Answered

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Air is compressed from 100 kPa, 300 K to 1200 kPa in a two-stage compressor with intercooling between the stages. The intercooler pressure is 400 kPa. The air is cooled back to 300 K in the intercooler before entering the second compressor stage. Each compressor stage is isentropic. For steady-state operation and negligible changes in kinetic and potential energy from inlet to exit, determine (a) the temperature at the exit of the seco

Sagot :

batolisis

Answer:

410.7 K

Explanation:

Given data:

P1 ( initial air pressure ) = 100 kPa

Pi ( intercooler pressure ) = 400 kPa

P2( compressed air pressure )  = 1200 kPa

T1 = 300 k

Ti = 300 k

Calculate for the temperature at the exit of the second stage ( T2 )

we can calculate this using the relation below

T2 / Ti = ( P2/Pi )^(r-1/r)

∴ T2 = 300 ( 1200 / 400 )^( 1.4 - 1 / 1.4 )

         = 300 ( 3 )^0.286

         = 300 * 1.369  = 410.7 K

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