CHEG231-F22 Homework Solution 2

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CHEG 231, Fall, 2021 Homework Assignment #1 – Solutions & Rubrics Problem 1: Use the Steam Tables to answer the following questions. Make sure to indicate which values you used for interpolation. Find (i) the specific internal energy of steam at 3.5 MPa and 300°C. 2738.0 kJ/kg (J/g) (ii) the specific density of steam at 2.75 MPa and 450°C. 0.13014m3/kg @ 2.50MPa and 450 deg C, 0.10787 @ 3.00 MPa and 450 deg C 0.13014+((0.13014-0.10787)/(2.50-3.00))*(2.75-2.50)=0.11901m3/kg Specific Density = 1/specific volume = 1/0.119005= 8.403kg/m 3 (iii) the molar enthalpy of steam at 74 bar and 553°C. 7.0MPa 8.0MPa 550 deg C 3530.9J/g 3521.0J/g 600 deg C 3650.3J/g 3642.0J/g double interpolation (3530.9-3521.0)*0.4+3521.0=3524.96J/g @7.4 MPa 550 deg C (3650.3-3642.0)*0.4+3642.0=3645.32J/g @7.4 MPa 600 deg C (3645.32-3524.96)*(3/50)+3524.96=3532.2 J/g@7.4 MPa 553 deg C 3532.2J/g *(18.015g/1mol)= 63,600J/mol=63.6kJ/mol (iv) the vapor pressure of (saturated) steam at 262°C. Pv=4MPa @ 250.4 deg C , Pv=5MPa @ 263.99 deg C (5-4)/(263.99-250.4))*(262-250.4)+4= 4.853 MPa (v) the molar enthalpy of vaporization at 180°C in units of kJ/mol. 2015.3J/g @ 179.91 deg C, 2000.4J/g @ 184.09 deg C (2015.3-2000.4)/(179.91-184.09)*(180-179.91)+2015.3= 2014.97J/g*(18.01g/mol)= 36290J/mol=36.29kJ/mol (vi) the specific internal energy of liquid water at 20 MPa and 135°C. 496.76@120 deg C 20MPa,580.69@ 140 deg C 20 MPa ((580.69-496.76)/(140-120))*(135-120)+496.76= 559.7J/g (vii) the specific enthalpy energy of liquid water at 17.5 atm and 85°C. Pressure has very little effect on enthalpy of sub cooled liquids. Neglect pressure dependence, use 5 MPa (as it’s in the tables) from the sub cooled liquid. 338.85J/g @ 80 deg C, 422.72J/g @ 100 deg C, ((422.72-338.85)/(100-80))*(85-80)+338.85= 359.82J/g (viii) the change in molar density on sublimation for water at 263 K. Vs=1.0891*10^-3m3/kg @ -10 deg C, Vv=466.7m3/kg @-10 deg C density=1/specific volume, solid density=1/1.0891e-3=918.19kg/m3 vapor density=1/466.7=0.0021427 change on sublimation=vapor density-solid density=-918.19kg/m3*(1mol/0.018kg)= -50,980mol/m 3
b) Plot should be similar to below. Interpolated values from superheated steam tables.
CHEG 231, Fall, 2021 Homework Assignment #1 – Solutions & Rubrics Problem 2: (a) Search online for the enthalpy change upon fusion for water and then use the Steam Tables to show that this value is consistent with the relationship. ΔĤ subl H2O,tp - ΔĤ vap H2O,tp = ΔĤ fus H2O,tp ΔĤ vap H2O,tp =2501.3J/g, ΔĤ subl H2O,tp =2834.8 2501.3-2834.8= -333.5J/g -334kJ/kg according to Young, Hugh D., University Physics, 7th Ed., Addison Wesley, 1992. (b) Search for the enthalpy changes for vaporization, fusion, and sublimation for your project compound at the triple point and show that ΔĤ subl tp - ΔĤ vap tp - ΔĤ fus tp =0 if all three values are available. If the information needed is not all available, which will be the case for most or all of the project compounds, you may need to use predicted values and/or make some assumptions to approximate the values. If you can find just two of the three enthalpy changes, you can use those to determine the missing one by using the relationship above. It may be useful to see that the triple point temperature is usually very close to the normal melting point. You may find that the enthalpy of vaporization is reported at room temperature and perhaps some other temperatures, but not at the triple point; if so, note that the largest value occurs at the triple point. You may need to extrapolate to obtain an estimate. These results will be useful for your project report! Using ammonia as the reference compound and the same logic as previously, ΔĤ fus tp of ammonia at the triple point is 332.16 Kj/Kg and the ΔĤ vap tp is 1369.8 Kj/Kg from knovel.com. By understanding that sublimation = vaporization -fusion, we can get ΔĤ subl tp of ammonia at the triple points as 1701.96 Kj/kg which is similar to literature reported values of 1718 Kj/Kg.
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CHEG 231, Fall, 2021 Homework Assignment #1 – Solutions & Rubrics Problem 3 : Study the Mollier Diagrams for Steam represented as Figures 3.3-1 and 3.3-2 in your textbook and find as many of the answers to Problem 1 as can be determined using these diagrams. The values will be consistent if the reference states match between the two data sources; is this the case? Some properties may not be obtainable. If this is the case, explain why. Comment as clearly as you can on the level of precision at which the values can be determined from these graphical representations. You will do much more with statistical issues in CHEM 220 and next term when you take CHEG 304. With practice, you may be able to obtain approximate values more rapidly from a graphical source than from tables, though the reverse may be true in the absence of practice, so use this opportunity to gain experience. (i) the specific internal energy of steam at 3.5 MPa and 300°C. specific enthalpy =~2975kJ/kg specific volume =~75*10^-3 m3/kg U=H-P*V U=2975-(3.5*10 3 )(75*10 -3 )=2712kJ/kg which is close to 2738kJ/kg from part one. (ii) the specific density of steam at 2.75 MPa and 450°C. ~120*10^-3m3/kg=8.3kg/m3 (iii) the molar enthalpy of steam at 74 bar and 553°C. H=3540kj/kg(18.01)=63,720J/mol=63.7kJ/mol (iv) the vapor pressure of (saturated) steam at 262°C. 4.9MPa (v) the molar enthalpy of vaporization at 180°C in units of kJ/mol. 2775kJ/kg for vapor,780kJ/kg for liquid (2775-780)=1995kJ/kg*0.018=35.91kJ.mol (vi) the specific internal energy of liquid water at 20 MPa and 135°C. U=H-PV H=550 V=1*10^-3m3/kg 550-(20*10^3)*(1*10^-3)=530kJ/kg (vii) the specific enthalpy energy of liquid water at 17.5 atm and 85°C. 375kj/kg (viii) the change in molar density on sublimation for water at 263 K. Cant do this one, No info on solids in diagrams (ix) Is it consistent with tables? It will be consistent as long as reference states are the same. In this case they are consistent. Differences are due to error in reading graph. (x) Comment on precision Cannot predict very accurately. 1/4-1/2 of the difference between line values can be used. Anything above that is not specific enough. Anything smaller and you are assuming too much precision.
CHEG 231, Fall, 2021 Homework Assignment #1 – Solutions & Rubrics Problem 4. Steam at 150 bar and 360°C undergoes J-T expansion. For the following questions, make sure to use the “Procedure” Procedure should be written out somewhere on page including steps: 1. Think 2. Declare system 3. Write balances 4. Simplify 5. Find missing properties 6. Solve 7. Think a)What is the specific enthalpy (in kJ/kg) of the inlet steam? (2975.5-2692.4)*(10/50)+2692.4= 2749.0kJ/kg b)What is the range of outlet temperatures for which condensation is possible? minimum 140-160 deg C, max of around 300 deg C. + or - 10 C c)What is the corresponding outlet pressure range? 0.3-0.5MPa for lower bound 8-10MPa for upper bound d)If the outlet steam is at 0.2 bar, what is its temperature? From Moiller diagram 125-135 deg C From steam tables, Double integrate to get 133 deg C e)What would be the change in enthalpy for the isobaric heating of the outlet steam of part d) back to 360°C? specific enthalpy = 3198 kJ/kg @ 360 deg C and 0.2 bar difference would be 3198-2749= 450kJ/kg