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Material Balance On Timber Dryer

material balance process flow sheet

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#1 naverrak

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Posted 20 August 2014 - 07:15 AM

My process flow sheet is with the question.
How do I go from here???.

Total balance= F=P+B

 

the answer is 0.125kg/kg(wt% of timber)

Attached Files



#2 Art Montemayor

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Posted 20 August 2014 - 08:02 AM

Ryu:

 

Your questions on Basic Stoichiometry have been so simple and basic, that I was concerned with your problems in understanding what you should be doing.  That is why I sent you the following message on one of your previous Posts:

 

"Ryu:
For my personal knowledge, I would like to know where you are studying (chemical engineering?) and what courses you are now taking. Are you in university or still in high school? How old are you and have you taken any previous chemical engineering courses before? Perhaps by knowing this information I can understand why you are having trouble in solving some very simple chemical engineering problems. Knowing this, we can help you better and more efficiently.

I hope to hear from you and trying to help."

 

I presume that you haven't read the message and that is why you haven't responded to my questions.  We want to help students on our Forum,  But we need your help to better understand why you are having such difficulty with basic problems that are pre-requisites to your future Chemical Engineering courses.  Can you help us help you?



#3 naverrak

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Posted 20 August 2014 - 08:15 AM

Hi, I am studying at the university of new south wales, I'm in my first year majoring chemical engineering, and this is the first chemical engineering course (MATS1101). I am following all the lectures/tutorials but none of the questions my lecturer went through cover this sort of question. In terms of material balance, I have only learnt ones that is given a flow (e.g. 50 lb/min) as the feed.

 

Also, the other subjects I do are: MATH1241, CHEM1021 and CEIC1001


Edited by naverrak, 20 August 2014 - 08:19 AM.


#4 Art Montemayor

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Posted 20 August 2014 - 08:46 AM

Ryu:

 

This problem is easier to comprehend than a steady state problem.  In my opinion, your instructor is being very kind in assigning this easy problem.

 

Follow the excellent advice and recommendation that Breizh gave you in your prior post:  Establish a Feed basis of 100 kg.

 

The questions you posted are:

  1. Draw process flowsheet and put the known stream data
  2. Determine the kg of water removed per kg of timber that enters the process.

Why do you confuse matters and state: “the answer is 0.125 kg/kg(wt% of timber)”.  What is this the answer to?  That is NOT what was asked of you.  Please be specific and identify what you mean by “answer”.

 

To obtain the answers requested, all you need to do is make a material balance around the dryer on the basis of 100 kg of wet timber feed.  You directly solve for the water removed and you can then answer both questions.  Furnish your calculations in detail (no text messaging or sloppy writing) and I am sure our members will gladly check out your solution and give you assistance.

 

This basic stoichiometry course is very, very important as a preparation for the real tough and complex studies that wait for you in future Chemical Engineering courses.  Therefore, you must thoroughly understand and dominate this basic stuff or you will have a very difficult time ahead in your university career.  Nobody – especially you – would want that to happen.  Apply our Forum comments and advice and make sure you communicate with your instructor about what it is that is giving you problems in his lectures.  It is YOUR career that is in jeopardy here, and you should be very pro-active in making it a success.

 

Good Luck.



#5 breizh

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Posted 20 August 2014 - 09:16 AM

Naverrak ,

 

Good start :   in =out ; F=P+B   Global mass balance ,  

Establish the same for Timber and water  , particular mass balances  and resolve the equations .

 

Or if you want to play directly with number , follow Art's advice :

in :

F = 100 Kg = mass of water + mass of timber >>>> mass of water = .. ; mass of timber =...

out :

mass of timber in = mass of timber out =....

 

 

then .....

 

waiting for your work.

 

Hope this helps .

 

Breizh



#6 naverrak

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Posted 20 August 2014 - 10:07 AM

assuming a basis of 100kg
Timber balance:
0.791*100=0.914*P+0*B
79.1=0.914P
P=86.92                            (1)

 

Water balance:
0.201*100=0.086*P+1.0( B )
20.1-0.086*P= B
B= 20.1 - 7.475
B=12.62

so is the answer to b ).
12.62 water removed per 86.92 tumber??

or 12.62 kg(water) / 86.92 kg (wet timber) = 0.145kg(water)/kg(wet timber).


Edited by naverrak, 20 August 2014 - 10:09 AM.


#7 Art Montemayor

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Posted 20 August 2014 - 12:58 PM

Naverrak:

 

Your answer is wrong.  But don't fret, so is everybody else's.  You did the proper logical steps in your algorithm (although you made a division mistake in your calculation).

 

Your problem - as stated and given - is either wrong (because of a typo error or lack of proper checking by your instructor) or is incomplete (because there might be another, third, component in the Feed.  Whichever is the cause, the problem cannot be resolved accurately because it is stated inaccurately.  The logic of the algorithm, however, is the important key to master because it will apply to all other similar problems.  Note the way I worked it in the attached workbook.  Always work your engineering problems involving calculations or diagrams in an Excel spread sheet.  The spread sheet will not make math errors and it will force you to organize your thoughts and algorithm in a logical, engineering manner.

 

This simple problem should teach you (and all other students reading this) that no engineering calculation is "sacred" and beyond challenge.   In the real world, ALL engineering work - especially calculations - must be peer checked for correctness and accuracy.  And this applies to problems given in any university as well!  Always check your data input in problems given you to make sure that it is correct, logical, and can be applied.  For example, all percentage compositions HAVE TO ADD UP TO 100%.

 

If I were you, I would return the problem to whoever gave it to you and complain about sloppy input data.  I hope this helps you out and you have had some key learnings.

 

Attached File  Naverrak Stoichiometry Problems.xlsx   40.57KB   30 downloads

 



#8 breizh

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Posted 20 August 2014 - 07:02 PM

Hi ,

my reading , correct me if I'm wrong .

 

Inlet drier  : wet timber ( water content : 20.1 % ) >>>>timber :100-20.1 = 79.9%

Outlet drier : Dry timber ( water content 8.6%) >>>>timber : 100-8.6 = 91.4%

 

F=P+B  , assume 100 Kg : >>>> 20.1 Kg water + 79.9 kg of timber

 

quantity of timber in = quantity of timber out : 79.9kg then based on water content at the outlet :

8.6/100 = water outlet drier /( timber +water) outlet drier  >>>> water outlet = 8.6*79.9/( 91.4)=7.5179 kg

 

>>>> P = 20.1-7.5179= 12.5824 kg

 

F= 100 kg

P=12.5824 kg

B= 79.9+7.5179=87.4179 kg

water in= 20.1kg

water out = 20.1 kg =12.5824 +7.5179

Timber in =Timber out = 79.9 kg

 

Answer : 12.5824/100 = 0.125824 kg/kg

 

Breizh



#9 naverrak

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Posted 20 August 2014 - 08:46 PM

Oh ok, I get it now, Thank you Art montemayor breizh. Also is the units for the answer kg(water)/kg(wet timber)?



#10 breizh

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Posted 20 August 2014 - 09:02 PM

Naverrak,

Refer to the calculation and you will find the answer : (P/W ) = Kg water /Kg wet Timber !

 

Breizh






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