# 8.5: Entropy and the Substance Models

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To apply the entropy accounting equation, we often need to determine the value of the specific entropy \(s\) of the substances found in or crossing the boundary of our system. The purpose of this section is to extend the two different substance models we studied previously—ideal gases and incompressible substances with room-temperature specific heats—to cover entropy.

## 8.5.1 Relating \(s\) to \(T\), \(P\), \(u\), \(h\), and \(v\) — The \(Tds\) relations

Before we give the working relationships for calculating changes in specific entropy as a function of other specific properties, we need to consider how, in fact, these properties are related.

Consider, if you will, a closed system that contains a *simple, compressible substance*. At this point you should be asking yourself, "What exactly is a simple, compressible substance?" A **simple, compressible substance** is a substance for which the only pertinent reversible (or quasiequilibrium) work mode is \(\text{PdV}\) work (compression and expansion work). There are substances (for example, a steel bar) that can undergo two different quasiequilibrium work processes—\(\text{PdV}\) work and elastic (or spring) work; however, in a given problem only one of the quasiequilibrium work modes may be important. (A steel bar where only elastic work is important would be called a *simple, elastic substance*). We will not belabor this point further except to say that both of our substance models apply to a simple, compressible substance.

Now consider the behavior of this system over a small time interval: \[\begin{array}{lll} \dfrac{d E_{sys}}{dt} = \dot{Q}_{\text{in}} + \dot{W}_{\text{in}} & \rightarrow & dE = \delta Q_{\text{in}} + \delta W_{\text{in}} \\ \dfrac{d S_{sys}}{dt} = \dfrac{\dot{Q}}{T_{b}} + \dot{S}_{gen} & \rightarrow & dS = \dfrac{\delta Q_{\text{in}}}{T_{b}}+\delta S_{gen} \end{array} \nonumber \] In addition we will make the following assumptions:

- the process is internally reversible, therefore all intensive properties are spatially uniform during the process,
- the substance temperature \(T\) and the boundary temperature \(T_{\mathrm{b}}\) are the same,
- the only possible work mode is PdV work, and
- the changes in kinetic and gravitational potential energy are negligible for the process.

Under these conditions, Eq. \(\PageIndex{1}\) simplifies as below: \[\begin{array}{l|l} \cancel{dE}^{=d U} = \delta Q_{\text{in}} + \cancel{\delta W_{\text{in}}}^{= -P d V\kern-0.5em\raise0.3ex-} & \\ dS = \dfrac{\delta Q_{\text{in}}}{\cancel{T_{b}}^{=T}} + \underbrace{ \cancel{\delta S_{gen}}^{=0} }_{\begin{array}{c} \text{Internally} \\ \text{Reversible} \end{array}} & \quad \rightarrow \quad dU = \cancel{\delta Q_{\text{in}}}^{=TdS} -P d V\kern-0.8em\raise0.3ex- \end{array} \nonumber \]

This finally gives us a relationship between the internal energy, temperature, entropy, pressure, and volume of the system: \[dU = T dS - P d V\kern-0.8em\raise0.3ex- \nonumber \] For our purposes, we will rewrite this in terms of intensive properties as follows: \[\begin{array}{l|l} dU = T dS-P d V\kern-0.8em\raise0.3ex- & \\ d(mu) = T d(ms)-P d(mv) & \quad \rightarrow \quad du = Tds - Pdv \\ m du = mTds - mPdv & \end{array} \nonumber \]

A related equation can be obtained by combining Eq. \(\PageIndex{3}\) with the defining equation for enthalpy as follows: \[ \begin{gather} \left. \begin{array} du = Tds - Pdv \\ h = u + Pv \end{array} \right| \quad \rightarrow \quad \underbrace{ d(h-Pv) }_{=u} = Tds - Pdv \nonumber \\ dh-[Pdv+vdP] = Tds - Pdv \\ dh = Tds + vdP \nonumber \end{gather} \nonumber \]

When rearranged in the "standard" form, Eqs. \(\PageIndex{3}\) and \(\PageIndex{4}\) are known collectively as the \(Tds\) relations for a simple, compressible substance: \[\begin{array}{lc} T ds = du + P dv \quad & \text { Tds relations for a } \\ T ds = dh - v dP \quad & \text { simple, compressible substance } \end{array} \nonumber \] These two equations are valid for any simple, compressible substance and are the means for relating the change in specific entropy to other more easily measured properties.

## 8.5.2 \(\Delta s\) — Ideal Gas Model

Starting with the first \(Tds\) relation, we find the following for an ideal gas: \[d s=\frac{du}{T} + \frac{P}{T} dv \quad \text { where } \quad du = c_{v} dT \quad \text { and } \quad \frac{P}{T} = \frac{R}{v} \nonumber \] After substitution, this gives up \[ds = c_{v} \frac{dT}{T} + R \frac{dv}{v} \quad\quad \text { Ideal gas } \nonumber \] This expression is good for *any* ideal gas. A similar expression can be obtained starting with the second \(Tds\) relation as follows: \[ds = \frac{dh}{T} - \frac{v}{T} dP \quad \text { where } \quad dh = c_{P} dT \quad \text { and } \quad \frac{v}{T} = \frac{R}{P} \nonumber \] After substitution, this finally gives us: \[ds = c_{P} \frac{dT}{T} - R \frac{dP}{P} \quad\quad \text { Ideal gas } \nonumber \] which is also good for any ideal gas.

If we restrict ourselves to the ideal gas model with room-temperature specific heats, Eqs. \(\PageIndex{6}\) and \(\PageIndex{7}\) can be integrated to give the following two equations: \[ \Delta s = s_{2}-s_{1} = c_{v} \ln \left(\frac{T_{2}}{T_{1}}\right) + R \ln \left(\frac{v_{2}}{v_{1}}\right) \quad \left| \begin{array}{c} \text { Ideal gas model } \\ \text { with room-temperature } \\ \text { specific heats } \end{array} \right. \nonumber \] and \[ \Delta s = s_{2}-s_{1} = c_{P} \ln \left(\frac{T_{2}}{T_{1}}\right) - R \ln \left(\frac{P_{2}}{P_{1}}\right) \quad \left| \begin{array}{c} \text { Ideal gas model } \\ \text { with room-temperature } \\ \text { specific heats } \end{array} \right. \nonumber \] Either of these equations can be used to calculate the change in specific entropy between two states. Note that unlike specific internal energy \(u\) and specific enthalpy \(h\), the specific entropy depends on *two* properties: \(T\) and \(v\), or \(T\) and \(P\). You will find both of these equations in the summary table for the ideal gas model (found in Section 7.4.3 of this text).

Under some operating conditions, we will find that the change in specific entropy is zero, i.e. \(\Delta s=0\). This will happen frequently enough that special equations are often developed for these conditions. Starting with Eq. \(\PageIndex{8}\) we have the following: \[\begin{aligned} &\cancel{\Delta s}^{=0} = c_{v} \ln \left(\frac{\cancel{T_{2}}^{=T_{2 s}}}{T_{1}}\right) + R \ln \left(\frac{\cancel{v_{2}}^{=v_{2 s}}}{v_{1}}\right) \quad \rightarrow \quad \ln \left(\frac{T_{2s}}{T_{1}}\right) = -\frac{R}{c_{v}} \ln \left(\frac{v_{2s}}{v_{1}}\right) \\ &\text { but } \frac{R}{c_{v}} = \frac{c_{p}-c_{v}}{c_{c}} = \frac{c_{p}}{c_{v}}-1 = k-1 \quad \rightarrow \quad \ln \left(\frac{T_{2s}}{T_{1}}\right) = -(k-1) \ln \left(\frac{v_{2s}}{v_{1}}\right) \end{aligned} \nonumber \]

This gives us finally \[\frac{T_{2s}}{T_{1}} = \left(\frac{v_{2s}}{v_{1}}\right)^{1-k} \quad \text { when } s_{2} = s\left(T_{2s}, \ v_{2s}\right) = s\left(T_{1}, \ v_{1}\right) = s_{1} \nonumber \] where \(T_{2 s}\) and \(v_{2 s}\) are the temperature and specific volume, respectively, at state 2 when \(s_{2}=s_{1}\). A similar expression can be developed from Eq. \(\PageIndex{9}\) and is presented here without development: \[\frac{T_{2s}}{T_{1}} = \left(\frac{P_{2s}}{P_{1}}\right)^{\frac{k-1}{k}} \quad \text { where } s_{2} = s \left(T_{2s}, P_{2s}\right) = s \left(T_{1}, P_{1}\right) = s_{1} \nonumber \] By combining Eqs \(\PageIndex{10}\) and \(\PageIndex{11}\), the following equation can be developed which is valid between any two states or along any process where \(s\) is a constant: \[ P v^{k} = \text { Constant } \quad\quad \begin{array}{c} \text{Ideal gas model} \\ \text {with room-temperature specific heats} \\ \text { and } \Delta s=0 \end{array} \nonumber \] If you were to remember one equation for conditions with \(\Delta s=0\), I would recommend Eq. \(\PageIndex{12}\). Combined with the ideal gas law, \(Pv = RT\), you can recover the other equations with a little algebra.

Any process in which the specific entropy remains constant is called an **isentropic process**. Eqs. \(\PageIndex{10}\), \(\PageIndex{11}\), and \(\PageIndex{12}\) are all valid for isentropic processes where the substance can be modeled as an ideal gas with room-temperature specific heats.

## 8.5.3 \(\Delta \mathbf{s}\) — Incompressible Substance Model

Starting with the first \(Tds\) relation, we find the following for an incompressible substance: \[ds = \frac{du}{T} + \frac{P}{T} dv \quad \text { where } \quad du=c dT \quad \text { and } \quad v = \text { constant } \nonumber \] After substitution, this finally gives us \[ds=c \frac{dT}{T} \quad \text { Incompressible substance } \nonumber \] This expression is good for any incompressible substance.

If we restrict ourselves further to the incompressible substance model with room-temperature specific heats we can integrate Eq. \(\PageIndex{13}\) giving us the following relationship: \[\Delta s = s_{2}-s_{1} = c \ln \left(\frac{T_{2}}{T_{1}}\right) \quad\quad \begin{array}{c} \text { Incompressible substance model } \\ \text { with } \\ \text { room-temperature specific heats } \end{array} \nonumber \] If you model something as an incompressible substance with room-temperature specific heats, use Eq. \(\PageIndex{14}\) to calculate the change in specific entropy. This equation can also be found in the summary table for the incompressible substance model (from Section 7.4.3 of this text).

Under some operating conditions, we will find that the change in specific entropy is zero, i.e. \(\Delta s=0\). Under these conditions, Eq. \(\PageIndex{14}\) reduces to the following: \[\begin{array}{cc} \cancel{\Delta s}^{=0} = c \ln \left(\dfrac{\cancel{T_{2}}^{=T_{2s}}}{T_{1}}\right) & \quad \begin{array}{c} \text { Incompressible substance model } \\ \text { with } \end{array} \\ T_{2 s}=T_{1} & \text { room-temperature specific heats } \\ & \text { and } \Delta s=0 \end{array} \nonumber \] Thus, for an isentropic process of an incompressible substance the temperature of the substance is constant.

## 8.5.4 Examples

In the following examples, the only thing that is new is the use of the ideal gas model and incompressible substance model as required to relate changes in specific entropy to other intensive properties.

Revisit the Pumping Kerosene example in Section 7.4.3 of this text. If the pumping occurs in an adiabatic, reversible process, determine (a) the temperature of the kerosene leaving the pump and (b) the power required to operate the pump, in \(\mathrm{ft} \cdot \mathrm{lbf} / \mathrm{lbm}\).

###### Solution

We will use exactly the same system as selected before and will also assume that kerosene can be modeled as an incompressible substance with room-temperature specific heats.

Applying the conservation of energy equation and the incompressible substance model to this system gives us the same result as before: \[\begin{aligned} \frac{\dot{W}_{\text {pump, in}}}{\dot{m}} &= \underbrace{\left(h_{2}-h_{1}\right)}_{=c\left(T_{2}-T_{1}\right) + v \left(P_{2}-P_{1}\right)} + \underbrace{ \cancel{\left(\frac{V_{2}^{2}}{2}-\frac{V_{1}^{2}}{2}\right)}^{=0}}_{\text {Same reasoning as before}} + g \left(z_{2}-z_{1}\right) \\ &= c \left(T_{2}-T_{1}\right) + v\left(P_{2}-P_{1}\right) + g \left(z_{2}-z_{1}\right) \end{aligned} \nonumber \] Now applying the entropy accounting equation to this system, we have

\[\underbrace{ \cancel{\frac{d S_{sys}}{dt}}^{=0} }_{\text{Steady-state}} = \underbrace{ \cancel{\sum_{j=1}^{N} \frac{\dot{Q}_{j}}{T_{b, j}}}^{=0} }_{\text{Adiabatic}} + \underbrace{\dot{m}_{1} s_{1} - \dot{m}_{2} s_{2}}_{\dot{m}_{1} = \dot{m}_{2} = \dot{m}} + \dot{S}_{gen} \quad \rightarrow \quad \underbrace{ \cancel{\dot{S}_{gen}}^{=0} }_{\begin{array}{c} \text{Reversible} \\ \text{process} \end{array}} = \dot{m} \left(s_{2}-s_{1}\right) \quad \rightarrow \quad s_{2}-s_{1} = 0 \nonumber \]

To go further we must apply the incompressible substance model and evaluate the change in specific entropy: \[\begin{array}{c|} \text { Substance model: } \quad \Delta s = c \ln \left(\frac{T_{2}}{T_{1}}\right) \\ + \\ \text { Process information: } \quad \Delta s=0 \end{array} \Rightarrow \quad \cancel{\Delta s}^{=0} = c \ln \left(\frac{T_{2}}{T_{1}}\right) \quad \rightarrow \quad T_{2}=T_{1} \nonumber \]

Thus the outlet temperature is the same as the inlet temperature and the work for the pump can be found as follows from our earlier efforts: \[\frac{\dot{W}_{\text {pump, in}}}{\dot{m}} = \underbrace{c \cancel{\left(T_{2}-T_{1}\right)}^{=0}}_{\begin{array}{c} T_{2}-T_{1} \text{ for this} \\ \text {process} \end{array}} + \frac{1}{\rho}\left(P_{2}-P_{1}\right) + g \left(z_{2}-z_{1}\right) = (126.6+15.0) \ \frac{\mathrm{ft} \cdot \mathrm{lbf}}{\mathrm{lbm}} = 142 \ \frac{\mathrm{ft} \cdot \mathrm{lbf}}{\mathrm{lbm}} \nonumber \]

**Comment:**

Note that if the process is internally reversible, it only takes approximately \(43 \%\) of the power input that was required when the kerosene temperature increased only \(0.5^{\circ} \mathrm{F}\).

If we revisit the entropy production equation, we can rearrange it so that the impact of entropy production on temperature change is easily seen: \[\begin{array}{c} \dot{S}_{gen} = \dot{m}\left(s_{2}-s_{1}\right) \\ s_{2}-s_{1} = c \ln \left(\dfrac{T_{2}}{T_{1}}\right) \\ T_{2} = T_{1} + \Delta T \end{array} \quad \rightarrow \quad \dfrac{\dot{S}_{gen}}{\dot{m}} = c \ln \left(1+\frac{\Delta T}{T_{1}}\right) \quad \text { or } \quad T_{2}-T_{1} = T_{1} \cdot \left[\exp \left(\frac{\dot{S}_{gen}}{\dot{m} c}\right)-1\right] \nonumber \]

Combining this with the general equation for the power into a pump gives the following result: \[\begin{aligned} \frac{\dot{W}_{\text {pump, in}}}{\dot{m}} &= c \left(T_{2}-T_{1}\right) + \frac{1}{\rho} \left(P_{2}-P_{1}\right) + \left(\frac{V_{2}^{2}}{2}-\frac{V_{1}^{2}}{2}\right) + g \left(z_{2}-z_{1}\right) \\ &= c T_{1} \cdot \left[\exp \left(\frac{\dot{S}_{gen}}{\dot{m} c}\right)-1\right] + \frac{1}{\rho}\left(P_{2}-P_{1}\right) + \left(\frac{V_{2}^{2}}{2}-\frac{V_{1}^{2}}{2}\right) + g\left(z_{2}-z_{1}\right) \end{aligned} \nonumber \] Now using the equation above, determine the theoretical minimum amount of power that must be supplied to the pump. [Just an equation is acceptable.]

For a pump, the "best" performance corresponds to minimizing the power required to operate the pump. What type of process gives you this "best" performance?

A rigid tank contains \(5 \mathrm{~kg}\) of nitrogen \(\left(\mathrm{N}_{2}\right)\) initially at a pressure of \(100 \mathrm{~kPa}\) and a temperature of \(27^{\circ} \mathrm{C}\). The tank also contains a paddle wheel that has negligible mass and volume. The walls of the tank are made of an insulating material. The paddle wheel operates for several minutes and does \(500 \mathrm{~kJ}\) of work on the gas inside the rigid tank. Determine the (a) final pressure and temperature of the gas, and (b) the amount of entropy produced during this process. Assume \(\mathrm{N}_{2}\) can be modeled as an ideal gas with room-temperature specific heats.

###### Solution

**Known:** Nitrogen gas is contained in a rigid tank and is stirred by a paddle wheel in the tank.

**Find:** (a) The final temperature and pressure of the gas.

(b) The entropy produced during the stirring process.

**Given: **

Initial conditions of \(\mathrm{N}_{2}\) gas:

\(T_{1}=27^{\circ} \mathrm{C}\)

\(P 1=100 \mathrm{~kPa}\)

\(m_{1}=5 \mathrm{~kg}\)

Work done by paddle wheel on the gas: \(W_{\text {Paddle-wheel, in}}=500 \mathrm{~kJ}\)

Paddle wheel mass and volume are negligible.

Tank is rigid and made of insulating material.

Figure \(\PageIndex{1}\): Tank contains a gas and a paddle wheel.

**Analysis:**

Strategy \(\rightarrow\) Assume we will need to use energy because we want to know the change in a property of gas when work is done on it. And we will need to use entropy because we are asked about the entropy produced.

System \(\rightarrow\) Treat the gas in the tank as a closed system.

Property to count \(\rightarrow\) Mass and energy.

Time interval \(\rightarrow\) Finite-time interval since beginning and end state are implied.

Starting with the conservation of energy equation for a closed-system, finite-time system we have the following: \[\underbrace{ \cancel{\Delta E}^{= \Delta U}}_{\begin{array}{c} \text { Neglecting changes in } \\ \text { kinetic and gravitational } \\ \text{ potential energy. } \end{array}} = \cancel{Q_{\text {in}}}^{=0} + \cancel{W_{\text{in}}}^{=W_{\text{PW, in}}} \quad \rightarrow \quad \Delta U = W_{\text{PW, in}} \nonumber \] (Don't despair if you cannot write down this closed-system, finite-time form immediately. Just start with the complete equation and apply the two assumptions — closed system and finite time. You should only start with a simplified form if you explicitly state the simplifications for the equation. Note that I clearly stated what I assumed before I wrote down the equation.)

Now if we apply the ideal gas model we have the following: \[\Delta U = m \Delta u = m c_{v} \left(T_{2}-T_{1}\right) \quad \text { where } c_{v}=0.743 \ \frac{\mathrm{kJ}}{\mathrm{kg} \cdot \mathrm{K}} \nonumber \] where the specific heat value is taken from the tables of thermophysical properties for \(\mathrm{N}_{2}\).

Substituting this back into the energy equation, we can solve for the final temperature as follows:

\[ \begin{gathered} m c_{v} \left(T_{2}-T_{1}\right) = W_{\text{PW, in}} \quad \rightarrow \quad T_{2} = T_{1} + \frac{W_{\text{PW, in}}}{m c_{v}} \\ T_{2} = \left(27^{\circ} \mathrm{C}\right) + \frac{500 \mathrm{~kJ}}{(5 \mathrm{~kg}) \left(0.743 \ \dfrac{\mathrm{kJ}}{\mathrm{kg} \cdot \mathrm{K}} \right)} = (27+134.6) ^{\circ} \mathrm{C} = 161.7 ^{\circ} \mathrm{C} \end{gathered} \nonumber \]

Solving for the final pressure, we can make use of the ideal equation and ratios: \[\begin{gathered} \begin{array}{c} P_{2} V\kern-1.0em\raise0.3ex-_{2} = m_{2} R T_{2} \\ P_{1} V\kern-1.0em\raise0.3ex-_{1} = m_{1} R T_{1} \end{array} \quad \rightarrow \quad \dfrac{P_{2} \cancel{V\kern-1.0em\raise0.3ex-_{2}}}{P_{1} \cancel{V\kern-1.0em\raise0.3ex-_{1}}} = \dfrac{\cancel{m_{2}} \cancel{R} T_{2}}{\cancel{m_{1}} \cancel{R} T_{1}} \quad \rightarrow \quad \dfrac{P_{2}}{P_{1}} = \dfrac{T_{2}}{T_{1}} \\ P_{2} = P_{1} \left(\dfrac{T_{2}}{T_{1}}\right) = (100 \mathrm{~kPa}) \left(\dfrac{161.7+273}{27+273}\right) = 145 \mathrm{~kPa} \end{gathered} \nonumber \]

Now to find the entropy produced, we must apply the entropy accounting equation to this same system. This time we will start with the general rate form and simplify it explicitly:

\[ \begin{gathered} \frac{d S_{sys}}{dt} = \underbrace{ \cancel{\sum_{j=1}^{N} \frac{\dot{Q}_{j}}{T_{b, \ j}}} }_{\text{Adiabatic}} + \underbrace{ \cancel{\sum_{\text{in}} \dot{m}_{i} s_{i} - \sum_{\text{out}} \dot{m}_{e} s_{e}}^{=0} }_{\text{Closed system}} + \dot{S}_{gen} \\ \frac{d S_{sys}}{dt} = \dot{S}_{gen} \quad \rightarrow \quad \int\limits_{t_{1}}^{t_{2}} \left( \frac{d S_{sys}}{dt} \right) dt = \int\limits_{t_{1}}^{t_{2}} \dot{S}_{gen} \ dt \quad \rightarrow \quad \underbrace{\Delta S_{sys}}_{\begin{array}{c} \text{Change in entropy} \\ \text{of the system} \end{array}} = \underbrace{S_{gen}}_{\begin{array}{c} \text{Entropy generated} \\ \text{inside the system} \end{array}} \end{gathered} \nonumber \]

Now to go further we must use the ideal gas model as follows: \[\begin{aligned} S_{gen} &= \Delta S = m \Delta s = m \underbrace{ \left[c_{v} \ln \left(\frac{T_{2}}{T_{1}}\right) + R \ln \cancel{\left(\frac{v_{2}}{v_{1}}\right)}^{v_{2}=v_{1}}\right]}_{\Delta s \text { from ideal gas model }} = m c_{v} \ln \left(\frac{T_{2}}{T_{1}}\right)\\[4pt] &= (5 \mathrm{~kg})\left(0.743 \ \frac{\mathrm{kJ}}{\mathrm{kg} \cdot \mathrm{K}}\right) \ \ln \left(\frac{161.7+273}{27+273}\right) = 1.378 \ \frac{\mathrm{kJ}}{\mathrm{K}} \end{aligned} \nonumber \]

**Comments:**

(1) The fact that entropy is produced in this process indicates that doing paddle wheel work on an ideal gas is an *irreversible* process. The source of the irreversibility is the fluid friction between the surface of the paddle wheel and the gas. If the gas was frictionless, the paddle wheel would just slide without friction through the gas and thus do no work on the gas.

(2) Could you have used the other equation for \(\Delta s\) to calculate the entropy change? Absolutely and you should get exactly the same result: \[S_{gen} = m \Delta s = m \left[c_{P} \ln \left(\frac{T_{2}}{T_{1}}\right) - R \ln \left(\frac{P_{2}}{P_{1}}\right)\right] \nonumber \]

\[\begin{aligned} S_{gen} &= (5 \mathrm{~kg}) \left[\left(1.04 \ \frac{\mathrm{kJ}}{\mathrm{kg} \cdot \mathrm{K}}\right) \ \ln \left(\frac{161.7+273}{27+273}\right)-\left(0.2968 \ \frac{\mathrm{kJ}}{\mathrm{kg} \cdot \mathrm{K}}\right) \ \ln \left(\frac{145}{100}\right)\right] \\[5pt] &= (5 \mathrm{~kg})\left[\quad 0.385709 \quad-\quad 0.110280 \quad \right] \frac{\mathrm{kJ}}{\mathrm{kg} \cdot \mathrm{K}}=1.377 \ \frac{\mathrm{kJ}}{\mathrm{K}} \end{aligned} \nonumber \] Note that this answer differs from the earlier calculation due to round off errors in the problem. Because entropy calculations often require taking the difference of two terms of equal magnitude, it is suggested that you carry several more significant figures in the intermediate calculations than you present in the final answer.

(3) How would your solution have changed if the mass and volume of the paddle wheel had not been negligible?

System \(\rightarrow\) must include paddle wheel that is modeled as an incompressible substance.

Conservation of energy \(\rightarrow\) would have to include internal energy change of paddle wheel in \(\Delta U_{sys}\)

Entropy accounting \(\rightarrow\) would have to include entropy change of the paddle wheel in \(\Delta S_{sys}\).

(4) How would the original solution change if an electric resistance heater of negligible mass had been used instead of a paddle wheel? A review of this problem should show you that nothing about the solution would change. All of the answers would be the same.

(5) How would the solution change if the energy had been added to the rigid tank by heat transfer instead of the paddle wheel?

Conservation of energy \(\rightarrow\) No change. Final \(P\) and \(T\) would be identical.

Entropy accounting \(\rightarrow \ S_{2}-S_{1} = \displaystyle \int\limits_{t_{1}}^{t_{2}} \left(\dfrac{\dot{Q}_{\text{in}}}{T_{b}}\right) dt + S_{g e n} \quad \rightarrow \quad S_{gen} = \underbrace{\left(S_{2}-S_{1}\right)}_{\begin{array}{c} \text {Same as as in problem} \\ \text {with paddle wheel} \end{array}} - \int\limits_{t_{1}}^{t_{2}} \left(\frac{\dot{Q}_{\text{in}}}{T_{b}}\right) dt\)

So the entropy production will be less than that found in the original problem since the change in entropy of the system only depends on the end states and the end states are the same in both processes. From an energy utilization standpoint, the fact that the paddle wheel and the electric resistor produce more entropy means that they have *destroyed more potential to do work* than would have been destroyed by the heat transfer process. If we can assume that the temperature of the gas is spatially uniform during the heating process and \(T_{b}=T\), then

\[\frac{dU}{dt} = \dot{Q}_{\text{in}} \quad \rightarrow \quad dU = \dot{Q}_{\text{in}} \ dt \quad \rightarrow \quad \frac{1}{T} dU = \frac{\dot{Q}_{\text{in}}}{T} dt \quad \rightarrow \quad m c_{v} \frac{dT}{T} = \frac{\dot{Q}_{\text{in}}}{T} dt \nonumber \]

and \[S_{gen} = m \left(s_{2}-s_{1}\right) - \underbrace{\int\limits_{T_{1}}^{T_{2}} \left(\frac{m c_{v}}{T}\right) dt}_{=m c_{v} \ln \left(\frac{T_{2}}{T_{1}}\right)} = m \left[\left(s_{2}-s_{1}\right) - c_{v} \ln \left(\frac{T_{2}}{T_{1}}\right)\right] = m \left\{ \underbrace{\left[c_{v} \ln \left(\frac{T_{2}}{T_{1}}\right) + R \ln \cancel{\left(\frac{v_{2}}{v_{1}}\right)}^{v_{2}=v_{1}}\right]}_{\Delta s \text { for an ideal gas }} - c_{v} \ln \left(\frac{T_{2}}{T_{1}}\right) \right\} = 0 \nonumber \]

Wow, what happened to the entropy production? It would seem that we have placed enough constraints on the process that it is now internally reversible. In practice, it would be *impossible* to heat the gas in such a fashion that there are no internal temperature gradients. With this in mind the answer represents a minimum estimate of the entropy that would be produced during the heat transfer process.

Experience has shown that anytime we use a work transfer of energy to accomplish something that could have been done by an equal heat transfer of energy, the process with work transfer of energy will produce more entropy. In fact, this is another way to determine if a given process is reversible or irreversible.

Revisiting "Mixing things up" in Section 7.4.3 of the text. We will now calculate the entropy production for the mixing process.

We will begin with the entropy accounting equation, since we already know the final temperature and pressure of the mixture.

Applying the entropy accounting equation to the closed, adiabatic system consisting of the two tanks gives the following: \[ \frac{d S_{sys}}{dt} = \underbrace{ \cancel{\sum_{j=1}^{N} \frac{\dot{Q}_{j}}{T_{b, \ j}}}^{=0} }_{\text{Adiabatic tanks}} + \underbrace{ \cancel{\sum_{\text{in}} \dot{m}_{i} s_{i} - \sum_{\text{out}} \dot{m}_{e} s_{e}}^{=0} }_{\text{Closed system}} + \dot{S}_{gen} \quad \rightarrow \quad \dot{S}_{sys}{dt} = \dot{S}_{gen} \quad \underbrace{\rightarrow}_{\begin{array}{c} \text{Integrating} \\ \text{for finite time} \end{array}} \quad \Delta S = S_{gen} \nonumber \]

Now to evaluate the entropy generation, we need to evaluate the change in entropy for the mixture. As in the past, you are advised to arrange the equation so that you can calculate the *change in specific entropy*. Here is an example of this approach.

\[\begin{aligned} S_{gen} &= \Delta S = S_{2}-S_{1} \\ &= m_{2} s_{2} - \left(m_{A, \ 1} s_{A, \ 1}+m_{B, \ 1} s_{B, \ 1}\right) = \underbrace{\left(m_{A, \ 1}+m_{B, \ 1}\right)}_{=m_{2}} s_{2} - \left(m_{A, \ 1} s_{A, \ 1} + m_{B, \ 1} s_{B, \ 1}\right) = m_{A, \ 1} \left(s_{2}-s_{A, \ 1}\right) + m_{B, \ 1}\left(s_{2}-s_{B, \ 2}\right) \end{aligned} \nonumber \]

To go further requires that we use the ideal gas model. \[\begin{aligned} & s_{2}-s_{A, \ 1}=c_{P} \ln \left(\frac{T_{2}}{T_{A, \ 1}}\right)-R \ln \left(\frac{P_{2}}{P_{A, \ 1}}\right)=\left(1.04 \ \frac{\mathrm{kJ}}{\mathrm{kg} \cdot \mathrm{K}}\right) \ln \left(\frac{367}{300}\right)-\left(0.2968 \ \frac{\mathrm{kJ}}{\mathrm{kg} \cdot \mathrm{K}}\right) \ln \left(\frac{236}{150}\right) = 0.0751338 \ \frac{\mathrm{kJ}}{\mathrm{kg} \cdot \mathrm{K}} \\ &s_{2}-s_{B, \ 1} = c_{P} \ln \left(\frac{T_{2}}{T_{B, \ 1}}\right)-R \ln \left(\frac{P_{2}}{P_{B, \ 1}}\right) = \left(1.04 \ \frac{\mathrm{kJ}}{\mathrm{kg} \cdot \mathrm{K}}\right) \ln \left(\frac{367}{400}\right)-\left(0.2968 \ \frac{\mathrm{kJ}}{\mathrm{kg} \cdot \mathrm{K}}\right) \ln \left(\frac{236}{300}\right) = -0.0183295 \ \frac{\mathrm{kJ}}{\mathrm{kg} \cdot \mathrm{K}} \end{aligned} \nonumber \]

Now substituting these values back into the entropy balance we have the following: \[\begin{aligned} S_{gen} &= m_{A, \ 1} \left(s_{2}-s_{A, \ 1} \right) + m_{B, \ 1}\left(s_{2}-s_{B, \ 1}\right) \\ &= (1 \mathrm{~kg}) \left(0.0751338 \ \frac{\mathrm{kJ}}{\mathrm{kg} \cdot \mathrm{K}}\right) + (2 \mathrm{~kg})\left(-0.0183295 \ \frac{\mathrm{kJ}}{\mathrm{kg} \cdot \mathrm{K}}\right) = 0.03847 \ \frac{\mathrm{kJ}}{\mathrm{K}} \end{aligned} \nonumber \] So as you might expect, entropy is produced as a result of the mixing process.

If you really want a challenge, prove that the final equilibrium temperature and pressure must be the values that you calculated earlier. Why exactly do these have to be *the* final equilibrium values? As you might guess, it has something to do with the entropy production. To do this you must assume do the following:

(1) Assume that each tank can have its own final temperature and pressure, i.e. \(P_{A, \ 2}\), \(P_{B, \ 2}\), \(T_{A, \ 2}\), and \(T_{B, \ 2}\).

(2) Reformulate the problem so that after you pick a final pressure and temperature for, say, Tank A, you can then calculate the corresponding pressure and temperature in Tank B.

(3) Now vary the final pressure and temperature for Tank A over all possible values and solve for \(S_{gen}\).

(4) Find the values for the final pressure and temperature in Tank A that maximizes the entropy generation.