1 00:00:00,080 --> 00:00:11,370 In 2 00:00:11,370 --> 00:00:20,750 the last class we had looked at cycle analysis of everything engines now let us look at rocket 3 00:00:20,750 --> 00:00:29,150 engines in the next few classes here we will be trying to derive expressions for specific 4 00:00:29,150 --> 00:00:38,410 impulse and how mass flow rate varies through a choke nozzle how do we get thrust how do 5 00:00:38,410 --> 00:00:44,370 we get exit velocity what are the assumptions that we make and as a consequence of these 6 00:00:44,370 --> 00:00:51,680 assumptions what do we have to deal with or what is the error of these assumptions we 7 00:00:51,680 --> 00:00:54,770 will look at it a little later in the course okay. 8 00:00:54,770 --> 00:01:10,790 So now let us look at rocket engines now rocket engines 9 00:01:10,790 --> 00:01:16,440 all the three kinds of rocket engines that we discover discussed in the earlier classes 10 00:01:16,440 --> 00:01:25,160 that is solid liquid and hybrid all three of them have the following that is they have 11 00:01:25,160 --> 00:01:45,950 propellant storage then they have a thrust chamber 12 00:01:45,950 --> 00:02:04,150 and lastly a convergent divergent nozzle 13 00:02:04,150 --> 00:02:10,440 in the solid rocket both these two are in the same physical location whereas in liquids 14 00:02:10,440 --> 00:02:16,770 and hybrids all three of them are separate but one thing common to all of them is this 15 00:02:16,770 --> 00:02:22,280 part. That has the nozzle part in the nozzle what 16 00:02:22,280 --> 00:02:30,120 you have as the terminal energy because of release because of chemical reactions is now 17 00:02:30,120 --> 00:02:48,550 converted into kinetic energy so in the nozzle you have thermal to kinetic energy conversion 18 00:02:48,550 --> 00:02:55,150 and as this is common to all three kinds of rockets we can study them exclusively okay 19 00:02:55,150 --> 00:03:00,380 and without having to pay any need to pay any attention to what kind of rocket to test 20 00:03:00,380 --> 00:03:05,330 whether it is a solid rocket or a liquid rocket or a hybrid rocket. So we will do that now 21 00:03:05,330 --> 00:03:13,230 we look at the converging diverging nozzle now before we go there what we are going to 22 00:03:13,230 --> 00:03:42,470 look at is quasi steady one dimensional analysis. That is we are going to assume that all changes 23 00:03:42,470 --> 00:03:47,709 happen along the axis only there are no changes that are happening in the radial or the azimuth 24 00:03:47,709 --> 00:03:53,010 direction R and ? directions we do not assume any changes to be happening so all changes 25 00:03:53,010 --> 00:03:58,170 are happening along the axis so it is one done which is not strictly true for rocket 26 00:03:58,170 --> 00:04:04,739 engine nozzle okay we will see what error this bring about a little later first as we 27 00:04:04,739 --> 00:04:23,759 are going to make this assumption then we are going to assume that the chamber pressure 28 00:04:23,759 --> 00:04:40,460 and temperature are given to us. That is let us say if this is the rocket motor 29 00:04:40,460 --> 00:04:54,159 we will assume that we know chamber pressure which is called PC and chamber temperature 30 00:04:54,159 --> 00:05:13,669 TC has known will also take these to be stagnation quantities 31 00:05:13,669 --> 00:05:20,320 that is if you look at velocities elsewhere in the nozzle and if you look at the velocity 32 00:05:20,320 --> 00:05:25,869 at the entry of the nozzle these velocities are very small compared to the velocities 33 00:05:25,869 --> 00:05:34,879 as you go along the nozzle so you can take them to be stagnation conditions at the entry 34 00:05:34,879 --> 00:05:55,000 of the nozzle okay. And we are also going to assume that you are 35 00:05:55,000 --> 00:06:04,139 also going to assume that as I said these are known to us firstly and we are also going 36 00:06:04,139 --> 00:06:21,879 to assume that fluid is an ideal gas 37 00:06:21,879 --> 00:06:45,159 with constant thermodynamic properties 38 00:06:45,159 --> 00:06:53,129 that is the CP ? thermal conductivity all these do not change as we go from this portion 39 00:06:53,129 --> 00:07:09,339 of the nozzle to the exit of the nozzle then we will also assume that the flow is in viscid 40 00:07:09,339 --> 00:07:30,749 that is viscosities are very small and will also assume fluid flow is isentropic. 41 00:07:30,749 --> 00:07:35,520 So with these assumptions let us see what we can derive and based on this we will also 42 00:07:35,520 --> 00:07:41,240 evaluate what are the short comings of some of these assumptions are these really valid 43 00:07:41,240 --> 00:07:52,949 and then we will try to estimate what is the error now from your basic gas dynamics you 44 00:07:52,949 --> 00:08:17,469 know that for a nozzle for a convergent. That was in nozzle I can write you buy you 45 00:08:17,469 --> 00:08:35,430 that is change in velocity as you move along the nozzle = 46 00:08:35,430 --> 00:08:43,080 now we know that if M is less than 1 what should be the dA / A if this has to flow it 47 00:08:43,080 --> 00:08:51,500 has to accelerate if M is less than 1 dA/ A has to be negative if fluid has to accelerate 48 00:08:51,500 --> 00:09:02,320 and if M is greater than 1 dA/ A has to be positive if fluid has to accelerate and we 49 00:09:02,320 --> 00:09:23,910 know that if at the throat at the throat of the C-D nozzle Mach number is 1 then we call 50 00:09:23,910 --> 00:09:35,560 it choke nozzle let us now find out what are the pressure ratio. 51 00:09:35,560 --> 00:09:47,200 If you have PC here what should be the PA or what should be this ratio of PC / PA for 52 00:09:47,200 --> 00:10:05,190 the nozzle to be choked okay so let us find the critical ratio of this 53 00:10:05,190 --> 00:10:21,800 pressure for doping 54 00:10:21,800 --> 00:10:33,180 in order to do this we will solve the energy equation from here to here okay from the entry 55 00:10:33,180 --> 00:10:52,070 to the root section and I can write one dimensional energy equation as one dimensional 56 00:10:52,070 --> 00:11:01,140 steady state and for in viscid flow with constant thermodynamic properties I can write the energy 57 00:11:01,140 --> 00:11:33,410 equation as HT + = UT. Where C were c there indicates our chamber 58 00:11:33,410 --> 00:11:56,370 conditions and T indicates throat conditions C indicates and T indicates throat that is 59 00:11:56,370 --> 00:12:10,510 we are looking at this to be see this to be okay now there is no heat addition that is 60 00:12:10,510 --> 00:12:17,200 taking place from here to here okay and the composition is the same so if the composition 61 00:12:17,200 --> 00:12:38,290 is the same then CP has to be the same so 62 00:12:38,290 --> 00:12:53,550 C to T that is from chamber to throat it means that CT is same and there is only change in 63 00:12:53,550 --> 00:13:07,350 sensible enthalpy and it also means that only there is change 64 00:13:07,350 --> 00:13:10,720 in enthalpy. Because there are no reactions that are taking 65 00:13:10,720 --> 00:13:22,000 place so if all the change from C to T is only because of sensible enthalpy change then 66 00:13:22,000 --> 00:13:51,560 I can rewrite this equation as right that is I have replaced HT / CpTt and HC / CpTc 67 00:13:51,560 --> 00:14:03,930 we also know that if the throat is if the flow is choked at the throat then add throat 68 00:14:03,930 --> 00:14:29,650 M = 1and therefore Ut = 80 that is the speed of sound which is given by here Rs nothing 69 00:14:29,650 --> 00:15:13,110 but R = Ru universal gas constant /molecular weight okay now using all this we can simplify 70 00:15:13,110 --> 00:15:33,380 this equation further and write it as. Because we know Ut = 80 and 80 is this I can 71 00:15:33,380 --> 00:15:44,820 write it bit like this = 72 00:15:44,820 --> 00:16:10,420 we know that CP / CV = ? and CP - CV = R so dividing the entire equation / CpTp that is 73 00:16:10,420 --> 00:16:45,130 dividing my CpTp we will get 1 + ?RTT / 2 okay now we know that R and CP we can get 74 00:16:45,130 --> 00:17:01,070 through this so let me write that down this will then become 1 + ? x 1 / 1 - ? this and 75 00:17:01,070 --> 00:17:39,690 this cancels out so you get 2 = TC/ DT or simplifying. 76 00:17:39,690 --> 00:17:55,879 I can write Tc / Tt = ? + 1 / 2 from this and knowing the connection between CP 2 temperatures 77 00:17:55,879 --> 00:18:19,179 to pressure for an isentropic flow I can write Pc / Pt as = so now we can get the critical 78 00:18:19,179 --> 00:18:31,190 ratio that is required for choking okay and if you substitute for ? =1.2 which is usually 79 00:18:31,190 --> 00:18:47,830 the value of ? for burnt gases you get Pc / Pt to be 1.7 that is if the chamber ratio 80 00:18:47,830 --> 00:18:54,129 to the exit pressure or the pressure outside is greater than this then the nozzle will 81 00:18:54,129 --> 00:19:14,350 be choked okay for any pressure ratio PC / or for any pressure 82 00:19:14,350 --> 00:19:24,389 ratio that is from Pc to ambient wear in it is greater than or = 1.7. 83 00:19:24,389 --> 00:19:44,529 Then the fluid flow is all choked flow as I had said in the previous class the advantage 84 00:19:44,529 --> 00:19:50,230 of chalk flow is now it becomes independent of downstream conditions and it is only a 85 00:19:50,230 --> 00:19:59,559 function of upstream conditions okay so let us see how we can use this to calculate what 86 00:19:59,559 --> 00:20:05,290 is the mass flow rate through a choke nozzle we have calculated what is the ratio that 87 00:20:05,290 --> 00:20:09,789 is required for the flow to be choked let us now calculate what is the mass flow rate 88 00:20:09,789 --> 00:21:00,210 through a choke nozzle. Now we know that m. which is the mass flow 89 00:21:00,210 --> 00:21:14,799 rate is given / ?tAt Vt where T as I said earlier indicates throat conditions. So this 90 00:21:14,799 --> 00:21:24,150 is the density into area x velocity now let me normalize ?T and VT with respect to chamber 91 00:21:24,150 --> 00:21:29,519 conditions why because I know for a choked flow that it is only dependent on the upstream 92 00:21:29,519 --> 00:21:36,169 conditions upstream of the throat of the chamber so if I normalize it with chamber conditions 93 00:21:36,169 --> 00:22:02,419 I will be able to derive some useful equations ?T and VT with ?C which is the density at 94 00:22:02,419 --> 00:22:23,249 in the chamber and At which is the acoustic speed or the speed of sound at the throat. 95 00:22:23,249 --> 00:23:05,149 So we will get m.= ?T / ?C x At xVtL / At x what is At I know that At = right so I will 96 00:23:05,149 --> 00:23:22,940 normalize this also with the chamber conditions and multiplied /under v TC what is this ratio 97 00:23:22,940 --> 00:23:28,529 is the ratio of local speed fluid speak to acoustic speed which is nothing but Mach number 98 00:23:28,529 --> 00:23:57,169 Mach number at throat for choked condition is 1 so this is 1 so we get m. = ?T / ?C x 99 00:23:57,169 --> 00:24:15,210 At okay so now we need to find this ratio and this ratio we already know for choked 100 00:24:15,210 --> 00:24:18,940 flow what is this ratio we also know pressure ratio. 101 00:24:18,940 --> 00:24:39,749 So density ratio is nothing but ?T / ?C is nothing but P t / Rt /tc so R cancels out 102 00:24:39,749 --> 00:25:11,809 you get Pt / Pc x Tc / Tt we have already derived them that is nothing but x ? + 1 / 2 103 00:25:11,809 --> 00:25:16,690 this is the temperature ratio this is the pressure ratio so now if we because these 104 00:25:16,690 --> 00:25:28,019 two are the same we can add the powers and we can simplify it as ?T / ?C I will get it 105 00:25:28,019 --> 00:25:59,179 as 106 00:25:59,179 --> 00:26:22,499 I also note Tt / Tc which is nothing but. I need under v so this would be- ½ now I 107 00:26:22,499 --> 00:26:29,250 know all the ratios that I wanted these two I can substitute them and get the relation 108 00:26:29,250 --> 00:26:41,080 for mass flow rate as M0 = I am sorry you do not remind me I needed to multiply/ if 109 00:26:41,080 --> 00:26:54,669 I divide this / ?c I need to multiply / ?c again so this is ?c x v y / ?c x this so I 110 00:26:54,669 --> 00:27:43,679 will get here ?c At okay now I can deal with them because their same base but different 111 00:27:43,679 --> 00:28:22,950 powers I can add the powers. So I will get 112 00:28:22,950 --> 00:28:34,149 and further this - I can replace it as + and this ?c I can write it in terms of pressure 113 00:28:34,149 --> 00:28:58,039 so I will get PC /RTc x At ? RTc because I have a - sign here I can write this as 2 / ? + 114 00:28:58,039 --> 00:29:16,279 1 x ? + 1 now I have RTc here and RTc here so I can find one of this and write it as 115 00:29:16,279 --> 00:29:57,029 okay this is very similar to something that you have already derived in gas dynamics where 116 00:29:57,029 --> 00:30:08,429 and you will not use Pc you will instead use P0 A* / T 0right this whole function this 117 00:30:08,429 --> 00:30:29,430 whole thing is a function of only ? so normally in rocket literature this is denoted as a 118 00:30:29,430 --> 00:30:46,409 ? of ? function. So you get m. = PcAC/ c* where C* is called 119 00:30:46,409 --> 00:31:09,340 the characteristic 120 00:31:09,340 --> 00:31:46,539 velocity and Z* is given / 1 /? of ? x Ru/ is nothing but R so C* is given / this where 121 00:31:46,539 --> 00:32:36,080 ? of ? is only a function of function of ratio specific heats that is okay 122 00:32:36,080 --> 00:32:42,279 so this is the typical expression that we use in Rockets when we are discussing Rockets 123 00:32:42,279 --> 00:33:01,590 for mass flow rate through a tube nozzle this ? of ? is also called as a random capture 124 00:33:01,590 --> 00:33:47,590 function and varies from 0.642, 0.67 for ? variation of 1.3. 125 00:33:47,590 --> 00:33:55,080 Now we know how to get the mass flow rate through a joke nozzle now if you are collect 126 00:33:55,080 --> 00:34:01,509 back we had derived expression for the thrust of the rocket motor when we were discussing 127 00:34:01,509 --> 00:34:08,150 about the thrust of the turbojet engine so or an air breathing engine. 128 00:34:08,150 --> 00:34:37,760 So 129 00:34:37,760 --> 00:34:58,310 we know that 130 00:34:58,310 --> 00:35:27,940 rocket engine f = m. Ue+ Ae Pe - Pa where e indicates exit conditions a indicates ambient 131 00:35:27,940 --> 00:35:42,740 condition so you have here Ue which is the exit velocity exit area exit pressure and 132 00:35:42,740 --> 00:35:55,270 ambient pressure now we in the thrust equation we already know how to calculate m. for a 133 00:35:55,270 --> 00:36:01,610 choked nozzles o this part we know we need to now find out how to get the exit velocity 134 00:36:01,610 --> 00:36:05,860 and then we need to get something about area ratios and other things we will look at it 135 00:36:05,860 --> 00:36:13,500 a little later firstly let us get an expression for the exit velocity. 136 00:36:13,500 --> 00:36:34,460 How do we get this expression we know that the energy equation is valid from the entry 137 00:36:34,460 --> 00:36:41,340 of the nozzle to the exit of the nozzle so from that we know we can write from energy 138 00:36:41,340 --> 00:37:07,790 equation he + Ue2 / 2 =hc we had seen a little earlier with respect to the throat this equation 139 00:37:07,790 --> 00:37:30,210 now we apply it to the exit of the nozzle okay so I can rewrite this as Ue = under v2 140 00:37:30,210 --> 00:37:48,330 hc – he since we know that there are no reactions that are taking place inside the 141 00:37:48,330 --> 00:37:53,460 nozzle and the thermodynamic properties are constant from the entry of the nozzle to the 142 00:37:53,460 --> 00:37:58,100 exit of the nozzle all this must be sensible enthalpy. 143 00:37:58,100 --> 00:38:20,650 So therefore we can write this in terms of Ue = 2CpTc x 1 - T/ tc + e / tc we can express 144 00:38:20,650 --> 00:38:57,480 it in terms of pressure as = under v ? – 1 this part CpTc part I can use e Ue CpTc I 145 00:38:57,480 --> 00:39:18,620 know = Cp / r x r Pc now this RTc what is C * C * is nothing but RuTc / molecular weight 146 00:39:18,620 --> 00:39:36,710 or RTC right this is nothing but = 1 / ? of ? x underv RTc so this part here RTc must 147 00:39:36,710 --> 00:39:54,980 be = C *2 ?2 okay so I can put this what is Cp / up this I can write it as ? / ? – 1 148 00:39:54,980 --> 00:40:59,790 x so if I substitute back this into the expression for u I will get V = because this has square 149 00:40:59,790 --> 00:41:09,020 here and this is in this v I can take it out and write it like this. 150 00:41:09,020 --> 00:41:14,530 Now what this tells us is that if we know the characteristic velocity and the pressure 151 00:41:14,530 --> 00:41:27,290 ratio we can get the exit velocity right so if Pe goes to 0 what happens if this will 152 00:41:27,290 --> 00:42:02,130 Ue will be a maximum when Pe goes to 0 so when e = 0 U is maximum and is called 153 00:42:02,130 --> 00:42:15,540 limiting exhaust velocity that is it will only become a function of the chamber temperature 154 00:42:15,540 --> 00:42:24,840 the becoming 0 here means in this case if P goes to 0T U also has to go to 0 so the 155 00:42:24,840 --> 00:42:39,540 Ue would this then become under v that is it becomes only a function of the chamber 156 00:42:39,540 --> 00:42:45,650 temperature but in reality. We cannot get a condition where in the exit 157 00:42:45,650 --> 00:42:53,040 velocity this is not the ambient pressure this is the exit of the nozzle being at zero 158 00:42:53,040 --> 00:42:58,210 pressure this is not possible this will be some finite number so therefore you cannot 159 00:42:58,210 --> 00:43:02,840 get this condition but this is the maximum that one can get with a convergent divergent 160 00:43:02,840 --> 00:43:28,580 nozzle. Now 161 00:43:28,580 --> 00:43:38,540 we need to also get one more parameter that is how does we still do not know how a and 162 00:43:38,540 --> 00:43:46,310 PE are connected that is if we know the area ratio there from the throat to the exit how 163 00:43:46,310 --> 00:43:54,590 does how does one obtain PE from if knowing PC how can we obtain PT. 164 00:43:54,590 --> 00:44:20,880 So let us do that so let us relate Pe / Pc to Ac/ At to start this what we know is we 165 00:44:20,880 --> 00:44:26,791 know that mass flow-rate once the nozzle is choked and the flow is steady the mass flow 166 00:44:26,791 --> 00:44:42,210 rate at any cross-section is the same and cannot vary so we know m. = Pc At / C* and 167 00:44:42,210 --> 00:44:57,920 is also = ?e Ue Ae where e indicates exit conditions so from this I can get an expression 168 00:44:57,920 --> 00:45:40,210 for Ae / At as Pc okay now I know ?e is nothing but ?e Pc/RTE and if I substitute for that 169 00:45:40,210 --> 00:45:47,440 here I will I also know the expression for Ue which we just derived okay. 170 00:45:47,440 --> 00:46:15,010 So if we substitute for both of them here ?e and Ue we will get an expression for A/ 171 00:46:15,010 --> 00:47:07,170 At eA this is the expression that we get once we substitute for Ue and ?e now here what 172 00:47:07,170 --> 00:47:17,550 is again C* x if we take a ? of ? here so you get C* x ? of ? 2 which is nothing but 173 00:47:17,550 --> 00:47:51,320 RTC right so we get Pc /Pc x RTe divided / RTc that is we need to have a ? of ? in the numerator 174 00:47:51,320 --> 00:48:16,670 to account for this x 2 ? / ? – 1 okay I can cancel the R’s and Te / Tc I can again 175 00:48:16,670 --> 00:48:23,180 express in terms of pressures I already have a Pc / Pe. 176 00:48:23,180 --> 00:48:37,620 So I know that Te / Tc is nothing but Pe / Pc to the power of ? - 1 / ? so if I substitute 177 00:48:37,620 --> 00:49:44,500 for all this for this here and simplify or I can. 178 00:49:44,500 --> 00:50:33,520 Because I can put them together and rewrite this as now if you notice this is a relation 179 00:50:33,520 --> 00:50:42,590 connecting area ratios to pressure ratios mostly we will know what is the geometric 180 00:50:42,590 --> 00:50:49,200 area ratio of the nozzle that we are taking and to get the pressure ratio from knowing 181 00:50:49,200 --> 00:50:55,010 the nozzle nice nozzle area ratio using this equation is very difficult so therefore we 182 00:50:55,010 --> 00:51:00,620 have gas dynamic tables which will give you this or there are plots that will tell you 183 00:51:00,620 --> 00:51:29,010 how T / Pc varies if you vary Ae / At okay we look at it and then next class thank you.