Thermodynamics by Van Ness Flashcards
A reversible process:
∙ Can be reversed at any point by an infinitesimal change in external conditions
∙ Is never more than minutely removed from equilibrium
∙ Traverses a succession of equilibrium states Is frictionless
∙ Is driven by forces whose imbalance is infinitesimal in magnitude
∙ Proceeds infinitely slowly
∙ When reversed, retraces its path, restoring the initial state of system and surroundings
Example 2.5
A horizontal piston/cylinder arrangement is placed in a constant-temperature bath.
The piston slides in the cylinder with negligible friction, and an external force holds it
in place against an initial gas pressure of 14 bar. The initial gas volume is 0.03 m^3. The
external force on the piston is reduced gradually, and the gas expands isothermally
as its volume doubles. If the volume of the gas is related to its pressure so that PV^t
is constant, what is the work done by the gas in moving the external force?
the reversible work W= −29,112 J
P2 = 700,000 Pa or 7 bar
If efficiency of such processes known to be about 80%, we could
multiply the reversible work by this figure to get an estimate of the irreversible
work, namely −23,290 J.
Calculate ΔU and ΔH for 1 kg of water when it is vaporized at the constant tempera-
ture of 100°C and the constant pressure of 101.33 kPa. The specific volumes of liquid
and vapor water at these conditions are 0.00104 and 1.673 m^3·kg^−1, respectively. For
this change, heat in the amount of 2256.9 kJ is added to the water.
ΔH = Q = 2256.9 kJ
ΔU = 2087.5 kJ
Example 2.7
Air at 1 bar and 298.15 K is compressed to 3 bar and 298.15 K by two different
closed-system mechanically reversible processes:
(a) Cooling at constant pressure followed by heating at constant volume.
(b) Heating at constant volume followed by cooling at constant pressure.
Calculate the heat and work requirements and ΔU and ΔH of the air for each path. The
following heat capacities for air may be assumed independent of temperature:
CV = 20.785 and CP = 29.100 J·mol−1·K−1
Assume also that air remains a gas for which PV/T is a constant, regardless of the changes
it undergoes. At 298.15 K and 1 bar the molar volume of air is 0.02479 m3· mol−1.
For the overall process:
Q = −5784 + 4131 = −1653 J
W = 1653 + 0 = 1653 J
ΔU = −4131 + 4131 = 0
ΔH = −5784 + 5784 = 0
For the two steps combined,
Q = 12,394 − 17,352 = −4958 J
W = 0 + 4958 = 4958 J
ΔU = 12,394 − 12,394 = 0
ΔH = 17,352 − 17,352 = 0
EXAMPLE 2.8
Calculate the internal energy and enthalpy changes resulting if air changes from an initial state of 5°C and 10 bar, where its molar volume is 2.312 × 10−3 m3·mol−1, to a final
state of 60°C and 1 bar. Assume also that air remains a gas for which PV/T is constant
and that CV = 20.785 and CP = 29.100 J·mol−1·K−1.
ΔU = 1144.0 J
ΔH = 1601.2 J
Open systems are characterized by flowing streams; there are four common measures of flow:
∙ Mass flow rate, m ̇
∙ Molar flow rate, n ̇
∙ Volumetric flow rate, q ∙
Velocity, u
The area for flow A is the_____ of a conduit, and ρ is specific or molar
density.
cross-sectional area
EXAMPLE 2.9
In a major human artery with an internal diameter of 5 mm, the flow of blood, averaged
over the cardiac cycle, is 5 cm^3·s^−1. The artery bifurcates (splits) into two identical
blood vessels that are each 3 mm in diameter. What are the average velocity and
the mass flow rate upstream and downstream of the bifurcation? The density of blood
is 1.06 g·cm^−3.
Average velocity = 25.5 cm⋅s^−1
m ̇up = 5.30 g⋅s−1
for each downstream vessel:
m ̇down = 2.65 g⋅s−1
The region of space identified for analysis of open systems is called a control volume; it is
separated from its surroundings by a__
control surface.
The term “steady state” does not necessarily imply that flow rates are constant, merely
that the inflow of mass is exactly matched by the outflow of mass.
When there is a single entrance and a single exit stream, the mass flow rate m ̇ is the
same for both streams; then,
m ̇ = const = ρ2 u2A2 = ρ1 u1A1
Example 2.11
An insulated, electrically heated tank for hot water contains 190 kg of liquid water
at 60°C. Imagine you are taking a shower using water from this tank when a power
outage occurs. If water is withdrawn from the tank at a steady rate of m= 0.2 kg·s^−1, how long will it take for the temperature of the water in the tank to drop from 60 to
35°C? Assume that cold water enters the tank at 10°C and that heat losses from the
tank are negligible. Here, an excellent assumption for liquid water is that Cv = Cp = C,
independent of T and P.
t = 658.5 s or 11 mins
