Solution pumped into a tank problem.

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In summary, the problem involves a large mixing tank initially holding 400 gallons of water with 65 pounds of dissolved salt. Another brine solution is pumped into the tank at a rate of 6 gal/min and pumped out at a faster rate of 8 gal/min. The concentration of the solution entering is 3 lb/gal. The differential equation for the amount A(s) of salt in the tank at time t can be determined by considering the flow rates and concentration of salt in the tank.
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derail
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Homework Statement


Suppose that a large mixing tank initially holds 400 gallons of water in which 65 pounds of salt have been dissolved. Another brine solution is pumped into the tank at a rate of 6 gal/min, and when the solution is well stirred, it is pumped out at a faster rate of 8 gal/min. If the concentration of the solution entering is 3 lb/gal, determine a differential equation for the amount A(s) of salt in the tank at time t.

Homework Equations


Flow rate is Volume times cross sectional area, however, I do not know if that is relevant, it has been a while since I have done something like this.
 
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  • #2
Flow rate is Volume times cross sectional area, however, I do not know if that is relevant, it has been a while since I have done something like this.
Geometry does not matter here.

If the concentration of salt is x and 8 gallons/min flow out, can you determine the rate of salt flowing out?
If 6gallons/min with a salt concentration of 3pounds/gallon (imperial units are weird) flows in, can you determine the rate of salt flowing in?

This should help to get your differential equation.
 
  • #3
So wouldn't the 8 gallons/min be the flow rate?

I would think i can find the rate of salt flowing out by taking the 6 gallons/min and the 3 pounds/gallon and trying to get pounds/minute by multiplying them to get 18 pounds/minute.
 
  • #4
derail said:
So wouldn't the 8 gallons/min be the flow rate?
There are two flow rates, one in, one out.
I would think i can find the rate of salt flowing out by taking the 6 gallons/min and the 3 pounds/gallon and trying to get pounds/minute by multiplying them to get 18 pounds/minute.
You mean the rate of salt flowing in, right? You know the flow rate of water out, so you need the concentration of salt in that. You have the variable A(t) for the amount of salt in the tank at time t, so you need to know how much water that's dissolved in. Can you find an equation for the volume of water in the tank at time t?
 
  • #5
I am not sure how to incorporate the flows of the liquid into the equation when I am ultimately looking for an equation for the amount of salt.
 
  • #6
You know the flow rates of water in and out, so how much water is in the tank at time t?
 
  • #7
I see, we should be able to use the same logic to make is so that the units of the A(t) are lbs per minute by using the concentration of salt water. Do you have a suggestion on how to write the equation?

I figure the way we are doing this, we are basically adding two separate equations, one for the flow rate going in and one for the rate going out.
 
  • #8
Forget the salt for the moment. There's 6g/m flowing in, 8g/m flowing out. What's the rate of change of volume of water? The tank started with 400 gallons. How many after t minutes?
 
  • #9
So the rate of change is 2g/m flowing out. The tank started with 400 gallons.
 
  • #10
So after t mins we should have, 2*t+400 gallons of water.
 
  • #11
derail said:
So after t mins we should have, 2*t+400 gallons of water.
Is the volume of water increasing or decreasing?
 
  • #12
Decreasing my bad, so we actually get -2*t+400, now to write that equation.
 
  • #13
So the water flow rate =(-2*t)+400, and we know that 3 pounds/gallon is being pumped in so we should be able to write: =((-2*t)+400)*3 to correct the units for Pounds per minute?
 
  • #14
derail said:
So the water flow rate =(-2*t)+400,
No, that's not a flow rate at all. That's the volume of water in the tank at time t.
If the salt in the tank at time t is A(t), what's the concentration in the tank at time t?
 
  • #15
I am trying to find a differential equation for the amount A(s), meaning the amount of salt. I miss wrote that last equation, you are right. However, I am trying to get the amount of salt, so I figure if I know the amount of water then I can use the concentration to get the amount of salt? like I said in reply #13


haruspex said:
If the salt in the tank at time t is A(t), what's the concentration in the tank at time t?

Well the concentration of the solution entering is 3lb/g.
 
  • #16
So my thought process was, we can take the flow rate equation and incorporate the concentration of salt to water?
 
  • #17
derail said:
I am trying to find a differential equation for the amount A(s),
Yes, I know, and I'm trying to lead you to that by asking a series of very simple questions. So please, try to answer them, not a question I didn't ask.
You have calculated the volume of water in the tank at time t. The amount of salt in the tank at time t is A. (Which one writes as A=A(t), or if you want to call the variable As then As=As(t). I don't know why you keep writing A(s). There is no variable s.) So, very simple question, what is the concentration of salt in the tank at time t?
 

FAQ: Solution pumped into a tank problem.

1. What is a "solution pumped into a tank" problem?

A "solution pumped into a tank" problem refers to a scenario in which a liquid solution is being pumped into a tank or container at a certain rate. This type of problem is commonly used in fluid mechanics and thermodynamics to calculate factors such as flow rate, pressure, and volume.

2. How is the rate of pumping calculated for a solution pumped into a tank problem?

The rate of pumping for a solution pumped into a tank problem can be calculated by dividing the volume of solution being pumped by the time it takes to pump that volume. This is known as the flow rate and is measured in units of volume per time, such as liters per second or gallons per minute.

3. What factors can affect the solution pumped into a tank problem?

There are several factors that can affect a solution pumped into a tank problem, including the pump's flow rate, the properties of the solution being pumped (such as density and viscosity), and any changes in the tank's volume or shape. Additionally, external factors such as temperature and pressure can also impact the problem.

4. How is the pressure of the solution calculated in a tank problem?

The pressure of the solution in a tank problem can be calculated using the ideal gas law, which states that pressure is equal to the product of the gas's density, temperature, and the gas constant. In this case, the gas constant is replaced with the specific gas constant for the solution being pumped.

5. What are some real-world applications of solution pumped into a tank problems?

Solution pumped into a tank problems have various real-world applications, such as in chemical and pharmaceutical industries, where precise control of solution flow and pressure is crucial. They are also used in hydraulic engineering to design systems for water supply and distribution. Additionally, these types of problems are commonly encountered in daily life, such as when filling a car's gas tank or a water tank for a household.

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