show answers

new problem

Directions

Details

About

In this simulation, you can perform mass balances on an evaporative crystallization process. Perform mass balances to determine the unknown values, then enter those values into the blue input boxes. Press the green "show answers" button above the canvas to check your work. Press "new problem" to reset the simulation with a new set of conditions.

$$ [1] \quad m_{feed} = m_{2} + m_{4} + m_{5} $$

$$ [2] \quad z_{w} \, m_{feed} = m_{2} + x_{w,5} \, m_{5} $$

$$ [3] \quad z_{k} \, m_{feed} = m_{4} + x_{k,5} \, m_{5} $$

\( m_{feed} = \) mass flow rate of fresh feed (kg/h)

\( m_{2} = \) mass flow rate of water exiting evaporator (kg/h)

\( m_{4} = \) mass flow rate of solid KCl crystals exiting the crystallizer and filter (kg/h)

\( m_{5} = \) mass flow rate of slurry exiting the crystalizer and filter (kg/h)

\( z_{k} = \) mass fraction of KCl in fresh feed

\( z_{w} = \) mass fraction of water in fresh feed

$$ [4] \quad m_{3} = m_{4} + m_{5} + m_{R} $$

$$ [5] \quad x_{k,3} \, m_{3} = m_{4} + x_{k,5} \, m_{5} + x_{k,R} \, m_{R} $$

\( m_{3} = \) mass flow rate of solution exiting evaporator (kg/h)

\( m_{R} = \) mass flow rate of recycle (kg/h)

\( x_{k,3} = \) mass fraction of KCl in stream exiting evaporator

\( x_{k,5} = \) mass fraction of KCl in solution exiting crystallizer and filter

\( x_{w,5} = \) mass fraction of water in solution exiting crystallizer and filter (this is equivalent to \( x_{k,R} \) as the exit stream from the crystallizer is split).

\( x_{k,R} = \) mass flow rate of solution exiting evaporator (kg/h)

$$ [6] \quad m_{feed} + m_{R} = m_{1} $$

$$ [7] \quad z_{k} \, m_{feed} + x_{k,R} \, m_{R} = x_{k,1} \, m_{1} $$

\( m_{1} = \) mass flow rate entering evaporator (kg/h)

\( x_{k,1} = \) mass fraction of KCl in stream entering evaporator

$$ [8] \quad m_{1} = m_{2} + m_{3} $$

$$ [9] \quad s_{3} = 0.2881 \, T_{evap} + 28.123 $$

$$ [10] \quad x_{k,3} = \frac{s_{3}}{s_{3} + 100 \; \mathrm{kg}} $$

$$ [11] \quad s_{5} = 0.2881 \, T_{crys} + 28.123 $$

$$ [12] \quad x_{k,3} = \frac{s_{5}}{s_{5} + 100 \; \mathrm{kg}} $$

\( s_{3} = \) solubility of KCl in 100 kg H_{2}O

\( T_{evap} = \) evaporator operating temperature (°C)

\( s_{5} = \) solubility of KCl in 100 kg H_{2}O

\( T_{crys} = \) crystallizer and filter operating temperature (°C)

Note that \( z_{k} + z_{w} = 1 \) and \( x_{k,i} + x_{w,i} = 1 \) where \( i \) is the stream. Recovery rate \( R = m_{4} / ( z_{k} \, m_{feed} ) \).

View this screencast for more practice with material balances on a crystallizer.

This simulation was created in the Department of Chemical and Biological Engineering, at University of Colorado Boulder for LearnChemE.com by Neil Hendren under the direction of Professor John L. Falconer. Address any questions or comments to learncheme@gmail.com. All of our simulations are open source, and are available on our LearnChemE Github repository.

Is your screen too small to fit this application? Try zooming-out on the web page (CTRL+"-" and CTRL+"=" on Windows, or ⌘+"-" and ⌘+"=" on Mac), then refreshing the page. This application is not compatible with Internet Explorer or web browsers that do not support WebGL and HTML5. Recommended browsers are the latest versions of: Chrome, Safari, Firefox, Edge, and Opera.