Determine the radius of a rising bubble

[Sabra_a]

Homework Statement

A small air bubble at the bottom of an open 4-m depth water tank
has a radius of 0.5 mm. Due to some reason the bubble comes off
the bottom. Determine the radius of the bubble when it is 0.1 m
below the surface. Assume that the pressure inside the bubble is
2/r above the pressure outside the bubble, where  =0.073 N/m –
the surface tension of water-air and r – the radius of the bubble. The
pressure and volume of the air in the bubble are related by the
expression pV=const.

Relevant Equations

ρ=2 σ/r
P=ρgh+P

Homework Statement: A small air bubble at the bottom of an open 4-m depth water tank
has a radius of 0.5 mm. Due to some reason the bubble comes off
the bottom. Determine the radius of the bubble when it is 0.1 m
below the surface. Assume that the pressure inside the bubble is
2/r above the pressure outside the bubble, where  =0.073 N/m –
the surface tension of water-air and r – the radius of the bubble. The
pressure and volume of the air in the bubble are related by the
expression pV=const.
Homework Equations: ρ=2 σ/r
P=ρgh+P

I calculated the pressure inside the bubble multiplying 2 into the given surface tension and the reduce I'm not sure if I should convert the mm to m for the radius or its not required.
ρ=2 σ/r
ρ=2 0.073/0.5
ρ= 0.292

the substituted that in P=ρgh+P
= (0.292)(9.81)(4) + 101300
=101311.4

this is what I did so far

 

[BvU]

Good recovery ! back .

Now what about this PV = Const equation ?
And: what you did for h = 4 you can also do for h = 0.1

Help me understand the units you use. 0.292 stones/cubic inch ?

 

[Sabra_a]

Thank you!
I think I should convert he radius unit to m from 0.5mm to 0.0005 m.
ρ=2 σ/r
ρ=2 (0.073N/m/0.0005m)
ρ= 292 N/m2 or Pascal

the substituted that in P=ρgh+P
= (292)(9.81)(4) + 101300
=112758.08

when h = 0.1
= (292)(9.81)(0.1) + 101300
= 101586.452

PV= constant
maybe rearrange the equation to calculate volume! PV=nRT but we don't have the n and T

 

[BvU]

No, but we do have p and V at 4 m depth ... so we have the constant nRT !

 

[Sabra_a]

so we calculate the volume of sphere by V=

V= 3/4 * 3.14 * (0.0005)3
= 2.94x10^-10

PV=constant
112758.08 * 2.94x10^-10
3.32x10^-5

 

[BvU]

Explain, please ...
Tip: use symbols, not numbers. And at the end: first check dimensions, then finally substitute to get a number + dimension for the answer. Oh, and don't let the number of significant digits explode. One or, tops, two -- like in the given information.

p at a depth of 4 m is bigger than 1.1 x 10⁵ N/m²

 

[Sabra_a]

because we have pressure we need to calculate volume by calculating the volume of sphere V=

radius = 0.0005 m
V= (3/4) * (3.14) * (0.0005)^3
= 2.94x10^-10 m^3

substituting the volume value in PV=constant
P and V at 4 m depth
PV=constant
(112758.08 * 2.94x10^-10) = constant
3.32x10^-5 = constant

as we have pressure and volume at 4 m depth but we have only pressure at 0.1 m we can use the following equation
P1V1=P2V2
V2 = P1V1/P2
V2 = 3.32x10^-5 / 101586.452
= 3.26x10^-10

 

[Chestermiller]

because we have pressure we need to calculate volume by calculating the volume of sphere V= View attachment 252197
radius = 0.0005 m
V= (3/4) * (3.14) * (0.0005)^3
= 2.94x10^-10 m^3

substituting the volume value in PV=constant
P and V at 4 m depth
PV=constant
(112758.08 * 2.94x10^-10) = constant
3.32x10^-5 = constant

as we have pressure and volume at 4 m depth but we have only pressure at 0.1 m we can use the following equation
P1V1=P2V2
V2 = P1V1/P2
V2 = 3.32x10^-5 / 101586.452
= 3.26x10^-10

Your equation for the volume is incorrect. It is 4/3, not 3/4. In addition, you did not take into account the effect of surface tension in getting the pressure within the gas bubble at the 0.1 m depth.

 

[Sabra_a]

volume of sphere
= 4/3 * 3.14 * r^3
= (4/3) * (3.14) * (0.0005)^3
= 5.23x10^-10 m3
I don't really get how to solve this question
by take into account the effect of surface tension at 0.1 m depth do you mean adding it to the equation
P=ρgh+P+ surface tension

 

[Chestermiller]

volume of sphere
= 4/3 * 3.14 * r^3
= (4/3) * (3.14) * (0.0005)^3
= 5.23x10^-10 m3
I don't really get how to solve this question
by take into account the effect of surface tension at 0.1 m depth do you mean adding it to the equation
P=ρgh+P+ surface tension

$$P_1=\rho g h_1+\frac{2\sigma}{r_1}$$
$$P_2=\rho g h_2+\frac{2\sigma}{r_2}$$$$P_2V_2=P_1V_1$$
$$V=\frac{4}{3}\pi r^3$$

 

[BvU]

Homework Equations: ρ=2 σ/r
P=ρgh+P

There are some things utterly wrong here !
I recognize the second one by its form but not in the way presented (leads to $\rho g h = 0$) . But that is definitely NOT the $rho$ in the first.

@Sabra_a : urgent advice to make good use of the tips in #6. Observe how much clearer things become when you use symbols, as @Chestermiller demonstrates in #10 ! You can count knowns, unknowns and number of equations to check if you have enough to start calculating, you can check dimensions, strike common factors, etc. etc. Much less prone to errors and easier to fix should you discover a mistake on the way.

 

[Sabra_a]

I hope this is clear now
$P=\frac {2σ} {r}$
r = 0.5mm →m = 0.0005m

$P=\frac {2*0.073N} {0.0005m}$

$P=292 N/m^2$
density of water = 1000 $kg/m^3$

$P_1=ρgh_1+\frac {2σ} {r_1}$
$= 1000*9.81*4+292$
$P_2=39532$

$P_2=ρgh_2+\frac {2σ} {r_2}$
$= 1000*9.81*0.1+292$
$P_2=1273$

$V_1=\frac {4} {3} πr^3$
$V=\frac {4} {3} π(0.0005)^3$

$V_1=5.23*10^{-10}$

$P_2V_2=P_1V_1$

$V_2=\frac {P_1V_1} {P_2}$

$V_2=1.62*10^{-8}$

​

 

[Chestermiller]

I hope this is clear now
$P=\frac {2σ} {r}$
r = 0.5mm →m = 0.0005m

$P=\frac {2*0.073N} {0.0005m}$

$P=292 N/m^2$
density of water = 1000 $kg/m^3$

$P_1=ρgh_1+\frac {2σ} {r_1}$
$= 1000*9.81*4+292$
$P_2=39532$​

You forgot to include atmospheric pressure in this calculation. If you are using the ideal gas law later, you need to use absolute pressures.

$P_2=ρgh_2+\frac {2σ} {r_2}$
$= 1000*9.81*0.1+292$
$P_2=1273$

You again forgot to include atmospheric pressure here. And the surface tension contribution is not 292 since you don't even know r2 yet.

$V_1=\frac {4} {3} πr^3$
$V=\frac {4} {3} π(0.0005)^3$

$V_1=5.23*10^{-10}$

This is correct.

$P_2V_2=P_1V_1$

$V_2=\frac {P_1V_1} {P_2}$

$V_2=1.62*10^{-8}$

This is incorrect because you need to use absolute pressures.​

 

[Sabra_a]

$P=\frac {2σ} {r}$
r = 0.5mm →m = 0.0005m

$P=\frac {2*0.073N} {0.0005m}$

$P=292 N/m^2$
density of water = 1000 $kg/m^3$

$P_1=ρgh_1+\frac {2σ} {r_1}+P$
$= 1000*9.81*4+292+101300$
$P_1=140832$

$V_1=\frac {4} {3} πr^3$
$V=\frac {4} {3} π(0.0005)^3$

$V_1=5.23*10^{-10}$

$P_1V_1=constant$
$140832 * 5.23*10^{-10}=const.$
$7.36*10^{-5}=const.$
​

 

[Chestermiller]

$P=\frac {2σ} {r}$
r = 0.5mm →m = 0.0005m

$P=\frac {2*0.073N} {0.0005m}$

$P=292 N/m^2$
density of water = 1000 $kg/m^3$

$P_1=ρgh_1+\frac {2σ} {r_1}+P$
$= 1000*9.81*4+292+101300$
$P_1=140832$

$V_1=\frac {4} {3} πr^3$
$V=\frac {4} {3} π(0.0005)^3$

$V_1=5.23*10^{-10}$

$P_1V_1=constant$
$140832 * 5.23*10^{-10}=const.$
$7.36*10^{-5}=const.$
​

OK so far. How much do you think the surface tension term contributes to the final answer?

 

[Sabra_a]

as the bubble move upwards it grows in size, so as the volume increase pressure outside the bubble drops therefore surface tension increases!

 

[Chestermiller]

as the bubble move upwards it grows in size, so as the volume increase pressure outside the bubble drops therefore surface tension increases!

Actually, as the radius of the bubble increases, the contribution of surface tension to the pressure of the gas inside the bubble decreases. And you have already shown that for the depth of 4 m, the surface tension contribution to the pressure is only 292 Pa out of 140832 Pa, or much less than 1%. You will find that the surface tension contribution at the shallower depth will be even less. So, you might as well neglect surface tension. But it's very minor effect can still be included if you really want to.

 

[Sabra_a]

I will probably include it, but how I will be able to calculate the radius at 0.1 m?

 

[Chestermiller]

You have a non-linear algebraic equation to solve, right? What methods do you know of for solving a non-linear algebraic equation?

 

[Sabra_a]

$ax+b=0$
$x=-b/a$

 

[Chestermiller]

$ax+b=0$
$x=-b/a$

I said non-linear.

 

[Chestermiller]

Let me summarize what you have so far: $$P_1\left(\frac{4}{3}\pi r_1^3\right)=P_2\left(\frac{4}{3}\pi r_2^3\right)$$ or, equivalently $$P_1r_1^3=P_2r_2^3$$with $$P_1=P_{atm}+\rho g h_1+\frac{2\sigma}{r_1}$$and$$P_2=P_{atm}+\rho g h_2+\frac{2\sigma}{r_2}$$Combining these equations gives:
$$(P_{atm}+\rho g h_1+\frac{2\sigma}{r_1})r_1^3=(P_{atm}+\rho g h_2+\frac{2\sigma}{r_2})r_2^3$$Next, defining $$x=\frac{r_2}{r_1}$$ we have:$$(P_{atm}+\rho g h_1+\frac{2\sigma}{r_1})=(P_{atm}+\rho g h_2+\frac{2\sigma}{r_1x})x^3$$Our objective is to solve for x. Do you agree with this so far?

 

[Sabra_a]

yes I do agree

 

[Chestermiller]

It is possible to solve this cubic equation in x analytically, but it is much easier to solve numerically. Any ideas on how to solve it numerically? Have you heard of the "method of successive substitutions?" What level of math have you had so far in your courses?

 

[Sabra_a]

I have stopped uni 3 years ago, don't remember much but trying my best. is it fine like this

 

[Chestermiller]

I have stopped uni 3 years ago, don't remember much but trying my best. is it fine like this
View attachment 252449

No. Your algebra is incorrect.