Modern Engineering Thermodynamics

5.2 Sealed, Rigid Containers 149
In this case, nothing in the problem statement leads us to believe that the vessel undergoes any change in specific
kinetic or potential energy during the heating process, so we assume that ke 2 =ke 1 and pe 2 =pe 1 . Also, since the container
is rigid, V = constant and dV = 0, the mechanical moving boundary work is zero (i.e., R pdV ¼ 0). Since no
other work modes are suggested in the problem statement, we assume that 1 W 2 ¼ 0. Applying these results to the previous
general energy balance yields the following simplified energy balance equation as the governing equation for
this problem:
1Q 2 = mðu 2 − u 1 Þ
Write any relevant auxiliary equations. Since we now know the values of two independent properties in each state
(T 1 , v 1 , and T 2 , v 2 ), we can find the values of any other properties in those states by use of thermodynamic tables,
charts, or equations of state. In particular, the qualities can be determined from the saturation tables and the auxiliary
equations
x 1 = v 1− v f 1
v fg1
= v 1 − v f ð20:0°CÞ
v fg ð20:0°CÞ
and
x 2 = v 2 − v f 2
v fg2
= v 2 − v f ð95:0°CÞ
v fg ð95:0°CÞ
Since we are not given enough information to use the conduction, convection, or radiation heat transfer equations,
we must find 1 Q 2 from this energy balance. The values of u 1 and u 2 can be found by using the saturation
tables and the following auxiliary equations:
u 1 = u f 1 + x 1 u fg1 = u f ð20:0°CÞ + x 1 u fg ð20:0°CÞ
and
u 2 = u f 2 + x 2 u fg2 = u f ð95:0°CÞ + x 2 u fg ð95:0°CÞ
Step 6. Algebraically solve for the unknown(s). At this point, we have algebraic equations for all the unknowns and we
know where all the numbers in these equations are to be found.
Step 7. Calculate the value(s) of the unknowns. We are now ready to make the calculations. From Table C.1b of
Thermodynamic Tables to accompany Modern Engineering Thermodynamics, we find that
a. At 20.0°C, v f 1 = 0:001002 m 3 /kg, v g1 = 57:79 m 3 /kg, and v fg1 = v g1 − v f 1 = 57:789 m 3 /kg. Also, u f 1 = 83:9 kJ/kg,
u g1 = 2402:9 kJ/kg, and u fg 1 = u g1 − u f 1 = 2319:0 kJ/kg.
b. At 95.0°C, v f 2 = 0:00104 m 3 /kg, v g2 = 1:982 m 3 /kg, and v fg2 = v g2 − v f 2 = 1:981 m 3 /kg. Also, u f 2 = 397:9 kJ/kg,
u g2 = 2500:6 kJ/kg, and u fg2 = u g2 − u f 2 = 2102:7 kJ/kg.
So the unknowns can now be determined as
a. x 1 = 0:500 − 0:001002 = 8:63 × 10
57:789
− 3 = 0:863%.
b. x 2 = 0:500 − 0:00104 = 0:252 = 25:2%.
1:981
c. u 1 = 83:9 + ð8:63 × 10 − 3 Þð2319:0Þ = 103:9 kJ/kgÞ and u 2 = 397:9 + ð0:252Þð2102:7Þ = 927:8 kJ/kg, so that
1Q 2 = mðu 2 − u 1 Þ = ð2:0kgÞð927:8 − 103:9 kJ/kgÞ = 1650 kJ.
Step 8. A check of the algebra, calculations, and units shows that they are correct.
1 We want three significant figures in the answer, so the data need to be given to three significant figures.
DO THERMO COMPUTER PROGRAMS GIVE THESE SAME ANSWERS?
Suppose you use a computer program like EES or CATT2 or NIST to solve this problem. Would the answer be the same?
Using EES you get (a) x 1 = 0:864%, (b)x 2 = 25:2%, and(c) 1 Q 2 = 1650 kJ; using CATT2 you get (a) x 1 = 0:864%,
(b) x 2 = 25:2%, and (c) 1 Q 2 = 1650 kJ; and using NIST you get (a) x 1 = 0:864%, (b) x 2 = 25:2%, and (c) 1 Q 2 = 1650 kJ (all
to three significant figures). So the source of the thermodynamic properties has no significant effect on the answer.