Username or E-mail:
Forget your password?
Young Writers Society
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?
Want 3 Reviews?
Tue Sep 20, 2011 1:20 am
Duuuuude. We had fun doing it!
The dreams were fun.
Ubi caritas est vera, Deus ibi est.
"The mark of your ignorance is the depth of your belief in injustice and tragedy. What the caterpillar calls the end of the world, the Master calls the butterfly." ~ Richard Bach
Moth and Myth
<- My comic!
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
Want 3 Reviews?
Sun Nov 13, 2011 3:26 am
I understood none of that!!!!!!!!!!!! Dang Algebra1!
We were made to corageous,
We're taking back the fight.
We were made to be corageous,
And it starts with us tonight.
And the only way we'll stand,
Is on our knees with lifted hands.
Make us corageous,
Lord make us corageous. - Casting Crowns
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. ^^
"Married to music - 'nuff said."
"Freedom is everything to me."
"Have you any idea why a raven is like a writing desk?"
"I shall futterwacken vigorously"
~ Tarrant Hightop, Alice in Wonderland.
Copyright © 2015
Young Writers Society
YWS logo created by Jordan Bobo
Header images ©
About / Info
Become a Supporter
Forums & RPG
Cover Art Creator
Poetic Lines Gen
Story Theme Gen
83,223 Literary Works • 430,438 Reviews