Help with Bernoulli's principle application

[Theelectricchild]

Hello everyone, I am having some difficulty with the following problem about bernoullis principle:

THe problem says that the level of liquid $$h = y_2 - y_1$$ drops at a rate

$$\frac{dh}{dt} = -\sqrt\frac{2gh{A_1}^2}{{A_2}^2-{A_1}^2}$$

where $$A_1$$ and $$A_2$$ are the areas of the opening (water spilling out) and the top of the surface respectively. Viscosity is ignored...

So the problem asks to solve this differential equation for h(t), letting $$h = h_0$$ at $$t = 0$$

So I notice from my diff eq class that this equation is seperable.

Thus I wrote
$${(\frac{dh}{dt})}^2 = \frac{2gh{A_1}^2}{{A_2}^2-{A_1}^2}$$

Then I isolate h:

$$\frac{(dh)^2}{h} = \frac{2g{A_1}^2}{{A_2}^2-{A_1}^2}(dt)^2$$

$$\frac{dh}{\sqrt h} = \sqrt\frac{2g{A_1}^2}{{A_2}^2-{A_1}^2}dt$$

$$2\sqrt h = \sqrt\frac{2g{A_1}^2}{{A_2}^2-{A_1}^2}t + C$$

$$\sqrt h = \frac{1}{2}\sqrt\frac{2g{A_1}^2}{{A_2}^2-{A_1}^2}t + C_1$$

So finally:

$$h = (\frac{1}{4})\frac{2g{A_1}^2}{{A_2}^2-{A_1}^2}t^2 + C_2$$

Would anyone be willing to confirm if I did indeed do this correctly? And how do I solve for the arbitrary constant?

In addition, the 2nd part asks how long it would take to fill a 9.4cm tall cylinder filled with 1.0 L of water if the opening is at the bottom and has a 0.50 cm diameter.

Would i just use my newly acquired h(t) equation and simply isolate t, plug in constants and solve?

Thanks a lot!

 

[Theelectricchild]

ok ok, so for "In addition, the 2nd part asks how long it would take to fill a 9.4cm tall cylinder filled with 1.0 L of water if the opening is at the bottom and has a 0.50 cm diameter."

I realize the 0.50cm will be used to determin A1, but what do i use to determine A2 using 1.0 L? Liters are in m^3 ! not m^2 ! that's my trouble for that part!

Thanks.

 

[Clausius2]

ok ok, so for "In addition, the 2nd part asks how long it would take to fill a 9.4cm tall cylinder filled with 1.0 L of water if the opening is at the bottom and has a 0.50 cm diameter."

I realize the 0.50cm will be used to determin A1, but what do i use to determine A2 using 1.0 L? Liters are in m^3 ! not m^2 ! that's my trouble for that part!

Thanks.

Taking into account what I've read in your last posts, either you have a fluid mechanics exam or you are getting in a fluid mechanics course. Your two last posts were about fluid mech. If so, God bless you!

I think your differential equation for h(t) is well posed. (I've not checked the solution :yuck: ). The constant is worked out considering the height at t=0.

About your second problem, you have the volume of the reservoir and its height (1.0 L and 9.4cm). So that, the top transverse area A2 is available (isn't it?).

 

[Theelectricchild]

OH DUH, volume divided by the height for a cylinder gives area pi*r^2, but today, pi*r justice.

Thanks, I hope my h equation is right. BTW i have not taken fluid mechanics yet, but instead this class is titled Thermal Physics, which deals with Hydrodynamics, Thermodynamics and theory of gases and kinetics. But I will take Fluid mechanics sometime in the near future...

 

[Theelectricchild]

anyone else still interested in checking? Hehehe i feel kinda bad cause it took be forever to type all that TeX out, so Id like to make the post worth it!

 

[Divergent13]

I believe your equation is right, but there's a lot going on here with squareroots, and I am not sure if the negative sign is there or not. Anyone know?

 

[Clausius2]

I believe your equation is right, but there's a lot going on here with squareroots, and I am not sure if the negative sign is there or not. Anyone know?

The negative sign is neccesary because of:

$$\frac{dh}{dt}<0$$ to have a physical meaning (outflowing).

Hmmm... your solution is wrong. The more the time the more the height, that's wrong, isn't it?.

$$\frac{dh}{\sqrt h} =- \sqrt\frac{2g{A_1}^2}{{A_2}^2-{A_1}^2}dt$$ (see the minus sign, you have missed to take the negative root).

$$2\sqrt h-2\sqrt{h_o} = -\sqrt\frac{2g{A_1}^2}{{A_2}^2-{A_1}^2}t$$ (here I've used the initial condition t=0; h=ho);

$$h = \Big(-\frac{1}{2}\sqrt\frac{2g{A_1}^2}{{A_2}^2-{A_1}^2}t + \sqrt{h_o}\Big)^2$$


Now I think that's the correct solution.

 

[Clausius2]

It would be an interesting problem worth of being thought by you (and me) to add another third part to your original problem:
"Answer the next questions, reasoning physically and mathematically what is happening if:

a)$$A_2>>>>A_1$$

b)$$A_2=A_2$$

c)$$A_2<A_1$$

Let me know what do you think about this.

 

[Theelectricchild]

Buit wait your h(t) equation when sqaured will take away that negative sign leading back to the h(t) i solved above right?

 

[Clausius2]

Buit wait your h(t) equation when sqaured will take away that negative sign leading back to the h(t) i solved above right?

The sequence you made in your original post was:

i) $$\frac{dh}{\sqrt{h}}=-\sqrt{something}$$

ii)$$\Big(\frac{dh}{\sqrt{h}}\Big)^2=\Big(-\sqrt{something}\Big)^2$$

iii)$$\Big(\frac{dh}{\sqrt{h}}\Big)^2=+something$$

iv)$$\frac{dh}{\sqrt{h}}=+\sqrt{something}$$

the pass of iii) to iv) is wrong. You should have chosen the negative root of the square root again.

 

[Theelectricchild]

Clausius I forgot to say thank you--- you really helped me out alot

 

[Clausius2]

Just thanks not needed, only think about the possible three more questions you would have as I posted three posts above. It seems rather interesting to think of it.