
#1
Sep1906, 02:21 PM

P: 17

This problem has been perplexing me all week, it doesn't look hard but somehow I can't get the right answer. The question is 
A salt tank of capacity 500 gallons contains 200 gallons of water and 100 gallons of salt. Water is pumped into the tank at 3 gallon/min with salt of 1 lb/gallon. This is uniformly mixed, on the other end end water leaves the tank at 2 gallon/min. Setup a equation that predicts the amount of salt in the tank at any time up to the point when the tank overflows. What I wrote was dy/dx = rate in  rate out dy/dx = 3*1  2*Q(t)/(200+t) Where Q(t) is the amount of salt in the tank. And 200+t represents the increasing volume of water in the tank. This equation doesn't give the correct answer, I have tried tacking 100 lbs of salt so that it becomes dy/dx = 3*1  2*(Q(t)+100)/(200+t) but nothing seems to work. Can someone help me setup the correct equation? 



#2
Sep1906, 04:20 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,881

Perhaps explicitly writing "y(x) is the amount of salt in the tank after x minutes" would help you remember what you are doing! If y(x) is the amount of salt in the tank, in pounds, after x minutes, then, yes, dy/dx= rate in rate out. Since you know "Water is pumped into the tank at 3 gallon/min with salt of 1 lb/gallon", you know that salt is coming at 3 gallons/min*1lb/gal= 3 lb/gal. If the amount of salt in the tank is y(x) and the amount of water is 200+ (32)x = 200+ x then the density of salt is y/(200+x) pounds/gal and so the rate out is (y/(200+x))*2 gal/min= 2y/(200+x) lbs/min. dy/dx= 3 2y/(200+x) 



#3
Nov3008, 05:45 AM

P: 1

I think this is the right ODE setting and its solution:
Y’= inflow of salt – outflow of salt Y’= 3 – 0.01y Y’= .01 (y –300) dy/dt = –.01 (y–300) dt dy/(y–300)= –.01 dt Integrate both sides yields: ∫dy/(y–300)= ∫ –.01 dt ln Iy300I= –.01t + c Take exponential for both sides: Y–300= ℮^(0.01t + c) Y = 300 + ce^(0.01t) At Y(0)=100 , at t=0 there was 100 lb of salt (initial points) Substitute to find C: 100= 300 + Ce^(0.01(0)) So c= –200 Y = 300 – 200℮^(0.1t) When salt (y) is 250 lb (as 100 lb is in 200 gallon, so in 500 gallons (capacity of the tank) there is 250 lb of salt, when this is reached tank overflows: 250 = 300 – 200℮^(0.01t) 50 = 200℮^(0.01t) 0.2= ℮^(0.01t) Ln 0.2= .01t 1.609=.01t t = 160.94 minutes, So after 160.94 minutes (2.68 hours) the tank will overflow. 



#4
Nov3008, 09:33 AM

P: 38

Salt Tank  Differential Equation
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