Water tank conservation of energy approach

[gneill]

ok so dR =(dr*dm)/(M+m-dm)

But then how would I integrate that??
doesnt it have to many terms in it?

You didn't take advantage of the fact that L = dr + dR in order to eliminate dr. Take the initial expression and replace dr right away.

 

[sg001]

You didn't take advantage of the fact that L = dr + dR in order to eliminate dr. Take the initial expression and replace dr right away.

Ok so I have dR= ((L-dR)dm)/(M+m-dm)

So integrating

R=∫((L-dR)dm)/(M+m-dm)

But how do I integrate that?
I still have to many terms?

 

[gneill]

Ok so I have dR= ((L-dR)dm)/(M+m-dm)

You haven't isolated dR. It still appears on both sides of the expression.

 

[sg001]

You haven't isolated dR. It still appears on both sides of the expression.

I am aware but from the eqn's

dR +dr = L & (M+m-dm)dR=dr(dm)

You can't isolate dR without introducing dr...
am I missing something?

 

[gneill]

I am aware but from the eqn's

dR +dr = L & (M+m-dm)dR=dr(dm)

You can't isolate dR without introducing dr...
am I missing something?

It's just algebra. Replace dr using dr = L - dR. Move everything to the L.H.S. Expand and collect all the dR terms together. Rearrange with dR alone on the L.H.S.

 

[gneill]

Rethinking it a bit, a simpler approach would be to let M be the total original mass of the tank plus water. You want to move a quantity m of that total mass (representing the water) through the pipe. Then

$(M - dm)dR = dm\;dr ~~~~$ and as before, $~~dr = L - dR$

$dR = L \frac{dm}{M}$

Integration is trivial, with limits 0 → m

$R = \frac{m}{M}L$

 

[sg001]

Rethinking it a bit, a simpler approach would be to let M be the total original mass of the tank plus water. You want to move a quantity m of that total mass (representing the water) through the pipe. Then

$(M - dm)dR = dm\;dr ~~~~$ and as before, $~~dr = L - dR$

$dR = L \frac{dm}{M}$

Integration is trivial, with limits 0 → m

$R = \frac{m}{M}L$

Ok so now the eqn for the distance traveled x = R=∫m/M *L with limits 0 &m

and final velocity =0.

Thanks!

 

[gneill]

Ok so now the eqn for the distance traveled x = R=∫m/M *L with limits 0 &m

That's R=∫(L/M)*dm

and final velocity =0.

Thanks!

 

[sg001]

Thanks for the continuing help
Greatly appreciated.

 

[gneill]

Thanks for the continuing help
Greatly appreciated.

We'll see...
I've decided to add some confusion back into the mix. Upon rethinking the rethinking, I've decided that the previous form of the integration makes more sense

At some time partway through the emptying of the tank, let's say that an amount x of the water has already gone down the pipe. That leaves the tank plus remaining water with mass M-x. It's this remaining mass for which we need to find the dR associated with the next dx that's moved.

$((M - x) - dx) dR = dx\;dr$

$((M - x) - dx) dR - dx (L - dR) = 0~~~~~$ because $dr = L - dR$

$dR = L \frac{dx}{M - x}$

Integrate that for x going from 0 to m, the mass of water drained. M is still the mass of the tank plus the initial amount of water.

EDIT: Changed variable dm to dx in order to have it reflect that it's affecting x, the water that's being shifted.

 

[sg001]

We'll see...
I've decided to add some confusion back into the mix. Upon rethinking the rethinking, I've decided that the previous form of the integration makes more sense

At some time partway through the emptying of the tank, let's say that an amount x of the water has already gone down the pipe. That leaves the tank plus remaining water with mass M-x. It's this remaining mass for which we need to find the dR associated with the next dx that's moved.

$((M - x) - dx) dR = dx\;dr$

$((M - x) - dx) dR - dx (L - dR) = 0~~~~~$ because $dr = L - dR$

$dR = L \frac{dx}{M - x}$


Integrate that for x going from 0 to m, the mass of water drained. M is still the mass of the tank plus the initial amount of water.

EDIT: Changed variable dm to dx in order to have it reflect that it's affecting x, the water that's being shifted.

ok so I have R = ∫L(dx/M-x) = ∫L (M-x)^-1 dx.

But then I have R= -L

Or could it be,,

R = -L(ln(M-x) + c

??

 

[gneill]

ok so I have R = ∫L(dx/M-x) = ∫L (M-x)^-1 dx.

But then I have R= -L

Or could it be,,

R = -L(ln(M-x) + c

??

You need to plug in both limits (0 and m, the total mass of water drained).

 

[sg001]

ok...
so x = R = (-L(ln(M-m)) - (-L(ln(M-0))

 

[gneill]

ok...
so x = R = (-L(ln(M-m)) - (-L(ln(M-0))

Pull out the L and combine the ln's.

 

[sg001]

Finally...
x = R= -L(ln(M-m)-(M-0))

 

[gneill]

Finally...
x = R= -L(ln(M-m)-(M-0))

You need to check your log manipulation math. ln(x) - ln(y) ≠ ln(x - y).

 

[sg001]

You need to check your log manipulation math. ln(x) - ln(y) ≠ ln(x - y).

Ohh yeah My bad simple yet stupid mistake.

x= R = -L(ln(M-m)-ln(M-0))

 

[gneill]

Ohh yeah My bad simple yet stupid mistake.

x= R = -L(ln(M-m)-ln(M-0))

Still review your log math operations; Those logs can be combined into a single log.

 

[sg001]

woops

R = -L(ln((M-m)/(M-0)) = x

 

[gneill]

woops

R = -L(ln((M-m)/(M-0)) = x

It can still be cleaned up some. Drop the zero and use the fact that -ln(b/a) = +ln(a/b).

 

[sg001]

It can still be cleaned up some. Drop the zero and use the fact that -ln(b/a) = +ln(a/b).

Ok so I have now

R = L (ln(M/M-m)) = x

 

[gneill]

Ok so I have now

R = L (ln(M/M-m)) = x

That looks much neater.

 

[sg001]

That looks much neater.

Thanks again gneil!

 

[sg001]

sorry gneill!