Nitrogen gas enters a diffuser at 100 kPa and 110^∘ C with a velocity of 205 m / s and leaves a

step 1 Here we have nitrogen entering a diffuser. So we have nitrogen and the ideal gas constant. It's

Nitrogen gas enters a diffuser at 100 kPa and 110^∘ C with a...

Key Concepts

Exergy Destruction

Exergy destruction quantifies the amount of useful work potential that is irreversibly lost due to entropy generation within a system. This concept is important in second-law analyses to measure the inefficiencies in processes, such as those occurring in diffusers where friction, heat loss, and other irreversibilities reduce the system's performance.

Heat Transfer in Open Systems

Heat transfer in an open system, such as a diffuser, affects the enthalpy of the fluid. This concept covers the assessment of heat exchange between the system and its surroundings, linking temperature changes to energy losses or gains. It is essential to correctly incorporate heat losses when performing energy and exergy analyses.

Second-Law Efficiency

Second-law efficiency, or exergetic efficiency, is the ratio of the actual performance of a device (in terms of minimum penalty due to irreversibility) to the ideal performance under reversible conditions. This concept is used to evaluate how effectively a system utilizes the available energy potential, taking into account losses due to irreversibility.

Diffuser Analysis

A diffuser is a device that slows down a moving fluid while typically increasing its pressure. The analysis involves tracking changes in velocity, pressure, and temperature while accounting for energy losses. Understanding a diffuser means applying both momentum and energy principles to see how fluid properties evolve in the presence of losses.

Open System Energy Balance

This concept involves applying the first law of thermodynamics to control volumes where mass and energy cross the boundaries. It accounts for changes in kinetic, potential, and internal energies as well as work and heat transfer. In diffuser problems, the energy balance is key to linking inlet and exit conditions, especially when kinetic energy differences are significant.

Variation of Specific Heats

In many thermodynamic problems, the specific heats (cp and cv) are not constant but vary with temperature. This concept emphasizes the importance of evaluating thermodynamic properties accurately over a range of temperatures, which is crucial for precise energy and exergy calculations in processes involving significant temperature changes.

Transcript

00:01 Here we have nitrogen entering a diffuser.

00:05 So we have nitrogen and the ideal gas constant.

00:08 It's entering at 100 kilopascals and 110 degrees c at a velocity of 205 meters per second.

00:17 And it leaves at a slightly higher pressure of 110 kilopascals, but a much lower velocity of 45 meters per second, which is what a diffuser is for.

00:28 Let's see here.

00:29 We're told that we estimate a heat, heat loss of 2 .5 kilojoules per kilogram.

00:35 And the surroundings are 27c and 100 kofascals.

00:40 And the diffuser, the exit area is 0 .04 square meters.

00:48 So we want to figure out the, accounting for the variation of the specific heats with temperature.

00:55 So need to be a little careful here.

00:58 Determint a, the exit temperature, the rate, b, the rate of extra temperature.

01:01 Destruction and the second law efficiency.

01:04 Okay then, well, since we're gonna, we're not gonna, we can't use the simple analysis of just using constant heat capacity, we've got to go to the tables, and so, or the software.

01:23 So we know these properties here, so we can look up or have the software spit out the entropy and the entropy at the inlet.

01:34 And then we also have the enthalpy and the entropy of nitrogen at the surrounding conditions.

01:46 So, yeah, they wanted to know something about extragy at the inlet.

01:50 So we needed that at some point or to get the second -lac efficiency.

01:55 So what we can do then is energy balance.

02:00 So this is a steady flow process.

02:03 So the energy flow in, the specific energy flow in, the specific energy flow out, and then the heat transfer out.

02:13 And in this equation, we know everything but h2.

02:15 So that allows us to figure out what the enthalpy at the exit is.

02:19 And we also know the pressure at the exit.

02:22 So now we have two properties of the exit.

02:24 So we can use the software or look at home tables to figure out what the temperature and the entropy are at the exit.

02:33 We can figure out the mass flow rate because we know what the area is and we know what the, let's see here, we know what the area is, and we know what the area is, and we know what the, and i messed up, i made these, these should be velocity means.

03:01 Not maybe not in red, velocity at two and velocity at two.

03:11 All right, so that's that's much better.

03:14 So we know that, again, this is cross -sectional area times the velocity that's flowing through there.

03:21 So this would be the volume flow rate and then we can divide through by the specific volume to get the mass flow rate.

03:28 All right.

03:32 We can then, you know, just use the ideal gas law to get the specific volume in terms of the pressure and the temperature.

03:40 And then now we know all of this stuff.

03:42 We could have calculated this, but just plugging this all in.

03:47 We know all of this stuff.

03:48 And we get a mass flow rate of about 1 .67 kilograms per second...