Question

Steady-state operating data for a simple steam power plant are provided in figure. Kinetic and potential energy effects can be ignored. Determine the a) the mass flow rate of the steam, (10 pts.) b) lin, (5 pts.) c) the mass flow rate of the cooling water, (5 pts.) d) thermal efficiency, (5 pts.) e) What happens to the thermal efficiency if we increase the turbine inlet pressure while keeping the inlet temperature (T1) constant? Explain it clearly. (10 pts.) PI T P1 4 MPa Turbine Power out wo out T1 610°C mini P2 0.2 bar P2 saturated vapor T4 Steam generator 50°C Cooling water in at Ts P4 4 MPa T5 15°C Condenser T6 35°C P4 TA Cooling water out at To Wout 3 MW Pump 11 kW Win + 3 in

Answer 1

From Steam table fore Su perncapead Condilion ar P = 4HPa T= 610°c We have; thi = 3697. 8145 ks/kg) (buy interpolation) Since, The that given figure suggens Point % is galwared Voupowr. so at P2 = 0.2 bar we have following prosperios. - 251.3997 K3/Kg hg = 2608.9414 Kalkg So, the = hq = 2608-9474 K3/kg] (given Point 2 is saturated vapower) he ie, Work curbet frem turbine Wout = melha - ha) >> 3X103 = mg (3697-8445 - 2608-944 > mig = 2.15508 kg/s mass flow rate of steam is 2.15508 Kgle. Any at by from steam table Pa = Pz = 0.2 bave ha = hf = 251-3997 kalkg as Work in puput in pump can be given Win = mg (14-93) 11 = o 2.15508 44-251-3997) h4 = 255-3923 K3/Kg!

C7 ie Qin - 9484-2312 kl. here, CPW is specific heat of halen Heat Inbet in boiler (Pin) = mgho-ha) = 2.75508( 3697.8445-255-3923) Cin= 9484.2312 KW Ans In Condenser, Heat transfer from bot steam to Convert in Inater is equal to hear transfer of Cooling water: ie, mw xCpu (T6-T6) → 2.75508(2608.9474- 251.3997) = mw84-187 (35-15) mw = 77.56427 K81% which is equal to 4.187 63 4.187 kalye ise, Hau How sease of cooling Water (mw) = 77.56427 kg 18. Whet 3 dy Thermal efficiency is given by (hr) mg (h2-23) Ay, Pin where What Wout-Win = 3000 - Il - 2989 Kul Pine 9484.2312 kW. 2989 т = 0.31515 9484.2312 lie 11 = 31.5151 Thermal Efficiency is 31:5157

ey let ug suppoge after increasing twebine inlet presence to P1 = 15 H Pa f Ti = 610°C (Constant) so, again from gleam table Corresponding to given Condition -h = 3690.01133 Kulka the = 2608-94 74 Kalkg) { salwcased vapowe Condition of B-02 brand wout = ng (hi-ha) 3x103 mg (3690-01783 -- 2608.9474) mg = 2.77487 kg/2 Tha: 251-3997 kJ/kg! { that the @o-2 bar ] so, Win - Il mg ( ne-h3) = 2.77487 ( 64 -25).3997) tha = 255-3638 kJ/kg! Hear input in boiler (Pin) = mg (hi-h4) = 2.71487 ( 3690.07733-255-3638) Min = 9530.8834 K lal Thermal Efficiency (hr) Wout-lin Qin 9530-8834 ht: 31-3617 3000 -11 --0-3136 80, clearly we can see that efficiency of Cycle decreases Slightly 94 we increase the inlet pressure While keeping the Anier temperature Constant,