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My Science Experiment

by Phrak
Tags: experiment, science
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Phrak
#1
Nov1-11, 05:22 AM
P: 4,512
1. The problem statement, all variables and given/known data

A small balloon was inflated with air at 1460 ft above sea level at a temperature of about 75 F. After inflating it was attached to a nanometer and depressed 18.25 inches of water. It was then tied-off. It's mass, with the nipple removed, measured about 156 milligrams. The volumetric displacement of the balloon was then measured by submerging in a 5 cup glass graduated kitchen measuring-thing. The volume of displaced water measured 360 ml.

Placed in a 7 inch deep kettle over 1 inch of boiling water, it sank to the bottom (as expected) rather than floating.

What was the minimum wetness of the steam as measured in grams of suspended water droplets per total grams consisting of both suspended water droplets and vapor? Better said, what minimum wetness of steam, measured in grams per total grams, would be required to achive boyancy in this experiment?

2. Relevant equations
...
3. The attempt at a solution

I have a CRC handbook with a table for standard atmosphere at given elevation, but it doesn't correct for actual temperature. Without correction, I get an air density in the balloon of 1.105 grams per liter. Is that correct?

I obtained a boiling point of 205 F for steam at 1460 feet alt. (right or wrong?) Using an online steam density calculator I got .528 grams per liter for dry steam. How's this number?
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Simon Bridge
#2
Nov1-11, 10:53 AM
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We can check your physics - the constants are up to you.
How would you check these values?

If only you had access to a large computer network and the tools to search it!
Phrak
#3
Nov2-11, 05:24 AM
P: 4,512
Yes sir. Here's some phyics, then.
[tex]v_B \rho_B = v_L\rho_L + v_V\rho_V [/tex]
[tex]v_B = v_L + v_V[/tex]
[itex]v[/itex] is volume, [itex] \rho [/itex] is mass density, and B, L, and V reference the balloon, water vapor and liquid water respectively

With m referencing mass, for neutral boyancy, the parts mass required for the liquid and vapor states of water require that,

[tex]\frac{m_V}{m_B} = \frac{\rho_V}{\rho_B} \left( \frac{\rho_B - \rho_L}{\rho_V - \rho_L} \right)[/tex]
and
[tex]\frac{m_L}{m_B} = \frac{\rho_L}{\rho_B} \left( \frac{\rho_B - \rho_V}{\rho_L - \rho_V} \right)[/tex]

How's that?

Simon Bridge
#4
Nov3-11, 02:24 AM
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My Science Experiment

That not physics, that's a bunch of equations. What was your reasoning?

For eg.
A small balloon was inflated with air at 1460 ft above sea level at a temperature of about 75 F.
... so you have a small container of air
... if you have it's pressure and temp can you estimate it's volume? The mass of air involved?

After inflating it was attached to a nanometer and depressed 18.25 inches of water.
Don't you mean a manometer? (A liquid column hydrostatic device for measuring small pressure differences?)
What does it mean that 18.25 inches of water were depressed in this device?
Would you expect the air pressure inside the elastic balloon to be higher, lower, or the same as the pressure of the air outside it?

It was then tied-off. It's mass, with the nipple removed, measured about 156 milligrams.
Wow - is this a lot? That's 0.156g. Mass of what? The whole balloon, the manometer, the nipple?
How does this relate to the air inside the balloon?
What does this measurement tell you?

The volumetric displacement of the balloon was then measured by submerging in a 5 cup glass graduated kitchen measuring-thing. The volume of displaced water measured 360 ml.
What is the mass of 360ml of water?
Is this displacement consistent with the situation? (How would you check?)

See ... physics.
If all you are doing is substituting numbers into equations you are doing accounting, not physics.
(Actually I know some accountants who would take issue with that statement.)


Also, you have only followed one third of the advise... <puzzled>
http://www.csgnetwork.com/h2oboilcalc.html
BP(1490ft=445m) = 98.6degC =209.4degF
... are there ways you could check your other numbers?
Phrak
#5
Nov4-11, 03:22 AM
P: 4,512
Quote Quote by Simon Bridge View Post
That not physics, that's a bunch of equations. What was your reasoning?
Oh. I thought it was physics. I used the Additive Law of Mass.

Also, you have only followed one third of the advise... <puzzled>
http://www.csgnetwork.com/h2oboilcalc.html
BP(1490ft=445m) = 98.6degC =209.4degF
... are there ways you could check your other numbers?
I got 98.4 C instead cause I used Engineeringtoolbox.com WTF?
Simon Bridge
#6
Nov4-11, 03:48 AM
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Oh. I thought it was physics. I used the Additive Law of Mass.
The additive law is relevant but it does not describe what is happening.
The equations are not the same thing as the physics. If you start with the physics, then describe it in mathematics, you'll automatically get the right equations.

I used an example (post #4) of what I meant when I said I wanted the physics you used ... giving you guiding questions. Try answering the other questions. The answers contain some of the mistakes you have been making and will help you understand what is going on. C'mon, you can do this!

I get 98.4 C using Engineeringtoolbox.com
So you would check some figures you looked up in one source by looking them up in another source? What if you are looking up the wrong figure? How would you know? (How you figure out why two sources disagree so much is a related matter.) How about seeing if the figures make sense in terms of the problem?

I strongly suspect the reason you are having trouble with this problem is that you have been too focused on looking up numbers and finding the right equations instead of understanding what the problem's description is telling you. I could just tell you the answer, and how to go about it, but you won't learn anything that way. All anyone can do otherwise is try to point you in the right direction.

If I thought you'd just got confused, I'd just point out the confusion and you'd go "oh yeah <headslap>" and that's it but I think it's deeper than that. Someone else can do that if they want but I hope they won't because it won't benefit you as much.
Phrak
#7
Nov5-11, 12:49 AM
P: 4,512
Quote Quote by Simon Bridge View Post
The additive law is relevant but it does not describe what is happening.
Correct. I am debunking a conjecture that would have the balloon float. The additive law is for neutral boyancy rather than a floating or sinking balloon. An inequality would describe what actually occurred. You sound hesitant as if you are unfamiliar with the additive law.

The equations are not the same thing as the physics. If you start with the physics, then describe it in mathematics, you'll automatically get the right equations.
I did that, starting with the assumption that for two masses m1 and m2, the combined mass is m1+m1, the additive law of masses.

I used an example (post #4) of what I meant when I said I wanted the physics you used ... giving you guiding questions. Try answering the other questions. The answers contain some of the mistakes you have been making and will help you understand what is going on. C'mon, you can do this!


So you would check some figures you looked up in one source by looking them up in another source? What if you are looking up the wrong figure? How would you know? (How you figure out why two sources disagree so much is a related matter.) How about seeing if the figures make sense in terms of the problem?

I strongly suspect the reason you are having trouble with this problem is that you have been too focused on looking up numbers and finding the right equations instead of understanding what the problem's description is telling you. I could just tell you the answer, and how to go about it, but you won't learn anything that way. All anyone can do otherwise is try to point you in the right direction.

If I thought you'd just got confused, I'd just point out the confusion and you'd go "oh yeah <headslap>" and that's it but I think it's deeper than that. Someone else can do that if they want but I hope they won't because it won't benefit you as much.
I see the problem. You got the altitude wrong. I'm at 1460 feet over sea level, not 1490.

But I'm confused. If an equation relating densities that no accountant would understand, isn't physics, what is it?

By the way, I appreciate you motivational help, but you should not direct folks to goofy web sites for reference data via some conversion calculator generated by some programmer with a degree in music appriciation, or computer science, or whatever.
Simon Bridge
#8
Nov5-11, 02:17 AM
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Quote Quote by Phrak View Post
Correct. I am debunking a conjecture that would have the balloon float. The additive law is for neutral boyancy rather than a floating or sinking balloon. An inequality would describe what actually occurred. You sound hesitant as if you are unfamiliar with the additive law.
I'm trying not to do your work for you ... so I have to be a bit evasive. sry. :)
I [started] with the assumption that for two masses m1 and m2, the combined mass is m1+m1, the additive law of masses.
But you did not say that was what you did, you just plonked the equations down, and you still have not related it to the situation described.

How did you use this to help solve the problem?

Note: the problem presented was to find the wetness of the steam, but you seem to have a different objective in mind.

I see the problem. You got the altitude wrong. I'm at 1460 feet over sea level, not 1490.
Fair enough - I read the 6 upside down ... it happens.

But I'm confused. If an equation relating densities that no accountant would understand, isn't physics, what is it?
I tried to illustrate what I meant by example. Work out the answers to the list of questions I provided and compare what you get from that with the bunch of equations.

Possibly it is just a cultural difference with language so I won't argue semantics with you. Instead I'll try to communicate ... here's another example:

If a student comes to you with a kinematics problem and you ask that worthy to show you the physics of the situation described in the problem, and that student responds by reproducing the kinematic equations ... to what extent can you confidently conclude that the student has done any physics?

There is nothing wrong with the equations - but what if he's chosen the wrong ones because he has misunderstood the physics? See? The equations are not the physics - they are just a description of some physics that may not be the physics.

There is clearly physics in the equations - but to actually do physics is something more right? When I wanted you to show me the physics I was looking for you to tell me about your understanding of the problem and the situation in more detail.

Answering those questions would have helped.

By the way, I appreciate you motivational help, but you should not direct folks to goofy web sites for reference data via some conversion calculator generated by some programmer with a degree in music appriciation, or computer science, or whatever.
I'm sorry, I thought I'd made myself clear - you have to come up with your own sources ;)

Well whatever - good luck.


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