Argon enters a turbine at a rate of 80.0 kg / min, a temperature of 800^∘ C, and a pressure of 1

Argon enters a turbine at a rate of 80.0 kg / min, a...

Argon enters a turbine at a rate of $80.0 \mathrm{kg} / \mathrm{min},$ a temperature of $800^{\circ} \mathrm{C},$ and a pressure of $1.50 \mathrm{MPa}$. It expands adiabatically as it pushes on the turbine blades and exits at pressure 300 kPa.(a) Calculate its temperature at exit. (b) Calculate the (maximum) power output of the turning turbine. (c) The turbine is one component of a model closed-cycle gas turbine engine. Calculate the maximum efficiency of the engine.

Argon enters a turbine at a rate of $80.0 \mathrm{kg} / \mathrm{min},$ a temperature of $800^{\circ} \mathrm{C},$ and a pressure of $1.50 \mathrm{MPa}$. It expands adiabatically as it pushes on the turbine blades and exits at pressure 300 kPa.(a) Calculate its temperature at exit.
(b) Calculate the (maximum) power output of the turning turbine. (c) The turbine is one component of a model closed-cycle gas turbine engine. Calculate the maximum efficiency of the engine.

Key Concepts

Carnot Efficiency

Carnot efficiency represents the maximum theoretical efficiency that any heat engine operating between two given temperature reservoirs can achieve. It is determined solely by the temperatures of the hot and cold reservoirs and serves as an ideal benchmark to measure the performance of real engines.

Power Calculation in Turbines

Power calculation in turbines involves determining the rate of work output by considering the mass flow rate of the working fluid and the change in its enthalpy as it passes through the turbine. This concept is fundamental in designing and analyzing turbine performance in energy conversion systems.

Ideal Gas Behavior

Ideal gas behavior assumes that the gas molecules do not interact significantly except through elastic collisions and that the volume of the molecules is negligible compared to the volume of the container. This assumption allows the use of simplified equations of state and specific heat relationships when analyzing thermodynamic processes involving gases.

Isentropic Process

An isentropic process is a special type of adiabatic process that is also reversible, meaning that entropy remains constant throughout the process. This concept is often used to estimate the maximum possible work output or minimum work input in turbines and compressors under idealized conditions.

Adiabatic Process

An adiabatic process is one in which no heat transfer occurs between a system and its surroundings. In thermodynamics, this concept is applied when a working fluid expands or compresses so rapidly that heat exchange is negligible, which simplifies energy balance considerations to only work interactions.

Steady-Flow Energy Equation

The steady-flow energy equation is a statement of the first law of thermodynamics for control volumes where mass flows continuously. It relates the changes in enthalpy, work done, and kinetic and potential energy of a fluid, which is essential for analyzing energy extraction in devices like turbines.

Transcript

00:01 Okay, in order to solve this question, we have to know some things about argon.

00:03 We know that the gamma value for argon is going to be 5 over 3, right, because argon is a monotomic molecule.

00:09 And we know the molar mass is about 40 grams from mole, and the flow rate, we're told, is 80 kilograms per minute.

00:14 And then we have some more stuff.

00:15 I already sort of converted these units to kelvin, the temperature, and the pressure to kilo -pascals.

00:22 I mean, you're setting 273 here and multiplying this by 1 ,000.

00:25 It's pretty simple.

00:26 And we also know that the energy is going to be equal to 3 over 2 nrt.

00:31 Now, importantly, since it is an adiabatic process, we have these two formulas, right? these two relations, but we don't have one that's for relating p and t.

00:42 So how do we do this? well, the first step is to use the p1 and p2 and solve for a ratio v1 over v2.

00:51 So let's do that.

00:52 If we rearrange this formula, we'll get v1 over v2 is going to be equal to p2 over p2.

00:56 P1 to the power of 1 over gamma, right? this is just rearranging this equation.

01:04 So if we get this answer, it's going to be 300 kpa, divided by 1 ,500 kpa, to the power of 3 over 5, right? and so if we put this in the calculator, we'll get an answer of 0 .381 about.

01:25 Cool.

01:26 So now we know v1 over v2 is equal to 0 .381.

01:28 Let's look at the first equation.

01:30 Let's solve for v2.

01:31 Or, sorry, t2.

01:32 T2 is going to be equal to t1 v1 gamma -minus 1 over v2 gamma -minus 1.

01:39 Now let's take the t1 out, and let's take the gamma -minus 1 out of the brackets...

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