Determining Dryness Fraction SUPERHEATED EVAPORATION 100 C WATER LINE 0 C g Temperature/Entalpy Dryness Fraction Tis andout assumes te student is amiliar wit Steam Tables and te terms ound terein. I required, please review Steam Tables beore continuing. Te eavy solid line in te above cart represents te transormation o 1 kg o water at 0 C and atmosperic pressure into steam, wit te addition o eat. From 0 C to 100 C, energy is added to te water in te orm o sensible eat. Tis causes an increase in temperature wile its state remains te same. Tis energy is reerred to in steam tables as. Te cange rom 1 kg o water at 0 C into 1 kg o water at 100 C requires te addition o 419.04 kj o energy. Tereore, at atmosperic pressure: = 419.04 kj/kg Te saturation temperature o water at atmosperic pressure is 100 C. Te addition o more eat will not cause a temperature cange but will, instead, cause a cange o state. In tis case, evaporation into steam at 100 C. Te cange rom 1 kg o water at 100 C
into dry saturated steam at 100 C requires te addition o 2257.0 kj o energy. Tereore, at atmosperic pressure: = 2257.0 kj/kg. Reerring to te diagram, you will notice tat g = + I not all o is added to te water, ten not all o te water can cange into steam. I, say, 50% o is added ten only 50% o te water will be canged into steam and te steam will be reerred to as 50% dry. Tis is known as te dryness raction o te steam. Expanding on tis, ten, it can be seen tat te entalpy content o steam o a certain dryness raction, and made rom water at 0 C, can be ound out by ling and tat portion o wic as been added ( multiplied by te dryness raction). I te steam is 100% dry, ten te entire amount o as been added and te dryness raction o te steam would be 100% or 1. Putting tis into a ormula we get: entalpy = dryness raction Using tis ormula and te inormation ound in Steam Tables, one can determine te dryness raction o steam at any pressure or temperature. Example 1: One kilogram o steam at 1400 kpa as a entalpy content o 2202.09 kj. Determine te dryness raction o te steam. From te Steam Tables: = 830.30 kj/kg = 1959.7 kj/kg g = 2790.0 kj/kg Since it is given tat te entalpy o te steam is less tan tat o dry saturated steam at 1400 kpa ( g ), one knows tat te steam is wet. dryness raction dryness raction Ten, substituting in te values,
2202.09 830.30 1959.7 2202.09 830.30 1959.7 0.7 Tereore te dryness raction o te steam is 0.7 or 70% Example 2 Consider 5 kg o steam at 2000 kpa wit a entalpy o 12571 kj. Wat is te dryness raction? First, te Steam Tables give igures tat are per kilogram. So we ave to determine te entalpy o 1 kg o te steam. 12571 5 2514.2 kj / kg From te Steam Tables Steam at 2000 kpa: = 908.79 kj/kg = 1890.7 kj/kg g = 2799.5 kj/kg Comparing te actual entalpy o te steam and g or steam at 2000 kpa, we know te steam is wet. Using te ormula, substitute te values in and solve or te dryness raction. dryness raction dryness raction Ten, substituting in te values, 2514.2 908.79 1890.7 2514.2 908.79 1890.7 0.849 Tereore te dryness raction is 0.849 or 84.9%
Example 3 How muc eat must be supplied to 200 kgs o water at 20C to make steam at 850 kpa wic is 87% dry? Tis is a two part question. You irst ave to determine wat te entalpy o te steam (at 850 kpa) is. D. F. Reerring to te steam tables, at 850 kpa: 732.22 kj / kg 2039.4 kj / kg D. F..87 Substituting tese into te ormula, we get 732.22 0.87 2039.4 2506.498 kj / kg Tis is te eat content o te steam. All te igures in te steam tables are assuming tat te steam is being created rom water at 0 degrees C. In tis question, owever, te water is warmer tan 0 degrees. So we ave to take tat into account. It's eat we don't ave to add again. So, rom Table II, we ind tat te water as an entalpy o: 83.96 kj/kg We subtract tis rom te eat content o te steam and we get: 2422.538 kj/kg Tis is ow muc eat we ave to add to 1 kg o water at 20 degrees to produce 1 kg o steam at 850 kpa and 87% dry. Multiply tis by 200 to get your inal answer: 484507.6 kj/kg
Example 4 Steam enters a turbine at 1100 kpa, dry and saturated. Te steam is exausted rom te turbine at 361.3 kpa and is 15% wet. Determine te quantity o eat used to do work in te turbine. In tis question, you need to determine te eat content o te steam entering te turbine and again wen it leaves te turbine. Te dierence is te amount o eat used to do work. From te steam tables, or steam at 1100 kpa, dry and saturated, g 2781.7 kj / kg Reerring to Table II For steam tat is at 361.3 kpa (or 140 degrees C) and 15% wet (85% dry) 589.13 2144.7 Inserting tese igures into te dryness raction equation we get: 589.13 0.85 2144.7 2412.125 kj / kg Subtracting tese two igures we get: 2781.7 2412.125 369.575 kj/kg Wic is te amount o eat used to do work.