00:00:28.039 or two problems on condensers then I'll 00:00:30.82900:00:30.839 introduce refrigerant evaporators so the 00:00:33.47000:00:33.480 specific objectives of this particular 00:00:34.70000:00:34.710 lecture hard to present worked out 00:00:38.45000:00:38.460 examples for thermal design of water 00:00:40.16000:00:40.170 cooled and air cooled condensers 00:00:42.10000:00:42.110 introduced have operators present 00:00:45.22900:00:45.239 classification of evaporators describe 00:00:48.07900:00:48.089 salient features of some of the 00:00:49.43000:00:49.440 important types of evaporators at the 00:00:51.95000:00:51.960 end of the lesson you should be able to 00:00:53.61900:00:53.629 carry out thermal design of air cooled 00:00:56.06000:00:56.070 and water cooled condensers using 00:00:57.71000:00:57.720 suitable correlations and formula 00:01:00.34000:01:00.350 classified refrigerant evaporators based 00:01:02.60000:01:02.610 on various criteria describe important 00:01:06.35000:01:06.360 features of some of the important types 00:01:07.70000:01:07.710 of evaporators so let me begin with a 00:01:10.42900:01:10.439 worked out example this is an example on 00:01:12.67900:01:12.689 shell and tube condenser and the problem 00:01:14.89900:01:14.909 statement is like this you have to find 00:01:17.24000:01:17.250 the length of tubes in a to pass 10 ton 00:01:20.06000:01:20.070 capacity shell and tube are 22 base 00:01:23.35900:01:23.369 water cooled condenser with 52 tubes 00:01:26.24000:01:26.250 arranged in 13 columns the heat 00:01:29.27000:01:29.280 rejection ratio is given as 1 point 2 7 00:01:32.00000:01:32.010 4 7 the condensing temperature is 45 00:01:34.61000:01:34.620 degree centigrade water inlet and outlet 00:01:36.83000:01:36.840 temperatures are 30 degree centigrade 00:01:38.36000:01:38.370 and 35 degree centigrade respectively 00:01:40.06000:01:40.070 the tube outer and inner diameters are 00:01:42.64900:01:42.659 14 and 16 mm respectively 00:01:44.93000:01:44.940 so the problem measure must be clear to 00:01:47.75000:01:47.760 you it's an R 22 based condenser water 00:01:50.53900:01:50.549 cooled condenser it has 52 tubes and 00:01:52.67000:01:52.680 thirteen columns 52 tubes are original 00:01:55.21900:01:55.229 13 columns and the other data is give as 00:01:57.46900:01:57.479 per the problem so now let us look at 00:02:00.08000:02:00.090 how to solve the problem so this is the 00:02:03.38000:02:03.390 schematic in fact I have last class have 00:02:05.45000:02:05.460 shown this schematic of it to pass shell 00:02:08.86900:02:08.879 and tube type of a condenser and as I 00:02:10.46000:02:10.470 have explained to you or here 00:02:12.38000:02:12.390 refrigerant is there on the shell side 00:02:14.36000:02:14.370 okay you can see that the salinity is 00:02:16.30900:02:16.319 entering at the top and it condenses as 00:02:18.55900:02:18.569 it comes in contact with the tubes and 00:02:20.36000:02:20.370 the liquid refrigerant goes out from the 00:02:22.03900:02:22.049 bottom and water enters from the bottom 00:02:24.53000:02:24.540 like this coolant is nothing but the 00:02:26.00000:02:26.010 water and it flows through these tubes 00:02:28.17900:02:28.189 since it is a to pass the same water 00:02:30.97900:02:30.989 again will flow through the condenser 00:02:34.61000:02:34.620 tubes okay so it comes here and again it 00:02:36.44000:02:36.450 goes back to the 00:02:37.40000:02:37.410 now convince' tube that's what you call 00:02:38.84000:02:38.850 it as a to pass condenser that means the 00:02:40.85000:02:40.860 water makes two passes as it flows to 00:02:43.55000:02:43.560 the condenser and in the problem it is 00:02:46.22000:02:46.230 mentioned that there are 52 tubes okay 00:02:48.71000:02:48.720 and they are arranged in thirteen 00:02:50.27000:02:50.280 columns now the first thing we have to 00:02:57.80000:02:57.810 do now to find out the properties of our 00:02:59.81000:02:59.820 22 in water because we have to evaluate 00:03:01.49000:03:01.500 the heat transfer coefficients and all 00:03:03.08000:03:03.090 so these are the properties at average 00:03:05.56000:03:05.570 temperature for water viscosity is given 00:03:09.56000:03:09.570 as seven point seven three into ten to 00:03:10.94000:03:10.950 the power of minus four kg per meter 00:03:12.44000:03:12.450 second thermal conductivity is given as 00:03:14.78000:03:14.790 point six one seven watt per meter 00:03:16.22000:03:16.230 Kelvin and density is 995 kg per meter 00:03:18.47000:03:18.480 cube specific heat of water is given as 00:03:20.66000:03:20.670 four point one nine kilo Joule per kg K 00:03:22.46000:03:22.470 and Prandtl number is given as five 00:03:24.41000:03:24.420 point two five PR is Prandtl number and 00:03:26.81000:03:26.820 for r22 the average properties are the 00:03:29.39000:03:29.400 viscosity is one point 8 into 10 to the 00:03:31.10000:03:31.110 power of minus 4 kg per meter second 00:03:33.38000:03:33.390 thermal conductivity KF is point zero 00:03:36.05000:03:36.060 seven seven nine baht per meter Kelvin 00:03:38.33000:03:38.340 density is eleven thousand one 18.9 kg 00:03:41.51000:03:41.520 per meter cube and latent heat of 00:03:43.93000:03:43.940 vaporization is 160 point nine kilo 00:03:46.94000:03:46.950 joule per kg this is at a condensing 00:03:48.71000:03:48.720 temperature and it's also given in the 00:03:50.96000:03:50.970 problem that the fouling resistance on 00:03:52.64000:03:52.650 water side and thermal conductivity of 00:03:54.77000:03:54.780 copper are following resistance is zero 00:03:57.95000:03:57.960 point zero zero zero one seven six meter 00:04:00.80000:04:00.810 squared Kelvin per watt and the thermal 00:04:02.96000:04:02.970 conductivity of water is given as three 00:04:04.37000:04:04.380 90 watt per meter Kelvin so this is how 00:04:07.25000:04:07.260 the problem is stated so now we have to 00:04:10.27000:04:10.280 find out the required length of each 00:04:12.80000:04:12.810 tube okay 00:04:16.87000:04:16.880 so first the step I used to find out 00:04:20.93000:04:20.940 what is the required heat transfer rate 00:04:22.46000:04:22.470 in condenser you know that required heat 00:04:24.74000:04:24.750 transfer rate in condenser is nothing 00:04:26.15000:04:26.160 but QC is equal to product of heat 00:04:28.88000:04:28.890 rejection ratio HR R into refrigeration 00:04:31.46000:04:31.470 capacity and the heat rejection rate in 00:04:33.59000:04:33.600 a ratio is given as one point two seven 00:04:35.39000:04:35.400 four seven and the refrigeration 00:04:37.16000:04:37.170 capacity is given as 10 tonnes so we 00:04:39.29000:04:39.300 convert that into kilowatts so this is 00:04:41.63000:04:41.640 10 into three point five one six seven 00:04:43.25000:04:43.260 kilowatt 00:04:44.00000:04:44.010 so from this equation you find that the 00:04:46.58000:04:46.590 condenser heat transfer rate is forty 00:04:48.86000:04:48.870 four point eight three kilowatt 00:04:51.38000:04:51.390 then we let us find the required mass 00:04:53.82000:04:53.830 flow rate of water so from energy 00:04:55.32000:04:55.330 balance we can also write QC is equal to 00:04:58.26000:04:58.270 M WC pw into TW o minus Twi where as you 00:05:03.45000:05:03.460 know MW is the mass flow rate of water 00:05:05.12000:05:05.130 CP W is the specific heat of water T wo 00:05:08.49000:05:08.500 and Twi are exit and Inlet temperatures 00:05:11.16000:05:11.170 of water and we know here we know the 00:05:14.49000:05:14.500 values of QC we also know the value of 00:05:17.22000:05:17.230 CP we also know the temperature so if 00:05:19.29000:05:19.300 you substitute those values you find 00:05:20.88000:05:20.890 that the total mass flow rate of water 00:05:22.83000:05:22.840 is 2 point 1 4 kg per second now this 2 00:05:26.52000:05:26.530 point 1 4 kg per second is distributed 00:05:28.77000:05:28.780 in many tubes okay since it is a two 00:05:32.37000:05:32.380 pass condenser it's 52 tubes 00:05:34.59000:05:34.600 water flow through each tube is given by 00:05:36.96000:05:36.970 this formula this is M subscript WI is 00:05:40.98000:05:40.990 the water flow through each tube that is 00:05:42.75000:05:42.760 equal to M W by 26 why do we get 26 00:05:46.50000:05:46.510 there are 52 tubes but it is a two pass 00:05:48.48000:05:48.490 to pass connection that means and one in 00:05:52.98000:05:52.990 one pass the water flows through 26 - so 00:05:56.07000:05:56.080 the total flow rate is given by 2 point 00:05:57.90000:05:57.910 1 4 kg per second 00:05:59.13000:05:59.140 so through individual tube the flow rate 00:06:01.77000:06:01.780 is or 2 point 1 4 divided by 26 so that 00:06:05.22000:06:05.230 is zero point zero 8 - 3 kg per second M 00:06:09.50000:06:09.510 now once we know the water flow rate and 00:06:11.88000:06:11.890 once we know the water properties we can 00:06:13.56000:06:13.570 calculate the Reynolds number for water 00:06:15.57000:06:15.580 side so as you know Reynolds number for 00:06:17.70000:06:17.710 water side is Rho VD by mu this can also 00:06:20.10000:06:20.110 be written in terms of the mass flow 00:06:21.54000:06:21.550 rates so in terms of mass flow rate it 00:06:23.79000:06:23.800 is 4 into M WI divided by PI D mu W 00:06:28.02000:06:28.030 where m WI as you have seen is the mass 00:06:30.54000:06:30.550 flow rate through each tube and di is 00:06:32.82000:06:32.830 the internal diameter of the tube and nu 00:06:35.01000:06:35.020 W is the viscosity of water so if you 00:06:37.53000:06:37.540 substitute these values because 00:06:38.67000:06:38.680 everything is known to you if you 00:06:40.11000:06:40.120 substitute these values you find that 00:06:41.49000:06:41.500 the Reynolds number for water side is 4 00:06:43.98000:06:43.990 6 8 8 82.6 since this is greater than 00:06:48.30000:06:48.310 2300 we can curve this is turbulent flow 00:06:51.36000:06:51.370 so we have to use the turbulent flow 00:06:53.19000:06:53.200 correlations for finding the heat 00:06:54.71900:06:54.729 transfer so let us find the heat 00:06:58.38000:06:58.390 transfer coefficient or water set in 00:06:59.96900:06:59.979 fact in last class I have mentioned that 00:07:01.35000:07:01.360 if it is turbulent flow we can use 00:07:03.83000:07:03.840 detestable tore equation or see data 00:07:05.48000:07:05.490 equation so in this problem let us use 00:07:07.70000:07:07.710 detest both of equation let us bolt to 00:07:09.92000:07:09.930 equation as you know is nothing but 00:07:12.02000:07:12.030 nusselt number is equal to point zero 00:07:13.76000:07:13.770 two three in also number to the power of 00:07:15.23000:07:15.240 point a tarantula number to the power of 00:07:17.12000:07:17.130 point four so we know Reynolds number we 00:07:19.43000:07:19.440 also know the Prandtl number so if we 00:07:20.78000:07:20.79000:07:22.61000:07:22.620 the Reynolds are nestled number four 00:07:23.96000:07:23.970 water side is sixty eight point nine six 00:07:26.18000:07:26.190 and so once you know the nestlé's number 00:07:27.95000:07:27.960 you can find out the heat transfer 00:07:29.06000:07:29.070 coefficient because heat transfer 00:07:30.29000:07:30.300 coefficient is nothing but the self 00:07:31.94000:07:31.950 number into thermal conductivity of 00:07:33.71000:07:33.720 water divided by the diameter so if we 00:07:35.72000:07:35.73000:07:37.24000:07:37.250 heat transfer coefficient and water side 00:07:39.56000:07:39.570 is 3039 watt per meter square Kelvin now 00:07:44.81000:07:44.820 let us find the heat transfer 00:07:46.49000:07:46.500 coefficient on the condensation side so 00:07:48.71000:07:48.720 condensation heat transfer coefficient 00:07:50.24000:07:50.250 since condensation is this is a 00:07:52.49000:07:52.500 shell-and-tube type of condenser so 00:07:54.74000:07:54.750 obviously water is flowing through the 00:07:55.96900:07:55.979 tube so condensation is taking place 00:07:57.26000:07:57.270 outside the tubes and this is since 00:08:00.29000:08:00.300 nothing is mentioned way let us take 00:08:01.76000:08:01.770 this as a horizontal shell and tube type 00:08:04.04000:08:04.050 of condenser so let us use the 00:08:05.39000:08:05.400 correlation for horizontal condensation 00:08:08.12000:08:08.130 on horizontal tubes so in fact in the 00:08:10.27900:08:10.289 last class I mentioned that we can use 00:08:11.99000:08:12.000 the classical correlation given by 00:08:14.51000:08:14.520 nestled in fact nestles correlation is a 00:08:16.90900:08:16.919 valid for laminar flow so here but this 00:08:20.69000:08:20.700 is valid in fact you can cross check 00:08:22.27900:08:22.289 that you will find that it is valid here 00:08:24.50000:08:24.510 so let us apply the nusselt correlation 00:08:27.29000:08:27.300 for finding the condensation heat 00:08:28.46000:08:28.470 transfer coefficient so what in a cell's 00:08:30.65000:08:30.660 correlation 00:08:39.46000:08:39.470 so this is the nacelles correlation for 00:08:42.11000:08:42.120 laminate film condensation outside 00:08:43.55000:08:43.560 horizontal tube I have explained this in 00:08:45.35000:08:45.360 the last lecture okay here as you know 00:08:49.13000:08:49.140 this is the thermal conductivity of 00:08:51.19000:08:51.200 saturated refrigerant this is the 00:08:54.23000:08:54.240 density of saturated refrigerant at 00:08:56.15000:08:56.160 condenser temperature and this is 00:08:58.22000:08:58.230 acceleration due to gravity 00:08:59.48000:08:59.490 these are latent heat of vaporization at 00:09:01.46000:09:01.470 that temperature and pressure and needs 00:09:03.86000:09:03.870 the total number of Stoops in a row and 00:09:06.38000:09:06.390 D naught is the outer diameter of the 00:09:09.65000:09:09.660 tube on which condensation is taking 00:09:11.03000:09:11.040 place mu F is the viscosity of the 00:09:14.36000:09:14.370 saturated liquid and delta T is nothing 00:09:16.94000:09:16.950 but the temperature difference between 00:09:18.80000:09:18.810 the refrigerant and the wall okay 00:09:22.58000:09:22.590 surface let us say okay so this is the 00:09:25.61000:09:25.620 nusselt correlation and here H naught is 00:09:29.27000:09:29.280 in watt per meter square Kelvin and all 00:09:31.46000:09:31.470 other units are in Si so you have to use 00:09:33.41000:09:33.420 the SI units this you have to be careful 00:09:35.18000:09:35.190 while using the unit cell so we know 00:09:37.76000:09:37.770 from the problem you know everything 00:09:39.92000:09:39.930 except delta T so let us find H naught 00:09:45.01000:09:45.020 okay so if you substitute the number of 00:09:47.90000:09:47.910 cubes per under a row we have to find 00:09:49.64000:09:49.650 out that is capital n sincerely 00:09:51.65000:09:51.660 mentioned that there are total number of 00:09:53.12000:09:53.130 there are fifty-two tubes and they are 00:09:55.40000:09:55.410 distributed in 13 rows so number of 00:09:57.50000:09:57.510 cubes per row is nothing but 52 by 13 00:09:59.84000:09:59.850 that is 4 so if we substitute this for 00:10:02.51000:10:02.520 and the other property values you will 00:10:03.98000:10:03.990 find that the nusselt correlation from 00:10:06.29000:10:06.300 the nestlé's correlation H naught is 00:10:08.03000:10:08.040 given as 2 175 divided by delta T to the 00:10:11.21000:10:11.220 power of point 2 5 1 00:10:12.71000:10:12.720 okay so here delta T as I have already 00:10:14.78000:10:14.790 mentioned is nothing but the temperature 00:10:16.85000:10:16.860 difference between the size of 00:10:18.47000:10:18.480 condensing refrigerant and the surface 00:10:20.69000:10:20.700 this is not known to us so what we have 00:10:23.33000:10:23.340 to do is we have to use a trial and 00:10:25.01000:10:25.020 error method okay trial and error method 00:10:27.29000:10:27.300 means initially we you use a guess value 00:10:29.21000:10:29.220 of delta T find out H naught find out u 00:10:32.54000:10:32.550 naught and all and finally you have to 00:10:34.64000:10:34.650 cross check whether the guess value is 00:10:36.98000:10:36.990 correct value or not I will show you 00:10:38.96000:10:38.970 that procedure now before that for water 00:10:44.63000:10:44.640 cooled condensers without fins the 00:10:46.85000:10:46.860 overall heat transfer coefficient is 00:10:48.20000:10:48.210 given by this formula this also I have 00:10:49.97000:10:49.980 explained in the last class 00:10:51.95000:10:51.960 as you know this is the convective 00:10:54.18000:10:54.190 resistance of the inside this is the 00:10:56.76000:10:56.770 fouling resistance on water side and 00:10:58.71000:10:58.720 this is the resistance offered by the 00:11:00.36000:11:00.370 wall and this is the external resistance 00:11:03.54000:11:03.550 okay now everything is known to us 00:11:06.33000:11:06.340 because a nought and I a I and all can 00:11:08.70000:11:08.710 be expressed in terms of diameters so 00:11:10.44000:11:10.450 diameters are given to us properties are 00:11:12.51000:11:12.520 also given and the fouling resistance is 00:11:14.88000:11:14.890 also mentioned in the problem statement 00:11:16.02000:11:16.030 so we know everything okay 00:11:18.42000:11:18.430 so if you substitute everything you find 00:11:20.67000:11:20.680 that the overall heat transfer 00:11:21.93000:11:21.940 coefficient is like this 1 by u naught 00:11:25.41000:11:25.420 is equal to 0.05 7/8 1 plus 1 by H 00:11:29.10000:11:29.110 naught H naught is not fully known to us 00:11:31.53000:11:31.540 because we do not know what is the delta 00:11:33.12000:11:33.130 T okay so as I said we have to go for a 00:11:40.20000:11:40.210 trial and error method so first let us 00:11:42.03000:11:42.040 take a guess the initial guess value 5 00:11:44.52000:11:44.530 degree centigrade there so once you take 00:11:46.71000:11:46.720 a guess value of 5 degree centigrade 00:11:48.14000:11:48.150 condensation heat transfer coefficient H 00:11:50.16000:11:50.170 naught is equal to 2 175 divided by 00:11:52.74000:11:52.750 delta T to the power of 0.25 so delta T 00:11:55.29000:11:55.300 is 5 degrees so from this we find that 00:11:57.33000:11:57.340 heat transfer coefficient is fourteen 00:11:59.67000:11:59.680 fifty four point five watt per meter 00:12:01.11000:12:01.120 square Kelvin on the refrigerant side 00:12:03.92000:12:03.930 once you know this you can substitute 00:12:05.91000:12:05.920 this in the expression for overall heat 00:12:07.68000:12:07.690 transfer coefficient and you find that 1 00:12:09.75000:12:09.760 by u naught is equal to point zero zero 00:12:12.15000:12:12.160 one two six five six meter squared 00:12:13.92000:12:13.930 Kelvin per bat that means the overall 00:12:17.67000:12:17.680 heat transfer coefficient u naught is 00:12:18.99000:12:19.000 equal to seven ninety point to write per 00:12:20.76000:12:20.770 meter square Kelvin so we have found the 00:12:23.46000:12:23.470 unit and we know that for the condenser 00:12:26.37000:12:26.380 we can write this equation QC is equal 00:12:28.29000:12:28.300 to u naught a naught into L MTD which is 00:12:31.47000:12:31.480 equal to forty four point eight three 00:12:32.76000:12:32.770 kilowatt QV is forty four point eight 00:12:34.86000:12:34.870 three kilowatt now you know so we have 00:12:37.32000:12:37.330 to power two in order to find out a 00:12:38.70000:12:38.710 naught we have to find L MTD because u 00:12:40.74000:12:40.750 naught is known to us them so LM today 00:12:43.32000:12:43.330 as you know is for a condensation 00:12:45.12000:12:45.130 process LM ttvo can be written like this 00:12:47.64000:12:47.650 T naught T wo minus TW I divided by 00:12:51.15000:12:51.160 natural log of TC minus TW I divided by 00:12:54.36000:12:54.370 TC minus TW o whereas you know TC is the 00:12:57.93000:12:57.940 temperature of the refrigerant T WI and 00:13:00.51000:13:00.520 TW or the inlet and outlet temperatures 00:13:02.97000:13:02.980 of water okay so all these 00:13:04.98000:13:04.990 things are known to us so if you 00:13:06.03000:13:06.040 substitute these value if you find that 00:13:07.44000:13:07.450 the LM TT is twelve point three three 00:13:09.48000:13:09.490 Kelvin okay so if we substitute the 00:13:11.94000:13:11.950 values of L MTD and you note in the 00:13:14.22000:13:14.230 expression for QC then you find that the 00:13:18.00000:13:18.010 area required area that is outer area is 00:13:21.36000:13:21.370 four point six meter square now as I 00:13:25.82900:13:25.839 have already mentioned you should not 00:13:28.11000:13:28.120 stop the this thing here because this 00:13:30.03000:13:30.040 value we got by taking an initial guess 00:13:32.73000:13:32.740 value of five degree Kelvin per for 00:13:36.06000:13:36.070 delta T okay so now we have to see 00:13:37.86000:13:37.870 whether delta T is really five degree 00:13:39.54000:13:39.550 Kelvin or not if it is not five Kelvin 00:13:41.94000:13:41.950 we have to go for the next trial okay so 00:13:46.29000:13:46.300 far let us calculate what is the delta T 00:13:48.62000:13:48.630 now delta T can also be written in this 00:13:51.75000:13:51.760 manner delta T is equal to QC divided by 00:13:54.09000:13:54.100 H naught into a naught because you can 00:13:56.85000:13:56.860 write QC in terms of QC can be written 00:14:05.91000:14:05.920 in terms of outer convective heat 00:14:08.31000:14:08.320 transfer coefficient outer area into 00:14:12.38000:14:12.390 refrigerant temperature minus surface 00:14:15.09000:14:15.100 temperature okay this is nothing but 00:14:16.92000:14:16.930 delta T right so it's not is nothing but 00:14:19.65000:14:19.66000:14:21.44900:14:21.459 which is known to us a not just we have 00:14:23.40000:14:23.410 computed okay 00:14:24.51000:14:24.520 so if you substitute everything you can 00:14:25.82900:14:25.839 find out what is the calculated value of 00:14:28.05000:14:28.060 delta T that's what we are doing now 00:14:29.84000:14:29.850 okay so if you substitute the values for 00:14:32.81900:14:32.829 QC it's not and they are not you find 00:14:34.56000:14:34.570 that delta T calculated is six point 00:14:38.43000:14:38.440 seven Kelvin so you find that the guess 00:14:40.38000:14:40.390 value is we have started the solution by 00:14:43.41000:14:43.420 taking an initial guess value of five 00:14:45.54000:14:45.550 degree Kelvin but when at the end when 00:14:47.28000:14:47.290 you calculate delta T you find that it 00:14:49.29000:14:49.300 is six point seven Kelvin since there is 00:14:50.81900:14:50.829 a difference of two degree two degrees 00:14:52.76900:14:52.779 between the gas value and the calculated 00:14:54.56900:14:54.579 value we have to go for a next trial 00:14:57.09000:14:57.100 then okay so in the next tale what we do 00:14:59.37000:14:59.380 is let us assume a delta T of seven 00:15:01.56000:15:01.570 Kelvin now okay and repeat the problem 00:15:03.15000:15:03.160 repeat the calculations so what we do is 00:15:07.05000:15:07.060 as I said since the calculate value is 00:15:08.69900:15:08.709 not equal to the assumed value we have 00:15:10.17000:15:10.180 to repeat the calculation with delta T 00:15:11.79000:15:11.800 is equal to seven Kelvin there is a 00:15:13.43900:15:13.449 second trial so once you assume delta T 00:15:16.92000:15:16.930 as seven Kelvin again 00:15:18.31900:15:18.329 you have to find out H not because it's 00:15:19.87900:15:19.889 not easy expressed in terms of delta T 00:15:22.10000:15:22.110 so once you find the H naught you find 00:15:24.76900:15:24.779 the overall heat transfer coefficient 00:15:26.11900:15:26.129 once you find the overall heat transfer 00:15:27.53000:15:27.540 coefficient you find out area and from 00:15:30.19900:15:30.209 the area again you find out what is 00:15:31.72900:15:31.739 delta T calculated and you compare delta 00:15:34.16000:15:34.170 T calculated with again the guess value 00:15:36.28900:15:36.299 okay so this process has to be repeated 00:15:37.93900:15:37.949 till you get the converse values right 00:15:41.15000:15:41.160 so from there if you repeat the 00:15:42.59000:15:42.600 calculation with seven Kelvin guess 00:15:44.98900:15:44.999 value you find that the delta T 00:15:46.93900:15:46.949 calculated will be six point nine six 00:15:49.51900:15:49.529 kelvin now okay so this is almost equal 00:15:52.72900:15:52.739 to seven Kelvin so you can stop here but 00:15:55.42900:15:55.439 if you want more accuracy of course you 00:15:57.04900:15:57.059 can again repeat the calculation by 00:15:59.50900:15:59.519 taking a third trial value right 00:16:02.62900:16:02.639 but since the difference is very less we 00:16:05.41900:16:05.429 need not go for third trial okay so 00:16:11.29900:16:11.309 since this value has it sufficiently 00:16:12.67900:16:12.689 close to second guess value of seven K 00:16:14.21000:16:14.220 we may stop here okay so for seven K or 00:16:19.30900:16:19.319 temperature difference we obtain the 00:16:20.72000:16:20.730 value of you not to be 754 watt per 00:16:23.17900:16:23.189 meter square Kelvin so once you know the 00:16:25.63900:16:25.649 value of u naught and LM TT and QC we 00:16:28.15900:16:28.169 find the value of a naught equal to four 00:16:30.16900:16:30.179 point eight two meter square okay 00:16:31.86900:16:31.879 ultimately we have to find the length of 00:16:33.73900:16:33.749 the tube okay so a naught is nothing but 00:16:36.82900:16:36.839 there are 56 tubes okay so my total area 00:16:40.15900:16:40.169 is 56 into PI D into L okay where PI D 00:16:45.07900:16:45.089 dy is the outer diameter of the tube 00:16:46.69900:16:46.709 right there are 56 tubes so we use 56 L 00:16:49.97000:16:49.980 is the length of this thing which is 00:16:51.37900:16:51.389 unknown to us but B naught is the outer 00:16:53.50900:16:53.519 diameter which is known to a 16 mm okay 00:16:55.72900:16:55.739 so if you substitute those values you 00:16:57.47000:16:57.480 find that the required length is one 00:16:59.68900:16:59.699 point seven one three meter okay this is 00:17:02.78000:17:02.790 how you have to do the design of a shell 00:17:07.73000:17:07.740 and tube condenser of course this is not 00:17:09.16900:17:09.179 a complete design this is only a thermal 00:17:11.00000:17:11.010 design of shell and tube condenser okay 00:17:13.34000:17:13.350 so you have to proceed in a systematic 00:17:15.52900:17:15.539 manner first by getting the required 00:17:17.36000:17:17.370 properties right and then using the 00:17:19.96000:17:19.970 correct formula and then calculate the 00:17:22.49000:17:22.500 various quantities and then you have to 00:17:25.75900:17:25.769 go as I said you have to go for or trial 00:17:27.76900:17:27.779 and error method okay and when you are 00:17:30.13900:17:30.149 using the trial and error method you 00:17:31.64000:17:31.650 have to take use intelligent guess 00:17:33.98000:17:33.990 values okay you should not use 00:17:35.24000:17:35.250 unrealistic gas values then it may take 00:17:38.12000:17:38.130 a long time you have to do the calculus 00:17:40.40000:17:40.410 for many try you have to take many 00:17:41.72000:17:41.730 trials okay so use the value judiciously 00:17:44.42000:17:44.430 then it will convert within two three 00:17:46.01000:17:46.020 steps right now let me explain the 00:17:50.24000:17:50.250 design of an air-cooled condenser again 00:17:52.49000:17:52.500 let me read the problem you have to 00:17:55.04000:17:55.050 determine the required phase area of an 00:17:57.38000:17:57.390 are twelve condenser for five ton 00:17:59.66000:17:59.670 refrigeration plant the condensing 00:18:01.91000:18:01.920 temperature is 40 degree centigrade the 00:18:03.98000:18:03.990 system Co P is four point nine and 00:18:05.84000:18:05.850 refrigeration effect is one ten point 00:18:08.39000:18:08.400 eight kilo Joule per kg here at an inlet 00:18:11.30000:18:11.310 temperature of 27 degrees centigrade 00:18:13.58000:18:13.590 flows to the condenser with a phase 00:18:15.20000:18:15.210 velocity of two point five meter per 00:18:17.18000:18:17.190 second the inside and outside diameters 00:18:19.55000:18:19.560 of the tubes are eleven point two six 00:18:21.68000:18:21.690 and twelve point six eight millimeter 00:18:23.51000:18:23.520 respectively fin efficiency is given as 00:18:25.94000:18:25.950 point seven three and the other 00:18:27.47000:18:27.480 dimensions these are the spacings and 00:18:29.30000:18:29.310 thickness of fins okay a facing spacing 00:18:33.20000:18:33.210 between two tubes in a row B is given by 00:18:38.39000:18:38.400 43 mm and C 38 mm D is three point one 00:18:42.77000:18:42.780 seven five mm and the thickness of the 00:18:44.39000:18:44.400 fin is 0.25 four mm okay so what is 00:18:48.44000:18:48.450 given here capacity is given and 00:18:50.39000:18:50.400 condensing temperature is given co P is 00:18:52.22000:18:52.230 given and the refrigeration effect is 00:18:53.57000:18:53.580 given okay and then air inlet 00:18:55.73000:18:55.740 temperature is given outlet temperature 00:18:57.11000:18:57.120 is not given and phase velocity of air 00:18:59.57000:18:59.580 is given as two point five meter per 00:19:00.83000:19:00.840 second and die inner and outer diameter 00:19:02.96000:19:02.970 of the tubes are given and other heat 00:19:05.60000:19:05.610 exchanger dimensions are given okay so 00:19:11.21000:19:11.220 as you know this we have discussed in 00:19:12.71000:19:12.720 the last class I have shown you typical 00:19:14.33000:19:14.340 I said this is the typical fin and tube 00:19:16.55000:19:16.560 type of condenser and as I said this is 00:19:19.52000:19:19.530 the face area okay this is the face area 00:19:22.57000:19:22.580 we have to find out what is the face 00:19:24.62000:19:24.630 area right and we have also defined 00:19:26.77000:19:26.780 certain parameters for example as I was 00:19:29.24000:19:29.250 telling you just now B is nothing but 00:19:32.24000:19:32.250 the center to Center distance between 00:19:34.16000:19:34.170 two consecutive tubes in a row right 00:19:37.16000:19:37.170 this value is given C is this is a side 00:19:40.40000:19:40.410 view okay C is the center to Center 00:19:42.14000:19:42.150 distance between two rows right 00:19:45.16900:19:45.179 so that is C and D is the center to 00:19:48.10900:19:48.119 Center distance between two consecutive 00:19:49.33900:19:49.349 scenes and T as I said is the thickness 00:19:53.20900:19:53.219 of the film right so all these 00:19:55.43000:19:55.440 parameters are given to us so we have to 00:19:57.64900:19:57.659 ultimately find out the phase area and 00:19:59.95900:19:59.969 we also know the phase velocity okay 00:20:03.22900:20:03.239 phase velocity is mentioned as two point 00:20:05.20900:20:05.219 five meter per second M now first thing 00:20:09.73900:20:09.749 we do here is let us find out various 00:20:11.38900:20:11.399 heat transfer areas all these areas have 00:20:13.60900:20:13.619 been defined in the last lecture so you 00:20:15.58900:20:15.599 can refer to the last lecture for these 00:20:17.26900:20:17.279 formulae okay so I am just giving the 00:20:18.76900:20:18.779 formula here now first is the bare tube 00:20:21.49900:20:21.509 area a B a B is given by D minus T by BD 00:20:25.99900:20:26.009 into PI D naught and D is nothing but 00:20:29.11900:20:29.129 fin to fin spacing so that is three 00:20:30.91900:20:30.929 point one seven five millimeter minus 00:20:33.52900:20:33.539 0.25 4 is the thickness of the fin and B 00:20:36.49900:20:36.509 is the spacing the center to Center 00:20:38.26900:20:38.279 spacing between two tubes in a row that 00:20:40.48900:20:40.499 is given as 43 mm and so if you 00:20:43.07000:20:43.080 substitute these values or D naught is 00:20:45.32000:20:45.330 the outer diameter of the tube so if you 00:20:46.87900:20:46.889 substitute all these values you get this 00:20:48.76900:20:48.779 value for a B and remember that this is 00:20:51.73900:20:51.749 defined as meter square per row per 00:20:54.95000:20:54.960 meter square face area this is very 00:20:56.57000:20:56.580 important okay the unit so everything 00:20:58.82000:20:58.830 all the areas we are expressing as meter 00:21:00.79900:21:00.809 square per row per meter square face 00:21:02.50900:21:02.519 area okay then let us calculate the fin 00:21:05.53900:21:05.549 area AF fin area F switch this is the 00:21:07.72900:21:07.739 formula again everything is known to us 00:21:10.66900:21:10.679 D is known C is known D naught is known 00:21:12.52900:21:12.539 B is known so if we substitute 00:21:13.94000:21:13.950 everything you get the fin area to be 00:21:15.73900:21:15.749 twenty-two point zero eight seven meter 00:21:18.07900:21:18.08900:21:20.06000:21:20.070 area okay now let us find out what is 00:21:22.63900:21:22.649 the minimum flow area minimum flow area 00:21:24.85900:21:24.869 is as I said that is nothing but the 00:21:26.74900:21:26.759 area between the two tubes okay the 00:21:29.38900:21:29.399 where the velocity becomes maximum and 00:21:32.93000:21:32.940 the flow area becomes minimum okay so 00:21:35.06000:21:35.070 that is defined as this D minus T by D 00:21:37.19000:21:37.200 into 1 minus D naught by B and again 00:21:39.70900:21:39.719 everything is known so if you substitute 00:21:40.94000:21:40.950 that you find that minimum flow area is 00:21:42.82900:21:42.839 0.6 for 87 meter square per row per 00:21:45.76900:21:45.779 meter square face area then total area a 00:21:51.04900:21:51.059 naught is nothing but a B plus F that is 00:21:53.59900:21:53.609 bare tube area plus fin tube area so 00:21:55.78900:21:55.799 that is found to be twenty-two point 00:21:57.70900:21:57.719 nine four meter square 00:21:58.90900:21:58.919 per meter square so I said yeah okay and 00:22:00.83000:22:00.840 the internal area internal area is AI 00:22:02.93000:22:02.940 and the formula a is 4 ei is PI di by B 00:22:06.34900:22:06.359 and this works out to be point 8 2 2 6 6 00:22:09.56000:22:09.570 meter square per row per meter square 00:22:11.33000:22:11.340 face area okay then we have to find out 00:22:13.36900:22:13.379 the hydraulic diameter because we want 00:22:14.86900:22:14.879 to find out the nusselt number and then 00:22:16.70000:22:16.710 also numbers hydraulic diameter is 00:22:18.68000:22:18.690 defined if you remember as the 4 into 00:22:22.27900:22:22.289 minimum flow area divided by the wetted 00:22:24.10900:22:24.119 perimeter okay and the formula for that 00:22:26.26900:22:26.279 is given like this 00:22:33.15900:22:33.169 okay so this is the formula for minimum 00:22:37.36900:22:37.379 flow area I mean a hydraulic diameter D 00:22:39.68000:22:39.690 H and C is known to us they see the 00:22:41.96000:22:41.970 minimum flow area and a naught is the a 00:22:44.47000:22:44.480 node is also known where these a total 00:22:46.70000:22:46.710 area okay that is 22.9 4 so if you 00:22:49.15900:22:49.169 substitute everything here okay this 00:22:51.13900:22:51.149 should be 22.9 4 I rounded off it okay 00:22:54.32000:22:54.330 it is actually 22.9 393 but I rounded it 00:22:57.01900:22:57.029 off to 22.9 4 so you find that the 00:23:00.10900:23:00.119 hydraulic diameter is given as 4 point 2 00:23:02.18000:23:02.190 9 8 4 into 10 to the power of minus 3 00:23:03.88900:23:03.899 meters and we also need these area 00:23:07.00900:23:07.019 ratios for calculating the overall heat 00:23:09.16900:23:09.179 transfer coefficient okay so this area 00:23:11.21000:23:11.220 ratio a naught by AI is the total area 00:23:13.39900:23:13.409 divided the internal area that works out 00:23:16.39900:23:16.409 to be twenty seven point eight eight 00:23:17.50900:23:17.519 four three and a B by a F that is a beta 00:23:20.29900:23:20.309 beta divided by screened area is point 00:23:22.27900:23:22.289 three seven one five so now we have 00:23:28.78900:23:28.799 found all the required areas now let us 00:23:31.15900:23:31.169 find out the condenser heat rejection 00:23:32.62900:23:32.639 rate so in the last problem for water 00:23:34.75900:23:34.769 cooled condenser the heat rejection 00:23:36.49900:23:36.509 ratio is given directly okay but in this 00:23:38.65900:23:38.669 problem it is not given directly but the 00:23:41.06000:23:41.070 co p value is specified and we know that 00:23:43.31000:23:43.320 heat rejection ratio is nothing but one 00:23:45.32000:23:45.330 plus one by few P okay so we have to 00:23:47.14900:23:47.159 find the heat rejection ratio first and 00:23:48.52900:23:48.539 from the heat rejection ratio we can 00:23:50.26900:23:50.279 find out the heat transfer rate at the 00:23:52.70000:23:52.710 condenser okay so that's what I have 00:23:54.91900:23:54.929 done here the condenser heat rejection 00:23:57.28900:23:57.299 rate QC is equal to H R R into Q II 00:23:59.41900:23:59.429 where Q is the refrigerant capacity that 00:24:01.75900:24:01.769 is five tons and hey char R is one plus 00:24:04.12900:24:04.139 one by COP and co-op is given as 4 point 00:24:06.28900:24:06.299 9 + QE we know 5 10 convert that into 00:24:09.01900:24:09.029 kilowatts by multiplying into three 00:24:11.06000:24:11.070 point five one six seven 00:24:12.41000:24:12.420 if you multiply that and substitute the 00:24:14.36000:24:14.370 values you find that the required heat 00:24:15.83000:24:15.840 rejection date condenser that is QC is 00:24:18.59000:24:18.600 equal to twenty one point one seven 00:24:20.33000:24:20.340 kilowatts in now we have to find out 00:24:22.73000:24:22.740 what is the mass flow rate of 00:24:23.63000:24:23.640 refrigerant to find out the mass flow 00:24:25.67000:24:25.680 rate of refrigerant we know the 00:24:27.38000:24:27.390 refrigeration capacity and also the 00:24:29.33000:24:29.340 refrigeration effect is specified okay 00:24:31.13000:24:31.140 so we know that if you do an energy 00:24:32.93000:24:32.940 balance for the evaporator neglecting 00:24:36.79000:24:36.800 the kinetic and potential energy changes 00:24:39.02000:24:39.030 we know that the refrigeration capacity 00:24:40.96000:24:40.970 is nothing but mass flow rate of 00:24:43.82000:24:43.830 refrigerant into refrigeration effect 00:24:45.80000:24:45.810 okay so refrigeration effect is 00:24:47.45000:24:47.460 specified so substitute the value of 00:24:49.58000:24:49.590 refrigeration effect in the refrigerant 00:24:51.05000:24:51.060 capacity you find that the mass flow 00:24:52.64000:24:52.650 rate of refrigerant is equal to 0.15 869 00:24:55.88000:24:55.890 kg per second okay so now first let us 00:24:59.96000:24:59.970 find out the condensation heat transfer 00:25:01.46000:25:01.470 coefficient for that we need the 00:25:04.04000:25:04.050 properties of r12 so there are 00:25:06.26000:25:06.270 properties of r12 I have not mentioned 00:25:09.08000:25:09.090 here but the properties have to be 00:25:11.06000:25:11.070 evaluated at condensing temperature and 00:25:13.10000:25:13.110 condensing temperature is specified as 00:25:15.41000:25:15.420 forty degree centigrade so you have to 00:25:16.70000:25:16.710 find out the saturated the properties of 00:25:18.98000:25:18.990 r12 liquid and vapour at a forty degree 00:25:22.55000:25:22.560 centigrade okay so I have used those 00:25:24.23000:25:24.240 values and using those values have found 00:25:26.30000:25:26.310 this non dimensional numbers okay first 00:25:29.06000:25:29.070 I have found the Prandtl number of a 00:25:30.56000:25:30.570 refrigerant that is CP a CP mu by K and 00:25:33.41000:25:33.420 this is the Prandtl number for the 00:25:35.15000:25:35.160 saturated liquid 00:25:36.05000:25:36.060 okay F stands for liquid then okay so I 00:25:41.87000:25:41.880 have you have to use all the liquid 00:25:43.64000:25:43.650 values alright so if you use these 00:25:45.47000:25:45.480 values you find that Prandtl number for 00:25:47.30000:25:47.310 the liquid refrigerant is three point 00:25:48.62000:25:48.630 two six four and the Reynolds number for 00:25:50.90000:25:50.910 the gas okay jeez for the vapor or the 00:25:53.24000:25:53.250 gas okay that is again can be written in 00:25:56.00000:25:56.010 terms of the mass flow rates 4 into m 00:25:57.92000:25:57.930 dot divided by PI D mu Z and this 00:26:00.56000:26:00.570 Reynolds number is calculated when all 00:26:03.17000:26:03.180 the mass is in vapor form okay so you 00:26:06.02000:26:06.030 have to use the total mass flow rate 00:26:07.40000:26:07.410 here right so this Reynolds number is 00:26:09.74000:26:09.750 the Reynolds number when all the 00:26:11.51000:26:11.520 refrigerant is in vapor form similarly 00:26:13.88000:26:13.890 Reynolds number of the fluid is 00:26:15.92000:26:15.930 calculated assuming that all the 00:26:17.75000:26:17.760 refrigerant is in liquid form okay that 00:26:20.15000:26:20.160 is our years from so if you substitute 00:26:22.01000:26:22.020 the values you find that these the 00:26:23.27000:26:23.280 Reynolds number of the refrigerant vapor 00:26:24.97000:26:24.980 similarly 00:26:26.03000:26:26.040 Reynolds number of the refrigerant 00:26:27.14000:26:27.150 liquid is this okay so we got the 00:26:33.68000:26:33.690 reynolds number and prandtl number then 00:26:37.67000:26:37.680 to find the condenser heat transfer 00:26:39.50000:26:39.510 coefficient inside tubes we use diene 00:26:41.69000:26:41.700 occurs and crosses correlation which 00:26:43.73000:26:43.740 assumes complete condensation in fact 00:26:45.89000:26:45.900 this is the new correlation in the last 00:26:47.15000:26:47.160 lecture I have shown two other 00:26:49.40000:26:49.410 correlation chart on Charles correlation 00:26:51.62000:26:51.630 and other correlation okay but here I am 00:26:54.02000:26:54.030 using a different correlation okay this 00:26:56.45000:26:56.460 is because I want to present as many 00:26:57.86000:26:57.870 Cordova correlations as possible right 00:26:59.54000:26:59.550 so let us calculate the condenser heat 00:27:01.58000:27:01.590 transfer coefficient using there are 00:27:03.77000:27:03.780 this D in a courser correlation and this 00:27:06.50000:27:06.510 correlation is valid for under the 00:27:08.87000:27:08.880 exemption that the condensation is 00:27:10.58000:27:10.590 complete okay so this correlation 00:27:13.73000:27:13.740 defines a modified Reynolds number REM 00:27:16.13000:27:16.140 and this modified Reynolds number is 00:27:17.96000:27:17.970 defined like this okay this correlation 00:27:21.53000:27:21.540 if you notice correlates nusselt number 00:27:24.29000:27:24.300 in terms of reynolds number and prandtl 00:27:26.21000:27:26.220 number so this is the basic correlation 00:27:29.33000:27:29.340 diene occurs and crossers correlation so 00:27:31.16000:27:31.170 you can see that this looks like a 00:27:32.42000:27:32.430 deters Volturi question okay you have a 00:27:34.19000:27:34.200 constant here re to the power of point 8 00:27:36.08000:27:36.090 and PR to the power of 1 by 3 okay only 00:27:38.78000:27:38.790 difference is the Reynolds number that I 00:27:40.46000:27:40.470 am using here is a modified Reynolds 00:27:42.80000:27:42.810 number okay either Reynolds number of 00:27:45.35000:27:45.360 the vapor nor the Reynolds number of gas 00:27:47.30000:27:47.310 it is a modified Reynolds number and 00:27:49.19000:27:49.200 modified Reynolds number formula is like 00:27:51.08000:27:51.090 this it is written in terms of the 00:27:52.88000:27:52.89000:27:54.47000:27:54.480 liquid and the density of the saturated 00:27:58.10000:27:58.110 liquid and density of the saturated 00:27:59.51000:27:59.520 vapor Rho a friend Rosie 00:28:00.80000:28:00.810 so densities can be written in terms of 00:28:02.39000:28:02.400 specific volumes and we know the 00:28:04.19000:28:04.200 specific volumes from the property data 00:28:06.05000:28:06.060 that is at 40 degree centigrade and re F 00:28:08.33000:28:08.340 also we have found so you can find out 00:28:10.22000:28:10.230 what is the modified Reynolds number and 00:28:12.53000:28:12.540 Prandtl number is also known to us so if 00:28:14.27000:28:14.280 you substitute all these values we can 00:28:16.19000:28:16.200 find out what is the nusselt number so 00:28:22.49000:28:22.500 substituting various property values and 00:28:23.99000:28:24.000 Reynolds number we find that our 00:28:27.26000:28:27.270 Reynolds number modified Reynolds number 00:28:28.97000:28:28.980 is 431 383 and nusselt number is 12 00:28:34.40000:28:34.410 sixty five point nine and condenser heat 00:28:37.40000:28:37.410 transfer coefficient H I 00:28:39.29000:28:39.300 is the 8200 6.7 watt per meter squared 00:28:42.59000:28:42.600 Kelvin um okay for that so we have found 00:28:44.51000:28:44.520 the kind of a teeny transfer coefficient 00:28:47.06000:28:47.070 now let us find the eight side heat 00:28:49.19000:28:49.200 transfer coefficient because for 00:28:50.29900:28:50.309 calculating the overall heat transfer 00:28:51.35000:28:51.360 coefficient you require heat transfer 00:28:53.72000:28:53.730 coefficient on the refrigerant side heat 00:28:55.16000:28:55.170 transfer coefficient on the a side okay 00:28:57.83000:28:57.840 so let us find out the heat transfer 00:28:59.36000:28:59.370 coefficient on the eight side remember 00:29:00.86000:29:00.870 that there is a fiend plate fin type of 00:29:02.69000:29:02.700 a condenser so you have to use suitable 00:29:05.48000:29:05.490 correlations for plate fin type of heat 00:29:07.73000:29:07.740 exchanger okay in fact if you remember 00:29:09.71000:29:09.720 last time I was mentioning that case and 00:29:12.14000:29:12.150 London have given several correlations 00:29:14.48000:29:14.490 for different types of condensers 00:29:16.79000:29:16.800 so we will be using the general 00:29:17.96000:29:17.970 correlation suggested by case in London 00:29:19.91000:29:19.920 okay so in fact if you remember the case 00:29:24.95000:29:24.960 in London correlation is the Reynolds 00:29:28.16000:29:28.170 number is defined in terms of the 00:29:29.45000:29:29.460 maximum velocity and maximum velocity 00:29:32.24000:29:32.250 takes place where the flow area is 00:29:33.77000:29:33.780 minimum okay for maximum velocity is 00:29:35.81000:29:35.820 nothing but the phase area if you can 00:29:37.58000:29:37.590 show that it is nothing but the phase 00:29:38.90000:29:38.910 area divided by I am sorry the phase 00:29:41.84000:29:41.850 velocity divided by minimum flow area 00:29:44.26000:29:44.270 this comes from your continuity equation 00:29:47.18000:29:47.190 okay how did you get this let us say 00:29:51.16900:29:51.179 that you have two tubes here okay and 00:29:54.79900:29:54.809 this is the minimum flow area AC okay 00:29:59.57000:29:59.580 and at this point of the velocity is U 00:30:02.93000:30:02.940 max right and let us say this is the at 00:30:07.07000:30:07.080 the face at the face you have the face 00:30:09.89000:30:09.900 area okay or Radia between two tubes and 00:30:15.98000:30:15.990 you also know the velocity phase 00:30:18.08000:30:18.090 velocity okay and from mass balance we 00:30:20.36000:30:20.370 know that au is constant 00:30:23.62000:30:23.630 okay so u max is you'll find that phase 00:30:28.01000:30:28.020 velocity divided by AC because we are 00:30:31.70000:30:31.710 writing AC in terms of per meter square 00:30:34.93000:30:34.940 face area okay so that is why a the face 00:30:37.31000:30:37.320 area term doesn't come here because we 00:30:39.16900:30:39.179 are doing all the calculation per meter 00:30:41.15000:30:41.160 square face area okay so you find that 00:30:46.29900:30:46.309 using that we find that the maximum 00:30:51.26000:30:51.270 velocity is three point eight five for 00:30:52.97000:30:52.980 me 00:30:53.18000:30:53.190 per second so the faithful are the 00:30:54.56000:30:54.570 velocity is two point five meter per 00:30:56.00000:30:56.010 second but by the time the air comes 00:30:58.19000:30:58.200 between the two tubes it gave its area 00:31:00.95000:31:00.960 of cross-section it gets reduced so it's 00:31:02.66000:31:02.670 velocity increases to three point eight 00:31:04.25000:31:04.260 five one 00:31:05.15000:31:05.160 okay then we find the Reynolds number 00:31:10.64000:31:10.650 Reynolds number is au max into hydraulic 00:31:13.22000:31:13.230 diameter divided by nu so whatever we 00:31:15.29000:31:15.300 have calculated the hydraulic diameter 00:31:16.49000:31:16.500 and all will be used now okay so we know 00:31:19.37000:31:19.380 the hydraulic diameter value we have 00:31:22.37000:31:22.380 computed this this is the kinematic 00:31:24.05000:31:24.060 viscosity and u max is the maximum 00:31:26.71000:31:26.720 velocity that a 3.85 four meter per 00:31:29.15000:31:29.160 second so we substitute that you find 00:31:30.53000:31:30.540 that Reynolds number is nine eighty 00:31:32.39000:31:32.400 three point six then we you as I said we 00:31:37.61000:31:37.620 use the general correlation suggested by 00:31:39.59000:31:39.600 a case in London which is given as 00:31:41.75000:31:41.760 nusselt number is H naught D H by K that 00:31:44.42000:31:44.430 is equal to 0.1 170 knots number to the 00:31:46.55000:31:46.560 power of 0.65 Prandtl number to the 00:31:48.14000:31:48.150 power of 1 by 3 so we know Reynolds 00:31:50.54000:31:50.550 number we know Prandtl number so we can 00:31:52.22000:31:52.230 substitute this so we find that the 00:31:54.14000:31:54.150 nusselt number is seven point eight 00:31:55.79000:31:55.800 three five one and heat transfer 00:31:58.37000:31:58.380 coefficient H naught is equal to 51 00:32:00.23000:32:00.240 point seven seven watt per meter square 00:32:02.36000:32:02.370 Kelvin now overall heat transfer 00:32:06.26000:32:06.270 coefficient for plate fin type of heat 00:32:09.92000:32:09.930 exchanger the formula is like this this 00:32:13.61000:32:13.620 formula I mentioned in the last class 00:32:14.81000:32:14.820 also so here you have the fouling 00:32:17.63000:32:17.640 resistance on the refrigerant side you 00:32:21.14000:32:21.150 don't have any fouling resistance on the 00:32:22.64000:32:22.650 a side that means outside following 00:32:24.44000:32:24.450 resistance is not there okay and you 00:32:29.72000:32:29.730 know various for this thing a naught by 00:32:31.28000:32:31.290 I and these are the area ratios are IRI 00:32:34.19000:32:34.200 and are not are the inner and outer 00:32:36.35000:32:36.360 radius of the tubes and etf's be the fin 00:32:39.62000:32:39.630 efficiency which is known to us H naught 00:32:41.75000:32:41.760 is the external heat transfer 00:32:42.83000:32:42.840 coefficient H ie the internal heat 00:32:44.63000:32:44.640 transfer coefficient okay so if you all 00:32:47.99000:32:48.000 these things are known to us we have 00:32:49.28000:32:49.290 already computed all these things so if 00:32:51.02000:32:51.03000:32:52.25000:32:52.260 that overall heat transfer coefficient 00:32:54.05000:32:54.060 is thirty one point two two nine watt 00:32:55.91000:32:55.920 per meter squared Kelvin okay and now we 00:32:59.18000:32:59.190 have to find out the LM TD because we 00:33:00.77000:33:00.780 would like to we have to find out the 00:33:01.94000:33:01.950 area so for that we have to find the LM 00:33:03.71000:33:03.720 TD okay again we have to do some trial 00:33:06.11000:33:06.120 and error 00:33:07.01000:33:07.020 method here because we do not know what 00:33:09.71000:33:09.720 is the outlet temperature of air neither 00:33:11.72000:33:11.730 we know the outlet temperature of Wales 00:33:13.27900:33:13.289 nor we know the mass flow rate of a room 00:33:15.50000:33:15.510 okay now since we do not know these two 00:33:17.81000:33:17.820 parameters 00:33:18.52900:33:18.539 we cannot calculate the LM TD directly 00:33:21.59000:33:21.600 so what we are to do is we have to 00:33:23.41900:33:23.429 assume either mass flow rate or the 00:33:25.43000:33:25.440 outlet temperature so it is easier to 00:33:27.82000:33:27.830 okay so now it is all the same but let 00:33:30.62000:33:30.630 us assume the outlet temperature of 8 00:33:32.65900:33:32.669 okay the outlet temperature of the a 00:33:35.57000:33:35.580 load I assume in such a way that it 00:33:36.86000:33:36.870 should not be greater than the 00:33:37.82000:33:37.830 condensing temperature obviously because 00:33:39.38000:33:39.390 heat transfer has to take place from 00:33:40.61000:33:40.620 refrigerant to a okay so in this problem 00:33:43.85000:33:43.860 let us assume a value of 35 degrees for 00:33:46.25000:33:46.260 the outlet temperature of a so once you 00:33:50.18000:33:50.190 assume the outlet temperature of the air 00:33:51.52900:33:51.539 inlet temperature is given as 27 degrees 00:33:53.38900:33:53.399 condensing temperature is given as 40 so 00:33:55.46000:33:55.470 we can calculate what is the LM TT so LM 00:33:58.15900:33:58.169 TD is eight point three seven two five 00:33:59.60000:33:59.610 degree centigrade okay 00:34:01.65900:34:01.669 so once you know the LM TD we can find 00:34:04.76000:34:04.770 out what is the total external area okay 00:34:07.70000:34:07.710 a total left how do you know this 00:34:13.60900:34:13.619 because QC is nothing but U naught into 00:34:15.56000:34:15.570 total external area multiplied by L MTD 00:34:18.02000:34:18.030 so LM TD is computed eight point three 00:34:21.05000:34:21.060 seven two five and QC is a twenty one 00:34:23.81000:34:23.820 point one seven and u naught is thirty 00:34:26.38900:34:26.399 one point two two nine okay I'm 00:34:27.88900:34:27.899 multiplying here because this twenty one 00:34:29.54000:34:29.550 point one seven is kilowatt so I am 00:34:31.07000:34:31.080 converting everything into vats because 00:34:32.99000:34:33.000 unit is in watt per meter square Kelvin 00:34:34.82000:34:34.830 now so if you do that you find that the 00:34:36.91900:34:36.929 total external area is eighty point nine 00:34:39.26000:34:39.270 seven six meter square okay so this is 00:34:42.16900:34:42.179 the total area and remember that this is 00:34:44.32900:34:44.339 a diff can have many rows right the 00:34:47.00000:34:47.010 total area means total area of the 00:34:48.98000:34:48.990 condenser that means area of all the 00:34:50.54000:34:50.550 rows right and so far nothing has been 00:34:52.84900:34:52.859 mentioned about the rows right and then 00:34:55.25000:34:55.260 the last class have mentioned that the 00:34:56.65900:34:56.669 number of rows can vary anywhere between 00:34:58.07000:34:58.080 two to eight since this is not a very 00:35:00.65000:35:00.660 large capacity system it is a five ton 00:35:02.35900:35:02.369 capacity system let us take the number 00:35:05.18000:35:05.190 of rows to be four okay so I am assuming 00:35:10.25000:35:10.260 the number of rays to be four once you 00:35:12.07900:35:12.089 assume the number of rows to be four 00:35:13.99000:35:14.000 total area is nothing but a face area it 00:35:17.78000:35:17.790 a number of rows into a naught Y is 00:35:20.96000:35:20.970 why are we writing like this because a 00:35:22.55000:35:22.560 naught is nothing but the total area per 00:35:24.89000:35:24.900 meter square face area per number of 00:35:27.56000:35:27.570 rows okay so that is how you get the on 00:35:29.42000:35:29.430 this kind of formula and a naught is 00:35:31.28000:35:31.290 known to us number of rows are assumed 00:35:33.38000:35:33.390 to be four and a safe area is not known 00:35:36.62000:35:36.630 for but eight total is just now we have 00:35:39.38000:35:39.390 calculated 80 point nine seven six so we 00:35:41.51000:35:41.520 substitute all these values you find 00:35:42.56000:35:42.570 that face area is 0.882 meter square but 00:35:46.52000:35:46.530 this is may not be the correct face area 00:35:48.32000:35:48.330 you have to check because you have 00:35:49.46000:35:49.470 assumed that the outlet temperature to 00:35:50.87000:35:50.880 be thirty five degree centigrade okay 00:35:52.40000:35:52.410 whether that is right under wrong you 00:35:54.23000:35:54.240 have to check now how do we check that 00:35:55.88000:35:55.890 we calculate the mass flow rate of air 00:35:58.01000:35:58.020 mass rate of the aid is nothing but if 00:36:00.80000:36:00.810 you apply the continuity equation is 00:36:02.54000:36:02.550 nothing but Rho into a phase into V 00:36:05.12000:36:05.130 where V the phase velocity that is two 00:36:07.31000:36:07.320 point five meter per second and a phase 00:36:09.47000:36:09.480 phase computed to be 0.882 four meter 00:36:11.66000:36:11.670 square and Rho is the density of air 00:36:13.76000:36:13.770 which is one point one seven seven four 00:36:15.47000:36:15.480 kg per meter cube these are mean density 00:36:17.66000:36:17.670 right so if you substitute these values 00:36:19.46000:36:19.470 you find that this is the mass flow rate 00:36:20.81000:36:20.820 of a right once you know the mass flow 00:36:23.51000:36:23.520 rate of air you can easily calculate 00:36:24.74000:36:24.750 what is the outlet temperature how can 00:36:27.32000:36:27.330 you calculate that we know that for the 00:36:30.17000:36:30.180 condenser you can also write QC as mass 00:36:34.16000:36:34.170 flow rate of air into CP into delta T so 00:36:37.60900:36:37.619 delta T is nothing but QC divided by 00:36:39.50000:36:39.510 mass flow rate of air into CP CP I have 00:36:42.05000:36:42.060 taken here as one point zero zero five 00:36:45.13000:36:45.140 okay so now and the mass flow rate of a 00:36:50.18000:36:50.190 we are computed and this is a Q seen 00:36:51.68000:36:51.690 right here you need not kind of multiply 00:36:53.87000:36:53.880 in 2000 because this is in kilo Joule 00:36:55.46000:36:55.470 per kg Kelvin so you find that delta T 00:36:57.95000:36:57.960 we have obtained we have obtained it to 00:36:59.54000:36:59.550 be eight point one one degree centigrade 00:37:01.13000:37:01.140 right now what is the delta T is eight 00:37:03.85900:37:03.869 point one one degree centigrade means 00:37:05.15000:37:05.160 what is the outlet temperature outlet 00:37:06.80000:37:06.810 temperature is nothing but the outlet 00:37:09.05000:37:09.060 for a of error is T Inlet of a plus 00:37:13.84900:37:13.859 delta T okay T Inlet is given as twenty 00:37:16.43000:37:16.440 seven plus eight point one one this is 00:37:19.28000:37:19.290 thirty five point one one degree 00:37:21.34900:37:21.359 centigrade so you find that 00:37:23.29000:37:23.300 coincidentally of the water you have 00:37:25.46000:37:25.470 calculated is very close to whatever you 00:37:27.62000:37:27.630 have guessed is the guessed value and 00:37:29.66000:37:29.670 calculated values are coming very close 00:37:31.49000:37:31.500 okay so if this accuracy is sufficient 00:37:34.37000:37:34.380 you can 00:37:34.78900:37:34.799 top the trial and error procedure at 00:37:36.79900:37:36.809 this point and you can take this face 00:37:38.26900:37:38.279 area is the required face area but if 00:37:40.78900:37:40.799 you want very high precision then you 00:37:43.06900:37:43.079 can go for a second trial by taking the 00:37:45.79900:37:45.809 outlet temperature to be that is a 00:37:47.47900:37:47.489 thirty five point one okay so you can 00:37:49.40000:37:49.410 continue this but sometimes if you get 00:37:51.64000:37:51.650 accuracies of this kind of accuracy 00:37:53.90000:37:53.910 there is no point in really going for 00:37:55.78900:37:55.799 more and more number of trials because 00:37:57.79900:37:57.809 remember that there is always an element 00:38:01.45900:38:01.469 of uncertainty in the calculation of 00:38:03.82900:38:03.839 heat transfer coefficients okay 00:38:05.83900:38:05.849 the themselves may give uncertainties 00:38:08.71900:38:08.729 could be as high as about plus or minus 00:38:10.54900:38:10.559 twenty five percent okay so there is no 00:38:12.10900:38:12.119 point in really looking for very close 00:38:14.16900:38:14.179 accuracies in areas right so that is 00:38:20.59900:38:20.609 what I have mentioned here so tp8 out as 00:38:22.75900:38:22.769 calculate is thirty five point one one 00:38:24.19900:38:24.209 degree centigrade 00:38:24.79900:38:24.809 since the guess value is close to the 00:38:26.29900:38:26.309 calculated value we may stop here for 00:38:28.72900:38:28.739 better accuracy calculations have to be 00:38:30.22900:38:30.239 repeated with second guess value of 00:38:31.60900:38:31.619 thirty four point one degree centigrade 00:38:32.87000:38:32.880 okay and there is one thing you must 00:38:35.15000:38:35.160 keep in mind the values obtained will be 00:38:37.30900:38:37.319 slightly different if other correlations 00:38:38.92900:38:38.939 are used for H I as I said a large 00:38:41.71900:38:41.729 number of correlations are available for 00:38:43.54900:38:43.559 estimating the heat transfer 00:38:45.34900:38:45.359 coefficients for example on the 00:38:46.64000:38:46.650 condenser side or on the a side had used 00:38:49.45900:38:49.469 some other heat transfer correlations 00:38:51.16900:38:51.179 you might have got slightly different 00:38:53.35900:38:53.369 results okay that's what I mean by 00:38:55.81900:38:55.829 saying there is a lot of uncertainty 00:38:57.04900:38:57.059 right now if you want you can check by 00:38:59.90000:38:59.910 using other correlations and see what 00:39:01.99900:39:02.009 value you are getting them okay so this 00:39:05.39000:39:05.400 is the procedure for estimating for 00:39:07.40000:39:07.410 design thermal design of condensers okay 00:39:10.42900:39:10.439 you I should proceed in a systematic 00:39:12.34900:39:12.359 manner the problem is very simple okay 00:39:14.53900:39:14.549 so at this point I stop my lecture on 00:39:17.23900:39:17.249 condensers and let us go to the next 00:39:19.78900:39:19.799 important component that is evaporator 00:39:21.67900:39:21.689 so I will give a brief introduction and 00:39:23.29900:39:23.309 classification of evaporators in this 00:39:25.33900:39:25.349 lecture okay 00:39:28.96900:39:28.979 so as you know evaporator like condenser 00:39:31.54900:39:31.559 is also a heat exchanger in an 00:39:34.03900:39:34.049 evaporator the refrigerant boils or 00:39:35.66000:39:35.670 evaporates and in doing so absorbs heat 00:39:37.99900:39:38.009 from the substance being refrigerated 00:39:41.22900:39:41.239 the name evaporator refers to the 00:39:43.51900:39:43.529 evaporation process occurring in the 00:39:45.01900:39:45.029 heat exchanger 00:39:47.06000:39:47.070 now let us look at the classification of 00:39:48.80000:39:48.810 evaporators in fact you can classify 00:39:50.30000:39:50.310 operators in many ways there are several 00:39:53.63000:39:53.640 ways of classifying the operators 00:39:55.13000:39:55.140 depending upon the heat transfer process 00:39:56.84000:39:56.850 or depending upon the refrigerant flow 00:39:58.73000:39:58.740 type or depending upon the condition of 00:40:00.65000:40:00.660 free transfer surface etc ok for example 00:40:03.77000:40:03.780 you can classify them as either force 00:40:06.86000:40:06.870 convection type or natural convection 00:40:08.63000:40:08.640 type now as you know involve what is a 00:40:11.75000:40:11.760 force convection type evaporator in 00:40:13.28000:40:13.290 force convection type evaporator just 00:40:15.20000:40:15.210 like force convection type condenser a 00:40:17.00000:40:17.010 fan or pump is used to circulate the 00:40:19.31000:40:19.320 external fluid X name so it could be 00:40:20.90000:40:20.910 water or air or any other media okay so 00:40:24.92000:40:24.930 you need a pump or fan for circulating 00:40:27.35000:40:27.360 this external fluid okay and make the 00:40:31.61000:40:31.620 make it flow or the heat transfer 00:40:32.93000:40:32.940 surface which is cooled by evaporation 00:40:34.82000:40:34.830 of refrigerant this is the force 00:40:36.17000:40:36.180 convection type evaporate ever natural 00:40:38.51000:40:38.520 convection type as you know a natural 00:40:40.49000:40:40.500 convection type we don't use either a 00:40:41.84000:40:41.850 fan or pond and the flow circulation 00:40:44.42000:40:44.430 takes place because of by in sea effects 00:40:46.52000:40:46.530 the baiioons effects are induced due to 00:40:48.29000:40:48.300 density differences which are caused by 00:40:50.45000:40:50.460 temperature difference okay this is 00:40:52.61000:40:52.620 nothing but the natural convection type 00:40:53.84000:40:53.850 evaporator just like natural convection 00:40:55.43000:40:55.440 type condenser okay in natural 00:40:58.01000:40:58.020 convection type evaporator refrigerant 00:41:00.41000:41:00.420 always boils inside tubes and evaporator 00:41:02.45000:41:02.460 is located at the top you have to locate 00:41:04.91000:41:04.920 the evaporator at the top because you 00:41:06.65000:41:06.660 are relying on the natural convection 00:41:08.06000:41:08.070 natural convection means what happens is 00:41:10.73000:41:10.740 when you are keeping at the top a vom 00:41:13.01000:41:13.020 air comes in contact with the evaporator 00:41:14.33000:41:14.340 it becomes cold once it becomes cold its 00:41:17.39000:41:17.400 density increases once it density 00:41:19.25000:41:19.260 increases because of the buoyancy effect 00:41:21.05000:41:21.060 it tries to settle down when it settles 00:41:24.14000:41:24.150 down or the vame from the bottom Rises 00:41:26.84000:41:26.850 hub and wom air goes to the evaporator 00:41:30.40000:41:30.410 it gets cooled and again it comes down 00:41:33.02000:41:33.030 so this cycle is repeated okay so to 00:41:35.36000:41:35.370 continue maintain this cycle you have to 00:41:37.04000:41:37.050 keep the evaporator at a height the 00:41:41.36000:41:41.370 temperature of fluid which is cooled by 00:41:42.80000:41:42.810 by decreases and I am just explaining 00:41:44.93000:41:44.940 what is the mechanism temperature of 00:41:46.52000:41:46.530 fluid which is cooled by it decreasing 00:41:48.32000:41:48.330 decreases and it's density decreases as 00:41:51.71000:41:51.720 a result the fluid moves downwards due 00:41:53.93000:41:53.940 to buoyancy and the bomb fluid rises up 00:41:55.61000:41:55.620 to replace it term okay you can also 00:41:58.64000:41:58.650 classify the evaporators based on the 00:42:00.59000:42:00.600 president flow whether it is taking 00:42:02.12000:42:02.130 place inside the tubes are outside tubes 00:42:05.20000:42:05.210 the heat transfer the this is very 00:42:07.49000:42:07.500 important because the heat transfer 00:42:08.90000:42:08.910 phenomena okay just like condensation is 00:42:11.45000:42:11.460 entirely different if it operational 00:42:14.39000:42:14.400 boiling is taking place inside the tubes 00:42:16.19000:42:16.200 or if it is staying outside the tubes 00:42:18.47000:42:18.480 okay the phenomena is different the 00:42:19.88000:42:19.890 correlations will be different and the 00:42:21.26000:42:21.270 values of heat times the coefficients 00:42:22.73000:42:22.740 also will be different 00:42:23.66000:42:23.670 so it's very very important to keep this 00:42:25.46000:42:25.470 in mind and use the suitable 00:42:27.11000:42:27.120 correlations okay so this is another way 00:42:29.66000:42:29.670 of for classifying the third way of 00:42:32.12000:42:32.130 classifying is by classifying them 00:42:35.21000:42:35.220 either as slider type of evaporators or 00:42:38.27000:42:38.280 dry expansion type of evaporators in 00:42:41.42000:42:41.430 flooded type evaporators liquid 00:42:44.12000:42:44.130 refrigerant covers the entire heat 00:42:45.74000:42:45.750 transfer surface okay this is known as a 00:42:47.72000:42:47.730 flooded type evaporator and the 00:42:49.01000:42:49.020 refrigerant leaves evaporated as liquid 00:42:51.74000:42:51.750 vapor mixture okay and what is the dry 00:42:55.07000:42:55.080 expansion type in a dry expansion type 00:42:57.07000:42:57.080 the refrigerant leaves the evaporator in 00:42:59.75000:42:59.760 vapour form and not the entire heat 00:43:02.24000:43:02.250 transistor is a surface area is covered 00:43:03.92000:43:03.930 with liquid okay there are some area is 00:43:06.08000:43:06.090 covered with a vapor okay so this is 00:43:08.06000:43:08.070 known as dry type now let let me explain 00:43:12.92000:43:12.930 the salient features of some of the 00:43:14.99000:43:15.000 important types of evaporators let me 00:43:17.84000:43:17.850 begin with natural convection type of 00:43:19.85000:43:19.860 evaporator okay natural convection type 00:43:22.61000:43:22.620 of evaporators are mainly used in 00:43:24.92000:43:24.930 domestic refrigerators and cold storages 00:43:28.24000:43:28.250 since flow is Banshee driven evaporator 00:43:30.89000:43:30.900 has to be kept at the top you might have 00:43:35.03000:43:35.040 a for example let me everybody must have 00:43:36.86000:43:36.870 seen it in old type refrigerators not 00:43:41.09000:43:41.100 Frost fitted in conventional type 00:43:42.77000:43:42.780 refrigerators the evaporator is kept at 00:43:45.05000:43:45.060 the top okay 00:43:48.26000:43:48.270 in fact this is nothing but evaporator 00:43:51.26000:43:51.270 comes freezer box okay operator and it 00:43:56.30000:43:56.310 also acts as a freezer compartment or 00:43:58.58000:43:58.590 freezer box 00:43:59.36000:43:59.370 okay so evaporated tubes are kept here 00:44:03.14000:44:03.150 it can be a roll bond type or cube and 00:44:05.03000:44:05.040 plate type and you store the food 00:44:08.54000:44:08.550 products everywhere the frozen food is 00:44:10.22000:44:10.230 kept here and other food foodstuffs are 00:44:12.17000:44:12.180 kept here vegetables fruits etcetera 00:44:14.48000:44:14.490 okay so the gravitation is in this 00:44:16.52000:44:16.530 direction this is the top so since you 00:44:18.92000:44:18.930 are keeping the Opera at the top of the 00:44:21.10900:44:21.119 air close to the evaporator becomes cold 00:44:23.42000:44:23.430 and since its density increases there it 00:44:26.00000:44:26.010 will come down okay so if you see from 00:44:27.77000:44:27.780 the side the operator will be something 00:44:32.00000:44:32.010 like this okay so cold air comes down 00:44:34.45000:44:34.460 right and as it comes down it comes in 00:44:37.07000:44:37.080 contact with the food products kept at 00:44:38.87000:44:38.880 the bottom and it takes the heat from 00:44:41.12000:44:41.130 the food products and it becomes mom 00:44:42.62000:44:42.630 once it becomes warm again it rises up 00:44:44.66000:44:44.670 okay once you try this up again it comes 00:44:47.15000:44:47.160 in contact with the evaporator surface 00:44:49.04000:44:49.050 it becomes cold and again it comes down 00:44:51.14000:44:51.150 so this is a natural circulation is 00:44:52.55000:44:52.560 maintained by keeping the operator at 00:44:54.26000:44:54.270 the top okay normally these are unfilled 00:45:01.49000:45:01.500 when used in cold storages the as I said 00:45:04.01000:45:04.020 these are mainly used in domestic 00:45:05.12000:45:05.130 refrigerators in cold storage in 00:45:06.80000:45:06.810 domestic refrigerators fins are added 00:45:09.41000:45:09.420 but in cold studies fins are not used 00:45:11.51000:45:11.520 okay a fabrication space should be 00:45:14.66000:45:14.670 provided all around the operators 4/8 00:45:16.58000:45:16.590 flow this is very important just like 00:45:18.34000:45:18.350 condensers the flatback condensers are 00:45:20.59900:45:20.609 Virant you type of condensers you are 00:45:22.76000:45:22.770 relying on by on sea effects for no air 00:45:25.40000:45:25.410 flow okay so the Delta T's are not 00:45:27.34900:45:27.359 normally very high so the potential for 00:45:30.02000:45:30.030 the air flow is not generally small so 00:45:32.66000:45:32.670 if there is large resistance then the 00:45:34.73000:45:34.740 air flow gets affected adversely okay so 00:45:37.01000:45:37.020 if you want to have good air flow you 00:45:38.96000:45:38.970 have to provide sufficient space all 00:45:41.06000:45:41.070 around the evaporator so that it can 00:45:43.13000:45:43.140 flow with minimum resistance them okay 00:45:45.44000:45:45.450 that's why you might have seen in the 00:45:46.97000:45:46.980 domestic refrigerator that they don't 00:45:48.10900:45:48.119 put the evaporator right at the top that 00:45:50.66000:45:50.670 means there will be some space all 00:45:51.89000:45:51.900 around the evaporator so that air can 00:45:53.21000:45:53.220 flow all around the evaporator okay and 00:45:56.53000:45:56.540 baffles are provided to separate the 00:45:58.70000:45:58.710 vomit and cold air plumes now what are 00:46:04.40000:46:04.410 the advantages of for natural 00:46:05.63000:46:05.640 circulation evaporators evaporator 00:46:08.63000:46:08.640 occupies less force per floor space this 00:46:10.82000:46:10.830 is very important especially in cold 00:46:13.16000:46:13.170 storages cinder because floor space if 00:46:15.65000:46:15.660 the system occupies a lot of floor space 00:46:17.35900:46:17.369 then valuable floor space is lost 00:46:19.49000:46:19.500 because you could have stored some 00:46:20.93000:46:20.940 products in that floor space and by 00:46:22.76000:46:22.770 storing more products you could have 00:46:24.41000:46:24.420 earned more money okay for floor space 00:46:26.54000:46:26.550 is very valuable 00:46:27.51000:46:27.520 so when you are using a natural 00:46:29.01000:46:29.020 convection type of evaporators in cold 00:46:31.95000:46:31.960 storage they are kept at the right near 00:46:34.02000:46:34.030 the ceiling okay so it doesn't occupy 00:46:35.94000:46:35.950 any floor space right this is one of the 00:46:38.22000:46:38.230 advantages of as insulation type 00:46:40.05000:46:40.060 evaporators second advantage is they can 00:46:42.87000:46:42.880 operate for longer periods without 00:46:44.46000:46:44.470 defrosting and they are simple to make 00:46:48.27000:46:48.280 and easy to maintain 00:46:49.23000:46:49.240 because especially in cold storages and 00:46:51.33000:46:51.340 all they are playing tubes okay normally 00:46:53.76000:46:53.770 they are unfilled so they are they'll be 00:46:55.26000:46:55.270 simply welded at the site so there are 00:46:57.60000:46:57.610 no fields nothing just you have to take 00:46:59.16000:46:59.170 the pipes and well the pipes at the side 00:47:01.02000:47:01.030 okay and the maintenance is also easy 00:47:03.66000:47:03.670 and very useful when low air velocities 00:47:07.17000:47:07.180 and minimum dehumidification of the 00:47:08.67000:47:08.680 product is required the since you are 00:47:11.46000:47:11.470 relying on natural convection and the 00:47:13.74000:47:13.750 delta T available for natural convection 00:47:15.83000:47:15.840 obviously you won't find any eight 00:47:17.70000:47:17.710 blasts or anything okay so the ideal 00:47:19.59000:47:19.600 velocity will be very very small once 00:47:21.35900:47:21.369 the air velocity is small there is no 00:47:22.83000:47:22.840 danger of products getting dried up too 00:47:25.89000:47:25.900 much okay these are one typical problem 00:47:27.84000:47:27.850 with forced convection type of 00:47:30.06000:47:30.070 evaporators because my velocity is high 00:47:31.98000:47:31.990 high air velocity means high heat and 00:47:33.78000:47:33.790 mass transfer rate so drying of a 00:47:35.43000:47:35.440 products take place whereas in natural 00:47:37.59000:47:37.600 convection type this problem is not 00:47:39.18000:47:39.190 there okay however there are certain 00:47:45.05000:47:45.060 disadvantages the disadvantage obviously 00:47:49.26000:47:49.270 is that you require very long lengths 00:47:51.24000:47:51.250 because the overall heat transfer 00:47:52.10900:47:52.119 coefficient is typically small this is 00:47:54.21000:47:54.220 small because you are relying on natural 00:47:56.19000:47:56.200 convection hence higher refrigerant 00:47:59.84900:47:59.859 inventory and higher pressure drops once 00:48:01.65000:48:01.660 the required to blend becomes large you 00:48:03.90000:48:03.910 have to put large amount of refrigerant 00:48:06.12000:48:06.130 inside the system okay once you put a 00:48:08.07000:48:08.080 large amount of refrigerant inside 00:48:09.09000:48:09.100 system there are other problems the cost 00:48:11.58000:48:11.590 will be more okay and if the refrigerant 00:48:13.62000:48:13.630 is a toxic and flammable it will have 00:48:15.75000:48:15.760 safety problems okay in addition to that 00:48:18.42000:48:18.430 there are other problems like pressure 00:48:21.15000:48:21.160 equalization takes a long time and 00:48:23.06000:48:23.070 defrosting also takes a long time okay 00:48:25.59000:48:25.600 so these are some of the disadvantages 00:48:26.79000:48:26.800 of having long lengths okay in addition 00:48:29.67000:48:29.680 to this if you have long refrigerant 00:48:31.98000:48:31.990 tubing pressure drop also will be large 00:48:33.81000:48:33.820 so if you want to minimize the pressure 00:48:35.09900:48:35.109 drop you may have to go for parallel 00:48:36.84000:48:36.850 circuits okay 00:48:39.82000:48:39.830 now let me quickly explain the second 00:48:41.68000:48:41.690 type that is the flooded type 00:48:42.88000:48:42.890 evaporators they are typically used in 00:48:45.91000:48:45.920 large ammonia systems the refrigerant 00:48:48.85000:48:48.860 enters a surge tank through a flow type 00:48:50.65000:48:50.660 expansion valve the compressor directly 00:48:54.01000:48:54.020 draws the flash flash vapor which 00:48:55.54000:48:55.550 improves the performance as you know 00:48:57.10000:48:57.110 that once you flash vapor is not allowed 00:48:58.84000:48:58.850 to go to the evaporator performance 00:49:00.64000:49:00.650 improves we have seen this in multistage 00:49:02.77000:49:02.780 systems and the liquid refrigerant 00:49:04.54000:49:04.550 enters the evaporator from the bottom of 00:49:06.31000:49:06.320 the surge tank the mixture of liquid and 00:49:08.95000:49:08.960 vapor flows along the evaporator tubes 00:49:11.01000:49:11.020 the vapor is separated as it enters the 00:49:13.75000:49:13.760 surge drum or surge tank okay another on 00:49:16.03000:49:16.040 evaporated liquid is recirculated so let 00:49:18.22000:49:18.230 me quickly explain this so this is the 00:49:21.82000:49:21.830 flooded type of evaporator okay so being 00:49:24.64000:49:24.650 the evaporator portion right that is the 00:49:29.56000:49:29.570 timing let us assume that if you use for 00:49:31.75000:49:31.760 let us say cooling area you have a fan 00:49:35.68000:49:35.690 and fins another it is blowing over this 00:49:37.36000:49:37.370 so we have a the component surge tank 00:49:40.00000:49:40.010 here and refrigerant from the condenser 00:49:42.33000:49:42.340 enters the surge tank through the float 00:49:44.98000:49:44.990 valve these float wall acts as an 00:49:47.02000:49:47.030 expansion device here okay these 00:49:49.66000:49:49.670 expansion device used here and what is 00:49:51.88000:49:51.890 the purpose of this float type of valve 00:49:53.80000:49:53.810 it always maintains a required level of 00:49:56.07000:49:56.080 refrigerant in the surge tank okay when 00:49:58.66000:49:58.670 the level Falls the valve will open more 00:50:01.12000:50:01.130 more refrigerant will come here so 00:50:03.13000:50:03.140 normally this is connected to a 00:50:04.60000:50:04.610 condenser or receiver condenser okay and 00:50:07.54000:50:07.550 you can see that when the refrigerant 00:50:09.97000:50:09.980 enters at this point it is vapor plus 00:50:12.52000:50:12.530 liquid because due to flashing across 00:50:14.50000:50:14.510 that expansion device some vapour would 00:50:16.21000:50:16.220 have been generated so you have both 00:50:18.01000:50:18.020 vapor and liquid but what you are doing 00:50:19.93000:50:19.940 is by separating vapor and liquid in the 00:50:21.67000:50:21.680 surge tank you are allowing only liquid 00:50:23.68000:50:23.690 to go to the evaporator and vapour 00:50:25.42000:50:25.430 instead of going to the evaporator and 00:50:26.92000:50:26.930 it directly goes to the compressor okay 00:50:28.69000:50:28.700 so that is how you can improve the 00:50:30.04000:50:30.050 efficiency of the evaporator so as you 00:50:32.77000:50:32.780 can see the from the picture that only 00:50:34.60000:50:34.610 liquid refrigerant goes through the 00:50:35.98000:50:35.990 evaporator and as the it flows to the 00:50:38.20000:50:38.210 evaporator it takes heat from the 00:50:39.85000:50:39.860 surroundings and vapour is generated 00:50:41.23000:50:41.240 then okay so you can see the vapour 00:50:44.47000:50:44.480 bubbles so you find that at the outlet 00:50:47.08000:50:47.090 of the evaporator again you have vapor 00:50:51.10000:50:51.110 liquid mixture this is the feature of a 00:50:53.80000:50:53.810 flooded 00:50:54.25000:50:54.260 type of operator not all that 00:50:55.84000:50:55.850 refrigerant that goes to the evaporator 00:50:57.55000:50:57.560 will evaporate okay there will be a 00:50:59.41000:50:59.420 large amount which is uh Nev operated 00:51:01.00000:51:01.010 then now another operated amount will 00:51:02.77000:51:02.780 simply recirculate so whatever vapor is 00:51:05.38000:51:05.390 there again that way / + water vapor is 00:51:07.45000:51:07.460 generated here both will be compressed 00:51:09.46000:51:09.470 by the compressor Ram okay so this is a 00:51:11.71000:51:11.720 flooded type evaporator 00:51:13.15000:51:13.160 you can see that there are a lot of 00:51:14.20000:51:14.210 liquid in the evaporator and the 00:51:15.61000:51:15.620 evaporator surfaces are always wet that 00:51:17.89000:51:17.900 will give you higher heat transfer 00:51:19.09000:51:19.100 coefficient on the refrigerant side so 00:51:21.19000:51:21.200 these are advantage of fluid type of 00:51:22.63000:51:22.640 evaporator okay and mass flow rate 00:51:26.59000:51:26.600 through evaporator is not same as the 00:51:28.15000:51:28.160 mass flow rate through compressor and 00:51:29.68000:51:29.690 the ratio of these two mass flow rates 00:51:31.51000:51:31.520 is called as recirculation factor s and 00:51:34.15000:51:34.160 if you apply the mass balance for steady 00:51:36.25000:51:36.260 state whatever mass is leaving the surge 00:51:38.86000:51:38.870 tank must enter the surge tank okay then 00:51:42.70000:51:42.710 that means what you have to do you have 00:51:44.26000:51:44.270 to apply mass balance across the surge 00:51:47.38000:51:47.390 tank okay from that you can easily show 00:51:49.48000:51:49.490 that the recirculation factor s is 1 00:51:53.86000:51:53.870 minus X 4 divided by X what are X 4 and 00:51:57.43000:51:57.440 X X 4 is a quality of refrigerant at the 00:52:04.21000:52:04.220 inlet to the surge tank and X is the 00:52:06.43000:52:06.440 quality of the refrigerant at the outlet 00:52:08.62000:52:08.630 of the evaporator okay so this is by 00:52:11.62000:52:11.630 applying the mass balance across the 00:52:13.12000:52:13.130 surge tank okay so once you know the 00:52:19.03000:52:19.040 qualities you can calculate what is a 00:52:20.35000:52:20.360 circulation factor circulation factor 00:52:22.12000:52:22.130 will be greater than 1 that means more 00:52:24.93000:52:24.940 refrigerant circulates through the 00:52:26.71000:52:26.720 evaporator than is evaporated okay now 00:52:32.08000:52:32.090 since liquid what are the advantages 00:52:33.61000:52:33.620 since liquid refrigerant is in contact 00:52:35.29000:52:35.300 with whole of operated surface the 00:52:37.30000:52:37.310 refrigerant side heat transfer 00:52:38.62000:52:38.630 coefficient will be very high sometimes 00:52:41.17000:52:41.180 the liquid refrigerant pump may also be 00:52:42.82000:52:42.830 used to further increase the heat 00:52:44.11000:52:44.120 transfer coefficient term the 00:52:46.51000:52:46.520 lubricating oil tends to accumulate in 00:52:48.19000:52:48.200 the flooded evaporator hence an 00:52:49.42000:52:49.430 effective oil separator must be used 00:52:50.98000:52:50.990 immediately after the compressor okay 00:52:55.80000:52:55.810 right I'll stop the lecture here and I 00:52:59.14000:52:59.150 will continue this lecture in the next 00:53:00.94000:53:00.950 class okay in the next class I will give 00:53:03.88000:53:03.890 you a formula for calculating the weight 00:53:06.73000:53:06.740 transfer coefficients and 00:53:07.88000:53:07.890 of evaporators okay thank you
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