 Hello students, I am Ganesh B. Agulavi working as an assistant professor in department of mechanical engineering, vulture used of technology, Solapur. In this session we will see theory and design of carburetor per second, learning outcome. At the end of this session students will be able to design the simple carburetor and also will be able to evaluate different parameters of simple carburetor. Now in the previous session we have seen the introduction of simple carburetor. So there we have seen the air is flowing in the downward direction as this is the downward type of the carburetor and the fuel injection is from this tube. So we will consider the section 11 at the inlet, section 22 at the throat and you can consider this as the section CC or 3. This is the throttle valve and this is the difference between the height, the nozzle tip is getting opened over here. So this is considered as a reference line and the petrol level in the float chamber will assume it is to be constant. So this is the jet. Now we will use the different relations and then we will try to get the air fill ratio. For that we required mass of air and mass of air considering compressibility of air. Compressibility of air means there is change in the density. For air applying steady flow energy equation at section 11 and section 22. So for that we can use the relation which we have studied in previous subjects. So try to recall steady flow energy equation. You might have studied in the basic mechanical engineering in the first year. Yes. So did the steady flow energy equation which is also called in the short form ESFEE. So H1 plus C1 square by 2 plus Q is equal to H2 plus C2 square by 2 plus W. So this is the section 11, this is the section 22. So this is your steady flow energy equation. This is the enthalpy at the inlet, this is the velocity of the fluid at inlet and this is the heat supplied which is equal to enthalpy at section 22 that is throat velocity at the throat and this is the work developed by the fluid. Now if we pass the fluid through the pipe, it will pass the fluid through pipe. It will not develop any work done. So for that purpose W can be neglected over here and as I have told you the complete carburetor is insulated one. So there is no heat transfer. So from this equation I can neglect Q and W. Then you just solve for so Q and W will be neglected. Then what is remaining? H1 plus C1 square by 2 which is equal to H2 plus C2 square by 2 ok. Now in this case C1 is to be neglected once again because it is very C1 is having very small quantity that is why usually it is neglected. So what is remaining? C2 square by 2 is equal to H1 minus H2 that is written here. So C2 will be square root of 2 in the Cp in the bracket T1 minus T2 because enthalpy H1 could be replaced by CpT and H2 also CpT2. So what will happen? The H1 minus H2 CpT1 minus CpT2. So Cp is taken common then thus the velocity at throat can be calculated if we know the temperature of the surrounding air inlet air and the throat temperature of the air ok next. Now I can take a T1 common. So equation will become C2 is equal to square root of 2 CpT1 in the bracket 1 minus T2 by T1 ok. Now try to recall the Cp of air if not given where assuming 1.005 ok next equation is square root of 2 CpT1 in the bracket 1 minus now T2 by T1 could be replaced by P2 by P1 because flow through carburetor is once again assumed to be isentropic. In that case we can replace T2 by T1 by P2 by P1 raise to gamma minus 1 by gamma ok there. You just go on reducing it you just go on reducing it then you will be getting the mass of air mass of air relation is equal to P1A2 by R square root of T1 under square root of main square root of 2 Cp in the bracket P2 by P1 raise to 2 by gamma minus P2 by P1 raise to gamma plus 1 by gamma bracket complete. Now this is the theoretical relation whenever any fluid flows through tube pipe then its coefficient of discharge is to be considered because theoretical flow cannot be equal to practical flow practical flow depends upon the material used for that pipe or tube and its friction factor. So, the manufacturer is supplying you this coefficient of discharge that is why the actual mass flow rate of air is the multiplication of Cd into Ma. So, you can calculate this one now. Now for fuel we can apply the Bernoulli's theorem. So, this is the Bernoulli's theorem ok. So, now here we are going to calculate for the fuel. So, the fuel pipe and the you are a float chamber in the float chamber atmospheric pressure section Cc or section 3 3 we assumed and at the throat the pressure is 2 ok. So, here this is the P3 P3 is nothing but Cc pressure and what pressure is acting in the float chamber atmospheric pressure. So, P1 and P3 are same. So, I can replace P3 by P1 also. So, this is P3 by rho f plus C1 square by 2 is equal to P2 by rho f plus C2 square by 2 plus Gz. Now this C2 is nothing but velocity of a fuel injected at the throat ok. Now as C1 is very very small can be neglected then P1 is also equal to P3. So, this equation I can reduce for C2 then we will get velocity of fuel injected at the throat is equal to square root of 2 by rho f in the bracket delta P minus rho f Gz. Now what is the delta P? Delta P is the difference in the pressure that is P1 minus P2 P1 minus P2 for delta P is equal to 0 mass of fuel injection is going to be 0 with decrease in the P2 with decrease in the P2 mass of fuel increases ok just keep in mind that point. So, this is the velocity of fuel injected at the throat from continuity equation we can write. So, this continuity equation recall yes m is equal to rho AC ok. Now so, mass of fuel after substituting in the rho AC here rho f A 2f and C2. So, A f as it is written Z. Now if they do not give in the numerical the value of the Z you can neglect this one. So, remaining parameter will be 2 rho f delta P ok. Now here also take now here what you can do in the previous equations we have derived M A and here M F. So, you can take M A by M F you will be getting the air field ratio. Now we will consider the second case neglecting means the density will remain constant then I can apply the Bernoulli's theorem. So, P1 by rho A plus C C1 square by 2 is equal to P2 by rho A plus C2 square by 2. Now this equation is for air that is why here suffix AA is written neglecting A1 and writing the equation for C2 square root of 2 P2 P1 minus P2 by rho A from the continuity equation I can write M A is equal to A2 square root of 2 rho A delta P. So, here also neglecting the compressibility. Obviously, actually the air is compressible, but if you are interested to find approximate air field ratio then you can adopt neglecting compressibility. Then actual air flow rate as usual is equal to coefficient of discharge into theoretical air field, theoretical mass flow rate of air. So, this is the third equation. So, we will be having the two air field ratios. First air field ratio based on compressibility, second air field ratio neglecting the compressibility ok. So, depending upon your accuracy you can select it. Now you can take once again the air field ratio and you can compare it. It is always recommended to use the accurate or compressibility of the air, accurate method and approximate method. For the accurate method you have to assume the air to be compressible. For more study you can refer IC engine Fundamentals by Hewitt and second book Fundamentals of IC engine by H N Gupta. And from these two books you can solve the numerical cells. Thank you all.