second law of thermodynamics

1. Ahmed Medhat - Thermodynamics 1 - ahmedhatfa@hotmail.com
Second law
of
thermodynamics

2. Ahmed Medhat - Thermodynamics 1 - ahmedhatfa@hotmail.com
1.Kelvin-Planck statement
It is impossible for any device that operates on a cycle to receive heat from a single
reservoir and produce an equivalent amount of work.
EFFICIENCY 100%

3. Ahmed Medhat - Thermodynamics 1 - ahmedhatfa@hotmail.com
EFFICIENCY 100%
Useful Output
Required Input
Net work
Thermal energy
in fuel
EFFICIENCY

4. Ahmed Medhat - Thermodynamics 1 - ahmedhatfa@hotmail.com
EFFICIENCY Heat engine
Net work
Thermal energy
in fuel
𝑾𝒏𝒆𝒕 = 𝑾𝒐𝒖𝒕 - 𝑾𝒊𝒏
𝑸𝒊𝒏 = 𝑾𝒏𝒆𝒕 + 𝑸𝒐𝒖𝒕
∴ 𝑾𝒏𝒆𝒕 = 𝑸𝒊𝒏 - 𝑸𝒐𝒖𝒕
𝜼 =
𝒘𝒏𝒆𝒕
𝑸𝒊𝒏
=
𝑸𝒊𝒏−𝑸𝟎𝒖𝒕
𝑸𝒊𝒏
= 𝟏 −
𝑸𝟎𝒖𝒕
𝑸𝒊𝒏

5. Ahmed Medhat - Thermodynamics 1 - ahmedhatfa@hotmail.com
2.Clausius statement
It is impossible for any device to operate on transferring heat from a cold reservoir to a high
temperature reservoir without receiving any work.

6. Ahmed Medhat - Thermodynamics 1 - ahmedhatfa@hotmail.com
Refrigerator & AC Heat Pump
(𝑪𝑶𝑷)𝑹=
𝑸𝒍
𝒘𝒏𝒆𝒕,𝒊𝒏
Coefficient of performance
(𝑪𝑶𝑷)
Useful Output
Required Input
(𝑪𝑶𝑷)HP =
𝑸𝑯
𝒘𝒏𝒆𝒕,𝒊𝒏
(𝑪𝑶𝑷)HP = (𝑪𝑶𝑷)𝑹 +1

7. Ahmed Medhat - Thermodynamics 1 - ahmedhatfa@hotmail.com
Carnot cycle
Maximum efficiency or COP
𝑸𝒊𝒏
𝑸𝑶𝑼𝑻
∝
𝑻𝒊𝒏
𝑻𝒐𝒖𝒕
∴
𝑸𝒊𝒏
𝑸𝑶𝑼𝑻
=
𝑻𝒊𝒏
𝑻𝒐𝒖𝒕
All temperatures must be in (K)

8. Ahmed Medhat - Thermodynamics 1 - ahmedhatfa@hotmail.com
∴
𝑸𝒊𝒏
𝑸𝑶𝑼𝑻
=
𝑻𝒊𝒏
𝑻𝒐𝒖𝒕
Heat engine Refrigerator & AC Heat Pump
𝜼𝒄 = 𝟏 −
𝑸𝟎𝒖𝒕
𝑸𝒊𝒏
= 𝟏 −
𝑻𝟎𝒖𝒕
𝑻𝒊𝒏
(𝑪𝑶𝑷)𝑹,Carnot =
𝑸𝒍
𝒘𝒏𝒆𝒕,𝒊𝒏
∴ 𝑾𝒏𝒆𝒕 + 𝑸𝒍 = 𝑸𝒉
=
𝑻𝒍
𝑻𝑯 − 𝑻𝑳
(𝑪𝑶𝑷)HP, Carnot =
𝑸𝑯
𝒘𝒏𝒆𝒕,𝒊𝒏
=
𝑻𝑯
𝑻𝑯 − 𝑻𝑳
∴ 𝑾𝒏𝒆𝒕 + 𝑸𝒍 = 𝑸𝒉