh/dt$. Therefore, when considering only the air volume, and considering the air as incompressible, $$ \rho_{air}\,A_0\,\frac{dh}{dt}=\rho_{air}\,A_4\,v $$ as you wrote (though I would use $v$ instead of $V$ for the velocity, since $V$ is often associated with volume).

Using $dh/dt\approx 0.1910\,\mathrm{ft/s}$ yields $v\approx 27.5\, \mathrm{ft/s}$. Using instead the 0.1484 (ft/s) for $dh/dt$ that given in the solution for (b), $v$ calculates to 21.4 ft/s.

PS: Please, unlike the author of the problem, use units in your calculations.

PPS: You might check if there is a newer version of the textbook available, or if there is an errata.


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