The gas inside a piston-cylinder experiences heat transfer, work transfer, and change in internal energy.
The Non Flow Energy Equation relates work, heat transfer, and internal energy, and one requires insight into each of these terms. Engines and compressors frequently employ piston-cylinders wherein the internal movement of gas generates or absorbs mechanical work. This behaviour is complex but we shall simplify our analysis by taking the "best case", quasi – equilibrium (or reversible) paths. The gas-to-solid interface constitutes the system boundary.
Figure 1 Progression of a piston-cylinder between two states (expansion). The red lines indicate the system boundary. (a) Initial
state 1, showing addition of heat and production of work (b) Final state 2 - the gas in the
cylinder has expanded, thereby moving the boundary.
Consider the expansion (Figure 1) from State 1 to State 2, under quasi-equilibrium such that pressures and temperatures are the same at any point inside the system boundary. Consider an infinitely small stroke displacement,
\( \delta s \), along which p is constant and (reversible) work done by the moving boundary is,
$$ \delta W = -F \delta s $$
Then taking the cross sectional area of the air mass as A we are normally interested in the thermodynamic
properties included in \(F=p A\) and \( \delta s = \delta V / A \). On substitution we get,
\begin{equation} \delta W = -p \delta V \end{equation}
It is important to understand the implications of Quasi-Equilibrium (or reversibility). If force on one side of the piston is F, then the opposing force differs by an infinitesimally small amount; it is F + dF and the piston is very close to equilibrium and follows a path that comprises a set of equilibria. (Such a path is also termed “reversible” because the system can move from state 1 to state 2 and back to state 1 leaving no trace on its surroundings. That is, no net heat or work has been transferred to the surroundings.) The theme of reversibility will be developed further in a separate section concerning the Second Law .
Equation 1 can be integrated across the complete volume change (V2-V1) to give, total (reversible) moving boundary work, \(W_b\) . Thus \(W_b\) is the area under the p-versus-V curve, and this curve constitutes a sensible choice of path diagram.
Figure 2 Path diagram for an expansion from (1) to (2). The shaded area
equates to moving boundary work, \(-W_b\)
We obtain,
\begin{equation} W_b = - \int^2_1 p dV \end{equation}
A compression process could well operate along the reverse direction (2) to
(1) on the above p-V diagram, changing the sign of the computed \(W_b\) in conformity with the IUPAC
sign convention. In an engine, the moving boundary work must tackle (1) moving the piston
against the atmosphere (2) all forms of friction (3) transmitting work to the crank - the
express intention of the piston-cylinder.
There are many possible paths between State 1 and State 2 and therefore many
possible integral areas and therefore many values of moving boundary work. (We shall see
later that this is a pre-condition for the successful operation of cyclic machines).
Fig. 3 Two alternative paths from State 1 to State 2 (a
compression). The shaded area equates to the additional work required
by path (B)
Path B does more moving boundary work than path A; the difference equates to the
shaded area. Work depends on the path and is not a state property
Example 3.010. Pressures are measured during the power stroke of an Otto Cycle with a piezo-electric sensor. The pressures are tabulated against the intantaneous gas volume below. Estimate the work done in this stroke.
Table 1 Hypothetical pressures measured in an engine (example 3.010)
Volume, litres
0.04
0.08
0.12
0.16
0.20
0.24
0.28
0.32
Pressure, bar
35.0
15.2
9.4
6.6
5.1
4.1
3.4
2.9
Problem Statement: Pressure from mass, temperature and volume.
Discussion - the Polytropic process will be investigated later. I have set the
problem with a polytropic equation (index n = 1.2) for which the analytical solution is \(W_b = -0.236kJ \) - a discrepancy of roughly 6%. In the context of engines, the moving work done over a complete cycle is termed indicated work . This excludes the frictional losses in mechanisms, unlike brake work.
2.) Constant Volume Processes (Isochoric Process)
An immobile system boundary does no work.
No work is done upon the surroundings in a constant volume process since there are no moving boundaries. Hence the NFEE becomes,
$$Q=U_2-U_1 \qquad ; \qquad (W=0)$$
The supply of heat is converted to a change in internal energy. If the heat is supplied reversibly (that is, in a quasi-equilibrium manner) then the process path is represented by a vertical line on the pV diagram.The zero area below such a line confirms that no work is done. The one special, irreversible, instance of a constant volume process where W ≠ 0 and concerns paddle wheel work. We shall ignore any such friction effects in constant volume processes.
Figure 4. Constant volume (isochoric) process. (a) System held in a piston cylinder, with no
piston movement (b) Process shown on a pV diagram (c) Process shown on a Ts diagram.
The process sketch for the pV curve is self-evident. With regards to the shape of the T-s curve recall that the specific entropy of an ideal gas follows generally,
$$ s = c_v ln (\frac{T}{T_o}) + R ln (\frac{V}{V_o} ) $$
The constant terms in the above are \(c_v, T_o, V_o \) and, at constant volume, V. With algebraic manipulation temperature (T) shows an exponential dependence on entropy,
$$ T = T_o \frac{ exp(s/c_v ) }{ (\frac{V}{V_o})^{R/c_v} } $$
The area underneath the Ts curve is equal to (reversible or quasi-equilibrium) heat transfer, \(Q_{rev}\).
For an ideal gas internal energy is a function of temperature only as discussed in
Topic 2.
Example 3.020 A rigid vessel contains 5 g of air. The temperature is raised from 473K to 573K. Estimate the required heat input to the air.
To answer this type of question, start by finding \( \Delta U \). Only then tackle heat addition, Q.
Problem Statement: Heat addition to an ideal gas at constant volume (hence W=0).
Schematics: - see Figure 4. above.
Assumptions: Ideal gas - internal energy is a linear function of temperature, and only temperature; one can use heat capacity to get \( \Delta U \); stationary fluid - ignore KE and PE changes and allows use of NFEE; quasi-equilibrium process (will be known as "reversible").
Physical Laws: First Law of Thermodynamics - gives NFEE. Note that for air \(R = 0.287 kJ kg^{-1} K^{-1} \), \( c_v = 0.718 kJ kg^{-1} K^{-1} \) and \(c_p = 1.005 kJ kg^{-1} K^{-1} \).
Calculation:
Let us identify the three contributions to the Non-flow Energy Equation, viz,
$$ U_2 - U_1 = W + Q$$
The change in internal energy is,
$$U_2-U_1 = m c_v (T_2-T_1) = 0.005\,kg \times 0.718\,kJ/kg\,K \times (573-473)K = 0.359 kJ$$
The work is \(W=0\). Therefore the NFEE is manipulated to give the heat addition as
$$ Q = U_2-U_1-W = 0.359-0 = 0.359kJ$$
The positive sign indicates heat addition.
Discussion: We are not told volume, and therefore know only a single property of state, temperature. The state postulate tells us that we cannot specify the complete equilibrium at 1 and 2. Term \(\Delta U\) is calculable only because the gas is ideal, and U(T) is a function of temperature only. Between 623K and 723K the same \(\Delta U\) would be found for any volume along the x-axis of part (b).
In Example 3.020 temperature is the only intensive property to be specified. For a non-ideal gas (i.e. steam) we need two properties. The next question
can be tackled with steam tables, such as the sample given in Table 3 of
Topic 2.
Example 3.030 A rigid vessel contains 5 g of superheated steam at a pressure of 5 bar and a temperature of 473K. The temperature is to be raised from 473K to 573K. Estimate the required heat input to the steam.
Problem Statement: Heat addition to steam (a non-ideal gas) at constant volume (hence W=0).
Schematics: - see Figure 4. above.
Assumptions: Stationary fluid - ignore KE and PE changes and allows use of NFEE; Quasi-equilibrium process (will be known as "reversible").
Physical Laws: First Law of Thermodynamics - gives NFEE. .
Calculation:
Let us identify the three contributions to the Non-flow Energy Equation, viz,
$$ U_2 - U_1 = W + Q$$
The specific energy, u, is found from steam tables, for instance in section 3, Topic 2. A the start condition (1) one finds specific internal energy using two other properties, \(p = 5bar, \; T=473K \). The specific volume is also noted
$$ u_1 = 2644 kJ/kg \qquad v_1 = 0.4252 m^3/kg $$
At the end condition the temperature, \(T_2 = 573K\) is specified but a second property is needed. The process is isochoric so \(v_2 = v_1 0.4252 m^3/kg\). The internal energy is to be found in the \(T_2 = 573K \) column if one applies
linear interpolation to the function u(v).
$$u_2 = 2801 + (2800-2801) \times \frac{0.4252-0.4344}{0.3714-0.4344} = 2800.85kJ/kg = 2801kJ/kg (4s.f.) $$
The change in internal energy is,
$$U_2-U_1 = m (u_2-u_1) = 0.005\,kg \times (2801-2644) = 0.785 kJ$$
The work is \(W=0 \). Therefore the NFEE is manipulated to give the heat addition as
$$ Q = U_2-U_1-W = 0.785-0 = 0.785kJ$$
The positive sign indicates heat addition.
Discussion: In this instance it was important to specify the gas pressure. In practice the influence of pressure/ volume on internal energy was weak and would have caused an error of 0.6%. Bigger errors are incurred at higher steam pressures.
Constant pressure heat addition is equivalent to change in enthalpy. Constant pressure work is the product of pressure and swept volume.
The constant pressure expansion is shown on pV and Ts plots below. The pV plot is a horizontal line.
Figure 5. Constant pressure (isobaric) process. (a) a) System held in a piston cylinder, with piston movement allowing constant pressure (b) Process shown on a pV diagram (c) Process shown on a Ts diagram.
The work is the area under the pV curve, simply
\begin{equation}
W = -p (V_2-V_1) \qquad at \;constant \; pressure
\end{equation}
\begin{equation}
Q = H_2-H_1 \qquad at \;constant \; pressure
\end{equation}
One recalls from
Topic 2
that for an ideal gas one can substitute the ideal gas law and the definition of isobaric heat capacity.
$$ W = -m R (T_2-T_1) \qquad ideal \; gas \; at \;constant \; pressure$$
$$ Q = m c_p (T_2-T_1) \qquad ideal \; gas \; at \;constant \; pressure $$
The area under the Ts curve indicates the (reversible) heat transfer. The shape is exponential, and for an ideal gas can be deduced from,
$$ s = c_p ln ( \frac{T}{T_o} ) - R ln (\frac{p}{p_o}) \qquad ideal \; gas $$
The constant terms in the above are \(c_p, T_o, p_o \) and, at constant pressure, p. Upon algebraic manipulation,
$$ T = T_o (\frac{p}{p_o}) ^ {R/c_p} exp (s/c_p) \qquad ideal \; gas \; at \; constant \; pressure $$
Example 3.040 A piston-cylinder contains 5g of air at 5 bar and at 473K. At the end of an isobaric process the pressure is still 5 bar and the temperature is 673 K. Estimate the work done and the heat input.
Problem Statement: Energy inputs for a constant pressure expansion (ideal gas).
Schematics: see Figure 5 above.
Assumptions: quasi-equilibrium (reversible) process (enables integration under pV curve). Ideal gas - simple laws relate pressure and volume to temperature, likewise internal energy follows from \( c_v \).
Physical Laws: Ideal gas law relates p, V, T. Work definition - area under pV curve is equal to work. First Law leads to Non Flow Energy Equation.
Calculation:
Work is defined as
$$ W = -\int_1^2 pdV = p (V_1-V_2) $$
Substitute the Ideal gas Law ( \(pV=mRT\) ) to get,
$$ W = m R (T_1-T_2) $$
$$ W = 0.005 kg \times 0.287 kJ/kgK \times (473-673) = -0.287kJ $$
The change in internal energy is,
$$ U_2 - U_1 = m c_v (T_2 - T_1) = 0.005 \times 0.718 \times (673-473) = 0.718 kJ $$
From the Non Flow Energy Equation,
$$Q = U_2-U_1-W = 1.005 kJ$$
Discussion: The heat transfer is also equal to the change in enthalpy. The above answer is confirmed by,
$$ Q = m c_p (T_2-T_1) = 1.005kJ $$
Example 3.050 A piston-cylinder contains 5g of steam at 5 bar and at 473K. At the end of an isobaric process the pressure is still 5 bar and the temperature is 673 K. Estimate the work done and the heat input.
Problem Statement: Energy inputs for a constant pressure expansion (steam, a none ideal gas).
Schematics: - see Figure 5. above.
Assumptions: quasi-equilibrium (reversible) process (enables integration under pV curve).
Physical Laws: Tabulated physical properties. First Law leads to Non Flow Energy Equation.
Physical properties: These are available from tables. I used the "number input" button on my
online interactive chart .
Physical Properties: At 5 bar and 473 K,
$$ u_1 = 2643 kJ/kg \qquad s_1 = 7.060 kJ/kgK \qquad v_1= 0.42488 m^3/kg $$
At 5 bar and 673 K,
$$ u_2= 2963 kJ/kg \qquad s_2 = 7.795 kJ/kg K \qquad v_2 = 0.61715 m^3/kg $$
Calculation:
The work done is
$$ W = p (V_1-V_2) = m (v_1-v_2) = 0.005 kg \times 5 bar \times 100 kN/m^2 bar \times (0.42488- 0.61715)m^3/kg=-0.4807kJ$$
The change in internal energy is,
$$ U_2-U_1 = m (u_2-u_1) = 0.005 kg \times (2963 kJ/kg - 2643 kJ/kg) = 1.60kJ$$
From the Non Flow Energy Equation,
$$ Q = U_2-U_1 - W = 2.081 kJ $$
Discussion: If we look up enthalpies at 473K and 673K( both at 5bar),
$$ h_1 = 2856 kJ/kg \qquad h_2 = 3272 kJ/kg $$
Then
$$ Q = m (h_2-h_1)= 0.005 \times ( 3272 - 2856 ) = 2.08 kJ $$
which confirms the above.
4.) Constant Temperature Processes (Isothermal Process)
When an ideal gas maintains a constant temperature transfers of heat and work transfers are equal .
This forms part of the Carnot Cycle (the most efficient engine cycle proposed). The pV and Ts diagrams are sketched below.
Figure 6. Constant temperature (isothermal) process. (a) System held in a piston-cylinder, with heat transfer to the surroundings such as to allow constant temperature (b) Process shown on a pV diagram (c) Process shown on a Ts diagram.
The theoretical treatements of ideal and non-ideal gases differ somewhat from each other. For an ideal gas pressure, volume and temperature are related by a simple relationship, namely the Ideal Gas Law,
$$ p = \frac{mRT}{V} $$
where temperature T is a constant. This can be substituted into the work definition, thus
$$ W_b = -\int^2_1 p dV = -mRT \int^2_1 \frac{dV}{V} $$
Carry out the integration and obtain,
\begin{equation}
W= -mRT ln(\frac{V_2}{V_1}) \qquad ideal \; gas \; at \; constant \; temperature
\end{equation}
For an ideal gas internal energy depends on temperature only, \(U=U(T)\), and hence is constant. By virtue of the Non-Flow Energy Equation,
$$ Q = -W = +mRT ln(\frac{V_2}{V_1})\qquad ideal \; gas \; at \; constant \; temperature $$
For steam and other non-ideal gases one can make use of tabulated properties, and exploit the constant temperature. Thus the heat transfer is,
\begin{equation}
Q = \int^2_1 T dS = T(S_2-S_1) \qquad quasi-equilibrium \; process \; at \; constant \; temperature
\end{equation}
and work follows from the Non-Flow Energy Equation,
$$ W = U_2-U_1 - Q $$
Example 3.060 A piston-cylinder contains 3.6 g of air at 7 bar and at 473K. At the end of an isothermal process the air has expanded to a volume 1.417 times the initial value (\(V_2=1.417V_1\)). Estimate the work done and the heat input.
Problem Statement: expansion of an ideal gas (air) at constant temperature.
Schematics: - see Figure 6. above.
Assumptions: stationary fluid - ignore KE and PE changes and allows use of NFEE; quasi-equilibrium process (will be known as "reversible"); ideal gas - so no change in internal energy and Q+W=0.
Physical Laws: First Law of Thermodynamics - gives NFEE. Ideal gas law and work definition are used in
an equation for work done.
Calculation: The expression for moving boundary work has already been given as,
$$ W_b = -mRT ln \frac{V_2}{V_1} $$
Substitute known values,
$$ W_b = -0.0036kg \times 0.287 kJ/kgK \times 473 K\times ln(1.417) = -0.170 kJ $$
An isothermal, ideal gas experiences no change in internal energy. Thus (from NFEE),
$$ Q = -W_b = +0.170 \times 10^{-3} kJ $$
Discussion: I checked this with
the interactive chart. In view of the limitations of setting parameters by eye the agreement ( \(W_b = -0.169kJ \) ) is
encouraging. A screenshot is shown below.
(I set up the 7 bar isobar and found thereon by trial and error the point where T1=473 K.
The interactive chart displays p1, V1, T1 and I used a pocket calculator to get V2 (see isochor), moved the limit lines over the isochors, highlighted the isotherm and pressed the "integrate" button.
Example 3.070 A piston-cylinder contains 5g of steam at 7 bar and at 473K. At the end of an isothermal process the steam has expanded to a volume 1.417 times the initial value (\(V_2=1.417V_1\)). Estimate the work done and the heat input.
Problem Statement: expansion of a non-ideal gas (steam) at constant temperature.
Schematics: - see Figure 6. above.
Assumptions: stationary fluid - ignore KE and PE changes and allows use of NFEE; quasi-equilibrium process (will be known as "reversible").
Physical Laws: First Law of Thermodynamics - gives NFEE. Ideal gas law and work definition are used in
an equation for work done. Data are extracted from steam tables, a sample is in
section 3, Topic 2
Calculations: At 7 bar, \( 200^oC = 473K \), the specific volume at initial state is
$$ v_1 = 0.3001 m^3/kg $$
And from the problem specification,
$$ v_2 = 1.417 \times 0.3001 = 0.4252 m^3/kg \qquad T_2=T_1 = 200^oC $$
In the tables, the above values of temperature and specific volume correspond to a pressure \( p_2 = 5 bar \). Importantly, values of internal energy (u) and entropy(s) are also tabulated. The
change in internal energy is,
$$U_2-U_1 = m (u_2-u_1) = 0.0036 \times (2644-2636) = 0.029 kJ$$
The heat transfer is,
$$ Q = \int^2_1 T dS = m T (s_2-s_1) = 0.0036kg \times 473K (7.060-6.888)kJ/kgK = 0.293 kJ $$
From the NFEE the work is
$$ W=U_2-U_1-Q = -0.264kJ$$
Discussion: The small change in internal energy is roughly 10% of the heat transfer ( \( \Delta U = 0 \) in an ideal gas.)
The expansion of steam in a piston-cylinder would now seldom be seen in practice; it would have been applicable to
steam locomotive engines in the past.
Moving boundary work is extracted uniquely from the internal energy of the gas. There is no heat transfer (adiabatic) and no gas friction.
The isentropic process forms another part of the Carnot Cycle (the most efficient engine cycle proposed). Within a piston-cylinder
there are no friction effects within the gas and no heat is transferred to the gas . Noting that entropy is defined by \(dS = dQ_{rev}/T \) one thus finds no change in entropy throughout the process. The pV and Ts diagrams are sketched below.
Figure 7. Constant entropy (isentropic) process. (a) System held in a piston cylinder, with no heat transfer during piston movement (b) Process shown on a pV diagram (c) Process shown on a Ts diagram with constant entropy.
Figure 7 shows an expansion . Note the reduction in temperature. It indicates the conversion of gas internal energy into work.
Isentropic Relationships for Ideal Gases
The theoretical treatements of ideal and non-ideal gases differ somewhat from each other. For an ideal gas pressure, the isentropic relationship is famously,
\begin{equation}
p_1 V_1^{\gamma} = p_2 V_2^{\gamma} \qquad constant \; entropy, \; ideal \; gas
\end{equation}
An alternative albegraic description is,
\begin{equation}
p V^{\gamma} = constant
\end{equation}
where term \( \gamma= c_p/c_v \) is the ratio of isobaric and isochoric heat capacities, and is very close to 1.4 for air. In Otto cycles it is more convenient to work with (T,V) pairs than with (p,V) pairs. The Ideal Gas Law eliminates pressure with ( \(p = mRT/V\) ) so,
In Brayton cycles it is more convenient to work with (p,T) pairs that with (p, V) pairs. The Ideal Gas Law eliminates volume with ( \(V = mRT/p\) ) so,
Proof of the isentropic relationship serves to emphasise to the student that work is converted to internal energy, with no heat transfer (Q=0). The proof is long-winded, so I would suggest that it is tackled only those driven by curiosity and with good capabilities in mathematics.
Start with the NFEE, noting zero heat transfer (Q=0).
$$ W=U_2 - U_1 \qquad isentropic $$
Consider an infinitessimally small change in system volume whereby,
$$ dW = dU $$
For an ideal gas, use work definition to replace the left hand side and the isochoric heat capacity to replace the right hand side. Thus,
$$ dW = -p dV $$
$$ dU = m c_v dT $$
Combine the three preceeding equations to obtain,
$$ -p dV = m c_v dT $$
From the above, eliminate temperature by means of the Ideal Gas Law . Substitute into the above equation ...
$$ T = \frac{p V}{m R}$$
To obtain,
$$ -p dV = m c_v d (\frac{p V}{m R})$$
The mass m cancels and the RHS and the heat capacity ratio, \( \gamma = c_p/c_v \) is introduced.
$$ -p dV = \frac{c_v}{R} d(p V) = \frac {d(p V)}{\gamma-1} $$
(Here, \( c_v/R = c_v/(c_p-c_v) = 1/(\gamma-1) \) .)
Now let us multiply both sides by \( \gamma - 1 \) and apply the
product rule to remaining terms on the right hand side.
$$ p dV - \gamma p dV = p dV + Vdp $$
Equal p dV terms on the LHS and RHS cancel. Now, separate variables and integrate,
$$ - \gamma \int^2_1 \frac{dV}{V} = \int^2_1 \frac{dp}{p} $$
Noting the
integration rules for reciprocals ,
$$ -\gamma ln (\frac{V_2}{V_1} ) = ln (\frac{p_2}{p_1}) $$
One applies the power rule of logarithms in a rearrangement to obtain the required form,
$$ p_1 V_1^{\gamma} = p_2 V_2^{\gamma} = constant; \qquad constant \;entropy, \; ideal \; gas $$
Work Done by Isentropic Ideal Gas
On integration, the isentropic relationship gives the moving boundary work.
\begin{equation}
W_b = \frac{p_2 V_2 - p_1V_1}{\gamma-1}
\end{equation}
One can substitute the Ideal Gas Law to obtain,
\begin{equation}
W_b = mR \frac{T_2-T_1}{\gamma-1}
\end{equation}
The same equation follows from manipulation of the NFEE (because work equals internal energy change and Q=0).
To prepare for integration, start with the isentropic relationship
$$ p V^{\gamma} = constant $$
and then write pressure as a function of volume.
$$ p = (p_1 V_1^{\gamma})\frac {1}{V^{\gamma}} $$
(I have used brackets to group the constant terms.) The integration of pressure with respect to volume is,
$$ W_b = -\int^2_1 p dV = - ( p_1 V_1^{\gamma} ) \int^2_1 \frac{dV}{V^{\gamma}}
= - p_1 V_1^{\gamma} \frac{V_2^{1-\gamma}- V_1^{1-\gamma}}{1-\gamma} $$
Note the multiplier \(p_1 V_1^{\gamma} = p_2 V_2^{\gamma} \). Also multiply numerator and denominator by (-1). Then
$$ W_b = \frac{p_2 V_2^{\gamma} \times V_2^{1-\gamma} - p_1 V_1^{\gamma} \times V_1^{1-\gamma}}{\gamma-1} $$
This simplifies to the required form,
$$ W_b = \frac{p_2 V_2 - p_1V_1}{\gamma-1} \qquad constant \; entropy, \; ideal \; gas $$
Example 3.080 A piston cylinder contains 3.6g of air at a pressure of 4 bar and a volume of
1.2 litres.The air expands isentropically to a volume of 2.4 litres. Estimate the moving boundary work.
Problem Statement: Expansion of an ideal gas (air) at constant entropy.
Schematics: - see Figure 7. above.
Assumptions: Stationary fluid - ignore KE and PE changes and allows use of NFEE; Quasi-equilibrium process (will be known as "reversible"). Ideal gas - so internal
energy is a linear function of temperature.
Physical Laws: First Law of Thermodynamics - gives NFEE (used in deriving isentropic relationship). Ideal gas law and work definition are used in
an equation for work done.
Calculation: The pressure at end of process is,
$$p_2 = p_1 (\frac{V_1}{V_2})^{\gamma} = 4 \times (\frac{1}{2})^1.4 = 1.516 bar $$
The work follows from
$$ W_b = \frac{p_2 V_2 - p_1 V_1}{\gamma-1} = \frac{1.516bar \times 2.4 litres - 4 bar \times 1.2 litres}{0.4} \times [0.1 \frac{m^3 kN/m^2}{litre bar}] $$
$$ W_b = -0.290 kJ$$
Discussion: Let us exploit an alternative route to check the calculation. Use the Ideal Gas Law to obtain the
initial temperature,
$$T_1 = \frac{p_1 V_1}{m R} = \frac{4 bar \times 1.2 litres \times [100kN/m^2bar] \times [0.001m^3/litre]}{0.0036kg \times 0.287kJ/kgK } = 465 K$$
$$ T_2 = T_1 (\frac{V_1}{V_2})^{\gamma-1 } = 465 \times (1.2/2.4)^{0.4} = 352 K) $$
Noting Q=0, apply the NFEE to obtain,
$$ W_b= U_2-U_1-Q = m c_v (T_2-T_1)-0 = 0.0036 \times 0.718 \times (352-465) = -0.292 kJ $$
(The difference in the third signficant figure can be attributed to rounding errors.) I also used the
interactive chart. In view of the limitations of setting parameters by eye the agreement \( W_b \approx -0.295 bar \) is
encouraging. A screenshot is shown below.
6.) Polytropic Processes
The equation \( pV^n \) follows from observation. It is not a physical law.
In internal combustion engines, in particular, it is frequently observed that the pressure-volume relationship is close to,
\begin{equation}
pV^n = constant
\end{equation}
where n is the polytropic index of compression or expansion. The value of n is in the range from zero to infinity, depending on conditions. Although \(n=\gamma\) is only one of many cases, the mathematics of polytropic and isentropic processes are identical. In equations, term n simply replaces term \(\gamma\). In terms of (T,V) and (p,T) datasets,
The polytropic relationships describe many processes. They do not describe all processes. (I used to set examination questions in which a piston-cylinder acted against a spring. About 30% of responses incorrectly attempted to fit the resulting linear pV relationship to a polytropic function.) If \( n \neq \gamma \) then the heat transfer is non-zero. For ideal gases special values of n are listed below:
\(n= 1.2, 1.3 \) are typical compression indices observed for internal combustion engines.
When compared with isothermal processes (in ideal gases), the polytropic process is less steep
if \(n < 1\) and steeper if \(n >1 \). The pV plot is a horizontal line if \(n=0\) and a vertical line if \(n=\infty\).The polytropic process describes processes that involve both heat and
work transfer.
Example 3.090 A piston cylinder contains 3.6g of air at a pressure of 4 bar and a volume of 1.2 litres. The air expands polytropically with an expansion index of n=1.2 to a volume of 2.4 litres. Estimate the moving boundary work and the heat transfer to the air.
Problem Statement: polytropic expansion of an ideal gas (air) with expansion index n=1.2.
Schematics: - see Figure 7. above (same shape as isentropic, albeit less steep).
Assumptions: stationary fluid - ignore KE and PE changes and allows use of NFEE; quasi-equilibrium process (will be known as "reversible"); ideal gas - internal energy is
a linear function of temperature.
Physical Laws: First Law of Thermodynamics - gives NFEE (used in deriving isentropic relationship); Ideal Gas Law and Work Definition are used in
an equation for work.
Calculation: The pressure at end of process is,
$$p_2 = p_1 (\frac{V_1}{V_2})^{n} = 4 \times (\frac{1}{2})^{1.2} = 1.741 bar $$
The polytropic moving boundary work follows from
$$ W_b = \frac{p_2 V_2 - p_1 V_1}{n-1} = \frac{1.741bar \times 2.4 litres - 4 bar \times 1.2 litres}{0.2} \times [0.1 \frac{m^3 kN/m^2}{litre \; bar}] $$
$$ W_b = -0.311 kJ$$
To obtain the internal energy change we require start and end temperatures. From the Ideal Gas Law,
$$T_1 = \frac{p_1 V_1}{m R} = \frac{4bar \times [100 kN/m^2 bar]\times 1.2 litres \times [0.001 m^3/litre]}{0.0036kg \times 0.287kJ/kgK } = 465K$$
To find the end temperature we can use either a polytropic relationship ( \( T \propto V^{n-1} \) ) or the Ideal Gas Law. From the latter,
$$ T_2 = \frac{p_2 V_2}{m R} = \frac{1.741bar \times [100 kN/m^2 bar]\times 2.4 litres \times [0.001 m^3/litre]}{0.0036kg \times 0.287kJ/kgK } = 404K $$
Find change in internal energy
$$ U_2-U_1 = m c_v (T_2-T_1) = 0.0036 \times 0.718 \times (404-465) = -0.158 kJ$$
Apply NFEE,
$$ Q = U_2-U_1-W_b = -0.158 -(-0.311) = +0.153kJ $$
Discussion: I also used the
interactive chart. In view of the limitations of setting parameters by eye the agreement to three significant Figures is
encouraging. A screenshot is shown below. It will be seen that \( \Delta U \) is exactly half the work \( \Delta U =W/2\). This can be checked
by comparing the multipliers \( c_v \) (to obtain \( \Delta U \)) and \( R/(n-1)\) (to obtain work).
(I set up the 7 bar isobar and found thereon by trial and error the point where T1=473 K.
The interactive chart displays p1, V1, T1 and I used a pocket calculator to get V2 (see isochor), moved the limit lines over the isochors, highlighted the isotherm and pressed the "integrate" button.
