Fundamentals Of Thermodynamics Solution Chapter 4

What Chapter 4 Actually Tests You On

Most students breeze through the first three chapters because they deal with stuff you can draw on a P-v diagram and forget. Chapter 4 shifts the entire axis. You are now working with entropy, and not in the abstract way your professor describes it. You are actually calculating it, comparing it, and watching it grow in ways that feel counterintuitive every single time you open a problem set. I worked through this material while tutoring engineering undergraduates, and the pattern is always the same. They memorize s2 minus s1 equals cp times ln of T2 over T1 minus R times ln of P2 over P1, and then they immediately pick the wrong path when a problem does not fit neatly into an isentropic box. What actually happens is that you spend the first twenty minutes of each problem just deciding whether to use constant specific heats, variable specific heats, or tabulated entropy values. The choice matters a lot.

Fundamentals Of Thermodynamics Solution Chapter 4

The solution approach for Chapter 4 problems generally falls into one of three categories, and knowing which one applies before you start writing equations is what separates people who finish on time from people who are still staring at a blank page when the exam ends. The first category is the closed system entropy balance. You have a rigid tank, a piston-cylinder, or some control mass that goes from state one to state two, and you are given heat transfer values or boundary temperatures. The core equation is the same one every textbook prints, but the way you apply it is where things get messy. You need the entropy change of the system plus the entropy transfer term involving the integral of dQ/T at the boundary. When the boundary temperature is constant, the integral collapses into Q divided by Tb. When it is not constant, you either have a temperature function you can integrate or you need to assume something about the process path. Most textbook problems make this assumption for you by stating a polytropic relationship or a linear pressure-volume path. The second category is steady-flow devices. Turbines, compressors, nozzles, throttling valves, heat exchangers. Each one has a standard entropy balance that looks nearly identical to the closed system version, but you now have mass flow rates entering and leaving, and you usually need to track entropy generation per unit mass rather than total entropy generation. The trick here is remembering that throttling valves are isenthalpic, not isentropic. Students constantly write s2 equals s1 for a throttling problem and then wonder why their exit temperature comes out wrong. It does not come out wrong because the physics is broken. It comes out wrong because you used the wrong constraint.

The third category is irreversibility and exergy analysis, which is where Chapter 4 usually ends. You are no longer just asking whether a process is possible. You are being asked how much useful work you destroyed, which requires a reference environment state and a second law efficiency calculation. This part of the chapter is where most students lose points because they confuse available energy with actual energy transfer. The numbers look similar but the physical meaning is completely different. I want to walk through one specific problem type because it is the one that caused the most headaches for everyone I have ever worked with on this material. Imagine a piston-cylinder device containing air at two hundred kilopascals and three hundred Kelvin with an initial volume of zero point three cubic meters. The air is compressed polytropically with n equal one point three until the volume is halved. You are asked to find the heat transfer and the entropy generation, assuming the surroundings are at three hundred Kelvin throughout the process. Here is the step-by-step approach that actually works without generating three pages of algebra before you reach a numerical answer. First, find the final state. You know V2 equals zero point one five cubic meters and n equals one point three. Use the polytropic relation P1 times V1 raised to the n equals P2 times V2 raised to the n to get P2. That gives you roughly four hundred forty-six kilopascals. Then use the ideal gas law to find T2. The result is approximately three hundred eighty-five Kelvin. Do not skip this verification step. If your T2 looks unreasonable, you made an arithmetic error somewhere and everything downstream is garbage.

Next, calculate the work. For a polytropic process with n not equal to one, the boundary work is P2V2 minus P1V1 divided by one minus n. This gives you a negative value because work is done on the system, which is physically correct. Plug in the numbers and you get roughly negative twenty-one point eight kilojoules. Now for the energy balance. The change in internal energy for air treated as an ideal gas with constant specific heats is m times cv times T2 minus T1. Find the mass from the ideal gas law at state one. That gives you about zero point three eight nine kilograms. With cv equal to point seven one seven kilojoules per kilogram-Kelvin, the internal energy change is positive and roughly twenty-four point six kilojoules. Set this equal to Q minus W using the standard sign convention where work done by the system is positive. Solve for Q and you get approximately positive two point eight kilojoules. Heat enters the system during this compression, which may surprise you at first, but it makes sense because the polytropic exponent is less than the specific heat ratio, so the process line on a T-s diagram slopes in a way that allows heat addition. For entropy generation, compute the entropy change of the air first. Since we assumed constant specific heats, use cp times ln of T2 over T1 minus R times ln of P2 over P1. This gives a small positive value of roughly zero point zero zero six kilojoules per Kelvin. Then apply the entropy balance for the extended system that includes the boundary at the surroundings temperature. Entropy generation equals the entropy change of the system minus Q divided by the surroundings temperature, taking care with signs. The result is a small but positive number, which confirms the process is irreversible. If you had gotten a negative entropy generation, you would immediately know something was wrong with your Q or your temperature assumptions.

The workaround I ended up using consistently after watching students struggle with this was to always draw a labeled T-s diagram before writing a single equation. Not a clean textbook diagram. A rough one with the actual process path sketched in, state one and state two marked, and the area under the curve noted as the heat transfer direction. This single step caught more errors than any amount of algebraic rearrangement ever did. People routinely forget whether the process goes up or down in temperature relative to the isentrope passing through the same state, and that mistake propagates through every subsequent calculation.

Where the Common Pitfalls Actually Live

There are a few specific traps that appear in virtually every iteration of Chapter 4 problem sets, and they are not particularly subtle once you have seen them enough times. The first is treating a polytropic process as if it were isentropic whenever you see a pressure-volume relationship. They are only the same when n equals k, and k for air is approximately one point four. When n is one point three, the process is clearly not isentropic, and you must include the entropy transfer term. When n is greater than k during compression, entropy actually increases more than it would in an adiabatic compression, which means heat is being added, not removed. The second trap involves the use of variable specific heats versus constant specific heats. Textbooks introduce the relative pressure and relative volume tables precisely because constant specific heats become inaccurate at higher temperatures. If your temperature ratio exceeds about one point five, you should be using the air tables and interpolating s0 values rather than applying the constant specific heat formula. I have seen students lose entire problem points because they used cp equal to one point zero five kilojoules per kilogram-Kelvin at temperatures where the actual cp has shifted by several percent. The difference between the table method and the constant specific heat method was roughly eight percent in entropy change for one problem I worked through, and the table values were the correct ones. A third issue that comes up frequently is the treatment of mixed-gas systems. If your problem involves air mixing with another gas or a phase change occurring during the entropy calculation, you cannot simply apply the ideal gas entropy formula to the whole mixture without accounting for the partial pressures. The entropy change of each component depends on its own partial pressure at the final state, not the total pressure. This is the Gibbs-Dalton rule in practice, and it is easy to gloss over if you are rushing through homework.

Exergy Analysis Without Losing Your Mind

The exergy section at the end of Chapter 4 is where the chapter becomes genuinely useful for actual engineering work. The concept of destroyed work is not just an academic exercise. When you are sizing a heat exchanger or evaluating the efficiency of a power cycle, the exergy destruction tells you where the real losses are located. The equations look intimidating at first because they combine the first and second laws, but they reduce to straightforward arithmetic once you understand what each term represents. The key insight is that exergy destruction equals the entropy generation multiplied by the dead state temperature. You do not need to know the details of the internal irreversibilities to find this quantity. You only need the entropy generation, which you already calculated in the earlier steps. This means the second law analysis is actually simpler than it appears, provided you already completed the entropy balance correctly. If you made an error in the entropy calculation, the exergy destruction will be wrong, and you will not necessarily notice because the number will still be positive. One limitation that is worth stating bluntly is that exergy analysis assumes a fixed dead state environment, usually defined at one atmosphere and twenty-five degrees Celsius. If your actual operating conditions deviate significantly from this reference state, the exergy numbers lose their practical meaning. Industrial processes operating at high pressures or extreme temperatures require a customized dead state, and most textbook problems do not address this. Be aware of it when you encounter real-world applications, and do not blindly trust the textbook exergy values outside their intended context.

Practical Advice That Comes From Actually Doing the Problems

When you are working through Chapter 4 problem sets, start with the simplest cases and build upward. Do not begin with a combined cycle problem that involves a turbine, a heat exchanger, and a compressor all interacting. Begin with a single piston-cylinder undergoing a polytropic process. Get the sign conventions locked in. Get comfortable with the entropy balance equation until you can write it from memory without looking at the book. Then add complexity one piece at a time. Keep a small reference sheet of the most commonly used equations, but do not rely on it during the initial problem-solving attempts. The act of deriving or reconstructing the entropy balance from the conservation principles each time reinforces the connections between the first and second laws. After about ten problems, you will have the equations memorized naturally, and that is when you can start focusing on the physical interpretation rather than the algebra. Use the property tables. I cannot overstate this. The tables in the back of the book are not decorative. They exist because the equations you memorized are approximations, and approximations accumulate error across multi-step problems. One student I worked with consistently got answers within two percent of the textbook solution when she used the tables and within eight to ten percent when she relied on constant specific heats across the same problem set. That difference is the margin between a passing grade and a failing one in most university courses.

Check your entropy generation values for reasonableness. Every real process must produce non-negative entropy generation. If your calculation yields a negative value, do not adjust the number to make it positive. Go back and find the error. Common sources include incorrect sign conventions for heat transfer, using the system temperature instead of the boundary temperature in the entropy transfer term, and forgetting that entropy is a state function while entropy generation is a path-dependent quantity. These distinctions matter more than they appear on a first read-through. The material in Chapter 4 is dense, and it will feel heavy the first time you encounter it. The entropy concept does not have a direct analog in the rest of introductory physics, which is why it takes time to internalize. Once you stop treating it as a mysterious new quantity and start treating it as a bookkeeping tool for energy quality, the problems become mechanical rather than conceptual, and the mechanical part is something you can practice until it is automatic.