  beispiel iverso yondई जी मेंँया ठाव मैंजिन न शोत मेंटर फाची आ��crany glio gression yet ceremony in evidencer is about 我iami Cute Discovery ilst subsequently today i am going to deliver a video session on performance of steam condenser two at the end of this video session You will be able to calculate the vacuum efficiency ॐ and condenser efficiency ॐ also you will be able to calculate quality of steam entering the Debatte the contents of this video session are the numericals on performance of steam condenser we are going to study two numericals related to performance of steam condenser the first problem is the following observations were recorded during trial on steam condenser barometric reading equals 765 mm of mercury condenser vacuum 710 mm of mercury mean condenser temperature given as 35 degree Celsius temperature of hot well 28 degree Celsius condensate collected equals 2000 kg per hour quantity of cooling water collected is 60000 kg per hour temperature of cooling water at inlet is 8 degree Celsius and temperature of cooling water at outlet is equal to 24 degree Celsius this is the given data and we are supposed to determine corrected vacuum to standard barometer vacuum efficiency of condenser under cooling of condenser condenser efficiency and quality of steam entering condenser now let us see how we can find out each term one by one the first step in the solution is first we need to calculate the absolute pressure in the condenser and already we have discussed in my previous sessions the absolute pressure in condenser is given by barometric pressure minus vacuum gauge reading now in the problem it is given that barometric pressure is 765 and vacuum gauge reading in the condenser is 710 so 765 minus 710 gives us 55 mm of mercury this is the vacuum in the condenser in mm of mercury which we can convert into bar which will be equivalent to 0.074 bar we know that 1 bar is equal to 750 mm of mercury therefore 55 mm of mercury can be converted into bar by dividing by 750 mm of mercury so we get absolute pressure in bar as 0.074 bar now let us see how to find out the corrected vacuum in the condenser corrected vacuum is given by an equation barometric pressure minus absolute pressure in the condenser now let us pause the video for movement and let us recall why do we find corrected vacuum as we know that the barometric pressure varies from location to location and condenser will be located at a particular point where barometric pressure will be different than standard barometric pressure standard barometric pressure is given by 760 mm of mercury hence we can find the corrected vacuum by using standard barometric pressure as 760 mm of mercury minus absolute pressure in the condenser 55 mm of mercury which we have decided in our previous step and this will be obtained as 705 mm of mercury now let us see how to find vacuum efficiency vacuum efficiency is given by an equation it is the ratio of actual vacuum to the ideal vacuum actual vacuum in the condenser is given by the relation barometric reading minus absolute pressure in the condenser barometric reading given in the problem is 765 mm of mercury and absolute pressure in the condenser we have obtained as 55 mm of mercury so difference of these two will give us the actual vacuum in the condenser equal to 710 mm of mercury now let us see how to find out the ideal vacuum ideal vacuum is nothing but the vacuum in the condenser when air is totally absent which is given by barometric reading minus saturation pressure of steam now barometric reading already we know that it is given as 765 mm of mercury whereas saturation pressure of steam is obtained from the steam table corresponding to mean condenser temperature of 35 degree Celsius so by using the temperature tables in the steam table we get corresponding to 35 degree Celsius absolute pressure of steam equal to 0.056 bar which we can convert into mm of mercury as we know that 750 mm of mercury equal to 1 bar so 0.056 bar will be equal to 42 mm of mercury therefore ideal vacuum is obtained by taking difference of barometric reading of 765 mm of mercury minus saturation pressure of steam corresponding to mean condenser temperature of 35 as 42 mm of mercury which is obtained as 723 mm of mercury therefore using an equation for vacuum efficiency which is ratio of actual vacuum on ideal vacuum we can put actual vacuum as 710 mm of mercury and ideal vacuum of 723 mm of mercury we get it as 98.20% now we are going to find out the under cooling of condenser means when the condenser steam is cooled below mean condenser temperature it is called as under cooling now under cooling of condenser heat is given by the formula mean condenser temperature minus hot well temperature that is the temperature of condenser heat in the hot well which will be equal to 35 minus 28 equivalent to 7 degree Celsius now let us see how to find out condenser efficiency condenser efficiency is given by an equation actual rise in temperature of cooling water divided by maximum possible rise in temperature of cooling water now actual rise in temperature of cooling water is given by TW2 minus TW1 which is nothing but the outlet cooling water temperature minus inlet cooling water temperature which is equivalent to 24 minus 8 ... only when the temperature of the water is given is as high as possible, the saturation temperature of the steam corresponding to the absolute pressure in the condenser which is obtained as that. We can see that a higher saturation temperature of the steam corresponding to the absolute pressure in the condenser, of 0.074 bar is equal to 39.95 °C. उसी आई� trust the condenser efficiency,jin उदविय Money , you mindls , 150 नुआ तो ं explicar the actual heard is in temperature of the cooling water, mayor । किचिन एक बाई किचिन खिलिग गभा movimiento ज़ीरhm । किचिन भी नान्गट equations , i find न�giggling which redef maker ==== उ को skipping heedcm वहार नछ time ज़ीर home विस धो � Veter gucken उऊठका서도 थ दोंं कून ऎन C now let us see how to find out the quality of steam entering the condenser for finding out quality of steam entering condenser we will be using the energy balance equation and heat lost by the steam equal to heat gained by the cooling water now heat lost by the steam is given by an equation mass of the steam multiplied by enthalpy of steam entering the condenser minus enthalpy of condenser leaving the condenser which will be equal to heat gained by cooling water will be equal to mass of cooling water into specific heat of cooling water into rise in temperature of cooling water now the enthalpy of steam entering steam is given by an equation h is equal to hf plus x into hfg since the steam entering the condenser is weight we have to use the equation for enthalpy of weight steam formula where hf and hfg are obtained from steam table corresponding to absolute pressure of steam in the condenser which is equal to 0.00704 bar therefore hf and hfg are taken from steam table corresponding to pressure table in the steam table with the reference to pressure of 0.0704 bar and hc is enthalpy of saturated liquid at hot well temperature of 28 degree Celsius which is obtained as 117.3 kJ per kg so using the values of hf plus x into hfg and hc as well as mass of steam entering condenser mass of cooling water supplied and specific heat of steam and rise in temperature of cooling water when we put these values we can obtain by solving this equation the value of x that is nothing but the dryness fraction of steam entering the condenser which is obtained as x is equal to 0.819 now let us see the second problem in which the problem statement is the vacuum efficiency of condenser is 95% and the temperature of condenser is 40 degree Celsius if the barometric reading is 75.4 cm of mercury find the vacuum gauge reading of the condenser so here we are given with the vacuum efficiency mean condenser temperature, barometric pressure and we are supposed to find out the vacuum gauge reading of the condenser that is vacuum in the condenser let us see how to find out the solution now already we have defined that vacuum efficiency is actual vacuum upon ideal vacuum actual vacuum we can denote it in terms of head of mm of mercury or cm of mercury as hv and ideal vacuum is nothing but barometric pressure barometric reading hv minus saturation pressure of the steam in terms of cm of mercury therefore we can write this as the saturation pressure of steam corresponding to mean condenser temperature of 40 degree Celsius is obtained from steam table corresponding to temperature table at 40 degree Celsius we get saturation pressure of 0.0738 bar and converting this into cm of mercury as 5.535 cm of mercury we can write now the value as vacuum efficiency is equal to given as 95% that is 0.95 equal to the actual vacuum that is vacuum gauge reading which we are interested in to find divided by barometric pressure given in the problem is 75.0 cm of mercury 75.4 cm of mercury minus this saturation pressure as 5.535 using these values and solving this we get the vacuum gauge reading in the condenser as 66.37 cm of mercury which we can convert as 0.88 bar these are the references thank you