A tank contains 400 liters of brine made by dissolving 30 kg of salt in water. Saltwater containing 0.1 kg of salt per liter runs into the tank at the rate of 8 liters per minute. A well – stirred mixture runs out of the tank at the rate of 12 liters per minute. Determine the percentage of the salt in the mixture at the end of an hour.
A tank contains 400 liters of brine made by dissolving 30 kg of salt in water. Saltwater containing 0.1 kg of salt per liter runs into the tank at the rate of 8 liters per minute. A well – stirred mixture runs out of the tank at the rate of 12 liters per minute. Determine the percentage of the salt in the mixture at the end of an hour.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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The topic is about Applications of First Order Differential Equations. The problem given is about Mixture and flow. The first image attached is how I want to answer the given problem which can be seen on the second image attached. Please provide a complete solution. (of how I want to answer it) Thanks! Will give you a thumbs up
Transcribed Image Text:A tank contains 400 liters of brine made by dissolving 30 kg of salt in water. Saltwater containing 0.1 kg of salt per liter runs into
the tank at the rate of 8 liters per minute. A well – stirred mixture runs out of the tank at the rate of 12 liters per minute.
Determine the percentage of the salt in the mixture at the end of an hour.
Transcribed Image Text:Cout
3Q
OUT =
100-t
100-t
dą
dą
30
= 2.
100-t
= 2
A tank contains 100 gallons of brine made by dissolving 60 lbs
of salt in water. Salt water containing 1 lb of salt per gallon
runs in at the rate of 2 gal/min., and the mixture kept uniform P(t) :
by stirring, runs out at the rate of 3 gal/min. Find the amount
of salt in the tank at the end of 1 hour.
dt
dt
100-t
3
W (t) = 2;
%3D
%3D
100-t
Ø = (100 – t)-3
%3D
Given:
Solution:
Q(100 – t)-3 = 2 | (100 – t)-3dt
V(0) = 100 gal
Q(0) = 60 lbs
lb
= 1
gal
gal
dQ
= IN – OUT
dt
Q(100 – t)-3 = (100 – t)-2 + K
Cin
lb
Q = 100 – t + K(100 – t)³
gal
IN = RinCin = 2;
(1
gal
%3D
min
Rin
= 2
min
Q(0) = 60:
= -4.000x10-5
%3D
K =
lb
: IN = 2
тin
Rout
gal
= 3
min
Q(t) = 100 – t – 4x10-5(100 – t)3
OUT = RoutCout; Cout
Required:
Q(60)
v(t)
Q(60)
= 100 – 60 – 4x10-5(100 – 60)3
V(t) = 100 gal + 2t – 3t = 100 – t
%3D
:: 0(60) = 37.44 lbs|
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