Reflection
Mid sem holiday had ended In a blink of eye, now it’s time to continue our lesson. Basically, right after holiday, we had covered second law of thermodynamic and entropy chapter using smith book.
From the first law of thermodynamic, we know that Q – W = ∆H/∆U + ∆KE + ∆PE. If we neglect kinetic and potential energy, and the system has no enthalphy/internal energy change, basically Q = W.
But the thing is the same amount of work done can be converted into same amount of heat (W = Q) but same amount of heat cannot produce same amount of work (Q = not W), thus it’s against the first law. This situation leads to second law equation which is to make Q = W equation work, we need special device that called heat engine. Heat engine operates in single way only.
Kelvin-planck statement says that,
“ it’s impossible for any device that operate on cycle to receive heat from single reservoir and produce net amount of work”
QH
W net out
QL
Kelvin-planck says again that,
“ no heat engine can have 100% thermal efficiency as for power plant to operate working fluid must be exchange heat with environment.”
Thermal efficiency of heat engine =
Wnet = QH – QL
To transfer heat from hot T to cold T is a natural process but to transfer heat from cold T to hot T, we need refrigerator and heat pump system to operate. Refregiretor system is used to keep refregirated space at low temperature by removing heat while heat pump is used to maintain heated space at high temperature.
QH
W net in
QL
Coefficient of performance (COP) refregirator =
COP heat pump =
Clausius statements says that,
“it’s impossible to construct device that operate in cycle and produce no effect other than transfer of heat from lower T to higher T”
Reversible/ideal process is assumed to neglect any irreversibilities in process. Many devices such as turbine, pump, and compressor are work in ideal process to get the maximum efficiency that can be obtained from the process. Carnott says,
“ efficiency of irreversible heat engine is less than reversible heat engine (carnott cycle) that operate at same two reservoir”
“ efficiency of all reversible heat engine are same between same two reservoir at the same time. “
Carnot heat engine efficiency =
Carnott COP refrigerator =
Carnott COP heat pump =
For chapter entropy, the objective of the study is to get ∆s net, ∆s surrounding and ∆s system. Entropy can be define as and entropy equation, ∆s net = ∆s system + ∆s surrounding.
To get ∆s system,
- From steam table in smith/cengel book, ∆s = S2-S1
- For solid and liquid state, ∆s = cp ln (T2/T1)
- Ideal gas, cv ln (T2/T1) + R ln (V2/V1) or cp ln (T2-T1) – R ln (P2-P1)
∆s surrounding = -Q system / T
Mid sem holiday had ended In a blink of eye, now it’s time to continue our lesson. Basically, right after holiday, we had covered second law of thermodynamic and entropy chapter using smith book.
From the first law of thermodynamic, we know that Q – W = ∆H/∆U + ∆KE + ∆PE. If we neglect kinetic and potential energy, and the system has no enthalphy/internal energy change, basically Q = W.
But the thing is the same amount of work done can be converted into same amount of heat (W = Q) but same amount of heat cannot produce same amount of work (Q = not W), thus it’s against the first law. This situation leads to second law equation which is to make Q = W equation work, we need special device that called heat engine. Heat engine operates in single way only.
Kelvin-planck statement says that,
“ it’s impossible for any device that operate on cycle to receive heat from single reservoir and produce net amount of work”
QH
W net out
QL
Kelvin-planck says again that,
“ no heat engine can have 100% thermal efficiency as for power plant to operate working fluid must be exchange heat with environment.”
Thermal efficiency of heat engine =
Wnet = QH – QL
To transfer heat from hot T to cold T is a natural process but to transfer heat from cold T to hot T, we need refrigerator and heat pump system to operate. Refregiretor system is used to keep refregirated space at low temperature by removing heat while heat pump is used to maintain heated space at high temperature.
QH
W net in
QL
Coefficient of performance (COP) refregirator =
COP heat pump =
Clausius statements says that,
“it’s impossible to construct device that operate in cycle and produce no effect other than transfer of heat from lower T to higher T”
Reversible/ideal process is assumed to neglect any irreversibilities in process. Many devices such as turbine, pump, and compressor are work in ideal process to get the maximum efficiency that can be obtained from the process. Carnott says,
“ efficiency of irreversible heat engine is less than reversible heat engine (carnott cycle) that operate at same two reservoir”
“ efficiency of all reversible heat engine are same between same two reservoir at the same time. “
Carnot heat engine efficiency =
Carnott COP refrigerator =
Carnott COP heat pump =
For chapter entropy, the objective of the study is to get ∆s net, ∆s surrounding and ∆s system. Entropy can be define as and entropy equation, ∆s net = ∆s system + ∆s surrounding.
To get ∆s system,
- From steam table in smith/cengel book, ∆s = S2-S1
- For solid and liquid state, ∆s = cp ln (T2/T1)
- Ideal gas, cv ln (T2/T1) + R ln (V2/V1) or cp ln (T2-T1) – R ln (P2-P1)
∆s surrounding = -Q system / T
Alhamdulillah, we've reach our study week :) that's mean we've already answered all the test questions (test 1, 2 and 3) throughout the semester. here, i want to summarized the achievement that i got along the way here.
in test 1, i have slightly misunderstanding on how to read psychometric chart, thus my result is kind of bad, but it's a way better than what i expect. hehe despite that i can't answer the questions well, i'm just grateful for the marks that i get.
for test 2, i've improved a bit, but not that well too~ thank you to dr for teaching us how to calculate mass balance in the easier way. however there's one question that i have answer it correctly but i lost so many mark because i've misunderstand the concept of the question. it's kind of loss, i know i can get more marks than what i've got :(
for test 3, i think i can do question 1 and 2 well, but for question 3 and 4, it's SO-SO. i thought bring cengel book was enough for the test because most of the assignment questions that dr gave us was easily being done using steam table from cengel book, unfortunately for question 4, it's REALLY difficult to use steam table from cengel, and i end up regret for not bringing my smith book along~ huhu this is the best lesson that i've learn, i should bring ALL my books to exam, in case anything happen.
thank you dr for being patient with us, for taught us so much new lesson. please pray for our final :)
It's been about 3 months since we started our PCP2 class with our lecturer, PM Dr. Ramli Bin Mat. Alhamdulillah, i've succesfully passed my PCP1 last semester. for this sem, we discovered how to do energy balances calculation on non-reactive and also in reactive system. As for energy balances, writing reference state is the basis that ‘must have’ in order to solve energy balance problem.
The objective of nonreactive process balances is to calculate enthalphy,H or internal energy,U from heat capacity equation using steam table, hypothethical path, psychrometric chart and other tables with the change in T, P, phase and chemical reaction mixing.
H = U + PV
∆U = U - U ref
∆H = H - H ref
Hypothetical path is use to calculate enthalpy in different T,P, phase or mixing process of energy balances. In this cases, boiling point, cv and cp are very important to develop the path. Based on the first law of thermodynamics for closed and open system,
Closed system : Q – W = ∆KE + ∆PE + ∆U
Open system : Q – W = ∆KE + ∆PE + ∆H
Normally, Q = ∆H in the absence of work shaft, potential and kinetic energy.
Q = n out . H out – n in . H in
While H = cp (T2-T1)
Energy balance table will be developed with the reference stated, to get the heat absorbed/released, Q.
We also learn how to read psychometric chart based on dry T/wet T/Absolute humidity/relative humidity/humid volume given. From the graph, we can get enthalphy value and saturation T.
For energy balance on reactive process, we have learn standard heat of reaction, heat of formation and heat of combustion.
Standard heat of reaction is the heat of reaction (enthalphy change when reactants consumed completely at temperature T and pressure P to form product at the same T and P ) at specified reference temperature and pressure, usually at 1 atm, 25°c. when ∆Hr° > 0, the reaction is exothermic (heat is released during process) while ∆Hr° < 0, the reaction is endorthermic (heat absorbed during process). Standard heat of reaction can also be calculated using hess law, which can be obtain as a linear combination of equations for other reactions. We also can get Hr from table B.1. if the product produced at different state/temperature/pressure than it’s reactant, hyphothethical path is needed to solve the problem.
While standard heat of formation is enthalpy change (heat of reaction) associated with the formation of 1 mole of the compound at a reference temperature and pressure (usually 25oC and 1 atm). To calculate heat of reaction from heat of formation using hess law is
∆Hr° = Hf°i = ( Hf°)product - Hf°)reactant
where v is stoichiometric coefficient of reactant/product species I and ∆Hf° is standard heat of formation for that species. standard heat of formation of substances also listed in table B.1.
The last one is the standard of heat of combustion, ∆Hc°. standard heat of combustion, ∆H°c, is the heat released when one mole of a substance reacts with oxygen to yield specified products in a specified phase. standard heats of reactions that involve only combustible substances and combustion products can be calculated from tabulated standard heats of combustion, in another application of Hess’s law which is
∆Hr° = Hc°i = ( Hc°)reactant - Hc°)product
Heat of combustion of substances also listed in table B.1.
To solve energy balance in reactive process, there are 2 common approach to choose reference state/condition.
- heat of reaction method
Reactants and products are considered as molecular species at To, which is the reference temperature for the heat of reaction (usually 25 oC)
Example of reference state : n-C4H12 (g), i-C4H12 (g) , HCl (g) at 25oC and 1 atm
This method is generally preferable when where is a single reaction for which ∆Hor is known
For multiple reaction,
- Heat of formation method
reference is taken as the elements that make up the reactants and products at reference temperature To.
example of reference state : N2 (g), O2 (g), H2 (g) at 25oC and 1 atm
This method is generally preferable for a single reaction and multiple reactions where ∆Hor is not readily available.
Comparison of the hypothetical process paths that correspond to the heat of reaction and heat of formation methods :
salam alaik!
i wish to share my thoughts and how i felt after nearly a sem, i've been learning PCP subject. first and foremost, i would like to thanks my lecturer, PM Dr. Azeman for letting me in his class. huhu it's long story of why we (me n my roommate) had register this subject under FPREE instead of FKK (my faculty). but really i'm glad that i've done that. thank you for teaching us, being patient with our attitude (sorry for being late to your class!) and for your motivation. it's inspired me a lot! :)
i do love this subject! eventhough doing energy balance is quite complicated, but the fact that i can solve every question correctly makes me feel really good. i'm not the type that can easily understand something new, but through a lot of exercises, it makes me improve my understanding better. for single and multiple mass balance, i think it's quite simple, as we learn futher into reactive process and energy balance, it's quite hard to understand. sometimes i do not know how to start answering the question and just staring at the paper for like 10 minutes? huh
i fail in my second test :(
as for the third test, i can sure it's simpler than test 2. eventhough the results are still unknown, i realllllyyy hope it's going out well, or else i have to work super hard on final. oh no!
i'm also enjoy being in the same class with the gas engineering students. they're veryy friendly and awesome. eventhough we're from different course, rarely meet in the college nor faculty, but still when we're bummed into one another, they will say hello and smile. it's wonderful how being in the same class for only one subject can turns strangers into friend. hihi i love you guys!
it's nearly the end of the sem.
i wish i can stay in sn 3 for PCP 2 next sem. can i? :o