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My Science Experiment 
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#1
Nov111, 05:22 AM

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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 tiedoff. 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 measuringthing. 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? 


#2
Nov111, 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! 


#3
Nov211, 05:24 AM

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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? 


#4
Nov311, 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 tiedoff. 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 measuringthing. 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? 


#5
Nov411, 03:22 AM

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#6
Nov411, 03:48 AM

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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 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. 


#7
Nov511, 12:49 AM

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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. 


#8
Nov511, 02:17 AM

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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. 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. Well whatever  good luck. 


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