1 00:00:11,370 --> 00:00:17,180 In the last class we were discussing about hybrid rockets or in this class let us look 2 00:00:17,180 --> 00:00:26,320 at how the burn rate relationship AGOX to the power of n is obtained so firstly to do 3 00:00:26,320 --> 00:00:48,530 that we need to understand how the burning takes place inside the hybrid rocket motor. 4 00:00:48,530 --> 00:01:02,749 Now if this is the hybrid rocket motor then oxidizer is injected here right so a boundary 5 00:01:02,749 --> 00:01:38,180 layer develops and the combustion takes place inside the boundary layer 6 00:01:38,180 --> 00:01:49,830 okay 7 00:01:49,830 --> 00:01:56,100 so this is how the combustion takes place inside a hybrid rocket motor now if we look 8 00:01:56,100 --> 00:02:04,870 closely into this combustion zone then we will be able to look at what are all the processes 9 00:02:04,870 --> 00:02:24,880 that are taking place in that zone. So let us say this is the solid fuel that 10 00:02:24,880 --> 00:02:35,040 we have then let us say we have taken somewhere in this region right where the boundary layer 11 00:02:35,040 --> 00:03:07,219 is already established so this is the boundary layers and let us say this is the 12 00:03:07,219 --> 00:03:15,890 flames own right now what is happening because of this boundary layer there is a heat transfer 13 00:03:15,890 --> 00:03:23,549 to the propellant which is convective in nature right and that heat transfer is being taken 14 00:03:23,549 --> 00:03:31,319 up partly for maintaining the temperature profile in the condensed phase and then partly 15 00:03:31,319 --> 00:03:41,319 to evaporate the fear. So if we were to draw the temperature profile 16 00:03:41,319 --> 00:03:52,480 inside the boundary layer it would go something like 17 00:03:52,480 --> 00:04:08,319 this is Ts this is Tc and this is Te the combustion that is achieved here I mean the temperature 18 00:04:08,319 --> 00:04:13,040 that is a trade here is the highest simply because there is combustion and after which 19 00:04:13,040 --> 00:04:27,710 it is diluted by oxidizer right now the propellant burns in this direction and there is because 20 00:04:27,710 --> 00:04:38,620 of that there is a mass addition that is given by ? Pr. okay mass is getting in and there 21 00:04:38,620 --> 00:04:43,110 is something known as blowing effect that we had discussed when we were discussing about 22 00:04:43,110 --> 00:04:57,060 erosive burning in solid Rockets. Now so there is heat that is being transferred 23 00:04:57,060 --> 00:05:11,509 let me call it as Qc okay that is heat that is being converted into the propellant surface 24 00:05:11,509 --> 00:05:32,319 now this Qc I can write it as h into Tc - Ts okay this is the convective heat transfer 25 00:05:32,319 --> 00:05:52,720 coefficient 26 00:05:52,720 --> 00:05:58,479 and this is the adiabatic flame temperature or the combustion temperature and this is 27 00:05:58,479 --> 00:06:06,979 the surface temperature now this heat as I said is a part of it goes to maintaining a 28 00:06:06,979 --> 00:06:23,300 temperature profile and part of it is used for evaporation so Qc must be equal to 29 00:06:23,300 --> 00:06:30,139 let me call this direction as X and this direction is y. 30 00:06:30,139 --> 00:06:48,879 So 31 00:06:48,879 --> 00:06:55,560 okay this is the temperature there is the part of the heat that is taken up to maintain 32 00:06:55,560 --> 00:07:03,319 the temperature profile and this is the Qs get seen earlier is the heat of evaporation 33 00:07:03,319 --> 00:07:09,330 at the surface so this we are seeing with regards to solid propellant that we can rewrite 34 00:07:09,330 --> 00:07:21,919 this part as ?P r . CPTS - T? let me call the temperature 35 00:07:21,919 --> 00:07:33,539 here as T8. So I can rewrite this as Qc must be equal 36 00:07:33,539 --> 00:08:04,759 to ?pr. okay now for ease of our terminology and other things to follow let me call this 37 00:08:04,759 --> 00:08:30,229 part as hv hv is nothing but Cp TS - T8 + Qs now we know that this problem is a boundary 38 00:08:30,229 --> 00:08:38,950 layer driven problem okay so what we are going to use is something known as Reynolds analogy 39 00:08:38,950 --> 00:08:46,410 what it says is there is a similarity between the momentum boundary layer and the thermal 40 00:08:46,410 --> 00:08:52,750 boundary layer right and we use that it is easier to calculate the momentum boundary 41 00:08:52,750 --> 00:08:59,430 layer there are correlations for it and we use that to calculate the thermal boundary 42 00:08:59,430 --> 00:09:03,280 layer. This can happen when only what Prandtl number 43 00:09:03,280 --> 00:09:12,779 is close to 1 which in carry gases it is always close to 1 so we can make this assumption 44 00:09:12,779 --> 00:09:50,800 so using approximate 45 00:09:50,800 --> 00:10:05,550 Reynolds analogy. That is there is similarity between 46 00:10:05,550 --> 00:10:27,449 momentum and heat or momentum and energy 47 00:10:27,449 --> 00:10:53,800 so we can define something known as a stanton number 48 00:10:53,800 --> 00:11:05,709 ue is nothing but the velocity here okay so h is the heat transfer coefficient ?e is the 49 00:11:05,709 --> 00:11:11,010 density at the beyond the boundary layer and Ue is the velocity beyond the boundary layer 50 00:11:11,010 --> 00:11:20,190 CP is the specific heat this is a non-dimensional number it is also known as diamond dimensionless 51 00:11:20,190 --> 00:11:30,260 heat transfer coefficient okay now if we use the Reynolds analogy what it says is that 52 00:11:30,260 --> 00:11:49,970 this stanton number indicated by Ch = Cf/2 where Cf is 53 00:11:49,970 --> 00:12:11,089 nothing but skin friction coefficient okay. And we know that rundle number is close to 54 00:12:11,089 --> 00:12:28,890 1 for gases so this term becomes merely Ch – Cf/2 now we have defined what a stanton 55 00:12:28,890 --> 00:12:37,240 numbers and we have found out that by Reynolds analogy we can equate it to this skin friction 56 00:12:37,240 --> 00:12:45,269 coefficient if we multiply the numerator and denominator of the stanton number by a ? T 57 00:12:45,269 --> 00:13:24,660 that is given by Ts - Tc - Ts okay we will do that and see what we will get. 58 00:13:24,660 --> 00:13:36,839 So we had stanton number Cf = h/?e ue Cp so we will multiply both numerator and denominator 59 00:13:36,839 --> 00:13:48,550 by 60 00:13:48,550 --> 00:14:01,140 Tc - Ts so it does not change anything to the left hand side but what is H into TC - TS 61 00:14:01,140 --> 00:14:19,310 that is the convective heat transfer right so Qs = h into Tc - Ts so what we will end 62 00:14:19,310 --> 00:15:00,630 up getting is sorry this is not Cf Ch Ch is equal to this which one oh okay QC fine now 63 00:15:00,630 --> 00:15:16,269 what we need to this part is the QC so what is this part we can define it as some change 64 00:15:16,269 --> 00:15:39,120 in enthalpy so we will call the ?h = CP TC - TS so what we will get this we also had 65 00:15:39,120 --> 00:15:50,600 the other term that is the heat. That was taken up by the solid let me call 66 00:15:50,600 --> 00:16:09,790 that as q s that was is equal to ?p r. hv right but we know that these two must be equal 67 00:16:09,790 --> 00:16:16,520 right that was the heat balance at the surface for the heat balance at the surface these 68 00:16:16,520 --> 00:16:48,770 two must be equal. So I can now replace in my stanton number 69 00:16:48,770 --> 00:16:59,839 definition this here right so and stanton number this also I know is nothing but Cf 70 00:16:59,839 --> 00:17:33,960 / 2 by two right so I will use that so I will get ?p or Cf / = ?p r . hv right or in other 71 00:17:33,960 --> 00:18:06,390 words I can rewrite a expression for r . as r . = Cf / 2 x ?e ue ?h/ hv / ?P now we have 72 00:18:06,390 --> 00:18:16,660 been able to get this expression for r. in terms of ?V x ue is nothing but in some sense 73 00:18:16,660 --> 00:18:27,560 mass flux and Cf /2 what we need to further determine is what is this Cf for the particular 74 00:18:27,560 --> 00:19:27,110 geometry that we have considered. So skin friction coefficient 75 00:19:27,110 --> 00:19:41,780 are without blowing 76 00:19:41,780 --> 00:20:02,309 we can write it as Cf0/ 2 is equal to in terms of Reynolds number 77 00:20:02,309 --> 00:20:21,000 this x indicates the location along the axis this is for a turbulent flow 78 00:20:21,000 --> 00:20:27,740 now what happens because of blowing is the boundary layer in essence tends to get thicken 79 00:20:27,740 --> 00:20:49,750 okay so the skin friction with blowing will be different 80 00:20:49,750 --> 00:21:28,659 and it can be given as Cf /Cf0 where ß varies from p less than 1ß is nothing but the blowing 81 00:21:28,659 --> 00:21:45,810 coefficient. And we can define ß as follows ß is defined 82 00:21:45,810 --> 00:22:04,770 as ?p r. that is the mass flux that is coming in from the surface to the mass that is going 83 00:22:04,770 --> 00:22:21,900 through the port okay. It is also known as non-dimensional 84 00:22:21,900 --> 00:22:39,140 so 85 00:22:39,140 --> 00:22:47,549 we have defined Cf in terms of Cf0 x ß and we have ß so we can now get the expression 86 00:22:47,549 --> 00:23:05,850 that we were looking for that is we had said that 87 00:23:05,850 --> 00:23:22,190 Ch is equal to this is nothing but ?p r . hv / ue that is this expression right. 88 00:23:22,190 --> 00:23:28,250 So we will treat in use this expression if you look at this expression and the expression 89 00:23:28,250 --> 00:23:43,760 for ß right there is some similarity between the two and we can write ?h \hv I it is nothing 90 00:23:43,760 --> 00:24:06,679 but ?pr .that is equal to ß right so now I have this expression for r .here wherein 91 00:24:06,679 --> 00:24:13,230 I have defined Cf I have defined all other terms so I can get plug them in and get my 92 00:24:13,230 --> 00:25:00,030 expression for r .which we will do r. = Cf / 2?e ue / ?p right now we know that Cf is 93 00:25:00,030 --> 00:25:33,880 nothing but Cf 0 x ß so we will use that here and get our expression for r. 94 00:25:33,880 --> 00:26:07,809 So r. =1.27 /2 xCf0 x ß0.77 x ?e ue / ?a x ß right because we know that ? h/h y is 95 00:26:07,809 --> 00:26:23,490 nothing but ß so Cf0 also we know in terms of Reynolds number as this expression here 96 00:26:23,490 --> 00:26:58,480 so if we substitute that we will get r . = ß is – 0.77 and one so I can write it as ß0.23 97 00:26:58,480 --> 00:27:44,149 Reynolds number is nothing but ?e ue x x/ µ so I get r.= so we have ?e ue here and 98 00:27:44,149 --> 00:27:54,799 here so we can Club those two. And rewrite our expression as r. = and I can 99 00:27:54,799 --> 00:28:07,029 take the ?p that is here this ? p I can multiply it back with the r. so I will get ?p r . is 100 00:28:07,029 --> 00:28:37,480 equal to 101 00:28:37,480 --> 00:29:07,190 what is ? ue is nothing but the mass flux in the put right so mass flux let us define 102 00:29:07,190 --> 00:29:30,070 G as m . /Ap so that will be nothing but ?e ue 103 00:29:30,070 --> 00:29:54,690 so you have we can replace this with G right and we will get - 104 00:29:54,690 --> 00:30:07,380 s thank you 105 00:30:07,380 --> 00:30:18,840 okay so we come to this expression there is 0 here. 106 00:30:18,840 --> 00:30:26,750 So we get this expression where in the burn rate is related to the mass flux through the 107 00:30:26,750 --> 00:30:36,710 port right and if you look at when we do experiments it is easier to characterize this in terms 108 00:30:36,710 --> 00:30:43,650 of the oxidizer mass flux itself because that is something that we have control of and we 109 00:30:43,650 --> 00:30:49,360 know what it is so which is why you will find that if you take a look at any hybrid related 110 00:30:49,360 --> 00:31:31,899 literature you will find that the burn rate is expressed as a function of G ox okay. 111 00:31:31,899 --> 00:31:42,740 So 112 00:31:42,740 --> 00:31:50,240 we can recast this in terms of zero X and that is what you will find in most literature 113 00:31:50,240 --> 00:32:01,799 as a dot is a function of 0 X to the power of n and if you look at this expression you 114 00:32:01,799 --> 00:32:09,902 will find that it depends very weakly on X the axial location and beta but it is a very 115 00:32:09,902 --> 00:32:18,809 strong function of the mass flux right now let us look at the problem that we had said 116 00:32:18,809 --> 00:32:27,630 yesterday with regards to what happens to burn rate as we proceed in from the head end 117 00:32:27,630 --> 00:32:33,120 to the nozzle length. If you look at this expression here it looks 118 00:32:33,120 --> 00:32:44,679 like what should happen to the burn rate as you increase X it should decrease right set 119 00:32:44,679 --> 00:32:54,700 the opposite yesterday right is that correct order is that wrong 120 00:32:54,700 --> 00:32:59,409 if you look at this yes you are saying but you are forgetting that there is a greater 121 00:32:59,409 --> 00:33:06,960 dependence on G if you look at what is happening as you move from the head end of the port 122 00:33:06,960 --> 00:33:12,780 to the nozzle end there is continuous mass addition and this is increasing as you go 123 00:33:12,780 --> 00:33:17,419 from heading to the nozzle end and this is raised to a power of 0. 8. 124 00:33:17,419 --> 00:33:22,210 So therefore you will find that the burn rate as you go from heading to the nozzle length 125 00:33:22,210 --> 00:34:17,070 will always increase 126 00:34:17,070 --> 00:34:33,620 okay now people have done people have conducted experiments and found out that if you take 127 00:34:33,620 --> 00:34:59,950 the oxidizer mass flux the dependence of burn rate varies this n varies from 0.42 to 0.7 128 00:34:59,950 --> 00:35:05,590 one of the things that we have kind of made an assumption right at the beginning was we 129 00:35:05,590 --> 00:35:10,870 had assumed something like a flat plate if we assume a pipe flow kind of a situation 130 00:35:10,870 --> 00:35:16,630 and redo the calculations. We will probably get closer to what we what 131 00:35:16,630 --> 00:35:24,590 is observed experimentally, experimentally what we can get is in terms of the oxidizer 132 00:35:24,590 --> 00:35:31,560 mass flux and not the overall flux k overall flux involves that you take across-section 133 00:35:31,560 --> 00:35:37,510 and find out what is the fuel plus oxidizer that is coming which is a lot more difficult 134 00:35:37,510 --> 00:35:43,180 to do if you are doing an experiment, experiment you are better off knowing what is your mass 135 00:35:43,180 --> 00:35:51,780 flow rate of oxidizer and then calculating the oxidizer mass flux there was another problem 136 00:35:51,780 --> 00:35:58,320 that I said hybrid Rockets suffer from because of this n. 137 00:35:58,320 --> 00:36:05,320 That is if you look at holding even in even when you hold the oxidizer mass flux constant 138 00:36:05,320 --> 00:36:14,840 we find that the fuel mass flow rate was changing now let us look at is there any condition 139 00:36:14,840 --> 00:36:22,170 at which the oxidizer mass flux if you hold constant or if you fold the oxidizer mass 140 00:36:22,170 --> 00:36:29,600 flow rate constant when the fluid flow rate also become constant. 141 00:36:29,600 --> 00:36:50,400 We know that the mass flow rate of fuel is given ?p Ab*I .so if you are taking a port 142 00:36:50,400 --> 00:37:07,780 configuration ?p into this is the diameter of the port 143 00:37:07,780 --> 00:37:20,860 and this is the length of the port this is the burning surface area into our 144 00:37:20,860 --> 00:37:33,490 dot is nothing but A into okay now let us determine what value of n will give us something 145 00:37:33,490 --> 00:37:39,110 that when you hold the oxidizer mass flow rate constant you can also get a few mass 146 00:37:39,110 --> 00:37:45,400 flow rate constant if you have to engineer such a hybrid then that is a very good situation. 147 00:37:45,400 --> 00:38:34,550 Because you can operate it at a set specific impulse even though the diameters are thinking 148 00:38:34,550 --> 00:38:57,920 so Gio X we know is nothing but fine this is geo X so if we plug it into this equation 149 00:38:57,920 --> 00:39:47,780 we will get m dot Fo know this is you have 2 into n right in the denominator and if you 150 00:39:47,780 --> 00:39:55,010 look at this equation you have D to to the part of one end d to the power of 2n as very 151 00:39:55,010 --> 00:40:11,040 obvious that if you have a value of n that is equal to 0.5 for n equal to 0.5 m dot F 152 00:40:11,040 --> 00:40:30,330 becomes 153 00:40:30,330 --> 00:40:39,510 independent of the pore diameter so therefore you can operate it at a set o by F ratio so 154 00:40:39,510 --> 00:40:49,460 that you can choose that over F ratio such that it gives the maximum is B. 155 00:40:49,460 --> 00:41:10,300 Now what happens if n is greater than 0.5 so it is desirable to have 156 00:41:10,300 --> 00:41:24,700 in equal 2.5 now for n greater than .5 what will happen to the mass of here as diameter 157 00:41:24,700 --> 00:41:35,120 increases it will reduce because you will have 2 N and you will have DP here so the 158 00:41:35,120 --> 00:41:41,760 denominator will be greater than the numerator so therefore as diameters increased mass flow 159 00:41:41,760 --> 00:41:58,990 rate of here will decrease 160 00:41:58,990 --> 00:42:29,900 and similarly or for the other case that is for n less than .5 okay this finishes our 161 00:42:29,900 --> 00:42:39,601 discussions on hybrid rockets so in a sense we have learnt three, three kinds of rocket 162 00:42:39,601 --> 00:42:45,720 engines in this course that is one is solid propellant the other one is liquid propellant 163 00:42:45,720 --> 00:42:54,810 and then lastly the hybrid propellant okay thank you.