Excel calculator for LMTD

LMTD, Log mean temperature difference is the temperature driving force for heat transfer in flow systems. When you are calculating heat transfer by overall heat transfer coefficient (U), you should not simply multiply temperature difference of hot stream or cold stream. Multiplying this temperature difference doesn't account the overall driving force.

Pressure drop in pipe lines and fittings. Part-2

Pressure drop due to valves and fittings

In the earlier post, tutorial goes around pressure drop in pipe lines. Pressure drop arises due to skin friction and form friction. Non separated boundary layers cause skin friction whereas separated boundary layers cause form friction. In plain pipe pressure drop is due to skin friction. But in a pipe line with valves and fittings pressure drop is  mainly due to form friction. the pipe lines may be connected to equipment and again pressure drop will be there in that equipment. This tutorial is restricted to pressure drop in valves and fittings only.

Pressure drop in pipe lines and fitting. Part-1

Pressure plays a prominent role in driving and arresting the fluid flow from one place to another. What exactly is this pressure? One classical definition is

Pressure is force acted per unit area

Excel Tutorial - 1 : Mass balance calculation using Microsoft Excel

We know that out there in this free information world, lot of soft-wares are available which ease the complexity of work. What if we can design our own mini super ninja software ( actually Excel files) to reduce our repetitive actions while calculating . Lets get started how to do simple calculations in excel.

In this tutorial neutralization of Sulphuric acid ( H2SO4 )   by Sodium hydroxide ( NaOH )was considered. Go through the problem statement given in the below image.

At first mention the Chemical reaction in one of the cells (A1 or B5 etc.,). Mention the molecular weights separately in a cell. This is important to put the value in a single cell ( L6 for H2O, L7 for NaOH, L8 for H2SO4, L9 for Na2SO4) it should not contain any alphabets. If you enter alphabets in the molecular weights column it will result in error.
Now we have given the H2SO4 quantity that should be neutralized. Put this value in a separate cell (D11) and convert it in to moles by dividing the mass with molecular weight mentioned in L8 cell.

Now use this moles and multiply this moles with 2 , as 1 mole of H2SO4 requires 2 moles of NaOH for Neutralization which would result in formation of 1 mole of Na2SO4 and 2 moles of H2O.

Multiply the moles by molecular weight in cell E17 to get the mass of NaOH required for neutralization.

Similarly do it for H2O & Na2SO4

Mass used for reaction should be equal to mass produced in the reaction.

Please feel free to put up your doubts and queries below.

To download the Excel file click the link

Mass Balance

First law of thermodynamics and sign convention of heat, work.

First law of thermodynamics

First law of thermodynamics states that

"Energy can be neither created nor destroyed but one form of energy can be converted to another form."

For example consider a ball is placed on the top of a table initially. It will have certain potential energy ( Energy possessed by virtue of its height ) as it is at a height from the ground. When it is allowed to fall from the table this potential energy will be converted into kinetic energy ( Energy possessed by virtue of its motion ). This kinetic energy will be converted to heat , sound etc., when it touches the ground.
In application of the first law to a given process, the sphere of influence of the process is divided into two parts namely system and surroundings. The region in which process occurs is the System and everything which the system interacts is the surroundings. First law of thermodynamics applies to both system and surroundings. In general,

Δ Energy of system  +  Δ Energy of surrounding  = 0

 For the above example if you consider ball as a system  initial energy is potential and final energy is kinetic,but the energy is gained by surroundings as heat and sound.

Systems are of two types.

• Open = System which exchange both mass and energy with surroundings.
• Closed = System which exchange only energy with surroundings.

For simplification here we are considering closed systems only. In general system contains some internal energy ( in the form of attractions and vibrations ) and this tend to change when the heat is added or removed, when work is done on the system or delivered by the system.For closed systems energy transfer between system and surroundings takes place in the form of work and heat. ( where as in open systems internal energy will be associated in transit also i.e., at entry and exit of the system ). For closed systems energy changes mostly occur in internal energy. So,

Δ Energy of system = Change in internal energy  = ± Q ± W 

Only change in internal energies can be found as it is hard to know the energy associated with  attractions and vibrations. Q is heat and W is work.

Sign Convention for heat and work.

Q and W always refer to system. 

• Heat given by the system, Heat produced by the system = -Q
• Heat given to the system, Heat supplied to the system = +Q

• Work done by the system, work produced by the system = -W
• Work done on the system, work given to the system = +W

Example :

Δ Internal energy  =  Q - W 

Heat is given to the system and work is done by the system.

How to solve excess air problems in Chemical Engineering?

In learning Chemical process calculations you might have encountered with the excess air or excess oxygen problems. This is more common in combustion type of problems, where oxygen is used as a aid to combustion. One can easily solve this kind of problems by simply noting down the four points. they are,

1. How many compounds are there which can be combustible ?
2. How much of oxygen is already there ?
3. How much actual oxygen is to be supplied ?
4. What is the percentage excess given ?

Consider a example where methane ( CH₄ ) undergoes combustion with air, according to the reaction,

CH₄     +         2O₂      =         CO₂     +          2H₂O

CH₄     +          Air (O₂ + N₂)                        =         CO₂  +          2H₂O  +          N₂

One mole of methane burns in the presence of two moles of oxygen giving one mole carbon dioxide and two moles water. If the above combustion occurs at 100% conversion rate the output stream contains only Carbon dioxide and water, if the conversion rate is less than 100%  then the output contains unreacted methane ( CH₄ ), unreacted oxygen, carbon dioxide and water. ( If we are considering air as the source for oxygen then output contains nitrogen also, because air contains 21% Oxygen and 79% Nitrogen and this 21% oxygen should be sufficient to achieve the given case ).

Let us consider the case where 2 Lb mole methane is burnt with 20% excess air in order to ensure complete combustion of methane. then what is the output gas composition ?

 Now apply the four steps.

1. Methane is the only compound which undergoes combustion.
2. Along with methane there is no oxygen , so oxygen already present is zero.
3. For 2 Lb mole methane 4 Lb mole oxygen is to be supplies as per stoichiometry.
4. But 20% excess air is supplied so indirectly it means that 20% excess oxygen is considered, so total oxygen moles = 4 Lb moles + 4*20/100  Lb moles = 4 *1.2 Lb moles.

Output gas composition :

Methane ( CH₄ ) is completely burnt so for 2 Lb moles methane 2 Lb moles carbon dioxide will form along with 4 Lb moles water as per stoichiometry. But the unreacted oxygen = 4*1.2 - 4 Lb moles = 0.8 Lb moles. Nitrogen in output is calculated by

Total air in = total oxygen/0.21 = 4*1.2/.21 = 22.857 Lb moles

Total nitrogen in inlet air  =  total air * 0.79 =22.857*.79 = 18.057 Lb Moles.

( If there is oxygen along with methane then net oxygen required for combustion = oxygen required for combustion as per stoichiometry - oxygen already present . Now apply percent excess to this quantity and proceed further )

Now start your own calculation. All the best.

Use of Microsoft Office Excel in Chemical engineering calculations.

In Chemical engineering, people need to solve lot of problems in the process calculations. If done by manual method these calculations consume more time and often creates confusion. On the other hand we can use simulation software to solve small material balance  to mighty design calculations, but they are too expensive to install for personal use. Even though some free simulation software are available, they are restricted to some calculations only. Then what is the solution? Here comes Microsoft Office Excel. In Microsoft Excel you can perform small arithmetic operations like addition, subtraction, multiplication, blah blah . to complex problems like iterations for a required case.

Microsoft Excel is equipped with some predefined functions along with charts (includes line,bar,pie charts,etc.,) which are very helpful to solve the process calculations. For example cost of production can be compared between different process by using charts for a given compound. You can also define your own function in Excel if you know VBA (Visual Basic Application) programming, which can reduce your hectic work considerably. Microsoft Excel can handle lot of information which can be compiled for the given instructions. This is particularly useful in doing the material balance calculations in stream wise across different sections like reactor, columns, strippers, etc.,
A lot of predefined and calculated excel sheets are available in internet. Here are some useful websites.

1. http://www.chemsof.com/
2. http://excelcalculations.blogspot.in/p/chemical-engineering-calculations.html
3. http://www.mycheme.com/category/spreadsheets/

Invest your time in learning excel and  you can easily solve complex problems in Microsoft Office Excel.
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