College maths.....lame

[killa]

I don't have notes for some of my assignment (they havent even covered it) and i'm stuck.

Question:- The ball of a float valve is a sphere of 200mm diameter. If the ball is immersed to a depth of 150mm, calculate
[i] It’s wetted surface area
[ii] The volume of liquid displaced in liqud

I need the working as well.

Thanks!
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[bob-a-job]

I assume it means the most bottom point is at 150 mm... not the centre point?

[killa]

I assumed it meant submerged to a depth of 150mm.
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[bob-a-job]

Well if it is totally submerged the calulations are easy (assuming there is no compression) or if it is not completely submerged the calcualtion is slightly (alot) harder as you have to remove part of the sphere...

[bob-a-job]

If fully submerged:

area = 4 * pi * r^2
area = 4 * pi * 100^2

= 125663.68 mm^2

= 0.126 Metres^2 (3sf)

[bob-a-job]

If fully submerged:

vol of liquid displaced = volume of sphere

= (4/3) * pi * r^3

= 1.3333333 * pi * 100^3

= 4188789.3 (.3 is recurring) mm^3

= 4188.8 ML (1dp)
or
= 4.2 Litres (1dp)

[bob-a-job]

I think!!!

[shorty]

ermmm before you follow bob-a-jobs advice too far as far as I'm aware this is a fairly straightforward calculus exercise you may find that this site offers some help.

And the volume of water displaced is equal to the volume of the submerged part of the sphere which will be another integration.

Worth bearing in mind that you will certainly get marks for mentioning this along with the standard surface area and volume formulae.

There are plenty of examples on the web for doing these type of calculations.

Shorty
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[bob-a-job]

Here is a diagram trying to explain my first post above...

[bob-a-job]

I found all my stuff here:
SPHERE
- https://en.wikipedia.org/wiki/Sphere

DOME (used to removed unsubmerged part)
- https://en.wikipedia.org/wiki/Dome_%28mathematics%29

Assuming that the sphere gets fully submerged (i believe the calcs above are right), however that does seem a bit simple for college so my gues is that you will have to calc the detaisl assuming it is not fully submerged.. so as shorty says, you will have to use integration to help you calulate the non submerged part.....

[killa]

Ok, thanks…..I think.

I do think it means the ball is only 150mm in the water.
* = X, right?

Wish I had some notes on this, even with the figures I’m a bit lost.

Karma for you when I get some to give.
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[bob-a-job]

[shorty]

I would suggest that you do it twice (as once you have done it once it will be dead easier doing it again -

1) Sphere completely submerged so that the top is 150mm below the surface (v. easy as is just spherical volume and total surface)
2) Sphere submerged so that bottom is 150mm below the surface.
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[shorty]

How's this....

The integral for the surface area of a sphere looks like this:

b
/
A = 2*Pi*R | dx = 2*Pi*R(b - a)
/
a

In your case R is 100 and you want to start at 100-150mm, so a = -50
and b = 100. In the normal case of a whole sphere, a = -R and b = R.
So:

b
/
A = 2*Pi*R | dx = 2*Pi*R(b - a) = 2*Pi*100(100+50)
/
a


= 47123.89mm^2 (2 d.p)

to check total area (as bob said) is 125663.68 mm^2

Taken from here: https://mathforum.org/library/drmath/view/52133.html

Just the volume to do then.

Shorty
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[Mal]

Vol of a sphere = (4/3)*pi*r^3
= (4/3)*3.142*100*100*100
= 4189333 mm^3

Vol of a dome = pi*r*h^2 - (1/2)*pi*h^3
= 3.142*100*50*50 - 0.5*3.142*50*50*50
= 785500 - 196375
= 589125 mm^3

Vol of water displaced = vol of sphere below water
= vol of sphere - vol of dome
= 4189333.3 - 589125
= 3600208.3 mm^3

Surface area of a sphere = 4*pi*r^2
= 4*3.142*100*100
= 125680 mm^2

Surface area of a dome = 2*pi*r*h
= 2*3.142*100*50
= 31420 mm^2

Wetted area = area of sphere - area of dome above water
= 125680 - 31420
= 94260 mm^2

Hope this helps.

Mal

[shorty]

Mal - am I missing something with the radius of the dome or are you just assuming that it is the same as that of the sphere? It isn't - that would only be the case if the sphere was submerged 100mm. The radius of the dome is unknown.

The point of this exercise, I am sure, is to use calculus - otherwise it is very simple arithmetic.

Shorty
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[Annabella]

[killa]

Now im confused
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[shorty]

OK Killa, I shall try my best to dis-confuse you. Bear in mind it's 10 odd years since i've done this so bear with me...

What Mal has said is entirely correct as far as the logic and workings go apart from the fact that the radius of the "dome" (i.e. the part of the sphere above the water) is unknown.

You can work out the radius of the dome as it changes depending on how far up the sphere you go. That means at the "equator" of the sphere the radius of the dome is equal to the radius of the sphere.

What you need to do is work out the radius of the dome using calculus (integration isn't it guys?).

My example uses calculus to work out the area of the sphere - what you could do is just work out the radius of the bottom of the dome and plug that into Mal's equations to give you the answers.

Hope that's a bit clearer.

Incidentally you will almost certainly get marks for mentioning all of the above as it shows that you understand in principle what you need to do.

HTH

Shorty
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[killa]

I gotcha

Thanks
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[Mal]

[instigator]

what course you doing at college and where killa?

[syl]

[syl]