MECH 213 - 10.10.07

# MECH 213 - 10.10.07 - MECH 213 Isentropic o S = 0 o With...

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MECH 213 – October 10, 2007 Isentropic o ΔS = 0 o With constant Specific heat 1TDS 0 = ΔS = c v *ln(T 2 /T 1 ) + R*ln(v 2 /v 1 ) T 2 /T 1 = (v 2 /v 1 ) -R/c_v T 2 /T 1 = (v 2 /v 1 ) -(k-1) = (v 1 /v 2 ) k-1 2TDS T 2 /T 1 = (p 2 /p 1 )^(R/c P ) = (p 2 /p 1 )^(k - 1/k) (v 1 /v 2 ) k = p 2 /p 1 p 1 v 1 k = p 2 v 2 k o Example Air at 25 C and 1 atm Compressed isentropically to 6 atm Find T 2 , constant c p T 2 = T 1 (p 2 /p 1 )^(k-1/k) k = 1.4 and temp must be absolute T 2 = (298)(6)^(1.4-1/1.4) = 497.2K o 0 = S 0 2 – S 0 1 – R*ln(p 2 /p 1 ) ln(p 2 /p 1 ) = (S 0 2 – S 0 1 )/R p 2 /p 1 = exp{(S 0 2 – S 0 1 )/R} p 2 /p 1 = exp(S 0 2 /R)/ exp(S 0 1 /R) = p R2 /p R1 o plug above into past equations v 2 /v 1 = (p R1 /T 1 )/(p R2 /T 2 ) = v R2 /v r1 o Example: 25C, 1atm, 6atm p 2 /p 1 = 6 = p r2 /p R1 p R1 = 1.3543 from air table p R2 = 8.1258 calculated T = 490 T = 500 p R interpolated to get a Temperature of 495.14K

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MECH 213 - 10.10.07 - MECH 213 Isentropic o S = 0 o With...

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