Mechanical and Aerospace Engineering

Office of Industrial Productivity and Energy Assessment 5 Michael R. Muller, Director Solution: Let the main flow be 0 m & , the mass flow drained off in the separator be 1 m & , and the mass flow through the calorimeter be 2 m & ; then 2 1 0 m m m & & & + = If h 1 is the specific enthalpy of the saturated liquid drained off in the separator and h 2 is the specific enthalpy of the superheated steam leaving the calorimeter, then, for an overall adiabatic system of separator and calorimeter, 2 2 1 1 0 0 h m h m h m & & & + = or 2 0 2 1 0 1 0 h m m h m m h & & & & + = Steam tables give h(20 psia, 228 ° F)=1156 Btu/lbm and h(20 psia, 250 ° F)=1167 Btu/lbm; so, at 20 psia and 230 ° F, h 2 =1156+[(230-228)/(250-228)](1167-1156)=1157 Btu/lbm. At the separator drain (which is normally upstream of the valve), h 1 = (h at 200 psia for the saturated liquid) = 355.5 Btu/lbm. Substituting these data, ( ) ( ) lbm Btu h o / 1061 1157 0 . 5 5 . 4 3 . 196 0 . 5 5 . 0 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = https://www.coursehero.com/file/6123276/thermorevsteamQ-A/ This study resource was shared via CourseHero.com Office of Industrial Productivity and Energy Assessment 6 Michael R. Muller, Director and the moisture fraction, y o , of the main flow is given by 1061=(y o )(355.5)+(1-y o )(1198.8), or y o ≈ 16%. 7. Can you determine the actual moisture contained in the steam that went into the calorimeter of the previous problem? Solution Since the steam was throttled and since the specific enthalpy of the main-flow steam at the calorimeter exit was determined to be 1157 Btu/lbm, then h 2 =(1-x)h cf +xh cg , where c refers to the calorimeter inlet. The inlet pressure is 200 psia, so that 1157=(y)(355.5)+(1-y)(1198.8), whence y=5%. 8. A hydroelectric turbine is supplied with 25m 3 /s of water. At the turbine inlet the water is at 5 bar and 25 ° C, with an elevation above datum of 100 meters and a flow velocity of 1m/s. At the turbine exit the water is at 1.2 bar and 25.1 ° C, with zero elevation and a flow velocity of 11m/s. The turbine loses 5J of heat per kg of water flowing through it. Assuming that the water is an incompressible liquid with specific heat 4.178kJ/kg ⋅ K, calculate (a) the potential energy change (kJ/kg), (b) the kinetic energy change (kJ/kg), (c) the internal energy change (kJ/kg), (d) the power output (MW). Solution: (a) Δ pe=0.00981(0-100)=-0.981kJ/kg. (b) Δ ke=(11 2 -1 2 )/(2 x 1000)=0.060kJ/kg. (note that a joule is a kg-m/s 2 ) (c) Δ u=c Δ T=(4.178)(25.1-25)=0.418kJ/kg. (d) The governing power equation is W & = m & ( Δ h+ Δ ke+ Δ pe-q). One finds: Δ h= Δ u+ νΔ P=(0.418kJ/kg)+(1.003x10 -3 m 3 /kg)[(120-500)kN/m 2 ]=0.037kJ/kg; also, m & = V & / ν =25/(1.003x10 -3 )=24925kg/s. It is given that q=-0.005kJ/kg. Therefore, W & =(24925)(0.037+0.060-0.981+0.005)=-21909kW=-21.9MW (negative because output). 9. A well-insulated turbine intakes steam at 75 psia, 1600 ° R. The exhaust occurs at 15 psia. A mass flow of 10 lbm/sec delivers 1750 hp. How efficient is the turbine? Solution: https://www.coursehero.com/file/6123276/thermorevsteamQ-A/ This study resource was shared via CourseHero.com

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