Engines & Refrigerators Flashcards
Heat engine
Any device that absorbs heat and converts part of that energy into work.
Reservoir
Anything that is so large that its temperature doesn’t change noticeably when heat enters or leaves.
Efficiency
Benefit/Cost Ratio
e= Wnet/Qh
What does first law of thermodynamics tell us about heat engines?
How can we use this to rewrite the efficiency equation?
The energy the engine absorbs must be precisely equal to the energy it expels.
Qh = Qc + Wnet
Thus
e = Qh-Qc/ Qh = 1- Qc/Qh
What does the second law of thermodynamics tell us about heat engines?
What can we conclude about efficiency?
Qc/Tc >= Qh/Th or Qc/Qh >= Tc/Th
where Qh/Th is entropy extracted from hot reservoir and Qc/Tc is entropy expelled to cold reservoir.
So we can conclude that
e <= 1 - Tc/Th
How does efficiency relate to the temperature of the reservoirs?
The smaller the ratio Tc/Th, the more efficient the engine will be.
Summary of what first two laws of thermodynamics tell us about heat engines.
First Law: Efficiency can’t be greater than one. (Can’t get more work out than the amount of heat being put in)
Second Law: Can’t achieve efficiency of 1 unless Tc = 0 or Th = inf.
Carnot Cycle
Theoretical thermodynamic cycle that provides maximum possible efficiency for heat engine.
T gas =~ Th
Steps:
- Isothermal Expansion at Th while absorbing heat.
- Adiabatic expansion to Tc.
- Isothermal compression at Tc while expelling heat.
- Adiabatic compression back to Th.
Refrigerators
Heat engine operated in reverse.
Coefficient of Performance (COP)
Efficiency of a refrigerator.
COP = Qc/W
We can rewrite using first law (Qh = Qc + W) as:
COP = Qc/ (Qh-Qc) = 1/ (Qh/Qc) -1
What does second law of thermodynamics tell us about refrigerators?
What does this tell us about the COP?
Entropy dumped into the hot reservoir must be at least as much as entropy absorbed from the cold reservoir.
Qh/Th >= Qc/Tc or Qh/Qc >= Th/Tc
Therefore,
COP <= 1 / (Th/Tc) -1 = Tc/ (Th-Tc)
Relationship between COP and temperature of reservoirs.
COP is largest when Th and Tc are not very different.
How can refrigerator have maximum possible COP?
Carnot Cycle in Reverse:
For heat to flow in opposite direction, working substance must be slightly hotter than Th while heat is being expelled and slightly colder than Tc while heat is being absorbed.
Efficiency of Carnot Cycle
e = 1- Tc/Th
Otto Cycle
Thermodynamic cycle that describes the processes in a four-stroke internal combustion engine, such as those used in most gasoline-powered vehicles.
Efficiency of Otto Cycle
e = 1 - (V2/V1)^ γ-1
where V1/V2 compression ratio and γ
is Adiabatic exponent.
Knowing that TV^ γ-1 is constant in Adiabatic process:
e = 1- T1/T2 = 1- T4/T3
Otto engine is less efficient than Carnot engine.
Rankine Cycle
Thermodynamic cycle that models the operation of a steam engine. It describes how heat is converted into mechanical work using a working fluid, typically water or steam.
Efficiency of Rankine Cycle
Heat is absorbed at constant pressure in boiler and expelled at constant pressure in condenser.
Since Enthalpy is equal to truth heat absorbed under constant-pressure conditions:
e = 1 - Qc/Qh = 1 - (H4 - H1)/ (H3 - H2) = 1 - (H4 - H1) / (H3 - H1)
Throttling Process (Joule-Thomson process)
Fluid is pushed through a porous plug and then expands into a region of lower pressure.
Since there is no head flow during process:
Uf- Ui = Q + W = 0 + Wleft + Wright
Wleft is positive, Wright is negative
Rewrite as: Uf - Ui = PiVi - PfVf
so Uf + PfVf = Ui + PiVi —> Hf = Hi
Purpose is to cool the fluid to below the temperature of cold reservoir so it can absorb heat as required.
If fluid is an ideal gas:
H = U + PV = f/2 NkT + NkT = (f+2)/2 NkT
Since Enthalpy is constant:
H = U potential + U kinetic + PV
and
COP = H1 - H3 / H2 - H1
