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8.3: Entropy Accounting Equation

  • Page ID
    81512
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    The recommended starting point for any problem that requires the application of the Second Law of Thermodynamics, the determination of the entropy production, or the change in entropy of a system, is the rate form of the entropy accounting equation derived earlier: \[\frac{d S_{sys}}{d t} = \sum_{j=1}^{N} \frac{\dot{Q}_{j}}{T_{b, \ j}} + \sum_{\text {in}} \dot{m}_{i} s_{i} - \sum_{\text {out}} \dot{m}_{e} s_{e} + \dot{S}_{gen} \nonumber \] In applying this equation to describe the behavior of a system, there are several modeling assumptions that are commonly used. These are described in detail in the following paragraphs. As always, you should focus on understanding the physics underlying the assumption and how they are used. Do not just memorize the simplified equations.

    Typical modeling assumptions

    Steady-state system: If a system is operating under steady-state conditions, all intensive properties and interactions are independent of time. Thus, the entropy of the system is constant, \(S_{sys}=\) constant. When applied to the entropy accounting equation you have the following \[\begin{aligned} \underbrace{\cancel{ \frac{d S_{sys}}{dt} }^{=0}}_{\text{Steady-state}} &= \sum_{j=1}^{N} \frac{\dot{Q}_{j}}{T_{b, \ j}} + \sum_{\text {in}} \dot{m}_{i} s_{i} - \sum_{\text {out}} \dot{m}_{e} s_{e} + \dot{S}_{gen} \\ 0 \ &=\sum_{j=1}^{N} \frac{\dot{Q}_{j}}{T_{b, \ j}} + \sum_{\text {in}} \dot{m}_{i} s_{i} - \sum_{\text {out}} \dot{m}_{e} s_{e} + \dot{S}_{gen} \end{aligned} \nonumber \] In words, the net heat transfer rate of entropy into the system plus the net mass transport rate of entropy into the system plus the net generation (production) rate of entropy must equal zero.

    Closed system: A closed system has no mass flow across its boundary. With this constraint, the entropy accounting equation simplifies as follows: \[\begin{aligned} &\frac{d S_{sys}}{dt} = \sum_{j=1}^{N} \frac{\dot{Q}_{j}}{T_{b, \ j}} + \underbrace{ \cancel{\sum_{\text {in}} \dot{m}_{i} s_{i}-\sum_{\text {out}} \dot{m}_{e} s_{e}}^{=0} }_{\text {Closed system} \rightarrow \text{no mass flow}} + \dot{S}_{gen} \\ &\frac{d S_{sys}}{dt} = \sum_{j=1}^{N} \frac{\dot{Q}_{j}}{T_{b, \ j}}+\dot{S}_{gen} \end{aligned} \nonumber \]

    Finite-time, closed system: For a closed system over a finite-time interval, you first apply the closed system assumption and then integrate the equation over the specified time interval: \[\begin{aligned} \frac{d S_{sys}}{dt} = \sum_{j=1}^{N} \frac{\dot{Q}_{j}}{T_{b, \ j}} &+ \cancel{\sum_{in} \dot{m}_{i} s_{j}-\sum_{\text {out}} \dot{m}_{e} s_{e}}^{=0} + \dot{S}_{gen} \\ {\frac{d S_{sys}}{dt}} &= \sum_{j=1}^{N} \frac{\dot{Q}_{j}}{T_{b, \ j}} + \dot{S}_{g e n} \\ \int\limits_{t_{1}}^{t_{2}} \left(\frac{d S_{sys}}{dt}\right) dt &= \int\limits_{t_{1}}^{t_{2}} \left(\sum_{j=1}^{N} \frac{\dot{Q}_{j}}{T_{b, \ j}}\right) dt + \int\limits_{t_{1}}^{t_{2}} \dot{S}_{gen} \ dt \\ S_{sys, \ 2}-S_{sys, \ 1} &= \sum_{j=1}^{N} \left[\int\limits_{t_{1}}^{t_{2}} \frac{\dot{Q}_{j}}{T_{b, \ j}} d t\right] + S_{gen} \end{aligned} \nonumber \] In words, this says the change in the entropy of the system equals the net heat transfer of entropy into the system plus the amount of entropy produced inside the system.

    Internally reversible process: When a process is internally reversible there is no entropy production and \(\dot{S}_{gen} \equiv 0\). Under these conditions the entropy accounting equation reduces to the following: \[\begin{aligned} &\frac{d S_{sys}}{dt} = \sum_{j=1}^{N} \frac{\dot{Q}_{j}}{T_{b, \ j}} + \sum_{\text {in}} \dot{m}_{i} s_{i} - \sum_{\text {out}} \dot{m}_{e} s_{e} + \underbrace{\cancel{\dot{S}_{gen}}^{=0}}_{\begin{array}{c} \text {Internally reversible} \\ \text{process} \end{array}} \\ & \frac{d S_{sys}}{dt} = \sum_{j=1}^{N} \frac{\dot{Q}_{j}}{T_{b, \ j}} + \sum_{\text {in}} \dot{m}_{i} s_{i} - \sum_{\text {out}} \dot{m}_{e} s_{e} \end{aligned} \nonumber \] This raises an interesting point. Unless you can assume that a process is internally reversible, all you know is that the entropy production rate is greater than or equal to zero, \(\dot{S}_{gen} \geq 0\). This means that when you are looking for equations to help solve a problem, the entropy accounting equation brings with it a built-in unknown. For example, if you have three equations and four unknowns before you apply the entropy accounting equation, after you apply the entropy accounting equation you will have four equations and five unknowns unless you can assume an internally reversible process. Actually, the situation is not quite as bleak as this appears. Although you may not know its exact value, you do know that the entropy production rate cannot be negative. In addition, as we will show shortly, any real system must have a value that is positive and will approach zero as we approach the best or optimum behavior for the specified conditions.

    Assumptions about heat transfer and work transfer of entropy:

    Heat transfer of entropy — In this course, we will usually make one of three assumptions about the heat transfer of entropy for a system:

    1. There is no heat transfer and thus there is no heat transfer of entropy.
    2. The heat transfer rate \(\dot{Q}_{j}\) and/or the surface temperature \(T_{b, \ j}\) may be unknown and thus the rate of entropy transfer with heat transfer is also unknown. In this case, the entropy accounting equation may provide an additional equation that relates these two variables.
    3. The heat transfer rate \(\dot{Q}\) and the surface temperature \(T_{b, \ j}\) are both specified and thus the rate of entropy transfer with heat transfer is known.

    Please remember that entropy transfer with heat transfer can only be defined with respect to a boundary. If you move the boundary you may change the heat transfer rate or the surface temperature at which it occurs. Without clearly indicating your system boundary, it is impossible to apply any of these assumptions.

    In applying the conservation of energy equation, you routinely calculated the net heat transfer rate for a system by summing the heat transfer rates over the system boundary. Because the entropy transfer rate with heat transfer depends on both the heat transfer rate and the boundary temperature where the heat transfer occurs, you must be careful to only sum heat transfer rates that occur at the same temperature when determining the heat transfer of entropy. For example, \[\sum_{j=1}^{N} \frac{\dot{Q}_{j}}{T_{b, \ j}} = \frac{1}{T_{b}} \sum_{j=1}^{N} \dot{Q}_{j} = \frac{\dot{Q}_{\text {net, in}}}{T_{b}} \quad\quad \text { only if } T_{b,\ j} = T_{b} \text { for all surfaces. } \nonumber \] Another common problem is assigning a boundary temperature when a system exchanges energy by heat transfer with a fluid-convection heat transfer. To help us answer this question, we need to examine the temperature variation within the system and the fluid near the system boundary. Figure \(\PageIndex{1}\) shows the temperature distribution normal to the boundary in the system and the surrounding fluid. If the convection heat transfer is occurring from the system to the surrounding fluid, the temperature distribution appears as shown in the figure. The temperature decreases inside the system as we approach the boundary. At the boundary (the interface between the system and the surrounding fluid), the temperature is \(T_{\text {surface}}\). Immediately adjacent to the system is a layer of air commonly referred to as the boundary layer. The boundary layer is a layer of air across which the temperature changes from \(T_{\text {surface}}\) to \(T_{\infty}\), the fluid temperature away from the wall.

    Graph of temperature vs position y. In the system, where y is less than or equal to 0, T decreases at a constant rate. For the boundary layer, with y between 0 and a positive value, T exponentially decays. For the fluid, which includes all y greater than the maximum y of the boundary layer, T is nearly constant at the value T_infinity.

    Convection heat transfer through the boundary layer: \(\dot{Q}_{\text {convection}}=h A_{\text {surface}}\left(T_{\text {surface}}-T_{\infty}\right) \)

    Entropy production rate within the boundary layer: \(\underbrace{\cancel{\dfrac{d S_{sys}}{dt}}^{=0}}_{\text{Steady-state}} = \dfrac{\dot{Q}_{\text {conv, in}}}{T_{\text {surface}}} - \dfrac{\dot{Q}_{\text {conv, out}}}{T_{\infty}} + \left.\dot{S}_{\text {gen}} \quad \rightarrow \quad \dot{S}_{\text {gen }}\right|_{\begin{array}{c} \text {boundary} \\ \text{layer} \end{array}} = \dot{Q}_{\text {conv, in}}\left[\dfrac{1}{T_{\infty}}-\dfrac{1}{T_{\text {surface}}}\right] \)

    Figure \(\PageIndex{1}\): Entropy production at a system boundary with convection heat transfer

    If our system does not include the boundary layer, then the correct temperature to use when calculating the heat transfer of entropy for the system is \(T_{b}=T_{\text {surface}}\). If the system does include the boundary layer, then the correct temperature to use when calculating the heat transfer of entropy is \(T_{b}=T_{\infty}\). However, you should note that because the latter system contains additional material (the boundary layer) it may have a different entropy production rate and different entropy transfer rates. As shown in Figure \(\PageIndex{1}\), the process of steady-state heat transfer through the boundary results in an entropy production rate.

    The reason this discussion is important is that frequently in applying the entropy accounting equation you will not know a surface temperature and you will only know (or can reasonably assume) an ambient fluid temperature, say room temperature. In these cases, it is perfectly acceptable to include the boundary layer inside your system so that you know a boundary temperature to calculate the heat transfer of entropy. However, please remember that the entropy production you calculate for a system that includes the boundary layer will be different and greater that what you would calculate for a smaller system that did not include the boundary layer.

    Work transfer of entropy — I have included this here to reinforce a point that there is no transport of entropy with work. Let me say that one more time: there is no transport of entropy with work.

    This leads us to another way to distinguish between a work transfer of energy and a heat transfer of energy. A heat transfer of energy always carries with it an amount of entropy; while a work transfer of energy never does.

    Assumptions about the substance:

    As we discovered with conservation of energy, we will need to evaluate thermophysical properties — \(u\), \(h\), \(s\), \(T\), \(P\), \(\rho\), and \(\upsilon\). This requires empirical knowledge about the behavior of the material within the system. In the last chapter we introduced two different substance models and shortly we will extend these to include the calculation of entropy changes.

    Problem Solving with the Entropy Accounting Equation

    In practice, we will typically apply the entropy accounting equation to three different types of problems:

    • problems where we are explicitly asked to determine the entropy production rate,
    • problems where we are asked to determine if a given process is possible, i.e. the entropy production rate is within acceptable limits, \(\dot{S}_{gen} \geq 0\), and
    • problems where we are asked to determine the "best" performance that is theoretically possible.

    For the first type of problem where we are asked explicitly to determine the entropy production rate, we will typically apply both the conservation of energy and the entropy accounting equations. In addition, we must be able to determine values for every term in the entropy accounting equation except for the entropy production rate.

    For the second type of problem where we are asked to determine if a given process or device is possible, we again must be able to determine the entropy production rate. Questions about whether a given process or device is possible almost always require that we find the entropy generation rate for the process. Not every process that satisfies conservation of energy also satisfies the entropy accounting equation. Failure to satisfy either of these physical laws means that a given process is impossible.

    For the third type of problem, we are asked to determine the "best" possible performance for the device or process. Again, the key to answering these questions involves the entropy accounting equation. To answer these questions you need to combine the conservation of energy and the entropy accounting equation so that the desired performance can be studied as a function of the entropy generation rate. (Recall that the entropy generation rate is the only quantity in either the conservation of energy or the entropy accounting equation that has any physical limitation placed on its value.) To find the "best" performance possible, you must study the desired performance of the device over the range of possible entropy-generation values and see what constitutes the "best" performance for the specified operating conditions, e.g. maximum power out or minimum power in for a system. Although you are encouraged to investigate specific cases to test this conclusion, experience has shown that the "best" performance that is theoretically possible always occurs when the device or process is internally reversible.

    The following examples serve two functions. First, they demonstrate the mechanics of how to apply entropy accounting with the various modeling assumptions. Second, they are designed to help you learn more about the importance of the property entropy and its production. Please read the problems carefully and, where requested, answer the various questions to the best of your ability.

    Example — The case of the hot resistor!

    A 9-volt battery energizes a \(200 \text{-} \Omega\) resistor. The resistor operates at steady-state conditions. Convection heat transfer occurs between ambient air at \(25^{\circ} \mathrm{C}\) and the resistor. The resistor surface area is \(2.5 \mathrm{~cm}^{2}\), and the convection heat transfer coefficient is \(10 \mathrm{~W} /\left(\mathrm{m}^{2} \cdot { }^{\circ} \mathrm{C}\right)\).

    Determine the (a) magnitude and direction of the heat transfer rate, (b) the entropy generation rate for the resistor alone, and (c) the entropy generation rate for an enlarged system that includes the resistor and the boundary layer of air surrounding the resistor.

    Solution

    Known: A resistor is energized by a battery and operates at steady-state conditions.

    Find: (a) Heat transfer rate, in \(\mathrm{W}\).

    (b) Entropy generation rate for the resistor alone, in \(\mathrm{W}/\mathrm{K}\).

    (c) Entropy generation rate for an enlarged system that includes the convection boundary layer

    Given:

    Current passes through a resistor. A boundary layer of air around the resistor is marked.

    Figure \(\PageIndex{2}\): Current passes through a resistor, which is surrounded by a boundary layer with the air.

    Ambient air temperature: \(T_{\infty}=25^{\circ} \mathrm{C}\)

    Convection heat transfer coefficient: \(h=10 \mathrm{~W} /\left(\mathrm{m}^{2} \cdot { }^{\circ} \mathrm{C}\right)\)

    Surface area: \(A=2.5 \mathrm{~cm}^{2}=2.5 \times 10^{-4} \mathrm{~m}^{2}\)

    DC voltage across resistor: \(\Delta V = 9 \mathrm{~volts}\)

    Resistor resistance: \(R=200 \ \Omega\)

    Steady-state system

    Analysis:

    Strategy \(\rightarrow\) Questions about heat transfer rates usually require application of conservation of energy.

    Questions about entropy production always require application of the entropy accounting equation.

    System \(\rightarrow\) Begin with just the resistor.

    Property to count \(\rightarrow\) Energy and then entropy.

    Time interval \(\rightarrow\) Because it described the system as steady-state, we can assume an infinitesimal time interval.

    A system consisting of only the resistor has heat entering at a steady rate and electric work entering at a steady rate.

    Figure \(\PageIndex{3}\): Closed system consisting only of the resistor.

    (a) To solve for the heat transfer rate we will treat the resistor alone as a closed system and write the conservation of energy equation for this system:

    \[ \begin{aligned} & \underbrace{ \cancel{\frac{dE_{sys}}{dt}}^{=0} }_{\text{Steady-state}} = \dot{Q}_{\text{net, in}} + \dot{W}_{\text{net, in}} + \underbrace{\cancel{ \sum_{\text{in}} \dot{m}_{i} \left(h_{i} + \frac{V_{i}^{2}}{2} + gz\right) - \sum_{\text{out}} \dot{m}_{e} \left(h_{e} + \frac{V_{e}^{2}}{2} + gz\right) }^{=0}}_{\text{Closed system}} \\ & 0=\dot{Q}_{\text {in}} + \dot{W}_{\text {electric, in}} \quad \rightarrow \quad \dot{Q}_{\text {in}} = -\dot{W}_{\text {electric, in}} \end{aligned} \nonumber \] As might be expected, we discover that the heat transfer rate in equals the negative of the electric power in. Alternatively, we could say the electrical power in equals the heat transfer rate out.

    Assuming that the resistor obeys Ohm's Law, we can calculate the electrical power and the heat transfer rate as follows: \[\dot{Q}_{\text {in}} = -\underbrace{\dot{W}_{\text {electric, in}}}_{=i \cdot \Delta V} = -i \cdot \Delta V = -\left(\frac{\Delta V}{R}\right) \cdot \Delta V = -\frac{(\Delta V)^{2}}{R} = -\frac{(9 \mathrm{~V})^{2}}{(200 \ \Omega)} = -0.405 \mathrm{~W} \nonumber \] Thus, the heat transfer rate out of the resistor is \(0.405 \mathrm{~W}\).

    (b) To find the entropy generation rate we will use the same system as above and apply the entropy accounting equation:

    \[ \underbrace{ \cancel{\frac{d S_{sys}}{dt}}^{=0} }_{\text{Steady-state}} = \sum_{j=1}^{N} \frac{\dot{Q}_{j}}{T_{b, \ j}} + \underbrace{ \cancel{\sum_{\text{in}} \dot{m}_{i} s_{i} - \sum_{\text{out}} \dot{m}_{e} s_{e}}^{=0} }_{\text{Closed system}} + \dot{S}_{\text{gen}} \quad \rightarrow \quad 0 = \frac{\dot{Q}_{\text{in}}}{T_{\text{surface}}} + \dot{S}_{\text{gen}} \quad \rightarrow \quad \underbrace{ -\frac{\dot{Q}_{\text{in}}}{T_{\text{surface}}} }_{\begin{array}{c} \text{Heat transfer of entropy} \\ out \text{ of the system} \end{array}} = \dot{S}_{\text{gen}} \nonumber \]

    This says that the entropy transfer rate out of the system with heat transfer equals the entropy generation rate inside the system. Notice that even though our original guess for the direction of the heat transfer rate was incorrect, we are still using the same assumptions for applying the entropy balance. (Changing signs and directions in the middle of a problem is a frequent source of errors in problem solving.)

    To solve for the entropy generation rate we need to find the surface temperature of the resistor, \(T_{b}\), because the system boundary coincides with the surface of the resistor. To do this we can make use of the convection heat transfer relationship as follows: \[\begin{aligned} &\dot{Q}_{\text {in}} = h \cdot A \cdot \left(T_{\infty}-T_{\text {surface}}\right) \quad \rightarrow \quad T_{\text {surface}}-T_{\infty} = \frac{-\dot{Q}_{\text {in}}}{h \cdot A} = \frac{-(-0.405 \mathrm{~W})}{\left(10 \dfrac{\mathrm{W}}{\mathrm{m}^{2} \cdot{ }^{\circ} \mathrm{C}}\right)\left(2.5 \times 10^{-4} \mathrm{~m}^{2}\right)} = 162^{\circ} \mathrm{C} \\[4pt] &T_{\text {surface}} = \left(T_{\text {surface}}-T_{\infty}\right)+T_{\infty} \quad \rightarrow \quad T_{\text {surface}} = 162^{\circ} \mathrm{C} + 25^{\circ} \mathrm{C} = 187^{\circ} \mathrm{C} \end{aligned} \nonumber \] From a practical standpoint, this surface temperature is unacceptably high. However, it does demonstrate a significant problem in miniaturizing electronic components-maintaining acceptable operating temperatures.

    Now that we know the surface temperature, we can calculate the entropy generation rate for the resistor: \[\left. \dot{S}_{\text{gen}}\right|_{\text {Resistor}} = \frac{-\dot{Q}_{\text {in}}}{T_{\text {surface}}} = \frac{-(-0.405 \mathrm{~W})}{(187+273) \mathrm{K}} = \frac{0.405 \mathrm{~W}}{460 \mathrm{~K}} = 0.880 \times 10^{-3} \frac{\mathrm{W}}{\mathrm{K}} \nonumber \] This is the entropy production rate inside the resistor and is the direct result of the irreversible conversion of electrical energy into thermal energy. For resistors, this is sometimes referred to as Joule heating. Note that the surface temperature value was converted to the Kelvin temperature scale, a thermodynamic temperature scale, before it was used to calculate the entropy transfer rate with heat transfer. Had we used the temperature in Celsius degrees, we would have obtained a different and incorrect value for the entropy transfer rate!

    (c) To solve for the entropy generation rate of the enlarged system that includes the boundary layer we need to redefine our system. (See the figure below.) If we apply conservation of energy to this system, we find the same result that we found for the resistor alone: \(-\dot{Q}_{\text{in}} = \dot{W}_{\text {electric, in}}\).

    System consisting of the resistor and its surrounding boundary layer. Heat enters the system at a constant rate, and electrical work enters the system at a constant rate.

    Figure \(\PageIndex{4}\): Closed system consisting of the resistor and its surrounding boundary layer.

    The entropy accounting equation will also look the same with one significant exception—the boundary temperature at which the heat transfer occurs. For the enlarged system, the entropy generation rate for this enlarged system is as follows: \[\left.\dot{S}_{\text{gen}}\right|_{\begin{array}{l} \text {Resistor}+ \\ \text{Boundary Layer} \end{array}} = \frac{-\dot{Q}_{\text {in}}}{T_{\infty}} = \frac{-(-0.405 \mathrm{~W})}{(25+273) \mathrm{K}} = \frac{0.405 \mathrm{~W}}{298 \mathrm{~K}} = 1.359 \times 10^{-3} \ \frac{\mathrm{W}}{\mathrm{K}} \nonumber \] where the boundary temperature was taken as the ambient air temperature. Notice that as one might expect the entropy production rate for the enlarged system is greater than the entropy production rate for just the resistor.

    Comment:

    Although we have answered all the questions, this is such a rich problem and we've already invested time getting started so let's see what else we can learn.

    Can we explain precisely where the additional entropy production is coming from? YES!

    Consider the convection boundary layer as a system by itself. Notice that the wires carrying the electrical power to the resistor pass through this system with no net transfer of electrical power; however, there are two heat transfers. At the inner surface of the boundary layer, the system exchanges energy by heat transfer \(\dot{Q}_{\text {in, surface}}\) with the resistor at \(T_{\text {surface}}\), and at the outer surface of the boundary layer, the system exchanges energy by heat transfer \(\dot{Q}_{\text {out, } \infty}\) with the surroundings at \(T_{\infty}\).

    System consisting only of the boundary layer. Outer edge of the boundary layer is at temperature T_infinity, and inner edge of the boundary layer is at temperature T_surface. Heat from the boundary layer moves out of the system at rate dot Q_out, infinity. Heat from the resistor moves into the boundary layer surface at rate Q_in, surface.

    Figure \(\PageIndex{5}\): System consisting only of the convection boundary layer.

    Now if we write the conservation of energy and the entropy accounting equation for this closed system we have

    \[ \begin{aligned} & \underbrace{ \cancel{\frac{d E_{sys}}{dt}}^{=0} }_{\text{Steady-state}} = \dot{Q}_{\text{in, surface}} - \dot{Q}_{\text{out, } \infty} \quad &\rightarrow \quad \dot{Q}_{\text{out, } \infty} = \dot{Q}_{\text{in, surface}} \\ & \underbrace{ \cancel{\frac{d S_{sys}}{dt}}^{=0} }_{\text{Steady-state}} = \frac{\dot{Q}_{\text{in, surface}}}{T_{\text{surface}}} - \frac{\dot{Q}_{\text{out, } \infty}}{T_{\infty}} + \dot{S}_{\text{gen}} &\rightarrow \quad \frac{\dot{Q}_{\text{out, } \infty}}{T_{\infty}} = \frac{\dot{Q}_{\text{in, surface}}}{T_{\text{surface}}} + \dot{S}_{\text{gen}} \end{aligned} \nonumber \]

    Note that from the entropy balance, the heat transfer of entropy out of the system cannot be less than the heat transfer of entropy into the system and it increases by the entropy production in the system.

    Combining these results gives us the entropy production rate for the boundary layer:

    \[\begin{aligned} \left.\dot{S}_{\text {gen}}\right|_{\begin{array}{l} \text {Boundary} \\ \text{Layer} \end{array}} &= \frac{\dot{Q}_{\text {out, } \infty}}{T_{\infty}}-\frac{\dot{Q}_{\text {in, surface}}}{T_{\text {surface}}} \\ &=\dot{Q}_{\text {in, surface}}\left[\frac{1}{T_{\infty}}-\frac{1}{T_{\text {surface}}}\right] = (0.405 \mathrm{~W}) \left[\frac{1}{298 \mathrm{~K}}-\frac{1}{460 \mathrm{~K}}\right] = 0.479 \times 10^{-3} \ \frac{\mathrm{W}}{\mathrm{K}} \end{aligned} \nonumber \]

    Notice that this entropy production is the result of a steady-state heat transfer across a layer of air with a finite-temperature difference. (Recall that heat transfer across a finite difference was on the list of dissipative effects.) Also notice that the three entropy production rates have a well-defined relationship: \[\underbrace{ \left.\dot{S}_{\text {gen}}\right|_{\begin{array}{l} \text {Resistor}+ \\ \text {Boundary Layer} \end{array}} }_{= 1.359 \times 10^{-3} \ \frac{\mathrm{W}}{\mathrm{K}} } = \underbrace{\left.\dot{S}_{\text {gen}}\right|_{\text {Resistor}}}_{=0.880 \times 10^{-3} \frac{\mathrm{W}}{\mathrm{K}}} + \underbrace{\left.\dot{S}_{\text {gen}}\right|_{\text {Boundary Layer}}}_{=0.479 \times 10^{-3} \ \frac{\mathrm{W}}{\mathrm{K}}} \nonumber \] because the enlarged system is just the sum of the two subsystems. Note that entropy cannot be generated on the boundary of a system, as a boundary is an infinitesimally thin surface with no mass.

    Would it be possible to reverse the direction of the heat transfer of energy for the resistor and get electrical power out of this hot resistor? NO!

    So why not? Let's begin by revisiting our original system that consisted only of the resistor and reverse the direction of the heat transfer rate and the electrical power.

    System consisting only of a resistor, with heat transfer in at a constant rate and electrical work outputted from the system at a constant rate.

    Figure \(\PageIndex{6}\): System from Figure \(\PageIndex{3}\) with the direction of electrical power reversed.

    Again we can apply conservation of energy and entropy accounting to give the following results:

    \[ \left. \begin{array}{ll} & \underbrace{ \cancel{\dfrac{d E_{sys}}{dt}}^{=0} }_{\text{Steady-state}} = \dot{Q}_{\text{in}} - \dot{W}_{\text{electric, out}} \quad &\rightarrow \quad \dot{W}_{\text{electric, out}} = \dot{Q}_{\text{in}} \\ & \underbrace{ \cancel{\dfrac{d S_{sys}}{dt}}^{=0} }_{\text{Steady-state}} = \dfrac{\dot{Q}_{\text{in}}}{T_{\text{surface}}} + \dot{S}_{\text{gen}} & \rightarrow \quad \dot{Q}_{\text{in}} = -\left( T_{\text{surface}} \cdot \dot{S}_{\text{gen}} \right) \end{array} \right| \quad \rightarrow \quad \dot{W}_{\text{electric, out}} = - \underbrace{\left(T_{\text{surface}} \cdot \dot{S}_{\text{gen}}\right)}_{\begin{array}{c} T_{\text{surface}} > 0 \\ \dot{S}_{\text{gen}} \geq 0 \end{array}} \leq 0 \nonumber \]

    Thus the maximum electric power out of this resistor is no power. [So I guess you need to give that kid at the science fair from Section 8.1.1 the prize for the best hoax.]

    Surely with modern technology we can build a steady-state device that receives energy by heat transfer at a single temperature \(\mathbf{T}_{\mathbf{b}}\) and completely converts that energy into a work transfer of energy out of the system. NO!

    Wow, this seems like a pretty strong restriction. Are you sure? Review the development above. Although our system diagram is labeled as a resistor with electrical power out, we could easily rewrite it for any steady-state system and for net power out. The only other restriction is that we have a net heat transfer into the system and that all heat transfer occurs at a single boundary temperature. Under these conditions we get the same result.

    A power cycle receives heat transfer at temperature T_H, converts part of this to work, and outputs the rest as heat transfer at the lower temperature T_L.

    Figure \(\PageIndex{7}\): Basic structure of a power cycle.

    Remember the Kelvin-Planck Statement in Section 8.1.1 that said it was impossible to have a power cycle that is \(100 \%\) efficient. If we had a steady-state device, like a closed-loop, steady-state power cycle that received a heat transfer of energy \(\dot{Q}_{H, \text { in}}\) at a single temperature \(T_{\mathrm{H}}\) and converted all of it to a net work out of the system \(\dot{W}_{\text {net, out}}\), its thermal efficiency would be \(\eta=\dot{W}_{\text {net, out}} / \dot{Q}_{H, \text { in}}=100 \%\). But we just showed that this was impossible! So I guess Kelvin and Planck were right. If you think the key is in whether there is some heat transfer out of the system, you are correct. (We'll investigate this later, or you can pursue it on your own now by trying to determine the minimum heat transfer rate out of a system when the two boundary temperatures and the heat transfer rate into the system are fixed.)

    So what have we learned here?

    First, it seems that work and heat transfer, while both being energy transfer mechanisms, are not interchangeable. Clearly we can build a steady-state device that completely converts work to heat transfer at a single temperature, but doing the reverse is impossible.

    Second, we have seen how some fairly straightforward applications of the entropy accounting equation have demonstrated that a whole class of devices is impossible. If we are given sufficient information to evaluate the entropy production, then we can determine whether a certain process is possible, internally reversible, or impossible. If we do not have sufficient information to calculate the entropy production, we can still use the entropy accounting equation and the restriction on entropy production to determine a range of physically possible solutions. As we will show shortly, the entropy accounting equation will also help us determine the theoretically "best" possible performance of a device.

    Example — Entropy and the "perfect" motor

    An electric motor operating at steady state draws a current of 10 amps with a voltage of 220 volts. The power factor is one. The output shaft rotates at \(1800 \mathrm{RPM} =188.5 \mathrm{rad} / \mathrm{s}\) with a torque of \(10 \mathrm{~N} \cdot \mathrm{m}\) applied to the external load. The heat transfer from the motor occurs by convection to the surroundings. The convection heat transfer coefficient, \(h_{\text{conv}}\), is \(20 \mathrm{~W} /\left(\mathrm{m}^{2} \cdot \mathrm{K}\right)\) and the surface area of the motor is \(A=0.300 \mathrm{~m}^{2}\).

    Determine

    (a) the entropy production rate for the motor, in \(\mathrm{W} / \mathrm{K}\), and

    (b) the maximum theoretically possible shaft power output for this device, i.e. what's the best possible performance?

    Solution

    Known: A motor operates at steady-state conditions.

    Find: (a) The heat transfer rate from the motor, in \(\mathrm{W}\).

    (b) The entropy production rate for the motor, in \(\mathrm{W} / \mathrm{K}\).

    (c) The maximum theoretically possible shaft power output for this device, in \(\mathrm{W}\).

    Given:

    Shaft Information

    \(\tau=10 \mathrm{~N} \cdot \mathrm{m}\)

    \(\omega=188.5 \mathrm{rad} / \mathrm{s}\)

    Electrical Information

    \(\Delta \mathrm{V}_{\text {effective}}=220 \mathrm{~volts}\)

    \(\mathrm{i}_{\text {effective}}=10 \mathrm{~amps}\)

    Power factor = 1

    Steady state operation

    \(\mathrm{h}_{\text {conv }}=20 \mathrm{~W} /\left(\mathrm{m}^{2} \cdot{ }^{\circ} \mathrm{C}\right)\)

    System consists of a motor with electric work entering, shaft work exiting, and transfer of heat out. The motor is at temperature T_motor and the surrounding air is at a temperature of 20 degrees C.

    Figure \(\PageIndex{8}\): System consiting of the motor only.

    Analysis:

    Strategy \(\rightarrow\) Calculating heat transfer may require using conservation of energy and/or the convection heat transfer equation. Calculating entropy production rate always requires the entropy balance

    System \(\rightarrow\) Take just the motor as a closed, non-deforming system.

    Property to count \(\rightarrow\) Entropy and energy

    Time interval \(\rightarrow\) Steady-state behavior as per problem statement.

    (a) To begin the analysis we refer to the closed system containing the motor shown in the figure above by the dashed line. Writing the energy balance for this closed system to find the heat transfer rate, we have the following: \[\underbrace{ \cancel{\frac{d E_{sys}}{dt}}^{=0} }_{\text{Steady-state}} = \dot{W}_{\text {electric, in}} - \dot{W}_{\text {shaft, out}} - \dot{Q}_{\text {out}} \quad \rightarrow \quad \dot{Q}_{\text {out}}=\dot{W}_{\text {electric, in}} - \dot{W}_{\text {shaft, out}} \nonumber \]

    To go further requires that we make use of the defining equations for electric and shaft power: \[\begin{gathered} \dot{W}_{\text {electric, in}} = i_{\text {effective}} \cdot \Delta V_{\text {effective}} \cdot \left(\begin{array}{c} \text {Power} \\ \text {Factor} \end{array}\right) = (10 \mathrm{~A}) \cdot (220 \mathrm{~V}) \cdot(1) = 2200 \mathrm{~W} \\ \dot{\mathrm{W}}_{\text {shaft, out}} = \tau \cdot \omega = (10 \mathrm{~N} \cdot \mathrm{m}) \cdot \left(188.5 \ \dfrac{\mathrm{rad}}{\mathrm{s}}\right) = 1885 \mathrm{~W} \end{gathered} \nonumber \] Substituting this back into the energy balance gives the heat transfer rate as follows: \[\dot{Q}_{\text {out}} = \dot{W}_{\text {electric, in}} - \dot{W}_{\text {shaft, out}} = (2200-1885) \mathrm{W}=315 \mathrm{~W} \nonumber \] Thus the heat transfer rate out of the system is \(315 \mathrm{~W}\), or \(14.32 \%\) of the electrical power input. (If we had started to solve for the heat transfer rate by first writing the convection heat transfer relation, we would have quickly realized that the motor temperature was unknown. Requiring another equation, we would then have turned to the conservation of energy.)

    (b) Now to find the entropy production rate, we use the same closed system but write the entropy accounting equation: \[\underbrace{ \cancel{\frac{d S_{sys}}{dt}}^{=0} }_{\text{Steady-state}} = -\frac{\dot{Q}_{\text {out}}}{T_{\text {motor}}} + \dot{S}_{\text{gen}} \quad \rightarrow \quad \dot{S}_{\text {gen}} = \frac{\dot{Q}_{\text {out}}}{T_{\text {motor}}} \nonumber \]

    To proceed, we need to find the temperature of the motor surface. We will do this using the convection heat transfer relationship as follows: \[\begin{gathered} \dot{Q}_{\text {out}} = h_{\text {conv}} \cdot A \cdot\left(T_{\text {motor}}-T_{\text {air}}\right) \quad \rightarrow \quad T_{\text {motor}} = \frac{Q_{\text {out}}}{\left(h_{\text {conv}} \cdot A \right)} + T_{\text {air}} \\ T_{\text {motor}}=\frac{(315 \mathrm{~W})}{\left(20 \ \dfrac{\mathrm{W}}{\mathrm{m}^{2} \cdot{ }^{\circ} \mathrm{C}}\right) \cdot \left(0.300 \mathrm{~m}^{2}\right)} + 20^{\circ} \mathrm{C} = 52.5^{\circ} \mathrm{C} + 20^{\circ} \mathrm{C} = 72.5^{\circ} \mathrm{C} \end{gathered} \nonumber \] Now to compute the entropy production rate, we have the following: \[\dot{S}_{\text{gen}} = \frac{\dot{Q}_{\text {out}}}{T_{\text {motor}}} = \frac{315 \mathrm{~W}}{(72.5+273) \ \mathrm{K}} = \ \frac{315 \mathrm{~W}}{345.5 \mathrm{~K}} = 0.912 \ \frac{\mathrm{W}}{\mathrm{K}} \nonumber \]

    (c) The final question asks us to consider the maximum shaft power output that is theoretically possible under the specified operating conditions-steady-state, adiabatic operation. Based on your personal experience, you might conclude that the maximum shaft output would occur when the heat transfer rate goes to zero. But what makes this the maximum value possible? Why couldn't you transfer energy into the system by heat transfer, i.e. \(\dot{Q}_{\text {out}} < 0\), and increase the shaft power out of the motor?

    To find the answer to this question we will make use of both the conservation of energy and the entropy accounting equations developed previously: \[\dot{W}_{\text {shaft, out}} = \dot{W}_{\text {electric, in}} - \dot{Q}_{\text {out}} \quad \text { and } \quad \frac{\dot{Q}_{\text {out}}}{T_{\text {motor}}} = \dot{S}_{\text{gen}} \nonumber \] We will assume that only the electrical power input is fixed. It then appears that shaft power output only depends on the heat transfer rate, and it in turn depends on the entropy production rate. The only one of these terms that we can say anything about is the entropy production rate. So our goal should be to relate the shaft power out to the entropy production rate.

    To do this, we combine these two equations by eliminating the heat transfer rate from as follows: \[\dot{W}_{\text {shaft, out}} = \dot{W}_{\text {electric, in}} - \underbrace{\dot{Q}_{\text {out}}}_{=T_{\text {motor}} \cdot \dot{S}_{\text {gen}}} = \dot{W}_{\text {electric, in}} - \underbrace{\left(T_{\text {motor}} \cdot \dot{S}_{\text {gen}}\right)}_{\begin{array}{c} T_{\text {motor}}>0 \\ \dot{S}_{\text{gen}} \geq 0 \end{array}} \quad \rightarrow \quad \dot{W}_{\text {shaft, out }} \leq \dot{W}_{\text {electric, in}} \nonumber \] Thus, the shaft power output will always be less than or equal to the electrical power input. (This proves that those of you who wanted to build a fire and make \(\dot{Q}_{\text {out}}<0\) to increase the shaft power output are out of luck.)

    Comment:

    Why doesn't a perfect motor, one with \(\dot{W}_{\text {shaft, out}} = \dot{W}_{\text {electric, in}}\), violate Kelvin-Planck's prohibition against a 100% efficient power cycle?

    The Kelvin-Planck prohibition only applies to the steady-state conversion of a heat transfer of energy completely into a work transfer of energy. For the motor there is no heat transfer of energy into the system.

    I'm still confused. What exactly is entropy production?

    Equation \(\PageIndex{2}\) gives us some additional insight into the meaning of entropy production. For this motor operating under the specified conditions, the theoretically possible maximum shaft power out of the system is equal to the electric power into the system. Using this, we can rewrite Eq. \(\PageIndex{2}\) as follows:

    \[ \left. \begin{array}{c} \left. \dot{W}_{\text{shaft, out}} \right|_{\text{actual}} = \dot{W}_{\text{electric, in}} - \left(T_{\text{motor}} \cdot \dot{S}_{\text{gen}}\right) \\ \dot{W}_{\text{shaft, out}} \left|_{\begin{array}{l} \text{max} \\ \text{possible} \end{array}} \right. \equiv \ \dot{W}_{\text{electric, in}} \end{array} \right| \quad \rightarrow \quad \left. \dot{W}_{\text{shaft, out}} \right|_{\text{actual}} = \dot{W}_{\text{shaft, out}} \left|_{\begin{array}{l} \text{max} \\ \text{possible} \end{array}} \right. - \left(T_{\text{motor}} \cdot \dot{S}_{\text{gen}}\right) \nonumber \]

    Now solving for the entropy production rate within the motor, we have the following relation: \[\left.\dot{S}_{\text{gen}}\right|_{\text {motor}} = \frac{\left[ \dot{W}_{\text{shaft, out}} \left|_{\begin{array}{l} \text{max} \\ \text{possible} \end{array}} \right. - \left. \dot{W}_{\text{shaft, out}} \right|_{\text{actual}} \right]}{T_{\text {motor}}} \geq 0 \nonumber \] Now, what does this tell us about entropy production?

    • First, the entropy production rate is a measure of how far a process deviates from ideal behavior.
    • Second, ideal behavior corresponds to an entropy production rate of zero and this can only occur for an internally reversible process.
    • Third, when viewed as the difference between the maximum possible power output and the actual power output, the entropy production is also a measure of the irreversible loss of the potential to do work.

    To grasp this last point, think about the flow of energy through the motor. Initially, the energy enters the motor as electrical work and leaves the system as shaft work and heat transfer. Experience has shown that a work transfer of energy is clearly more valuable than an equal heat transfer of energy. Why? Because we can do anything with a work transfer of energy, including driving an ideal DC generator that converts the shaft power back into electrical power and supplying the electrical power to an electric resistor which converts electrical work back into a heat transfer of energy. However, as will be shown in the next example, it is impossible to convert all of the heat transfer from the motor (or any other system) completely into a work transfer of energy.

    Exercise — Work is more valuable than heat transfer? I don't believe it!

    To investigate the relative "value" of work transfers of energy and heat transfers of energy, consider two steady-state closed systems shown in the diagrams:

    • A "work converter" that receives a work transfer of energy and then "converts" that into a work transfer and a heat transfer of energy out of the system. Heat transfer occurs by convection heat transfer between the work converter at \(T_{\text {surface}}\) and the surroundings at temperature \(T_{\mathrm{o}}\).
    • A "heat converter" that receives and rejects energy by heat transfer at boundary temperatures \(T_{\text {surface}}\) and \(T_{\mathrm{o}}\), respectively, and has a net work output.

    In a work converter, work enters at rate dot-W_in and exits at rate dot-W_WC, out. Heat is transferred away from the system from a boundary at temperature T_surface, at the rate dot-Q_surface. In a heat converter, heat enters the system at rate dot-Q_surface through a boundary at temperature T_surface, and exits the system at rate dot-Q_o through a boundary at temperature T_o. Work exits the system at a rate dot-W_HC, out.

    Figure \(\PageIndex{9}\): Structure of a work converter and a heat converter.

    Answer the following questions:

    (a) Starting with the conservation of energy and entropy accounting equations shown below, develop an expression for the power out of the work converter as a function of the surface temperature, the entropy generation rate, and the power input, i.e. \(\dot{W}_{\text {WC, out}} = f\left(T_{\text {surface}}, \ \dot{S}_{gen, \text{ WC}}, \ \dot{W}_{\text {in}}\right)\)

    \[\frac{d E_{sys}}{dt} = \dot{W}_{\text{in}} - \dot{W}_{\text{WC, out}} - \dot{Q}_{\text {surface}} \quad\quad\quad \frac{d S_{sys}}{dt} = -\frac{\dot{Q}_{\text {surface}}}{T_{\text {surface}}} + \dot{S}_{gen, \text{ WC}} \nonumber \]

    Answer

    \( \dot{W}_{\mathrm{WC}, \text { out }}=\dot{W}_{i n}-\left(T_{\text {surface }} \cdot \dot{S}_{g e n, \mathrm{WC}}\right) \)

    (b) What fraction of the power input to the work converter can in theory be returned to the surroundings as power out from the work converter, i.e. what's the maximum value of the ratio \(\dot{W}_{\text{WC, out}} / \dot{W}_{\text {in}}\) ?

    (c) Starting with the conservation of energy and entropy accounting equations, develop an expression for the power out of the engine as a function of the two boundary temperatures, the entropy generation rate, the heat transfer rate into the engine, i.e. \(\dot{W}_{\text{HC, out}} = f\left(\dot{Q}_{\text {surface}}, \ T_{\text{o}}, \ T_{\text {surface}}, \ \dot{S}_{gen, \text{ HC}}\right)\)

    \[\frac{d E_{sys}}{dt}=\dot{Q}_{\text{surface}} - \dot{Q}_{\text{o}} - \dot{W}_{\text{HC, out}} \quad\quad\quad \frac{d S_{sys}}{dt} = \frac{\dot{Q}_{\text{surface}}}{T_{\text{surface}}} - \frac{\dot{Q}_{\text{o}}}{T_{\text{o}}} + \dot{S}_{gen, \text{ HC}} \nonumber \]

    Answer

    \( \dot{W}_{\text {HC, out}} = \dot{Q}_{\text {surface}} \left[1-\frac{T_{\text{o}}}{T_{\text {surface}}}\right] - \left(T_{\text{o}} \cdot \dot{S}_{gen, \text { HC}}\right) \nonumber \]

    (d) What fraction of the heat transfer input to the heat converter can in theory be returned to the surroundings as power out from the heat converter, i.e. what's the maximum value of the ratio \(\dot{W}_{\text {HC, out}} / Q_{\text {surface}}\) assuming the \(T_{\text {surface}}\) and \(T_{\mathrm{o}}\) are fixed ?

    From parts (b) and (d) above we learn two things:

    • given a work transfer of energy and an ideal (internally reversible) work converter, we can completely convert all of the work transfer of energy into the system into an equal amount of work out of the system. Under the worst possible conditions, all of the energy into the work converter would leave the system as heat transfer, and
    • given a heat transfer of energy and an ideal (internally reversible) heat converter, we can at best only turn a fraction of the heat transfer of energy into the system into a work transfer of energy out of the system.

    Now let's consider what happens when energy flows through a non-ideal work converter. The figure below gives a graphical interpretation of the energy flow through a work converter combined with a heat converter.

    Work enters a system at rate dot-W_in, going into a work converter. The output of the work converter is split between dot-W_HC, out, which exits the system and is less than or equal to the input to the system, and dot-Q_surface, which enters a heat converter. The outputs of the heat converter, which both exit the system, are dot-W_WC, out and dot-Q_o.

    Figure \(\PageIndex{10}\): Energy flow through a system consisting of a work converter and a heat converter, whose input is the heat output of the work converter.

    With an ideal work converter, \(\dot{S}_{gen, \text{ WC}}=0\) and all of the work transfer of energy into the system leaves the system as an equal work transfer of energy. With a non-ideal work converter, \(\dot{S}_{gen, \text{ WC}}>0\) and some of the energy leaves the system by heat transfer. To convert this heat transfer back into work we feed it into the heat converter. Even under the best of conditions only a fraction of the heat transfer of energy entering the heat converter can be converted into a work transfer of energy out of the system.

    The work that could be recovered by combining the work output from both converters is

    \[ \begin{aligned} \dot{W}_{\text{combined}} &= \dot{W}_{\text{WC, out}} + \dot{W}_{\text{HC, out}} = \underbrace{\dot{W}_{\text{HC, out}}}_{\begin{array}{c} \text{Actual power out} \\ \text{of the work converter} \end{array}} + \underbrace{ \dot{Q}_{\text{surface}} \left[1 - \left(\frac{T_{\text{o}}}{T_{\text{surface}}}\right)\right] - T_{\text{o}} \dot{S}_{gen, \text{ HC}} }_{\begin{array}{c} \text{Actual power out} \\ \text{of the heat converter} \end{array}} \\ &= \underbrace{ \left[\dot{W}_{\text{WC, out}} + \dot{Q}_{\text{surface}}\right] }_{=W_{\text{in}}} - \underbrace{ \dot{Q}_{\text{surface}} }_{= T_{\text{surface}} \dot{S}_{gen, \text{ WC}}} \cdot \left(\frac{T_{\text{o}}}{T_{\text{surface}}}\right) - T_{\text{o}} \dot{S}_{gen, \text{ HC}} = \dot{W}_{\text{in}} - \left(T_{\text{surface}} \dot{S}_{gen, \text{ HC}}\right) \left(\frac{T_{\text{o}}}{T_{\text{surface}}}\right) - T_{\text{o}} \dot{S}_{gen, \text{ HC}} \\ &= \dot{W}_{\text{in}} - T_{\text{o}} \left(\dot{S}_{gen, \text{ WC}} + \dot{S}_{gen, \text{ HC}}\right) \leq \dot{W}_{\text{in}} \end{aligned} \nonumber \]

    Now, what's the impact of entropy production in the work converter? Any entropy production in the work converter results in a heat transfer out of the work converter. If we could convert all of the heat transfer back into work, there would be no problem. Unfortunately this is not the case. Even if we assume an ideal heat converter to recover the maximum amount of work from the heat transfer, we only recover part of the work transfer of energy supplied to the work converter: \[\dot{W}_{\text {combined}}\left|_{\begin{array}{l} \text {Ideal} \\ \text {Heat Converter} \end{array}} \right. = \dot{W}_{\text {in}}-T_{\text{o}} \left(\dot{S}_{gen, \text{ WC}} + \underbrace{ \cancel{\dot{S}_{gen, \text { HC}}}^{=0} }_{\text {Ideal Heat Converter}}\right) = \dot{W}_{\text{in}} - T_{\text{o}} \dot{S}_{gen, \text{ WC}} < \dot{W}_{\text{in}} \nonumber \]

    Thus, any entropy production within the work converter will reduce our potential to do work. This reinforces the point that it pays us at least thermodynamically to minimize entropy production because anytime entropy is produced we lose the capacity to do some work. Economically, this may not be the best approach; however, as the cost of energy increases there is greater economic incentive to reduce entropy production. Thermoeconomics is the discipline that attempts to assign the true value to various forms of energy. By combining conservation of energy and the entropy accounting principle, engineers have developed a new extensive property called exergy or availability that describes the work potential of any quantity or transfer of energy. Thus, it is possible to price energy and energy transfers based on its work potential.

    Exercise — Steady-state Heat Transfer Through a Wall

    Energy flows steadily through a cylindrical "plug" with diameter \(D=0.5 \mathrm{~m}\) and length \(L=0.25 \mathrm{~m}\). The heat transfer rate and surface temperature at Surface 1 are \(500 \mathrm{~kW}\) and \(300 \mathrm{~K}\), respectively. The surface temperature at Surface 2 is \(400 \mathrm{~K}\). Assume steady-state behavior.

    A steel cylinder is oriented so that its central axis is horizontal. Energy flows from left to right, from Surface 1 to Surface 2.

    Figure \(\PageIndex{11}\): Energy passes through the axis of a cylindrical steel plug.

    (a) Determine the heat transfer rate at Surface 2, in \(\mathrm{W}\).

    Answer

    \(500 \mathrm{~kW}\)

    (b) Using the entropy balance, determine the entropy production rate within the steel plug, in \(\mathrm{W} / \mathrm{K}\).

    Answer

    \(0.4167 \mathrm{~kW} / \mathrm{K}\)

    (c) What would happen to the rate of entropy production as the temperature difference across the steel plug gets very small? What is the entropy production rate as the temperature difference goes to zero? [Hint: Replace \(T_{2}\) by the expression \(T_{2}=T_{1}-\Delta T\). Then examine the entropy production as \(\Delta T\) gets small.]

    (d) If the boundary temperatures remained unchanged, would it be possible for the heat transfer to flow in the opposite direction? Yes or no? Why?

    Example — Behavior of a closed system with no energy transfer

    (a) Sketch a closed system that has no energy transfers with the surroundings.

    (b) Simplify the rate form of the conservation of mass, conservation of energy, and the entropy balance for this system and write the resulting equation in the blank column of the table:

    Mass \[\frac{\mathrm{dm}_{\text {sys}}}{\mathrm{dt}} = \sum_{\text {in}} \dot{m}_{i}-\sum_{\text {out}} \dot{m}_{e} \nonumber \]  
    Energy \[\frac{\mathrm{dE}_{\text {sys}}}{dt} = \dot{Q}_{\text {net, in}}+\dot{W}_{\text {net, in}} + \sum_{\mathrm{in}} \dot{m}_{i}\left(h+\frac{V^{2}}{2}+gz\right)_{i} - \sum_{\text {out}} \dot{m}_{e} \left(h+\frac{V^{2}}{2}+gz\right)_{e} \nonumber \]  
    Entropy \[\frac{dS_{\mathrm{sys}}}{dt} = \sum_{j} \frac{\dot{Q}_{j}}{T_{\text{b, j}}} + \sum_{\mathrm{in}} \dot{m}_{i} s_{i} - \sum_{\text {out}} \dot{m}_{e} s_{e} + \dot{S}_{gen} \nonumber \]  

    (c) Now using this information, plot the system energy \(E_{\mathrm{sys}}\), the system mass \(m_{\mathrm{sys}}\), and the system entropy \(S_{\mathrm{sys}}\) as a function of time on the graphs below. (The small dot at \(t=0\) on each graphs represents the initial value of \(m_{\mathrm{sys}},\) \(E_{\mathrm{sys}}\), and \(S_{\mathrm{sys}}\).)

    First-quadrant axes for m vs t, E vs t, and S vs t. At t=0, an arbitrary point is marked on each of the vertical axes.

    Figure \(\PageIndex{12}\): Axes to plot system mass, energy, and entropy.

    What happens for very long times, i.e. as time goes to infinity? (For example, if you assumed that the system finally reached a steady-state, equilibrium value, what must happen to \(S_{\text {sys}}\)? What restriction does this place on the shape of your curve for \(S\) vs. \(t\)?)


    This page titled 8.3: Entropy Accounting Equation is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Donald E. Richards (Rose-Hulman Scholar) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.