Carnot Cycle Efficiency & COP

The Performance Ceiling for Heat Engines

We've already established that the most efficient engine possible is a reversible one, the Carnot engine. Its efficiency depends only on the absolute temperatures of the hot reservoir ($T_H$) it takes heat from and the cold reservoir ($T_L$) it dumps waste heat into.

${\eta }_{max}={\eta }_{Carnot}=1-\frac{T_L}{T_H}$

This formula gives us the maximum possible efficiency for any heat engine operating between those two temperatures. It doesn't matter if it's a steam engine, a car engine, or a futuristic fusion-powered engine. None can beat this limit.

This leads to a simple but powerful way to evaluate any claim about a heat engine:

1. Calculate the Carnot efficiency for the given operating temperatures.
2. Compare the claimed efficiency to the Carnot efficiency.

• If ${\eta }_{claimed}<{\eta }_{Carnot}$, the engine is possible but irreversible (like all real engines).
• If ${\eta }_{claimed}={\eta }_{Carnot}$, the engine is theoretically possible and reversible.
• If ${\eta }_{claimed}>{\eta }_{Carnot}$, the engine is impossible. It violates the Second Law of Thermodynamics.

This video provides a nice recap of this idea and sets the stage for some examples.

Now, let's put this into practice.

Worked Problem 1: Evaluating an Inventor's Claim

An inventor presents a heat engine that operates between a geothermal source at 150°C and the ambient air at 20°C. The engine produces 10 kW of work while taking in 40 kW of heat. Is this engine possible?

• Step 1: Calculate the actual thermal efficiency of the inventor's engine.
  The definition of thermal efficiency is what you get divided by what you pay for.
  ${\eta }_{th}=\frac{W_{net}}{Q_H}=\frac{10\text{ kW}}{40\text{ kW}}=0.25\text{ or }25\%$

• Step 2: Calculate the maximum possible (Carnot) efficiency for these temperatures.
  First, and most importantly, we must convert the temperatures to Kelvin.

  • $T_H=150°C+273.15=423.15$ K
  • $T_L=20°C+273.15=293.15$ K

  Now, apply the Carnot formula:
  ${\eta }_{max}=1-\frac{T_L}{T_H}=1-\frac{293.15\text{ K}}{423.15\text{ K}}=1-0.6928\approx 0.307$
  So the maximum possible efficiency is 30.7%.

• Step 3: Compare the actual efficiency to the maximum efficiency.
  The engine's actual efficiency (25%) is less than the maximum theoretical efficiency (30.7%).
  $25\%<30.7\%$

• Conclusion: Yes, the engine is possible. Since its efficiency is less than the Carnot limit, it's an irreversible engine, just like every real engine in the world.

Flipping the Engine Around: Refrigerators and Heat Pumps

So far, we've focused on using heat to create work. But what if we want to do the opposite? What if we use work to move heat from a cold place to a warm place? This is exactly what your refrigerator and air conditioner do.

Imagine taking a heat engine and running it backwards.

• Instead of producing work, we input work ($W_{in}$).
• Instead of taking heat from a hot source, we dump heat to a hot source ($Q_H$).
• Instead of dumping heat to a cold sink, we take heat from a cold sink ($Q_L$).

This is the fundamental principle behind refrigeration.

Notice that refrigerators and heat pumps are physically the same device; their "purpose" is just different.

• A refrigerator or air conditioner's goal is to keep the cold space cold. Its desired effect is the heat it removes from the cold reservoir, $Q_L$.
• A heat pump's goal is to keep the warm space warm. Its desired effect is the heat it delivers to the hot reservoir, $Q_H$. (Think of it as an air conditioner running in reverse in winter to heat your house).

A New Metric: Coefficient of Performance (COP)

For these devices, the term "efficiency" doesn't quite fit. Why? Because the amount of heat you move can be much larger than the work you put in. An efficiency greater than 100% doesn't make sense.

So, we use a different metric called the Coefficient of Performance (COP). The principle is the same: what you get divided by what you pay for.

For a Refrigerator/AC:
You want to remove heat ($Q_L$). You pay for the work ($W_{in}$).
$CO{P}_R=\frac{\text{Desired Output}}{\text{Required Input}}=\frac{Q_L}{W_{in}}$

For a Heat Pump:
You want to deliver heat ($Q_H$). You pay for the work ($W_{in}$).
$CO{P}_{HP}=\frac{\text{Desired Output}}{\text{Required Input}}=\frac{Q_H}{W_{in}}$

From the First Law of Thermodynamics, the energy must balance. For one cycle of the device, the heat we dump to the hot side ($Q_H$) must equal the heat we took from the cold side ($Q_L$) plus the work we put in ($W_{in}$).
$Q_H=Q_L+W_{in}\text{or}{W}_{in}=Q_H-Q_L$

Using this, we can write the COP formulas in terms of just the heat transfers:
$CO{P}_R=\frac{Q_L}{Q_H-Q_L}$
$CO{P}_{HP}=\frac{Q_H}{Q_H-Q_L}$

And look at this neat relationship:
$CO{P}_{HP}=\frac{Q_L+W_{in}}{W_{in}}=\frac{Q_L}{W_{in}}+\frac{W_{in}}{W_{in}}=CO{P}_R+1$
This makes perfect sense! The heat delivered by a heat pump is always exactly one unit of work more than the heat removed by the equivalent refrigerator.

The Best Possible Refrigerator: The Carnot COP

Just as no engine can be more efficient than a Carnot engine, no refrigerator or heat pump can have a higher COP than one based on a reversed Carnot cycle.

How do we find this maximum COP? We use the same magic trick as before! For a reversible (Carnot) cycle, we know:
$\frac{Q_L}{Q_H}=\frac{T_L}{T_H}$

Let's substitute this into our COP equations. To make it a bit cleaner, we can divide the top and bottom of the $CO{P}_R$ equation by $Q_H$, and the $CO{P}_{HP}$ equation by $Q_L$. Let's do the refrigerator first:

$CO{P}_{R,max}=\frac{Q_L}{Q_H-Q_L}=\frac{Q_L/Q_H}{1-Q_L/Q_H}$
Now, substitute the temperature ratio:
$CO{P}_{R,max}=\frac{T_L/T_H}{1-T_L/T_H}=\frac{T_L/T_H}{(T_H-T_L)/T_H}$
$\mathbf{CO}{\mathbf{P}}_{\mathbf{R},\mathbf{max}}=\frac{{\mathbf{T}}_{\mathbf{L}}}{{\mathbf{T}}_{\mathbf{H}}-{\mathbf{T}}_{\mathbf{L}}}$

And for the heat pump, we know $CO{P}_{HP,max}=CO{P}_{R,max}+1$, but let's derive it for practice:
$CO{P}_{HP,max}=\frac{Q_H}{Q_H-Q_L}=\frac{1}{1-Q_L/Q_H}$
Substitute the temperature ratio:
$CO{P}_{HP,max}=\frac{1}{1-T_L/T_H}=\frac{1}{(T_H-T_L)/T_H}$
$\mathbf{CO}{\mathbf{P}}_{\mathbf{HP},\mathbf{max}}=\frac{{\mathbf{T}}_{\mathbf{H}}}{{\mathbf{T}}_{\mathbf{H}}-{\mathbf{T}}_{\mathbf{L}}}$

Crucial Point: Just like with Carnot efficiency, you MUST use absolute temperatures (Kelvin or Rankine) for these formulas to work.

These formulas tell us something very important about the real world. The performance of a refrigerator or heat pump gets worse as the temperature difference ($T_H-T_L$) gets bigger.

• Your kitchen fridge runs more on a hot summer day (large $T_H-T_L$).
• A heat pump works great in a mild climate but struggles to heat a house when it's freezing outside (large $T_H-T_L$).

This article provides an excellent explanation of these concepts, especially for heat pumps.

Worked Problem 2: Maximum Performance of a Freezer

A freezer is designed to keep its contents at -18°C while operating in a room that is 22°C.
(a) What is the maximum possible COP for this freezer?
(b) For every 1 kJ of electrical energy (work) the freezer consumes, what is the maximum amount of heat it can remove from its contents?

• Step 1: Identify temperatures and convert to Kelvin.
  The "cold" space is inside the freezer, and the "hot" space is the room.

  • $T_L=-18°C+273.15=255.15$ K
  • $T_H=22°C+273.15=295.15$ K

• Step 2 (a): Calculate the maximum COP for a refrigerator.
  $CO{P}_{R,max}=\frac{T_L}{T_H-T_L}=\frac{255.15\text{ K}}{295.15\text{ K}-255.15\text{ K}}=\frac{255.15}{40}=6.38$
  The absolute best this freezer could ever be is a COP of 6.38. A real freezer would have a lower COP, maybe around 2.5 to 4.

• Step 2 (b): Use the definition of COP to find the heat removed.
  We know $CO{P}_R=Q_L/W_{in}$. To find the maximum heat removed, we assume the freezer is perfect and has the maximum COP.
  $Q_{L,max}=CO{P}_{R,max}\times W_{in}$
  $Q_{L,max}=6.38\times 1\text{ kJ}=6.38\text{ kJ}$
  This is amazing when you think about it. For every 1 unit of high-quality electrical energy we use, we can pump up to 6.38 units of low-quality thermal energy out of the freezer. This is not a violation of energy conservation! The extra 5.38 kJ is extracted from the cold freezer interior. The total energy dumped into the room would be $Q_H=Q_L+W_{in}=6.38+1=7.38$ kJ.

Key Takeaways for Today

What a productive session! We've taken the theoretical foundation of the Carnot cycle and applied it to a full range of real-world devices.

Here are the main points:

• The Carnot efficiency (${\eta }_{th,rev}=1-T_L/T_H$) is the absolute performance ceiling for any heat engine.
• Refrigerators and heat pumps are reversed heat engines that use work to move heat from cold to hot.
• Their performance is measured by the Coefficient of Performance (COP), which is (what you get) / (what you pay for).
• The maximum possible COP for a refrigerator is $CO{P}_{R,rev}=\frac{T_L}{T_H-T_L}$.
• The maximum possible COP for a heat pump is $CO{P}_{HP,rev}=\frac{T_H}{T_H-T_L}$.
• In all these formulas, it is absolutely critical to use absolute temperatures (Kelvin or Rankine).

Preview of the Next Lesson

We've now fully explored the limits of performance set by the Second Law. But we're left with a deeper question. We keep saying that real processes are "irreversible" and that's why they can't reach the Carnot limit. But what is irreversibility, really? Why does friction generate heat that you can't undo? Why does heat always flow from hot to cold spontaneously?

To answer these questions, we need to introduce one of the most profound and fascinating properties in all of science: Entropy. In our next lesson, we will explore what entropy means intuitively and how it gives us a way to measure the "direction of time" and the irreversibility of processes. It's the concept that underpins everything we've just learned. See you then