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Algebra & Calculus

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Sun Sep 18, 2011 8:21 am
seeminglymeaningless says...



D: I didn't want you guys to spend ages reading up on it D: I feel bad now... but it really will help me, and if I can figure out the 2nd one myself, then I'm all set :)

Thanks again, the two of you, for helping me out ^^

lol, were the dreams any good? :P




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Tue Sep 20, 2011 1:20 am
Snoink says...



Duuuuude. We had fun doing it! :)

The dreams were fun. :)
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Tue Sep 20, 2011 1:19 pm
seeminglymeaningless says...



Tank Problem flow in and flow out with a salt concentration

Find formula’s for V(T) and C(T)

For V(t) first find for the changing volume state is dV/dT

Therefore dV/dT = Q1 - Q2 integrate both side with respect to “t” time

So V(t) = Q1t - Q2t + constant , constant found as start condition called Vo so that at

t = 0 V(t) = Vo

Therefore V(t) = Q1t - Q2t + Vo

For C(t) the formula for changing salt concentration in the tank is dependent

On both the changing mass of salt and the changing volume with respect to time

As C(t) = M(t) / V(t) and we already have the volume formula, the mass formula needs defining

dM/dT = mass in – mass out = Q1C1 - Q2C2

integrate both sides with respect to “t” gives

M(t) = Q1C1t - Q2C2t + constant , constant found by making t = 0 and is called the start mass Mo is found from Vo and Co for ease of formulation will leave as Mo

Therefore M(t) = Q1C1t - Q2C2t + Mo

Now rearrange the equation M(t) = Mo + Q1C1t - Q2C2t
Remember that C2 is the same concentration as the tank so that

C2 = M(t)/ V(t) so rewrite the equation you get

M(t) = Mo + tQ1C1 - tQ2 M(t)/ V(t) need to get M(t) by itself

M(t) + tQ2 M(t)/ V(t) = Mo + tQ1C1

M(t)( 1 + tQ2/ V(t) ) = Mo + tQ1C1

M(t)(( V(t) + tQ2)/ V(t) ) = Mo + tQ1C1

M(t)(( Q1t - Q2t + Vo + tQ2)/ V(t) ) = Mo + tQ1C1 remember V(t) = Q1t - Q2t + Vo

M(t)( Q1t + Vo )/ V(t) = Mo + tQ1C1 now cross multiple to obtain M(t) by itself

M(t) = V(t) (Mo + tQ1C1 )/( Q1t + Vo ) the formula for mass in the tank

But remember we want the C(t) which is equal to M(t) / V(t)

Therefore C(t) = (V(t) (Mo + tQ1C1 )/( Q1t + Vo ))/ V(t) as there is a V(t) term on

the top and the bottom they can be cancelled out.

Therefore C(t) = (Mo + tQ1C1 )/( Q1t + Vo ) to check if correct plug in the time t = 0

Which gives C(t) = Mo / Vo which from definitions is start mass of salt and start

volume so that C(t) = Co

thus to find salt concentration at overflow first find time taken to fill tank

From V(t) = Q1t - Q2t + Vo where V(t) = 200 l , Vo = 100 l , Q1 = 20 l/m

Q2 = 10 l/m

There for 200 = 20t – 10t + 100 gives 100 = 10t therefore t = 10 minutes

Put into C(t) = (Mo + tQ1C1 )/( Q1t + Vo ) given that Mo = VoCo

You get C(t) = ( 100 x 0.1 + 10 x 20 x 0.4 )/( 20 x 10 + 100 )

C(t) = ( 10 + 80 )/( 200 + 100 ) = 90/300 = 0.3 g/l at t = 10 minutes




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Sun Nov 13, 2011 3:26 am
youngwolf1105 says...



I understood none of that!!!!!!!!!!!! Dang Algebra1!
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Wed Jan 11, 2012 7:13 am
xXmusicaXx says...



I thought I'd check this out since it would be good practice for Maths (my pre-board is in a week's time), and.. i just want to know if you're in college seeminglymeaningless.
Because we have the same stuff for Math in grades 11 and 12. (The Indian system is crazy after all.) *sweatdrops*
Though, Differential equations is actually kinda fun. ^^
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