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