- #1
bacon
- 68
- 0
Suppose that two tanks, 1 and 2, each with a large opening at the top, contain different liquids. A small hole is made in the side of each tank at the same depth 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 \rho_{1}/\rho_{2} 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: Av\rho= constant
Since the mass flow rates are the same,
A_{1}v_{1}\rho_{1} = A_{2}v_{2}\rho_{2}
and since A_{1}=A_{2}/2 the above equation becomes, with a little algebra,
\rho_{1}/\rho_{2}=2v_{2}/v_{1}
And setting v_{1}=v_{2} I got
\rho_{1}/\rho_{2}=2
(b) The volume flow rate equation: Av= R a constant
A_{1}v_{1}=R_{1}
A_{2}v_{2}=R_{2}
Using the same substitutions as in (a) I end up with,
R_{1}/R_{2}=1/2
Both the answers are correct(from the back of the book), but going back through the problem I cannot justify v_{1}=v_{2}. 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,
A_{1}v_{1}=A_{2}v_{2} and A_{1}=A_{2}/2
This gives v_{1}=2v_{2}
I'm suspecting Bernoulli's Equation may come into play here but I'm fuzzy as to how.
P_{1} + 1/2\rho_{1}v_{1}^{2}+\rho_{1}gh= a constant
P_{2} + 1/2\rho_{2}v_{2}^{2}+\rho_{2}gy_{2}= a constant
But not necessarily the same constant. Solving for y_{2} is what I would want to do but there are too many unknowns.
Thanks for any help.
Here is how I first went through parts (a) and (b):
(a) The mass flow rate equation: Av\rho= constant
Since the mass flow rates are the same,
A_{1}v_{1}\rho_{1} = A_{2}v_{2}\rho_{2}
and since A_{1}=A_{2}/2 the above equation becomes, with a little algebra,
\rho_{1}/\rho_{2}=2v_{2}/v_{1}
And setting v_{1}=v_{2} I got
\rho_{1}/\rho_{2}=2
(b) The volume flow rate equation: Av= R a constant
A_{1}v_{1}=R_{1}
A_{2}v_{2}=R_{2}
Using the same substitutions as in (a) I end up with,
R_{1}/R_{2}=1/2
Both the answers are correct(from the back of the book), but going back through the problem I cannot justify v_{1}=v_{2}. 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,
A_{1}v_{1}=A_{2}v_{2} and A_{1}=A_{2}/2
This gives v_{1}=2v_{2}
I'm suspecting Bernoulli's Equation may come into play here but I'm fuzzy as to how.
P_{1} + 1/2\rho_{1}v_{1}^{2}+\rho_{1}gh= a constant
P_{2} + 1/2\rho_{2}v_{2}^{2}+\rho_{2}gy_{2}= a constant
But not necessarily the same constant. Solving for y_{2} is what I would want to do but there are too many unknowns.
Thanks for any help.
