Biological Physics/Heat Engines, Refrigerators, and Efficiency

Efficiency

A general definition of efficiency is the benefit over cost.

So, since efficiency = benefit / cost, to be 100% efficient, the benefit / cost needs to be equal to 1.

We know that ${\displaystyle q_{H}=w+q_{C}}$  for a heat engine (see figure below). Rearranging give us ${\displaystyle w=q_{H}-q_{C}}$ . The efficiency is the work we get out of the engine in regards to the heat put into the engine. efficiency ${\displaystyle efficiency={\frac {w}{q_{H}}}={\frac {(q_{H}-q_{C})}{q_{H}}}={\frac {1-q_{C}}{q_{H}}}}$ . Well, we would also like to show that ${\displaystyle efficiency={\frac {1-q_{C}}{q_{H}}}\leq {\frac {1-T_{C}}{T_{H}}}.}$  This means that the efficiency is going to be related to the hot and cold temperatures, as well as the heat, ${\displaystyle q}$ .

Using the ideal gas equation we know that ${\displaystyle PV=Nk_{B}T}$ . And based on the Carnot Cycle (Figure #) we know that ${\displaystyle {\frac {q_{c}}{q_{H}}}={\frac {(Nk_{B}T_{C}ln(V_{4}/V_{3}))}{(Nk_{B}T_{H}ln(V_{2}/V_{1}))}}}$ . The ${\displaystyle Nk_{B}}$  term cancels leaving only ${\displaystyle {\frac {q_{C}}{q_{H}}}={\frac {T_{C}ln(V_{4}/V_{3})}{T_{H}ln(V_{2}/V_{1})}}}$ .

To show that this relationship lets us relate ${\displaystyle {\frac {q_{C}}{q_{H}}}={\frac {T_{C}}{T_{H}}}}$  we look at ${\displaystyle \Delta U=w={\frac {f}{2}}Nk_{B}\Delta T}$ . This is true for adiabatic because Q = 0. So ${\displaystyle \Delta U=-P\Delta V}$ . There are ΔT and ΔV so it is necessary to make small incremental changes which leads to integration. (The numbers, #.), indicate steps in the derivation).

1.) ${\displaystyle {\frac {f}{2}}Nk_{B}\Delta T=-{\frac {Nk_{B}T}{V}}\Delta V}$

2.) ${\displaystyle {\frac {f}{2}}{\frac {\Delta T}{T}}=-{\frac {\Delta V}{V}}}$

Then, we integrate both sides:

3.) ${\displaystyle \int _{T_{i}}^{T_{f}}{\frac {f}{2}}{\frac {1}{T}}dT=-\int _{V_{i}}^{V_{f}}{\frac {1}{V}}dV}$

4.) ${\displaystyle {\frac {f}{2}}ln(T_{f}/T_{i})=ln(V_{i}/V_{f})}$  (Note the sign change that leads to the flipping of the terms inside the natural log on the left side of the equation).

If we exponentiate both sides, then the terms in the ln() are freed.

5.) ${\displaystyle e^{{\frac {f}{2}}ln(T_{f}/T_{i})}=e^{ln(V_{i}/V_{f})}}$

6.) ${\displaystyle e^{ln(T_{f}/T_{i})^{\frac {f}{2}}}=e^{ln(V_{i}/V_{f})}}$

7.) ${\displaystyle {\frac {T_{f}^{\frac {f}{2}}}{T_{i}^{\frac {f}{2}}}}={\frac {V_{i}}{V_{f}}}}$

Group the final and initial terms:

8.) ${\displaystyle T_{f}^{\frac {f}{2}}V_{f}=T_{i}^{\frac {f}{2}}V_{i}}$

We know that ${\displaystyle T^{\frac {f}{2}}V}$  is a constant for an adiabatic processes. We can divide the sides because they are equal and cancel ${\displaystyle T_{H}}$  and ${\displaystyle T_{C}}$ .

9.) ${\displaystyle {\frac {T_{H}^{\frac {f}{2}}V_{2}}{T_{H}^{\frac {f}{2}}V_{1}}}={\frac {T_{C}^{\frac {f}{2}}V_{3}}{T_{C}^{\frac {f}{2}}V_{4}}}}$

Which gives:

10.) ${\displaystyle {\frac {V_{2}}{V_{1}}}={\frac {V_{3}}{V_{4}}}}$

If we put these fractions into the original ${\displaystyle {\frac {q_{C}}{q_{H}}}={\frac {T_{C}ln(V_{4}/V_{3})}{T_{H}ln(V_{2}/V_{1})}}}$ , we know that ${\displaystyle {\frac {ln(V_{4}/V_{3})}{ln(V_{2}/V_{1}}}=1}$  based on the equality we just derived.

And, thus, we show that ${\displaystyle {\frac {q_{C}}{q_{H}}}={\frac {T_{C}}{T_{H}}}}$ .

Now we are back to where we began with: ${\displaystyle efficiency=1-{\frac {q_{C}}{q_{H}}}=1-{\frac {T_{C}}{T_{H}}}}$

Example Problems

We can do some examples with these equations. Let's look at the efficiency of boiling water.

T_H = 373 K

T_C = 295 K

${\displaystyle efficiency=1-{\frac {295}{373}}}$

${\displaystyle =1-0.791}$

${\displaystyle =0.21}$

${\displaystyle =21\%}$

So for a steam engine, the very best we could do is 21% efficiency when drawing the heat from steam.

How about an example of a gasoline engine:

T_C = 295 K (An average outdoor temperature) T_H = 1500 K (The combustion temperature of gasoline)

${\displaystyle efficiency=1-{\frac {295}{1500}}}$

${\displaystyle =1-02}$

${\displaystyle =0.8}$

${\displaystyle =80\%}$