Saturated vapor steam at 100 Psia at a rate of 1 lbm/s enters an adiabatic mixer

Question

Saturated vapor steam at 100 Psia at a rate of 1 lbm/s enters an adiabatic mixer and is mixed with 5 lbm/s saturated liquid water at 100 Psia pressure. The mixture leaves the mixer at the same pressure. Calculate: a) The mixture enthalpy and quality at the exit, and b) The entropy generation for this process.

Explanation / Answer

adiabatic mixture ==> Q = 0 (insulated) also no work is done in the process of mixing W = 0 therefore energy conservation yields: mi1= mass rate inlet stream1; hi1 = specific enthalpy inlet stream 1 ; he = specific enthalpy of exit stream; me= mass rate of exit ; mi1(hi1+ kei1 + pei1) + mi2(hi2+kei2+pe2)+ Q -W = me(he+ ke+ pe)........(i) change in ke,pe from inlet to exit is very small compared to change in h ? (i) reduces to mi1(hi1) + mi2(hi2)= me(he) ..............................(ii) using mass conservation: mi1+ mi2 = me he = [mi1(hi1) + mi2(hi2)] / [mi1+ mi2] hi1= (hg at Psat=100Psia )= 1187.5 hi2 = (hf at 100 Psat=Psia) = 298.51 mi1 = 1 ; mi2= 5 he= 446.675 at P=100Psia a) xe = quality at exit stream = (he-hf) / (hg-hf) = 1/6 b) adiabatic process Q=0 ? mi1*si1 + mi2*si2 + Sgen + integral(Q/T)= me*se from values from table se = sf + (1/6)*(sg-sf) si1 = 1.6032 ; si2 = 0.47427 Sgen = 0 ( reversible adiabatic process )

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