heat and thermodynamics - i / dr. mathivanan velumani
TRANSCRIPT
SECOND LAW OF THERMODYNAMICS
It is impossible for a self acting machine
unaided by any external agency to transfer
heat from a body at a lower temperature to
another body at a higher temperature.
ENTROPY
A measure of the randomness or measure of disorder
of the system of the microscopic constituents of a
thermodynamic system. Symbol: S.
Examples of increase in entropy:
Solid converting into liquid
Liquid converting into gas
Examples of decrease in entropy:
Liquid converting into solid
Gas converting into liquid
The mixing decreases the entropy
of the hot water but increases the
entropy of the cold water by a
greater amount, producing an
overall increase in entropy. ... that
could have been used to run a heat
engine is now unavailable to do
work.
ENTROPY OF HOT WATER MIXED WITH COLD
WATER
FORMULA TO FIND CHANGE IN ENTROPY
Working : The Carnot engine has
the following four stages of
operations. It is a reversible
engline which converts heat
energy into mechanical work.
1. Isothermal expansion
2. Adiabatic expansion
3. Isothermal compression
4. Adiabatic compression.
From A to B -
The cylinder is placed over the source which is at the temperature T1.
The piston is moved upward –Pressure decreases and volume increases at constant temperature T1.
The heat energy absorbed by the system is H1.
Gain in entropy of the working
substance =
From B to C -
The cylinder is placed over the
insulated stand
The piston is moved little upward –
Pressure decreases and volume
increases at a decreased temperature
T2.
There is no heat energy absorbed by
the system.
There is no change in entropy
From C to D -
The cylinder is placed over the sink which is at the temperature T2.
The piston is moved downward –Pressure increases and volume decreases at constant temperature T2.
The heat energy rejected by the
system is H2.
Loss in entropy of the working
substance =
From D to A -
The cylinder is placed over the
insulated stand
The piston is moved little downward–
Pressure increases and volume
decreases at an increased temperature
T1.
There is no heat energy absorbed by
the system.
There is no change in entropy
𝑑𝑠 = 𝐻1
𝑇1−
𝐻2
𝑇2 = 0
Total change in entropy of the
working substance in a
complete reversible process
Carnot theorem:
The efficiency of a reversible engine does not
depend on the nature of the working substance.
It merely depends upon the temperature limits
between which the engine works.
"All the reversible engines working between the
same temperature limits have the same efficiency.
No engine can be more efficient than a carnot's
reversible engine working between the same two
temperatures."
Quantity of heat taken from the sink = H2’- H2
Quantity of heat given to the source = H1’- H1
Entropy changes of Vander-waals gas:
(P + 𝒂
𝑽𝟐 ) (v-b) = RT
A modification of the ideal gas law (PV = RT) was
proposed by Johannes D. van der Waals in 1873 to
take into account molecular size and inter molecular
forces. It is usually referred to as the van der Waals
equation of state. Here, P – Pressure, V- Volume, a -
correction for intermolecular force of attraction, b –
correction for
molecular size and gas constant R = 8.3145 J/mol K.
Two most important corrections neglected
in the ideal gas are included.
During an expansion, both the temperature and
the volume may change. To calculate the change
in entropy, we need
_____________1
and
-------------------2
Cv – represent the specific heat capacity at constant volume
we can calculate the total entropy change by
integrating, first at constant T and then at
constant V
For isothermal expansion b = 0
and T1 = T2
Therefore, the change in entropy
becomes
∆s = R ln 𝑉2
𝑉1
∆𝑠 = Rln 𝑉2
𝑉1 + Cv ln (
𝑉1
𝑉2)𝛾−1
= Rln 𝑉2
𝑉1 + Cv ln (
𝑉2
𝑉1)−𝛾+1
= Rln 𝑉2
𝑉1 - Cv
𝐶𝑝
𝐶𝑣 ln (
𝑉2
𝑉1) + Cv ln (
𝑉2
𝑉1)
= ln ( 𝑉2
𝑉1)(R-Cp +Cv)
= ln ( 𝑉2
𝑉1)(R-(Cp-Cv))
= ln ( 𝑉2
𝑉1)(R-R)
=0
Therefore change in entropy for adiabatic process: Let
b =0 and
𝑇2
𝑇1 =
𝑉1𝛾−1
𝑉2𝛾−1
NUMERICAL PROBLEM 1
NUMERICAL PROBLEM 2
NUMERICAL PROBLEM 3
NUMERICAL PROBLEM 4
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