Calculating Water Volume in a Tank Using Pressure Measurements

[Regla]

You are supposed to be integrating from 0 to h, not from 0 to 4. You are supposed to be integrating x, not 4.

the whole problem was stated as a volume of 64 dm³... so I thought that since 4³=64 doesn't that mean h=4?

 

[jbriggs444]

the whole problem was stated as a volume of 64 dm³... so I thought that since 4³=64 doesn't that mean h=4?

You are trying to find h. It will not be equal to 4.

h in this context refers to the height (or depth) of the water in the tank. The tank is not full.

 

[Regla]

You are trying to find h. It will not be equal to 4.

like this?

 

[jbriggs444]

like this?

That appears to be a correct result. However, as has been suggested previously, it would be better to leave it in symbolic form and to avoid replacing $\rho$, $g$ and $L$ with numeric values:

$$F_{side} = \rho g L \frac{h^2}{2}$$

Now can you write down a formula for the total force on the bottom of the tank in terms of $\rho$, $g$, $L$ and $h$?

 

[Regla]

That appears to be a correct result. However, as has been suggested previously, it would be better to leave it in symbolic form and to avoid replacing $\rho$, $g$ and $L$ with numeric values:

$$F_{side} = \rho g L \frac{h^2}{2}$$

Now can you write down a formula for the total force on the bottom of the tank in terms of $\rho$, $g$, $L$ and $h$?

Like this? or does the 4 have to be an x?

 

[jbriggs444]

Like this? or does the 4 have to be an x?

You can include "L" or you can include "4". Using both is nonsense. As I had suggested in #34, the correct equation is:

$$F_{side} = \rho g L \frac{h^2}{2}$$

You had previously come up with a formula for the force on the base:

I've got the formula for the base of the tank bF=(ro)gxL2

When you wrote that you were apparently using the variable "x" to denote depth. Let us use "h" for depth and reformat that result:

$$F_{base} = \rho g L^2 h$$

Now then. We have formulas for the force on the side and for the force on the base. The problem statement tells us the ratio of those two forces. Can you write down an equation expressing that fact?

 

[Regla]

Now then. We have formulas for the force on the side and for the force on the base. The problem statement tells us the ratio of those two forces. Can you write down an equation expressing that fact?

Fbase/Fside=3 you mean something like this? ρgL2h/ρgL(h²/2)=3

 

[jbriggs444]

Fbase/Fside=3 you mean something like this? ρgL2h/ρgL(h²/2)=3

Yes. You should be able to simplify that significantly.

 

[Regla]

​

Yes. You should be able to simplify that significantly.

so how do we find the volume or height of the water level? since that only tells the ratio.

 

[jbriggs444]

More calculating. Fewer questions.

 

[Regla]

More calculating. Fewer questions.

so what does the integral in post #34? h? the height of the water level? or the number that I have to put in the post #37 formula?
but we still get the h in the integral.

 

[jbriggs444]

In #37 you had an equation:

Fbase/Fside=3 you mean something like this? ρgL2h/ρgL(h2/2)=3

Solve that equation for h. Or at least simplify it as I had suggested in #38.

 

[Regla]

More calculating. Fewer questions.

I get h=0.00991988 or 0.01

 

[Regla]

In #37 you had an equation:
Solve that equation for h. Or at least simplify it as I had suggested in #38.

if I try to find the pressure to the bottom and side by putting in h I get always the same thing as 2. (ρgL²h/ρgL(h²/2)) It's a bit counter intuitive since I was stated that that should be 3, is the formula for the side equation right?

 

[Nidum]

ρgL2h/ρgL(h2/2)=3

Let's do a tidy re-arrangement of that equation :

ρgL²h = (3/2)ρgLh²

the ρg's cancel so :

L²h = (3/2)Lh²

Divide both sides by hL :

L = (3/2)h

Can you go from there ?

 

[Regla]

ρgL2h/ρgL(h2/2)=3

Let's do a tidy re-arrangement of that equation :

ρgL²h = (3/2)ρgLh²

the ρg's cancel so :

L²h = (3/2)Lh²

Divide both sides by hL :

L = (3/2)h

Can you go from there ?

that simplifies a lot of things, thanks, but when I calculate h (if L=40) it I get h=26.6667 or 27. the problem is that if I try to calculate the pressures they don't equal to the statement give in the equation (Pbase/Pside=3)
Pbase=(1000*9.8*40²*27)/40*40=264600
Pside=(1000*9.8*40*(27²/2)/40*27=132300
264600/132300=2. It doesn't match with the statement given. I feel like there is something wrong with the side force equation.

 

[Nidum]

L is not 40 . Simple numerical mistake I think ?

 

[Regla]

L is not 40 . Simple numerical mistake I think ?

If I know that the volume of the tank is 64 dm³ 4³=64 since that 4 is dm I change it into cm which is 40cm, isn't that right?

 

[Nidum]

Cube root of 640 is roughly 8.62

8.62cm x 8.62 cm x 8.62 cm = 640 cm³ .

.

 

[Regla]

Cube root of 640 is roughly 8.62

8.62cm x 8.62 cm x 8.62 cm = 640 cm³ .

.

even if I change L with 8.62 and h changes to roughly 5.75 I still get 2 when I calculate the Pressures in the way stated in post #46

 

[Nidum]

even if I change L with 8.62 and h changes to roughly 5.75 I still get 2 when I calculate the Pressures in the way stated in post #46

That is correct . The pressure on the base is twice the average pressure on the side .

Anyway you don't need to work that out except possibly for your own interest . You just need to work out the volume of the water .

 

[Regla]

That is correct . The pressure on the base is twice the average pressure on the side .

Anyway you don't need to work that out except possibly for your own interest . You just need to work out the volume of the water .

ohhhhhhh, so that's AVERAGE pressure, makes some sense. But is there any way to make sure that h is actually 5.75?

 

[Nidum]

Work out the actual areas and pressures and pressure forces . If pressure force on the bottom is three times pressure force on the side then you have the right answer for h .

To complete this problem finally what is the actual volume of the water ?

 

[Regla]

Work out the actual areas and pressures and pressure forces . If pressure force on the bottom is three times pressure force on the side then you have the right answer for h .

To complete this problem finally what is the actual volume of the water ?

if the height of the water is 5.75 the the volume is 427 cm³ and the weight is 0.427kg. I also have to convert the weight into kilomoles, but I think I can handle it. thanks. If I'll have any problems while going over the whole solution I'll ask here, is that ok?

 

[Nidum]

That's the correct answer for the water volume - good result .

Certainly ok to ask new questions anytime .

 

[Regla]

That's the correct answer for the water volume - good result .

Certainly ok to ask new questions anytime .

since you're an regular visitor to this forum, what do I do that his post or problem would be shown as solved?

 

[Regla]

That's the correct answer for the water volume - good result .

Certainly ok to ask new questions anytime .

you made a small mistake (as did I) L is 40cm, 64dm³ is not 640cm³ it's 64000cm³. It's because we're dealing with volume not length.

 

[Nidum]

We are told that the total volume of the cube is 640 cm³ . If that's the volume then length of side is 8.62 cm . I don't know where you get all the other numbers from ?

 

[haruspex]

64dm3 is not 640cm3

In post #1 you wrote 640cm³. Are you now saying that was wrong, that it is 64dm³?

 

[Regla]

In post #1 you wrote 640cm³. Are you now saying that was wrong, that it is 64dm³?

It was my mistake. I was stated that the volume is 64dm³ and stupidly stated that it's 640cm³ since I wanted it to be more applicable to people with imperial system to understand, I admit that I made a mistake, but the solution as far as I know doesn't change. it was a stupid mistake.