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

kuhse4461a 2021-12-30 Answered
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?
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Pansdorfp6
Answered 2021-12-31 Author has 27 answers
Solution:
The salt enters the tenk at rate
R=(4 galmin)×(3 lbgal)
R=12 lbs/min
Since the tenk loses liquid at a rate of 2 gal/min
=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
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Heather Fulton
Answered 2022-01-01 Author has 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
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karton
Answered 2022-01-09 Author has 439 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

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