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can you derive the formula for efficiency regarding the Brayton Cycle

          can you derive the formula for efficiency regarding the Brayton Cycle
        

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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Transcript

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00:01 So as for the statement of this question, we need to find the cycle efficiency for the cycle that is in the problem 9 .41, right? so let's start with it.
00:15 Now we are going to get all the values that we're obtaining the previous problem, right? so because it would be used.
00:23 Now, to obtain the risk, first the specific volume at state 3 is determined from the given pressure.
00:29 And the saturated liquid using the table b .2 .1, right? so the specific volume of state 3 is equal to 0 .056 meter, cubic per kilogram, right? now we need to determine the enthalpy at state 4, right? so this is determined from the energy balance on the second form, so that is a sub 3 plus alpha sub 3 times the difference of the pressures, right? we have the values just plug them in and we would get that the enthalpya state for it's equal to 172 .17 .17 kilojoules per kilogram right now we need to determine the quality the quality at state seven and this is determined from the condition that the intropia state seven is equal to the intropia state of five and the intropies of all the components are the given condensions.
01:28 Temperature right so we have quality at the state seven is equal to the entropy as state seven negative the entropy of all the liquids at negative 15 temperature divided by entropy evaporation and negative 15 right again you have the values you just need to plug them in and we would get the required value of the quality at state seven now that we have the quality we need to calculate the enthalpy at the state 7 right so the enthalpy at state 7 is equal to the enthalpy of the liquid at negative 15 temperature plus the quality at 7 times the enthalpy of evaporation negative 15 again you have the values you just plug them in and obtain the value of so ithalpia stage 7, that is equal to 1 ,414 .06 kilojoules per kilogram, right? now that we have all the required parameters, we just need to plug them in in the expression for the thermal efficiency.
02:34 That is equal to omega divided by the input quality, right? so now i'm going to plug in the expression for this variable...
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