Required information problem 11004 a steady flow carnot...

! Required information Problem 11.004 - A steady flow Carnot cycle A steady-flow Carnot cycle uses water as the working fluid. Water changes from saturated liquid to saturated vapor as heat is transferred to it from a source at 240.0\deg C. Heat rejection takes place at a pressure of 10 kPa . Problem 11.004.d - Work output from the Carnot cycle Determine the net work output. The net work output is | kJ/kg.

Transcript

00:01 Now in this question we are given that a steady flow carnot cycle uses water as a working field now i've also drawn this t's diagram for the carnot cycle now water changes from saturated liquid to saturated vapor this is the saturated liquid and it converts to saturated vapor point four is saturated vapor and the source temperature we are given to be 250 degrees celsius so this is this process is happening at 250 degrees celsius now heat rejection takes place at a pressure of 20 kioskle.

00:33 So this pressure is 20 kioskl and these lines are the constant pressure lines.

00:42 Now first we need to draw this diagram on the ts diagram that i have already drawn and these are the saturation lines.

00:52 Now in the first part we have to find the thermal efficiency.

00:55 Now in case of karnot cycle, how we can we know that the efficiency for the karnot cycle is given by 1 minus t l divide by the t h t l is the lower temperature and t h is the higher temperature now this would be the higher temperature this is t h here and this would be the t l now t h we are already given in the equation to be 250 degrees celsius and tl now to find this tl we can refer the saturation table and we can find the saturation temperature corresponding to this saturation pressure which is 20 kilo pascal now from the table stream table will get this lower temperature to be 60 .06 degrees celsius now we have obtained all this th and tl we can plug it here so one minus now one thing you should know that the values of tl and thh should be put in kelvin not in degrees celsius so we're going to convert it into kelvin so 60 point 06 we'll go to add 273 to it to convert it into kelvin divide by th we have 250 plus 273 and this is also in kelvin now from here we'll get this efficiency to be 0 .363 which can be written as 36 .3 percentage so this is the efficiency for the car note cycle or the thermal efficiency now next in the b part we have to calculate the amount of heat rejected.

02:44 Now let me show it on the figure.

02:48 Now here heat is added.

02:52 This is q in and here heat is rejected.

02:56 This is qr.

02:57 We have to calculate this qr...

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