Hello, and welcome to lecture number 11 of this lecture series on jet propulsion. We have been discussing about the various aspects of jet engine cycles. We have discussed about the ideal cycles, the real cycles and also the component performance parameters based on which we were able to derive the real cycles. So, based on our discussion during the last several lectures, you must have had some idea of what is involved in carrying out a cycle analysis. And as I have been discussing earlier, it is also being assumed that you have undergone some sort of training in the form of the ideal cycle analysis, especially if you have undergone the previous course on introduction to aerospace propulsion. So, in today's class as promised in the last class, we shall be discussing a few problems. We will basically be having a tutorial session. We will solve a few problems in today's class and we will see how we can carry out real cycle analysis for a few configurations of the jet engine. So, we will basically be having a problem on a simple turbojet and then I have configured a problem on a turbojet, the same problem with afterburning. We will subsequently solve a problem on a turbofan, one of the configurations of a turbofan and then I also have a problem for you on turboprop engines. And towards the end of the class, I will also give you a few exercise problems which you can try and solve based on our discussion in the previous classes as well as based on the tutorial session that we have today. So, today's class is basically a tutorial session. We will solve a few problems from the jet engine cycles. We will basically be solving real cycle problems and not the ideal cycle. So, the first problem that we have for today's discussion is a simple turbojet. So, let us take a look at the problem statement. So, problem number one states that an aircraft using a simple turbojet engine flies at max 0.8 where the ambient temperature and pressure are 223.3 Kelvin and 0.265 bar respectively. The compressor pressure ratio is 8 and the turbine inlet temperature is 1200 Kelvin. The isentropic efficiencies of the compressor are 0.87, the turbine is 0.90, the intake is 0.93, the nozzle is 0.95, the mechanical efficiency is 0.99, the combustor efficiency is 0.98. The pressure loss in the combustor is 4 percent of the compressor delivery pressure. Determine the thrust and specific fuel consumption. So, this is the first problem that we shall be attempting to solve today. It is basically on a simple turbojet engine. A simple turbojet as you know is a turbojet engine which does not have any form of after burning. So, which means that there is no reheating in this particular cycle. So, we have been given a lot of data corresponding to this particular turbojet engine. We have the compressor pressure ratio, the turbine inlet temperature and all the efficiencies of the components like the compressor, the turbine, diffuser, nozzle and so on. We also have been given the ambient conditions and the flight Mach number. So, based on this data we should be able to carry out a cycle analysis and determine what is the thrust and specific fuel consumption of such an engine. So, given this particular problem statement, where do we first begin to solve this problem? So, as you must have already realized by now, the first step towards solving such a problem is to get the cycle diagram first that is on either a temperature entropy plot or on a pressure volume plot. So, once we get the cycle diagram right and also we mark the points where we have the data with us and we would therefore be able to determine which are those points where we need to find the properties of temperature and pressure and therefore we will help us in determining the exhaust velocity and hence the thrust and fuel consumption. So, the first step towards solving this problem is to draw the cycle diagram. So, this is a turbojet cycle without after burning as this problem does not have this particular jet engine that we have does not have any after burning. So, this is the real turbojet cycle that we had discussed in the last class on a temperature entropy plot. So, I will just quickly review the different processes involved in this particular turbojet cycle. The ambient conditions are denoted by a subscript a, station 2 corresponds to the intake exit or the diffuser exit which is also the compressor inlet, station 3 is the compressor outlet, station 4 is turbine inlet, 5 is turbine outlet and 7 is the nozzle exit. So, process between a and 2 is the compression in the intake or the diffuser and the line shown by this dotted line is the real process of the actual process and because it is an actual process and because they have been given a diffuser efficiency it means that that process is not isentropic. So, process a to 2 is non isentropic the compression begins at process from station a between 2 and 3 is the compressor again this is also a non isentropic process which is why we have this dotted line which indicates the compression process. Between process states 3 and 4 we have the combustion chamber or the heat addition and we have been given the combustion efficiency as well as the pressure loss in the combustion chamber. So, this process is no longer a constant pressure process. So, there is a pressure loss occurring in the heat addition process or the combustion process. Between stations 4 and 5 is the turbine and the turbine process expansion process is non isentropic and that is why the expansion process is indicated by this dotted line. Between stations 5 and 7 is the nozzle and this again is a non isentropic process and basically we get the exhaust thrust as a result of expansion through the nozzle. So, these are the various processes that are that basically constitute the turbojet cycle and what we shall try to do now is that we will identify those points where we have been given the data and then we will get some idea of which are those points where we need to find out temperature and pressure and because that is what will be required for calculating the exhaust velocity. So, we have been given the ambient velocity ambient temperature and pressure and the Mach number and then the diffuser efficiency is given the compressor pressure ratio is given then we have the turbine inlet temperature. So, these are the different property or parameters which have been specified in the problem besides of course, the efficiencies of all the components. So, based on the ambient temperature and pressure and the Mach number we can calculate the ambient stagnation conditions and also because intake efficiency is given we can calculate the intake exit conditions like it is stagnation temperature and pressure and then we have the compressor pressure ratio. Therefore, compressor inlet conditions being known we can find out the compressor exit conditions from the pressure ratio and the efficiency. Then we arrive at the combustion chamber where again we have been given the efficiency, the pressure loss and the exit conditions of the combustor that is the turbine inlet temperature. So, from the combustor analysis we can find out the fuel to air ratio and then we have the turbine. Turbine inlet condition is known and so we can find out by equating the turbine work to the compressor work the turbine exit conditions and also the exhaust nozzle which will be solved. So, these are the steps that will be that are involved in solving this particular problem. So, let us begin with the intake where we have the intake inlet conditions that is the ambient temperature pressure Mach number and the intake efficiency. So, for the given ambient conditions that is the temperature and pressure and the Mach number we can calculate the flight speed because flight speed will be equal to Mach number into the speed of sound that is square root of gamma RT. So, from that we can calculate the flight speed which comes out to be 239.6 meters per second. So, this is calculated because the Mach number is known and the ambient temperature is known. So, we can calculate the speed of sound which is square root of gamma RT that multiplied by the Mach number gives us this the flight speed. And so once we know the flight speed or the Mach number we can calculate the intake exit stagnation conditions that is stagnation temperature which is the ambient temperature plus V square by 2 C p. And so we the ambient temperature is known the velocity is known and the C p for air we will be assuming as 1.005 kilo joules per kilogram Kelvin. And for the exhaust products or the combustion products we will assume it to be 1.147 kilo joules per kilogram Kelvin. So, based on these when we substitute these values and simplify we get the stagnation condition as 251.9 Kelvin. Now, similarly this stagnation pressure is given by this stagnation is related to the stagnation temperature through the isentropic efficiency of the diffuser. So, P02 by P a is equal to 1 plus diffuser efficiency multiplied by T02 by T a minus 1 raise to gamma by gamma minus 1. So, diffuser efficiency is given to us it is 0.93 stagnation temperature we have just now calculated ambient temperature is known and gamma for air is 1.4. So, if we substitute these values we get P02 by P a is equal to 1.482. And since ambient pressure static pressure is known stagnation pressure can be calculated by P a multiplied by 1.482 that is 0.265 into 1.482 that is 0.393 bar. So, what we have calculated now are the exit conditions of the intake which will basically be the inlet conditions of the compressor. So, based on these conditions that correspond to the compressor inlet we can now proceed towards calculating the compressor exit conditions. And how do we calculate that? Compressor exit pressure is straight forward because the pressure ratio is specified. So, given the pressure ratio we can calculate the compressor exit stagnation pressure which is basically the stagnation pressure multiplied at the inlet of the compressor multiplied by the pressure ratio. So, P03 will be equal to pi c which is the compressor pressure ratio multiplied by P02. And how do you calculate the temperature? Temperature is basically calculated using the compressor isentropic efficiency definition which is eta c is equal to T03 minus T02 T03 s minus T02 divided by T03 minus T02. And if we simplify this we get an expression in terms of the pressure ratio and therefore, we can calculate T03. We have discussed that during the last lecture when we took up the cycle analysis of the turbojet. So, compressor exit pressure is basically just the product of the pressure ratio and the inlet stagnation pressure which is already calculated as 0.393 bar. And so that multiplied by the exit by the pressure ratio that is 8 gives us the exit stagnation pressure. So, compressor exit stagnation pressure is 3.144 bar and exit stagnation temperature is from the efficiency definition T03 is equal to T02 multiplied by 1 by eta c which is the isentropic efficiency of the compressor multiplied by pi c raise to gamma minus 1 by gamma minus 1 plus 1. So, this basically gives us the stagnation temperature at the compressor exit. So, all these parameters on the right hand side are known to us now T02 we have calculated in the previous expression in previous slide eta c is specified pi c is the pressure ratio gamma is 1.4. So, we substitute for all these values in this expression on the right hand side we should be able to get an expression for we should be able to find out the compressor exit stagnation temperature. So, substituting for all these values we get T03 is equal to 486.8 Kelvin. So, this is the stagnation temperature at the compressor exit. So, compressor exit stagnation temperature we have related that to the pressure ratio and the efficiency based on which we can calculate the compressor exit conditions. So, the next component. So, we are now solved or carried out the cycle analysis up to the compressor exit and. So, we now have data up to the compressor exit in terms of the stagnation pressure and the stagnation temperature. We also have which is basically the combustion chamber inlet conditions and for the combustion chamber we have the combustion efficiency which has been specified and the pressure loss taking place in the combustion chamber. And what else we have we also have the combustor exit temperature stagnation temperature which is the turbine inlet temperature. So, based on this much data that we have we should be able to calculate the fuel to air ratio and how do we calculate that if you recall in the last lecture we had discussed to calculate the fuel to air ratio we just carry out an energy balance from between the inlet and exit. So, energy balance will be able to help us in calculating the fuel to air ratio. So, enthalpy at inlet which is the an enthalpy of the air coming in from the compressor plus the enthalpy or heat of reaction of the fuel is equal to enthalpy at the exit of the combustion chamber. So, we solve this and we should be able to get the fuel to air ratio. So, h 0 4 which is the enthalpy at the exit of the combustion chamber which is the turbine inlet is equal to h 0 3 which is the compressor exit plus the burner efficiency or combustion chamber efficiency multiplied by the fuel to air ratio and the heat of reaction of the fuel we should be able to solve this and get the fuel to air ratio. So, C p times T 0 4 is equal to C p a times T 0 3 plus eta b into f times q dot f. So, C p g here corresponds to the specific heat of the combustion products C p a corresponds to specific heat of air. So, all these numbers are known to us now. So, f we can simplify and that would be equal to C p g by into T 0 4 divided by C p a into T 0 3 minus 1 divided by burner efficiency or combustion efficiency into q dot f divided by C p a T 0 3 minus C p g T 0 4 by C p a T 0 3. Now, we will assume a value of q dot f because it is not explicitly stated in the problem. Usually for jet engines the fuel that is used is known as aviation turbine fuel and that is a fuel which is similar to kerosene and for that particular fuel the q dot f or heat of reaction is about 44 mega joules per kilogram. So, we will assume that value here and the rest of the values are already known to us. So, we substitute for all these values and if we substitute for all of them T 0 4 is basically the turbine in the temperature which is given T 0 3 is compressor exit stagnation temperature which we have calculated and C p a and C p g we have known eta b burner efficiency is also given. So, f is equal to if you simplify substitute for all these values and simplify we get 0.0198. Now, so this is the fuel to air ratio what else is required to be calculated we need to now calculate the combustion chamber exit stagnation pressure exit stagnation temperature is known that is the turbine inlet temperature exit stagnation pressure is equal to the pressure loss in the combustion chamber multiplied by the inlet stagnation pressure. It is given in the problem that there is 4 percent loss occurring in the combustion chamber. So, pi b is basically equal to 0.96. So, p 0 4 will be pi b multiplied by p 0 3 that is 0.96 into 3.144. So, that is 3.018 bar. So, this is how we have estimated and calculated properties all the way from the inlet then proceeding towards the compressor then the combustion chamber. The next component is the turbine we have the turbine inlet stagnation temperature and pressure and we also have the mass flow rate in the sense that we have mass flow rate of air plus the mass flow rate of fuel we have just calculated the fuel to air ratio. So, we should now be able to calculate the fractional mass flow rate that is actually going into the turbine. So, all these conditions are known for the turbine. Now, what we need to calculate are the exit conditions of the turbine and how do we calculate that remember in a simple turbojet engine the basic function of a turbine is only to drive a compressor that means the turbine is generating just enough work to drive the compressor. So, turbine work the work output of a turbine will be equal to the work input of the compressor and of course, there is an if there is a mechanical efficiency involved that means the work output of the turbine is getting diminished by a certain fraction which is equal to the mechanical efficiency. So, mechanical efficiency multiplied by the turbine work will be equal to the compressor work. So, thus if you simplify we should be able to calculate the compressor well the turbine exit stagnation temperature and then from the efficiency definition we can calculate the stagnation pressure and. So, that is how we will be proceeding towards calculating turbine exit conditions. So, let us do that and see about the turbine exit conditions from all three. Now, since we know that the turbine exit as well the work done by the turbine is equal to that of the compressor we know that the mechanical efficiency multiplied by m dot which is m dot mass flow rate of air plus mass flow rate of fuel this sum of this multiplied by C p of the gases into the temperature difference across the turbine that is T 0 4 minus T 0 5 is equal to m dot which is mass flow rate of air into C p of air into T 0 3 minus T 0 2 which is the differential temperature across the compressor. So, left hand side is the work done by the turbine right hand side is the work done by work required by the compressor. So, let us simplify this once we simplify this we in this equation the only unknown is T 0 5. So, if you simplify this and we get an expression in terms of T 0 5. So, T 0 5 on simplification is equal to C p into well C p g into T 0 4 minus C p a into T 0 3 minus T 0 2 divided by mechanical efficiency into m 1 plus f because we cancel out m dot. So, we get 1 plus f. So, the entire all the terms which are involved on the right hand side are known from our cycle analysis so far. So, we have C p g which is 1 1 4 7 into T 0 4 which is 1200 minus C p a which is 1005 multiplied by T 0 3 that is 486.8 minus T 0 2 251.9 divided by mechanical efficiency is 0.99 into 1 plus f that is 1 plus 0.0198. So, if you calculate this we will get the exit temperature as 992.3 Kelvin. So, this is this stagnation temperature at the turbine exit. Now, how do we calculate this stagnation pressure at the exit? We will now make use of the efficiency definition of the turbine. Well efficiency of a turbine, isentropic efficiency of a turbine is defined as eta t is equal to the inlet temperature stagnation temperature T 0 4 minus T 0 5 which is the actual temperature divided by T 0 4 minus T 0 5 isentropic. Now, if we divide this and convert them to temperature ratios in the denominator we have the isentropic temperature ratio T 0 5 s divided by T 0 4 which is equal to T 0 5 by T 0 4 raise to the gamma minus 1 by gamma. So, from this simplification we should be able to get an expression for T 0 5 which is the turbine exit stagnation pressure. We have already derived that in the last lecture and so T 0 5 can be expressed in terms of the efficiency and the temperature ratios. So, if we do that we get T 0 5 is equal to T 0 4 multiplied by 1 minus 1 by eta t which is the turbine efficiency into 1 minus T 0 5 by T 0 4 this raise to gamma by gamma minus 1. So, all these values are known to us T 0 4 is known turbine efficiency is known and these two temperatures are also known. And remember we are going to use gamma here as 1.33 and not 1.4 because we are now dealing with the combustion products and so gamma is going to assume an average gamma for all the combustion products and that is going to be assumed as 1.33 and for the average gamma for air is 1.4. So, if you substitute for all these values here and simplify we get T 0 5 which is the turbine exit stagnation pressure is equal to 1.284 bar. So, this is the pressure of at the exit of the turbine. So, we have data now all the way up to turbine exit we also have now the turbine exit stagnation pressure and the temperature. And in this particular engine configuration there is no after burner and since there is no mention of any other losses after the turbine except the nozzle efficiency we can assume that all these parameters which we have calculated for the turbine exit is valid for the nozzle in net also. That means T 0 5 and P 0 5 will be true or will be the same at the nozzle entry. So, we will use these same values at the nozzle entry to calculate the nozzle exit conditions and therefore, the velocity of the jet. But, before that we have to make one check which is basically to see if this nozzle is operating under choked conditions or not. Because if it is choking then we the exit conditions get fixed by the critical parameters that is the exit temperature will be the critical temperature exit pressure will be the critical pressure and density will be critical density and so on. So, based on which we need to now calculate the other conditions. If it is un choked then the nozzle exit conditions are to be calculated using the enthalpy drop across the nozzle something we have done in the last class during the cycle analysis discussion. So, let us check for the choking of the nozzle. So, we first check for nozzle choking and how do you do that if you are familiar with this ideal cycle analysis you must have had some experience of calculating the nozzle pressure ratio and to check whether it is indeed choking. The nozzle pressure ratio is basically P05 by P a P05 is known from our previous calculation that is 1.284 and P a is given as 0.265 bar. So, P05 by P a is equal to 4.845 and that is it nozzle pressure ratio and what is the critical pressure ratio that is P05 by P c how do we calculate this critical pressure ratio this is basically calculated by equating Mach number is equal to 1 and so, we can from isentropic relations we should be able to calculate the critical pressure ratio. But in this case we have been given a nozzle efficiency as well. So, there is a nozzle efficiency that also comes into picture here. So, this basically can be derived when we either from the nozzle efficiency definition or from the pressure ratio definition. So, critical pressure ratio is 1 by 1 minus 1 by eta n into gamma minus 1 by gamma plus 1 this raise to gamma by gamma minus 1. So, if you substitute for all these values we have the efficiency and gamma which is 1.33 then we get a critical pressure ratio of 1.914. So, here we have now two pressure ratio one is the actual nozzle pressure ratio which is 4.845 and the critical pressure ratio which we have calculated as 1.914. So, what we see here is that the nozzle pressure ratio is greater than the critical pressure ratio and what does that mean? That is nozzle pressure ratio being greater than nozzle the critical pressure ratio means that the nozzle is choking. And why should that be because from the pressure ratio that we have it means that the nozzle entry pressure is much higher than the critical pressure that is it will expand only up to the critical pressure ratio. If you have a pressure ratio higher than that it will still mean that the exit conditions of the nozzle are choked or fixed because it is a convergent nozzle. So, because it happens to be convergent nozzle then the exit conditions are fixed that is Mach number will become 1 mass flow rate will be maximum mass flow rate and therefore, it is choking. And so, if it is choking then the nozzle exit conditions are fixed and they are fixed because those conditions will be equal to the critical parameters that is the critical temperature and the critical pressure and the density. That will also fix the exhaust velocity because since Mach number is equal to 1 the exhaust velocity will now be equal to square root of gamma r t which is this period of sound based on the critical temperature. So, in this problem we now have a scenario where the nozzle is choking and therefore, we now will calculate the nozzle exit conditions based on the critical parameters. So, let us see how we can calculate the critical parameters. So, we will calculate the critical temperature pressure and density. So, nozzle exit conditions will get fixed by critical parameters. So, the critical temperature that is static temperature T 7 will be equal to critical temperature T c which is 2 by gamma plus 1 into T 0 5 and it becomes 2 by gamma plus 1 because of the temperature ratios that is T 0 5 by T c is 1 plus gamma minus 1 by 2 m square and if we put m is equal to 1 we get this expression. So, T c is 2 by gamma plus 1 into T 0 5. So, if we substitute for all these values we get 850.7 Kelvin. Similarly, P 7 or P c is equal to P 0 5 multiplied by 1 by P 0 5 by P c therefore, this comes out to be 0.671 bar. The density rho 7 is equal to P 7 by R T 7 this on simplification we get 0.275 kilograms per meter cube. So, we have temperature pressure and density at the exit. So, exhaust velocity V exit is equal to square root of gamma R T 7 and we substitute and we find out that the exhaust velocity is 570.5 meters per second. So, this is the exit velocity we can also calculate from the data that is known to us this particular parameters that is the area to mass flow rate ratio because this will be required for calculating thrust. So, exit area by mass flow rate is equal to 1 by rho 7 into V exit because mass flow rate is rho into area into velocity. So, this parameter that is area divided by mass flow rate is 0.006374 meter square second per kilogram. Therefore, we now have the exit velocity the flight Mach number and the pressures that is P c minus P f critical parameter pressure is known that is exit pressure is known ambient pressure is also known and they are not equal. And so, if we substitute for all these values the specific thrust we can calculate as 1 plus f into V exit minus V plus a e by m dot into P c minus P a. So, if you substitute all the values we get 596.25 Newton second per kilogram. So, this will be the thrust specific thrust which is Newton's per that is thrust per unit mass flow that is 596.25 Newton second per kilogram. So, we have calculated the thrust what is the next parameter to be calculated that is fuel consumption specific fuel consumption. So, specific thrust we have calculated and that get that is basically 596.25 Newton second per kilogram. And the next parameter to be calculated is the fuel consumption the specific fuel consumption we have already calculated the fuel to air ratio. So, that divided by the specific thrust will give us the specific fuel consumption. So, 0.0198 is the fuel to air ratio that divided by 596.25. So, specific fuel consumption comes out to be 3.32 into 10 raise to minus 5 kilograms per Newton second. And sometimes this is also converted to kilograms per Newton hour that is multiplied by 3600. And so, but basically I have expressed it in the SI units here. So, we have now completed the cycle analysis for this first problem which was basically a simple turbojet engine without any after burning. And the given a certain amount of data we were able to calculate the different temperatures and pressure across various components finally, leading to the nozzle exit conditions and the nozzle exhaust velocity. And so, using that we should be we were able to calculate the specific thrust and also the specific fuel consumption. So, this is how we would be carrying out real cycle analysis based on the data that is provided to us. Of course, with certain assumptions like we have assumed a constant specific heat for air all the way up to compressor exit. And we have also assumed an average specific heat for the combustion products. And similarly, the ratio of specific heats and so on. And so, the next problem that we are going to take up is basically on the same turbojet, but if the turbojet were to be operating in an after burning mode. That is if this particular turbojet is to operate with an after burner then what would be the corresponding thrust and fuel consumption. So, the problem statement for the second problem statement is the determine the thrust and specific fuel consumption in the above problem if the engine operates with an after burner. The nozzle inlet temperature in this case is limited to 1800 Kelvin. All other parameters and operating conditions remain unchanged. That is if we assume that other parameters can be fixed as it is in the previous case. The only change is that the after the turbine exit we now have an after burner that is additional fuel is added in the after burner taking the temperature to 1800 Kelvin. In which case what will be the new value of the thrust as well as the fuel consumption. So, before we start solving this problem we will first take a look at the cycle diagram like we did in the previous case. So, we will look at turbojet with after burning. And then we shall attempt to solve this problem and find out the thrust or the increase in thrust and fuel consumption. So, for an after burning turbojet real turbojet cycle with after burning the cycle diagram up to 5 is the same as what we discussed in the previous problem because it is the same turbojet engine. Now, at the exit of the turbine instead of a nozzle now we have an after burner. After burner will increase the temperature and so the temperature which was at T 0 5 at the exit of the turbine will now increase and reach T 0 6 A that is temperature after the exit of the after burner or the nozzle entry. And after the after burner we have the nozzle which expands the combustion products and generates the thrust. So, we can see that there is a substantial increase in the temperature that is expected in the case of an after burning turbojet. The turbine only temperature in that problem was 1200 Kelvin whereas, the temperature at the exit of the after burner is specified as 1800 Kelvin. So, we can afford to have higher temperatures in the after burner and therefore, we should be able to get a substantial increase in thrust. So, let us see how much increase in thrust we should be able to achieve. So, all other it is also mentioned that the operating conditions or operating parameters remain unchanged up to the turbine exit which means that the cycle analysis of up to the turbine exit is exactly the same. And so we will not repeat the same calculations here and so we will now need to see if the increase in temperature what kind of how would it affect the performance. So, the nozzle will continue to be choked because we had calculated the nozzle pressure ratio and so we have we had seen in the previous case that the nozzle was choking which means that even if you use an after burner the nozzle will continue to be choked. But we will have an increase in this stagnation temperature, but we will see if that makes a difference in the thrust. So, what else will change we will have an additional fuel flow rate in the after burner because we will have to add additional fuel in the after burner which is what will lead to this increased temperature. Therefore, the fuel consumption is going to change and the total fuel flow rate will now be equal to the fuel you added in the combustion chamber that is main combustor plus the fuel added in the after burner. So, this will directly impact the fuel consumption, but besides this there will also be a change in the thrust. So, to calculate the fuel flow rate which is what we will calculate first in the after burner because the rest of the parameters up to the turbine exit remains unchanged. So, if we will use the same principle as we did for the main combustor that is basically an energy balance across the after burner. So, an enthalpy of the after burner exit is h 0 6 which is equal to h 0 5 plus the efficiency or burner efficiency into fuel flow rate multiplied by the heat of reaction or the calorific value of the fuel. So, this on simplification we get f 2 which is the fuel added in the after burner is equal to C p into t 0 6 by C p g into t 0 5 minus 1 divided by the burner efficiency into calorific value divided by C p into t 0 5 minus C p t 0 6 by t 0 5. So, all these numbers are known to us t 0 6 is specified as 1800 Kelvin other parameters are known from our previous cycle analysis. So, if you substitute all these values we get f 2 which is fuel flow rate in the after burner is equal to 0.0 2256 and therefore, the total fuel flow rate f is equal to f 1 plus f 2 where f 1 is the fuel to air ratio in the main combustor that plus f 2 in the after burner. So, the total fuel flow rate is 0.04236. So, there is as you can see more than 50 percent more than 100 percent increase in the fuel to air ratio in this turbojet with after burning. Now, the nozzle exit conditions we will now calculate we have already calculated or and seen that the nozzle is operating under choked condition. So, the nozzle exit temperature critical temperature will now be equal to 2 by gamma plus 1 into t 0 6 where t 0 6 is given as 1800 Kelvin. So, this is 1545.06 Kelvin similarly, the exit pressure which is the critical pressure is equal to p 0 5 by 1 by p 0 5 by p c which is 0.671 bar which is the same as what we calculate in the last problem. The density will now change because the temperature has changed that is p 7 by r t 7 that is 0.151 kilograms per meter cube. Therefore, the exhaust velocity will be equal to square root of gamma r t 7 where t 7 is something we have already calculated. So, this is equal to 787.9 meters per second. So, at the nozzle exit we now have the exhaust velocity and also the critical parameters. So, we now calculate the specific thrust in the same way as we calculated in the previous case. We have the specific thrust is equal to 1 plus f into v exit minus v plus a m dot into p c minus p a. Here f is different from what we calculated in the previous problem and so is the exhaust velocity. So, if you substitute these values we get the specific thrust as 912.56 Newton second per kilogram. Similarly, the specific fuel consumption is f divided by specific thrust this comes out to be 4.64 into 10 raise to minus 5 kilograms per second Newton second. So, if you compare these values with what we had calculated in the first problem we will see that after burning indeed leads to a very substantial increase in thrust. We get about 35 percent increase in the thrust and this is also accompanied by a corresponding increase in the fuel consumption. It is with a cost of 28 percent increase in the fuel consumption. So, this is just to highlight that after burning is something that is used to increase the thrust or to achieve momentary increase in thrust and also to sustain high Mach numbers in supersonic flights. So, after burning turbo jets are usually used in these applications when there is a need to increase thrust momentarily. So, since we have seen that after burning can also lead to an increase in fuel consumption it is not something that is used in civil aviation turbo fans normally do not operate with after burning. So, after burning is usually used or limited to military engines. So, we have now solved two problems related to turbo jet engines. One was a simple turbo jet engine without any after burning and the second problem was an extension of the first problem with after burning and we have seen how we can calculate and carry out a cycle analysis realistic cycle analysis because we have efficiencies of all the components and how the cycle analysis can be carried out in a systematic manner. So, we will now take up another version of a jet engine we will now discuss about a turbo fan engine. We will solve one problem on turbo fan engine followed by another problem on a turbo prop engine. So, let us take a look at the third problem. Problem number three states that at twins pool unmixed turbo fan engine has the fan driven by the LP turbine and the compressor driven by the HP turbine. The overall pressure ratio is 25 and the fan pressure ratio is given as 1.65. The engine has a bypass ratio of 5 and the turbine in the temperature of 1550 Kelvin. The fan turbine and the compressor have polytropic efficiencies of 0.9. The nozzle efficiency is 0.95 and the mechanical efficiency for each spool is 0.99. The combustor pressure loss is 1.5 bar and the total air mass flow rate is 215 kilograms per second. Find the thrust under the sea level static conditions where ambient pressure and temperature are 1 bar and 288 Kelvin. So, in this problem which is to do with an unmixed turbo fan which means that there are two separate nozzles which are used a cold nozzle or the bypass nozzle or the secondary nozzle and the primary nozzle which is like that used in a turbo jet engine. So, we will solve these two streams separately and then the total thrust will be equal to some of the thrust generated by the primary nozzle and this thrust developed by the secondary nozzle. So, the steps that are going to be followed are very similar to what we had done in the case of a turbo jet engine. So, I will go through this problem little more quickly little faster than what we had discussed for the turbo jet because the steps are identical. So, we will begin with the fan and it is mentioned in the problem that we are required to calculate the thrust under static conditions which means that the engine is stationary Mach number or flight speed is 0. Therefore, P 0 1 will be equal to ambient temperature P 0 1 will be equal to ambient pressure and therefore, the fan exit temperature T 0 2 prime is equal to the inlet temperature multiplied by the fan pressure ratio raise to gamma minus 1 divided by in this problem we have been given the polytropic efficiency of the fan. And if you recall during our discussion little earlier when we were talking about the component performance we had discussed about polytropic efficiency of the compressor and we have seen that we can relate the polytropic efficiency to the isentropic efficiency and through the ratio of specific cage. So, we are going to use the polytropic efficiency here. So, we have pi f which is the fan pressure ratio raise to gamma minus 1 divided by the polytropic efficiency of the fan multiplied by gamma. So, this comes out to be 337.6 Kelvin. Now, it is given that the overall pressure ratio is 25 and the fan pressure ratio is known which is 1.65. Therefore, the compressor pressure ratio will be equal to overall pressure ratio divided by the fan pressure ratio. So, we have 25 divided by 1.65 that is 15.15. So, the compressor exit stagnation temperature T 0 3 will be equal to T 0 2 into the fan at the compressor pressure ratio that is pi c raise to gamma minus 1 by polytropic efficiency of the compressor into gamma. So, this temperature comes out to be 800.1 Kelvin. So, we will first calculate the secondary nozzle properties we will calculate the thrust developed by the secondary nozzle and then we will proceed towards the primary nozzle and calculate the thrust developed by that and then add up the two to get the final thrust generated by the turbofan. Now, in the case of nozzle as we have seen we will first need determine whether the nozzle is choking or not. So, we will calculate the nozzle pressure ratio and the critical pressure ratio and see if check if the nozzle is indeed choking. So, the cold nozzle pressure ratio is the fan pressure ratio because that is what is available at the inlet. So, it is 1.65 and critical pressure ratio is P 0 2 prime by P c which is 1 by 1 minus 1 by eta n into gamma minus 1 by gamma plus 1 raise to gamma by gamma minus 1. So, for the cold nozzle we will take gamma as 1.4 whereas for the second nozzle we will take gamma as 1.33. So, if you substitute the nozzle efficiency and the value of gamma we get the critical pressure ratio as 1.965. Now, since the pressure ratio of the nozzle is less than the critical pressure ratio it means that the nozzle is not choking. So, if the nozzle is not choking then we have the nozzle exhaust velocity which will be equal to square root of 2 C p into nozzle efficiency into the temperature T 0 2 prime into 1 minus P a by P 0 2 prime raise to gamma minus 1 by gamma. So, how we have derived this expression we have discussed in earlier lecture on cycle analysis this basically comes from the enthalpy drop across the nozzle and if we simplify we get the exhaust velocity. So, this is equal to 2 into the C p which is 1005 nozzle efficiency temperature and the pressure ratio. So, we get the exhaust velocity at the fan on the secondary nozzle is 293.2 meters per second. Now, it is given that the bypass ratio is 5 therefore, the cold mass flow m dot C is equal to mass total mass flow into bypass ratio divided by B plus 1 that is bypass ratio plus 1 which we get as 179.2 kgs per second. Therefore, the thrust developed by the secondary nozzle is mass flow rate in the exhaust velocity that is 52.532 kilo Newtons. Now, we now move on to the core nozzle or the core engine. So, for the HP turbine we have the work done work done by turbine equated to the work done by work required by the compressor. So, HP turbine drives the compressor. So, the temperature across the HP turbine T 0 4 minus T 0 5 prime is C p a divided by mechanical efficiency into C p g into T 0 3 minus T 0 2. So, we simplify this and we get the HP turbine exit stagnation temperature which is 1141 Kelvin. Similarly, we can calculate the LP turbine exit conditions because LP turbine drives the fan and the fan mass flow rate as you can see here is different from the core mass flow because fan drives a larger amount of mass flow. So, T 0 5 that is LP turbine exit temperature is 877.8 Kelvin. Similarly, we proceed towards finding the exit pressures. We have these pressure ratio P 0 4 by P 0 5 prime which is temperature ratio raised to gamma minus 1 by gamma which is 3.902. Similarly, for the LP turbine and the turbine inlet pressure is equal to compressor exit pressure minus delta P pressure loss in the burner. So, that is 25 minus 1.5 23.5. So, P 0 5 is basically equal to P 0 4 divided by these pressure ratios and so we get 1.878 bar and the hot therefore, hot pressure ratio of pressure ratio of the nozzle hot nozzle is 1.878. The critical ratio pressure ratio for this nozzle is different from that of the primary nozzle from that of the secondary nozzle because gamma is different here gamma is 1.33. Critical pressure ratio is actually 1.914. So, we find that since the pressure ratio is less than the critical pressure ratio this nozzle is also not choking. So, we proceed to find out the exit velocity exhaust velocity here in the same manner as we calculated for the secondary nozzle and that we get 528.3 meters per second and the mass flow rate through the hot nozzle is 35.83. Therefore, thrust developed is 35.83 times the exhaust velocity which is 18.931. Therefore, the total thrust is equal to primary thrust which is developed by the primary nozzle plus that of the secondary nozzle that is 71.5 kilo Newtons. So, this is a problem which we which involved a turbofan engine and an unmixed turbofan engine which was consisting of 2 separate streams each of them generating a thrust and contributing towards the total thrust. The last problem that we will solve today corresponds to that of a turboprop engine where in an aircraft operating on a turboprop engine flies at 200 meters per second while ingesting a primary mass flow of 20 kgs per second. The propeller of the engine having an efficiency of 0.8 generates a thrust of 10000 Newton while the jet thrust is 2000 Newton. The power turbine and the nozzle have efficiencies of 0.88 and 0.92. If you remove the power turbine and nozzle what would be the thrust developed by the engine while operating under the same conditions. So, here we have a turboprop and we have been given all the efficiencies of the propeller the turbine and the nozzle. So, we have required to find out that if we remove the power turbine and nozzle what will be the thrust developed by the same engine operating under the same conditions. So, in this case we will basically be calculating the enthalpy drop across the power turbine plus the nozzle combination and then we see that if you remove the power turbine then the entire enthalpy drop is available for exhaust through the nozzle and that is how we will be able to calculate the thrust developed by this particular engine. So, thrust power that is developed by the propeller we have already discussed this in our last class. Thrust power by the propeller is basically equal to the thrust developed by the propeller into the velocity. This is equal to the propeller efficiency into the power turbine efficiency into alpha delta H into m dot. Therefore, if you simplify this we get alpha delta H as the thrust by the propeller which is given as 10000 Newton into velocity 200 divided by propeller efficiency 0.8 power turbine efficiency 0.88 into m dot which is 20. So, alpha delta H is 142045 joules per kilogram and the nozzle thrust is also given it is 2000 Newton. So, nozzle thrust 2000 is equal to m dot into exit velocity minus v therefore, we get the exit velocity as 300 meters per second. We also know that exit velocity is equal to square root of 2 into nozzle efficiency into 1 minus alpha into delta H. So, from this we can calculate what is delta H because alpha delta H we have already calculated. So, delta H comes out to be 190958.49 joules per kilogram. Now, it is mentioned that now if you remove the power turbine and propeller what will be the thrust which means that this entire delta H will now be available for expansion through the nozzle. Therefore, the exit velocity now will change it was earlier 300 now the exit velocity is not equal to 1 minus alpha delta H, but it is just equal to delta H. So, we get exit velocity is equal to square root of 2 into efficiency of the nozzle into delta H which is equal to 592.76 meters per second. Therefore, the thrust the new thrust will now be equal to 20 that is mass flow rate into v exit minus v that is 592.76 minus 200 that is 7855.19 Newton. So, this is the thrust that will be developed by the nozzle if we were to remove the power turbine and the propeller. So, we have today discussed 4 problems we have solved 4 problems 2 of them related to turbojet one of them to turbofan and another one with turboprop. So, I have now 4 exercise problems for you and the first one is to do with the turbojet engine which is operating with the compressor pressure ratio of 8 turbine let temperature of 1200 and mass flow rate of 15 kg per second the altitude of the aircraft and the speed is given. So, assuming these efficiencies that is polytropic efficiencies of turbine and compressor as 0.87 intake and nozzle efficiency as 0.95 mechanical efficiency as 0.99 combustion efficiency 0.97 pressure loss as 6 percent of compressor delivery we need to calculate the net thrust the area nozzle area required net thrust and specific fuel consumption. So, the answer to that is 0.0713 meter cube meter square 0.7896 Newton which is the thrust and fuel consumption is 0.126 kilograms per Newton hour the second problem is the same problem with the after burning. So, at the turbine exit the gases are reheated 2000 Kelvin and pressure loss is 3 percent calculate the percentage increase in nozzle area required if the mass flow rate is to remain unchanged and also the percentage increase in net thrust that is we need to calculate how much area increase of the nozzle is required if we have to operate this engine keeping the same mass flow. So, the answer to that is 48 percent 3 percent increase in area and 64.5 percent increase in net thrust. Third problem is to do with a turbofan engine a twin spool turbofan engine which is again unmixed turbofan we have overall pressure ratio the fan pressure ratio bypass ratio is given as 3, turbine inlet temperature is 1300 pressure loss is given as 1.25 in the combustor total air mass flow is 115 kgs per second. We need to find out the thrust when the ambient pressure and temperature are 1 bar and 288 Kelvin the fan and compressor turbine efficiencies are 0.9 nozzle has an efficiency both the nozzle have efficiency of 0.95. So, thrust in this case is 47.6 kilo newtons and the last problem is to do with a turbo prop turbo prop operating under sea level conditions with flight speed is 0 air flow entering the compressor is 1 kgs per second compressor pressure ratio is 12 all the efficiencies are given turbine inlet temperature is 1400 Kelvin stagnation pressure leaving the second turbine is 4.6 bar and we need to calculate the horsepower delivered to the propeller that is by the power turbine and the thrust developed by the gases passing through the engine. So, the power developed by the propeller is 632 kilo watts and the thrust developed by the nozzle is 875 Newton. So, these are four different problems that I have sorted out here for you basically to do with turbo two of them to do with turbo jet one to do with a turbo fan and another one with turbo form turbo prop engine all these problems are exactly in line with what we have solved with slight differences here and there. So, I am sure you would be able to solve these problems based on what we had discussed during today's tutorial session. So, that brings us to the end of today's tutorial session and the next class we will be taking up analysis of compressors axial flow compressors we will begin a simplistic analysis of axial flow compressors.


Myself Piyusha Shedgar, Assistant Professor, Walchand Institute of Technology, Solapur. In today's session we will discuss the modes of helical antenna. These are the learning outcomes for today's session. At the end of this session, students will be able to differentiate between different modes of helical antenna. They will be able to list the applications of helical antenna. These contents will be discussed in today's session. So before going to start actually the modes of a helical antenna that is actual working of the helical antenna, you must know what is the helical antenna. So you can pause video here for a second and recall that what is helical antenna. Yes, the helical antenna or it is also called as a helix antenna is the antenna in which the conducting wire is wound in helical shape and connected to the ground plate with a feeder line. Now this is the structure of a helix antenna or helical antenna. So the helical antenna which having the shape of a helix that is the number of turns can be used in shape of a helix with the spacer are provided between each turn. The helical aerial element is used to make the number of turns. The central support is provided at the centrally which is also called as a axis of the helix and the base is used at the left side. So it is also called as a reflector or the ground plate with the feeding line. So here the coaxial cable is used. The inner conductor is connected to the helix whereas the outer conductor is connected to the ground plane. There are the different types of the ground planes. So according to the structure of the helix for designing of the helical antenna to get a proper radiation pattern these elements are considered these elements are nothing but the spacing between the turn, length of the turn or the length of the helix and the circumference. So circumference depend upon the diameter of the helix. Diameter of the helix is given by the capital D and the circumference is given by this capital C. C is given by the equation pi times of the D whereas alpha is nothing but the peach angle. Peach angle made between this normal line of the circumference and its perpendicular to the consideration with the length of the helix. Here the spacing D and the C that is circumference and the diameter of the helix antenna decides the alpha value. So by using the Pythagoras theorem you can write the equation for the length of the helix. It is given by under root s square plus c square whereas the alpha can be calculated by taking the tangent value of the alpha tan alpha is equal to the opposite side by adjacent side. So s upon c where c is given by the pi times of D. So alpha is equal to tan inverse of s by pi D or it is also can be written as tan inverse of s upon c. So this is the structure of the helix antenna. Now consider the different modes of operation of the helical antenna. So there are the different modes of helical antennas but generally the two main modes of operations are there. So one is the normal mode of operation whereas the axial mode of the radiation of helical antenna. So normal mode it is also called as the perpendicular mode of radiation whereas the axial mode is also called as a end fire or beam mode of radiation. Now what is actually the normal mode? In normal mode the radiation field is normal to the helix axis. So as you know that the helix axis is provided at the centre to the number of turns the major radiation is perpendicular to this helix axis with the minor lobes are also present. Radiated waves are circularly polarized. It is the property of the helical antenna. It is having the circular polarization that is the EM wave which having the horizontal as well as the vertical components. Mode of the radiation is obtained if the dimension of the helix are small as compared to the wavelength. So for obtaining the proper radiation this should be the condition for the normal mode of operation for the helical antenna. For the normal mode of radiation dimension of the helix are must be small as compared to the wavelength whereas the radiation pattern of this helical antenna is the combination of short dipole antenna and the loop antenna. So actually how the radiation for the normal mode of operation is shown in this figure. So suppose the helical antenna which having the number of turns in this direction with the axis of the helix is provided at the centre then for the normal mode of operation the major radiation pattern which is perpendicular to the axis of the helix. Therefore at the 90 degree and 270 degree you are getting the major loop. Obviously the minor lobes are also present here. So from this you can says that the radiation pattern depends upon the values of the diameter of the helix denoted with the capital letter D and the spacing between the turn that is capital S. So what is actually the drawback of this mode of operation? So it having the low radiation efficiency along with the narrow bandwidth therefore very hardly it is used. Next mode of operation is the axial mode. In axial mode of operation the radiation is in the direction of the helical axis whereas in case of the normal mode it is perpendicular. So the major radiation is in the direction of the helical axis itself. Waves are again same the circularly or nearly circularly polarized in case of the axial mode. In axial mode the dimension is obtained by raising the circumference in order to get one wavelength and the spacing of approximately should be lambda by 4. So this is how the pattern for the axial mode is defined. So here the major radiation is in the direction of the axis of the helix. So antenna can be designed for the right-handed or the left-handed circularly polarized wave. So whenever the antenna is designed for right-handed circularly polarized well it will not receive the left-handed circularly polarized wave and vice versa. So mode of operation is generated with greater is and therefore more practically this type of a mode of operation is used. So for the proper radiation the circumference of the helix should be greater than 3 lambda by 4 and it should be less than 4 lambda by 3. So the condition for the proper radiation for this type of a mode of operation the circumference should lie between 3 lambda by 4 to 4 lambda by 3. Some of the advantages of the helical antennas, the helical antennas are simple to design. It having the highest directivity along with the wider bandwidth. It can achieve the circular polarization. It can be used for the ultra high frequency very high frequency and high frequency bands also. Along with the advantages some of the disadvantages of the helical antennas. As you know that the helical antennas are having the larger shape so it requires the most space and the efficiency decreases with the number of turns of the helical antenna. These are the applications the single helical antenna or its array is used to transmit and receive the very high frequency signal. It can be frequently used for satellite and the space communication purpose. It is used for the telemetry links with ballistic missiles and satellites at earth stations whereas it also used to establish communication between the moon and the earth. Again one of the application in radio astronomy thus the helical antennas can be used for the different applications. These are the references used for today's session. Thank you.


Hey, this is Brett, and this 2023 Jeep Grand Cherokee Limited, with the Black Appearance Group, is stock number 23J7. I am here at Summit Automotive in Fond du Lac, Wisconsin, your new and used Jeep and Jeep Grand Cherokee headquarters. This is our very first 2023 Jeep Grand Cherokee L that we've gotten on our lot. Really a nice package on this one. I'm going to go all the way around in this video, give you all the information that I can on this particular one. We're going to check out the original window sticker, go inside under the hood, turn all the lights on at the end of the video, and that's coming pretty quick. So stay tuned for that. First off, I'll say that this one has the 3.6 liter Pentastar V6 engine, puts out 293 horsepower and 260 foot pounds of torque. This one is capable of towing up to 6,200 pounds and has a max payload of 1,410 pounds capability. It's going to get you 25 miles per gallon on the highway, 18 city for an average of 21 miles per gallon with the four wheel drive system. Diamond black is the color and PXJ is the paint code on this one. This one also has a 23 gallon fuel tank. As we come around to the front here, the redesign from two years ago, still looking really fresh and good. The seven slotted grill of course for the seven continents that Jeep has driven on. The black appearance group appears to give you the blacked out Jeep lettering, the blacked out bezels on there and the blacked out front fascia and bumper. Blacked out headlight bezels as well. This one does have LED headlamps, LED running lights and LED fog lamps. And like I said, we're going to turn all those on at the end of the video.  These are the Scorpion Verdez and the size on these is 265 50 R20s. Kind of a highway tread pattern on there. Four wheel disc brakes of course on the Grand Cherokees and diamond black has just a little bit of metal flake to it. It's got a little bit of a gold flake, which I really like when it's out in the sunlight. It's really spectacular to check out. You also get the blacked out Grand Cherokee and American flag lettering on the side. We'll take a look at the original window sticker. Feel free to pause this. I'm not going to go through everything, but most of it, Grand Cherokee L Limited, diamond blacks to color, global black interior, Capri leather seats, 3.6 liter with the eight speed automatic transmission. Everything on the left there is your standard equipment including the Quadra Track 1 Jeep Wave, select terrain, remote start, power lift gate, adaptive cruise control, active lane management, all that stuff is standard equipment, interior features, those are all standard as well. And then the optional equipment starts with the diamond black paint, which is 395. This one has a trailer till prep group. So it's kind of an interesting thing because it doesn't actually have the receiver hitch or wiring, but it has the prep group, which gives you the trailer hitch zoom, automatic headlamp leveling system, the rear load leveling suspension and the full size spare tire. So it's all set up, all it means is the receiver hitch and wiring installed and then it's all ready to tow up to 6,200 pounds. Black appearance package is 2095 because of the 20 inch rims and the gloss black exterior accents. This one also has the rear entertainment group, which is 1995, this is the 2, 2 point or 10.1 inch rear entertainment screens and Amazon Fire TV built in. Total MSRP on this Grand Cherokee L is 56,795, three year 36 bumper to bumper, five year 60 powertrain. And if you wanna get your summit price in the upper right hand part of the screen is a link to our website to get all the pricing and information on this particular Grand Cherokee. There's your miles per gallon, 18 city, 25 highway for an average of 21 and your average fuel cost is 2,100 bucks. Does not have any safety ratings yet. Although I'm pretty sure the Grand Cherokees are known for their safety ratings being very high. As we come around to the back, you get the LED tail lamps as well, the rear bumper parking sensors and you'll see that it does have the hitch cover that does come off, but it's just a bar behind there. So it's all ready to have the receiver hitch installed. This one also has the power lift gate, which comes standard on the limited package, back storage area, get a nice amount of space back here. And I believe your jack tools are in there. So these seats do fall down for even more storage and then it's kind of like the two row Grand Cherokee, which is nice. But you have the options for the option for the third row seating, which is a nice thing to have if you have to haul around extra people. Notice you get the blacked out four by four and L and Jeep and limited badge on the back. This one has the capitalist fuel fill, which is a really nice feature to have. Never get gas. So you notice you get the roof rack rails up there. They are gloss black as well and all the door trims are gloss blacked as well. You do get blind spot monitoring built in directional signals and heated mirrors.  Take a quick look at the back seats. So you get the second row quad seating in this one with the captain's chairs, nice arm rest. There are the 10.1 inch screens with the Amazon Fire TV built into them. You get rear air controls, second row heated seats, 115 volt, 150 watt plug-in, two USBs and two USB Cs. So these seats go down a couple different ways. The first way is just the backrest goes down and then the other way is that the whole seat kind of goes up and forward like that for access to that third row. And then they even kind of fold up a little bit more, which makes it pretty easy to get to that third row seat. That's just the two-seater back there. It goes down like that and then these are adjustable as you can see. You get the nice wood grain trim on the back doors, child safety locks as well. We'll hop inside, check out the miles radio, well, instrument cluster radio. It's only got 38 miles on it. So you can see rear seat alert active. This one has the driver assist system. You have adaptive cruise controls on the steering wheel, the lane departure of course, normal cruise controls. You get the heated leather wrap steering wheel, information center controls, Bluetooth controls, nice wood grain trim on there as well. And then you can select all your different options from the screen there and just a really nice system to have. And the steering wheel is trimmed out very nicely. I like the Jeep right in the middle there. This one comes with the touchscreen radio that I believe this is a UConnect 5 radio. So you get the factory navigation system and you have all your different media that you can do here, AM, FM, Sirius, connected to your Bluetooth device. You have your radio controls here and your climate controls, including your heated seats and heated steering wheel controls there or up on the screen. Notice the dual climate controls. You have all your different apps here. If you go to all, you can see, you can pretty much get to everything in here and it's all touchscreen, it's all scroll. So a lot of really nice stuff that you can do. I wonder if I can get to the backup camera. And I love these backup cameras. They are crystal clear HD cameras. And like I said, you can zoom in on the receiver hitch once you get one installed. So that's really cool too. You can put those third row headrest down. I'll press the button and show you that happened. So that's pretty cool. Don't do it when anybody's sitting back there. They'll get mad at you. Then you have all your different other apps here including the off-road pages which I'll show you real quick. You have your steering angle, whether the transfer case is locked or unlocked. You have your accessory gauges, your pitch and roll and your select terrain, latitude, longitude, altitude, all that good stuff. So really cool stuff that you can do with this radio. So then the terrain settings, you have snow or sand, mud, snow, auto and sport. And if you change that, it changes it up there. That's snow, sand and mud. And then auto just kind of configures to whatever terrain you're on. And then you have sport mode with the Jeep drag strip there. So pretty cool stuff. 8-speed automatic transmission. Everything's all trimmed out in the gloss black. USBs, two USB C's and AUX jack. This is gloss black as well. Period, your start, stop, your lane departure, stability control, your parking sensors. Love the wood grain trim in the Grand Cherokee. Really good looking. And then up here, you get SOS and assist buttons, map lights and home link buttons for your garage door security systems and lane systems. There is the LED lights there. And like I said, we'll start it up. Take a look at all those lights in the front. And yeah, we'll go from there. Show you the back lights as well. We'll take a look under the hood. Grand Cherokees look pretty sharp from the back. I would personally like to thank you for checking out the video today and hopefully from this HD video, you've been able to verify the looks, styling and options on this particular Grand Cherokee. Hope you enjoyed it. There are those LED lights, both the fogs, the running lights and the headlights are ultra bright. And under the hood here, we have the 3.6 liter V6 dual overhead cam, Pentastar engine, 293 horsepower, 260 foot pounds of torque. Remember, it's gonna get you 25 highway, 18 city, 21 average with the four wheel drive has a max tow of 6,200 pounds and a max payload of 1,410 pounds and a 23 gallon fuel tank as well. There is the mission sticker. And that's pretty much it. Thanks for checking out the video. Remember to comment on it, like, subscribe, share. Let me know what you think of the 2023 Grand Cherokee L limited especially with the black appearance group. I think the black appearance group looks great. And to see more pictures of this Jeep or one of our other 550 new and used cars, trucks, SUVs, minivans, Wranglers, Cherokees, Grand Cherokees, Grand Cherokee Ls, you name it, we gotta go to that website right there summitauto.com full pictures and descriptions of every single vehicle from two locations all at summitauto.com. And if you'd like to check out more HD videos, you can go to youtube.com slash summitauto. Remember to like, subscribe and share on this video and all the videos that you see there. Click the bell notifications and you'll get updates on the videos I do each and every day here at Summit Auto. In fact, in a second you will see a link to subscribe to my YouTube channel in the upper left. The link to more Grand Cherokee videos like this one in the upper right. A link to this vehicle on our website in the lower left. And a link to one of our latest YouTube videos in the lower right. Click those, check us out. And we're super excited to be offering the brand new 2023 Jeep Grand Cherokee L limited with the black appearance group in Diamond Black Metallic. Thanks again for checking out the video. Remember to like, subscribe and share on the YouTube channel. I really appreciate it. Thanks again.


The conclusion is that using p equals 0.05 to suggest a discovery will result in being wrong at least 30% of the time, and experiments underpowered will increase this error rate. This conclusion is supported by tree diagrams, simulations, and evaluations of effect size overestimation. To keep false discovery rate below 5%, a 3 sigma rule or p less than or equal to 0.001 should be used, and the word significant should be avoided. This article was authored by David Kahoon.


What is up guys and welcome back to the Johnny Q daily vlog Thanks so much for watching today's video. You guessed it more spin art and the first piece a heart Let's get this. All right guys. Here's our heart-shaped canvas.  Let it kind of just spread on itself just like this And just like All right, let's get this Final result is Look at that. Let's go So good All right next piece. We're gonna do another heart-shaped piece, but with galaxy colors Let's get this galaxy heart piece. Let's get it first.  We got some blue because blue We got some purple. Oh, oh some galaxy. Let's go bro. Oh Oh Some white because stars Just a little bit just like that. Just like that and then again, we'll finish it with some black Oh, yes, sir Let it kind of just drool down, you know, just let it drool Only one thing to do. Let's get it. Here we go. Oh baby Final result is ah next piece. We got a small heart-shaped canvas. Let's get it next piece We have a large heart-shaped canvas. Let's get her ship canvas just like this Next one we're gonna do another heart ship canvas another large one with sunset colors with the glitter twist Let's get it. Well, I hope you guys like this video. I Enjoyed making the heart shape canvases honestly something different It's a cool variety to make and I do have them in large medium and small And if you guys want to email me or check out the website go down All the info will be in the description below and as always Thank you so much for watching and I'll catch you in the next Johnny Q. Daddy vlog


So hey guys, welcome to come again if you're new make sure you hit the subscribe button and the little bell Count below hit the like button. I mean even share with your friends Good job. We have cookies and milk might be seen in that shirt All right guys, welcome back come again today. I'm gonna be reviewing Funko pop pop heroes number 127 from DC Comics superheroes the Harley Quinn in popster For those of you who don't know what the impopsters are it is Mainly members of Batman's rogues gallery who did their own take on his classic costume As you can see with Harley Quinn hers is the red and black and because of the black on part of her costume She does have the white eyes Now let's take a look around This does look more like the Batgirl costume rather than the Batman costume because of The nose cutout Batgirl classically has that nose cut out on hers As you can see here this one looks more like the Batman impopster Whereas like I said before this is more like the Batgirl impopster and here's a few of the Collectibles in the wave we got the pink one impopster number 122 To face impopster number 123 Harley Quinn Batman impopster number 124 Scarecrow impopster number 125 Riddler impopster number 126 Harley Quinn Batgirl impopster number 127 and poison ivy Batgirl impopster number 128 So let's go ahead and take her out of the box and take a look First things first as usual with all Funko Pop characters. It does come with a little mini catalog Join today at legionofcollectors.com Every box 100% exclusive item the debut box featured items from Batman vs Superman where you got a Batman Funko Pop Little Superman and Aquaman I call them Chibis because that's what they look like Batman vs Superman shirts Then we've got the postcard legion of collectors for only $25 you'll receive 100% Exclusive DC Comics collectibles created by Funko each box contains exclusive pop vinyl guaranteed Box ships every other month six boxes per year no fluff every box features at least $50 of value Join a community of fans and get news direct from DC Comics join today at legionofcollectors.com Let's go and put that over to the side go ahead and take Batgirl Harley Quinn out of the package now first things first, let me adjust this real quick She does appear to be a little bit top heavy Or back heavy. I should say because of her hair on the back This did not come with a stand for her so You may have to slightly bend it a little bit To get her to stand All right, as you can see her right half is red her black half of the cow is black No mouth on this Funko Pop vinyl You have her hair flowing out the back which This is the red hair so more than likely they just took the Batgirl Funko Pop and repainted her they should have probably added Harley Quinn's special multicolored hair But you know or at least painted it blonde to signify that it was Harley Quinn under there then you've got The cape which matches the cow and then opposite of that her shirt Which is black on the right side red on the left side with black diamonds and red diamonds red glove black glove and Her pants which once again match up with her cowl With the black diamonds and red diamonds, and then her boots red and black Now like with all Funko Pops her head does turn side to side however it You will have a hard time getting her to turn all the way around because of her hair being in the way The cape is non-removable and it looks like she's only got eyelashes on one side because The black on the other side. They could have probably did red eyelashes on this side Got the white utility belt and then on the inside of her cape it looks to be a little bit gray Almost like they couldn't paint that part So I'm not sure why they wouldn't just sculpt them sculpt them separately and paint them separately and then attach them But you know what beggars can't be choosers Let's see if the arms on this one no the arms do not move, but it does look like they do It does look like there's a joint, but they are glued on there. See if I can get her to stand back up, right? So there you have it guys Funko Pop Harley Quinn Batgirl in popster Number 127 from pop heroes DC Comics superheroes collection My I believe my brother found this one out at Walmart in Springfield, so Hey no one out find yourself a Harley Quinn Batgirl in popster and add her to your collection.


So here's the Matatalad little robot and hi hi so who are you hi and and here you have a little this is the kids you sell this together and this is a educational robot how does it work okay it's educational coding robots and we put the coding blocks into this board and to the start button it will set the signal to the robot so the robot can see the instruction, the child, the case, the command size, the basics like coding like coding instructions and the sequence. Does it start here? It starts here. How does it work? This is the coding block. Which one is the first one? The top right left corner? What do you... It means go forward. Go forward, turn left? Turn left. And we have all kinds of blocks. Let's take them all out. What do you have here? We have 60 degrees? 60 degrees. The real time 60 degrees. And then what? This is building blocks. We have loop blocks. We put the loop blocks between the loop block and the loop block which we will do these two blocks for two times. Let's do it. Can you show? Can you push the button? Let's do a little bit more. Can you do the music one? And then what? Okay let's try it. Push the button. Turn that first. Go forward. Nice. That's all. And nice. Can you articulate the arms now? I can put the ball into this basket. Okay. So you're adding front, left, basket. Okay. Let's try it. Cool. So this was sale? 209 dollars for this set. Do you have different versions? Yeah, we have sales 109 dollars. It has three modes. The first mode is a remote control. A model control. A remote control. And it runs just like a remote control car. Is it connected now or not? Yeah. Okay. And we have motion control. The second mode is coding mode. We can set instructions from the buttons. Push the button. Push the button. Here are the instructions. The third mode is combined with the coding blocks. We have a sensor box. And it will... How does that change? Much complex. Use the blocks and the sensor. You can add sensor. You can add stuff to it. It's just a magnetic. Or how do you add? You add stuff and then it's more and more? Like this? No problem. So it just connects? No problem. I don't want to interrupt. Yeah, no problem. You have all this stuff here with this. And you get the price. You got the medal. Innovation award 2019. It's like the Olympics. You win. Where are you based? Shenzhen. And this is also the pack of all the different things. Is it ready for sale? Yeah. We have entered 40 countries around the world. All right. So the distributor can contact you and sell more. And then in the future millions of new programmers will exist. They will all learn to be software engineer. Is it possible to go into software also or just with the blocks? We have a new application and you can use the app to control the robot too. On the phone? Yeah. Android, iOS? Yeah, Android and iOS. And then it's like a scratch? Yeah, it looks like scratch. All right.


Today we're opening the first Yu-Gi-Oh Packs ever made. What's up guys, we're back with another epic video. Today we're opening a ton of old school packs, which include the first four sets ever made, Legend of Blue Eyes, Metal Raiders, Spell Ruler, Pharaoh's Servant, and then some Invasion of Chaos, which are not the first five or whatever, but they're also very cool. But before that, we have a giveaway. I'll be giving away these two original print secretaries. This one has a little bit of a ding on, but we have Gaia, the Dragon Champion, and the Thousand Dragon. All you guys have to do is like this video, be subscribed, let me know what you wanna see pulled today. We have 25 packs to open today, so hopefully we can get some really nice pulls because last time we opened some Legacy Packs, it didn't go that well. So let's hope that we can rebound a little bit. These came out of Target Walmart Blisters. We've got a Yellow Luster Shield Inferno. We've got a Ryu Koki and Dimension Distortion. All right, Ultra Evolution Pill Chop Man and Prickle Fairy. So a slow start, but we have a long way to go. I kinda wanna do the pack trick today. Let's do it. Two, three, four, like that. I think that's the rare, no. Wait, is that, yeah. Okay, we're good, yeah. Okay, four will be good. Solemn Wishes, we got Burning Land, Rain of Mercy, Doha Ryaba, seven completed, Infinite Dismissal, Flame Champion, Bombardment Beetle, Monster Recovery, and nothing. We're 0 for two to start, but we have a long way to go. Let's go into Spell Ruler. All right, Spell Ruler. Let's do the pack trick again for a little bit of fun, a little bit of extra suspense. We've got an Eternal Rest, Performance of the Sword, Twin Long Rounds, number two, Umi Ruka, Whippetail Crow, Magical Labyrinth, Hungry Burger, and UFO Turtle and a Taylor of the Fickle. Okay, slow start, Metal Raiders. This is your chance, the first Yukio packs ever made. We have Legend of Blue Eyes, Metal Raiders, Spell Ruler, which is a magic ruler, and Pharaoh Server, the first four sets ever. We got Leo Gun, all 2004 sets. That's all of them, the main sets at least. Pharaoh Imp, Yadu Kuru, oh, we didn't do it. Wait, let's go like this. We've got the Aki Beam, Stem Pack, Wing Dragon, Guardian of the Fortress, number one, Paralyzing Potion, Yadu Kuru, Seven Tools of the Bandit, There's an Ultra Rare Pull. Wow, that's a bummer, that could have been something really good. All the Ultra Rares except Seven Tools, Magic Jammer, and the other one. What is that, Horn of Heaven? But I do like Seven Tools, I think it's a cool card, but it's definitely not one you're looking for. We only have one Legend of Blue Eyes pack, so we're gonna save that for the end, because I had all these other ones, but I only had the one Legend of Blue Eyes. So let's see, oh wait, let's do this for a little bit of suspense. You guys know, yeah, like that, that's good. We've got the Lord Poison, Terror King Salmon, the Salmon, Self-Destruct Button, Chaos Necromancer, Soul Tiger, Amphibious Burgoth, MK3, Oh, Jameela. Oh, Secret Rare Invader of Darkness. We're invading the darkness, rather than getting the Chaos Ember, we are getting Invader Darkness. But still really cool, it's a Secret Rare, we will take that, okay, definitely better opening than last time so far though. So now we have a Secret Rare and an Ultra Rare, not bad actually for how many packs we've opened. The pack trick seems to be paying off a little bit. We switched it over to the pack trick, it's actually working. We've got the Inspection, we've got the Wing Weaver, Salmon's Law Book. This way, all the commons, they come before the foil so you can respect them a little more. DC Warrior, the Mr. Volcano. We're not always gonna do this, but sometimes it's fun. Minor Goblin Official, Flame Champion, especially on these old school ones where we like the commons, Vibonic Vermin, and I think that's it. Yeah, Prohibition, okay. It's like maybe a $3 Rare, that's pretty good actually. Maybe $2 at this point, it just depends. All right, we'll take that, a decent Rare. Let's see what the next pack can give us. All right, Spell Ruler. What shall you provide for the opening? So far, pretty good. Ceremonial Bell, we can show each other our hands while we play, that's such a weird card. Electric Snake, also a weird card. I always thought that this like electric ball on it, it looks, I don't know, I always thought it looks kinda cool, but it also looks like 2000s of video games on a PC. So, Darkness Approach, we've got the Commencement Dance, Hungry Burger, Whiptail Crow, and Reliable Guardian and the Shining Angel. Okay, nothing in that one. But now we're back to the Metal Raiders, which is one for one so far. We have pulled, what was it, Seven Tools of the Bandit? Yeah, an amazing Ultra Rare. One, two, three, four. The best Ultra Rare you could possibly want. Watch, we're gonna pull Horn of Heaven now. Gazelle the King of Mythical Beasts, Armored Lizard, Mortal of Thunder, Aki Beam, Baby Dragon, Labyrinth Tank, Dream Clown, Mystic Horseman, and The Great, The Grande Moth. Very beautiful, The Grande Moth, such a great pull. We will take that all day of the week. We are back to Invasion of Chaos. The Chaos will be invaded. We already have the Invader of Darkness, so there's definitely some invasion going on here. We have the Recycle, Skullmark to Lady Bug, Energy Drain, Curse Seal of the Forbidden Spell, Dimension Distortion, Ceaser Room Warrior of Darkness, G and Insect Whistle. I don't know where the foil is coming either, which makes this also kind of exciting. Oh, the Stray Lambs, because I'm so used to it. I know it's like two from the back, but you can't really tell as easily when you're doing it through so many cards. Okay, Pharaoh's Servant. So that's a little bit of a change up, because now I'm so used to it being the fourth card, it makes it a little bit more suspenseful for me. So let's see. Gamble, so we're gambling on Yu-Gi-Oh Packs, we're opening them up and losing money. Keys to the Tide, but that's the way we do it. So all I'm wishing is Minor Goblin Official, Flame Champion, the Boobon, the Vermin. Oh, Gear Free, the Iron Knight, such a nice card, super rare. We seem to pull this guy quite a bit, but it is one of the better super rears. The Iron Knight, when an equip card is equipped to this card, destroy the equip card. Very exciting, Butterfly Dagger Elma, you guys know. Spell Ruler, the ruler of the spell. Here we go. Can we pull something out of Spell Ruler? Have we pulled anything out of Spell Ruler? I don't know, Twin Long Rods, number two, an Invibious Creature with two whip-like tails. Very beautiful. Dark Zebra, Eternal Rest, Spear Cratin. The Tyhone, number two, Turtle Oath. Can we get the Crab Turtle, Liquid Beast, a liquid life form that thrives on water? Not in water, on water, interesting. Minar, okay, and the Giant Germ, he thrives on water. So he's just slithering around on top of the water. That's even more scary now that I think about it. Metal Raiders, let's go. One, two, three, four, let's do this. We have Lady Defe. Cocuna Evolution, can we pull the Picanium off? Disc, not desk. Paralyzing Potion, Rainbow Flower, Ancient Lizard Warrior, Destroyer Golem, Castle of Dark Illusions, and the Elegant Egotist. Elegant Egotist, all right, that's now what we're looking for. Invasion of Chaos, the Chaos, Shall Invade. Let's go, 25 Legacy Packs. So far we've had pretty good pulls, I would say, but not the best we could possibly have, which would be, you know, we probably, you know, best we could possibly have, we've already not had them because we've had a pack without a foil. I guess you could technically have a Seeker Rare or something like that in every pack. Chaos Necromancer, we've got the Yellow Luster Shield, Inferno, Ryu, Koki, Dimension Distortion, and the Ultra Evolution Pill. Let me know what you guys think would be the best pack, or the best opening ever. Like, what would I have to pull? Like, 25 Ultras or like the best card in the set every time or something like that? Let me know what you guys think. We've got the Eye of Truth. We're back to Pharaoh's Sermon. Bite Shoes, The Driving Snow. Oh, oh, we forgot to do it. We got the Nobleman of Crossout. Okay, that's very good, Super. I think it's around $5, even unlimited. That's a very good one. It's a very cool artwork too. It's shiny sword, I like to see that. We've got the Morphin Dragon Number Two and a Darkfire Soldier Number Two. Honestly, I wanna know who's idea was to have like Number Two. So like have the same name, but just put in Number Two at the end. Like, is that not seem like the laziest thing ever? I don't know. It's sort of cool actually because it's like two different cards that just say Number Two, but it's also like really like just gonna put, like they could have made Slot Machine Number Two. Just made a different card. Isn't that just weird? Weather Report, ceremonial bell. Giant Turtle who feeds on flames. Psychic Cabo, what does his say? An amphibian with a myriad of powers to shield it from enemy attacks. A myriad of powers, very cool. We got a Curse of Fiend. The Crab Turtle, there he is. Rising Air Current and the Giant Rat. We have returned to the Raiders of Metal. The Metal Raiders. Inside Laze, one of the strongest cards ever made. The Pecainio Moth, which we now have our first edition PSA 10 Petite Moth. We're looking for that Pecainio Moth still though. We got the Oguchi. Love to see that guy, the Disc Magician. Queen's Double. Oh, and the Legendary. This small but deadly creature is better off avoided and you best avoid this sucker because this thing is amazing. Let's pull that guy out. Tremendous fire, that's an amazing pull. Dragon Piper, Tainted Wisdom and the Bladefly. So we got the best card in the set there. We cannot complain. We are back to the Invasion of Chaos. The Chaos, can it be invaded? We've been looking to do that. Well, we did invade with the of Darkness, the Invader of Darkness. The Neo Bug, Salvage, Sylphied, Recycle, Boganean, Primal Seed, Grenmaju, Deyiza. And the DD Scout Plane. We love to see the DD Scout Plane. Always shows its face in an Invasion of Chaos opening. Very odd how that always happens. Spell Ruler, we're skipping Pharoah Servant because we only have one left. We're gonna save that. We had less Pharoah Servant than the rest besides LOB. So one, two, three, four. Let's go to the Fickle. The Taylor of the Fickle, Performance of Sword. Hamburger Recipe, Eat Gaboon. Fairy's Hand Mirror. Umi Ruka, Flash of Salem. Invader of the Throne, a super rare Invader of the Throne. Snatch Steel if you are able to flip this sucker up but can't be activated in the battle phase. So not very good. It would have been pretty good, theoretically, if it could have been activated in the battle phase. Really good, actually. You just set that sucker and just take their monster. But yeah, they didn't want it to be that good. So they just made Snatch Steel the next set and made it even better. We got a King of Yami Makai. Why did I see an Inn there? We got the Gazelle. Armed Lazar, Immortal Thunder. We could pull another Pekanium off. Sword of Deep Seated. Rock Ogre Grotto, number one. So they made this guy number one before they even made number two, I guess. That's interesting. Leg hole and share the pain. Mukamuk. We're down to one of each pack left. Besides Invasion of Chaos, we have two of those. So we're gonna do two in a row here for Invasion of Chaos. Let's make the chaos. We've been looking to chaos the entire time. Ojama Black. The Prickle, Fairy, Griffin's Featherduster. Grenadoura, Gigantus. Jade Insect Whistle. The Chaos Sork, very good one. DD Borderline, I think that was it. And the Drillago. Five more packs in the opening. So far, pretty good. I would say pretty good opening. We've got a Seaguar. We've got an Ultra. We've got a bunch of Supers. It's gone fairly well. It hasn't gone crazy though. So let's see if we can really turn it up a notch here with the Torpedo Fish. Probably not gonna do it. Terri King Salmon. We got the Ojama Delta Hurricane. Soul Absorption. Boganian. Stealth Bird. Soul Tigray. And the Orca Mega Fortress of Darkness. Another one that tends to show up in Invasion of Chaos openings. DD Scoutplane and Orca Mega Fortress. I think DD is a little bit more popular, but that one's close. Now we are on to the Servant of Pharaohs. We've not pulled Jinzo in quite a while. We used to pull Jinzo like every single time and it has been a while. Let's see if we can do it here. The Burning Land. Thousand Eyes Idol. Our Fire Soldier. The number two. Solemn Wishes. Grouching Kuba Gata. Earth Shaker. Souls of the Forgotten. Is it the next one? Yeah, we get the Morphing Drone number two. And World Suppression. Okay, we're suppressing the world. We're not pulling anything. Onto the final three packs of the first packs ever made. These were all in 2002. So was Ferris Servant, but we forgot to mention that one. But this one, third set, main set I should say ever made. Spell Ruler was a magic ruler. Change of Spell Ruler. You guys know that. Two, three, four, because you know Magic Gathering got a little salty. We've got Dark Zibra. Guy Power. Luminous Spark. Chain Energy. Red Archery Girl. Got the Crab Turtle. Meta Fish. Reliable Guardian. And a Mother Grizzly. Okay, nothing in there, but that's okay because we have two more packs. We have Metal Raiders. Let's pull ourselves another Pakenio Moth. The Moth that is deadly and better off avoided. Here we go. King of Yami Makai. He became a King of Yami Makai. Bottom Dweller. We got the Pumpkin King of Ghost. Yeah, that was that panic voice in one grade. Soul Release. I need to practice that. Mushroom Man. Wait, is there a Mushroom Man number one? I don't think there is. Is there? Maybe it's in a tournament pack? I don't know. Trent. We've got a guy from Lord of the Rings. Block Attack. Elegant Egotist. Final pack. Legend of Blue Eyes. This is the only Legend of Blue Eyes pack we have in this opening. So if there is a Blue Eyes, White Dragon in here, we're going to lose it. Here we go. We did pull one the other day. It was pretty cool. So one, two, three, four, let's do that. Final pack of the opening M Warrior. Number one, he's the first M Warrior. We got the Remove Trap Dissolve Rock. Oh, we rocked the special, but that's okay. Mr. Rizame. Wait. Oh, we did it back, because we did that. Oh, never mind. There's an Yami. It'll have to be after this. Nope, root water unfortunate. The old school pack, so it'd be Yami. Oh, wait, here we go. Metal Dragon. And Yurabi, the second highest attack monster. That's a four star that you can actually play. So, okay, maybe not what we were looking for. Overall, it was a fun opening. If you guys enjoyed the video, make sure to subscribe for more epic old school openings as they will be happening on this channel. Shout out to TCG Trust, the trust cards, Tomefo Show, Tomato Juice, Noa J, Stanley, Mike, Nance, Mimic, Gekko, Duelist, Game Charts, Bank of it, Father, and G-rated. Thank you guys for supporting this channel, and I will see you guys next time. Peace.