 We're now going to work an example problem looking at the newtons or conservation of momentum for the case of rectilinear acceleration of the control volume and what we'll look at is that problem of a rocket. Okay so there is our problem and what I've drawn is I've drawn a control volume around our rocket and you can see it sketched here. I've drawn the control volume to be right across the nozzle exit and we can see that we have our inertial reference frame that is fixed and not moving and then we have our moving reference frame that in this particular case it turns out that being that it's on the rocket it's actually accelerating and consequently we have the rectilinear acceleration problem so what we are told to find other things we were told is that the rocket starts from rest so it starts from being stationary not moving. We're told to neglect the drag forces across or on the rocket being exerted by the error resistance. Fuel consumption we're given five kilograms per second and that's equal to the mass full rate leaving exhaust velocity 3500 meters per second and then finally the original mass of the rocket is 400 kilograms so we're asked to find a couple of things so we're asked to find the initial acceleration a0 and then the velocity is a function of time so what we have here this is an example problem involving rectilinear acceleration so we know in terms of a solution that the equations that we can apply one would be the y component of linear momentum for the the case of linear acceleration so let's look at that and if you recall in last lecture segment what we did we came up with an expression that was basically a modification of the form of the momentum equation where what we did is we modified it with this term that I'm writing right now pertaining to the acceleration of the reference frame and then everything else was the same however we had to be careful of measuring the velocities with respect to the moving reference frame and then finally we have the mass flux leaving term the last three terms give us that and then we have conservation of momentum or of mass the continuity equation okay so what we're going to do we're going to begin by looking at conservation of mass and we'll work with that first come up with an expression for the mass of the control volume and then we'll go back to the y momentum equation so this first term here can be expressed as being the mass of the control volume with respect to time and that is equal to the integral over the control surface I'm just using the continuity equation here and I basically moved one term to the left and kept the other on the right so with this if we're looking at the bottom of our rocket we have we have to be careful here with the dot product but essentially we have a control surface like this the vector leaving is in this direction the area of vectors in that direction and consequently this turns out to be a positive when we do the dot product so what we end up with and that is equal to minus m exiting and I'll put a dot over that's that's the mass flow rate exiting our control volume and with this what we come up with is an integral equation so we want to integrate that and the control volume mass is only changing with respect to time so that would be an ordinary differential equation and what we then get is mcv is minus the mass flow rate leaving times time plus some constant of integration we apply a boundary condition and we were told that the mass of the rocket or the mass of the control volume at time zero was equal to m zero and I think we were given 400 kilograms so with that we can get C is equal to the original mass and from that we get the mass the control volume is equal to the mass exiting multiplied by time the mass flow rate exiting times time plus the original mass that's kind of an obvious intuitive expression basically just saying the mass leaving you subtract that from the original mass and that gives you the mass of the control volume is a function of time so we get that expression from continuity now what we want to do is go on to momentum and take a look at the y component of momentum and for this the body force is just going to be the mass of the control volume times the gravitational vector and it's minus because it's acting down the term that modifies and enables us to apply control volume analysis for the accelerating reference frame and then on the right hand side okay now what we're going to do we're going to make a simplification here and this is a bit of an approximation but I'm going to get rid of the time rate of change term and I'm going to use two different reasons for justifying that one is that the unburnt fuel and the structure of the rocket is not moving with respect to the accelerating control volume and the other one is that the nozzle exhaust velocity is steady so even though the masses is very large and we have a steady velocity more or less and a large mass that really is not changing that much with time and and consequently with that we can then make the approximation that the timer to change that term is small so what that leaves us with then is this term here that we're going to take a look at and work a little bit more with so let's look at that now so here the only mass leaving the control volume or the only place crossing a control surface is the exhaust itself and so we have the exit velocity is V e and if you look at the vector V e is in the down direction so I would write minus V e j and that's why we have a minus for the first velocity term and then that's multiplied by the mass flux looking the area vectors in that direction so the dot product comes out to be a positive so we express that as being a positive here and so row V e area exit that is nothing more than just the mass flow rate exiting so m dot e which we expressed earlier from continuity so what we can do we can take this and plug it back up into our equation for the y momentum equation we have the gravity term or the body force the acceleration of our reference frame times the mass of the control volume and then on the right hand side V e times m dot e so what I'm going to do I'm going to isolate the reference frame acceleration and I'm going to rearrange this and also what I will do is I'm going to go back to the result that we got from the continuity equation right here and plug that in as well and so we get this equation here for the acceleration and we were asked to find the initial acceleration so for part a we have that time t equals zero so what you find plugging in the values is right at the beginning of the lift off the rocket the acceleration is three point four six G's and the second part we were asked to find was the velocity as a function of time and so for that we will integrate the acceleration expression and upon integrating this we end up with the following relationship or expression so that gives us an expression for the velocity as a function of time and plugging in the values that we were given for the problem so that becomes an expression we can do calculations at different times so let's say velocity at 10 seconds we would get 369.3 meters per second so assuming that the speed of sound is around it'd be the square root of gamma or t approximately 340 meters per second within 10 seconds we would have exceeded the speed of sound and so at that point the rocket would be going supersonic and then it would continue to accelerate if you plot out the velocity profile for this problem I don't have it plotted neatly however it would look something kind of like this I given the relationship that we've drawn we would have t on the horizontal u of t this way you get a function that would kind of do something like this and then it would start going up and climbing very very quickly so you can see it takes a while for the rocket to get going in terms of velocity but then when you get later on in time by time 80 seconds which I think is right in around here somewhere by then given the mass flow rate we have I think you've depleted all of the mass of the rocket and that's why the velocity is going very very high there because you don't have a lot of mass that you have to accelerate so anyways that's the case solving a problem with an accelerating control volume the biggest thing to remember is the fact that you have to deal with the modified version of the momentum equation where was it I think it was right here and the modification was due to this term here for the accelerating reference frame so that concludes this problem and this lecture