Before you tackle this tutorial, you might wish to look at Thermal Physics Tutorial 4.
The Second Law of Thermodynamics states that it is impossible for any heat engine to be 100 % efficient:
No process is possible which results in the extraction of an amount of heat from a reservoir and its conversion to an equal amount of mechanical work.
The theory behind this is that entropy increases. In other words all processes tend towards chaos (which might explain my physics lessons when I worked in schools). If you drop a pack of cards, they will scatter and the chances of their landing in a meaningful order are very small indeed.
Heat is work and work's a curse,
And all the heat in the Universe
Is gonna cool down.
That will mean no more work,
And there'll be perfect peace.
That's entropy, man!
[Michael Flanders and Donald Swan]
Most energy is lost to the surroundings as low grade heat. We can show this in the diagram below:
In this diagram, called a Sankey Diagram, we can see that of 72 kW of power from the fuel, only 9 kW are used in actually driving a car along a road. The rest is lost as low grade heat. As we said before, getting energy out of heat is remarkably difficult.
All heat engines work by extracting mechanical energy from a temperature gradient. A heat engine has to operate between the hot reservoir and the cold reservoir to satisfy the Second Law of Thermodynamics. Heat flows from hot to cold, never the other way round:
Heat won't pass from a cooler to a hotter.
You can try it if you like,
But you far better notta,
Because the cold in the cooler
Will get hotter as a ruler,
And that's a physical law!
[Michael Flanders and Donald Swan]
We can show the heat flowing from a hot reservoir through a heat engine to a cold reservoir.
All heat engines give up their energy to a cold reservoir. We can define the terms used on the diagram:
Qin = the heat flow from the hot reservoir to the engine
Qout is the heat flow from the engine to the cold reservoir.
The work done by the heat engine is the difference between Qin and Qout.
W = Qin - Qout
We can write down an efficiency relationships from this:
A car uses energy from the fuel at a rate of 72 kJ s-1. It uses 9 kJ s-1 to move along the road. How much heat is lost as waste? What is the efficiency?
An ideal heat engine takes a quantity of heat Qin from a hot reservoir of temperature TH and sends a quantity of heat Qout as waste to a cold reservoir of temperature TC. It can be shown that:
We can rewrite the efficiency equation:
This can be rearranged to give us a useful relationship:
The temperature must always be in Kelvin. If we set TC at 0 K, we could have a heat engine that was 100 % efficient, but as we can't get down to 0 K, forget it! However we can make heat engines more efficient by making the difference between that hot reservoir and the cold reservoir as big as possible. In a power station, the steam coming from the boiler is at about 400 oC, while for the cold reservoir, water at about 10 oC is used.
Note that some text books use the code h for efficiency. The strange looking symbol h is "eta", a Greek letter long 'ē'.
The picture below shows the condenser of a steam turbine:
In this case, steam enters the condenser at a temperature of 110 o C and the cooling water temperature is about 15 oC.
What is the maximum possible efficiency of an engine using steam at a temperature of 100 oC on a day when the temperature is 24 oC?
The dipping duck in the photograph below is a heat engine:
What do you think are the hot and cold reservoirs?
There are limitations to the theoretical efficiency of any heat engine.
TH cannot be too high, otherwise components could melt;
TC will be in the normal range of atmospheric temperatures.
Careful analysis of the cycle of an engine can help improve efficiency;
Careful design of ports so that gas can get in and out with the minimum resistance.
Friction cannot be eliminated. Lubrication reduces friction in bearings, but there is some viscous drag with the oils themselves.
A small geothermal power station in Iceland pumps cold water into hot rock strata far below the Earth’s surface to be heated and returned at a constant temperature of 87 °C. The power station uses the hot water as the heat source for a heat engine which rejects energy to the much colder sea water near the station.
(a) When the temperature of the sea water is 7 °C the power output from the heat engine is 5.0MW. Calculate:
(i) the maximum theoretical efficiency of the heat engine,
(ii) the rate at which heat energy must be transferred from the hot water if the engine works at the maximum theoretical efficiency,
(iii) the rate at which energy must be transferred to the sea water under these conditions.
(b) The power station produces electrical power with an overall efficiency which is much lower than the maximum theoretical efficiency of the heat engine. Give two reasons for this lower efficiency.
(c) The overall efficiency of an oil-fired power plant of similar size to the geothermal station is over four times as great. Suggest one reason, other than less pollution, why the geothermal source was preferred for the power station.
(AQA Past question)
A real engine does less work for a given heat transfer Qin. Additionally, if we do a job of work on the real engine, we would not get Qin back. The real engine is much less efficient than the reversible engine. If you are driving a car downhill in gear, the engine acts as a brake. It will not produce the same heat flow as it would if driving the car along a level road. Just as well, otherwise you would boil the engine going downhill. Not a good idea.
We can reverse the process, pumping energy from the cold side to the hot side:
Such a device is called a heat pump. The cold reservoir is the environment, which can be the ground, the air, or water. The hot reservoir is the house. In this case the heat pump is shifting the heat energy from the cold to the hot. The cold reservoir is very large; the hot reservoir is small. The idea is shown in the diagram.
The pump moves the compressed gas which is hot. It releases the heat into the hot reservoir, say a room. The gas continues around the loop, and expands cooling down. As it expands, heat flows from the cold reservoir, which may be the atmosphere, or the ground, or water. The gas is therefore warmed up by the heat flow into it. It is then compressed again, and the cycle continues.
Since the heat flow comes from the environment, which is the cold reservoir, a much greater amount of heat can be pumped into the room than the power of the compressor motor. The heat flow possible from the environment is almost limitless.
This is a good way of heating houses sustainably, but the equipment is expensive and planning permission may be needed.
An important quantity is the coefficient of performance. This is:
The ratio of the heat transferred to the hot reservoir to the input energy required to pump the heat from the cold to the hot.
For a heat pump, the coefficient of performance is given by:
The coefficient of performance is a ratio, so it has no units. For a heat pump, the coefficient of performance can never be less than 1.
The output temperature of a heat pump needs to be 30 o C, while the ground temperature is 0 oC.
(a) Calculate the coefficient of performance.
(b) The motor of the heat pump has a power of 1.5 kW. What is the power of such a heat pump heater?
Convert Celsius temperatures to Kelvin:
30 oC = 303 K
0 oC = 273 K
CoPHP = 303 K ÷ (303 K - 273 K) = 10.1
QH = 10.1 × 1.5 kW = 15.15 kW = 15 kW (2 s.f.)
In reality, the coefficient of performance of such a situation is rather less than the theoretical maximum.
The picture below shows a heat pump:
By Kristoferb at English Wikipedia, CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=10795550
The heat pump provides about 3 to 4 times as heat than a simple resistance heater. However the installations are rather expensive.
In a refrigerator, the cold reservoir is small, and heat is pumped from the cold reservoir to the hot, which is the environment (i.e. the kitchen where the fridge stands). While this may appear to contravene the Second Law of Thermodynamics, remember that the source of energy is the real hot reservoir which is the boiler in the power station that produces the energy to turn the turbine to generate the electricity.
The compressor motor pumps and compresses a coolant. The compressed coolant goes to a heat exchanger outside on the back, where heat is transferred by convection into the room. The coolant then is sprayed into an expansion chamber in the ice compartment of the fridge. The liquid evaporates, which requires energy as latent heat. The energy required is taken from the inside of the cabinet, which is heavily insulated to prevent heat flowing from the room. The coolant gas is taken back to the pump.
For a refrigerator, the coefficient of performance is given by:
A way of telling which of the coefficient of performance formulae to use is to consider the target reservoir. If it's a room being heated, we use the CoPHP formula. If it's space that's being cooled, then we use the CoPref formula.
We can work out the theoretical coefficient of performance for a refrigerator.
A refrigerator can maintain the contents at a temperature of 4 oC while the room has a temperature of 30 oC.
(a) Calculate the coefficient of performance.
(b) The motor has a power of 250 W. What is the rate at which heat is transferred from the cabinet to the room?
Convert the Celsius temperatures to Kelvin:
4 oC = 277 K;
30 oC = 303 K
CoPref = 277 K ÷ (303 K - 277 K) = 10.7
(b) QC = CoPref × W = 10.7 × 250 W = 2663 W = 2700 W (2 s.f.)
(a) Explain what is meant by the coefficient of performance of a heat pump.
(b) The box labelled E in Figure 3 shows a diagram of a combined heat and power scheme. The scheme provides electrical energy W from an engine-driven generator and heat Q1 for buildings situated near to the generator. Some of the electrical energy is used to drive the heat pump shown in the box labelled P. Output Q2 is also used to heat the buildings.
You may assume that the engine runs at its maximum theoretical efficiency and that the electrical generator is 100% efficient. The output power of the engine-driven generator is 80kW.
(i) The fuel used in the engine (E) is propane of calorific value 49 MJ kg–1. Calculate the rate of flow of propane into the engine. State an appropriate unit.
(ii) The heat pump has a coefficient of performance of 2.6. The power supplied by the electrical generator to the heat pump (P) is 16 kW. Calculate the total rate at which energy is available for heating from both the engine and heat pump.
(iii) The conversion of electrical energy to heat is nearly 100% efficient. Explain why the designer has proposed installing a heat pump rather than an electrical heater to provide the additional heat Q2.
(AQA past question June 2014 Q4)
Other Laws of Thermodynamics (Extension)
Thermal equilibrium happens when two bodies are are brought into contact with each other, there is no heat flow. This means that the temperature of the body remains the same. Consider the three bodies, A, B, and C below. The brown arrows show the heat flows between the three bodies.
According to the Zeroth Law of Thermodynamics, if A and B are in thermal equilibrium, and A and C are in thermal equilibrium, then B and C must be in thermal equilibrium.
The law was called this because the First and Second laws of Thermodynamics depend on it. It was discovered later than the First and Second Laws.
The Third Law of Thermodynamics states that:
The entropy of a perfect crystal is exactly zero when the temperature is absolute zero.