CHEG231-F22 Homework Solution 2

.pdf

School

University of Delaware *

*We aren’t endorsed by this school

Course

231

Subject

English

Date

Jan 9, 2024

Type

pdf

Pages

5

Uploaded by MateButterfly4469

Report
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.
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help