Brine containing 3lb/gal of salt enters a tank at the

kuhse4461a

kuhse4461a

Answered question

2021-12-30

Brine containing 3lb/gal of salt enters a tank at the rate of 4gal/min and the solution well-stirred leaves at 2gal/min. The tank initially contains 20 gal of water. What is the amount of salt in the tank after 5 minutes?

Answer & Explanation

Pansdorfp6

Pansdorfp6

Beginner2021-12-31Added 27 answers

Solution: 
At a rate, salt enters the tube.
R=(4 galmin)×(3 lbgal) 
R=12 lbs/min 
Due to the tenk's 2 gal/min loss of liquid
=4 galmin2galmin 
=2 galmin 
after to time: (minutes). The number of gallons of Brine in the tenk 
=20+2t 
So after 5 min 
=20+2×5 
=30 gallons 
and 2 gallons =.3454lbs 
So amount of salt =30×8.3454 
=250.362

Heather Fulton

Heather Fulton

Beginner2022-01-01Added 31 answers

Given that the initial amount of salt is zero( because of pure water)
Rate in 4gal/min
Rate out 2gal/min initial volume is 20 gal
Brine containing 3lb/gal of salt
The inteving rate 4gal/min
So intering concentration
=3×4
=12
Volume =20gal
Let the Amount of Brine is y(+)
Then rate change
y1=ratelnrateout
=122×y20+(42)t
y1=12y10+t,y(0)=0
Salving y+110+t y=12
IF=e1lot tdt=eln(10+t)=10+t
y×(10+t)=12×(10+t)dt+c=6(10+t)2+c
y(+)=6(10+t)+c10+t
By y(0)=00=60+c10c=600
y(t)=6(10+t)60010+t
The amount after 5 min
y(5)=6×(10+5)60010+5=9060015
y(5)=9040=50lb
karton

karton

Expert2022-01-09Added 613 answers

R=(4 gal/min)×(3 lb/gal)R=12lbs/min=4 gal/min2gal/min=2 gal/min=20+2t=20+2×5=30 gallonsand 2 gallons =.3454lbs=30×8.3454=250.362

xleb123

xleb123

Skilled2023-05-13Added 181 answers

Given information:
- Brine enters the tank at a rate of 4 gallons per minute, with a concentration of 3 pounds of salt per gallon.
- The solution is well-stirred, meaning the concentration remains uniform throughout the tank.
- The solution leaves the tank at a rate of 2 gallons per minute.
- The tank initially contains 20 gallons of water.
To determine the amount of salt in the tank after 5 minutes, we need to find S(5).
Let's break down the problem into two parts: the amount of salt entering the tank and the amount of salt leaving the tank.
1. Amount of salt entering the tank:
The rate of brine entering the tank is 4 gallons per minute, with a concentration of 3 pounds of salt per gallon. Therefore, the amount of salt entering the tank per minute is 4×3=12 pounds. Thus, the total amount of salt entering the tank in the first t minutes is 12t pounds.
2. Amount of salt leaving the tank:
The rate of solution leaving the tank is 2 gallons per minute. Since the solution is well-stirred, the concentration of salt in the tank is uniform. Therefore, the concentration of salt leaving the tank is also 3 pounds per gallon. Thus, the amount of salt leaving the tank per minute is 2×3=6 pounds. Therefore, the total amount of salt leaving the tank in the first t minutes is 6t pounds.
Now, let's calculate the amount of salt in the tank after 5 minutes.
At time t, the amount of salt in the tank is given by:
S(t)={initial amount of salt}+{amount of salt entering}{amount of salt leaving}
The initial amount of salt in the tank is 0 pounds since it contains only water initially.
Using the formulas we derived earlier, we have:
S(t)=0+12t6t
Simplifying the expression:
S(t)=6t
Finally, we can calculate S(5) to find the amount of salt in the tank after 5 minutes:
S(5)=6×5=30 pounds.
Therefore, the amount of salt in the tank after 5 minutes is 30 pounds.

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