 So, the next three examples that we are going to look at are all steady flow applications and the last example that we will look at involves a heat engine executing a cyclic process. So, the first example here reads like this air is compressed steadily in an 8 kilowatt compressor from ambient condition to 650 kPa and 187 degree Celsius. The mass flow rate is given to be 2.1 kg per minute. Neglecting the changes in KE kinetic and potential energy is determined the second law efficiency of the compressor, okay. So, basically we have situation like this. So, we have a compressor air enters at ambient condition then it is compressed to 650 kPa 187 degree Celsius. So, we apply steady flow energy equation to the compressor and we make use of the fact that air is an ideal gas and calorically perfect. So, we simply write H1 minus H2 as Cp times T1 minus T2. If we plug in the numbers we get Q dot to be minus 2.2711 kilowatts, okay. Remember the power of the compressor is given to be 8 kilowatts. So, Wx dot is actually minus 8. We need to keep that in mind. So, Q dot is known now and the rate of entropy generation and so exergy destruction may be evaluated like this. So, this is sigma dot. So, basically it is m dot times S2 minus S1 plus Q dot surroundings divided by T0 and you may recall that Q dot surrounding is equal to minus Q0 and S2 minus S1 here may be evaluated using the TDS relations because it is an ideal gas. So, if you go ahead with this you get the lost work expression for lost work reads like this and the lost work itself comes out to be 1.2102 kilowatts. So, now we can write the second law efficiency of the compressor as absolute value of Wx dot reversible divided by the absolute value of Wx dot. Remember this is a power absorbing device, okay. So, which means that so the lost work is added to reversible work. The actual work is more than the reversible work. So, the lost work is added to the reversible work to get the actual work. So, we may simplify this expression and evaluate the second law efficiency of the compressor as 85 percent. So, notice that here to get around the difficulty involving the negative sign for Wx dot we have used absolute values and we have treated lost work as what it is. So, lost work in this case adds to the reversible work to arrive at the actual work. So, the actual work is more than the reversible work because it is a power consuming device. The next example involves a steam turbine. So, here we have so steam enters the turbine steadily at 3 MPa 440 degree Celsius. So, we know that it is super heated based on the property values that are given at a rate of 8 kg per second and exits at 0.35 MPa and 160 degree Celsius. So, again we know that since pressure and temperature are given it is super heated at the exit also. Heat is lost to the surroundings at the rate of 500 kilowatts kilowatts. So, let us say so heat is lost to the surroundings at the rate of 500 kilowatts. Ke and Pe changes are negligible, determine the actual power output maximum possible power output and the second law efficiency. So, we can directly retrieve the values for specific enthalpy and specific entropy at the inlet and exit from the superheated table. And if we apply steady flow energy equation to the turbine we may get the actual power output as 3815.2 kilowatts. This is the actual power output. Now, if we add the lost work to this we get the maximum possible work or the reversible work, maximum possible power output or reversible power output. So, the rate of exergy destruction may be evaluated like this. So, remember exergy destruction is nothing but T0 times sigma dot. So, sigma dot itself is equal to m dot times S2 minus S1 plus Q dot surrounding over T0 and Q dot surrounding is minus Q0. So, if you plug in the values we get the rate of exergy destruction to be 679.5152 kilowatts. So, we add this rate of exergy destruction which is also lost work to WX2R actual to arrive at the maximum possible work or maximum possible power that we could have obtained from the turbine and that comes out to be 4494.72. Since this is a power producing device, the second law efficiency is simply WX dot actual divided by WX dot maximum. And the efficiency of the turbine second law efficiency comes out to be 85 percent or so, which is quite high. 85 percent second law efficiency is usually considered to be high, but typical of turbines. The next example that we will look at involves feed water heater or mixing chamber. So, steam at 1 mPa 300 degree Celsius. So, this is superheated steam enters at inlet 1 and saturated liquid water at 10 kilopascal whose pressure has been raised to isentropically to 1 megapascal enters the heater. So, steam at 1 megapascal and saturated liquid water whose temperature whose pressure has been raised to 1 mPa enters the heater. Heat is lost to the ambient at the rate of I am sorry heat loss to the ambient occurs at the rate of 100 kilowatts. If saturated liquid at 1 mPa is to exit the heater determine the required mass flow rate of steam for every unit mass flow rate of the exiting stream. So, that means m3 dot is 1 kg per second. So, we are asked to determine the amount of I am sorry the mass flow rate of steam that is required to accomplish the given conditions. Remember the condition is that saturated liquid at 1 mPa should leave the heater. So, we adjust the mass flow rate of steam until we are able to accomplish that. So, basically we assume the mass flow rate of steam to be x kg per second. So, this mass flow rate is then 1 minus x kg per second because this is at the exit the mass flow rate is 1 kg per second. We are asked to calculate the rate of x kg destruction and the second law efficiency of the feed water heater or mixing chamber. In the next module where we are going to discuss Rankine cycle you will learn this is usually referred to as a feed water heater and not as a mixing chamber in the next module. And this is although this is irreversible we know that mixing is highly reversible. This device is still very good device because it allows steam to be heated by mixing and not by adding heat to it directly. So, the water which comes in at 1 mPa is actually heated until it leaves as saturated liquid at 1 mPa. So, that means it is being heated to that temperature without directly adding heat and that reduces the external irreversibility associated with the heat addition process and actually increases the overall efficiency of the plant. We will see this in the next module. So, for the superheated steam that comes in through inlet 1 we may retrieve the property values like this and at state 3 we have saturated liquid at 1 mPa the state is given. So, we have hf h3 equal to hf and h3 equal to sf at 1 mPa. Now, state 2 is a compressed liquid state and temperature at state 2 is not known. Usually what we do is we approximate h of t comma p for a compressed liquid as uf of t plus p times vf of t but t is not known. So, here we have to resort to a slightly different tactic. We use the TDS relationship and write TDS equal to dh minus vdp. It is given that the water that is entering at inlet 2 is actually compressed isentropically from a pressure of 10 kPa to a pressure of 1 mPa. So, for that process we can take ds to be equal to 0 and dh then may be evaluated like this. So, h2 equal to hf at 10 kPa plus vf of 10 kPa times 1000 minus 10. So, the enthalpy of the liquid that is entering compressed liquid that is entering at inlet 2 works out to 192.72 and s2 equal to sf because it is an isentropic process. So, s2 equal to sf at 10 kPa and that is 0.6489. We also need the enthalpy or properties of the fluid at the reference state. So, at the reference state of 25 degree Celsius and 100 kPa it is a compressed liquid. So, we can get h0 as uf at 25 degree Celsius plus this. So, that is equal to this. And we may also approximate h0 as sf at 25 degree Celsius. So, now we have all the property values that we require. So, I suggest that you review this once more just to make sure that you are comfortable with what we have done. So, basically it is given that saturated liquid water at 10 kPa is compressed isentropically or is pressure is increased isentropically in a pump. So, that is the process that the water has undergone before entering inlet 2. So, basically if you recall, if I draw a TS diagram like this, let us say this is the isobar corresponding to 10 kilopascal and this is the isobar corresponding to 1 mPa. Then the saturated liquid at 10 kPa is pumped in a pump from 10 kPa to 1 mPa and that is an isentropic process. So, this is state 2. So, we apply TDS relation to this process, to this process the pumping process and we are able to calculate h2 and because it is an isentropic process, s2 is equal to sf at this state. So, that is what we have done. So, if we now apply steady flow energy equation to the feed water heater, we get x to be equal to 0.2345. So, for every kilogram per second of saturated liquid at 1 mPa that leaves the feed water heater, we need 0.2345 kg per second of steam. Now, specific exergy of the outgoing stream psi 3 may be evaluated like this after substituting the values here, psi 2 and psi 1 may also be evaluated in this in the same manner. So, the rate of exergy destruction is nothing but rate at which exergy comes in minus rate at which exergy leaves the feed water heater. So, this is the rate at which exergy comes in, this is the psi 3 is the rate at which exergy leaves the control volume. So, the rate of exergy destruction is 91.7135 kilowatts. Rate at which exergy enters, we already said that this is the rate at which exergy enters the feed water heater. So, this comes out to be 221.7 kilowatts. Therefore, we use the second law efficiency, this is not a work producing or a work absorbing device. So, you may recall that for such devices, we write the second law efficiency like this. So, rate at which or that we wrote it like this, exergy recovered divided by exergy supplied. So, rate at which exergy is recovered, so rate at which exergy is recovered is nothing but exergy supplied minus rate at which exergy is destroyed. So, this is equal to exergy supplied all on a rate basis because this is a steady flow problem minus exergy destroyed. So, that the final expression for each of two becomes 1 minus rate of exergy destruction divided by rate at which exergy is supplied and it comes out to be 58.6 percent. Now, as we said because it is a mixing chamber, its second law efficiency is poor, but in an overall sense because we are able to heat stream of water without directly supplying heat to it, the overall efficiency of the cycle still improves as we will demonstrate in the next module. The last example involves a cyclic process. So, here we have a direct heat engine. So, we have two reservoirs, one at TC. We have a direct heat engine which receives an amount of heat QH from a high temperature reservoir rejects heat to a low temperature reservoir at TC and we are asked to develop operates in a cycle and we are asked to develop an expression for the second law efficiency of the heat engine. So, the exergy recovered during each cycle basically is this plus the exergy associated with QC. You may recall that the exergy associated with QC may be written like this. Please go back and review the discussion on exergy transfer or racing from heat transfer and you should be able to write the exergy associated with heat rejection of QC to a reservoir at TC works out to be something like this. Now, the exergy supplied to the cycle is the equivalent of or is the exergy associated with QH and that works out to QH times 1 minus T0 divided by TH. So, we know exergy supplied, we know exergy recovered. So, second law efficiency because it is a power producing cycle, we may write second law efficiency as exergy recovered divided by exergy supplied and since the device operates in a cycle, W equal to QH minus QC. So, these two cancel out and if you divide through by QH, we may write this expression. So, here it has the actual efficiency of the cycle. So, eta is equal to and what we have done the denominator is to multiply and divide by a TC and then we have rewritten TC over TH by using the definition of the Carnot efficiency, eta Carnot equal to 1 minus TC over TH. So, this is the efficiency of a Carnot engine operating between the same two reservoirs. So, by using that we have rewritten the expression like this. So, this is the second law efficiency of the cycle expressed in terms of the first law efficiency and the Carnot efficiency. Now, if heat is rejected to the ambient instead of to the if heat is rejected to the ambient instead of to a reservoir at TC, basically we said TC equal to T0, then we get the second law efficiency of the cycle to be just eta over eta Carnot. Otherwise, if it is rejected to an intermediate reservoir at a temperature higher than the ambient temperature, this is the expression that we get. So, you can see that the second law efficiency gives us more perspective on the performance of the cycle and insights on the operations of components because we can define second law efficiencies for individual components which are executing a non-cyclic process. It gives us insights on the performance of the individual devices, the exergy destroyed and this will allow us. So, when we have let us say several designs for executing or accomplishing a certain process, this metric will allow us to make a comparative evaluation of these designs because they may be identical on a first law basis and which they will be usually. So, we need to discriminate against the poorer designs by using the second law efficiency. So, second law efficiency in the notion of exergy are thus very, very important in practical thermodynamic applications. So, what we are going to do in the next module is to discuss thermodynamic cycles. We will discuss three cycles. First one is the Rankine cycle or steam power cycle. Next one is the air standard cycle. Third one is the vapor compression refrigeration cycle, which is a power absorbing cycle. Now, the Rankine cycle and the vapor compression refrigeration cycle have as we showed in the previous course, they are genuine heat engines because there is a fixed quantity of mass of the working substance which executes the cyclic process throughout. So, whatever we have discussed, so it qualifies as a heat engine and the discussion proceeds very smoothly. Now, in the case of cycles that utilize air as the working substance, typically air is drawn from the atmosphere and then after passing through the engine, the air is then exhausted back to the atmosphere or clean air is taken from the ambient and usually combustion products are let out of the engine into the atmosphere. So, the air does not actually execute a cyclic process. So, what we will sort of try to do is to idealize the process by saying that a certain quantity of air executes a cyclic process just like in the case of Rankine cycle or vapor compression cycle and we will treat this as an idealized form of the actual process. So, this will be a cyclic process although the actual process involved in a gas turbine engine or a spark ignition engine or a diesel engine is not a cyclic process, but this will give us some ideas on the performance of such devices.