Thermodynamic Cycles

Definitions

Working substance (WS) = the WS is used as the carrier for heat energy. The state of the WS is defined by the values of its properties, e.g. pressure, volume, temperature, internal energy, enthalpy, entropy. These properties are also sometimes called functions of state.

Key facts

A thermodynamic cycle comprises a series of operations (e.g. expansion/compression of volume) carried out on the WS during which heat is supplied, and after which the WS is returned to its original state.

The work done during a thermodynamic cycle equals the area of the cycle when plotted either on a $PV$ plot (result is in $ft-lb$ ), or on a $TS$ plot (result is in $BTU$ ).

The thermal efficiency of any engine working between the temperatures of $T_1$ and $T_2$ is:

$\eta_{th} \leq 1 - \frac{T_1}{T_2}$

(1)

with the equality being satisfied only for the Carnot efficiency. Therefore, the Carnot cycle is the cycle with the highest possible thermal efficiency, while in order to increase the efficiency of all the other engines the $(T_2-T_1)$ value should be increased.

An introduction to thermodynamic cycles, and discussing the Carnot cycle.

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In order to introduce the notion of a thermodynamic cycle, consider a series of operations (e.g. expansion/compression of volume - see Figure 1) carried out on the working substance (WS) during which heat is supplied, and after which the WS is returned to its original state.

For example, imagine that initially there is an expansion of volume from $V_1$ to $V_2$ , corresponding to a work output of $W_1$ . If we are to illustrate such a process on a pressure ( $P$ ) - volume ( $V$ ) plot (see Figure 2A), then the work output $W_1$ will equal the area under curve $A$ (the blue shaded area in Figure 2B).
(For a more detailed discussion on the work done in a reversible process see Reversible Processes )

Next, consider that we are dealing with a compression of volume which returns the volume of the WS from $V_2$ back to $V_1$ , as a consequence of a work input of $W_2$ . If we are to plot this change on a $PV$ diagram (see Figure 3A), then the work $W_2$ will equal the area under curve $B$ (the blue shaded area in Figure 3B).

When we combine the two processes on a single $PV$ plot, we get a closed loop (see Figure 4A). This closed loop corresponds to a thermodynamic cycle.

As the net work output $W$ is given by:

$W = W_1 - W_2$

(2)

it will thus equal the area of the cycle (the blue shaded area in Figure 4B).

Such thermodynamic cycles can also be represented on temperature ( $T$ ) - entropy ( $S$ ) diagrams. For example, imagine that initially there is a certain amount of heat supplied ( $Q_S$ ) to the WS, leading to an increase in entropy. If we are to illustrate such a process on a $TS$ plot (see Figure 5A), then the heat energy supplied $Q_S$ will equal the area under curve $A$ (the blue shaded area in Figure 5B).

Next, consider that there is a certain amount of heat rejected in the WS and in losses ( $Q_R$ ), reverting the system back to its initial state. By plotting this process on a $TS$ diagram (see Figure 6A), the heat rejected $Q_R$ will equal the area under curve $B$ (the blue shaded area in Figure 6B).

If we are to illustrate both processes on a single $TS$ plot, then we will get again a closed loop, corresponding to the thermodynamic cycle (see Figure 7A).

By denoting the energy of the WS at the start of the cycle with $E_i$ , and the energy of the WS at the end of the cycle with $E_f$ , and by applying the law of conservation of energy, we can write that:

$E_i + Q_S = E_f + W + Q_R$

(3)

where $W$ is the work output.
(For a more detailed discussion on the law of conservation of energy see First Law of Thermodynamics )

However, as $E_i = E_f$ (the system is reverted back to its initial state), equation (#2) becomes:

$Q_S = W + Q_R$

(4)

from which:

$W = Q_S - Q_R$

(5)

Therefore, when plotting the thermodynamic cycle on a $TS$ diagram (as in Figure 7A), the work output will again equal the area of the cycle (the blue shaded area in Figure 7B).

It is important to note that although the work done can be calculated from the area of the thermodynamic cycle for both a $PV$ and a $TS$ plot, the results obtained are expressed in different units. For example, if calculating in imperial units, the area of the cycle on the $PV$ diagram gives the work done in $ft-lb$ , while the area of the cycle on the $TS$ diagram gives the work done in $BTU$ .

Carnot Cycle

In order to discuss the Carnot cycle, we first have to introduce the thermal efficiency of a cycle.

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:

$\eta_{th} = \frac{W}{Q_S}$

(6)

where $W$ is the work output, and $Q_S$ the heat energy supplied. Considering the expression of the work done $W$ from (#4), equation (#5) becomes:

$\eta_{th} = \frac{Q_S - Q_R}{Q_S}$

(7)

from which:

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

(8)

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

This cycle consists of a reversible adiabatic (i.e. isentropic) compression of the WS 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 WS from $T_2$ to $T_1$ (step $3-4$ ), and ended with an isothermal cooling with compression which reverts the system back to its initial state (step $4-1$ ).

The Carnot cycle can also be represented on a $TS$ diagram (see Figure 9A), and this representation is more useful in calculating the Carnot cycle efficiency.

In order to calculate the thermal efficiency of the Carnot cycle, we first have to calculate the $Q_S$ and $Q_R$ terms (see equation #7).

The heat supplied $Q_S$ during step $2-3$ of the Carnot cycle can be calculated on a $TS$ diagram as the area under the cycle beneath the $T_2$ line (the blue shaded area in Figure 9B).

The area of this rectangle can also be calculated as:

$Q_S = T_2 (S_3 - S_2)$

(9)

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).

The area of this rectangle is also given by:

$Q_R = T_1 (S_4 - S_1)$

(10)

Taking into account (#8) and (#9), the thermal efficiency of the Carnot cycle becomes:

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

(11)

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

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

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

(12)

or, written in a different form:

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

$\eta_{th} = \frac{T_2-T_1}{T_2}$

(13)

It should be noted that in equations (#11) and (#12) 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)"

As previously stated, the thermal efficiency of a Carnot 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 (#11), the thermal efficiency of any engine (other than a engine with Carnot efficiency) working between the temperatures of $T_1$ and $T_2$ is:

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

(14)

From equation (#13) 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:

$\eta_C = \frac{T_2-T_1}{T_2}$

(15)

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:

$T_1 = 150^\circ F = 338.70 \; K$

(16)

and

$T_2 = 350^\circ F = 449.82 \; K$

(17)

we get the Carnot efficiency of the steam engine:

$\eta_C = \frac{449.82 - 338.70}{449.82}$

(18)

from which we obtain:

$\eta_C = 0.247$

(19)

or, expressed as percentage:

$\eta_C = 24.7\%$

(20)

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

$\eta = \frac{52}{100} \cdot 0.247$

(21)

from which we obtain:

$\eta = 0.128$

(22)

or, expressed as percentage:

$\eta = 12.8\%$

(23)

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

$\eta = \frac{W}{Q_S}$

(24)

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:

$Q_S = \frac{W}{\eta}$

(25)

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:

$W = 1 \cdot 550 \cdot 60$

(26)

from which we obtain:

$W = 33000 \; ft-lb$

(27)

By using (#13) and (#8) in equation (#11), 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:

$Q_S = \frac{33000}{0.128\cdot 778}$

(28)

from which we obtain:

Solution

$Q_S = 331.38 \; BTU$

(29)

Thermodynamic Cycles

Carnot Cycle

Example 1 [imperial]

Example 2 [metric]