00:00:04.130 this example problems show how can we 00:00:07.61000:00:07.620 use the steam table to calculate the 00:00:10.07000:00:10.080 enthalpy and composition of partial to 00:00:12.53000:00:12.540 condense stream so on this problem you 00:00:16.18900:00:16.199 have been gave a stream of superheated 00:00:19.07000:00:19.080 steam a 2 kilo gram per second 10 bar 00:00:23.99000:00:24.000 and 210 Celsius degree as an energy from 00:00:31.13000:00:31.140 this stream is used in a heat exchanger 00:00:35.18000:00:35.190 to preheat a reactor field so the first 00:00:40.34000:00:40.350 thing we should do is to draw the 00:00:42.41000:00:42.420 diagram here is a heat exchanger hae 00:00:58.69000:00:58.700 here is the steam stream life coming 00:01:02.50000:01:02.510 with mass flow rate is equal to two 00:01:05.98000:01:05.990 kilogram per second pressure is equal to 00:01:11.23000:01:11.240 10 bar temperature is equal to 210 00:01:16.74900:01:16.759 Celsius degree of course F thambi is 00:01:20.68000:01:20.690 unknown in a heat exchanger reactor feed 00:01:26.46900:01:26.479 which come in and extract 450 kilowatt 00:01:36.76000:01:36.770 or 450 kilo joule per second of energy 00:01:45.41900:01:45.429 so because of this extraction we have a 00:01:49.27000:01:49.280 change in the stream stream so the mass 00:01:52.24000:01:52.250 flow rate will still equal to 2 00:01:55.51000:01:55.520 kilograms per second the pressure will 00:02:00.04000:02:00.050 still equal to 10 bar because a pressure 00:02:04.66000:02:04.670 drop across the heat exchanger is 00:02:07.06000:02:07.070 negligible and of course we don't know 00:02:11.22900:02:11.239 the temperature and the ophthalmic of 00:02:14.35000:02:14.360 this equation so the next step is to 00:02:20.47000:02:20.480 find the enthalpy of feed stream so I 00:02:23.56000:02:23.570 have this superheated steam table which 00:02:27.13000:02:27.140 will help us in managing this situation 00:02:30.99000:02:31.000 the first column is pressure with bars 00:02:35.41000:02:35.420 and here we have the temperature from 50 00:02:39.75900:02:39.769 to 350 Celsius degree our example is in 00:02:45.50900:02:45.519 10 part of pressure and the temperature 00:02:51.46000:02:51.470 is 210 which is between 200 and 250 we 00:02:59.25900:02:59.269 need to know the cross between the 00:03:01.12000:03:01.130 pressure and temperature as we go down 00:03:05.88000:03:05.890 to reach 10 bar we will stop here 00:03:14.25900:03:14.269 which is an cross with ten bar of 00:03:18.19900:03:18.209 pressure now I will put the law which is 00:03:25.03900:03:25.049 enthalpy n equal to temperature minus 00:03:32.03000:03:32.040 temperature low / temperature high minus 00:03:39.19900:03:39.209 temperature low multiplied by H high 00:03:50.31900:03:50.329 minus h lo + H low now we will do the 00:04:00.31900:04:00.329 calculation 00:04:30.08000:04:30.090 two hundred and eighty two thousand 00:04:33.08000:04:33.090 eight hundred fifty kilo Joule per 00:04:37.87000:04:37.880 kilogram so now we can do energy balance 00:04:43.58000:04:43.590 on a steam string so that says Q is 00:04:48.43000:04:48.440 equal to minus m dot delta h we can 00:04:55.10000:04:55.110 simplify this by minus m dot H out minus 00:05:03.02000:05:03.030 H n from this equation we know the Q 00:05:10.31000:05:10.320 which is 450 and also we know the flow 00:05:14.39000:05:14.400 or the mass flow rate and we know the M 00:05:18.74000:05:18.750 tell me out sorry the Afghan M this will 00:05:27.26000:05:27.270 be Q would equal to 450 minus 2 enthalpy 00:05:39.62000:05:39.630 out minus enthalpy and which is 2,850 00:05:50.83000:05:50.840 help me out 00:05:53.00000:05:53.010 what equal to 200 2625 kilo Joule per 00:06:01.15000:06:01.160 kilogram now let try to find the 00:06:06.40900:06:06.419 temperature of a steam coming out so the 00:06:09.26000:06:09.270 enthalpy which is two hundred to two 00:06:13.27900:06:13.289 thousand six hundred twenty-five which 00:06:16.04000:06:16.050 is decrease so we go to the left and 00:06:19.65900:06:19.669 this vertical line which represent the 00:06:24.68000:06:24.690 saturated temperature is in between 150 00:06:29.48000:06:29.490 and 200 Celsius degree as we go to the 00:06:33.35000:06:33.360 left to reach 10 bar which is our 00:06:35.99000:06:36.000 pressure the temperature is about 180 00:06:41.45000:06:41.460 Celsius degree 00:06:43.49000:06:43.500 the enthalpy of Tim of water and this 00:06:46.04000:06:46.050 temperature is 760 - and the enthalpy of 00:06:52.00000:06:52.010 steam is two thousand seven hundred 00:06:55.87000:06:55.880 seventy six so we know the enthalpy is 00:07:01.43000:07:01.440 between the pure vapor and pure liquid 00:07:04.55000:07:04.560 and saturation between them so we know 00:07:08.18000:07:08.190 the temperature of a steam coming out is 00:07:10.85000:07:10.860 a saturated steam or saturated liquid is 00:07:15.25000:07:15.260 one hundred seventy nine point nine 00:07:21.76000:07:21.770 which is one hundred seventy nine point 00:07:27.41000:07:27.420 nine and we already mentioned enthalpy 00:07:31.25000:07:31.260 coming out which is two thousand six 00:07:35.81000:07:35.820 hundred twenty five so the last 00:07:40.34000:07:40.350 information is to calculate how much 00:07:43.16000:07:43.170 vapor and how much liquid do we have so 00:07:46.31000:07:46.320 that means just apply the lever rule to 00:07:49.37000:07:49.380 determine how much each do we have so 00:07:53.86000:07:53.870 the operation fraction equal to H minus 00:08:02.80000:08:02.810 H saturation of liquid and its 00:08:08.93000:08:08.940 saturation of divided by its saturation 00:08:12.47000:08:12.480 of paper - its saturation of liquid now 00:08:19.37000:08:19.380 we will do the calculation 2625 minus 00:08:29.05000:08:29.060 seven hundred sixty three divided by two 00:08:34.19000:08:34.200 thousand seven hundred seventy six which 00:08:38.00000:08:38.010 is here - seven hundred sixty three and 00:08:43.24000:08:43.250 this will be equal to 93 percent
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