Heat Engines and Refrigerators
Ron Reifenberger Birck Nanotechnology Center Purdue University February 27, 2013
Exam II will be held on Tuesday April 30, 2013 TIME: 8:00-10:00 AM PLACE: ARMS 1021
Lecture 8 1 DEFINITION
Any device that has the primary purpose to convert heat to work is called a Heat Engine.
First, a quick review of different types of heat engines
2 Basic Idea ABCD
Mass = m
h
gas
Apply Heat! Remove Schematic: Heat!
Qin C P B
D PdV mgh A Qout V 3 EXAMPLE I: Early heat engines were steam engines (Savery in 1698, Newcomen in 1705, Watt in 1770)
Simplified Steam Engine*
Mechanical Piston Linkage
Turning Steam wheel
Boiler Water Cold Condenser 4 * REQUIRED A LICENSED OPERATING ENGINEER Modern steam engine
(Work comes from decrease of Eint of steam) High pressure Low pressure
flywheel
Steam condenses Clean water to water at low is recycled! pressure (invented ~1865)
An example of an external combustion engine Earliest steam engines were ~1% efficient (they burned enormous amounts of fuel!); by early 1900’s efficiencies reached ~10%; modern steam engines have efficiencies above 30%. 5 Classic Heat Engines
• Rankine (steam) engine – 1859 (theory) 1st generation • Stirling engine - 1816 • Internal Combustion (Otto) engine - 1876 2nd generation • Diesel engine - 1893
6 Steam (Rankine) Cycle
change water to steam
Isentropic (constant entropy) expansion
change steam Eint = 0 to water Isentropic Why? compression
isentropic = quasistatic plus adiabatic7 Example II: The Stirling Engine
(Invented by Robert Sterling in 1816)
cooling fins
two pistons
Heat Heat Heat Heat
Characteristics: Recycles same trapped gas, quiet, low vibration, higher efficiency (~40%) than steam engine or internal combustion engine, low power applications.
http://en.wikipedia.org/wiki/Stirling_engine8 Moveable gas tight piston Simplified Stirling Engine
3 Flywheel not shown 2 4 Moveable 1 5 (or 1) displacer Cold Cold Cold Cold Cold w. shaft
Loose- fitting P displacer (takes up P P space) P P
Tube, open at top end, Hot Hot Hot Hot Hot closed at bottom end Expansion Compression caused by caused by thermal downward 9 expansion of motion of gas red piston Stirling Thermodynamic Cycle
Qin
isothermal process Eint = 0 Why?
Qout
10 A commercial Stirling Engine
Floor model Table model
"The Lake Breeze Motor“ was marketed in 1919. Its original price was ~$25. It functions as a hot-air engine. A hot-air engine is a heat engine that uses the expansion and contraction of air under the influence of a temperature change to convert thermal energy into mechanical work. When you light the wick in the kerosene lamp, the hot air pushes through a turbine that then turns a shaft that ultimately turns the fan. Popular11 before the electrification of rural America which commenced in the mid 1930s. Example III: Second generation heat engines - Internal Combustion; William Barnet (1838) (takes advantage of the energy content of fossil fuels)
How is heat derived from fuel? Through two basic chemical reactions:
C + O2 CO2 + heat energy Exothermic or reactions: heat of O + H2 H2O + heat energy combustion
Example: The reaction of a hydrocarbon heptane with oxygen
6 C7H16 + 11O2 7CO2 + 8H2O + 4.8 x 10 J (per 100 gm of C7H16)
12 Internal Combustion Engine (enabled by the extraction of petroleum from the earth by Drake in 1859) converts ~20% of chemical energy to mechanical energy
Characteristics: • undesirable exhaust gas, • noisy, ~70 atm. gas pressure exerts • high vibration, force on piston • high power
Compression to 6-10 atmospheres T ~ 1000 C 13 The Otto Thermodynamic Cycle (close approximation to the internal combustion engine)
burning gasoline adiabatic process – rapid expansion
Eint = 0 Why? adiabatic exhaust to process – rapid environment compression
14 Example IV: Diesel Engines Rudolf Diesel (1893)
6135H (6 cylinder, 13.5L, Turbocharged and Aftercooled) Tier 3 engine at 448kW @ 2100rpm (rated speed), BTE (brake thermal efficiency) is around 39% +/- 0.6%.
4024T (4 cylinder, 2.4L, Turbocharged) Tier 2 engine at 49kW @ 2800rpm,15 BTE is about 32%. The Wartsila-Sulzer RTA96-C turbocharged two-stroke diesel engine is the most powerful and most efficient prime-mover in the world today (12 cylinder version) Cylinder diameter 38 “
• Length 89 ft ; Width 44 ft • Max. Power 108,920 hp at 102 rpm •. At maximum economy, the engine exceeds 50% thermal efficiency. One piston and rod • Consumes 1,660 gallons of heavy fuel oil per hour 16 Internal Combustion Engines
If an automobile engine (gasoline) were an ideal engine, it would have an efficiency of ~55%. Its actual efficiency is probably closer to 25%. A considerable amount of heat input is lost to the environment and NOT converted to work.
Diesel engines have a slightly higher efficiency than gasoline engines. Diesels do not mix fuel with air and do not require electric spark ignition. The combustion occurs because a high temperature is reached upon compression. The efficiency of a diesel engine is ~30%.
17 Schematic Representation of Generic Heat Engine
W = Qh -Qc
Every heat engine transfers some fraction of heat energy from a hot reservoir to the surroundings (i.e. a cold reservoir). 18 Efficiency of Engine Efficiency = work done/energy in
Don’t worry about previous sign conventions for Q (Qc is inherently negative; origin of minus sign?). Treat all Q’s as positive numbers!
The proper definition of efficiency is important for historical reasons. It showed that the efficiency of an engine does NOT depend on the type of fuel, the pressure of the gas, the volume of the piston, etc. 19 Rate of heat transfer
Rate of heat transfer: For a heat engine with ε =0.25, for every watt W = Qh -Qc (1W=1 J/s) of power generated by the engine, heat is removed from the hot reservoir at a rate of 4 watts (Qh) and heat is exhausted to the cold reservoir at a rate
of 3 watts (Qc).
1 horsepower = 745.7 watts 20 Typical efficiencies:
Liquid fuel rocket ~0.48 Steam turbine ~0.46 Diesel engine ~0.37 Automobile engine ~0.25 Steam locomotive ~0.08
21 Example During one cycle, an engine extracts 2 x 103 J of energy from a hot reservoir and transfers 1.5 x 103 J of energy to a cold reservoir.
a) What is efficiency of engine? Q = 1 -c = 1 – 0.75 = 0.25 Qh b) How much work is done in one cycle?
W = Qh-Qc = 500 J c) How much power is delivered if one cycle takes 0.4 s? W P = = 1250 W t 22 KEY IDEA: Knowing the shape and parameters of a thermodynamic cycle, it is possible to calculate the efficiency of a thermodynamic process SIMPLE EXAMPLE
Take 1 mole of ideal gas quasistatically through the Useful factors: reversible process shown below: 101.3 J= 1 ℓ•atm R=8.206 x 10-2 ℓ•atm/(mol•K) P (in atm) PTVPV/T 23 2 1 1 atm 300K 24.6 ℓ 0.082 2 2 atm 600K 24.6 ℓ 0.082 1 1 4 3 2 atm 1200K 49.2 ℓ 0.082 1 atm 600K 49.2 ℓ 0.082 V (in ℓ) 4
24.6 49.2 23 See Tipler and Mosa: Prob. 19:25 W1 2: PΔV=0
Q1 2: ΔEint=CVΔT=1.5nRΔT=3.74 kJ
W2 3: PΔV=4.99 kJ
Q2 3: CPΔT=2.5nRΔT=12.5 kJ
W3 4: PΔV=0
Q3 4: ΔEint=CVΔT=1.5nRΔT=-7.48 kJ
W4 1: PΔV=-2.49 kJ
Q4 1: CPΔT=2.5nRΔT=-6.24 kJ WkJkJkJWW 4.99 ( 2.49 ) 2.5 23 41 15.4% QQin 12 Q 23 3.74 kJkJ 12.5 16.24 kJ 24 Note that W= Qhc - Q (3.74 12.5) (7.48 6.24) 2.5kJ 4. High pressure liquid Refrigeration expands (and cools) through high-pressure expansion valve and returns to cooling coil
1. Refrigerant 3. Since temperature absorbs heat of compressed gas is and goes from higher than room liquid to gas temperature, heat is given off. Gas then condenses to a liquid under high pressure
2. Compressor increases pressure (and raises temperature) of gas
25 A refrigerator is a heat engine running backwards
Q COP = c W
Typical refrigerators have a COP 5 - 6
Rate of heat transfer: For a refrigerator with a COP = 5, for every watt of power used by the compressor (1W=1J/s), heat is removed from the cold reservoir at a rate of 5 watts (Q ) and heat is nd c The 2 Law of exhausted to the hot reservoir at Thermodynamics requires a rate of 6 watts (Qh). a COP < ∞ 26 After a party, 2 L (0.002 m3) of beer is placed in a refrigerator. The beer initially is at 323 K and is cooled to 283 K. If the COP of the fridge is 5, how much energy is needed to cool the beer?
Assume beer is mostly water: c = 4190 J/(kg K); = 1 x 103 kg/m3
m = V = 2.0 kg
5 Qc = mcT = -3.35 x 10 J (heat flows out)
Q 3.35 x 105 J COP = c = = 5 W = 6.7 x 104 J W W 27 Why can’t ε=1?
Mass = m
h
W= Q–q=mgh gas
n,T,V Apply T+ΔT, V+ΔV Heat! Q
some heat q is always deposited in the
gas: q=nCp ΔT =W/Q=(Q-q)/Q =1-[q/Q] < 1 since q>0
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