Question

One whom the commowe 4MP, SC, the conditions we 30 Pa 92 percent quality P = 4 MPa Ti = 500°C V = 80 m/s Steam m= 12 kg/s $. w out The other 12 De hecho, the power out, and the subtle area P= 30 kPa x=0.92 V = 50 m/s

Answer 1

Assumptions: Steady flow X adiabatis => =0 n = n2 = n = 12 kg/ neglect change in Potential ener99 P = 4 MPa Ti= sooc P2 = 30kPa x3 0.92 (Wet Steam) V2= som's, Ta=? V = somis oke=?, Wout = ? , ALE? turbine inlet area. from steady flow energy equation 902:37] z &-wam [Chath) + 12²_v2 - - » [Cha-h) + ] -w=in [ Cha-hi) + oke]

Data from steam tables Inlet : Pi = 4mPaz 4000 kea T = 500°c Tsat @ 4 MPa 25033 Tis Tsat. superheated vapour. &=0.08644 m3/kg h = 3446 kJ/kg outlet: wet steam Pe= 30 kPa 7 he = 289.27 Alt 20.92 hfg=2335.3 kJ/kg hz= ht+xz hag h2 = 289.27+0.92 x 2335.3 h2= 24 37.75 kJ/kg SKE= V2²_v2_s0²_802 302.3 -1.95kJ/ks 2 energy û =- m DKE = -1.95 kr/kg + change in kinetic - W = m [lha-bi) + AkE] m[Ch2-hi) +Ske] = -12 [(2437-75-3946.0 )+ (-1.95)] w = 12122kW [W = 12.1 MW Z The power output M= A.V. nih u V → A, 12 X0.08544 80 A1 = 0.013 m2 / The turbine inlet > A, a aseg