 So, all the examples that we have seen so far are steady state situations where there was no change in the inside the control volume. So, you may recall that we wrote down this requirement as like this. So, DECV DT was 0 in all the cases that we saw so far and DMCV DT was also equal to 0. So, whatever mass came in went out so that there is no accumulation or depletion of mass within the control volume these two conditions were satisfied so all the examples could be treated as steady flow examples. What we will do now is look at a couple of examples involving unsteady situation where there is some change inside the control volume from beginning to end of the process. So, that calls for an unsteady analysis. So, either DECV DT is not equal to 0 or DMCV DT is not equal to 0, one of the two or both. So, let us take a look at what we have here. So, an insulated rigid vessel of volume 500 liters which is initially evacuated is connected to a line in which steam at 20 bar 400 degree Celsius flows through a valve. So, we have seen this before so we have a line in which steam is flowing and we are connecting an initially evacuated vessel insulated rigid vessel so it is insulated so we show that like this. So, steam is flowing in this line and the control volume for this case may be taken like this. So, this is the control volume suitable for this analysis we have seen this before. So, we are asked to so the valve is open steam is allowed to slowly flow into the vessel until the pressure inside is the same as the line pressure at which point the valve is closed. We are asked to determine the final temperature inside the vessel and the mass that enters neglect HE and neglect heat loss and KE and PE changes. So, if you recall the unsteady energy equation looks like this in its full form let us just go back and take a look. So, this is the unsteady flow energy equation in its full form. So, for the problem under consideration it is given that there is no heat loss so Q dot may be taken as 0 and there is no external work so WX dot may also be taken to be 0, mass enters the control volume but no mass leaves the control volume so we take ME dot to be equal to 0. We are also been told to neglect KE and PE changes which means that we may take this to be 0 and E control volume you may recall that E control volume is equal to U plus KE plus PE in the control volume. Since there is no change in KE and PE in the control volume you may also write for this particular case ECV equal to UCV. So, with these simplifications the unsteady flow energy equation becomes DE CV DT equal to DU CV DT equal to MI dot times HI dot I am sorry MI dot times HI and the unsteady mass balance equation becomes DM CV DT equal to MI dot since ME dot is equal to 0. So, if I combine this or if I combine these two equations I get the following DU CV DT equal to HI times DM CV DT what is that HI is equal to H line in this case. So, HI is the enthalpy of the fluid at entry into the control volume which would be this point. So, at entry into the control volume the fluid enters with the same enthalpy as it had in the line so there is no thermodynamic process between here and here so it enters with the same enthalpy so that HI is equal to H line. And UCV is the total internal energy in the control volume you should recall that we are using since we are using an upper case letter here this is an extensive quantity. So, U is equal to mass within the control volume times the specific internal energy within the control volume so capital U equal to M times U and since H line remains constant with time we can integrate both sides of this equation to get this expression. And initially the vessel is evacuated so M1 is equal to 0 so if you rearrange you get U2 equal to H line which is the same expression that we obtained when we used a system approach for solving this problem. So, basically unsteady analysis proceeds in more or less the same manner as steady analysis you have the unsteady flow energy equation and the mass balance equation so you look at the equation simplified throwing out terms which are not relevant or which are equal to 0 and then simplify and then integrate using the information that is given. Let us move on to the next problem so here so the this example this example looked at filling a vessel with steam the so that is a filling process the next one looks at emptying a vessel you know through a certain process so this is an emptying process ok. So, rigid vessel initially contains a certain amount of saturated R134A vapor and a certain amount of saturated R134A liquid at pressure of 900 kilopascals so we are looking at a situation like this so there is a valve that is provided on top of the vessel so the pressure inside the vessel is maintained constant by this valve so we have liquid here and we have saturated vapor here so the valve allows the pressure to remain constant I am sorry the valve maintains the pressure inside the vessel constant by allowing saturated vapor to escape. So, we now heat the contents of the vessel and until all the liquid evaporates we are asked to calculate the mass of vapor that escapes and the heat that is supplied. Now, control volume analysis for this problem would require us to define a control volume that looks like this ok. So, using the given information the dryness fraction at the initial state may be evaluated as 0.0625 which means it is practically all liquid and the total mass is 16 kg 1 kg of vapor plus 15 kg of liquid so the total mass is 16 kg. So, 900 kPa is the pressure so from the pressure table we may retrieve Vf to be equal to this and Vg, Uf and Ug to be equal to this. So, specific volume at the initial state is equal to 2.223 times 10 raise to minus 3 meter cube per kilogram. So, the volume of the vessel may be evaluated since we know the mass that is initial there and we know the specific volume the volume of the vessel may be evaluated to be 0.0356 meter cube. Now, we supply heat starting from this state we start supplying heat until all the liquid evaporates which means there is only saturated vapor inside the vessel. So, the final state is saturated vapor ok. So, at the final state the specific volume of the contents of the vessel is simply equal to specific volume of the saturated vapor which is equal to this and the mass that is contained may be evaluated quite easily because we know the volume of the vessel we know the specific volume at the final state. So, the mass that remains finally is 1.57 kilogram. So, the mass of vapor that escapes is 16 minus 1.57 which is 14.43 kilogram. So, we defined the control volume to be like this. So, we simplify the unsteady flow energy equation to this case and we end up with an expression like this. Remember, Mi dot is equal to 0 for this case and Me dot is not equal to 0, Ke and Pe changes are neglected. The unsteady mass balance equation looks like this. So, we combine these two equations and we may write du Cv dt equal to q dot plus He times dm Cv dt. Now, if I look at my control volume, He is the enthalpy of the fluid as it leaves the control volume. So, you can see that at the location where it leaves the control volume the fluid has the same enthalpy as the enthalpy in the control volume itself. So, we can write He equal to Hg. Since we are allowing only saturated vapor to escape and it leaves at the same state as the state inside the control volume, we may write He equal to Hg and Hg is saturated specific enthalpy of saturated vapor at 900 kilopascal and it does not change with time or during the process which means that we can integrate this equation to get this expression finally. And if you substitute the known values, we get the heat supplied to be 2515.285 kilojoules. Notice how we evaluate He in this problem and Hi in the previous problem. So, He is the enthalpy of the fluid at entry to the control volume, Hi is the enthalpy of the fluid, I am sorry, He is the enthalpy of the fluid at exit to the control volume and Hi is the enthalpy of the fluid at entry to the control volume. That was how it was defined when we derived the unsteady flow energy equation. You must keep that in mind and define the control volume accordingly so that Hi and He are known. The next example involving unsteady analysis is this. So, we have a rigid vessel of volume 500 liter which contains air at a pressure of 1 bar. So, we have a valve and we connect this to a vacuum pump. So, the vacuum pump extracts air at a constant rate of 0.1 meter cube per minute. So, the volume flow rate leaving the vessel is constant and it is given to be 0.01. So, heat exchange also occurs between the vessel and the surroundings to keep the temperature of the air within the vessel constant, determine time taken to reduce the pressure in the vessel to one fourth of the initial value and the magnitude and side of the heat interaction between the vessel and the surroundings. Now, control volume that is suitable for this case would look like this. So, the unsteady flow energy equation simplified to this particular situation looks like this WX dot is 0, MI dot is 0 and ME dot is non-zero. So, we have something like this and this is the unsteady mass balance equation. So, this is the mass, MCV is the mass that is inside the control volume at any instant. So, this is nothing but PV over RT. You may recall that we have air inside the control volume. So, the equation of state gives us PV equal to MRT at any instant. So, the mass inside the control volume at any instant may be written as the pressure inside the control volume times the volume divided by RT where PV and T are evaluated at each instant. But you may also recall that the temperature remains constant in this particular case, heat exchange occurs to keep the temperature constant, the volume of the vessel is also constant. Now, ME dot in this particular case the volume flow rate leaving the control volume is a constant and it is equal to 0.01 meter cube per minute. So, the mass flow rate that leaves the control volume may be written as density times the volume flow rate, density as it leaves times the volume flow rate. Now, density as the fluid leaves the control volume may be rewritten using the equation of state like this where P is the pressure as it leaves the control volume, T is the temperature as it leaves the control volume, which is the same as the pressure inside the control volume and the temperature inside the control volume because of the manner in which we have drawn the control volume. So, we may write this equation like this and if you use the fact that temperature remains constant, this equation simplifies to something like this which may then be integrated to give this expression for the pressure at any instant in the vessel. So, the time taken to reduce the pressure to one for the initial value is 69.31 minutes. Now, if I combine these two equations, I get the following. So, q dot is equal to d u c v dt, remember u c v is equal to m c v times u c v and m c v may be rewritten in terms of I mean using the equation of state, u c v is nothing but c v times T and we have done the same thing for this expression here and h is equal to c p times T. So, if you go through a little bit of algebra, you get q dot to be equal to this and since the volume of the vessel remains constant, we can integrate both sides of the expression to get q to be equal to 37.5 kilojoules plus 37.5 kilojoules. That means, it is being supplied by the ambient to the vessel. So, again as I said before, the key steps in solving problems that involve unsteady situation are to identify a proper control volume, write down the full form of the unsteady flow energy equation, simplify according to the information given in the problem and then usually it involves integrating the unsteady equation to obtain the final values. Another key aspect to unsteady flow analysis problem is HI and HE. Be careful when defining the control volume so that we can evaluate HI and HE in a convenient manner. This is the last example on unsteady flow analysis that we will look at. This involves a rigid tank which contains air initially at a temperature P1 and T1 and instead of discharging the air directly to the atmosphere, it is actually discharged through a turbine into the atmosphere. So, there is a turbine that is connected here. The idea is to extract some work from the air before it is exhausted to the ambient. So, as you can see here, the tank is fully insulated. So, let us show it like this. Now, the air is always expanded to the atmospheric pressure. So, PE is always equal to P atmosphere and the valve is opened and the air is allowed to expand through the turbine until the pressure in the tank finally becomes equal to the atmospheric pressure. And for the purpose of simplifying the analysis, we are also told that the mass of air in the turbine at any instant and its energy may be neglected. Because as the air expands, there will be a finite amount of mass in the turbine. We have been asked to neglect it and also the energy that this contains because we have to apply the unsteady energy equation as well. So, that is the problem description. So, we take the tank and the turbine together as the control volume. So, our control volume is going to look like this. So, this is our control volume. And if you apply unsteady energy equation to this control volume, Q dot may be taken to be equal to 0 because the tank and the turbine are insulated. Mi dot equal to 0, no mass enters the control volume, this control volume. What is that? No mass enters this control volume and then but the mass leaves the control volume. So, Mi dot is not equal to 0 and we also neglect Ke and Pe terms. So, with these simplifications, the unsteady flow energy equation reduces to du Cv dt equal to minus Wx dot minus Mi dot times He dot, I am sorry, minus Mi dot times He. So, if we combine these two equations, we finally get d by dt of m times u. Notice that we have replaced Cv with the tank because the amount of mass in the turbine and its energy may be neglected. So, we can replace Cv with tank here. Now, we may assume the air in the tank to have undergone a process that obeys PV raise to gamma equal to constant because the tank and the turbine are insulated, there is no heat exchange in the surroundings and it is a fully registered process. So, we had shown earlier that for an ideal gas that undergoes an adiabatic process, the process may be written as PV raise to gamma equal to constant. So, we assume the same here that the process follows, expansion process follows this. So, if I apply this to the initial state at any instant and the exit here, I may write it like this after noting that the final pressure of the air in the tank is P atmosphere. So, this is in the initial state and this is at any instant and this is at the final state. At the final state, the pressure in the tank is equal to P atmosphere and the temperature of the air in the tank is also equal to Te. So, the exit temperature Te also does not change during this process. So, we integrate the previous expression that we derived here because He is equal to Cp times Te and since the air is always expanded to the atmospheric pressure, Te remains the same. So, Te is a constant. So, He is a constant which means we can integrate this equation and obtain an expression that looks like this. So, here we have used the equation of state PV equal to MRT for rewriting the mass in the tank at any instant. So, if you simplify this, you finally get an expression for the work done that looks like this. Typically, if the initial pressure is very high and if you expand the air to the atmospheric pressure, the temperature Te will be very, very low. So, a more practical situation for instance, if you are looking at using such a device in an automobile for developing energy or power in an automobile and normally the tank, I am sorry the air in the tank will remain at constant temperature which is equal to the ambient temperature. So, that the pressure with which it leaves will be slightly different that is more meaningful scenario. But here for the purpose of demonstrating how to carry out an unsteady analysis, we assume that the air is always expanded to the atmospheric pressure as a result of which the temperature here will be very, very low. Notice how we have combined two devices into a single control volume and simplified the analysis. We can also split this into two different control volumes and then combine the two. So, we come up with the two sets of equations, then we can combine the two equations and we will then end up with the same governing equations that we have here. It is as if we have taken two different control volumes and then combined them together. So, this concludes our discussion on first law analysis of systems and first law analysis of devices using control volume approach and what we will do next is to move on to second law of thermodynamics.