Drawing up a liquid by forward motion of a tanker

[annnoyyying]

Homework Statement

We are in a planet where the gravitational acceleration is 1 ms-2.
Tankers fly over the surface of a liquid ammonia lake (at 239K) at a height of 40m. A special, insulated, frictionless tube is lowered into the ammonia 1m below the surface.
We want to draw up ammonia through the tube into an insulated tank by the forward motion of the tanker without pumping.
The tank is initially full of hydrogen at 5 bar and 239K. The hydrogen is displaced through a vent at the top of the tanker by the incoming ammonia. The maximum depth of ammonia in the tank is 10m (the bottom of the tank forms the base).
What is the minimum speed the tanker must fly over the lake to fill its tank?

Homework Equations

1/2 mv^2=mgh? F=ma?

The Attempt at a Solution

the total height the ammonia has to travel to reach the tank is 41m. Something to do with the height and gravitational acceleration? And the maximum height of ammonia? And the pressure of the hydrogen in the tank initially?

 

[BvU]

Not much to go by, eh ?
And what would the three question mark equation contribute to solving this problem ?

Not much of an attempt at solution either: 40 + 1 = 41 m and that's all.
What v does that give when substituted in the four question mark equation ?

But the problem statement gives me the idea 40 + 1 + 10 m might be more appropriate !

The story made me think of Pitot tubes
Google led me to Bernoulli equation. But if that isn't in the preceding sections of your textbook or class notes, then perhaps all you have is Newton ?

I also wondered at the 1 m/s². But even more at the 5 Bar of hydrogen that has to be expelled. Does that mean that even at such low g the atmospheric pressure is 5 Bar ? Without further info I'd be inclined to assume that.

 

[mfb]

Energy conservation is a good approach. Moving a drop of ammonia from the tanker to the surface of the lake should not release energy.

 

[annnoyyying]

i know there's something to do with bernoulli's equation/energy balances
I'm just stuck on the fact that there's a forward motion involved. (force in x direction to push liquid in y direction)
According to the other parts of the question I guess I am safe to assume the ambient pressure is 1 bar.

 

[BvU]

What other parts of the question(s) ?
And why the whole 5 bar story if just opening the vent can get rid of 80% of the hydrogen ? Or do they want the NH3 to stay at 5 Bar ?

 

[annnoyyying]

The other parts are irrelevant to this question so i just extracted the required information. And yes, they want the NH3 to stay at 5 bar.
If i can differentiate between the excess information and the required information i probably wouldn't be here. Sadly.
im not given density or atmospheric pressure.

 

[haruspex]

Energy conservation is a good approach. Moving a drop of ammonia from the tanker to the surface of the lake should not release energy.

Did you mean, raising a drop from the lower end of the pipe to the surface should not require energy? If so, I agree.

Annnoyyying, I assume the lower end of the pipe points forwards, so acts as a scoop. You quoted a useful energy equation. Note that the m's cancel, turning it into a SUVAT equation.

 

[annnoyyying]

Did you mean, raising a drop from the lower end of the pipe to the surface should not require energy? If so, I agree.

Annnoyyying, I assume the lower end of the pipe points forwards, so acts as a scoop. You quoted a useful energy equation. Note that the m's cancel, turning it into a SUVAT equation.

Yes the lower end of the pipe points forwards to act as a scoop. I originally suspected that particular energy equation as it does not require pressure or density terms like in the energy/Bernoulli's equation.

raising a drop from the lower end of the pipe to the surface should not require energy

Do you mind expanding a bit on this? My mechanics side of my brain has never been well developed.

 

[BvU]

Means you add 40 m to 10 m and get the 1 m for granted haru was nice to me by not saying I totally overlooked that,

 

[annnoyyying]

Means you add 40 m to 10 m and get the 1 m for granted haru was nice to me by not saying I totally overlooked that,

So I just need to use 1/2 mv^2 = mgh, no bernoulli's? using that I get v = 10ms-1

 

[BvU]

And that's from 50 m. The 5 Bar is out of the picture ?

Actually, I have no idea, but: can planes fly at 36 km/h in a 1 m/s² environment ?

 

[mfb]

Did you mean, raising a drop from the lower end of the pipe to the surface should not require energy? If so, I agree.

Same thing?

Actually, I have no idea, but: can planes fly at 36 km/h in a 1 m/s² environment ?

They can. Birds can even do it on earth.

Where did the hydrogen pressure go now?

 

[haruspex]

Same thing?

No, the tanker is 40m above the lake surface; the lower end of the pipe is 1m below the lake surface. And moving it from tanker to lake would be opposite to the way it is flowing.

 

[mfb]

In equilibrium, nothing flows, and "the energy difference is zero" is equivalent to both statements.

 

[haruspex]

In equilibrium, nothing flows, and "the energy difference is zero" is equivalent to both statements.

Then I find your original comment most confusing:

Moving a drop of ammonia from the tanker to the surface of the lake should not release energy.

That seems to say there's no PE change.

 

[mfb]

Sure there is a change in potential energy. We still have the velocity => kinetic energy to take into account.