Mixing problems You are an industrial engineer, and you are tasked to produce a solution that contains about 1/4 grams of salt per liter. You are working with the following: • A tank with a mixture of 10 grams of salt dissolved into 200 liters of water • Pouring in a solution that contains 1/2 gram per liter at a rate of 4 liters per minute • The tank is being drained at 3 liters per minute. We can use the mixing model to define the rate of change of salt in the tank to the quantity of salt within the tank. The key is assuming that the salt mixture pouring into the tank is automatically mixed. The mathematical model is O' = rate in of salt - rate out of salt The "rate in for salt" is already given to you as 1/2 grams per liter times 4 liters per minute. But the "rate out for salt" is the concentration of salt times the amount being drained. rate out of salt = in Where Q is the quantity of salt in grams dissolved in the tank at time t, Q'is the rate of change to the quantity of salt, and W(t) is the volume of solution in the take. time t. Note the volume of fluid or solution changes as more fluid is added then drained. At time 0 there are 200 liters of fluid in the tank. You can refer to the text in section 4.2 for examples or in the modeling examples of lesson 1. Post the differential equation model 2. Solve the differential cquation 3. Determine how long it will take to reach a concentration of 1/4 graphs per liter.
Mixing problems You are an industrial engineer, and you are tasked to produce a solution that contains about 1/4 grams of salt per liter. You are working with the following: • A tank with a mixture of 10 grams of salt dissolved into 200 liters of water • Pouring in a solution that contains 1/2 gram per liter at a rate of 4 liters per minute • The tank is being drained at 3 liters per minute. We can use the mixing model to define the rate of change of salt in the tank to the quantity of salt within the tank. The key is assuming that the salt mixture pouring into the tank is automatically mixed. The mathematical model is O' = rate in of salt - rate out of salt The "rate in for salt" is already given to you as 1/2 grams per liter times 4 liters per minute. But the "rate out for salt" is the concentration of salt times the amount being drained. rate out of salt = in Where Q is the quantity of salt in grams dissolved in the tank at time t, Q'is the rate of change to the quantity of salt, and W(t) is the volume of solution in the take. time t. Note the volume of fluid or solution changes as more fluid is added then drained. At time 0 there are 200 liters of fluid in the tank. You can refer to the text in section 4.2 for examples or in the modeling examples of lesson 1. Post the differential equation model 2. Solve the differential cquation 3. Determine how long it will take to reach a concentration of 1/4 graphs per liter.
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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Transcribed Image Text:Mixing problems
You are an industrial engineer, and you are tasked to produce a solution that contains about 1/4 grams of salt per liter. You are working with the following:
• A tank with a mixture of 10 grams of salt dissolved into 200 liters of water
• Pouring in a solution that contains 1/2 gram per liter at a rate of 4 liters per minute
• The tank is being drained at 3 liters per minute.
We can use the mixing model to define the rate of change of salt in the tank to the quantity of salt within the tank. The key is assuming that the salt mixture pouring into the tank is automatically mixed. The mathematical model is
O' = rate in of salt – rate out of salt
The "rate in for salt"
already given to you as 1/2 grams per liter times 4 liters per minute. But the "rate out for salt" is the concentration of salt times the amount being drained.
rate out of salt=
W(t)
Where Q is the quantity of salt in grams dissolved in the tank at time t, Q'is the rate of change to the quantity of salt, and W(t) is the volume of solution in the take at time t. Note that the volume of fluid or solution changes as more fluid is added
then drained. At time 0 there are 200 liters of fluid in the tank. You can refer to the text in section 4.2 for examples or in the modeling examples of lesson
1. Post the differential equation model
2. Solve the differential equation
3. Determine how long it will take to reach a concentration of 1/4 graphs per liter.
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