• Support PF! Buy your school textbooks, materials and every day products Here!

Find the quantity of salt in the tank?

  • Thread starter Math10
  • Start date
  • #1
301
0

Homework Statement


A 200 gallon tank initially contains 100 gallons of water with 20 pounds of salt. A salt solution with 1/4 pound of salt per gallon is added to the tank at 4 gal/min, and the resulting mixture is drained out at 2 gal/min. Find the quantity of salt in the tank as it's about to overflow.

Homework Equations


None.

The Attempt at a Solution


(1/4)lbs/gal*(4)gal/min=1lbs/min
(Q)lbs/100 gal*(2)gal/min=Qlbs/50 min
dQ/dt=1-Q/50=(50-Q)/50
50/(50-Q)dQ=dt
-50ln abs(50-Q)=t+C
ln abs(50-Q)= -t/50+C
50-Q=Ce^(-t/50)
Q=50-Ce^(-t/50)
Q(0)=0
C=50
Q(t)=50(1-e^(-t/50))
Q(50)=31.606
But the answer is Q(50)=47.5
Where did I make a mistake?
 

Answers and Replies

  • #2
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,519
734

Homework Statement


A 200 gallon tank initially contains 100 gallons of water with 20 pounds of salt. A salt solution with 1/4 pound of salt per gallon is added to the tank at 4 gal/min, and the resulting mixture is drained out at 2 gal/min. Find the quantity of salt in the tank as it's about to overflow.

Homework Equations


None.

The Attempt at a Solution


(1/4)lbs/gal*(4)gal/min=1lbs/min
(Q)lbs/100 gal*(2)gal/min=Qlbs/50 min
dQ/dt=1-Q/50=(50-Q)/50
50/(50-Q)dQ=dt
-50ln abs(50-Q)=t+C
ln abs(50-Q)= -t/50+C
50-Q=Ce^(-t/50)
Q=50-Ce^(-t/50)
Q(0)=0
C=50
Q(t)=50(1-e^(-t/50))
Q(50)=31.606
But the answer is Q(50)=47.5
Where did I make a mistake?
I didn't check all your work, but I notice you used Q(0)=0. But the problem says the tank started with 20lbs of salt.
 
  • #3
301
0
So I used Q(0)=20,
C=30
Q=50-30e^(-t/50)
Now what?
 
  • #4
DrClaude
Mentor
7,163
3,306
(Q)lbs/100 gal*(2)gal/min=Qlbs/50 min
This line I don't understand. Why are you dividing by 100 gal?
 
  • #5
301
0
So what do I do? Can you guys correct me?
 
  • #6
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
32,738
5,034
This line I don't understand. Why are you dividing by 100 gal?
It's attempting to calculate the instantaneous rate at which salt is leaving the tank (in lb/min).
Math10, you've overlooked that fact that the volume is increasing too. It's only 100 gal at the start.
We shouldn't have to guess what you are doing. Explain as you go along.
 
  • #7
19,924
4,096
You need to do 2 material balances: (1) The volume of water in the tank and (2) the amount of salt in the tank.

Let V (gal) represent the volume of water in the tank at time t, and let C (lb/gal) represent the concentration of salt in the tank at time t. You need to solve for both of these. The balances should be of the form

(rate of input) - (rate of output) = (rate of accumulation).

Let qin represent the rate of water addition to the tank, and let qout represent the rate of water output from the tank. Let Cin represent the concentration of salt in the inlet stream. Assume that the tank is always well-mixed, so that the concentration of salt in the tank is always uniform.

To start with, show us your balance on the volume of water in the tank.

Chet
 

Related Threads on Find the quantity of salt in the tank?

Replies
1
Views
699
  • Last Post
Replies
0
Views
863
  • Last Post
Replies
3
Views
11K
Replies
1
Views
378
  • Last Post
Replies
1
Views
7K
  • Last Post
Replies
3
Views
23K
Replies
3
Views
2K
Replies
5
Views
1K
Top