Question
Steam enters a horizontal 6-in.-diameter pipe as a saturated vapor at 20 lbf/in.2 with a velocity of 30 ft/s and exits at 14.7 lbf/in.2 with a quality of 95%. Heat transfer from the pipe to the surroundings at 80degreeF takes place at an average outer surface temperature of 220degreeF. For operation at steady state, determine the velocity at the exit. in ft/s. the rate of heat transfer from the pipe, in Btu/s. the rate of entropy production, in Btu/s degree R, for a control volume comprising only the pipe and its contents. the rate of entropy production, in Btu/s degree R, for an enlarged control volume that includes the pipe and enough of its immediate surroundings so that heat transfer from the control volume occurs at 80degreeF. Why do the answers of parts (c) and (d) differ?
Explanation / Answer
a) mass flow rate at entrance m1=1 A1V1 take the at saturated vapour atP=20lbf/in2 then m=0.0497*/4*0.52*30 =0.2927lb/sec since mass flow rate remains constant m2=m1=2A2V2 take 2 at pressure P2 take 2 at pressure P2 0.2927=0.0392*/4*0.52*V2 V2=38.028ft/s b)at P1=20lbf/in2 from stem tables h1=hg=1156.4Btu/lb s1=sg=1.732Btu/lb R at P2=14.7lbf/in2 andx2=095 then h2=1102.03Btu/lb s2=1.6858Btu/lb R by using S.F.E.equation h1+V12/2+q=h2+V22/2 then substituting known values we will get q=-54.359Btu/lb c)rate of entropy production cv=-q/Ts+(s2-s1) =54.359/(220+460)+(1.6858-1.732) =0.037Btu/lbm.R d)sgen=s2-s1-q/T =1.6858-1.732-(-54.359/(80+460)) =0.0544Btu/lbm R