MECH 310 Problems Chapter 6 Solutions, Fall 2011

MECH 310 Thermodynamics I November 28, 2011 American University of Beirut, Fall 2011 Handout #5 Chapter 06 Solved Problems Problem 6.25 C.V. Nozzle, steady state, 1 inlet and 1 exit flow, insulated so no heat transfer. 2, 2, q+h,+ V;/2= h.+ V,/2, q=0, V;=0 Table B.2.2: h; = 1464.9 =h, + 4507/(2*1000) => h, = 1363.6 kI/kg Table B.2.1: P.=300kPa Sat. state at -9.2°C : h, = 1363.6 = 138.0 + x, = 1293.8, => x, = 0.947, vz = 0.001536 + x, « 0.4064 = 0.3864 m/kg = ih,v,/V_ = 0.01 = 0.3864 / 450 = 8.56 x 10-6 m2 Inlet Exit Low V av HiP.A Low P,A Problem 6.27 Continuity Eq.6.3: 1; = A;V;/v; = m, =A.V2/Ve. Energy Eq.(per unit mass flow)6.13: h; + av 2 =h,+ we he - h; =} «2007/1000 — 3 «20/1000 = 19.8 ki/kg Te = Tj + (he - hj)/Cp = 300 + 19.8/1.004 = 319.72 K Now use the continuity equation and the ideal gas law AN. AN. Ve=Vj AV, |= = (RT; Pi) | AN; =RT,/P. T 19. 72) 100 x 200 P. =P; Eales an 100 (230 (7 - we |= 123.92 kPa Exit Low V HiP,A Problem 6.35 Energy Eq. hy + 3 Vv; +9Z,=h, +5 : + gZ> Process: Z;=2Z and V3=V; State 1, Table B.5.1: hy = 234.59 kJ/kg. vy = v¢ = 0.000829 m/kg Use energy eq.: => hy =h, = 234.59 ki/kg State 2: P2>&h, = 2-phase and T, =T,, (165 kPa) =-15°C hy = h¢+ Xz hey = 234.59 ki/kg Xq = (hg - hg) / hg= (234.59 — 180.19) / 209 = 0.2603 V2 = Vg + X2 * Vfg = 0.000746 + 0.2603 x 0.11932 = 0.0318 m/kg Now the continuity equation with V>= Vj) gives m= pAV = AV/v = A\Vjy/¥ = (Az V)) / v2 (Ap /Ay)=v2/ vy =(D2/D) , , 0.5 0.5 (D>/D)) = (v2 /-¥,) = (0.0318 / 0.000829) ~ = 6,19 Problem 6.42 C.V. Throttle, Steady, q = 0 and w = 0. No change in kinetic or potential energy. The energy equation then reduces to Energy Eq.6.13: hy =h, = 2927.2 kJ/kg from Table B.1.3 . . 110 C.V. Turbine, Steady, no heat transfer, specific work: w = 025 440 kJ/kg Energy Eq.: hy = hy =h3 + w= 2927.2 kI/kg => hy = 2927.2 - 440 = 2487.2 kI/kg State 3: (P,h) Table B.1.2 h< hg 2487.2 = 191.83 + x3 x 2392.8 U T=45.8°C , x3 = 0.959 Problem 6.50 C.V. air compressor q=0 Continuity Eq: my = my Energy Eq.6.13: hy + We =h> Table A.7: We jn = hs - hy = 607.02 - 290.17 = 316.85 kJ/kg C.V. cooler w= 0 Continuity Eq.: m3 = my, Energy Eq. hp = dout * hs dout = hz - hy = 607.02 - 300.19 = 306.83 kJ/kg Problem 6.55 C.V.: boiler steady single inlet and exit flow, neglect KE, PE energies in flow Continuity Eq.: my =m, =m; 600 1 2\3 1 2 boiler Table B.6.1: hy =-81.469 kJ/kg, hy = 86.85 kJ/kg, Table B.6.2: h3 = 289.05 kI/kg Energy Eq. boiler = bz — hy = 86.85 - (- 81.469) = 168.32 kI/kg Qhoiler = 14 Uhojler = 0.005 x 168.32 = 0.842 kW C.V. Superheater (same approximations as for boiler) Energy Eq. dcup heater = 3 — hy = 289.05 — 86.85 = 202.2 kI/kg Qsup heater — ™>4sup heater = 9-005 x 202.2 = 1.01 kW Problem 6.72 C.V. Turbine Steady state, 1 inlet and 2 exit flows. Continuity Eq.: m, =m + m3 => m3 = my - mz = 80 kg/s Energy Eq: mh, a Wr + mph, + mh; Table B.1.3 hy = 3582.3 kJ/kg, hy = 3137 kI/kg Table B.1.2: h3 = h¢+ x3hfg = 384.3 + 0.95x2278.6 = 2549.1 kI/kg From the energy equation, => Wy = mh; — mzh — m3h; = 91.565 MW Problem 6.76 C.V. Heat exchanger Energy Eq. Megohz9 + My,0h309 = Meoolh3o + My, oh 10 kPa 300°C sat. liq. _ — ae Meool 30°C 20°C Table B.1.1: hy9 = 83.96 kJ/kg, hgg = 125.79 kI/kg Table B.1.3: hoo. 10kPa — 3076.5 kJ/kg, B.1.2: he 10 kPa — 191.83 kJ/kg h3o0-h¢ okpa_, 3076.5 - 191.83 Meool = ™MH0~ hy5-hyy | ~ 125.79 - 83.96 ©? Ke/s Problem 6.94 Turbine As = (7/4)(0.2)? = 0.031 42 m?, vs = 0.06163 m*/kg V5 = thvs/As = 25 kg/s = 0.061 63 m*/kg / 0.031 42 m* = 49 m/s he = 191.83 + 0.92 2392.8 = 2393.2 kI/kg 1 2 2 wr =hs ~hg +3 (V5 -VG) = 3404 - 2393.2 + (497 - 2007 )/(2 = 1000) = 992 kI/kg Wy =mwy = 25 « 992 = 24 800 kW Remark: Notice the kinetic energy change is small relative to enthalpy change. Condenser A7 = (x/4)(0.075)* = 0.004 418 m?, v7 =0.001 008 m*/kg V; = thv/A; = 25 = 0.001 008 / 0.004 418 = 5.7 m/s hg = 191.83 + 0.92 2392.8 = 2393.2 kI/kg 1 2 2 dconp = by - hg +3 (V5 - VG) = 168 — 2393.2 + (5.77 — 200 )/(2 1000) = -2245.2 kI/kg Qconp = 25 * (-2245.2) = -56 130 kW This rate of heat transfer is carried away by the cooling water so -Qcond = My,0(Hout — hin)#,0 = 56 130 kW 56 130 => 1H50 ~ 104.9 - 63.0 = 1339.6 kg/s Economizer Ajz= nD-/4 = 0.004 418 m*, v7 =0.001 008 m*/kg V>=V- =thv-/A; = 25 « 0.001 008/0.004 418 = 5.7 ms, V3 = (v3/v7)V> = (0.001 118 / 0.001 008) 5.7 = 6.3 m/s x V; so kinetic energy change unimportant dECON = 3 - hy = 744 - 194 = 550.0 kI/kg Qecon =lWdEcon = 25 (550.0) = 13 750 kW 2 2 Generator A4 = 72D,/4 = 0.031 42 my, v4 = 0.060 23 m°/kg V4 = tavy/Ay = 25 « 0.060 23/0.031 42 = 47.9 m/s GEN = 3426 - 744 + (47.9? - 6.37)/(2* 1000) = 2683 kJ/kg QcEN = MqGEN =25 x (2683) = 67 075 kW Problem 6.99 CV: Compressor Qooyyp = 1n(hy - hg) + Woop = 0.05 (377 - 284) - 5.0 = -0.35 kW CV: Condenser Qconp = m(h; - hy) = 0.05 kg/s (134 - 367) kJ/kg = -11.65 kW C.V. Valve: hy =h3 = 134 kJ/kg CV: Evaporator Qryap = m (hs- hy) = 0.05 kg/s (280 - 134) kI/kg = 7.3 kW | P condenser {f \ 21 Ww compressor 4 evaporator 5 Problem 6.100 The compressor, turbine and nozzle are all steady state single flow devices and they are adiabatic. We will use air properties from table A.7.1: h, = 260.32, hz = 800.28, h3; = 1635.80, hy = 933.15, hs = 649.53 kI/kg Energy equation for the compressor gives We in = hy — hy = 800.28 — 260.32 = 539.36 kJ/kg Energy equation for the turbine gives wr = h3 — hy = 1635.80 — 933.15 = 702.65 kJ/kg Energy equation for the nozzle gives 2 hy=hs + “V5 5 “ V; = hy - hs = 933.15 — 649.53 = 283.62 kJ/kg V5 = [2( hy — hs) ] /? = (2x 283.62 x1000 ) 7 = 753 m/s Problem 6.101 Separation of phases in flash-evaporator constant h in the valve flow so Table B.1.3: hy = 763.5 kJ/kg hy, = 763.5 = 604.74 + x « 2133.8 => x=0.07439 = m,/m, Table B.1.2: hy = 2738.6 kJ/kg: Sat. liquid out h3 = 191.83 + 0.9 x 2392.8 = 2345.4 kI/kg Energy Eq.6.12 for the turbine 1000 W=tho(hy-h3) => hy = 55596-93454 = 2.543 kg/s > my, =m)/x = 34.19 kg/s = 123 075 kg/h