 Hello, I am Ganesh Biyagalai working as an assistant professor in Department of Mechanical Engineering, Walsh Institute of Technology, Sallapur. In this session, we will see Weber compression cycle, the basics required, learning outcome. At the end of this session, students will be able to describe simple Carnot cycle, second describe reverse Carnot cycle and express ton of refrigeration contents. First, we will see simple Carnot cycle, then reverse Carnot cycle and definition and expression for ton of refrigeration, which is required for to study refrigeration and air conditioning. In the first part, that is simple Carnot cycle, in the simple Carnot cycle, there are 4 processes, I will show these 4 processes on T S diagram. So, process 1 to 2, 2 to 3, 3 to 4 and 4 to 1, already you have studied this Carnot cycle in basic mechanical engineering in first year. So, can you think over the 2 types of the process, which are involved in the Carnot cycle? Try to recall, yes, there are 2 isothermal processes and 2 isentropic processes. In the Carnot cycle, we assume the hot body is brought in contact with the piston and cylinder arrangement, then the heat will be supplied to the piston cylinder. Because of this heat supplied, we may get work done by the engine or the Carnot cycle will be during process 2 to 3, which is isentropic process, entropy is constant. In process 1 to 4, the heat will be rejected. Now, for to reject this heat from the piston cylinder, the cold body is required to be brought in contact with the cylinder head, so this will be heat rejected. Then process 4 to 3 is once again the isentropic process, isentropic process entropy is constant and work required to be done on the system. So, the cycle 1, 4, 3, 2, 1 is simple Carnot cycle. We have derived the efficiency for the simple Carnot cycle as output to the input that is heat supplied. The output can be written as heat supplied minus heat rejected divided by heat supplied. Now, though the numbers are not in flow with the clockwise direction of the cycle as this is simple Carnot cycle, we will see after this simple reverse Carnot cycle for that cycle I have written 1, 2, 3, 4 notations. You can change the notations instead of 3, you can write here 1, 2, 3, 4 also. Now, the efficiency of the Carnot engine will be equal to now for this I will take this as the high temperature source because T3 is equal to T2 and this as low temperature sink as T4 is equal to T1, both these processes are isothermal processes. Now, this can be represented by high temperature minus low temperature divided by high temperature. So, this is the simple Carnot cycle. In the reverse Carnot cycle, now we will see the reverse Carnot cycle. What I will do? I will just change the direction of the processes. Once again, I will show the rectangle that is the processes 4 processes on the TS plane 1, 2, 3, 4 now simply I will change the rotation or direction so 1, 2, 2, 2, 3, 3, 2, 4 and 4 to 1 this becomes a reverse Carnot cycle. In this cycle 1 to 2 is not the work developed process, the work is required to drive the compressor. So, the work will be consumed by the compressor during process 1 to 2 obviously it is isentropic compression process. Process 2 to 3 is the heat rejection process. Now, here we will consider condenser because in the refrigeration condenser is used to reject the heat. During process 3 to 4, there may be work development depending upon the type of the cycle here the work will get developed and process 4 to 1 is the heat absorption process. Here we are calling it as a refrigerating effect. Now, in the reverse Carnot cycle we cannot use the heat utter for efficiency so effectiveness is calculated using the term coefficient of performance and for refrigerator suffix is written as R. Now, so it is the COP, COP is the ratio of desired effect nothing but refrigerating effect here to the work done to achieve this effect. Now, how much work is required that is W c. So, this can be written as refrigerating effect divided by Q c minus Q r. Now, this Q can be replaced by a temperature once again I will use the notation T h and T l. Now, the heat is absorbed from the source at low temperature. So, here I can write T l divided by T c is high temperature sink. So, this is T h minus T l. So, this will be the Carnot COP for refrigerator. Similarly, I can write Carnot COP for heat pump as now in the heat pump the desired effect will be the heat addition into the room. So, Q c divided by same work will be required by the heat pump. So, it will be the ratio of T h divided by T h minus T l remember the Carnot COP of heat pump is always greater than Carnot COP of the refrigerator. Next is the third point definition of ton of refrigeration. So, ton of refrigeration is defined by the ASHRAE as 1 US ton water is taken which is taken at 0 degree Celsius required to be converted into ice at a same 0 degree Celsius. So, the ton of refrigeration is defined as the amount of heat removed from 1 US ton water at 0 degree Celsius to be converted into ice at a same temperature in 1 day or 24 hours. So, 1 ton of refrigeration will be equal to now here 1 US ton ice is nothing but 709 sorry 907 kg of water 907 kg of water. Now, this water is to be converted into ice means latent heat is to be removed from that water which is 335 kilo joule per kilogram. In 1 day or 24 hours are to be converted 24 hours are to be converted into seconds. If we will do the calculation then we will get the exact figure as 3.5 167 3.5 167 here kilogram kilogram will get cancelled the unit will be kilo joule per second nothing but 3.5 for all calculations in refrigeration and air conditioning we use will take 3.5 kilowatt. Now, this you can convert into kilo joule per minute or BTO per pound etcetera etcetera, but for all numerical calculations we will be consider we will be considering 1 ton of refrigeration as 3.5 for further study you can refer refrigeration and air conditioning by CP Aurora and refrigeration and air conditioning by WF Stoker. Thank you.