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.