How to Solve a Material Balance Problem for Two Liquids A and B?

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Homework Help Overview

The discussion revolves around a material balance problem involving two liquids, A and B, with specific solid content percentages. The original poster presents a scenario where these liquids are mixed, and additional solids are added, leading to a final mixture with a defined solid percentage. The goal is to determine the original amounts of liquids A and B.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the number of equations needed to solve for the unknowns A and B, with some attempting to formulate equations based on mass and solid balances. Others express confusion regarding the setup of the problem and the correct placement of terms in their equations.

Discussion Status

There is ongoing exploration of different formulations for the equations needed to solve the problem. Some participants have identified potential errors in their initial equations and are revisiting their calculations. Guidance has been offered regarding the placement of terms in the equations, but no consensus has been reached on the correct approach.

Contextual Notes

Participants are grappling with whether the process is a batch or rate process, which affects how they formulate their equations. There is also mention of specific percentages and total masses that need to be accounted for in the equations.

nahian taskin
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Two liquids, A and B, are mixed together. A is 5.00% solids, the rest water. B is 11.0% solids, the rest water. To the mixture is added 18.5 kg of bone-dry solids resulting in a mixture of 2 150 kg containing 8.25% solids. What were the original amounts of the liquids A and B?
 
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How many equations in how many unknowns?
 
2 unknown. A and B.
 
one example related to
Q.Anna's Purification Co. produces drinking water from saltwater by partially freezing the saltwater to create salt-free ice and a brine solution. Tf saltwater is 3.50% salt by mass and the brine solution is fow1d to be an 8.00% concentrate by mass, determine how many kilograms of saltwater must be processed to form 2.00 kg of ice.
Ans: Eqauation: input= output
Overall balance : S = B + 2.00
salt balance : 0.035S = 0.08B
water balance : 0.965S = 0.920B + 2.00
then solve for S and B.
 
Bystander said:
How many equations in how many unknowns?
one example related to
Q.Anna's Purification Co. produces drinking water from saltwater by partially freezing the saltwater to create salt-free ice and a brine solution. Tf saltwater is 3.50% salt by mass and the brine solution is fow1d to be an 8.00% concentrate by mass, determine how many kilograms of saltwater must be processed to form 2.00 kg of ice.
Ans: Eqauation: input= output
Overall balance : S = B + 2.00
salt balance : 0.035S = 0.08B
water balance : 0.965S = 0.920B + 2.00
then solve for S and B.
 
In terms of A and B, what is the total mass of the final mixture?

In terms of A and B, what is the mass of solids in the final mixture?

Chet
 
Chestermiller said:
In terms of A and B, what is the total mass of the final mixture?

In terms of A and B, what is the mass of solids in the final mixture?

Chet
initial amount of A n B in the total mixture. How much kg of A and B was added in the mixture
 
Chestermiller said:
In terms of A and B, what is the total mass of the final mixture?

In terms of A and B, what is the mass of solids in the final mixture?

Chet
Answers for this question is, A = 1260 kg, B = 872 kg
 
nahian taskin said:
initial amount of A n B in the total mixture. How much kg of A and B was added in the mixture
I have no idea what you are saying here. Can you please answer the questions I asked (in the form of equations).

Chet
 
  • #10
Chestermiller said:
I have no idea what you are saying here. Can you please answer the questions I asked (in the form of equations).

Chet
No equations are given. From this, We have to make our own equation to get the solution.
I have tried with this equation:
For solid : 0.05A+ 0.11B = 18.5+ 2150(0.08125)
For Water: 0.95A+ 0.89B = 2150(0.9175)
solve the equations.
but i wrong answer,which means something is wrong in my equation
 
  • #11
nahian taskin said:
No equations are given. From this, We have to make our own equation to get the solution.
I have tried with this equation:
For solid : 0.05A+ 0.11B = 18.5+ 2150(0.08125)
For Water: 0.95A+ 0.89B = 2150(0.9175)
solve the equations.
but i wrong answer,which means something is wrong in my equation
how many equations it can be here, that's the confusion because i don't know wether its a batch process or rate process.
 
  • #12
nahian taskin said:
how many equations it can be here, that's the confusion because i don't know wether its a batch process or rate process.
Batch process : total input = total output.
Rate process: rate of input- rate of output= rate of accumulation.
if the rate process is uniform then ,
rate of input= rate of output.
 
  • #13
nahian taskin said:
No equations are given. From this, We have to make our own equation to get the solution.
I have tried with this equation:
For solid : 0.05A+ 0.11B = 18.5+ 2150(0.08125)
For Water: 0.95A+ 0.89B = 2150(0.9175)
solve the equations.
but i wrong answer,which means something is wrong in my equation
This is what I was looking for. Your formulation is very close to being correct. There are just a couple of minor errors.

1. In the solid equation, are you sure that 18.5 belongs on the right hand side of the equation?
2. In the water equation, recheck that 0.9175. There's an arithmetic error.

Chet
 
  • #14
Chestermiller said:
This is what I was looking for. Your formulation is very close to being correct. There are just a couple of minor errors.

1. In the solid equation, are you sure that 18.5 belongs on the right hand side of the equation?
2. In the water equation, recheck that 0.9175. There's an arithmetic error.

Chet
if 8.25% is solid then (100-8.25)% is water which is 91.75%
and i tried by giving 18.5 on the left hand side, but answer didnt came. It was more far than the actual answer.
 
  • #15
nahian taskin said:
if 8.25% is solid then (100-8.25)% is water which is 91.75%
and i tried by giving 18.5 on the left hand side, but answer didnt came. It was more far than the actual answer.
in the first equation i made a mistake it is 2150(0.0825) not 0.08125
 
  • #16
nahian taskin said:
in the first equation i made a mistake it is 2150(0.0825) not 0.08125

I know. I just noticed that. Do you get the right answer now with the 18.5 on the left?

Chet
 
  • #17
Chestermiller said:
I know. I just noticed that. Do you get the right answer now with the 18.5 on the left?

Chet
No, i didnt. I am now trying with by corrcting that value in the equation
 
  • #18
nahian taskin said:
No, i didnt. I am now trying with by corrcting that value in the equation
Didnt came again :( with this value .
somewhere is wrong, i don't know where!
 
  • #19
Here are the equations I solved, and I got the correct answer:

For solid : 0.05A+ 0.11B + 18.5 = 2150(0.0825)
For Water: 0.95A+ 0.89B = 2150(0.9175)

If you didn't get the correct answer using these equations, then you need to check your algebra.

Chet
 
  • #20
i was checking by putting the actual value in the equation then i get for the 1st equation , 18.5 would on left side .
and my second equation is wrong.
Chestermiller said:
Here are the equations I solved, and I got the correct answer:

For solid : 0.05A+ 0.11B + 18.5 = 2150(0.0825)
For Water: 0.95A+ 0.89B = 2150(0.9175)

If you didn't get the correct answer using these equations, then you need to check your algebra.

Chet
Ohk, thank you :)
 

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