# Carnot cycle

1. May 18, 2015

### nocar

In a reversible carnot cycle the amount of heat added is always more than the amount of heat taken out. So why doesn't it explode? I'm wondering how heat is conserved mathematically as the area under the curve of a p-v diagram. u=q+w. If the heat during expansion is always more than the heat taken out during compression where does that heat go. Does the temperature going down mean that it lost heat?

2. May 18, 2015

### EM_Guy

The difference between the heat input and the heat output is the work performed by the Carnot engine.

w = qin - qout.

3. May 18, 2015

### nocar

If qin is always greater than qout then the heat is transformed into mechanical work? How does mechanical work reduce heat of the system if its isolated? Even if there is less heat per unit area because there are less particles per unit area there should still be the same heat overall in the system right? If your taking less heat out at a lower temperature and putting more heat in at a higher temperature wouldn't that mean the net heat in the system would always increase?

4. May 18, 2015

### Staff: Mentor

System is not isolated in the sense you think it is - mechanical work is "taken out" of the system (is done on the surroundings, not on the system itself).

5. May 18, 2015

### EM_Guy

First, let's establish the conservation of energy firmly:
Energy In = Energy Out.

So,

qin + win = qout + wout

Or equivalently,

qin = qout + wout (net).

The Carnot cycle consists of four reversible processes:
From state 1 to state 2: Reversible, isothermal expansion.
From state 2 to state 3: Reversible, adiabatic expansion.
From state 3 to state 4: Reversible, isothermal compression.
From state 4 to state 1: Reversible, adiabatic compression.

From state 1 to state 2: The working fluid of the engine is at some hot temperature TH, and it is brought into close contact with a thermal reservoir that is effectively the same temperature (or just a differential amount hotter). Heat is transferred in a reversible, isothermal manner (no change of temperature) slowly from the hot thermal reservoir to the working fluid, causing the gas to slowly expand. A total of QH amount of thermal energy is transferred from the hot thermal reservoir to the working fluid during this process.

From state 2 to state 3: The working fluid is no longer in contact with the thermal reservoir. Having been energized by the heat transfer to the system (and the work input during compression), the working fluid now undergoes a reversible, adiabatic expansion - performing work on the surroundings. (In a car engine, this would correspond to the power stroke). There is no heat transfer (adiabatic) during this process. But since the gas is expanding - performing work on the surroundings, energy is transferred from the gas to the surroundings by the process of work (expansion). Thus, the internal energy of the gas decreases, which causes the temperature of the gas to drop to some low temperature TL.

From state 3 to state 4: The working fluid is brought into contact with a thermal reservoir at a low temperature TL. Heat is transferred from the working fluid to the low temperature thermal reservoir in a reversible, isothermal manner (no change of temperature). The fact that this is an isothermal process (temperature remains constant) implies that work is being done on the working fluid - increasing the pressure and decreasing the specific volume of the fluid. A total of QL thermal energy is transferred out of the working fluid by the process of heat.

From state 4 to state 1: The insulated working fluid is now reversibly and adiabatically compressed back to its original state. Pressure increases; specific volume decreases; and temperature increases from TL to TH.

Each process is reversible. Note that reversible processes are idealized processes. Every real process has irreversibilities. Thus, no engine operating between two thermal reservoirs can be more efficient than the Carnot engine.

6. May 18, 2015

### EM_Guy

7. May 18, 2015

### nocar

so the expansion from step 2 to 3 causes the heat content to be reduced because the temperature went down?

8. May 18, 2015

### nocar

Entropy can be a state variable in a Carnot cycle because putting more heat in at a higher temperature is the same as taking less heat out at a lower temperature, but doesn't that still leave a net increase in heat for every cycle?

9. May 19, 2015

### EM_Guy

Be careful with the term "heat content." Some people use the term "heat content" synonymously with enthalpy (h = u + Pv). The enthalpy is reduced during a reversible, adiabatic expansion. But don't make this more complicated than what it is. During the cycle, the system gets energy in two ways. One way is by heat transfer to the system from the hot thermal reservoir. The other way is by work done on the system - compressing the working fluid. Also, during the cycle, energy leaves the system in two ways. One way is by work done by the system on the surroundings (during expansion), and the other is by heat transfer from the working fluid to the low temperature thermal reservoir.

So, qin + win = qout + wout

wout,net = wout - win

So, qin - qout = woutnet.

When a working fluid that is insulated from the surroundings expands, it does work on the surroundings. Work is an energy transfer. Thus, during an adiabatic expansion, the working fluid gives some of its energy to the surroundings. Having lost some of its energy, its temperature decreases.

10. May 19, 2015

### EM_Guy

It seems like you are mixing together several ideas here, and I'm trying to follow you.

Entropy is a thermodynamic property. The Carnot cycle is comprised of four reversible processes. According to the Clausius inequality, the cyclical integral of dq/T is always less than zero. The cyclical integral of dq/T equals zero for cycles completely comprised of reversible processes. It is less than zero for cycles in which there are irreversibilities. This is the 2nd law of thermodynamics. Equivalently, the Kelvin-Planck statement of the second law states that, "No system can produce a net amount of work while operating in a cycle and exchanging heat with a single thermal energy reservoir."

When Clausius realized that the cyclical integral of dq/T equals zero for cycles comprised completely of reversible processes, he realized that he had discovered a thermodynamic property. The cyclical integral of any property is zero. A quantity whose cyclical integral is zero depends only on the state and not the process path, and is thus a property. Clausius named this property entropy.

Specific entropy: ds = (dq/T), internally reversible processes.

A reversible, adiabatic process is an isentropic process; i.e., the entropy does not change. So, the expansion from state 2 to state 3 and the compression from state 4 to state 1 are both isentropic. The entropy of the system undergoes no change during these processes. However, when a system is heated, its entropy increases. And when a system is cooled (i.e. when a system loses thermal energy by the process of heat), it loses entropy. Since the Carnot cycle is comprised entirely of reversible processes, and since two of these processes are adiabatic, and since the cyclical integral of entropy is zero for cycles comprised of entirely reversible processes, it follows that the increase of entropy from state 1 to state 2 must equal the decrease of entropy from state 3 to state 4. This can be shown easily by integrating dq/T for those processes, noting that for isothermal processes, the 1/T factor can be taken out of the integral.

11. May 19, 2015

### EM_Guy

Correction: According to the Clausius inequality, the cyclical integral of dq/T is always less than or equal to zero.

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