 Let us now solve a few work examples which will hopefully illustrate these concepts ok. The first one reads like this which are the following two choices is the best way to increase the thermal efficiency of a reversible engine operating between T h and T c. Because it is given that it is a reversible engine I may write the efficiency as 1 minus T c over T h ok. Note that although I wrote here only for a corno engine remember the efficiency of all reversible engines operating between the same reservoir is the same. So, in fact you may actually want to write this as reversible engine ok. So, since the engine given engine here is reversible the idea is which is a better way to improve the efficiency of the engine. So, if I decrease T c the efficiency goes up if I increase T h the efficiency goes up ok which is a which is a better strategy decreasing T c or increasing T h. So, we take the differential of this expression which is nothing but partial eta partial T h times delta T h partial eta partial T c delta T c. What is that this partial derivative implies that T c is held constant and this partial derivative implies that T h is held constant. So, if I go through and compute the partial derivatives I end up with an expression like this ok. So, this coefficient here with the negative sign is the amount by which the efficiency will change when I reduce T c. Let us say that I decrease T c by 1 Kelvin ok. So, that means delta T c is negative. So, that means the negative sign multiplies this. So, this becomes positive. So, the efficiency will increase by 1 over T h if I decrease T c by 1 Kelvin. If I increase T h by 1 Kelvin that means delta T h is positive the efficiency will increase by this amount that is what this equation is trying to tell you. Now by comparing these two coefficients after ignoring the sign keeping in mind that delta T c is going to be negative because we want to increase the temperature. So, you are only comparing the magnitudes notice that we have 1 over T h here 1 over T h here and this is a quantity which is we just do the following and this quantity here is less than 1 since T c is less than T h. So, the best strategy for improving the efficiency of the reversible engine is to reduce the cold reservoir temperature. Ironically however in practice efficiency gains have been obtained only by increasing T h for instance I mentioned the ultra supercritical cycle which operates at a very very high temperature. So, all advances have come by pursuing an increase in the T h rather than decrease in the T c because the cold temperature the I am sorry the cold reservoir is at a temperature of 300 Kelvin or ambient temperature and it cannot be decreased any further. So, we really know I mean we really have no choice when it comes to this we cannot reduce the temperature of the ambient any more than this because we need an infinite reservoir we have to reject heat to the ambient and that temperature cannot be reduced any further than 300 Kelvin or thereabouts. So, the only practical strategy that we can pursue is to increase D h and thereby increase the efficiency of the corner engine. So, the next example reads like this a house is to be maintained at 27 degree Celsius when the outside ambient temperature is minus 5 degree Celsius two strategies are being considered one operate a 1 kilowatt space heater for 1 hour. So, that is simply a resistance heater that we keep in the room and it puts out heat. The other strategy is to supply the same amount of work. So, we are supplying a certain amount of electrical work to the space heater supply the same amount of work to a reversible heat pump that operates between the ambient and the house. So, here if you look at the block diagram that we add for a heat pump that is just here is a block diagram. So, in this case we wish to operate it as a heat pump. So, T c is actually the ambient and T h is the house which needs to be maintained at a reasonable temperature. So, that is what we are trying to do. So, the heat pump operates between the ambient and the house. So, let us go through and now when the 1 kilowatt heater is operated for an hour it is being supplied with the power of 1 kilowatt. So, the energy that is supplied is simply 3600 kilo joules. So, that is transferred to resist to heating and that is the amount of heat that will be available in the room under ideal circumstances. So, 3600 kilo joules is the amount of heat that we can actually put in the room if you operate the 1 kilowatt heater for 1 hour. So, the COP of the heater which is effect short divided by input effort input effort is 3600 kilo joules because we are operating for 1 hour and the amount of heat that we are putting in the house is also 3600. So, the COP of the space heater is 1 which is actually quite poor. Now, the COP of the heat pump is equal to Q H over W. Now, we are supplying the same amount of work which is 3600 kilo joules. Since the heat pump is reversible I may actually write the COP of the heat pump like this and if I substitute the numbers remember the temperatures have to be in Kelvin. So, TC is at minus 5 degree Celsius so minus 5 plus 273 the house has to be maintained at 27 degree Celsius. So, we convert this to Kelvin and we get a COP for the reversible heat pump to be 9.375. So, from this expression we can then see that 9.375 times 3600 which is 33750 kilo joules is can be put inside the house compared to 3600. So, it is almost a factor of 10 more because this COP is almost equal to 10 or 9 more. So, you can see that this is a much better way of maintaining a dwelling at a comfortable temperature than using the space heater. Although space heater is quite widely used it is probably much cheaper than a heat pump. In the long run from an energy perspective the heat pump is much much more efficient than the space heater. So, the next example that we are going to look at is this seawater desalination plant using solar energy is being proposed. The plant collects 9.9 times 10 raise to 9 joules of solar energy during a 10 hour period at an average temperature of 100 degree Celsius and rejects heat to the ambient at 25 degree Celsius. It is claimed that the plant can deliver anywhere between 1000 liters per second to 1000 liters per minute of fresh water evaluate this claim. What is the maximum possible yield for this plant? It is known that in an ideal process seawater has to be pumped through a pressure difference of 27 bar to obtain fresh water. This usually means that we are using the reverse osmosis process for desalinating the water. So, let us evaluate this claim. The power required for pumping water at the rate of m dot kg per second through a pressure difference of delta P is given as m dot times V times delta P. You may recall that we derived this expression in one of our earlier examples involving steady flow analysis for the steady flow analysis for a pump. So, we can see that we derived this expression W x dot equal to m dot times V times delta P in one of the steady flow examples. So, we make use of that expression now. So, the power required is m dot V times delta P or if you write it in terms of volume flow rate we may write it as V dot times delta P and delta P is given to be equal to 27 bar. Now, the engine receives an amount of heat equal to 2.9, 9.9 times 10 raise to 9 during an entire 24 hour period. Now, assuming that the engine is a corno engine or a corno engine is actually used to convert this heat into work that will then be used to drive the seawater desalination plan. So, we are collecting so much solar energy and we are going to use best circumstances. We are going to use a corno engine to convert this heat into work which will then be used to run the desalination plant. So, the efficiency of the corno engine is of course, 1 minus T C over T H and the temperatures are given average temperature of T H is average value of T H is 100 degree Celsius and the ambient is at 25 degree Celsius. So, which means that the work that the corno engine can produce per day is 1.991 times 10 raise to 9 joules. Therefore, the maximum possible volume flow rate of desalinated water we are saying maximum possible because we are using a corno engine for the purpose of converting the solar energy to work. So, the maximum possible volume flow rate of desalinated water that can be delivered is given as W divided by delta P. So, if you convert it into proper units we get this to be 512 liters per minute. Assuming that it is operating the desalination plant operates 24 hours a day. Although the energy itself is collected only for 10 hours a day, we assume that the plant operates 24 hours a day. So, we can then convert it into something like this. So, this is 512 liters per minute. So, even an ideal reversible engine can produce only 512 liters per minute of fresh water which is quite far from half of the claim that is made by the inventor. This of course is too high to be even plausible. This seems more reasonable and if you use a corno engine then you get half of that approximately. So, in real life it is probably likely that the yield will even be considerably less than 512 liters per minute. The next example reads like this two reversible power cycles labeled 1 and 2 are arranged in series. The first cycle receives energy from a reservoir at TH and rejects energy to a reservoir at an intermediate temperature T. So, let us try to draw this. So, the first engine is a reversible engine. Let us denote this using capital R. So, this is a reversible engine. It receives energy from a reservoir at TH and rejects energy to a reservoir at an intermediate temperature of T. So, this we take to be a T. The second cycle receives the energy rejected by the first cycle to the reservoir at T and rejects energy to a reservoir at temperature Tc lower than T. So, which means that the second engine operates like this. So, this in fact we have labeled this as 2 and labeled this as 1. So, this is W1, this is W2. So, this is Qh1, Qc1, Qh2, Qc2. It is given that the second cycle receives the energy rejected by the first cycle to the reservoir at T and then rejects to this. So, which means that these two are equal. Now, we are asked to derive an expression for the intermediate temperature T in terms of Th and Tc when the network of the two cycles are equal and the thermal efficiencies of the two cycle are equal. So, you must be able to from the given problem statement, you must be able to draw a schematic or a block diagram like this. It is very important for analysis of problems using the second law statements and the kind of things that we are doing. You should be able to sketch a block diagram with the information that is given and fill in as many details as possible. So, now let us actually look at a nicer illustration. So, what we have drawn there is illustrated very nicely here. So, let us proceed with the analysis. So, since the two cycles are reversible, we may write Qh over Th equal to Qc over T for the two cycles from which we get these expressions for Qh1 and Qc1. From first law, the network produced by each cycle is Qh minus Qc and in part A, these are given to be equal. So, simply we equate these two. And you may also recall that it was given that Qc1 is equal to Qh2. So, the amount of heat rejected by engine 1 is picked up by engine 2 and then after developing what some amount is rejected to reservoir at Tc. So, Qc1 equal to Qh2. So, if you substitute that, then we end up with the expression for this expression for T, which is simply that the intermediate temperature in this case with the work being equal is simply the arithmetic average of the two reservoir temperatures. Now, in case B, the efficiencies of the two cycles are given to be the same. So, the efficiencies of the two cycles may be written like this. Since these are equal, we can show from here that the intermediate temperature T is actually the geometric mean of Th and Tc. In the first case, it was the arithmetic mean of Th and Tc. In the second case, it is the geometric mean of Th and Tc. So, the important thing is being able to draw a block diagram like this helps the analysis immensely. So, that is very, very important. Let us look at the next example. A reversible engine operates between reservoirs at 1000 Kelvin and 500 Kelvin. The entire work delivered by this engine is used to operate a reversible heat pump that operates between the reservoir at 500 Kelvin and one at a temperature T, which lies in between 1000 Kelvin and 500 Kelvin. Heat removed by the pump and that received by the engine or equal, determine the temperature T. This is what we have been asked to do. So, if you try to draw a block diagram for this case, so let us start with reversible engine R. So, here we have reservoir at 1000 Kelvin. So, we have a reversible engine R. This rejects heat to the reservoir at 500 Kelvin. Now, we have a reversible heat pump, so which we will call R inverse. So, the reversible heat pump is driven by the reversible engine. So, it operates between the reservoir at 500 K. So, let us just erase this part and draw it like this. So, it operates between the reservoir at 500 K and another one at some temperature T. So, the heat removed by the pump, so let us call this QC prime, let us call this QC. So, the heat and so the heat removed by the reverse by the heat pump which is this one is equal to the heat that is received by the engine whereas, to determine the temperature T. So, here we have this block diagram drawn in a nice manner. So, you can see that it is from the problem statement it is given that QC prime equal to QA and we are asked to evaluate the intermediate temperature T. So, for engine A since it is reversible QA divided by 1000 is equal to QC divided by 500 which means that QC is one half of QA. The work developed by the engine per cycle is QA minus QC that is again QA over 2. For the heat pump since it is reversible QB over T divided is equal to QC prime over 500 QC prime. So, QB is equal to this. Since the work input for the heat pump notice that all the work that is developed by the engine is used to drive the heat pump. So, since the work developed by the engine is entirely used for running the heat pump you may equate the expression for work for both this is the work for the heat pump this is the work for the engine. So, we equate this and it is given further that QA is equal to QC prime. So, you can get the intermediate temperature T to be 750 Kelvin from this. Note that once the block diagram is drawn the analysis actually becomes easier. Once you draw the block diagram analysis becomes easier. The next example reads like this an internally reversible engine. Remember we talked about internal reversibility, internally reversibilities and externally reversibilities. So, this is an internally reversible engine that operates between a reservoir at 1000 k and one at 300 Kelvin. So, if you draw the diagram it will look something like this 1000 k 300 k this is an internally reversible engine. If the engine receives and rejects heat across a 20 Kelvin temperature difference with the reservoirs determine its efficiency. So, this means that the engine actually in this device we have externally reversibilities. So, remember temperature heat transfer across a finite temperature difference was classified as an external irreversibility. So, we can say that this is a device which operates with external external irreversibility. We are asked to calculate its efficiency. So, the engine receives QH from the reservoir at TH and we have shown that here in QC. The only thing is the engine operates at when the engine receives this it assuming that it operates in a corner cycle the engine is at 980 Kelvin 1000 minus 20 when it receives the heat. And the engine is at a temperature of 320 Kelvin when it rejects heat to the reservoir. So, since the engine is internally reversible we may write QH over TH minus delta TH. So, basically what is happening here is this if we draw the cycle. So, the cycle executed by the engine is like this 1, 2, 3, 4. So, this is 1000 Kelvin this is 880 Kelvin this is 320 Kelvin this is 300 Kelvin. So, we may write for the engine. So, we may write for the engine QH over TH minus delta TH this is the reversible isotherm along which the engine operates equal to QC over TC plus delta TC. Again this is the reversible isotherm along which the engine operates when it is rejecting heat. And the thermal efficiency of the engine is 1 minus QC over QH. So, we may write it like this and if you substitute the numbers we get the efficiency to be 0.6735. Had there been no external irreversibility the efficiency of this engine would have been 0.7. So, you can see that the external irreversibility both on the supply side and on the rejection side decreases the efficiency of the engine by almost let us say about 5 percent or so I think. So, you can see that external irreversibility caused by a temperature I am sorry heat transfer over a temperature difference can be accounted for in the expressions that we have derived. So, all these things are still very useful. What is important is to recognize the internal irreversibility and external irreversibility. So, an engine can be internally irreversible and operate like this what we have just said. But it can it may still have external irreversibility. So, we can still do a meaningful analysis and get and calculate values for efficiencies and other things.