Carnot Cycle

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The ideal thermodynamic cycle

Engineering Thermodynamics

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The Carnot Cycle is an entirely theoretical thermodynamic cycle utilising reversible processes.

Thermal Efficiency

The thermal efficiency of a cycle, also denoted by $\eta_{th}$, is a measure of the ability to convert heat energy into work. Therefore, the thermal efficiency can be defined as:

$$\eta_{th} = \frac{W}{Q_S}$$
(1)

where W is the work output, and Q\_S the heat energy supplied. Replacing W with the heat supplied minus the head rejected, then equation (1) becomes:

$$\eta_{th} = \frac{Q_S - Q_R}{Q_S}$$
(2)

from which:

$$\eta_{th} = 1 - \frac{Q_R}{Q_S}$$
(3)

The cycle with the highest possible thermal efficiency is the Carnot cycle (diagramed on a $PV$ plot in Figure 8).

Figure 8
Figure 1

This cycle consists of

  • a reversible adiabatic (i.e. isentropic) compression of the gas from temperature T\_1 to T\_2 (step 1-2),
  • followed by an isothermal heating with expansion (step 2-3),
  • then a reversible adiabatic (isentropic) expansion of the gas from T\_2 to T\_1 (step 3-4),
  • and ending with an isothermal cooling with compression which reverts the system back to its initial state (step 4-1).

The Carnot Cycle can be represented on a TS diagram (see Figure 9), which is useful for calculating the Carnot Cycle efficiency.

Figure 9
Figure 2

To calculate the thermal efficiency, first calculate the Q\_S and Q\_R terms (see Eq. 3). The heat supplied Q\_S during step 2-3 can be calculated on a TS diagram as the area under the cycle beneath the T\_2 line (the blue shaded area in Figure 9).

The area of this rectangle can also be calculated as:

$$Q_S = T_2 (S_3 - S_2)$$
(4)

On the other hand, the heat rejected Q\_R during step 4-1 of the Carnot cycle can be calculated on a TS diagram as the area under the cycle beneath the T\_1 line (the blue shaded area in Figure 9C).

Figure 9C
Figure 3: C

The area of this rectangle is also given by:

$$Q_R = T_1 (S_4 - S_1)$$
(5)

Taking into account (4) and (5), the thermal efficiency of the Carnot cycle becomes:

$$\eta_{th} = 1 - \frac{T_1 (S_4 - S_1)}{T_2 (S_3 - S_2)}$$
(6)

\calc{1-(T_C/T_H)} "Instant calc. Carnot eff. (K)"

However, we can see from Figure 9 that $S_1=S_2$ and $S_3=S_4$. Therefore, we obtain the Carnot cycle efficiency as:

$$\eta_{th} = 1 - \frac{T_1}{T_2}$$
(7)

or, written in a different form:

\calc{1-( (T_C+459.67)/(T_H+459.67) )} "Instant calc. Carnot eff. (F)"

$$\eta_{th} = \frac{T_2-T_1}{T_2}$$
(8)

It should be noted that in equations (7) and (8) the temperatures (also identified as T\_1=T\_C, the temperature of the cold reservoir, and T\_2=T\_H, the temperature of the hot reservoir) are expressed on an absolute scale, such as the Kelvin scale. On the right side we provide calculators for the Carnot efficiency where you can input the temperatures in degrees Fahrenheit or degrees Celsius as well (the conversions are computed automatically).

\calc{1-( (T_C+273.15)/(T_H+273.15) )} "Instant calc. Carnot eff. (C)"

The thermal efficiency of the cycle (and in general of any reversible cycle) represents the highest possible thermal efficiency (this statement is also known as Carnot's theorem - for a more detailed discussion see also Second Law of Thermodynamics ). This ultimate thermal efficiency can then be used to compare the efficiencies of other cycles operating between the same two temperatures. Thus, taking into account (7), the thermal efficiency of any engine working between the temperatures of T\_1 and T\_2 is:

$$\eta_{th} < 1 - \frac{T_1}{T_2}$$
(9)

From equation (9) it can be seen that in order to improve the thermal efficiency of an engine, we should basically increase the value of (T\_2 - T\_1), i.e. increase the temperature difference under which the engine works.

Example 1 [imperial]
Problem

Consider a steam engine for which the steam is supplied at $350^\circ F$ and condensed at $150^\circ F$. If the thermal efficiency of the steam engine is $52\%$ of the Carnot efficiency, find the heat required (expressed in $BTU$) to produce a work output of 1 horsepower ($HP$) for 1 minute.

Workings

We know that the Carnot efficiency of an engine working between temperatures $T_1$ and $T_2$ is given by:

$$\eta_C = \frac{T_2-T_1}{T_2}$$
(10)

where $T_1$ and $T_2$ are absolute temperatures expressed in kelvins ($K$).

Therefore, in order to calculate the Carnot efficiency of the steam engine from the hypothesis, we first have to convert the temperatures into absolute temperatures. As we have that:

$$T_1 = 150^\circ F = 338.70 \; K$$
(11)

and

$$T_2 = 350^\circ F = 449.82 \; K$$
(12)

we get the Carnot efficiency of the steam engine:

$$\eta_C = \frac{449.82 - 338.70}{449.82}$$
(13)

from which we obtain:

$$\eta_C = 0.247$$
(14)

or, expressed as percentage:

$$\eta_C = 24.7\%$$
(15)

As the actual efficiency of the steam engine is $52\%$ of the Carnot efficiency, we can calculate the actual efficiency $\eta$ as:

$$\eta = \frac{52}{100} \cdot 0.247$$
(16)

from which we obtain:

$$\eta = 0.128$$
(17)

or, expressed as percentage:

$$\eta = 12.8\%$$
(18)

We know that the thermal efficiency of an engine is given by:

$$\eta = \frac{W}{Q_S}$$
(19)

where $W$ is the work output, and $Q_S$ the heat supplied. Therefore, the heat supplied $Q_S$ in order to produce a work output of $W$, while working at an efficiency of $\eta$ can be written as:

$$Q_S = \frac{W}{\eta}$$
(20)

From the hypothesis we have that the required work output $W$ is $1\;HP$ for $60\;s$. Taking into account that $1\;HP=550\;ft-lb/s$, the work $W$ can also be expressed in $ft-lb$ as:

$$W = 1 \cdot 550 \cdot 60$$
(21)

from which we obtain:

$$W = 33000 \; ft-lb$$
(22)

By using (22) and (17) in equation (20), and also considering that $1\;BTU = 778 \; ft-lb$, we get the heat required (expressed in $BTU$) to produce $1\;HP$ for $1\; min$ as:

$$Q_S = \frac{33000}{0.128\cdot 778}$$
(23)

from which we obtain:

Solution
$$Q_S = 331.38 \; BTU$$
(24)