Problem 2 Air enters a compressor at a pressure of 1.0 bar...

Problem 2 Air enters a compressor at a pressure of 1.0 bar and exits at 8.5 bar. The compressor inlet temperature is 37 °C. The compressor is not insulated. Kinetic and potential energy effects can be neglected. The compressor outlet temperature is 650 K and the power input to the compressor is 500 kW. The compressor is operating at steady-state, and the air can be modeled as an ideal gas. The air mass flow rate through the compressor is 1.8 kg/s. a. Determine the direction and rate of heat transfer, in kJ/kg. b. Determine the compressor isentropic efficiency. c. Draw the process on a T-s diagram (clearly indicate the direction of the process and label the states). d. Determine the entropy generation rate, in kJ/kg/K, assuming the boundary temperature is 100 °C, and state the nature of the process (reversible, irreversible, or impossible). Summarize you results in the table below a. heat transfer, kJ/kg (include direction, so in or out?) b. isentropic efficiency d. entropy generation rate, in kJ/kg/K, and is it reversible, irreversible, or impossible?

Transcript

00:01 Hello students here we have a air compressor that has the input power as pressure as p1 that is equal to one bar and the outlet temperature p2 is 8 .5 bar and it has the temperature of the inlet that is d1 equal to 37 degrees celsius and the kinetic energy potential energy or here neglected the compressor outlet temperature d2 is 650 kelvin and the power input to the compressor is 500 kilowatt so here we consider only the ideal gas with the steady state so here the mass flow rate of air is air flow that is equal to 1 .8 kilogram per second so from this we have to determine the direction and rate of the heat transfer so here we consider the reaction is adiabatic so we can write w equal to work done that is equal to delhg equal to cp t2 minus t1 so here cp equal to 1 .005 kilojou per kg per kelvin so here t1 in terms of kelvin that is 310 kelvin so substituting the values we will get the work done as 1 .005 into 650 minus 310 that is equal to 3 4 5 6 kilojoule per kg so this is our work done so from this we can write the rate of the heat transfer us rate of heat transfer that is q equal to m tp into t 2 minus d1 that is equal to 1 .8 into 1 .005 into 650 minus 3 10 that is equal to you equal to 59 96 .7 below vote similarly we have to determine the is centrosophic efficiency of the system so that is written as eta equal to h2 as minus h1 divided by h2 minus h1 so here we can write that terms as h1 equal to tp t1 that is equal to 1 .005 into 310 that is equal to 311 .55 kilojure per kg and we have a h2 that is equal to d2 that is equal to 1 .005 into 850 that is equal to 653 .25 kilojoule per g so here we have to determine h h h h hd2s that is equal to that can be determined by the relationship that is p1 divided by b2s that is equal to p2s that is equal to p2s divided by t1 whole power k divided by k minus 1 this is our equation number 2 so from this we have to determine the t2s so t2s can be written as t 1 into b2 s divided by b1 into k power my k power minus 1 divided by k so here substituting the values we will get 310 into 8 .5 divided by 1 into 4 minus 1 1 divided by 1 .4 that is equal to 509 .38 kelvin so from this we can determine our h2 s that is equal to cp t 2s that is equal to 1 .005 into 509 .38 that is equal to 511 .92 below jules per kg so from this we can determine our efficiency see that is equal to 511 .92 minus 311 .55 divided by 653 .25 minus 311 .55 so by solving this we will get our theta that is equal to 0 .5863 and from in section b we have to draw the bv diagram for the system so here we calculated our b1 b1 that is equal to 1 comma 37 degree and v2 v2 that is equal to 8 .5 comma 650 so from this we can write in our bv diagram that is here we have a a axis is b and x axis p so from this we can write 1 .37 8 .5.

06:34 650 so from this we get a diagonal line of bb diagram for the system...

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