 Assistant Professor, Department of Mechanical Engineering, Valchin Institute of Technology, Solapur. Today we are going to study properties of pure substance. At the end of this session, students will be able to use MOLLEAR diagram. Students will be able to calculate the heat required for wet, dry and superheated stream for given conditions. These are the contents we are going to discuss today. That is MOLLEAR diagram. What are the concepts required to solve the numericals? MOLLEAR diagram. It is the representation of the condition of the steam on enthalpy versus entropy diagram. And this diagram is divided into two parts. This will be the, this is the line called as the saturated liquid line. So the saturated liquid line is the weight mixture and above this, this is the superheated. And this line is represented as the quality of X1, means the dryness fraction. If you are going to measure at any point on this line, that is 1. And if you are supposed to draw the parallel lines to this, these are having the constant dryness fraction lines. Means at this, this line will represent X of 1, this line will represent X of 0.9, this line will represent X of 0.8. These lines are representing the constant temperature lines and these lines are representing constant pressure lines. That is the use of what you can say MOLLEAR diagram to find for the given conditions. If two conditions are known to you, you can find the remaining conditions. Suppose for example, if you are having pressure and the temperature, if let us say pressure is given as 1 bar and temperature is given as 220. Now from this 220 you are going to draw this and whatever this pressure line is there, at which point it is going to meet, that is the given point, you can find out the remaining parameters. These are nothing but the enthalpy, entropy or specific volume. Means if you know two properties, then you can find out the remaining properties for the given MOLLEAR diagram. That is the use of MOLLEAR chart or MOLLEAR diagram. Now what are the concepts required to solve the numericals? First one that is important one that is dryness fraction that is denoted by X which is calculated by it is the ratio of how many, what is the mass of the steam divided by the total mass of the given mixture and that is calculated by MS divided by MS plus MW. For weight the value is in between 0 to 1, for dry the value is 1 and for superheated that is also 1. Weight steam, this is the formula that you are going to use to solve the numerical that is total heat contained by the weight steam for per unit mass that is equal to HF plus XHFG where HF is the sensible heat content of the that liquid, HFG is the latent heat of the liquid at a given pressure and X is the dryness fraction. For the weight steam the value of that X is in between 0 to 1, moving towards the next that is saturated steam. For the saturated steam same thing is there only the value of that X will become 1 so that the formula will become as total heat content by the saturated steam per unit mass is equal to HF plus HFG where X is 1, moving towards the next that is superheated steam means you are heating that steam above the saturated steam that the formula is given by HF plus HFG which is also known as HG plus MCP delta T superheated where that CP is the specific heat of saturated specific heat of the steam and delta T is called as the degree of superheated where that delta T is calculated as T superheated that is the temperature of superheated steam minus saturated saturation temperature with respect to the given pressure. These formulas are required to solve the numericals let us see the numericals what amount of heat would be supplied to produce 4.4 kg of steam means M is given as 4.4 at a pressure 6 bar pressure is 6 bar and at 250 degree Celsius means the output temperature of the steam must be 250 degree Celsius and the pressure is 6 bar from water at 30 degree Celsius means initially water is available at 30 degree Celsius and they have given tech specific heat for superheated steam as 2.2 kilo joule per kg Kelvin. Now first thing from the steam table we are going to find out the values of HF HFG at 6 bar at 6 bar from the same table HF is 670.4 kilo joule HFG 2805 kilo joule per kg T saturation temperature is 158.8 T steam is equal to 250 degree Celsius. Now I have a question for all of you what will be the type of steam means is that the output of the steam is wet or dry or superheated pause this moment for a while and think on that the answer is superheated why superheated because the output temperature of the steam is 200 degree Celsius and the saturation temperature with respect to given pressure given pressure is 6 bar and for a 6 bar the saturation temperature is 158.8 as the temperature of the saturation steam is greater than the sorry as the temperature of the steam is greater than the saturation temperature therefore definitely the quality of the steam is superheated means the value of x is 1. Now total heat contained by 1 kg of superheated steam is given by the equation of superheated that is HF plus HFG plus Cp delta T Cp by putting the values Cp as 2.2 we will get 2956 kilo joule per kg that is for per unit already the water is at 30 degree Celsius water this HF is there that is from 0 degree Celsius now I have to subtract this sensible heat therefore per unit sensible heat associated with 1 kg of water will be given as MCP delta T simply m is 1 4.18 that is the specific heat of water and delta T is 30 minus 0 the total will be 125.4 therefore net quantity of heat to be supplied per kg will be this minus this you will get 2830.6 that is for per unit mass but the total quantity is given as 4.44 kg therefore the final value that is total quantity of heat to be supplied to water is 4.44 multiplied by this per unit mass you will get 124.546 kilo joule. Moving towards the next numerical that is determine the mass of 0.15 meter cube of weight steam at a pressure of 4 bar and dryness fraction 0.8 also calculate the heat of 1 meter cube of steam. Now first thing at 4 bar what will be the values of HF HFG that we are going to find out by using the steam table these are the values now density is calculated by 1 upon x vg where vg is the specific volume and x will be the dryness fraction as they have mentioned that is weight steam and then they have given the value that is pointed therefore we will get the density 2.7056. Now once you know the density if you multiplied by that is mass you will get the total mass that is 0.4058 total heat contained by 1 meter cube of steam which is mass of 2.705 kg will be given by 2.7056 multiplied by H where H is the total heat contained of 1 kg that already you have calculated and as they have given that the quality of the steam is weight therefore HF plus x HFG by putting the values you will get the total heat contained by 1 meter cube of steam that value is 6252.9 these are the references thank you.