Reflection
Well this chapter cover about the entropy. Entropy owes its existence from the extension of second law of thermodynamics . In Entropy the formula is Sgeneration = S system + S sorrounding and if the S generation is > 0 it is a irreversible process , if the S generation is = 0 it a reversible process . In this chapter , theres also a process called Isentropic Process , which the S system is = 0 which simplified to S 2 = S 1 . Many engineering system or devices such as pump , turbines , nozzles are essentially adiabatic in their operation and it perform the best in irreverseable . Then in isentropic process enable us to compare the actual performance of theses devices to the performance under idealized conditions. To determined wether the process is Isentropic or not , it depend on the state. To achieve isentropic , the system must be adiabatic and it is reversible. .
Lastly entropy change of an ideal gases, which the formula is obtained by combining entropy and ideal gases formula. S2-S1 = Cp ln (T2/T1)- R ln (P2/P1)
To find effeciencies of a turbine , N= W isentropic/ W actual . The ratio of the actual work output of the turbine to the work output that would be achieved if the process between the inlet state and the exit pressure were isntropic.
In this chapter , it concerned with transformation of energy, and the laws of thermodynamics.The laws called Second Law of Thermodynamics. It cover the differences between two forms of energy, heat and work. In an energy balance, both work and heat are included as a simple addtive term, implying that one unit of heat.
In this chapter , It covers about heat engines , thermodynamics temperature scales, ideal-gas temperature scale : Carnot Equation and entropy changes of an ideal gas.
First we are going to talk about heat engines. Heat engines is a machines that produce work from heat in a cyclic process. An example is a steam power plant in which the working fluid periodacally return to its original state. The formula provided in this subtopics is W= Qh- Qc. and Thermal efficiency=1-Qh/Qc.
For the Carnot Equation the formula is Qh/Th=Qc/Tc. If the process is reversible and adiabatic Qrev=0 then S=0. Thus the entropy of a system is constant during adiabatic process and the process is said to be isentropic
Assalamualaikum,
Based on the previous chapter , I've learned on how to calculating the specifc heat of entalphy on non reactive process. I learned on how to applying hypothetical path , watson corellation and how to interpret data from the pyscometric chart .
Next , i've learned about the balances on reactive process where it required me to understand the concept heat of reaction , heat of reaction and HESS law and quite frankly , from all chapter in (PCP II ), this chapter is the chapter that i love the most because it play with a lot of molar things. .In other words , it mean that it required me to do a lot of atomic balance and i love to do atomic balances . So throughout this chapter , what i can say is , an energy balances on a reactor tell the process engineer how much heating or cooling the reactor requires in order to operate at the desired conditions. It show how entalphy changes that accompany chemical reaction are determined from tabulated phsical propeties of the reactant and products and how calculated entalpies if reaction are incoporated in energy balances on reactive processes. About the heat of formation and heat of reaction , both this reaction were a method that you can choose in order to finding the specific entalphy Q. What differentiate both of this method was the equation used and the formula. To put in another word , heat of formation basically is a formation reaction of a compound in which the compound is formed from its constituentselement as they normally occur . For example the formation of CO2 (C(s) + O2(g) --------> CO2(g) ) and the heat of formation of this reaction is -393.5 kj/mol ( from table B1) and the formula for heat of formation can be seen in page 447 (9.3-1). For the heat of reaction , it is about the entalphy changes for a process in which stoichiometric quantities of reactant at temperature T and pressure P react completely in a single reaction to product at the same temperature and pressure.
To summarize , when performing energy balances on a reactive process, two procedures( two method) maybe followed in the calculations of H . The choices is in your hand, you can choose in which method you prefer , but this doesnt mean you to specialize in only one method . Sometimes question may asked you to do for both method but don't worry as long you do the exercise given by Dr. Azeman , insyaAllah , everything gonna work out just fine .
Assalamualaikum,
So far i've learn till chapter 8 in the Principle of Chemical Processes II and yet still using the same red thick book .Balance On Non Reactive Process (chapter 8) , In this chapter i've learn about the important of reference state and hypothetical path. I also have learned using Watson corellation formula. Then i learned the Psychometric Chart which quite difficult to understand this .It is important to know that the chart is only for water and dry air.I use this chart when i want to find dry-bulb temperatue, absolute humidity, relative humidty, dew point, humid volume, wet-bulb temperature, specific enthaphy and enthalphy deviation.
In my opinion , PCP II (Principle of Chemical Processes II) was the best subject among the others subject because it requires me to think critically in finding the best solution and answer.
Thank you.
It is already toward the end of semester 2 . Throughout the semester , there's a lot of thing i've learned in Principle Chemical Process (PCP) teached by Dr. Azeman . Dr. Azeman always gave me words of motivation that I became eager to learn. Before this , my senior once said that this subject is tough and he said i've to give full concentration in class if i want to pass this subject (they said) . Alhamdullilah , so far i can concentrate in class with out any interference . Thanks to Dr. Azeman because the way he taught making me want to focus only on the whiteboard. With the full concentration , i've gain skill to perform mass balance easily and calculating molar composition . For the test 1 where held at N24 i've got 81% . I am very happy with my result . However my happiness was only for a short time because for test 2 i got only 50% only . I jus dont know why this is happening. Maybe because of the question for test 2 really tough and difficult compared with the question for test 1 . Few weeks ago , i've just done my test 3 and the question is quite simillar with test 2 . The questions given was so tough. The only thing that i can do is to wait and hoping for a miracle that i will gain a good result. For the final examination , i will ensure that i am fully prepared. The final exam is arround the corner . I hoped that i will pass the test with a flying colour . Last but not the least , i want to thank to my brilliant lecturer (Dr. Azeman) for teaching me and continously giving me motivation . Thank you .