2. (25 pts) The following data are for a simple steam power...

2. (25 pts) The following data are for a simple steam power plant as shown in figure. P1= 6.2 MPa T2=45 C P2=6.1 MPa T3= 175 C P3= 5.9 MPa T4= 500 C P4= 5.7 MPa T5= 490 C P5= 5.5 MPa x6= 0.92 P6=10 kPa T7= 40 C P7= 9 kPa V6= 200 m/s Rate= of steam flow= 25 kg/s Power input to pump= 300 kW Pipe diameters: Steam generator to turbine: 200mm Condenser to steam generator: 75 mm Calculate: a) Power output of the turbine b) Heat transfer rates in condenser, economizer and steam generator c) Flow rate of cooling water through condenser, if the temperature of the cooling water increases from 15 to 25 C in the condenser

Transcript

00:01 So here, let the given data, this is n -suff is this is 25 kilogram per second, right? and p2 is 6 .1 megapascals and t2 is the second temperature.

00:18 This is 45 degrees celsius, right? so by using the steam table, h2 will be 193 .75 kilojol per kilogram.

00:29 This is h2, right? this is enthalpy, right? now, let at p3, at third pressure, p3 is 5 .9 megapascals, and the third temperature, t3, is 175 degrees celsius, right? so here, from that table, h -sub -3 will be 743 .80 kilojol per kg, right? now at p poor at p 4 this is 5 .7 megapascals so at temperature port this will this is 5 hundred degrees celsius so this h4 will be from that table this is h4 will be this will be 3187 .5 kilojol per kilogram right now at p6 right at p6 this is 10 kilo pascal right so here is x sub 6 this is 0 .92 so here h 6 will be h sub f plus h sub 6 x 6 h g minus h f right this is the formula here and this is 8 sub 6 this is 191 .81 plus 0 .92 times 2583 .9 minus 191 .81 right now if we calculate this by using a calculator this will be 23992 .53 kilojoules per kg right we have calculated all associated inthalpies right now at p7 let at the 7 pressure this is p 7 this is 9 kilo pascal right at t 7 this is 40 degrees celsius this is 8 7 this is a 167 5 5 4 kilojou per kg right now here let it in part a let coming to the problem let in part a the power output is the m x times h subpip minus eight sub six these are enthalpyes right and this is let power output is this is 25 x right this is given this is 25 because this is the that 25 g per second right this is is so let this is h subpipe minus 2 ,392 .53 .3 so each pipe will be h pipe is 3 ,4005 .3 per kjoules per kg right so the power output will be this will be 25 times 3 ,4005 .3 minus 2 ,3192 .53.

04:01 And if we calculate this by a calculator, this will be power output and this is 25 ,319 .2 .25 kilowatts.

04:13 So this is the output power, right? now in part b, we are asked in part b we are asked the heat transfer rates in condenser, economizer and steam generator, right? so let the heat change is q sub c and this is that 25 and 8 sub 6 minus 8 7.

04:42 So this is 25 times 2 ,392 .53 minus 167 .54 and this enthalpy or heat transfer will be.

04:55 Be 55 ,624 .74 .7 .5 kilowatt, right? this is the answer, right? this is the heat transfer in the condenser, right? and now the heat transfer in the economizer is, this is a condenser, c represents condenser.

05:22 So the heat transfer in economizer, which is e, he represents that this is x.

05:31 This is h3 minus h2...

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