C Secure https://we work2. su.edu/webwo 2/Lopez MAI 2/5 Fal 201 //Section 1.1 In
ID: 1770457 • Letter: C
Question
C Secure https://we work2. su.edu/webwo 2/Lopez MAI 2/5 Fal 201 //Section 1.1 Introduction First Order Systems/5/ ey=pk56GdW/Macke/cmqOZ51x 52Ph5roHA user-emleung8 effectiv ::: Apps a Amazon.com Onlin G ASU Fana | APT DOtherWenwlark E RashmarkMercan N Nrtfi Tutoring a Grades Problems 1 point Problem 1 Problem 2 v Problem1 roblem Problam 8 Problem 8 Consider wo interconnected tanks as shosn in the tigure shove lank 1 intial contains 10 L (ltersi ot water and 4 g ot salt, whle tank 2 initialy contains 40 L at weter and 1/u g ot salt weter containing 25 gL of salt is poured into tenk1 ata rate of 3.5 Lmin while the mixture fowing nto tank 2 contains a salt concentration of 30 gL of salt nd is flowing at the You may assume that the sa utions in eath tank are hori hly m xed so that th concentration at the mixture ea ing any tan any of h utes has lhe same cannen rat n t salt as the tank as a wnole. (This is not completely realistc, but as in real physics, we are going to work with the approximete, rather than exact description. The resl equatons ot physics are often too complicated to even write down precisely, much less solve.) How does the water in each tank change over tme? 1 and respectively wrtedi rental equatons 1 r l se lhe Symbols and ratner than it and I ett ana git)) t helhe amount ot salt ing at tme t in tank and q s i Give the initial valus p(0) 465 170 3:39 PM O Type here to searchExplanation / Answer
Inital Volume of Tank 1 = 10L
Initial Volume of Tank 2 = 40L
First, you have to note the inflow and outflow of water for each tank:
Tank 1: In= 6 L/min Out= 6 L/min
Tank 2: In= 7.5 L/minOut= 7.5 L/min ---->(2.5L/min + 5L/min)
Now let p(t) = amount of salt, in grams, in Tank 1 as function of time per min
andq(t) = amount of salt, in grams, in Tank 2 as function of time per min
where at time = 0 --> p(0) = 465 grams of salt
q(0) = 170 grams of salt
Thus the derivative is:
p'(t) = the rate of change with respect to time of the amount of salt in Tank 1
q'(t) = the rate of change with respect to time of the amount of salt in Tank 2
p'(t) = (Incoming flow in Tank 1)(concentration fo salt in flow)
+ (Incoming flow from Tank 2)(Concentration of salt in Tank 2)
- (Outgoing flow from Tank 1)(concentration in Tank 1)
p'(t) = (Incoming flow)(concentration of salt in flow)
+ (Incoming flow from Tank2)[p(t)/(Volume of water in Tank 2)]
- (Outgoing flow from Tank 1)[q(t)/(Volume of water in Tank 1)]
p'(t) = (25g/L)(3.5L/min) + (2.5L/min)[p(t)/40L] - (6/min)[q(t)/10L]
p'(t) = 87.5 + (2.5/40)p(t) - (6/10)q(t) g/min
p'(t) = 87.5 + (1/16)p(t) - (3/5)q(t) g/min
q'(t) = (Incoming flow)(Concentration of salt in flow)
+ (incoming flow from Tank 1)[p(t)/(Volume of water in Tank 1)]
- (Outgoing flow from Tank2)[q(t)/(Volume of water in Tank 2)]
q'(t) = (30L)(1.5L/min) + (6L/min)[p(t)/10L] - (7.5L/min)[q(t)/40L]
q'(t) = 45 + (3/5)p(t) - (3/16)q(t) g/min
Therefore p' and q' is:
p' = 87.5 + (1/16)p - (3/5)qg/min for Tank 1
q' = 45 + (3/5)p - (3/16)qg/min for Tank 2
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.