In 1872, George Brayton applied for a patent for his "Ready Motor", a reciprocating heat engine operating on a gas power cycle. One type of Brayton cycle is open to the atmosphere and uses an internal combustion chamber and another type is closed and uses a heat exchanger. The reversed Joule cycle uses an external heat source and incorporates the use of a regenerator. It is also sometimes known as the Joule cycle. The engine cycle is named after George Brayton (1830–1892), the American engineer who developed it originally for use in piston engines, although it was originally proposed and patented by Englishman John Barber in 1791. Although the cycle is usually run as an open system (and indeed must be run as such if internal combustion is used), it is conventionally assumed for the purposes of thermodynamic analysis that the exhaust gases are reused in the intake, enabling analysis as a closed system.
The original Brayton engines used a piston compressor and piston expander, but modern gas turbine engines and airbreathing jet engines also follow the Brayton cycle. Thank you.The Brayton cycle is a thermodynamic cycle that describes the operation of certain heat engines that have air or some other gas as their working fluid. So this is the required answer to the question. Our answer to the question is option b, so our answer is here: option b, which is 95 kilo, joule per k g. So if we round off it will get 95 kilo joule per k g. Joule per k g now, let us see which option is correct, so our option b is correct, which is 95 kilo joule per k g. Actually, so it will be equals to 212.5 minus 117.647, so it will be 94.853 kilo. We know that the work done net actual is equal to work done by the turbine actual minus the work done by the compressor. So from here, work done by the turbine actual will come out equal to 212.5 kilo, joule per k g. So it will be equals to 0.85 is equal to work done by the turbine actual over 250 point. Tropic efficiency is equal to work done by the term when actual over the work done by the turbine ideal. Now the tropic efficiency of the turbine is also. Actual is equal to 117.647 kilo jule per kg. So from here we get work done by the compressor. Actually so here it will be equal to 0.85 is equal to 100 over work done actual. Joules per kg, the isentropic efficiency is 1.85, so centropic efficiency of the compressor is equal to work done by the compressor ideal over the work done by the compressor. Joule per kg and work done by the compressor ideal is equal to 100 kilo.
We will basically get work done by the turbine.
So it is our ideal and it is equal to 0.4 now solving this first equation and this second equation so from here we write from 1 and 2. W r is equal to basically w c over w t work done by the compressor over the turbine. W r of ideal is given to us as 115 point so b. So we know that w net, which is the net work done, is equal to work done by the turbine minus the work done by the compressor. So now here we have to find a net work output of the cycle. W r is equal to 0.4, and here we are given that the efficiency t is equal to the efficiency of c is equal to 0.85. W net is done, is equal to 150 kilo, joule per k, g and backwork ratio, which is represented by b. Hello students in this question, we are given an ideal batrons cycle where the net work done.
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