 Assistant Professor, Department of Mechanical Engineering, Volchian 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 weight, 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. Below 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 make, 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 M S divided by M S plus M W. 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 H F plus X H F G where H F is the sensible heat content of that liquid. H F G 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 H F plus H F G 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 H F plus H F G which is also known as H G plus M C P delta is superheated. Where that C P is the specific heat of saturated, specific heat of the steam. And delta T is called as the degree of superheat. Where that delta T is calculated as T superheated that is the temperature of superheated steam minus 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 K G 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 K G Kelvin. Now first thing from the steam table we are going to find out the values of H F, H F G at 6 bar. At 6 bar from the steam table H F is 670.4 kilo joule. H F G 2805 kilo joule per K G. 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? 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 one. A total heat contained by one cage of superheated steam is given by the equation of superheated. That is H F plus H F G plus C P delta T. C P by putting the values C P as 2.2 we will get 2956 kilo joule per K G. That is for per unit. Already the water is at 30 degree Celsius. Water this H F is there that is from 0 degree Celsius. Now I have to subtract this sensible heat. Therefore, per unit sensible heat associated with one cage 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 K G 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 K G. 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 H F H F G that we are going to find out by using the steam table. These are the values. Now, density is calculated by 1 upon X V G where V G 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 multiply it 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 content of 1 kg that already you have calculated. And as they have given that the quality of the steam is weight therefore, H F plus X H F G 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.