Lec 17: Carnot principles, entropy

For next time: Outline: Important points: Read: § 6-9 to 6-14 and 7-1 HW 9 due October 29, 2003 Outline: Carnot’s corollaries Kelvin temperature scale Clausius inequality and definition of entropy Important points: Do not forget the first law of thermodynamics and the conservation of mass – we still need these to solve problems Kelvin temperature scale helps us find maximum efficiencies for power cycles Understand how entropy is defined as a system property

Carnot’s first corollary The thermal efficiency of an irreversible power cycle is always less than the thermal efficiency of a reversible power cycle when each operates between the same two reservoirs.

Carnot’s first corollary So, WI  WR, and So th,I  th,R

Carnot’s second corollary All reversible power cycles operating between the same two thermal reservoirs have the same thermal efficiencies.

Carnot’s second corollary And so

Refrigerators and heat pumps We can show in a manner parallel to that for the Carnot corollaries: The COP of an actual, or irreversible, refrigeration cycle is always less than the COP for a reversible cycle operating between the same two reservoirs. 2. The COP’s of two reversible refrigerators or heat pumps operating between the same two reservoirs are the same.

Kelvin Temperature Scale The thermal efficiency of all reversible power cycles operating between the same two thermal energy reservoirs are the same. It does not depend on the cycle or the mechanism. What can  depend upon?

Kelvin Temperature Scale = (L, H) where the ’s are temperatures. It is also true that =1 -(QL/QH)rev, so it must be true that

Kelvin Temperature Scale The previous equation provides the basis for a thermodynamic temperature scale--that is, one independent of the working fluid’s properties, of the cycle type, or any machine. We are free to pick the function  any way we wish. We will go with the following simple choice:

Kelvin Temperature Scale So,

Kelvin Temperature Scale This only assigns the T ratio. We proceed by assigning Ttp 273.16 K to the triple point of water. Then, if that is one reservoir,

(Reversible) Power Cycles’ Efficiency With Carnot’s first corollary, the maximum efficiency one can expect from a power-producing cycle is that of a reversible cycle. Also, recall that for any cycle (reversible or irreversible)

(Reversible) Power Cycles’ Efficiency But So

TEAMPLAY Many power cycles for electricity supply operate between a steam supply reservoir of about 1,000 °F and a heat rejection reservoir of about 70 °F. What is the maximum thermal efficiency you can expect from such a system?

Carnot Efficiency This maximum thermal efficiency for a power cycle is called the “Carnot Efficiency.”

Efficiencies So, if an efficiency is obtained that is too large, it may be an impossible situation:

TEAMPLAY Problem 6-90E

Refrigerators, air conditioners and heat pumps Hot reservoir at TH System Cold reservoir at TL

Coefficient of Performance Refrigerators/Air conditioners

For a Carnot refrigerator or air conditioner Substituting for temperatures

Coefficient of Performance for Heat Pumps

For a Carnot heat pump Substituting for temperatures

TEAMPLAY Problem 6-104

This is going to seem pretty abstract..so hang on for the ride!

Clausius Inequality Another corollary of the 2nd Law. Now we will deal with increments of heat and work, Q and W, rather than Q and W. We will employ the symbol , which means to integrate over all the parts of the cycle.

Clausius Inequality The cyclic integral of Q/T for a closed system is always equal to or less than zero.

Look at a reversible power cycle Hot reservoir System Cold reservoir

Look at a reversible cycle: We know: And:

For the reversible cycle Look at Q/T: Since the heat transfer occurs at constant temperature, we can pull T out of integrals:

For the reversible cycle This allows us to write:

What were signs of irreversibilities? Friction unrestrained expansion mixing heat transfer across a temperature difference inelastic deformation

For an irreversible cycle Hot reservoir System Cold reservoir

For an irreversible cycle For the same heat input: For both cycles we can write: and Apply inequality: or or

Apply cyclic integral For the irreversible cycle:

Clausius Inequality And so, we have Where the  goes with an irreversible cycle and the = goes with a reversible one.

Clausius Inequality Why do we go through the “proof”?…The inequality will lead to a new property.

Let’s look at a simple reversible cycle on a p-v digram with two processes Let A and B both be reversible

Find the cyclic integral: We can then write:

What can we conclude? The integral is the same for all reversible Paths between points (states) 1 and 2. This integral is only a function of the end states and is therefore a property of the system. We’ll define a new property, entropy as:

Entropy s2 - s1 = Units are