An ideal gas is contained in a piston-cylinder device and undergoes a power cycle as follows: 1-

An ideal gas is contained in a piston-cylinder device and...

Key Concepts

Isentropic Process

An isentropic process is a thermodynamic process that occurs at constant entropy, meaning no entropy is generated. In ideal gas cycles, this process is idealized as reversible and adiabatic, so there is no heat transfer with the surroundings, and the work done solely changes the internal energy according to the compression or expansion of the gas.

Constant Pressure Process

A constant pressure process is one where the pressure remains fixed while other state properties, such as volume and temperature, change. In thermodynamic cycles, heat addition or rejection often occurs at constant pressure because it simplifies the analysis and reflects processes like combustion or heating in engines.

Constant Volume Process

A constant volume process, also known as an isochoric process, is characterized by a fixed volume where no work is performed since the system does not expand or contract. In such a process, any heat added or removed directly affects the internal energy and temperature of the system, which is particularly useful in analyzing heat rejection phases in cycles.

Thermodynamic Cycle Analysis

Thermodynamic cycle analysis involves evaluating a sequence of processes that return a system to its initial state, allowing the net work, heat transfer, and efficiency to be determined. This concept is essential in power cycles and engine analysis, where understanding each process's energy interactions helps optimize performance and energy utilization.

P-v and T-s Diagrams

Pressure-volume (P-v) and temperature-entropy (T-s) diagrams are graphical representations of thermodynamic processes and cycles. The P-v diagram illustrates the work interactions as changes in volume at varying pressures, while the T-s diagram provides insight into the heat transfer and entropy changes during the process, highlighting the efficiency and irreversibility of real processes.

Heat and Work Interactions

Heat and work interactions describe the energy transfers between a system and its surroundings via thermal energy and mechanical work. In thermodynamic cycle analysis, correctly accounting for heat added or removed and work done during each process is crucial to determining the overall energy balance and efficiency of the cycle.

Thermal Efficiency

Thermal efficiency is a measure of how well a thermodynamic cycle converts heat energy into work. It is defined as the net work output divided by the heat input and is a critical parameter in evaluating the performance of engines and power cycles. Analyzing the efficiency helps in understanding the limitations posed by the second law of thermodynamics.

Ideal Gas and Constant Specific Heats

Assuming an ideal gas with constant specific heats simplifies the analysis of energy transfers in thermodynamic processes. This assumption allows for the use of simple relationships between temperature, work, and heat, making calculations more straightforward while giving insight into the fundamental behavior of gases in cycle processes.

Compression Ratio

The compression ratio is the ratio of the volume at the start of the compression process to the volume at the end. It is an important parameter in thermodynamic cycles, as it influences the efficiency, work output, and temperature changes during compression and expansion processes. A higher compression ratio typically improves efficiency but may also increase the risk of engine knock or other practical limits in real systems.

Transcript

00:01 In this problem, we have a thermodynamic cycle where we have an ideal gas in a piston cylinder device and it's going a cycle.

00:13 Where from 1 to 2 here, we have isotropic compression and also it's adeptic there.

00:21 So we go from t1, which is 20 degree c, up to a different temperature as we compress the gas.

00:29 The entropy doesn't change, or the specific entropy doesn't change during that process.

00:36 And we go from a volume, we compress the gas by a factor of five.

00:43 So the volume at two is one -fifth the volume at one.

00:48 So we, and then this is a line of constant, constant entropy.

00:55 So i guess, put some arrows in here how the cycle goes.

00:58 The numbering is, should make it clear.

01:02 So from one to two, then, so that means we have this compression ratio of, there's a factor of five there.

01:10 From two to three, we have constant pressure heat transfer.

01:19 So we have heat being added to the gas, it's heating it up.

01:24 And we keep the pressure constant.

01:26 So again, we let the volume change, but we keep the pressure constant.

01:32 And then we let the volume get back up to what it was initially.

01:40 So this is another factor of five here.

01:45 Volume at three is five times that is the volume at two.

01:50 And so then we heat is as we drop the pressure back down, we have a at a constant volume.

02:04 So we drop the pressure and then we have heat transfer out.

02:09 And so over here we have a constant pressure line here and then a constant specific volume line here.

02:18 So we have work being done from here and then from this work positive and then the negative work is this value here.

02:29 So the network that we're going to wind up calculating is the volume of this.

02:35 So let's see here.

02:37 What can we do here? well, we're given some ideal gas constants.

02:42 We've given r and cv.

02:44 We can find cp because that's just r plus cv, and then we can find k because that's just cp over cv.

02:52 And now, for an ideal gas undergoing isotropic compression, we have this relationship between the temperature and the specific volumes.

03:01 We know what this ratio is.

03:04 This is just r1 -2, so this is a factor, five.

03:10 So v1 was five times v2.

03:13 And so we know everything here and we can plug in and we get that we raise the temperature from 20 degree c to 540, so about 397 kelvin to about 584 kelvin.

03:26 So we raise the temperature up here as we compress the gas to a fifth of its volume.

03:35 And in doing that, we put some work into the system and that's that work is given by the for an ideal gas undergoing uh let's see here um isotropic compression we have that it's cv times the change in temperature and in the end that winds up being 204 kilojoules per kilogram and we had no heat transfer during this first this first stage of the process.

04:13 Now, for ideal gas with a constant pressure from 2 to 3, we have t3 over t2 equals v3 over v2.

04:24 V3 is v1.

04:27 So we start back here.

04:29 And v1 over v2 is just r12 or 5.

04:33 So t3 is 5 times t2 or 2 ,922 kelvin.

04:39 So this, this, this, this, this, this, this, this, this, this, this, this scale is way off because this temperature should be five times this temperature.

04:50 So this is really, this is kind of going up a little bit here and then this way goes way up here and then comes back way down here.

04:58 So this is not to scale, but it's a representation of the cycle.

05:05 Now from two to three we have work being done, work coming out and that's just because it's a constant pressure it's just that pressure times a change in specific volume and because it's an ideal gas we have um that that winds up being the um ideal gas constant times of change in temperature and that winds up being 701 .3 kilojoules per kilogram so that's the work we get out there now we need the heat transfer that came in here um so as we were we were putting heat in we get, let's see here, we have the heat in, if we do basically an energy balance from between those two states, the heat in over that state is the workout plus the change in internal energy.

06:00 And again, just using the definition of enthalpy, we can read about that this is just actually the change in enthalpy.

06:07 And the change in enthalpy is the heat capacity, constant pressure times the change in temperature.

06:13 And we have everything, we know all of these values.

06:16 And that winds up being 2 ,338 kilojoules per kilogram.

06:20 So that's how much heat we put in here or here.

06:26 And now the heat that we dumped out of the system as we were letting the pressure back down to our initial state, that is just the change in internal energy because there was no work done here.

06:44 So that's just the change in internal energy in the gas from 3 to 1.

06:49 And that is the heat capacity constant volume times the change in temperature.

06:54 So the heat transferred, the heat that we got out was 1 ,840 .3 kilojoules per kilogram.

07:04 Now, let's see here.

07:06 We should have the, let's see here.

07:13 Then we have, so we have that and we have no work, no work in or out here because the volume didn't change.

07:20 So we know the net work out is this value here, 497 .3 kilojoules per kilogram...

Please give Ace some feedback