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Second law Chapter 18 of and

Describe what happens to the in the following situations. What is the reverse process?

1. A hot gas is thermally connected to a cold gas.

2. A block slides on a table and stops?

3. Two containers of an at p1 and p2 and the same are connected.

Why do the reverse processes not occur in nature? Experimentally the degree of disorder or randomness in a never decreases with time. The quantitative measure of randomness is called . The greater the randomness the greater the entropy. The most general statement of the 2nd law of thermodynamics is that entropy or randomness always increases. There are other equivalent statements having to do with heat engines (see below) In the 3 examples above how is the randomness (entropy) increased? What about ice freezing? It is more ordered after it freezes! Heat Engines

A heat is a device which converts heat into i.e. does . The working substance of a heat engine is what absorbs heat and does the work. e.g. In a the working substance is the steam.The steam absorbs the heat from burning , expands and does work on a piston. What is the working substance is a car engine?

Usually the working substance is carried through a cycle or closed thermodynamic path in which initial and final states are equal. Thus U=Q-W=0 and the net heat absorbed by the engine equals the work done per cycle.

In each cycle of an idealized heat engine an amount

of heat QH flows from a hot reservoir where the temperature is assumed to be constant at TC. Only some of the thermal energy can be converted to mechanical work W. The remainder of the heat

QC is exhausted to a cold reservoir at temperature TC Since QH is transferred ito the engine it is +ve whereas QC is -ve. st According to the 1 law W QH QC QH QC .

The efficiency of a heat engine is ratio of the work done to heat input: W Q Q e 1 C 1 C QH QH QH

Nuclear or Coal Plant

QH Boiler (nuclear or coal)

A 750 MW power plant with operating with an efficiency of 30% has a heat input water Turbine (W) pump W 750MW Q 2500MW H e 0.30 Q Q W 1750MW Condenser C H QC e.g river or lake and Heat Pumps

It is also possible to operate a heat engine in reverse and thereby take heat from the cold reservoir and move it to the hot reservoir. However one must also do some work for this to be consistent with the 2nd law of thermodynamics.

st applying 1 law QH QC W 0 and W are Since QH both negative QH QC W Q Q Coefficient of performance K C C W QH QC Equivalent Statements of 2nd Law of Thermodynamics

A It is impossible for any process to have its sole result to transfer heat from a cooler body to a hotter body. or B It is impossible for any process to absorb heat from a hot reservoir and convert it completely to work

These statements are equivalent as may be seen below. If A is not true and W=0 (what is K?). Such a workless may be combined with a regular heat engine and convert heat from a reservoir completely to work. Similarly if B is false then such an engine could be combined with a regular fridge to make A false. (after French Engineer Sadi Carnot (1796-1832))

The most efficient heat engine is based on the Carnot cycle

Vb QH Wab nRTH ln Va

Q=0 Q=0

Vd Vc QC Wcd nRTC ln nRTC ln Vc Vd Q T ln(V /V ) Taking the ratio C C c d QH TH ln(Vb /Va )

However we can eliminate know Vc and Vd using results from last chapter since :

1 1 For the b c THVb TCVc (1) 1 1 and for adiabatic process d a THVa TCVd (2)

1 1 V V V V Dividing (1) by (2) one obtains b c or b c Va Vd Va Vd

Q T TC Then the ratio C C or the efficiency e 1 carnot T QH TH H (Minister Robert Stirling 1816)

Regenerator R made of packed steel wool or metal baffles allows gas exchange while maintaining the temperature gradient. Expansion piston on right and compression piston on left both connected to same shaft but out of

1 2 isothermal compression, cold piston moves up and QC released to the cold reservoir. 2 3 hot piston goes down, cold piston so no change. regenerator transfers QR to the gas. 3 4 isothermal expansion, hot piston down and cold piston stationary as QH transferred in from hot reservoir. (working part of the cycle) 4 1 hot piston up, cold piston down so no volume change. Gas releases QR to the regenerator. Stirling Engine (Minister Robert Stirling 1816)

Efficiency of an idealized Stirling engine

QC nRTC ln(V2 /V1) TC estirling 1 1 1 ecarnot QH nRTH ln(V2 /V1) TH Four Stroke Automobile Engine heat of combustion for gasoline 4 Lc=5.0X10 J/g

Compression ratio

r=Vmax/Vmin typically less than 10 due to preignition problems due to the high temperature of the air fuel mixture for r>10 Efficiency of (an idealized Cycle for Internal Combustion Engine)

QH Qbc nCv (Tc Tb ) since Wbc 0

QC Qda nCv (Td Ta ) since Wda 0 Q T T e 1 C 1 a d QH Tc Tb Since for an adiabatic process

1 1 1 TcV Td (rV ) or T c Td r 1 and similarly Tb Tar

Ta Td 1 Thus e 1 1 1 1 1 Td r Tar r

Note higher improves efficiency r=8 gamma=1.4 gives e=0.56 This document was created with Win2PDF available at http://www.daneprairie.com. The unregistered version of Win2PDF is for evaluation or non-commercial use only.