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bacon

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*h*below the liquid surface, but the hole in tank 1 has half the cross-sectional area of the hole in tank 2. (a) What is the ratio [itex]\rho_{1}/\rho_{2}[/itex] of the densities of the liquids if the mass flow rate is the same for the two holes? (b) What is the ratio of the volume flow rates from the two tanks? (c) To what height above the hole in the second tank should liquid be added or drained to equalize the volume flow rates?

Here is how I first went through parts (a) and (b):

(a) The mass flow rate equation: [itex] Av\rho= constant[/itex]

Since the mass flow rates are the same,

[itex] A_{1}v_{1}\rho_{1} = A_{2}v_{2}\rho_{2}[/itex]

and since [itex]A_{1}=A_{2}/2[/itex] the above equation becomes, with a little algebra,

[itex]\rho_{1}/\rho_{2}=2v_{2}/v_{1}[/itex]

And setting [itex]v_{1}=v_{2}[/itex] I got

[itex]\rho_{1}/\rho_{2}=2[/itex]

(b) The volume flow rate equation: [itex] Av= R [/itex] a constant

[itex]A_{1}v_{1}=R_{1}[/itex]

[itex]A_{2}v_{2}=R_{2}[/itex]

Using the same substitutions as in (a) I end up with,

[itex]R_{1}/R_{2}=1/2[/itex]

Both the answers are correct(from the back of the book), but going back through the problem I cannot justify [itex]v_{1}=v_{2}[/itex]. It made sense at the time I did it but now it does not, so I need conceptual help here.

(c) I can't get very far here.

I know that the volume flow rates are equal so,

[itex]A_{1}v_{1}=A_{2}v_{2}[/itex] and [itex]A_{1}=A_{2}/2[/itex]

This gives [itex]v_{1}=2v_{2}[/itex]

I'm suspecting Bernoulli's Equation may come into play here but I'm fuzzy as to how.

[itex]P_{1} + 1/2\rho_{1}v_{1}^{2}+\rho_{1}gh= a constant[/itex]

[itex]P_{2} + 1/2\rho_{2}v_{2}^{2}+\rho_{2}gy_{2}= a constant[/itex]

But not necessarily the same constant. Solving for [itex]y_{2}[/itex] is what I would want to do but there are too many unknowns.

Thanks for any help.