03_Exercise_Set_3_Solutions_

PE 4023/6023- Fundamentals and Applications of Geothermal Energy Fall 2023/TU Exercise Set No. 3 - SOLUTIONS Given Date: 09-21-2023 Due Date: 09-26-2023 Subject: Derivation of time and space derivative of Enthalpy as a function of 𝑝 & 𝑇 , Computation of Geothermal Heat and Power, compressibility, saturation line with the effects of dissolved CO2 and salts. Problem 1: Specific enthalpy of a fluid is expressed as a function of 𝑝 and 𝑇 ; i.e., ℎሺ𝑝, 𝑇ሻ or inversely, temperature can be expressed as a function of specific enthalpy, ℎ , and 𝑝 ; 𝑇ሺℎ, 𝑝ሻ . Show that the time derivative and the gradient of specific enthalpy can be derived, respectively, as డ௛ డ௧ ൌ െ𝜇 ௃்,௙ 𝑐 ௣,௙ డ௣ డ௧ ൅ 𝑐 ௣,௙ డ் డ௧ ൌ 𝑐 ௣,௙ ቀ డ் డ௧ െ 𝜇 ௃்,௙ డ௣ డ௧ ቁ , (1.1) and ∇ℎ ൌ 𝑐 ௣,௙ ൫∇𝑇 െ 𝜇 ௃்,௙ ∇𝑝൯ , (1.2) Where 𝑐 ௣,௙ is the specific heat capacity of the fluid, 𝜇 ௃்,௙ is the Joule-Thomson coefficient of fluid, and gradient vector for given coordinate system. For a Cartesian gradient system, the gradient vector of a scalar function 𝑓ሺ𝑥, 𝑦, 𝑧ሻ is given by ∇𝑓 ൌ ቀ డ௙ డ௫ , డ௙ డ௬ , డ௙ డ௭ ቁ ൌ డ௙ డ௫ 𝐢 ൅ డ௙ డ௬ 𝐣 ൅ డ௙ డ௭ 𝐤 , (1.3) where i , j , and k are unit coordinate vectors in the 𝑥 , 𝑦 , and 𝑧 directions. Solution 1: As ℎ is treated as a function of state variables 𝑝 and 𝑇 , then its time derivative from the chain rule follows as డ௛ డ௧ ൌ ቀ డ௛ డ௣ ቁ ் డ௣ డ௧ ൅ ቀ డ௛ డ் ቁ ௣ డ் డ௧ (S1.1) By definition (see Aside A - Eqs. A5 and A7); ቀ డ௛ డ் ቁ ௣ ൌ 𝐶 ௣,௙ (S1.2) and ቀ ௗ௛ ௗ௣ ቁ ் ൌ െ𝜇 ௃்,௙ 𝐶 ௣,௙ (S1.3) Using Eqs. S1.2 and S1.3 in Eq. S1.1 gives: డ௛ డ௧ ൌ െ𝜇 ௃்,௙ 𝐶 ௣,௙ డ௣ డ௧ ൅ 𝐶 ௣,௙ డ் డ௧ ൌ 𝐶 ௣,௙ ቀ డ் డ௧ െ 𝜇 ௃்,௙ డ௣ డ௧ ቁ (S1.4) Similarly, the gradient of ℎ can be written as: ∇ℎ ൌ ቀ డ௛ డ௣ ቁ ் ∇𝑝 ൅ ቀ డ௛ డ் ቁ ௣ ∇𝑇 (S1.5)

Using Eq. S.1.2 and S1.3 in Eq. S1.5 gives: ∇ℎ ൌ െ𝜇 ௃்,௙ 𝐶 ௣,௙ ∇𝑝 ൅ 𝐶 ௣,௙ ∇𝑇 ൌ 𝐶 ௣ ൫∇𝑇 െ 𝜇 ௃்,௙ ∇𝑝൯ . (S1.6) --------------------------------------------------------------------------------------------------------------------- Aside A: Let’s consider, 𝑇 ൌ 𝑇ሺℎ, 𝑝ሻ . 𝑑𝑇 ൌ ቀ డ் డ௛ ቁ ௣ 𝑑ℎ ൅ ቀ డ் డ௣ ቁ ௛ 𝑑𝑝 (A1) Taking the derivative of Eq.A1 with respect to 𝑇 while ℎ remains constant gives: 1 ൌ ቀ డ் డ௣ ቁ ௛ ቀ ௗ௣ ௗ் ቁ ௛ (A2) Taking the derivative of Eq.A1 with respect to 𝑝 while 𝑇 remains constant gives: 0 ൌ ቀ డ் డ௛ ቁ ௣ ቀ ௗ௛ ௗ௣ ቁ ் ൅ ቀ డ் డ௣ ቁ ௛ (A3) From Eq. A3; ቀ ௗ௛ ௗ௣ ቁ ் ൌ െ ቀ ങ೅ ങ೛ ቁ ೓ ቀ ങ೅ ങ೓ ቁ ೛ or ቀ ௗ௛ ௗ௣ ቁ ் ൌ െ ቀ డ் డ௣ ቁ ௛ ቀ డ௛ డ் ቁ ௣ (A4) By definition: ቀ డ௛ డ் ቁ ௣ ൌ 𝐶 ௣ (A5) ቀ డ் డ௣ ቁ ௛ ൌ 𝜇 ௃் (A6) Using Eqs. A5 and A6 in Eq. A4 gives: ቀ ௗ௛ ௗ௣ ቁ ் ൌ െ𝜇 ௃் 𝐶 ௣ , (A7) where 𝐶 ௣ is the specific heat capacity and 𝜇 ௃் is the Joule-Thomson coefficient. Problem 2: For a geothermal Project 800 tons/hour waste water at 175 ˚C is disposed to a nearby river rather than reinjected into reservoir. What is the wasted energy in MWt ? Find the quantity of annual oil equivalent of this wasted energy ( Energy equivalent of 1 ton oil = 42x10 9 J ). Take a reference temperature of 15 o C. Solution 2: By assuming that the annual average temperature is 15 ˚C, wasted energy in MW t can be calculated from 𝑄 ሶ ௪௔௦௧௘ௗ ሺMW ୲ ሻ ൌ 𝑞 ሺkg/sሻ ൈ 𝐶 ௣,௪ ሺj/kg. ˚Cሻ ൈ ሺ𝑇 ௪௔௦௧௘ െ 𝑇 ௥௘௙ ሻ ሺ˚Cሻ ൈ 10 ି଺ (S2.1) Using the given input values of parameters in Eq. S2.1, we compute the wasted power as

- The outlet temperature = 40 ˚C - The maximum flow rate = 225 kg/s - The annual average flow rate = 110 kg/s a) Find the installed capacity in MWt b) Find the annual energy in use in TJ/year c) Find the quantity of oil required to provide the amount of thermal annual energy use calculated in part b ( Energy equivalent of 1 ton oil = 42 GJ). d) Find the capacity factor Solution 4: a) Installed capacity can be calculated by using the equation; I nstalled capacityሺMWtሻ ൌ Maximum flow rate ൬ kg s ൰ ൈ ሺInlet Temp. െ Outlet Tempሻሺ˚Cሻ ൈ 0.004184 ൌ 225 ൬ kg s ൰ ൈ ሺ95 െ 40ሻሺ˚Cሻ ൈ 0.004184 ൌ 𝟓𝟏. 𝟖 𝐌𝐖𝐭 b) To be able to calculate the annual energy, we can use following equation can be used; Annu al energy use ൬ TJ yr ൰ ൌ Annual average flow rate ൬ kg s ൰ ൈ ሺInlet Temp. – Outlet Temp. ሻሺ˚Cሻ ൈ 0.1319 ൌ 110 ሺkg/sሻ ൈ ሺ95 െ 40ሻሺ˚Cሻ ൈ 0.1319 ൌ 𝟕𝟗𝟖. 𝟎𝟎 𝐓𝐉/𝐲𝐫 c) Quantity of oil required ൌ Annual energy use ሺTJ/yrሻ/Energy equivalence of 1 ton oil ൌ 798 ൈ 10 ଷ GJ yr 42ሺGJሻ ൌ 𝟏𝟗, 𝟎𝟎𝟎 ሺ𝐭𝐨𝐧𝐬/𝐲𝐫ሻ d) Capacity Factor Capacity Factor ൌ Annual Average Use ቀ ୘୎ ୷୰ ቁ ൈ ଴.଴ଷଵ଻ଵ ୍୬ୱ୲ୟ୪୪ୣୢ Capacity ሺMWtሻ ൌ 𝟕𝟗𝟖 𝐓𝐉 𝐲𝐫 ൈ 𝟎.𝟎𝟑𝟏𝟕𝟏 𝟓𝟏.𝟖 𝐌𝐖𝐭 ൌ 𝟎. 𝟒𝟖𝟗 or 48.9% Problem 5: A geothermal field produces 770 tons/h water and 230 tons/h steam at saturated conditions with 150 ˚C. a) Find the total amount of energy produced in kJ/h and the fraction of energy produced with water with respect to the total amount of energy produced. b) Find the total amount of fluid volume produced in m 3 /h and the fraction of water volume for the total volume produced. Solution 5 a) It is convenient to convert flow rate data given in tons/hour to kg/h for simplicity; 770 ሺtons/hሻ ൈ ሺ1000 kg/tonሻ ൌ 77x10 ସ kg/h 230 ሺtons/hሻxሺ1000kg/tonሻ ൌ 23 ൈ 10 ସ kg/h The specific enthalpy of steam and water phases are determined for saturated conditions and temperature of 150 ˚C by using steam tables;

ℎ ௪ ൌ 632.18 kJ/kg ℎ ௦ ൌ 2745.9 kJ/kg Energy produced with water ሺ𝑞 ௪ ሻ ൌ 𝑤 ௪ ൈ ℎ ௪ ൌ ሺ77 ൈ 10 ସ ሻ ൈ ሺ632.18ሻ ൌ 𝟒𝟖𝟔𝟕𝟓. 𝟓𝟓 ൈ 𝟏𝟎 𝟒 kJ/hr Energy produced with steam ሺ𝑞 ௦ ሻ ൌ 𝑤 ௦ ൈ ℎ ௦ ൌ ሺ23 ൈ 10 ସ ሻ ൈ ሺ2745.9ሻ ൌ 𝟔𝟑𝟏𝟒𝟒. 𝟐 ൈ 𝟏𝟎 𝟒 kJ/hr Total amount of energy produced ሺ𝑞 ௧ ሻ ൌ 𝑞 ௪ ൅ 𝑞 ௦ ൌ 48677.86 ൈ 104 ൅ 63155.7 ൈ 10 ସ ൌ 𝟏𝟏𝟏𝟖𝟏𝟗. 𝟕𝟓 ൈ 𝟏𝟎 𝟒 kJ/hr 𝑞 ௪ /𝑞 ௧ ൌ ሺ48677.86 ൈ 10 ସ ሻ/ሺ 111833.56 ൈ 10 ସ ሻ ൌ 𝟎. 𝟒𝟑𝟓 (43.5 percentage of energy produced with (liquid) water phase) b) To be able to compute the volume flow rate of steam and water phases, specific volumes are needed to be determined from steam tables at specified pressure and temperature conditions; v w = 0.001091 m 3 /kg v s =0.39245 m 3 /kg Volume of water produced ሺ𝑉𝑤ሻ ൌ 𝑣 ௪ ൈ 𝑤 ௪ ൌ ሺ0.001091ሻ ൈ ሺ77 ൈ 10 ସ ሻ ൌ 𝟖𝟒𝟎. 𝟎𝟕 m ଷ /h Volume of steam produced ሺ𝑉𝑠ሻ ൌ 𝑣 ௦ ൈ 𝑤 ௦ ൌ ሺ0.39245ሻ ൈ ሺ23 ൈ 10 ସ ሻ ൌ 𝟗𝟎𝟐𝟕𝟎. 𝟒 m ଷ /h Total volume of fluid produced ሺ𝑉 ௧ ሻ ൌ 𝑉 ௦ ൅ 𝑉 ௪ ൌ 840.07 ൅ 90270.4 ൌ 𝟗𝟏𝟏𝟏𝟎. 𝟒𝟕 m ଷ /h 𝑉 ௪ / 𝑉 ௧ ൌ 840.07/ 91110.47 ൌ 𝟗. 𝟐𝟐 ൈ 𝟏𝟎 ି𝟑 (0.9 volume percentage of (liquid) water phase) Problem 6: The following data are available for a geothermal field used for district heating; - maximum flow rate = 180 kg/s -Annual average flow rate = 114 kg/s -Inlet temperature = 90 ˚C -Outlet temperature = 20 ˚C a) Determine the capacity in MW t , annual energy use in TJ/yr, and capacity factor as a fraction for this field. b) Repeat part a if half of the geothermal fluid produced is reinjected to the reservoir at 45 ˚C. c) Find your answers if all produced water is reinjected at 45 ˚C. Solution 6: (a) Installed capacityሺMWtሻ ൌ Maximum flow rate ൬ kg s ൰ ൈ ሺInlet Temp. െ Outlet Tempሻሺ˚Cሻ ൈ 0.004184 ൌ 180 ሺkg/sሻ ൈ ሺ90 െ 20ሻሺ˚𝐶ሻ ൈ 0.004184 ൌ 52.72 MWt

c) Since all water produced will be reinjected, the outlet temperature can be assumed to be the reinjection temperature which is 45 ˚C. By using outlet temperature as 45 ˚C, the installed capacity, annual energy use, and capacity factor can be calculated as in section a. Installed capacityሺMWtሻ ൌ Maximum flow rate ൬ kg s ൰ ൈ ሺInlet Temp. െ Outlet Tempሻሺ˚Cሻ ൈ 0.004184 ൌ 180 ሺkg/sሻ ൈ ሺ90 െ 45ሻሺ˚Cሻ ൈ 0.004184 ൌ 33.89 MWt Annual energy use ൬ TJ yr ൰ ൌ Annual average flow rate ൬ kg s ൰ ൈ ሺInlet Temp. – Outlet Temp. ሻሺ˚Cሻ ൈ 0.1319 ൌ 114 ሺkg/sሻ ൈ ሺ90 െ 45ሻሺ˚Cሻ ൈ 0.1319 ൌ 676.65 TJ/yr Capacity factor ൌ Annual energy use ሺTJ/yrሻ ൈ 0.03171/ınstalled capacityሺMWtሻ ൌ ሺ676.65 ൈ 0.03171ሻ/33.89 ൌ 0.633 ሺ% 63.3ሻ Problem-7 Following data are available for saturated liquid water containing geothermal reservoir and surface facilities; -Reservoir temperature =217 ˚C. -Production flow rate = 250 kg/s -Seperator temperature =150 ˚C.( Mixture of steam and hot water exists in the seperator). a) Find the amount of energy produced from the reservoir in terms of MW t . b) Neglecting the heat loss in the wellbore, determine the energy of steam to a turbine in MW t , find (Energy of steam to turbine /Energy produced) ratio. c) Assuming the annual energy power of the plant is 11 MW e , find (Annual energy power /energy of the steam to turbine) ratio. d) Determine (Annual average power/Energy produced) ratio. e) For 11 MWe of the annual average power of the plant, determine the amount of electricity generated (kJ.h/yr) in a year. Solution 7a) Since, in the reservoir, only water phase exists at specified pressure and temperature, the total energy produced can be calculated by using the mass flow rate of water and enthalpy of water and given temperature and pressure. Because pressure increment on the liquid water phase does not affect the specific enthalpy of water so much, the specific enthalpy of water can be chosen as the specific enthalpy of the saturated water phase at a given temperature which is ℎ ௪ ൌ 929.81 kJ/kg @217 ˚C Amount of energy produced ሺMWtሻ ൌ ℎ ௪ ൈ 𝑤 ௪ ൌ 250 ሺkg/sሻ 𝑥 929.81 ሺkJ/kgሻ ൈ 10 ିଷ ൌ 𝟐𝟑𝟐. 𝟒𝟓 𝐌𝐖𝐭

b) Since heat losses in the wellbore are neglected, the total specific enthalpy will remain constant. As two phases exist in the separator, steam quality should be determined by using total specific enthalpy and steam and water specific enthalpies at 150 ˚C ℎ ௦௪ ൌ 929.81 kJ/kg, ℎ ௪ ൌ 632.18 kJ/kg, ℎ ௦ ൌ 2745.93 kJ/kg 𝑥 ൌ ሺℎ ௦௪ െ ℎ ௪ ሻ/ሺℎ ௦ െ ℎ ௪ ሻ ൌ ሺ929.81 െ 632.18ሻ/ሺ2745.93 െ 632.18ሻ ൌ 𝟎. 𝟏𝟒𝟎𝟖 ሺ% 14.08ሻ Steam mass flow rate ሺw ௦ ሻ ൌ 250 ሺkg/sሻ ൈ ሺ0.14076ሻ ൌ 𝟑𝟓. 𝟏𝟗 kg/s Energy of the steam to turbineሺMWtሻ ൌ w ୱ ൈ ℎ ௦ ൌ ሺ35.19ሻ ൈ ሺ2746.5ሻ ൈ 10 ିଷ ൌ 𝟗𝟔. 𝟔𝟓 𝐌𝐖𝐭 ሺEnergy of steam to turbineሻ/ሺTotal energy produced from the reservoirሻ ൌ 96.65/232.45 ൌ 𝟎. 𝟒𝟏𝟔 ሺ% 𝟒𝟏. 𝟔 ሻ c) ሺAnnual average energy powerሻ/ሺEnergy of steam to turbineሻ ൌ 11ሺMWeሻ/96.65ሺMWtሻ ൌ 𝟎. 𝟏𝟏𝟒 ሺ% 11.4ሻ d) ሺAnnual average power of plantሻ/ሺEnergy produced from the reservoirሻ ൌ 11ሺMWeሻ/232.45ሺMWtሻ ൌ 𝟎. 𝟎𝟒𝟕 ( % 4.7) e) 11 MWe ൌ 11000 ሺkJ/sሻ ൈ ሺ24 h/dayሻ ൈ ሺ365 day/yrሻ ൌ 𝟗𝟔𝟑𝟔𝟎𝟎𝟎𝟎 ሺkW. hr/yrሻ Problem-8 A geothermal reservoir contains two-phase steam-water mixture at pressure and temperature of 2548 kPa and 225 ˚C; 𝜙 ൌ 0.15 𝐶 ௣,௦ ൌ 1 kJ/kg.˚C 𝜌 ௦ =2600 kg/m 3 𝑆 ௪ ൌ 0.02 Compute 2-phase compressibility and compare it with steam and water compressibilities at the same pressure and temperature by using the data given above. Solution 8: To compute the two-phase steam-water mixture compressibility at a given p and T , we will use 𝑐 ்ଶ∅ ൌ 0.42 ൈ 10 ିହ ሺ𝜌𝐶ሻ ோ 𝜙 𝑝 ିଵ.଺଺ We can compute volumetric heat capacity of the geothermal reservoir from: ሺ𝜌𝐶ሻ ோ ൌ ሺ1 െ 𝜙ሻ𝜌 ௦ 𝐶 ௣,௦ ൅ 𝜙𝑆 ௪ 𝜌 ௪ 𝐶 ௣,௪ ൅ 𝜙𝑆 ௦௧௘௔௠ 𝜌 ௦௧௘௔௠ 𝐶 ௣,௦௧௘௔௠ We find the water and steam properties at given temperature of 225 ˚C and 2548 kPa (25.48 bar) by using steam tables (H2O Software); 𝜌 ௪ ൌ 833.47 kg/m ଷ 𝜌 ௦௧௘௔௠ ൌ 12.74 kg/m ଷ 𝐶 ௣,௪ ൌ 4.6498 ୩୎ ୩୥ ౥ େ 𝐶 ௣,௦௧௘௔௠ ൌ 3.42579 kJ/ሺkg ୭ Cሻ

𝐾 ுௌ ൌ 5.36124 ൈ 10 2 ൅ ሺ80.1681582 ൈ 0.07 ൅ 0.33462 ൈ 200 ൈ 0.07ሻ ൌ 5.46421 ൈ 10 2 𝑝 ஼ை మ ൌ ଵ଼ ସସ ∗ 5.46421 ൈ 10 2 ∗ 0.015 ൌ 3.35304 MPa The compressibility of CO 2 with salt is then computed from 𝑐 ஼ை మ ൌ 1 𝐾 ுௌ 𝜌 ௪ 𝜌 ௦ 𝑝 ௦ 𝑝 ஼ை మ But, first, we need to find density of liquid water and steam at 200 o C and the saturation pressure 𝑝 ௦ at the saturation line from steam tables. I found 𝜌 ௪ ൌ 864.658 kg/m ଷ and 𝜌 ௦ ൌ 7.861 kg/m ଷ and 𝑝 ௦ ൌ 15.5493 bar ൌ 1.55493 MPa 𝑐 ஼ை మ ൌ ଵ ௄ ಹೄ ఘ ೢ ఘ ೞ ௣ ೞ ௣ ಴ೀ మ ൌ ଵ ହ.ସ଺ସଶଵ ൈ10 2 ଼଺ସ.଺ହ଼ ଻.଼଺ଵ ଵ.ହହସଽଷ 3.35304 ൌ 9.3349 ൈ 10 ିଶ MPa -1 The water compressibility at 200 o C and 69 bar is found from steam tables 𝑐 ்௪ ൌ 8.49376 ൈ 10 ିହ bar -1 = 8.49376 ൈ 10 ିହ ଵ ୠୟ୰ ቀ బ.భ ౉ౌ౗ భ ౘ౗౨ ቁ ൌ 8.49376 ൈ 10 ିସ MPa -1 Clearly, the compressibility of CO 2 is about two order of magnitude larger than the liquid water compressibility at 200 o C and 69 bar (subcooled region; see P-T diagram of pure water). 𝑐 ஼ைଶ ≫ 𝑐 ்௪ b. The solution of this question is similar to the solution of Problem 8 above. So, to compute the two-phase steam-water mixture compressibility at a given 𝑝 and 𝑇 , we will use 𝑐 ்ଶ∅ ൌ 0.42 ൈ 10 ିହ ሺ𝜌𝐶ሻ ோ 𝜙 𝑝 ିଵ.଺଺ We can compute volumetric heat capacity of the geothermal reservoir from: ሺ𝜌𝐶ሻ ோ ൌ ሺ1 െ 𝜙ሻ𝜌 ௦ 𝐶 ௣,௦ ൅ 𝜙𝑆 ௪ 𝜌 ௪ 𝐶 ௣,௪ ൅ 𝜙𝑆 ௦௧௘௔௠ 𝜌 ௦௧௘௔௠ 𝐶 ௣,௦௧௘௔௠ As we assume we are at the saturation line, saturation of Sstem can be taken as zero. We find the water and steam properties at given temperature of 200 ˚C by using steam tables (H2O Software); 𝜌 ௪ ൌ 864.658 kg/m ଷ and 𝜌 ௦ ൌ 7.861 kg/m ଷ and 𝑝 ௦ ൌ 15.5493 bar ൌ 1.55493 MPa 𝐶 ௣,௪ ൌ 4.49584 ୩୎ ୩୥ ౥ େ 𝐶 ௣,௦௧௘௔௠ ൌ 2.98955 kJ/ሺkg ୭ Cሻ ሺ𝜌𝐶ሻ ோ ൌ ሺ1 െ 0.15ሻ ∗ 2000 ∗ 1000 ൅ 0.15 ∗ ሺ1.00 ∗ 864.658 ∗ 4495.84 ൅ 0.0 ∗ 7.861 ∗ 2989.55 ሻ ൌ 2.2831𝑥10 ଺ J/(m 3 )

𝑐 ்ଶ∅ ൌ 0.42 ൈ 10 ିହ ሺఘ஼ሻ ೃ థ 𝑝 ିଵ.଺଺ ൌ 0.42 ൈ 10 ିହ ଶ.ଶ଼ଷଵ௫ଵ଴ ల ଴.ଵହ 1.55493 ିଵ.଺଺ ൌ 30.7216 MPa -1 From H 2 O tables, we find that the compressibility of liquid water and steam at the saturation pressure and temperature is given by: 𝑐 ்௪ ൌ 8.83197 ൈ 10 ିହ 𝑏𝑎𝑟 ିଵ ൌ 8.83197 ൈ 10 ିସ MPa ିଵ and 𝑐 ்௦௧௘௔௠ ൌ 7.24682 ൈ 10 ିଶ 𝑏𝑎𝑟 ିଵ ൌ 7.24682 ൈ 10 ିଵ MPa -1 Recall in part a, we computed 𝑐 ஼ை మ ൌ 9.3349 ൈ 10 ିଶ MPa -1 . So, 𝑐 ்ଶ∅ ൐ 𝑐 ்௦௧௘௔௠ ൐ 𝑐 ஼ைଶ ≫ 𝑐 ்௪ Problem-10. Investigate the effect of amount of salt and CO2 dissovled in geothermal water on saturation pressure. (This is a bonus 30 points to be added to one of your quizzes). Use the expressions given Chapter 5, make plots of ps versus Ts, where ps and Ts are saturation pressure and temperature, respectivly for xg = 0, 0.01, and 0.05, and ms = 0, 2. Compare your results with the saturation line of pure water (plotted ps versus Ts).