An air standard cycle with constant specific heats at room...

Name: an air standard cycle with constant specific heats at room temperature is executed in a closed system with 6000e 3 kg of air and consists of the following three processes 1 2 vconstant heat 71945
Uploaded: 2020-08-24T23:06:28-08:00
Duration: 12 min 27 s
Channel: Nathan Silvano
Description: an air standard cycle with constant specific heats at room temperature is executed in a closed system with 6000e 3 kg of air and consists of the following three processes 1 2 vconstant heat 71945

An air-standard cycle with constant specific heats at room temperature is executed in a closed system with 6.000E-3 kg of air and consists of the following three processes: 1-2 V=Constant heat addition from 95 kPa and 17°C to 380 kPa 2-3 Isentropic expansion to 95 kPa 3-1 P=Constant heat rejection to initial state The properties of air at room temperature are cp=1.005 kJ/kg⋅K , cv=0.718 kJ/kg⋅K , and k=1.4 . Determine the thermal efficiency. The thermal efficiency is %.

Transcript

00:01 Okay, well, in this problem we have an error standard cycle with three process.

00:08 And in the first part of this problem, in part a, we have to show the ts and pv diagram.

00:16 So let's begin putting here the axis for our pv diagram.

00:24 This is the vertical axis, this is the horizontal axis, and this is going to be the ts diagram, okay? okay, well, let's see.

00:44 The pv diagram has three processes.

00:51 The first process is an isentropic process.

00:56 An isentropic process, as we know, is described by, how can we say, hyperboles.

01:06 So we have a high, this is not a good hyperbole.

01:10 Let's put here the hyperbole of the isentropic process.

01:16 So this is from point one to point two.

01:22 Now the second one is a process with constant pressure.

01:28 And the process with constant pressure is described by this horizontal line and here.

01:39 So this is the horizontal line and this is point three.

01:46 And finally, the last process is a process with, that we're going to say, is a straight line.

01:55 Because it's a heat rejection that depends on the initial state constants.

02:04 So this is going to be represented as a straight line.

02:09 Oops, let's put in blue.

02:11 Straight line from three to one.

02:15 Okay.

02:16 And again, as we know, the heat enters the system in here and leave the system in here because an isentropic does not change exchange heat with the surroundings.

02:31 Okay, so that's the definition of an isentropic process.

02:36 So this is the pv diagram for this cycle.

02:42 Now let's see the ts diagram.

02:45 Always remember that isentropic process in ts diagram are represented by vertical lines.

02:54 So this is the vertical lines which is an isentropic vertical process.

03:02 Okay okay but this is point one this is point two now the constant pressure is going to be a hyperbole let's put this way from two to three and the final one also a hyperbole but a little strange one.

03:32 And again the heat enters the system here and leaves the system here.

03:40 Okay, so this is the t -s diagram and the left one is the pv diagram.

03:53 This is item a.

03:55 Now let's continue with item b.

03:58 In item b we want to find the output that is correct yes that is correct we want to find the output heat of the cycle so the output heat as we can see in the diagram is represented actually no we do not want the output we want the input no that is correct the heat rejected so the heat rejected is the output heat okay so let's see the output heat is going to be represented by the work done by by the system actually into the system plus the change in the internal energy of the system okay so first of all what is the work so the work as we know is just the area under the curve so if we look here, the area is going to be this trapezoid...

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