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Algebra & Calculus
Sun Sep 18, 2011 8:21 am
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?
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Tue Sep 20, 2011 1:20 am
Duuuuude. We had fun doing it!
The dreams were fun.
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Tue Sep 20, 2011 1:19 pm
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
I understood none of that!!!!!!!!!!!! Dang Algebra1!
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And the only way we'll stand,
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Wed Jan 11, 2012 7:13 am
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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