ChemE 260 Thermodynamic Temperature Scales & The Carnot Efficiency Dr. William Baratuci Senior Lecturer Chemical Engineering Department University of Washington TCD 6: F & G CB 5: 10 & 11 May 2, 2005

Thermodynamic Temperature Scales From Ch 5: 2nd Carnot Principle : Therefore : Thermodynamic Temperature Scales Operate a HE between two thermal reservoirs Measure QC, QH and W Calculate  Arbitrarily assign a value to either TC or TH If we know the function: we can calculate the other, unknown T. This is a Thermodynamic Temperature Scale In Ch 5 we learned that thermal efficiency depends only on the ratio of QC to QH. From the 2nd Carnot Principle, we learned that thermal efficiency depends only on TC and TH. We can conclude that QC/QH depends only on TC and TH. Our goal is to determine this functional relationship ! This will be a very important result. Why ? For starters, it will allow us to define thermodynamic temperature scales Thermodynamic temperature scales are not dependent on the properties of any material ! Baratuci ChemE 260 May 2, 2005

QC/QH = Fxn(TH, TC) Algebra tells us that : Hot Reservoir, TH QH HEA Cold Reservoir, TC HEA QH Q1 WA HEB HEC Reservoir, T1 QC WB WC Algebra tells us that : Because we are applying the 2nd Carnot Principle to these heat engines, the result of this work will only apply to REVERSIBLE cycles ! Notice that fxn(TH, T1) means plug TH and T1 into the function. It does NOT mean fxn multiplied by TH or T1. Ex: fxn(x,y) = ax + by + c, then fxn(TH=400, T1=300) = 400 a + 300 b + c The bottom line is: The function fxn must be a very special function in order for the boxed equation to be true Let’s see what kind of function would satisfy the relationship in the box. Baratuci ChemE 260 May 2, 2005

The Kelvin Relationship All of the T1 terms in fA and fB must drop out when they are multiplied together. Only true if the fxn has the form : Conclusion : Lord Kelvin : Result : There is only one way for all of the T1 terms to drop out. This leads to such a nice, simple result. It makes life much easier than all those other functions. We will use the Kelvin Relationships in most of the problems in the rest of this course ! Choices other than the one made by Kelvin are possible. Fxn2 must be a function that increases monotonically, so fxn2(T) = eT This is called a logarithmic temperature scale Temperature on this scale goes from – to + as opposed to 0 to + on the Kelvin scale. Now, back to temperature scales… Use water at its triple point as one reservoir for a HE and assign its temperature a numerical value of 273.16. Run a reversible HE between this reservoir and another reservoir at an unknown temperature. Measure Qtriple and Qunk. Tunk = 273.16 * ( Qunk / Qtriple ) This IS the Kelvin Temperature Scale ! The Kelvin Temperature Scale IS A THERMODYNAMIC TEMPERATURE SCALE ! This is a wonderful result. So, if we know the temperatures of two reservoirs in Kelvin, we can immediately and easily compute the thermal efficiency of a reversible HE operating between those reservoirs. What is the thermal efficiency of a reversible HE operating between a human body and the ambient air in the classroom ? What is the thermal efficiency of a reversible HE operating between a nuclear reactor at 2000 K and the cooling water at 300 K ? This is why we want to have a very hot reservoir to generate electrical power. Baratuci ChemE 260 May 2, 2005

The Ideal Gas Temperature Scale Apply the 1st Law to Steps 1-2 and 3-4 carried out in a closed system to determine QH and QC. P V QH QC 1 2 3 4 TH TC Use the definition of boundary work and CV for ideal gases. As usual, assume changes in kinetic and potential energies are negligible. The only tricky part here is converting from the 1st Law…which uses our sign convention… to QC and QH which are both positive and DO NOT use the sign convention. Here is the key: QH = Q12 QC = -Q34 The results come naturally when you realize that Steps 1-2 and 3-4 are both isothermal. Results : Baratuci ChemE 260 May 2, 2005

Adiabatic Steps in a Carnot Cycle Apply the 1st Law to Steps 2-3 and 4-1: Use the definition of boundary work and CV for ideal gases. Results : The only tricky part here involves T. In order to understand why this technique is valid, we must write the 1st Law in differential form. - W = dU Then, because we assume only boundary work occurs : - P dV~ = dU = CoV dT Ideal Gas EOS : - (RT/V~) dV = dU = CoV dT We cannot integrate the left-hand side of this equation at this point because T is NOT a constant ! We must get the T into the right-hand side of the equation where we will integrate with respect to T and take into account the fact that T changes during the adiabatic process. Divide by T : - R ( dV~ / V~ ) = ( CoV / T ) dT Now, we can integrate ! Conclusion : Baratuci ChemE 260 May 2, 2005

Ideal Gas & Kelvin Scales The isothermal steps : The adiabatic steps : Most of the algebra here is straightforward. The only trick is that – Ln[A] = Ln[1/A]. The results is very cool. It shows in a much clearer way that I could back in Ch 1 why the Ideal Gas and the Kelvin Temperature Scales are identical. They are both thermodynamic temperature scales. These results enable us to analyze a wide new area of thermodynamic cycle problems. Combining yields : The Ideal Gas & Kelvin Scales are identical ! Baratuci ChemE 260 May 2, 2005

Carnot Efficiency and COP Power Cycles Refrigeration Cycles Heat Pump Cycles Baratuci ChemE 260 May 2, 2005

Performance vs. Reservoir T All cycles shown here are reversible. Power Cycle  TH TC = 300 K COPR TH Refrigeration Cycle TC = 265 K COPHP TC TH = 300 K Heat Pump Cycle Power Cycle: TC = 300 K Efficiency  100 % as TH   Efficiency  0 % as TH  TC Refrigeration Cycles: TC = 265 K COP   as TH  TC COP  0 as TH  TH Heat Pump Cycles: TH = 265 K COP   as TC  TH COP  1 as TC  0 K Baratuci ChemE 260 May 2, 2005

The Relative Value of Energy Power Cycle  TH TC = 300 K 100 J of heat at 1000 K produces more work than 100 J of heat at 300 K Heat energy available at a higher T is more valuable or has a higher quality. You can always completely convert work into heat at ANY temperature using friction. We can conclude that work is a more valuable form of energy than heat ! The idea that energy has a certain quality depending on the temperature at which it is available leads to the concept of entropy that we will study in the next chapter ! No matter what the temperature, if you do some work, the resulting friction yields heat. So, work can be completely converted to heat at ANY temperature. Heat can NEVER be completely converted into work (K-P statement of the 2nd Law). Work is therefore more valuable than heat, regardless of the temperature at which the heat is available. Baratuci ChemE 260 May 2, 2005

Next Class … Next class … CIDR Learning Styles Survey Chapter 7: Entropy Here the 2nd Law finally boils down to an equation that we can use to analyze thermodynamic cycles. The Clausius Inequality leads to the definition of a new property…entropy, S. CIDR Learning Styles Survey Baratuci ChemE 260 May 2, 2005