Engines & Refrigerators

Question 1

Heat engine

Answer 1

Any device that absorbs heat and converts part of that energy into work.

Question 2

Reservoir

Answer 2

Anything that is so large that its temperature doesn’t change noticeably when heat enters or leaves.

Question 3

Efficiency

Answer 3

Benefit/Cost Ratio

e= Wnet/Qh

Question 4

What does first law of thermodynamics tell us about heat engines?

How can we use this to rewrite the efficiency equation?

Answer 4

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

Question 5

What does the second law of thermodynamics tell us about heat engines?

What can we conclude about efficiency?

Answer 5

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

Question 6

How does efficiency relate to the temperature of the reservoirs?

Answer 6

The smaller the ratio Tc/Th, the more efficient the engine will be.

Question 7

Summary of what first two laws of thermodynamics tell us about heat engines.

Answer 7

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.

Question 8

Carnot Cycle

Answer 8

Theoretical thermodynamic cycle that provides maximum possible efficiency for heat engine.

T gas =~ Th

Steps:

1. Isothermal Expansion at Th while absorbing heat.
2. Adiabatic expansion to Tc.
3. Isothermal compression at Tc while expelling heat.
4. Adiabatic compression back to Th.

Question 9

Refrigerators

Answer 9

Heat engine operated in reverse.

Question 10

Coefficient of Performance (COP)

Answer 10

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

Question 11

What does second law of thermodynamics tell us about refrigerators?

What does this tell us about the COP?

Answer 11

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)

Question 12

Relationship between COP and temperature of reservoirs.

Answer 12

COP is largest when Th and Tc are not very different.

Question 13

How can refrigerator have maximum possible COP?

Answer 13

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.

Question 14

Efficiency of Carnot Cycle

Answer 14

e = 1- Tc/Th

Question 15

Otto Cycle

Answer 15

Thermodynamic cycle that describes the processes in a four-stroke internal combustion engine, such as those used in most gasoline-powered vehicles.

Question 16

Efficiency of Otto Cycle

Answer 16

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.

Question 17

Rankine Cycle

Answer 17

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.

Question 18

Efficiency of Rankine Cycle

Answer 18

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)

Question 19

Throttling Process (Joule-Thomson process)

Answer 19

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