Question

A cogeneration system works with a water cycle and a refrigerant (ammonia) cycle combined. Superheated water vapor enters turbine 1 (efficiency of 85%) at a flow rate of 5 kg/sec, 50 bar and 500°C and expands to 1.5 bar. Half of the flow is extracted for industrial heating and the rest enters a heat exchanger. The condensate leaves the heat exchanger as saturated liquid at 1 bar and combines with the return flow from the industrial process, which comes back at 60oC and 1 bar. The combined flow is pumped (efficiency of 85%) to the boiler pressure. The refrigerant cycle is an ideal Rankine cycle. The ammonia enters turbine 2 at a pressure of 14 bar and a temperature of 100°C and leaves the condenser at 3 bar. Calculate: a) The amount of heat required by the boiler, in kW. b) The net power output of the cogeneration system, in kW. c) The heat transfer provided to the industrial process.

Answer 1

To handle such problems , first thing to do is making T-S diagram of all the state points given in the question.

After determining all the points we can proceed to solve for the required quantities.

So here we go...

T for water (H₂O) so bar 그 Sook 55 1.5 has . Ibar 111.4°C 99.624 4 60°c ba s T for Ammonia (NM) 14 bar 11 3 bore 10 9 s from steam Tables ! کو KJ 3434.7 h = 6.978) for point as in Ses si =) [S+ (49-8f)] = ? 1 1,434 + 2 (7.223-1.434) = 6.9881 =) 1.434) = 6.9981 -> X = 0.958

1. hes 11 [the + x (ty-the) Jet worden 467.1 + 0,958 ( 2694 - 467.1) has 2600.47 KJ/kg no Talline 1 85/ Turbine 1 h, he th, the 2725-6KT = 0.85 => ho = 417.4 KT 21.26 417.40 4118 the the - Gw (T3-T6) 417.4 4.16 99.62-60 hi - g 51.8 KJ Kg By Energy Babance! mz azt mohol my h4 -) 2.5(4174) + 2,51251.8) = 5 ( ha) > ha = 334.6 KT kg hes 11 ha t YA (PS - P4) 24 CVS EL E specific volume of Liquids Remains more or less Same

11.04 150-1) 100 he 334.6 + 1000 kPa kg KT hos 339,7 kg npump I 85% > hss - hq ha 17 O, &5 hy KT his 340.6 kg 1651.4 KJ for Aminonia Cycle; hoa Tables, ho (from 1431.7 KJ kg har 138.82 T kg the thing + Yo (or) = 138.827 TE (14-3) x100 Kla he =190.5 Ig

for Heat Exchanger By Heat Balance & Heat given By Steam Heat Taken By NH3 mm Na (ng-hop = m Cherha) - > 2.5 (272506 - 417.4) = mung (1851.4 – 190.5) mono 3.82 kg &2 kg Boiler !- (a) Amount of heat required By Channs) [ 3434,7 in Boilet 5 340.6] = 15470.5 kW And. 16) Net Power output ? Whet a Wsteam & Wanna steam = com Elly hy [lhi-hy-(s – ho] | (3434.7 – 27954)- (940,6 -334,6}] sream Ich it muna arry [thy thg) - Can-Ahre)] - 5 + 3.82 [11057.2 - 1312) (140.5-138.827 (3515.5 + 837.83) What a 4348.33 A KW Heat Transfer to Indrihstry in mua (ho-ho) to Industry (299506 - $518] 6184,5 kW
Hope you got the concept well...!!!