Dr. Mohammed H. Ahmed
College of Petroleum and Mining Engineering
University of Mosul
• Engineering Thermodynamics:
Work and Heat Transfer
Book by G. F. C. Rogers and Y. R. Mayhew
• Applied Heat for Engineers
Thermodynamics
• Applied Thermodynamics for
Engineering Technologists
(5th Edition)
T.D. Eastop , A. Mcconkey
Octave Sneeden, Samuel Vallance Kerr
Lecture 7,8
𝑄12 = 0
𝑊12 = −∆𝑢12
Ex.13/ A closed vessel having a volume of 0.4 m3 contains 2.0 kg
of a liquid water and water vapor mixture in equilibrium at a
pressure of 6 bar (0.6 MPa). Calculate:
a) The volume and mass of the liquid.
𝑚3
b) The volume and mass of the vapor. (Take 𝑣𝑓 = 0.001101 𝑘𝑔 , 𝑣𝑔
𝑚3
= 0.3156 𝑘𝑔 )
4. Adiabatic process
An adiabatic process occurs without transfer of heat or mass of substances
between a thermodynamic system and its surroundings. (Isolated)
𝑸𝟏𝟐 = 𝟎
Apply the N.F.E.E
𝑸𝟏𝟐 = 𝑾𝟏𝟐 + ∆𝒖𝟏𝟐
∴ 𝑾𝟏𝟐 = −∆𝒖𝟏𝟐
For this process that is a property has constant through reversible adiabatic
process which called “Entropy (s)”
𝟐
𝑷𝒅𝒗 =
𝟏
−𝒎𝑪𝒗 𝑻𝟐 − 𝑻𝟏
Take differentials,
𝑷𝒅𝒗 = −𝒎𝑪𝒗 𝒅𝑻
And we have; 𝑃𝑉 = 𝑚𝑅𝑇 ⟹
𝑷=
And we have;
𝑪𝒗 =
𝒎𝑹𝑻
𝑽
𝑹
𝜸−𝟏
𝑑𝑣
𝑅
𝑚𝑅𝑇
=𝑚
𝑑𝑡
𝑉
(𝛾 − 1)
𝑑𝑣
1 𝑑𝑡
=−
𝑉
(𝛾 − 1) 𝑇
2
1
𝑑𝑣
1
=−
𝑉
(𝛾 − 1)
2
1
𝑑𝑡
𝑇
1
2
ln 𝑉 1 = −
[ln 𝑇] 1
(𝛾 − 1)
2
𝑉2
𝑇2
−(𝛾 − 1)ln
= ln
𝑉1
𝑇1
𝑉2
ln
𝑉1
𝑉2
𝑉1
−(𝛾−1)
−(𝛾−1)
𝑇2
= ln
𝑇1
𝑇2
=
𝑇1
∴ 𝑇1 𝑉1 (𝛾−1) = 𝑇2 𝑉2 (𝛾−1)
or
𝑇1
𝑉2
=
𝑇2
𝑉1
(𝛾−1)
(i.e)
𝑇𝑉 (𝛾−1) = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑃1 𝑉1
𝑇1
as;
𝑉1
𝑉2
=
𝑃2 𝑉2
𝑇2
(𝛾−1)
=
⟹
𝑇2
𝑇1
=
𝑃2 𝑉2
𝑃1 𝑉1
𝑃2 𝑉2
𝑃1 𝑉1
𝑃1 𝑉1 . 𝑉1 (𝛾−1) = 𝑃2 𝑉2 . 𝑉2 (𝛾−1)
∴ 𝑃1 𝑉1
(i.e)
(𝛾)
= 𝑃2 𝑉2
𝑃𝑣 (𝛾) = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑃1
𝑉2
=
𝑃2
𝑉1
𝛾
(𝛾)
From the equation of adiabatic process;
∴ 𝑾𝟏𝟐 = −∆𝒖𝟏𝟐
= −𝑚𝐶𝑣 𝑇2 − 𝑇1 , let say per 1 kg
= 𝐶𝑣 𝑇1 − 𝑇2
And, 𝐶𝑣 =
𝑅
𝛾−1
𝑊12
𝑅 𝑇1 − 𝑇2
=
𝛾−1
𝑊12
𝑝1 𝑣1 − 𝑝2 𝑣2
=
𝛾−1
We have, 𝑝𝑣 = 𝑅𝑇,
And for reversible process,
𝑣2
𝑊12 =
𝑃𝑣 (𝛾) = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑃𝑑𝑣
𝑣1
Then,
𝑊12 =
𝑣2 𝐶.𝑑𝑣
𝑣1 𝑣 𝛾
= 𝐶.
𝑊12 = 𝐶.
𝑣2 𝑑𝑣
𝑣1 𝑣 𝛾
−𝛾+1 𝑣2
𝑣
−𝛾 + 1
𝑣2 −𝛾+1 − 𝑣1 −𝛾+1
= 𝐶.
= 𝐶.
1−𝛾
𝑣1
𝑣1 −𝛾+1
− 𝑣2 −𝛾+1
𝛾−1
And 𝑃𝑣 (𝛾) = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
∴ 𝑊12 =
𝑃1 𝑣1 𝛾 𝑣11−𝛾 − 𝑃2 𝑣2 𝛾 𝑣21−𝛾
𝛾−1
∴ 𝑊12 =
=
𝑃1 𝑣1 − 𝑃2 𝑣2
𝛾−1
𝑃1 𝑣1 − 𝑃2 𝑣2
𝛾−1
Ex.14/0.285 m3 of air at a pressure of 2.7 bar is expanded to
volume of 0.708 m3. find the work during expansion for each of
the following process:
- Constant pressure
- Isothermal
- adiabatic
Thank you for your attention
Next lecture: Application for open system