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### Unit8

1. 1. PROPERTIES OF STEAM J2006/8/1 UNIT 8PROPERTIES OF STEAM OBJECTIVESGeneral Objective : To define the properties of wet steam, dry saturated steam and superheated steam using information from the steam tables.Specific Objectives : At the end of the unit you will be able to:  define the word phase and distinguish the solid, liquid and steam phases  understand and use the fact that the vaporization process is carried out at constant pressure  define and explain the following terms: saturation temperature, saturated liquid, wet steam, saturated steam, dry saturated steam, , dryness fraction and superheated steam  determine the properties of steam using the P-v diagram  understand and use the nomenclature as in the Steam Tables  apply single and double interpolation using the steam tables  locate the correct steam tables for interpolation, including interpolation between saturation tables and superheated tables where necessary
2. 2. PROPERTIES OF STEAM J2006/8/2 INPUT8.0 Introduction In thermodynamic systems, the working fluid can be in the liquid, steam or gaseous phase. In this unit, the properties of liquid and steam are investigated in some details as the state of a system can be described in terms of its properties. A substance that has a fixed composition throughout is called a pure substance. Pure chemicals (H2O, N2, O2, Ar, Ne, Xe) are always pure substances. We all know from experience that substances exist in different phases. A phase of substance can be defined as that part of a pure substance that consists of a single, homogenous aggregate of matter. The three common phases for H2O that are usually used are solid, liquid and steam. When studying phases or phase changes in thermodynamics, one does not need to be concerned with the molecular structure and behavior of the different phases. However, it is very helpful to have some understanding of the molecular phenomena involved in each phase. Molecular bonds are strongest in solids and weakest in steams. One reason is that molecules in solids are closely packed together, whereas in steams they are separated by great distances.
3. 3. PROPERTIES OF STEAM J2006/8/3 The three phases of pure substances are: - Solid Phase In the solid phase, the molecules are; (a) closely bound, therefore relatively dense; and (b) arranged in a rigid three-dimensional pattern so that they do not easily deform. An example of a pure solid state is ice. Liquid Phase In the liquid phase, the molecules are; (a) closely bound, therefore also relatively dense and unable to expand to fill a space; but (b) they are no longer rigidly structured so much so that they are free to move within a fixed volume. An example is a pure liquid state. Steam Phase In the steam phase, the molecules; (a) virtually do not attract each other. The distance between the molecules are not as close as those in the solid and liquid phases; (b) are not arranged in a fixed pattern. There is neither a fixed volume nor a fixed shape for steam. The three phases described above are illustrated in Fig. 8.0 below. The following are discovered: (a) the positions of the molecules are relatively fixed in a solid phase; (b) chunks of molecules float about each other in the liquid phase; and (c) the molecules move about at random in the steam phase. (a) (b) (c) Source: Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles Figure 8.0 The arrangement of atoms in different phases
4. 4. PROPERTIES OF STEAM J2006/8/4 8.1 Phase-Change Process The distinction between steam and liquid is usually made (in an elementary manner) by stating that both will take up the shape of their containers. However liquid will present a free surface if it does not completely fill its container. Steam on the other hand will always fill its container. With these information, let us consider the following system: A container is filled with water, and a moveable, frictionless piston is placed on the container at State 1, as shown in Fig. 8.1. As heat is added to the system, the temperature of the system will increase. Note that the pressure on the system is being kept constant by the weight of the piston. The continued addition of heat will cause the temperature of the system to increase until the pressure of the steam generated exactly balances the pressure of the atmosphere plus the pressure due to the weight of the piston. STATE 1 STATE 2 STATE 3 STATE 4 W W W W Steam Superheated Liquid Steam Figure 8.1 Heating water and steam at constant pressure At this point, the steam and liquid are said to be saturated. As more heat is added, the liquid that was at saturation will start to vaporize until State 2. The two-phase mixture of steam and liquid at State 2 has only one degree of freedom, and as long as liquid is present, vaporization will continue at constant temperature. As long as liquid is present, the mixture is said to be wet steam, and both the liquid and steam are saturated. After all the liquid is vaporized, only steam is present at State 3, and the further addition of heat will cause the temperature of steam to increase at
5. 5. PROPERTIES OF STEAM J2006/8/5 constant system pressure. This state is called the superheated state, and the steam is said to be superheated steam as shown in State 4.8.2 Saturated and Superheated Steam While tables provide a convenient way of presenting precise numerical presentations of data, figures provide us with a clearer understanding of trends and patterns. Consider the following diagram in which the specific volume of H2O is presented as a function of temperature and pressure1: T, oC 300 4 Superheated steam Saturated 2 mixture 3 100 Compressed liquid 20 1 v, m3/kg Figure 8.2-1 T-v diagram for the heating process of water at constant pressure Imagine that we are to run an experiment. In this experiment, we start with a mass of water at 1 atm pressure and room temperature. At this temperature and pressure we may measure the specific volume (1/ρ = 1/1000 kg/m3). We plot this state at point 1 on the diagram. If we proceed to heat the water, the temperature will rise. In addition, water expands slightly as it is heated which makes the specific volume increase slightly. We may plot the locus of such points along the line from State 1 to State 2. We speak of liquid in one of these conditions as being compressed or subcooled liquid.1 Figures from Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
6. 6. PROPERTIES OF STEAM J2006/8/6 State 2 is selected to correspond to the boiling point (100 oC). We speak of State 2 as being the saturated liquid state, which means that all of the water is in still liquid form, but ready to boil. As we continue to heat past the boiling point 2, a fundamental change occurs in the process. The temperature of the water no longer continues to rise. Instead, as we continue to add energy, liquid progressively changes to steam phase at a constant temperature but with an increasing specific volume. In this part of the process, we speak of the water as being a saturated mixture (liquid + steam). This is also known as the quality region. At State 3, all liquid will have been vaporised. This is the saturated steam state. As we continue to heat the steam beyond State 3, the temperature of the steam again rises as we add energy. States to the right of State 3 are said to be superheated steam. Summary of nomenclature: Compressed or subcooled liquid (Between States 1 & 2) A liquid state in which the fluid remains entirely within the liquid state, and below the saturation state. Saturated liquid (State 2) All fluid is in the liquid state. However, even the slightest addition of energy would result in the formation of some vapour. Saturated Liquid-Steam or Wet Steam Region (Between States 2 & 3) Liquid and steam exist together in a mixture. Saturated steam (State 3) All fluid is in the steam state, but even the slightest loss of energy from the system would result in the formation of some liquid. Superheated steam (The right of State 3) All fluid is in the steam state and above the saturation state. The superheated steam temperature is greater than the saturation temperature corresponding to the pressure.
7. 7. PROPERTIES OF STEAM J2006/8/7 The same experiment can be conducted at several different pressures. We see that as pressure increases, the temperature at which boiling occurs also increases.2 T, oC P = 221.2 bar Critical point P = 150 bar P = 80 bar 374.15 P = 10 bar P = 5 bar P = 1.01325 bar Saturated Saturated liquid steam 0.00317 v, m3/kg Figure 8.2-2 T-v diagram of constant pressure phase change processes of a pure substance at various pressures for water. It can be seen that as pressure increases, the specific volume increase in the liquid to steam transition will decrease. At a pressure of 221.2 bar, the specific volume change which is associated to a phase increase will disappear. Both liquid and steam will have the same specific volume, 0.00317 m3/kg. This occurs at a temperature of 374.15 oC. This state represents an important transition in fluids and is termed the critical point.2 Figures from Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
8. 8. PROPERTIES OF STEAM J2006/8/8 If we connect the locus of points corresponding to the saturation condition, we will obtain a diagram which allows easy identification of the distinct regions3: Saturated liquid line T Dry saturated steam line Critical point P2 = const. P2 > P1 COMPRESS P1 = const. LIQUID REGION SUPERHEATED STEAM REGION WET STEAM REGION v Figure 8.2-3 T-v diagram of a pure substance The general shape of the P-v diagram of a pure substance is very much like the T-v diagram, but the T = constant lines on this diagram have a downward trend, as shown in Fig. 8.2-4. Saturated liquid line P Dry saturated steam line Critical point SUPERHEATED STEAM REGION COMPRESS LIQUID T2 = const. REGION T2 > T1 WET STEAM T1 = const. REGION v Figure 8.2-4 P-v diagram of a pure substance3 Figures from Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
9. 9. PROPERTIES OF STEAM J2006/8/9 Activity 8A TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…!8.1 Each line in the table below gives information about phases of pure substances. Fill in the phase column in the table with the correct answer. Statement Phase The molecules are closely bound, they are also relatively dense and unable to expand to fill a space. However they are i._____________ no longer rigidly structured so that they are free to move within a fixed volume. The molecules are closely bound, they are relatively dense and arranged in a rigid three-dimensional patterns so that they ii.____________ do not easily deform. The molecules virtually do not attract each other. The distance between the molecules are not as close as those in the iii.____________ solid and liquid phases. They are not arranged in a fixed pattern. There is neither a fixed volume nor a fixed shape for steam.8.2 Write the suitable names of the phases for the H2O in the P-v diagram below. ( ii ) P ( iv ) ( vi ) (v) (i) T2 = const. T2 > T1 ( iii) T1 = const. v
10. 10. PROPERTIES OF STEAM J2006/8/10 Feedback To Activity 8A8.1 i) Liquid Phase ii) Solid Phase iii) Steam Phase8.2 i) Compress liquid region ii) Saturated liquid line iii) Wet steam region iv) Dry saturated steam line v) Superheated steam region vi) Critical point
11. 11. PROPERTIES OF STEAM J2006/8/11 INPUT8.3 Properties of a Wet Mixture Between the saturated liquid and the saturated steam, there exist a mixture of steam plus liquid (wet steam region). To denote the state of a liquid-steam mixture, it is necessary to introduce a term describing the relative quantities of liquid and steam in the mixture. This is called the dryness fraction (symbol x). Thus, in 1 kg of wet mixture, there must be x kg of saturated steam plus (1 – x) kg of saturated liquid. x kg of steam total mass = 1 kg (1 - x ) kg of liquid Figure 8.3-1 Liquid-steam mixture The dryness fraction is defined as follows; mass of dry saturated steam dryness fraction = total mass msteam (8.1) x= mtotal where mtotal = mliquid + msteam
12. 12. PROPERTIES OF STEAM J2006/8/12 Sat. steam Sat. steam At point A, x = 0 Sat. liquid Sat. liquid At point B, x = 1 P x = 0.2 x = 0.8 Between point A and B, 0 < x < 1.0 Note that for a saturated liquid, x = 0; and that for dry saturated steam, x = 1. A B ts vf vg v Figure 8.3-2 P-v diagram showing the location point of the dryness fraction 8.3.1 Specific volume For a wet steam, the total volume of the mixture is given by the volume of liquid present plus the volume of dry steam present. Therefore, the specific volume is given by, volume of a liquid + volume of dry steam v= total mass of wet steam Now for 1 kg of wet steam, there are (1 – x) kg of liquid and x kg of dry steam, where x is the dryness fraction as defined earlier. Hence, v = vf(1 – x) + vgx The volume of the liquid is usually negligibly small as compared to the volume of dry saturated steam. Hence, for most practical problems, v = xvg (8.2) Where, vf = specific volume of saturated liquid (m3/kg) vg = specific volume of saturated steam (m3/kg) x = dryness fraction
13. 13. PROPERTIES OF STEAM J2006/8/13 8.3.2 Specific enthalpy In the analysis of certain types of processes, particularly in power generation and refrigeration, we frequently encounter the combination of properties U + PV. For the sake of simplicity and convenience, this combination is defined as a new property, enthalpy, and given the symbol H. H = U + PV (kJ) or, per unit mass h = u + Pv (kJ/kg) (8.3) The enthalpy of wet steam is given by the sum of the enthalpy of the liquid plus the enthalpy of the dry steam, i.e. h = hf(1 – x) + xhg h = hf + x(hg – hf ) ∴ h = hf + xhfg (8.4) Where, hf = specific enthalpy of saturated liquid (kJ/kg) hg = specific enthalpy of saturated steam (kJ/kg) hfg = difference between hg and hf (that is, hfg = hg - hf ) 8.3.3 Specific Internal Energy Similarly, the specific internal energy of a wet steam is given by the internal energy of the liquid plus the internal energy of the dry steam, i.e. u = uf(1 – x) + xug ∴ u = uf + x(ug – uf ) (8.5) Where, uf = specific enthalpy of saturated liquid (kJ/kg) ug = specific enthalpy of saturated steam (kJ/kg) ug – uf = difference between ug and uf
14. 14. PROPERTIES OF STEAM J2006/8/14 Equation 8.5 can be expressed in a form similar to equation 8.4. However, equation 8.5 is more convenient since ug and uf are tabulated. The difference is that, ufg is not tabulated.
15. 15. PROPERTIES OF STEAM J2006/8/15 8.3.4 Specific Entropy A person looking at the steam tables carefully will notice two new properties i.e. enthalpy h and entropy s. Entropy is a property associated with the Second Law of Thermodynamics, and actually, we will properly define it in Unit 9 . However, it is appropriate to introduce entropy at this point. The entropy of wet steam is given by the sum of the entropy of the liquid plus the entropy of the dry steam, i.e. s = sf(1 – x) + xsg s = sf + x(sg – sf ) ∴ s = sf + xsfg (8.4) Where, sf = specific enthalpy of saturated liquid (kJ/kg K) sg = specific enthalpy of saturated steam (kJ/kg K) sfg = difference between sg and sf (that is, sfg = sg - sf ) REMEMBER! These equations are used very often and are, therefore, important to remember! v = xvg h = hf + xhfg u = uf + x(ug – uf ) s = sf + xsfg
16. 16. PROPERTIES OF STEAM J2006/8/168.4 The Use of Steam Tables The steam tables are available for a wide variety of substances which normally exist in the vapour phase (e.g. steam, ammonia, freon, etc.). The steam tables which will be used in this unit are those arranged by Mayhew and Rogers, which are suitable for student use. The steam tables of Mayhew and Rogers are mainly concerned with steam, but some properties of ammonia and freon-12 are also given. Below is a list of the properties normally tabulated, with the symbols used being those recommended by British Standard Specifications. Table 8.4 The property of steam tables Symbols Units Description p bar Absolute pressure of the fluid o ts C Saturation temperature corresponding to the pressure p bar 3 vf m /kg Specific volume of saturated liquid 3 vg m /kg Specific volume of saturated steam uf kJ/kg Specific internal energy of saturated liquid ug kJ/kg Specific internal energy of saturated steam hf kJ/kg Specific enthalpy of saturated liquid hg kJ/kg Specific enthalpy of saturated steam hfg kJ/kg Change of specific enthalpy during evaporation sf kJ/kg K Specific entropy of saturated liquid sg kJ/kg K Specific entropy of saturated steam sfg kJ/kg K Change of specific entropy during evaporation These steam tables are divided into two types: Type 1: Saturated Water and Steam (Page 2 to 5 of steam tables) Type 2: Superheated Steam (Page 6 to 8 of steam tables) 8.4.1 Saturated Water and Steam Tables
17. 17. PROPERTIES OF STEAM J2006/8/17 The table of the saturation condition is divided into two parts. Part 1 Part 1 refers to the values of temperature from 0.01 oC to 100oC, followed by values that are suitable for the temperatures stated in the table. Table 8.4.1-1 is an example showing an extract from the temperature of 10oC. Table 8.4.1-1 Saturated water and steam at a temperature of 10 oC t ps vg hf hfg hg sf sfg sg 0 C bar m3/kg kJ/kg kJ/kg K 10 0.01227 106.4 42.0 2477.2 2519.2 0.151 8.749 8.900 Example 8.1 Complete the following table for Saturated Water and Steam: t Ps vg hf hfg hg sf sfg sg o C bar m3/kg kJ/kg kJ/kg K 0.01 206.1 0.02337 8.666 100 1.01325 Solution to Example 8.1 From page 2 of the steam tables, we can directly read: t Ps vg hf hfg hg sf sfg sg o C bar m3/kg kJ/kg kJ/kg K 1 0.006566 192.6 4.2 2498.3 2502.5 0.015 9.113 9.128 20 0.02337 57.84 83.9 2453.7 2537.6 0.296 8.370 8.666 100 1.01325 1.673 419.1 2256.7 2675.8 1.307 6.048 7.355
18. 18. PROPERTIES OF STEAM J2006/8/18 Part 2 Part 2 (Page 3 to 5 of steam tables) is values of pressure from 0.006112 bar to 221.2 bar followed by values that are suitable for the pressures stated in the table. Table 8.4.1-2 is an example showing an extract from the pressure of 1.0 bar. Table 8.4.1-2 Saturated water and steam at a pressure of 1.0 bar p ts vg uf ug hf hfg hg sf sfg sg o 3 bar C m /kg kJ/kg kJ/kg kJ/kg K 1.0 99.6 1.694 417 2506 417 2258 2675 1.303 6.056 7.359 Note the following subscripts: f = property of the saturated liquid g = property of the saturated steam fg = change of the properties during evaporations Example 8.2 Complete the missing properties in the following table for Saturated Water and Steam: p ts vg uf ug hf hfg hg sf sfg sg o 3 bar C m /kg kJ/kg kJ/kg kJ/kg K 0.045 31.0 2558 10 0.1944 311.0 5.615 Solution to Example 8.2 From page 3 to page 5 of the steam tables, we can directly read: p ts vg uf ug hf hfg hg sf sfg sg o bar C m3/kg kJ/kg kJ/kg kJ/kg K 0.045 31.0 31.14 130 2418 130 2428 2558 0.451 7.980 8.431 10 179.9 0.1944 762 2584 763 2015 2778 2.138 4.448 6.586 100 311.0 0.0180 1393 2545 1408 1317 2725 3.360 2.255 5.615 2
19. 19. PROPERTIES OF STEAM J2006/8/19 Example 8.3 For a steam at 20 bar with a dryness fraction of 0.9, calculate the a) specific volume b) specific enthalpy c) specific internal energy Solution to Example 8.3 An extract from the steam tables p ts vg uf ug hf hfg hg sf sfg sg 20 212.4 0.09957 907 2600 909 1890 2799 2.447 3.893 6.340 a) Specific volume (v), v = xvg = 0.9(0.09957) = 0.0896 m3/kg b) Specific enthalpy (h), h = hf + xhfg = 909 + 0.9(1890) = 2610 kJ/kg c) Specific internal energy (u), u = uf + x( ug -uf ) = 907 + 0.9(2600 - 907) = 2430.7 kJ/kg P bar x = 0.9 20 ts = 212.4 oC v m3/kg v vg uf u u g hf h h s g sf sg
20. 20. PROPERTIES OF STEAM J2006/8/20 Example 8.4 Find the dryness fraction, specific volume and specific enthalpy of steam at 8 bar and specific internal energy 2450 kJ/kg. Solution to Example 8.4 An extract from the steam tables, p ts vg uf ug hf hfg hg sf sfg sg 8 170.4 0.2403 720 2577 721 2048 2769 2.046 4.617 6.663 At 8 bar, ug = 2577 kJ/kg, since the actual specific internal energy is given as 2450 kJ/kg, the steam must be in the wet steam state ( u < ug). From equation 8.5, u = uf + x(ug -uf) 2450 = 720 + x(2577 - 720) ∴ x = 0.932 From equation 8.2, v = xvg = 0.932 (0.2403) = 0.2240 m3/kg From equation 8.4, h = hf + xhfg = 721 + 0.932 (2048) = 2629.7 kJ/kg P bar x = 0.932 8 ts = 170.4 oC v m3/kg v vg
21. 21. PROPERTIES OF STEAM J2006/8/21 Activity 8B8.3 The internal energy of wet steam is 2000 kJ/kg. If the pressure is 42 bar, what is the value of dryness fraction?8.4 Determine the specific volume, specific enthalpy and specific internal energy of wet steam at 32 bar if the dryness fraction is 0.92.8.5 Find the dryness fraction, specific volume and specific internal energy of steam at 105 bar and specific enthalpy 2100 kJ/kg.
22. 22. PROPERTIES OF STEAM J2006/8/22 Feedback To Activity 8B8.3 DRYNESS FRACTION (x): u = uf + x(ug -uf) 2000 = 1097 + x(2601 - 1097) x = 0.68.4 SPECIFIC VOLUME (v): v = xvg = 0.92(0.06246) = 0.05746 m3/kg SPECIFIC ENTHALPY (h): h = hf + xhfg = 1025 + 0.92(1778) = 2661 kJ/kg SPECIFIC INTERNAL ENERGY (u): u = uf + x( ug -uf ) = 1021 + 0.92(2603 - 1021) = 2476 kJ/kg8.5 DRYNESS FRACTION (x): h = hf + x hfg 2100 = 1429 + x(1286) x = 0.52 SPECIFIC VOLUME (v); v =xvg = 0.52(0.01696) = 0.00882 m3/kg Specific internal energy (u): u = uf + x( ug -uf ) = 1414 + 0.52(2537 – 1414) = 1998 kJ/kg
23. 23. PROPERTIES OF STEAM J2006/8/23 INPUT 8.4.2 Superheated Steam Tables The second part of the table is the superheated steam tables. The values of the specific properties of a superheated steam are normally listed in separate tables for the selected values of pressure and temperature. A steam is called superheated when its temperature is greater than the saturation temperature corresponding to the pressure. When the pressure and temperature are given for the superheated steam then the state is defined and all the other properties can be found. For example, steam at 10 bar and 200 o C is superheated since the saturation temperature at 10 bar is 179.9 oC. The steam at this state has a degree of superheat of 200 oC – 179.9 oC = 20.1 oC. The equation of degree of superheat is: Degree of superheat = tsuperheat – tsaturation (8.5) The tables of properties of superheated steam range in pressure from 0.006112 bar to the critical pressure of 221.2 bar. At each pressure, there is a range of temperature up to high degrees of superheat, and the values of specific volume, internal energy, enthalpy and entropy are tabulated. For the pressure above 70 bar, the specific internal energy is not tabulated. The specific internal energy is calculated using the equation: u = h – pv (8.6) For reference, the saturation temperature is inserted in brackets under each pressure in the superheat tables and values of vg, ug, hg and sg are also given. A specimen row of values is shown in Table 8.5.2. For example, from the superheated table at 10 bar and 200 oC, the specific volume is 0.2061 m3/kg and the specific enthalpy is 2829 kJ/kg.
24. 24. PROPERTIES OF STEAM J2006/8/24 Table 8.4.2 Superheated steam at a pressure of 10 bar p t 200 250 300 350 400 450 500 600 (ts) vg 0.1944 v 0.2061 0.2328 0.2580 0.2825 0.3065 0.3303 0.3540 0.4010 10 ug 2584 u 2623 2711 2794 2875 2957 3040 3124 3297 (179.9) hg 2778 h 2829 2944 3052 3158 3264 3370 3478 3698 sg 6.586 s 6.695 6.926 7.124 7.301 7.464 7.617 7.761 8.028 Example 8.5 Complete the missing properties in the following table for Superheated Steam: p t 300 350 400 450 (ts) vg 0.0498 v 0.0800 40 ug 2602 u 2921 (250.3) hg 2801 h 3094 sg 6.070 s 6.364 Solution to Example 8.5 From page 7 of the steam tables, we can directly read p t 300 350 400 450 (ts) vg 0.0498 v 0.0588 0.0664 0.0733 0.0800 40 ug 2602 u 2728 2828 2921 3010 (250.3) hg 2801 h 2963 3094 3214 3330 sg 6.070 s 6.364 6.584 6.769 6.935
25. 25. PROPERTIES OF STEAM J2006/8/25 Example 8.6 Steam at 100 bar has a specific volume of 0.02812 m 3/kg. Find the temperature, degree of superheat, specific enthalpy and specific internal energy. Solution to Example 8.6 First, it is necessary to decide whether the steam is wet, dry saturated or superheated. At 100 bar, vg = 0.01802 m3/kg. This is less than the actual specific volume of 0.02812 m3/kg. Hence, the steam is superheated. The state of the steam is at point A in the diagram below. P bar A 100 425 oC ts = 311.0 oC v m3/kg vg= 0.01802 v = 0.02812 An extract from the superheated table, p t 425 (ts) vg 0.01802 v x 10-2 2.812 100 hg 2725 h 3172 (311.0) sg 5.615 s 6.321 From the superheated table at 100 bar, the specific volume is 0.02812 m 3/kg at a temperature of 425 oC. Hence, this is the isothermal line, which passes through point A as shown in the P-v diagram above. Degree of superheat = 425 oC – 311 oC
26. 26. PROPERTIES OF STEAM J2006/8/26 = 114 oC So, at 100 bar and 425 oC, we have v = 2.812 x 10-2 m3/kg h = 3172 kJ/kg ∴ From equation 8.6, u = h – Pv = 3172 kJ/kg – (100 x 102 kN/m2)(2.812 x 10-2 m3/kg) = 2890.8 kJ/kg Note that equation 8.6 must be used to find the specific internal energy for pressure above 70 bar as the specific internal energy is not tabulated.
27. 27. PROPERTIES OF STEAM J2006/8/27 Activity 8C8.6 Steam at 120 bar is at 500 oC. Find the degree of superheat, specific volume, specific enthalpy and specific internal energy.8.7 Steam at 160 bar has a specific enthalpy of 3139 kJ/kg. Find the temperature, degree of superheat, specific enthalpy and specific internal energy.
28. 28. PROPERTIES OF STEAM J2006/8/28 Feedback to Activity 8C8.6 From the superheated table at 120 bar, the saturation temperature is 324.6 oC. Therefore, the steam is superheated. Degree of superheat = 500 oC – 324.6 oC = 175.4 oC So, at 120 bar and 500 oC, we have v = 2.677 x 10-2 m3/kg h = 3348 kJ/kg ∴ From equation 8.6, u = h – Pv = 3348 kJ/kg – (120 x 102 kN/m2)(2.677 x 10-2 m3/kg) = 3026.76 kJ/kg8.7 At 160 bar, hg = 2582 kJ/kg. This is less than the actual specific enthalpy of 3139 kJ/kg. Hence, the steam is superheated. From the superheated table at 160 bar, the specific enthalpy of 3139 kJ/kg is located at a temperature of 450 oC. The degree of superheat = 450 oC – 347.3 oC = 102.7 oC At 160 bar and 450 oC, we have v = 1.702 x 10-2 m3/kg ∴ From equation 8.6, u = h – Pv = 3139 kJ/kg – (160 x 102 kN/m2)(1.702 x 10-2 m3/kg) = 2866.68 kJ/kg
29. 29. PROPERTIES OF STEAM J2006/8/29 INPUT8.5 Interpolation The first interpolation problem that an engineer usually meets is that of “reading between the lines” of a published table, like the Steam Tables. For properties which are not tabulated exactly in the tables, it is necessary to interpolate between the values tabulated as shown in Fig. 8.5-1 below. In this process it is customary to use a straight line that passes through two adjacent table points, denoted by α and β. If we use the straight line then it is called “interpolation”. f(x) Interpolation x α β Figure 8.5-1 Interpolation The values in the tables are given in regular increments of temperature and pressure. Often we wish to know the value of thermodynamic properties at intermediate values. It is common to use linear interpolation as shown in Fig. 8.5-2. y y2 (x2 , y2) From Fig. 8.5.2, the value of x can be determined by: y (x , y) x − x1 x − x1 = 2 y1 (x1 , y1) y − y1 y 2 − y1 ( y − y1 )( x2 − x1 ) x= + x1 x ( y 2 − y1 ) x1 x x2 Figure 8.5-2 Linear interpolation
30. 30. PROPERTIES OF STEAM J2006/8/30 There are two methods of interpolation: i. single interpolation ii. double interpolation 8.5.1 Single interpolation Single interpolation is used to find the values in the table when one of the values is not tabulated. For example, to find the saturation temperature, specific volume, internal energy and enthalpy of dry saturated steam at 77 bar, it is necessary to interpolate between the values given in the table. Example 8.6 Determine the saturation temperature at 77 bar. Solution to Example 8.6 The values of saturation temperature at a pressure of 77 bars are not tabulated in the Steam Tables. So, we need to interpolate between the two nearest values that are tabulated in the Steam Tables. t s − 290.5 295 − 290.5 P = 77 − 75 80 − 75 80 t s − 290.5 295 − 290.5 77 = 2 5 75 2( 4.5) ts = + 290.5 ts 5 290.5 ts 295 ts = 292.3 oC
31. 31. PROPERTIES OF STEAM J2006/8/31 Example 8.7 Determine the specific enthalpy of dry saturated steam at 103 bar. Solution to Example 8.7 P hg − 2725 2715 − 2725 = 103 − 100 105 − 100 105 103 3( −10) hg = + 2725 5 100 hg hg = 2719 kJ/kg 2725 hg 2715 Example 8.8 Determine the specific volume of steam at 8 bar and 220oC. Solution to Example 8.8 From the Steam Tables at 8 bar, the saturated temperature (ts) is 170.4 oC. ∴The steam is at superheated condition as the temperature of the steam is 220oC > ts. An extract from the Steam Tables, p / (bar) t 200 220 250 o o (ts / C) ( C) 8 v 0.2610 v 0.2933 (170.4) P 250 v − 0.2610 0.2933 − 0.2610 = 220 − 200 250 − 200 220 200 v = 0.27392 m3/kg v 0.2610 v 0.2933
32. 32. PROPERTIES OF STEAM J2006/8/32 8.5.2 Double Interpolation In some cases a double interpolation is necessary, and it’s usually used in the Superheated Steam Table. Double interpolation must be used when two of the properties (eg. temperature and pressure) are not tabulated in the Steam Tables. For example, to find the enthalpy of superheated steam at 25 bar and 320oC, an interpolation between 20 bar and 30 bar is necessary (as shown in example 8.9). An interpolation between 300oC and 350oC is also necessary. Example 8.8 Determine the specific enthalpy of superheated steam at 25 bar and 320oC. Solution to Example 8.8 An extract from the Superheated Steam Tables: t(oC) 300 320 350 p(bar) 20 3025 h1 3138 25 h 30 2995 h2 3117 Firstly, find the specific enthalpy (h1) at 20 bar and 320 oC; At 20 bar, T h1 − 3025 3138 − 3025 350 = 320 − 300 350 − 300 320 300 h1 = 3070.2 kJ/kg h 3025 h1 3138
33. 33. PROPERTIES OF STEAM J2006/8/33 Secondly, find the specific enthalpy (h2) at 30 bar and 320 oC; At 30 bar, T h2 − 2995 3117 − 2995 = 320 − 300 350 − 300 350 320 h2 = 3043.8 kJ/kg 300 h 2995 h2 3117 Now interpolate between h1 at 20 bar, 320oC, and h2 at 30 bar, 320oC in order to find h at 25 bar and 320oC. At 320oC, P h − h1 h − h1 = 2 25 − 20 30 − 20 30 h − 3070.2 30438 − 3070.2 . 25 = 25 − 20 30 − 20 20 h = 3057 kJ/kg. h h1 h h2 Example 8.9 0.9 m3 of dry saturated steam at 225 kN/m2 is contained in a rigid cylinder. If it is cooled at constant volume process until the pressure drops to180 kN/m2, determine the following: a) mass of steam in the cylinder b) dryness fraction at the final state Sketch the process in the form of a P-v diagram.
34. 34. PROPERTIES OF STEAM J2006/8/34 Solution to Example 8.9 Data: V1 = 0.9 m3 , P1 = 225 kN/m2 = 2.25 bar, P2 = 180 kN/m2 = 1.80 bar a) Firstly, find the specific volume of dry saturated steam at 2.25 bar. Note that the pressure 2.25 bar is not tabulated in the steam tables and it is necessary to use the interpolation method. From the Steam Tables, vg at 2.2 bar = 0.8100 m3/kg vg at 2.3 bar = 0.7770 m3/kg ∴vg1 at 2.25 bar, v g1 − 0.8100 0.7770 − 0.8100 = 2.25 − 2.20 2.30 − 2.20 vg1 = 0.7935 m3/kg V1 ∴ Mass of steam in cylinder, m = (m3 x kg/m3) vg1 0.9 = 0.7935 = 1.134 kg b) At constant volume process, Initial specific volume = final specific volume v1 = v2 x1vg1 at 2.25 bar = x2vg2 at 1.8 bar 1(0.7935) = x2 (0.9774) 1(0.7935) x2 = 0.9774 P = 0.81 bar 1 v1 = v2 2.25 1.80 2 v m3/kg 0.7935 0.9774
35. 35. PROPERTIES OF STEAM J2006/8/35 Activity 8D8.8 Determine the specific enthalpy of steam at 15 bar and 275oC.8.9 Determine the degree of superheat and entropy of steam at 10 bar and 380oC.8.10 A superheated steam at 12.5 MN/m2 is at 650oC. Determine its specific volume.8.11 A superheated steam at 24 bar and 500oC expands at constant volume until the pressure becomes 6 bar and the dryness fraction is 0.9. Calculate the changes in the internal energy of steam. Sketch the process in the form of a P-v diagram.
36. 36. PROPERTIES OF STEAM J2006/8/36 Feedback to Activity 8D 8.8 T h − 2925 3039 − 2925 = 300 275 − 250 300 − 250 275 h = 2982 kJ/kg 250 h 2925 h 3039 8.9 Degree of superheat = 380oC – 179.9oC = 200.1oC T 400 s − 7.301 7.464 − 7.301 = 380 380 − 350 400 − 350 350 s = 7.3988 kJ/kg K s 7.301 s 7.464 8.10 An extract from the superheated steam table: t(oC) 600 650 700 p(bar) 120 3.159 x 10-2 v1 3.605 x 10-2 125 v -2 130 2.901 x 10 v2 3.318 x 10-2
37. 37. PROPERTIES OF STEAM J2006/8/37 Firstly, find the specific volume (v1) at 120 bar and 650 oC; At 120 bar, T 700 v1 − 3.159 x 10 −2 3.605 x 10 −2 − 3.159 x 10 −2 = 650 650 − 600 700 − 600 600 v1 = 3.382 x 10 −2 m3/kg v 3.159 x 10-2 v1 3.605 x 10-2 Secondly, find the specific volume (v2) at 130 bar and 650 oC; At 130 bar, T v 2 − 2.901 x 10 −2 3.318 x 10 −2 − 2.901 x 10 −2 700 = 650 − 600 700 − 600 650 v2 = 3.1095 x 10-2 m3/kg 600 v 2.901 x 10-2 v2 3.318 x 10-2 Now interpolate between v1 at 120 bar, 650oC, and v2 at 130 bar, 650oC in order to find v at 125 bar and 650oC. At 650oC, v − v1 v −v P = 2 1 125 − 120 130 − 120 130 125 v − 3.382 x 10 −2 3.1095 x 10 −2 − 3.382 x 10 −2 = 125 − 120 130 − 120 120 v = 3.246 x 10-2 m3/kg v1 v2 v v
38. 38. PROPERTIES OF STEAM J2006/8/38 8.11 Data: P1 = 24 bar T1 = 500oC P2 = 6 bar x2 = 0.9 Firstly, find the initial internal energy at 24 bar, 500oC. Note that the pressure 24 bar is not tabulated in the Superheated Steam Tables and it is necessary to use the interpolation method to find the changes in the internal energy of steam. At 500oC, P 30 u1 − 3116 3108 − 3116 = 24 24 − 20 30 − 20 20 u1 = 3112.8 kJ/kg u 3116 u1 3108 Secondly, find the final internal energy at 6 bar where x = 0.9, u2 = uf2 + x2( ug2 -uf2 ) = 669 + 0.9(2568 - 669) = 2378.1 kJ/kg ∴ The changes in the internal energy of steam is, (u2 – u1) = 2378.1 – 3112.8 = - 734.7 kJ/kg P bar 500oC v1 = v2 o 221.8 C 1 24 158.8oC 500oC 6 221.8oC 2 158.8oC v m3/kg v1 = v2
39. 39. PROPERTIES OF STEAM J2006/8/39 SELF-ASSESSMENT You are approaching success. Try all the questions in this self-assessment section and check your answers with those given in the Feedback to Self- Assessment on the next page. If you face any problem, discuss it with your lecturer. Good luck. 1. With reference to the Steam Tables, i. determine the specific volume, specific enthalpy and specific internal energy of wet steam at 15 bar with a dryness fraction of 0.9. ii. determine the degree of superheat, specific volume and specific internal energy of steam at 80 bar and enthalpy 2990 kJ/kg. iii. complete the missing properties and a phase description in the following table for water; P t x v u h s Phase o 3 bar C m /kg kJ/kg kJ/kg kJ/kg K description 2.0 120.2 6.4 12.0 1 2784 175 354.6 0.9 200 425 2. With reference to the Steam Tables, i. find the dryness fraction and specific entropy of steam at 2.9 bar and specific enthalpy 2020 kJ/kg. ii. determine the degree of superheat and internal energy of superheated steam at 33 bar and 313oC. iii. determine the enthalpy change for a process involving a dry saturated steam at 3.0 MN/m2 which is superheated to 600 oC and carried out at constant pressure.
40. 40. PROPERTIES OF STEAM J2006/8/40 Feedback to Self-Assessment Have you tried the questions????? If “YES”, check your answers now. 1. i. v = 0.11853 m3/kg h = 2600 kJ/kg u = 2419.8 kJ/kg ii. degree of superheat = 55 oC v = 2.994 x 10-2 m3/kg u = 2750.48 kJ/kg iii. P t x v u h s Phase o bar C m3/kg kJ/kg kJ/kg kJ/kg K description 2.0 120.2 0.87 0.7705 2267 2421 6.4 Wet steam 12.0 188 1 0.1632 2588 2784 6.523 Dry sat. steam 175 354.6 0.9 0.007146 2319.8 2448.1 5.0135 Wet steam 200 425 - 0.001147 2725.6 2955 5.753 Superheated steam 2. i. x = 0.68 s = 5.2939 kJ/kg K ii. Degree of superheat = 73.85oC u = 2769 kJ/kg iii. h2 – h1 = 879 kJ/kg