Understanding Thermal Efficiency and COP

The Big Idea: What You Get vs. What You Pay

Before we jump into any formulas, let's start with a simple, universal idea. Any time we talk about "efficiency" or "performance," we're really asking:

"How much of what I want did I get, for the amount of what I had to pay?"

Think about a car. The "performance" we often care about is its fuel economy.

• What you get: Miles driven.
• What you pay: Gallons of expensive fuel.
• Performance Metric: Miles per gallon (What you get / What you pay).

This simple ratio is the key to everything we're doing today. In thermodynamics, our "currency" is energy. The "cost" is always the high-value energy we have to provide (like the heat from burning fuel, or the electricity to run a compressor). The "desired effect" is the job we want done (getting work out, cooling a room, or heating a room).

This is a great framework to keep in mind. As a resource from Durham College puts it, all these performance measures follow the general formula: Performance Measurement = Desired Effect / Cost. Let's apply this to our three devices.

Heat Engine Performance: Thermal Efficiency (${\eta }_{th}$)

First up, the heat engine—the hero of power plants and car engines. Its job is to take heat and produce work.

• What you get (Desired Effect): Net work output, $W_{net}$. This is the energy that turns the wheels of your car or the generator in a power plant.
• What you pay (Cost): The heat supplied from the high-temperature source, $Q_H$. This is the energy released by burning gasoline or coal.

So, the performance metric, which we call thermal efficiency (using the Greek letter eta, $\eta$), is:

${\eta }_{th}=\frac{\text{Desired Output}}{\text{Required Input}}=\frac{W_{net}}{Q_H}$

Remember from our discussion on heat engines that not all the heat Q_H can be converted to work. Some heat, Q_L, must be rejected to the cold reservoir (the environment). The First Law (energy conservation) tells us:

$W_{net}=Q_H-Q_L$

Now we can substitute this into our efficiency formula:

${\eta }_{th}=\frac{Q_H-Q_L}{Q_H}=1-\frac{Q_L}{Q_H}$

This is a hugely important formula. It tells you the fraction of the heat you supplied that was successfully converted into useful work. Because the Kelvin-Planck statement guarantees that $Q_L$ must be greater than zero for any cycle, the thermal efficiency of a heat engine is always less than 1 (or 100%). You always lose some heat to the environment.

Refrigerator Performance: Coefficient of Performance (COP)

Now let's look at a refrigerator or an air conditioner. Here, our goal is different. We want to make a cold space even colder.

• What you get (Desired Effect): The heat removed from the cold space, $Q_L$. This is the "cooling effect."
• What you pay (Cost): The work you have to supply to the compressor, $W_{in}$. This is the electricity your fridge uses.

So, its performance metric is:

$\text{Performance}=\frac{\text{Desired Output}}{\text{Required Input}}=\frac{Q_L}{W_{in}}$

Now, we don't call this "efficiency." Why? Because the value can be, and often is, greater than 1! This would be confusing if we used the word "efficiency," which we associate with being less than 100%. So, engineers use a different name: the Coefficient of Performance, or COP. For a refrigerator, we denote it as $CO{P}_R$.

$CO{P}_R=\frac{Q_L}{W_{in}}$

Wait, how can it be greater than 1? Are we breaking the laws of physics?

Not at all! This is a beautiful point that's often misunderstood. A great article from the blog "Do the Math" explains it perfectly: a refrigerator doesn't create cold; it moves heat. It takes much less energy to move something that already exists than to create it from scratch.

Think of it like this: Imagine you have to get a huge boulder off your lawn.

• Option 1 (Like an inefficient heater): Get a giant hammer and smash the boulder into dust to get rid of it. This takes a massive amount of energy.
• Option 2 (Like a refrigerator): Use a lever and a bit of effort (work) to roll the boulder off your lawn and onto your neighbor's (the kitchen). You've achieved your goal (boulder is gone) with far less energy than it would take to destroy it.

The refrigerator is just moving the heat energy from inside the fridge to outside in the kitchen. The work you put in is just the "push" needed to roll that energy "uphill." Because you're just moving energy, not creating it, the amount of heat you move ($Q_L$) can easily be larger than the work you put in ($W_{in}$).

The following video gives a great summary of this concept.

Just like before, we can use the First Law for a fridge: $W_{in}=Q_H-Q_L$. Substituting this gives an alternative formula for $CO{P}_R$:

$CO{P}_R=\frac{Q_L}{Q_H-Q_L}$

Heat Pump Performance: Coefficient of Performance (COP)

A heat pump is the twin of a refrigerator. It's the exact same machine, but our goal is flipped. In winter, we use a heat pump not to cool the outdoors, but to heat our house.

• What you get (Desired Effect): The heat delivered to the hot space (your home), $Q_H$. This is the "heating effect."
• What you pay (Cost): The same as the fridge—the work supplied to the compressor, $W_{in}$.

So, the Coefficient of Performance for a heat pump, $CO{P}_{HP}$, is:

$CO{P}_{HP}=\frac{\text{Desired Output}}{\text{Required Input}}=\frac{Q_H}{W_{in}}$

Again, we can substitute $W_{in}=Q_H-Q_L$ to get:

$CO{P}_{HP}=\frac{Q_H}{Q_H-Q_L}$

Like $CO{P}_R$, the $CO{P}_{HP}$ is also typically greater than 1. For every 1 Joule of electricity you pay for, you might get 3 or 4 Joules of heat delivered to your house! The extra 2-3 Joules are "free" energy that the pump scavenged from the cold outside air.

A Neat Connection

Let's look at the formulas for $CO{P}_R$ and $CO{P}_{HP}$ side-by-side. Is there a relationship?

$CO{P}_R=\frac{Q_L}{W_{in}}$

$CO{P}_{HP}=\frac{Q_H}{W_{in}}$

We know that $Q_H=Q_L+W_{in}$. Let's plug this into the heat pump formula:

$CO{P}_{HP}=\frac{Q_L+W_{in}}{W_{in}}=\frac{Q_L}{W_{in}}+\frac{W_{in}}{W_{in}}$

Look at that! The first term is just $CO{P}_R$, and the second term is 1. So we get a very simple and powerful relationship:

$CO{P}_{HP}=CO{P}_R+1$

This makes perfect sense. The total heat you dump into the hot room ($Q_H$) is the heat you took from the cold space ($Q_L$) plus the work you paid for ($W_{in}$), which also gets converted to heat and dumped in the room. So the heating effect is always exactly one "work unit" bigger than the cooling effect.

Visual Summary

This diagram brings it all together. It's a fantastic visual reference for the energy flows and goals of each device.

Practice Problems

Let's solidify these concepts with a couple of problems.

Conceptual Questions

1. Why must the thermal efficiency of a heat engine be less than 1, while the COP of a heat pump is normally greater than 1?
2. You use a window air conditioner to cool your room on a summer day. Will its $CO{P}_R$ be higher on a mildly warm day (e.g., 27°C outside) or a very hot day (e.g., 37°C outside)? Why?

Numerical Problems

1. Problem: A small gasoline engine takes in 10,000 J of heat from the combustion of fuel and produces 2,500 J of net work.
   a) How much heat does it reject to the atmosphere?
   b) What is its thermal efficiency?

   Solution:

   • Identify the givens:
     • Heat input, $Q_H=10,000$ J.
     • Net work output, $W_{net}=2,500$ J.
   • a) Find the rejected heat, $Q_L$:
     • From the First Law, $W_{net}=Q_H-Q_L$.
     • Rearranging gives $Q_L=Q_H-W_{net}$.
     • $Q_L=10,000\text{J}-2,500\text{J}=7,500\text{J}$.
     • The engine rejects 7,500 J of heat.
   • b) Find the thermal efficiency, ${\eta }_{th}$:
     • The formula is ${\eta }_{th}=W_{net}/Q_H$.
     • ${\eta }_{th}=2,500\text{J}/10,000\text{J}=0.25$.
     • Answer: The thermal efficiency is 0.25, or 25%.

2. Problem: A residential heat pump is used to keep a house at 22°C. On a winter day when it's cold outside, the pump delivers 15,000 kJ/h of heat to the house while consuming 1.2 kW of electrical power.
   a) What is the $CO{P}_{HP}$ of the heat pump?
   b) At what rate is heat being absorbed from the cold outdoor air, in kJ/h?
   c) If this same device were used as an air conditioner in the summer, what would its $CO{P}_R$ be?

   Solution:

   • Identify the givens and standardize units:
     • Heat delivered to hot space, ${\dot{Q}}_H=15,000$ kJ/h. (The dot signifies a rate).
     • Work input, ${\dot{W}}_{in}=1.2$ kW.
     • We need consistent units. Let's convert kW to kJ/h.
     • Recall that 1 W = 1 J/s. So 1 kW = 1 kJ/s.
     • ${\dot{W}}_{in}=1.2\frac{\text{kJ}}{\text{s}}\times \frac{3600\text{s}}{1\text{h}}=4,320$ kJ/h.
   • a) Find the $CO{P}_{HP}$:
     • $CO{P}_{HP}=\frac{{\dot{Q}}_H}{{\dot{W}}_{in}}=\frac{15,000\text{kJ/h}}{4,320\text{kJ/h}}\approx 3.47$.
     • Answer: The Coefficient of Performance for heating is 3.47.
   • b) Find the rate of heat absorption, ${\dot{Q}}_L$:
     • The energy balance is ${\dot{Q}}_H={\dot{Q}}_L+{\dot{W}}_{in}$.
     • Rearranging gives ${\dot{Q}}_L={\dot{Q}}_H-{\dot{W}}_{in}$.
     • ${\dot{Q}}_L=15,000\text{kJ/h}-4,320\text{kJ/h}=10,680$ kJ/h.
     • Answer: The pump absorbs 10,680 kJ/h from the outside air.
   • c) Find the $CO{P}_R$:
     • We can use the simple relation: $CO{P}_{HP}=CO{P}_R+1$.
     • $CO{P}_R=CO{P}_{HP}-1=3.47-1=2.47$.
     • Answer: As an air conditioner, its COP would be 2.47.

Conclusion

Excellent work today! We've taken the abstract rules of the Second Law and attached concrete numbers to them. You now have the tools to grade the performance of any heat engine, refrigerator, or heat pump.

Here are the key takeaways:

• Performance is always measured as What You Get / What You Pay.
• For a heat engine, this is Thermal Efficiency, ${\eta }_{th}=W_{net}/Q_H$, which is always less than 1.
• For a refrigerator, this is the Coefficient of Performance, $CO{P}_R=Q_L/W_{in}$, which measures the cooling effect.
• For a heat pump, this is also the Coefficient of Performance, $CO{P}_{HP}=Q_H/W_{in}$, which measures the heating effect.
• These performance metrics are not just numbers; they tell a story about how effectively a device is using high-value energy to accomplish a task. And for the same device, $CO{P}_{HP}=CO{P}_R+1$.

Preview of the Next Lesson

We now know how to measure performance. But this raises a fascinating question: What is the best possible performance? Is there a theoretical limit, an ultimate speed limit for efficiency and COP that no engineer can ever beat?

In our next lesson, we will answer that question. We'll explore the idea of perfect, idealized processes called reversible processes and introduce a legendary theoretical cycle called the Carnot Cycle. This cycle represents the absolute pinnacle of performance that the laws of thermodynamics will allow.