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