TheCarnot heat engineis an imaginary model machine that Carnot devised in 1824 to represent a steam engine.Asimplesteam engineis depicted schematically in Figure 3.1a.
It has a cylinder with a piston connected to a crankshaft by a connecting rod. There is an intake valve through which a boiler can inject high-pressure steam into the cylinder and an exhaust valve through which spent steam can be exhausted into the atmosphere.
This steam engine operates with a two-stroke cycle. The cycle begins with the piston at top dead center (the position of minimum volume in the cylinder) and with the intake
(b)
Working fluid (system) Piston Hot reservoir Cold
reservoir
Connecting rod Crankshaft (a)
Intake valve Steam inlet Steam
exhaust Exhaust valve
Piston Connecting rod Crankshaft
Figure 3.1 Comparison of a Simple Steam Engine and a Carnot Engine (Schematic).
(a) A Simple steam engine. (b) A Carnot heat engine.
valve open. High-pressure steam from the boiler enters the cylinder through the intake valve and pushes on the piston, which turns the crankshaft. When the piston reaches bottom dead center (the position of maximum volume in the cylinder) the intake valve closes and the exhaust valve opens. The inertia of the crankshaft and flywheel pushes the piston back toward top dead center, expelling the spent steam through the exhaust valve. The exhaust valve closes and the intake valve opens when top dead center is reached, and the engine is ready to repeat its cycle.
The Carnot engine is depicted in Figure 3.1b. It operates reversibly, so there can be no friction. The cylinder contains a gaseous “working fluid,” which we define to be the system. The Carnot engine has no valves and the system is closed. To simulate passing steam into and out of the cylinder the Carnot engine allows heat to flow from a “hot reservoir” into its working fluid and exhausts heat into a “cold reservoir” by conduction through the cylinder walls or cylinder head.
The Carnot engine operates on a two-stroke cycle that is called theCarnot cycle. We begin the cycle with the piston at top dead center and with the hot reservoir in contact with the cylinder. We break the expansion stroke into two steps. The first step is an isothermal reversible expansion of the system at the temperature of the hot reservoir.
The final volume of the first step is chosen so that the second step, which is an adiabatic reversible expansion, ends with the system at the temperature of the cold reservoir and with the piston at bottom dead center. The compression stroke is also broken into two steps. The third step of the cyclic process is a reversible isothermal compression with
V1 V4 V2 V
P
P1
P2
P4 P3
V3
(b) Step 1:
Isothermal expansion Step 2:
Adiabatic expansion Step 3:
Isothermal compression Step 4:
Adiabatic compression
V1 V4 V2 V3
V
T
Tc Th
(a)
Step 2:
Adiabatic expansion
Step 3:
Isothermal compression Step 4:
Adiabatic compression
Step 1: Isothermal expansion
Figure 3.2 The Path of the State Point during a Carnot Cycle.(a) In theV–Tplane. (b) In theV–Pplane.
the cylinder in contact with the cold reservoir. This step ends at a volume such that the fourth step, a reversible adiabatic compression, ends with the piston at top dead center and the system at the temperature of the hot reservoir. The engine is now ready to repeat the cycle.
Figure 3.2a shows the path that the state point of the system follows as the engine undergoes one cycle, usingV andT as the state variables. The state at the beginning of each step is labeled with the same number as the step. Figure 3.2b shows the same cycle usingVandPas the state variables.
Since the second and fourth steps of the Carnot cycle are adiabatic,
q2q40 (3.1-2)
For the entire cycle,
qcycleq1+q2+q3+q4q1+q3 (3.1-3) SinceUis a state function and because the cycle begins and ends at the same state,
∆Ucycle0 (3.1-4)
From the first law of thermodynamics,
wcycle∆Ucycle−qcycleqcycle −q1−q3 (3.1-5) The efficiency,ηCarnot, of the Carnot engine is the work done on the surroundings divided by the heat input from the hot reservoir. The heat exhausted at the cold reservoir is wasted and is not included in the efficiency calculation.
ηCarnot wsurr
q1 −wcycle
q1 qcycle
q1 q1+q3
q1 1+q3
q1
(3.1-6) From the Kelvin statement of the second law, the efficiency must be less than unity, so thatq3must be negative. It is not possible to run a Carnot engine without exhausting some heat to a cool reservoir.
ACarnot heat pumpis a Carnot heat engine that is driven backwards by another engine. It removes heat from the cool reservoir and exhausts heat into the hot reservoir.
Figure 3.3 represents a Carnot heat pump cycle, which is the reverse of the cycle of Figure 3.2. The steps are numbered with a prime () and are numbered in the order in which they occur. Since we are considering the same Carnot engine run backwards,
q4 −q1 (3.1-7)
and
q2 −q3 (3.1-8)
V
T
Th
(a)
Step 3:
Adiabatic compression
Step 2:
Isothermal expansion Step 1:
Adiabatic expansion
Step 4: Isothermal compression
Tc
Figure 3.3 The Path of the State Point in theV–TPlane during a Carnot Heat Pump Cycle.
Because the cycles are reversible, the amount of work done on the system in the reverse (heat pump) cycle,w, is equal to the amount of work done on the surroundings in the forward (engine) cycle:
wcycle −wcyclewsurr (3.1-9)
For a heat pump, the output is the heat delivered to the hot reservoir and the input is the work put into the heat pump. The ratio of the output to the input is called thecoefficient of performanceof the heat pump and is denoted byηhp. The coefficient of performance equals the reciprocal of the Carnot efficiency, because the input and output are reversed in their roles as well as their signs.
ηhp q4
wcycle − q4
−q2−q4 q1
q1+q3
1
1+q3/q1 1 ηc
(3.1-10) The Carnot efficiency is always smaller than unity, so the Carnot heat pump coefficient of performance is always greater than unity. The amount of heat delivered to the hot reservoir is always greater than the work put into the heat pump because some heat has been transferred from the cold reservoir to the hot reservoir. There is no violation of the Clausius statement of the second law because the heat pump is driven by another engine. A real heat pump must have a lower coefficient of performance than a reversible heat pump but can easily have a coefficient of performance greater than unity.
No reversible heat engine can have a greater efficiency than the Carnot engine. We prove this by assuming the opposite of what we want to prove and then show that this assumption leads to a contradiction with experimental fact and therefore must be incorrect. Assume that a reversible heat engine does exist with a greater efficiency than a Carnot engine. We call this engine a “superengine” and label its quantities with the letter s and label the quantities for the original Carnot engine by the letter c. By our assumption,ηs> ηc, so that
1+q3(s)
q1(s) >1+ q3(c)
q1(c) (3.1-11)
Now use the superengine to drive the Carnot engine as a heat pump between the same two heat reservoirs that are used by the superengine. If there is no friction all of the work done by the engine is transmitted to the heat pump:
w(s) −w(c) (3.1-12)
From Eq. (3.1-10) the amount of heat put into the hot reservoir by the Carnot heat pump is equal to
−q4(c) w(c) 1+q3(c)/q1(c)
The amount of heat removed from the hot reservoir by the superengine is q1(s) − w(s)
1+q3(s)/q1(s) < w(c) 1+q3(c)/q1(c) Therefore
q1(s)<−q4(c)
A greater amount of heat has been put into the hot reservoir by the Carnot heat pump than was removed from this reservoir by the superengine. This conclusion is contrary to the Clausius statement of the second law. The only source of this contradiction is our assumption that a superengine exists with a greater efficiency than that of a Carnot engine, so the efficiency of the second reversible engine cannot be larger than that of a Carnot engine.
The second reversible heat engine also cannot have a smaller efficiency than the first Carnot engine. If it did its coefficient of performance as a heat pump, which is the reciprocal of its efficiency as a heat engine, would be larger than that of a Carnot heat pump, and the second law could be violated by using the first engine to drive the second engine as a heat pump. We have shown thatthe efficiency of a reversible heat engine operating with two heat reservoirs does not depend on the nature of the working fluid or on the details of its design, but depends only on the temperatures of the heat reservoirs.
Exercise 3.1
Carry out the proof that a reversible engine cannot have a smaller efficiency than a Carnot engine if it uses the same heat reservoirs.
A heat engine operating irreversibly can have a lower efficiency than a Carnot engine. If a heat engine operates irreversibly its coefficient of performance as a heat pump will not necessarily be the reciprocal of its engine efficiency since each step cannot necessarily be reversed. Therefore, driving the irreversible engine backward as a heat pump by a Carnot engine would not necessarily violate the second law if it has a lower efficiency than a Carnot engine.