2nd Law Thermodynamics

2nd Law Thermodynamics - BME100L:Topic2 SecondLawof...

This preview shows pages 1–6. Sign up to view the full content.

1 BME 100L: Topic 2 Second Law of Thermodynamics: Entropy Always Increase Entropy Always Increases From 1 st law to 2 nd law y E = q + w y If I can somehow reduce the energy of a system by E, I can then extract work. I can also do this continuously through a cyclic process (for each cycle E = 0) y Even if I can reduce heat loss (q) to zero, the work I extract will never exceed what’s reduced in the energy of the system (First law: I can’t win ). y Actually, even this cannot be achieved: the reduced energy can’t be 100% converted into useful work in a cyclic process – there is always gonna be some heat loss (Second law: I can’t break even).

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
2 From 1 st law to 2 nd law y First Law: Energy can change from one form to another as a result of changes in system states y Second Law: y Will a system change states? OR y Under what conditions will the system change states? T hot Consider an engine trying to extract engine w q total q cold work by operating between a hot reservoir and a cold reservoir. Two important points: 1) In order to do any work through a cycle, we must have T hot > T cold. T cold 2) w < q total; But, what’s the maximum work that can be extracted?
3 Step I, isothermal expansion : 22 hot 1 1h o t ln 0 VV dw PdV nRT V w PdV dV nRT =− = < ∫∫ 11 2 For ideal gas, Δ E = 0 along an isotherm. Thus, according to 1 st law: 2 111 1 1 h o t 1 0l n 0 V Eqw q wn R T V = Δ=+ ⇒= −= > The system acquires heat and converts it to an equal amount of work during the first step. Step II, adiabatic expansion : 2 c o l d h o t 0; () V q wE qE n C TT = = Thus, the system does work without acquiring any heat, resulting in a decrease in its temperature.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
4 Step III, isothermal compression : 4 3 3 3c o l d 4 ln 0 V V V wP d V n R T V =− = > 33 0 qw < The system will release heat while receiving equal amount of work from the surroundings (cold reservoir). Step IV, adiabatic compression : 0; 4 4 4 4 4 hot cold () V q wE qE n C T T = = The system recovers its temperature while acquiring work from surroundings 4 3 1 hot cold hot cold hot cold 1 24 ln ( ) ln ( ) VV i i V V w w nRT nC T T nRT nC T T = == + + + Total extracted work: 3 1 hot cold ln ln 0 (consistent with 1st law) V V nRT nRT E =+ Δ= 1. Overall, a Carnot cycle clearly obeys the 1 st law. The surroundings (hot reservoir and cold reservoir) supplies the system with heat (q), which is converted into equal amount of work (w).
5 2. However, there’s something permanent that has happened. Consider the amount of heat extracted from the hot reservoir (from step 1): 2 1h o t 1 ln V qn R T V = Total heat extracted from hot reservoir: Not all of this is converted to w. Instead, there’s a fraction that goes into the cold reservoir: V 4 3c o l d 3 ln R T V = Total heat “wasted” on the cold reservoir: Carnot’s discovery: We know the sum of q over the process is not 0 Consider the sum of q/T over the process.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/21/2011 for the course BME 100 taught by Professor Yuan during the Spring '07 term at Duke.

Page1 / 21

2nd Law Thermodynamics - BME100L:Topic2 SecondLawof...

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online