PURE SUBSTANCES DEFINITIONS: A pure substance is one which consists of only single species with characteristic feature of being of homogeneous and invariable composition irrespective of phase or phases in which it exists. A pure substance can exist in three different phases/states/conditions namely, solid, liquid and gas (vapour). Examples of pure substances are (1) a system composed of liquid and vapour phases of water (2) air (mixture of 79%-N and 21%-O by volume) if its composition does not change when undergoes a cooling/heating expansion/compression process (3) vapours of Fr-12, Fr-22 and NH3 used in refrigeration and air conditioning systems for cooling effect, etc. If some of the liquid is vaporized during a process, the system will be in chemically homogeneous but not in physical homogeneous because liquid density is more than that of vapour density. A phase is any homogeneous part of the system that is physically distinctive. It is essentially homogeneous, chemical and physical state of aggregation of molecules comprising the substance. Each phase of the system is separated from the other phases by the interfaces called phase boundaries. For example, considering a solid ice cube in liquid water, the two phases are separated by a phase boundary which is the surface of the cube. A change of phase is the transition from one phase to another phase of the substance. For example, melting/freezing is the change of phase between solid and liquid, vaporization/condensation is the change of phase between liquid and vapour and sublimation/deposition is the change of phase between solid and vapour. Whenever a change of phase occurs, all the mass of substance does not change its phase simultaneously. Melting is the process of changing the phase from solid to liquid. Freezing is the process of changing the phase from liquid to solid. Vaporization is the process of changing the phase from liquid to gas/vapour. Condensation is the process of changing the phase from gas/vapour to liquid. Sublimation is the process of changing the phase from solid to gas/vapour. Deposition is the process of changing the phase from gas/vapour to solid. Evaporation is the process of vapour generation from the surface of the liquid. It occurs at any temperature. Boiling is the process of formation of vapour from the whole mass of the liquid. When heat is added to a liquid, it starts boiling and vapours get generated both from the free surface as well as inside the liquid. Saturation temperature (ts) is the definite temperature at which all the liquid is transformed into vapour. At this constant temperature, the liquid is in equilibrium with its own vapour. For ex: ts for water is equal to 100C. Saturation pressure (ps) is the pressure of the liquid/vapour phase corresponding to saturation temperature at which the liquid boils or is in thermal equilibrium with its own vapour. For ex: ps for water is equal to 1.01325 bar corresponding to ts = 100C. Saturated liquid is the liquid which is in equilibrium with its own vapour at specified temperature/pressure at its boiling point. Saturated vapour is the vapour which is in equilibrium with its own liquid at specified temperature/pressure. Saturated vapour/steam may be wet or dry. Wet saturated steam is the saturated vapour which contains particles of liquid evenly distributed over the entire mass of the vapour. Dry saturated steam is the saturated vapour that contains no liquid particles. Saturation point (or) saturation temperature (or) boiling point is the temperature at which the molecules escaping from the liquid become equal to the molecules returning to it. Dryness fraction is the ratio of mass of dry saturated vapour to the total mass of the mixture. It is denoted by ‘x’. Let, mf = mass of water vapour (saturated liquid) in suspension with saturated dry steam. mg = mass of dry saturated steam = mf + mg m = total mass of the mixture, mass of dry saturated vapour mg mg then, Dryness fraction, x total mass of mixture m m f mg Wetness fraction is the ratio of mass of water or moisture in suspension with the total mass of the steam which contains it. It is denoted by ‘y’ mass of water m mf mg Wetness fraction, y f 1 1 x total mass of mixture m m f mg m f mg Degree of superheat is the difference between the temperature of superheated steam (tsup) and temperature of saturation liquid (ts). Higher the degree of superheat, greater is the volume of steam occupied. At very high superheat, the steam behaves like an ideal gas. STEAM GENERATION: The process of steam generation can be divided into three stages namely, 1. Heating of water up to boiling point 2. Evaporation of boiling water and its conversion into dry saturated steam, and 3. Transformation of dry saturated steam into superheated steam. Considering an example of 1kg of ice at -10C to be heated to 200C generate steam. Consider the following T-Q, T-v and P-v diagrams. When heat is added to ice at -10C (point 1), its temperature increases to 0C without any Step-1 change in its phase. This quantity of heat used to increase the temperature without change in phase is called sensible heat of ice. (point 2, process 1-2) When more heat is added to ice at 0C, the ice is completely transformed into water without any change in its temperature. This quantity of heat used to change the phase at constant Step-2 temperature is called latent heat of fusion of ice. At atmospheric pressure of 1.01325 bar (1 atm), its value is equal to 335 kJ/kg. There is a nominal decrease in volume for water and increase in volume for other substances. This is the peculiarity of water. (point 3, process 2-3) During supply of heat to water at 0C, it continues till boiling point is reached. This heat required to change from liquid state to vaporization is called sensible heat of water. This boiling point depends on pressure at which heat is supplied and is called saturation temperature. At Step-3 normal atmospheric pressure, its value is equal to 100C. There is a decrease in volume of water up to 4C and thereafter volume increases up to the commencement of boiling (point 4, process 3-4). Evaporation of water starts at 100C and once again another change of phase from liquid to vapour occurs due to supply of latent heat of vaporization at this constant temperature of 100C Step-4 whose value at constant pressure of 1 atm is equal to 2257 kJ/kg.. This process of vaporization which is both isothermal and isobaric, increases the volume of water. (point 4, process 4-5) At point 5, More heat is supplied to vapour at 100C to generate dry saturated steam at 200C. This quantity of heat is called sensible heat of vapour. Once the steam attains dry saturated Step-5 state, its temperature and volume start increasing with further supply of heat. The steam so obtained is called supersaturated steam and it behaves like a perfect gas. (point 6, process 5-6) The horizontal distance between saturated liquid line and dry saturated vapour line becomes less and less with increase of pressure and saturation temperature. The point at which this distance becomes zero and represents highest temperature and pressure above which liquid no longer exists is called Critical point. For water, the temperature, pressure and volume corresponding to critical point are called critical temperature (tcritical = 374.14K), critical pressure (pcritical = 220.9 bar) and critical volume (vcritical = 0.005155 m3/kg) SOLID-LIQUID-VAPOUR EQUILIBRIUM The figure shows the phase diagram drawn between pressure on y-axis and temperature on x-axis for water in which regions of solid, liquid and vapour are indicated. Single phase regions are separated from each other by saturation lines Equilibrium exists between (i) Solid & vapour phases along sublimation(1-T) or deposition(T-1) curves (ii) Solid and liquid phases along melting (2-T) or freezing (T-2) curves (iii) liquid and vapour phases along evaporation (T-C) or condensation (C-T) curves The slope of the melting/freezing curve is positive for most of the substances, but is negative for water. The evaporation line ends at critical point where the liquid and vapour phases are indistinguishable. The sublimation/deposition, melting/freezing and evaporation/condensation meet at a point called triple point at which all the three phases co-exist in equilibrium. This point exists at a definite temperature and pressure. For water, its values are ttp = 0.01C and ptp = 0.611 kPa The process where solid is directly transformed into vapour without touching liquid phase is called sublimation. The reverse of sublimation process is referred as deposition/ablimation. ENTHALPY CHANGE DURING FORMATION OF ICE The quantity of heat required to change from ice at -10C to steam at 250C is called change in enthalpy. Change in enthalpy of ice from -10C to 0C is called sensible heat of ice which is equal to h12 c pidt where c pi specific heat of ice = 2.09 kJ/kg-K h12 2.090 10 20.9kJ / kg K Change in enthalpy during melting of ice at 0C is called latent heat of fusion of ice, h23 335kJ / kg K Change in enthalpy during heating of water from 0C to 100C is called sensible heat of ice which is equal to h34 c pwdt where c pw specific heat of water = 4.186 kJ/kg-K h34 4.186100 0 418.6kJ / kg K Change in enthalpy during vaporization of water at 100C is called latent heat of vaporization of water, h45 2256.9kJ / kg K Change in enthalpy during vaporization from 100C to 250C is called sensible heat of vapour which is equal to h56 c psdt where c ps specific heat of steam = 2.1 kJ/kg-K h56 2.1250 100 315kJ / kg K Total change in enthalpy during change of ice at 0C to steam at 250C is equal to h h12 h23 h34 h45 h56 20.9 335 418.6 2256.9 315 3346.4kJ / kg K ENTROPY CHANGE DURING FORMATION OF ICE Change in entropy of ice from -10C to 0C is equal to s 1 c pi ln t2 t1 Change in entropy during melting of ice at 0C is equal to s 2 Qice Qice 335 1.23kJ / kg K t2 t3 273 Change in entropy during heating of water from 0C to 100C is equal to s 3 c pw ln t4 t3 Change in entropy during vaporization of water at 100C is equal to s 4 Qwater Qwater 2257 6.05kJ / kg K t4 t5 373 Change in entropy during heating of water from 100C to 250C is equal to s 5 c ps ln t6 t5 Total change in entropy during change of ice at 0C to steam at 250C is equal to t2 t t s s 1 s 2 s 3 s 4 s 5 c pi ln 1.23 c pw ln 4 6.05 c ps ln 6 t1 t3 t5 INDEPENDENT PROPERTIES OF PURE SUBSTANCE Let us consider for a pure substance, For liquid-vapour Property Symbol For liquid For vapour Mixture dryness When liquid starts boiling, When vaporization completes X 0<x<1 fraction x = 0 (saturated water) x = 1 (dry saturated vapour) specific volume V vl (or) vf (or) vw vg vfg pressure P psat = saturated pressure psup = superheated pressure pcr = critical pressure temperature T tsat = saturated temperature tsup = superheated temperature tcr = critical temperature specific h hl (or) hf (or) hw hg hfg enthalpy specific entropy s sl (or) sf (or) sw sg sfg mass m ml (or) mf (or) mw mg m (or) mfg specific internal u ul (or) uf (or) uw ug ufg energy If the pure substance exists at saturation points (i.e. points on saturated liquid or saturated vapour curves), all the independent properties can be extracted from the tables at different saturation temperature and saturation pressure. If the pure substance exists in sub-cooled region (i.e. wet region) or superheated region (i.e. dry region), the properties are to be evaluated as discussed below. Wet region: The latent heat of ice (hfg) is to be added to saturated liquid, the enthalpy increases from hf to hg. Therefore, hg h f h fg (1) If x% of latent heat is supplied, then hx h f xh fg (2) s x s f xs fg (3) u x u f xu fg (4) vx v f xv fg (5) Superheated region: The enthalpy of a saturated vapour (hg) can directly be taken from tables. The enthalpy of superheated vapour is equal to the energy of the vapour when it is at superheated temperature (tsup). So the change in enthalpy at constant pressure process is given by, hsup hg c p tsup t sat (or) hsup hg c p tsup t sat (6) tsup tsup ssup s g c p ln (or) ssup s g c p ln (7) t sat t sat usup u g cv tsup t sat (or) usup u g cv tsup t sat (8) vsup vs (or )v g (9) tsup t sat Sub-cooled region: In the sub-cooled region, the enthalpy is less than the enthalpy of a saturated liquid and adopting same procedure as that of superheated region, we have hsub h f c pl t sat t sub (10) t sat ssub s f c pl ln (11) t sub usub u f cvl t sat t sub (12) vsub vsat (13) EQUATION OF STATE FOR VAPOUR PHASE OF A SIMPLE COMPRESSIBLE SUBSTANCE Referring to the figure below, a-b-c-d represents the elementary Carnot cycle for dry saturated steam on p-v diagram. The point a lies on saturated liquid line and point b lies on dry saturated steam line. Let dtsup represents difference between two isothermal lines a-b and c-d. The efficiency of this elementary Carnot cycle is given by, t1 t 2 dt sat (1) t1 t sat The efficiency of this elementary Carnot cycle can also be given from the following relation, work done v v f dp g (2) heat supplied h fg where dp is the difference in pressures of two isothermal lines. Equating the above two equations, we get dt sat vg v f dp t sat h fg (or) v g vf h fg dt sat (3) t sat dp This equation (3) is called Clapeyron’s equation which gives specific volume at any required pressure and dt temperature. The value of sat is obtained from slope of temperature-pressure curve which is plotted dp from experimental data. PROBLEMS ON PURE SUBSTANCES 1. Calculate the dryness fraction of steam which has 1.25 kg of water in suspension with 40 kg of steam. Solution: Mass of steam, = 40 kg, Mass of water, m f = 1.25 kg Mass of mixture, m mg m f = 40 + 1.25 = 41.25 kg mg 40 Dryness fraction, x 0.9696 m 41.25 2. Using steam tables, determine the mean specific heat for superheated steam at 1 bar and between 150C and 200C. Solution: From steam tables of superheated steam and at 1 bar pressure, we have At 150C: hsup 2776.4kJ / kg At 200C: hsup 2875.3kJ / kg 2875.3 = 2776.4 + c ps (200 – 150) 2875.3 2776.4 c ps 1.978kJ / kgK 50 3. A spherical shell of 30 cm in radius contains saturated steam and water at 300C. Calculate the mass of each if their volumes are equal. 4 3 4 Solution: Total volume of the shell = r (0.3)3 0.113m3 3 3 0.113 Volume of vapour and that of water = V f Vg 0.0565m3 2 From steam tables, corresponding to saturation temperature of 300C, we have v f 0.001404m3 / kg , vg 0.0216m3 / kg Vg 0.0565 Mass of vapour, mg 2.616kg , and vg 0.0216 Vf 0.0565 Mass of water, m f 40.242kg vf 0.001404 4. A tank contains 100 kg of liquid water and 5 kg of water vapour under saturation conditions at 20C. If the specific volume of saturated vapour at that temperature is taken as 57.78 m 3/kg, calculate the volume of tank and moisture content of the mixture. Solution: From steam tables, corresponding to saturation temperature of 20C, we have, v f 0.001002m3 / kg , vg 57.79m3 / kg that gives Volume of water = V f m f v f 100 0.001002 0.1002m3 Volume of vapour = Vg mg vg 5 57.79 288.95m3 Total volume, V V f Vg 0.1002 288.95 289m3 Moisture content of the mixture refers to wetness fraction, mass of water mf 100 100 y 0.952 total mass of mixture m f mg 100 5 105 5. A closed container holds 1 kg of water at 153.3C temperature with the composition by volume 1/3 liquid and 2/3 vapour. Determine the pressure, quality, volume and enthalpy of the mixture. Solution: Since a mixture of water and vapour is considered, its pressure would be the same as the saturation pressure corresponding to 153.3C. Then from steam tables, psat 5.2bar Vf 1/ 3 1 mf vf 1 Given, (or) Vg 2/3 2 mg v g 2 mf 0.361 m 165 mg f mg 2 0.001094 165 mg m f / 165 1 Dryness fraction, x 0.00602 m f mg m f m f / 165 166 For wet steam with dryness fraction, x 0.00602 , v 1 x v f xv g 1 0.00602 0.001094 0.00602 0.361 3.2606 103 m3 / kg h h f xh fg 646.7 0.00602 2103.8 659.363.2606 103 kJ / kg 6. A vessel of 0.3 m3 capacity contains 1.5 kg of water and steam in equilibrium at a pressure of 5 bar. Calculate the volume and mass of liquid and water vapour. 0.3 Solution: The specific volume of the given mixture, v 0.2m3 / kg 1.5 From steam tables, the specific volumes of saturated liquid and saturated vapour at 5 bar pressure are noted as v f 0.001093m3 / kg and v g 0.375m3 / kg v xv g 1 xv f 0.2 x 0.375 1 x0.001093 x 0.532 mg Now by definition, x m g x m m g m mg f f Mass of vapour, mg x m mg 0.532 1.5 0.798kg f Mass of water, m f 1.5 0.798 0.702kg Volume of vapour, V g mg v g 0.798 0.375 0.2992m3 Volume of water, V f m f v f 0.702 0.001093 0.00135m3 7. A vessel of volume 0.04 m3 contains a mixture of saturated water and saturated steam at a temperature of 250C. The mass of the liquid present is 9 kg. Find the pressure, the mass, the specific volume, the enthalpy, the entropy and the internal energy. Solution: From steam tables, corresponding to saturation temperature of 20C, psat 39.73bar , v f 0.0012512m3 / kg vg 0.05013m3 / kg h f 1085.36kJ / kg h fg 1716.2kJ / kg s f 2.7927kJ / kgK s fg 3.2802kJ / kgK Volume of liquid, V f m f v f 9 0.0012512 0.01126m3 Volume of vapour, Vg V V f 0.04 0.01126 0.02874m3 Vg 0.02874 Mass of vapour, mg 0.575kg vg 0.05013 Total mass of mixture, m m f mg 9 0.575 9.575kg mg 0.575 Quality of mixture, x 0.06 (or) 6% m 9.575 v v f xv fg 0.0012512 0.06(0.05013 0.0012512) 0.00418m3 / kg h h f xh fg 1085.36 0.06 1716.2 1188.32kJ / kg s s f xs fg 2.7927 0.063.2802 2.9895kJ / kg K u h pv 1188.32 3.973 103 0.00418 1171.72kJ / kg Also at 250C, u f 1080.39kJ / kg u g 1522kJ / kg u u f xu fg 1080.39 0.06 1522 1071.71kJ / kg 8. Calculate the enthalpy and entropy of steam at 10 bar when it is (a) saturated (b) 0.95 dry (c) superheated to 260C. Take cp of steam = 2.09 kJ/kg-K. Solution: (a) When steam is saturated and is at 10 bar, the properties can be obtained directly from the saturated steam tables: at 10 bar, hg 2776.2kJ / kg and s g 6.5828kJ / kg K (b) When steam is 0.95 dry x 0.95 and is at 10 bar, hx h f x hg h h0.95 762.6 0.95 2776.2 762.2 2676kJ / kg f s x s f x s g s s0.95 2.1382 0.95 6.5828 2.1382 6.36kJ / kg K f (c) When steam is superheated at 260C tsup 260C , From superheated steam tables, t sat 179.9C hsup hg c pg tsup t sat 2776.2 2.09 260 179.9 2943.6kJ / kg tsup 260 273 ssup s g c pg ln 6.5828 2.09 ln 6.923kJ / kg K t sat 179.9 273 9. Calculate the amount of heat supplied to a boiler to generate 10 kg/s of steam at a pressure of 10 bar and 0.95 dry if the water is supplied to it at 30C and at the same pressure. Solution: From first law of thermodynamics, dQ dh dW dQ dh dW 0 in constant pressure process dQ m h2 h1 Where, h2 is enthalpy of steam at 10 bar and 0.95 dry and is given by, h2 h f 0.95 hg h f 762.6 0.95 2776 2762.2 2676kJ / kg Where h1 is enthalpy of steam at 30C and is equal to h1 125.8kJ / kg Heat supplied, dQ 10 2676 125.8 25502kJ / s 25.5MW 10. One kg of steam at 20 bar exits in the following conditions (a) 0.9 dry (b) dry and saturated (c) at a temperature of 250C. Find its enthalpy, specific volume, density, internal energy and entropy in each case. Assume cp of steam = 2.01 kJ/kg-K. Solution: (a) When steam is 0.9 dry, hx h f x hg h h0.9 908.6 0.9 2797.2 908.6 2608.3kJ / kg f s x s f x s g s s0.9 2.446 0.9 6.336 2.446 5.947kJ / kg K f v0.9 xv g 0.9 0.0995 0.08955m3 / kg 1 1 0.9 11.166m3 / kg v0.9 0.08955 u0.9 h pv0.9 2608.3 2000 0.08955 2428.6kJ / kg (b) When steam is dry saturated, hg 2797.2kJ / kg , s g 6.3366kJ / kg K , v g 0.0995m3 / kg , 10.05kg / m3 u h pv 2797.2 2000 0.0995 2598.2kJ / kg (c) When steam is superheated, tsup 250C , t sat 212.4C , hsup hg c pg tsup t sat 2797.2 2.01 250 212.4 2872.7kJ / kg tsup ssup s g c pg ln 6.4865kJ / kg K t sat v 250 273 0.1072m3 / kg , g 0.0995 v t sup t sat sup 212.4 273 1 1 sup 9.328kg / m3 vsup 0.1072 usup h pv sup 2872.7 2000 0.1072 2658.3kJ / kg 11. Ten kg of water at 45C is heated at constant pressure of 10 bar until it becomes superheated vapour at 300C. Find the changes in volume, enthalpy, internal energy and entropy. (Ans: 2.569 m3, 28627.5 kJ, 26047.6kJ, 64.842 kJ/K) 12. A rigid vessel of o.86 m3 capacity contains 1 kg of steam at 2 bar. Evaluate the specific volume, temperature, dryness fraction, internal energy, enthalpy and entropy of steam. (Ans: 0.86 m3/kg, 120.2C, 0.97, 2468.5 kJ/kg, 2640.5 kJ/kg, 6.959 kJ/kg-K) 13. A mass of wet steam at 165C is expanded at constant quality of 80% to a pressure of 3 bar. Use steam tables to determine the initial pressure of steam and changes in enthalpy and entropy during the expansion process. (Ans: 7 bar, 59 kJ/kg, 0.163kJ/kg-K) 14. Steam initially at 60 bar and 450C is allowed to cool to 200C at constant volume. Find the final condition of syeam and heat transfer if the mass of hte steam is 0.75 kg. (Ans: 0.4038, -1423.35 kJ) 15. Two kg of dry saturated steam at 11 bar is to be generated from water available at 45C. Determine the quantity of heat to be supplied for raising the steam. If subsequently the steam is to be superheated by 50C, what additional heat input is required. Take cp for superheated steam = 2.25 kJ/kg-K and cp for water = 4.18 kJ/kg-K. (Ans: 25714.8kJ/kg, 112.5kJ/kg) QUESTIONS ON PURE SUBSTANCES 1. Define pure substance, phase of a substance, critical point, triple point, dryness fraction. 2. Differentiate between evaporation, boiling and vaporization of steam. 3. Differentiate between saturation liquid and saturation vapour for a pure substance. 4. Expalin phase equilibrium diagram for water with neat diagram. 5. Explain the process of steam generation at constant pressure process. Show the stages on p-v, t-v and t-s diagrams. 6. Differentiate between sub-cooled region, wet region and superheated region for a pure substance. 7. Express various independent properties of a pure substance for a dryness fraction ‘x’ . 8. Explain equation of state for vapour phase of a simple compressible pure substance. 9. Explain Mollier diagram and its features with a neat sketch. 10. List out the advantages of using superheated steam.