Question

An industrial steam turbine shown in the figure below receives 20 [kg/s] of superheated steam at 10 [MPa) and 500 [°C] (State 1). Steam is extracted for another industrial process at a rate of 5 kg/s) ata location in the turbine where the pressure is 1 [MPa) and the temperature is 200 [°C] (State 21. The remaining steam continues to expand through the turbine and exit at State 3, where the pressure is 10 [kPa) and the quality is 90%. Assume that expansion of the steam through the turbine occurs adiabatically. The steam from the exit of the turbine then enters a condenser where it is converted to a saturated liquid (State 4). Calculate the following: (a) The mass flow rate at the exit from turbine, ne b) The power produced by the turbine, W, neglecting the effect of kinetic energy in the energy cquation If the diameter at the extraction pipe from the turbine (a: State 2) is 40 [cm), calculate the exit velocity from this pipe The rate of heat transfer removed from the steam in the condenser Draw a T-v diagram showing the state points and process path(s). Label the values of T and for cach state point and clearly label the constant pressure lines that pass through the state points. id es W. P, 10 MPa Ti = 500 °C m = 20 kg/s Turbine Py = 10 kPa 2) = 90% P, 1 MPa T = 200C m = 5 kg's D = 40 cm Condenser P = 10 kPa Saturated liquid

Answer 1

20 Impa, 200c I- h P = 10 mpa CP R=1mpa 500. 500°c 72019 2006 nga 200°ct ngga erkak P lokea saturated Liquid? д V V₂ hn- her do impa, 200°c = Solution From Thermodynamic DHB diagram BT Mijagung 13 lompa, 500°C = 3374.6K5/kg 2826.8 K5/164 Ad lokpa ar oolbar 191.8 kJ/kg skeople ya 2392.9 kJ/kg has hdz + x3 hlas lo rokupa 24-90% hz 191.8+0.962392-2) = 23 28:41K3/19 191.8 K5/kg @ 10lpa flow rate at turbine excut (ing) mau balance im = netim z = 20 = 3 +mz =) na + m 3 = 20 = 3 tmz -) | nz = is kg/s b) power produced by turbine (lit) We = 20chi - ha) +15 ha-ha) = 20 (3372.6-2826.8) +15(282618-23h5.hd) int 1817 6.8kW c) if Dz = toom. 0592 m²/kg mz = Az Vz from supocheat steamtable (The Coode) 2], vz =) V₂ = 8.193 m/ a ) man Var? = 0.20592m3 hee have 2 5 0 20592 d) create of heat removed in the condenser. QR = n3 (hz-ha) 15 (23h5.41- 191.8) Q R = 32304.15 KW