In a simple gas turbine air leaves the compressor at 445 k...

In a simple gas turbine air leaves the compressor at 445 K and leaves the combustor at 1000 K and leaves the turbine at 750 K. The thermal efficiency of the simple gas turbine engine is 20.6 %. The combustor has a burning efficiency of 98% with QR = 44,000 kJ/kg.K and Cp =1.07 kJ/kg.K. If a recuperator with effectiveness of 70% is fitted to the engine, determine: a) the specific power output of the engine. b) The improvement in the thermal efficiency of the engine c) The improvement in the specific fuel consumption in kg /kW.hr.

Transcript

00:01 Hello students in this question like given as that air enters the compressor of a gas turbine engine at and after that some data is given regarding the temperature pressure and the efficiency of turbine so here and we need to find the like a solution of two parts of this question first is the party and second is the part so first of all like if we draw the diagram of this problem then the diagram of this problem is looks like this this type of diagram okay here 300 kelvin is the initial temperature this is also given in qion 300 kelvin and 300 kelvin here is the initial temperature and here 500 atty kelvin is the like a compressor exit temperature and in like in q'son also given us that the compressor exit pressure p2 and the compressor initial pressure p1 is also given an according to question initial pressure even is equal to 100 kiosk and the compressor exit pressure which p2 is equal to it is equal to 700 kilo pascal okay and in question also given as the like heat input and the value of heat input incision given is 950 kilo joule per kege and in question also given us the turbine efficiency and just writing the turbine of change by nita so turbine exchange is given 86 % now we like solve the part of this question and but before solving the part of discussion first we need to like discuss something regarding this problem well first of all we already know that an actual gas turbine plant operates at specific conditions and the fraction of turbine work output used to drive the compressor and the thermal efficiency are to be determined okay and after that like this is the some small information in which we need to solve this question but for following discussion we also need to take some adeptions without any adaption we are not able to solve this case then what is the adeption like first adeption is just assume that steady operating conditions exist here and second just assume that the air standard adjunctions are applicable and third just assume that the kinetic and potential energy are negligible okay and fourth just assume that a air is an ideal guess with variable specific heats.

02:56 Okay, now we solve party of discussion and for solving the party of discussion we use the isentrophic relations so by using the as a tropic relations relations we will write that if temperature t1 is equal to 300 kelvin then h1 h1 this this is h1 okay h1 is equal to 300 point 3 hundred 1 .19 kilojoule per k and just assume that if temperature t2 is equal not assume if temperature t2 is equal to 580 kelvin then h2 is 2 so h2 is equal to 580 0 .04 kilojoule per keg okay now in question also given us the heat input that means q in and we already know that q in is equal to q in is equal to h3 minus h1 okay h3 sorry h3 minus h2 this is 2 and this is 3 so q in is equal to h 3 minus h 2 q in is equal to h3 minus h2 so from here we write that h3 is equal to where we transfer h2 on left side then this equation becomes h3 is equal to q in plus h2 and the value of q in is also given in question it is 950 kilojoule per kg plus h2 and what is the value of h2? the value of h2 is 586 0 .04 kilojoule per cage so when we solve this then we get the value of s3 and the value of s 3 is s 3 is equal to 1 ,500 36 .0404 040 kg okay now we calculate the pressure ratio so pressure ratio is equal to p2 divided by p here what is the p2 here p2 is the is the compressor exit pressure and here p1 is the initial pressure so just put the value of p2 and p1 and according to the the way we just write the ratio so pressure ratio is equal to p2 and the value of feet according to position is 700 kioskd divided by and the value of p1 according to basically 100 kilo pascal so when we solve this then we get the ratio of pressure is 7 okay now like if we find the liquor pr 3 so what is the fear 3 it is like 700 it's also we can just write that pr 3 is equal to 474 474 .11.

07:10 Okay, this is pressure ratio.

07:13 Yeah, now we also know that if we want to find pr4 then pr4 is equal to p4 divided by p3.

07:28 Okay, these are the conditions p4 p3.

07:33 Okay, so p4 divided by f3.

07:34 Okay, so p4 divided by f3...

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