Download Chapter 3. Energy and the First Law

Chapter 3.
Energy and the First Law
Soong Ho Um
Sungkyunkwan University
Chemical Engineering
Introduction
• Thermodynamic systems are characterized by an additional
state variable: temperature.
• Temperature is a measure of the energy stored inside matter
in various forms, which we collectively call internal energy. It
gives rise to another type of energy exchange that is not
encountered among purely mechanical systems, heat.
• The incorporation of heat effects into the energy balance
constitutes one of the fundamental principles of
thermodynamics known as the first law.
Instructional Objectives
• Formulate the mathematical statement of the first law for
a closed system and learn how to:
1. Do energy balances in closed system.
2. Distinguish between path and state functions.
3. Use the steam tables to calculate internal energy and enthalpy.
4. Apply the energy balance to systems undergoing vaporization
or condensation.
5. Perform calculations of internal energy and enthalpy in the
ideal-gas state.
Energy and Mechanical Work
• Heat, work and energy are measured in the same units but
they represent different physical entities.
• Energy is the ability of a system to produce work, namely,
ability to cause the displacement of a force. It is a property of
the state, storable quantity.
• Example: an object with mass m resting on the floor. Potential
energy -> Kinetic energy; if no change on the location at
certain time, the energy is preserved until relocated.
• Work takes place when a force is displaced. It is exchanged
during a process and characterizes, not the state of the
system, but the transition of the system between states.
Energy -- Mechanical work -- Energy --
A system initially in equilibrium state A undergoes a process that brings it to
final state B; during this process the system exchanges work with the surroundings
so that the energy change of the system is equal to the amount of work exchanged:
∆EAB = W
Work represents energy that passes from one system into another.
Energy vs. Work
• Energy is storable; it remains in the system for as long as the state
of the system is preserved. It is a state function.
• Work is energy in transit; it appears when energy is passed from
one system into another.
• Work is not a storable quantity as such. Work that enters a system
must be stored in some form of energy.
• Work is associated with a direction from one system in which it
originates to another, where it is transferred to.
• Sign convention for work: work is positive if it enters the system,
negative if it exits.
Shaft work and PV work
• Shaft work ≈ mechanical work
• Another more subtle form of work is associated with
movement of the system boundaries, PV work. Can you
discriminate these (shaft work vs. PV work)?
Pex = P + δP (External pressure =
System pressure + increment; It
is a positive increment for
compression but it is a negative
increment for expansion.)
(Mechancially reversible process)
If the process is conducted in
a quasi-static manner, then
δP - 0 and Pex - P.
For a process that moves
the direction of decreasing
(compression), the area
negative and the work
positive.
△?
in
V
is
is
Importantly,
PV
work
depends on the entire path
that connects the two states.
PV work is a path function
whose value depends not
only on the initial and final
states, but on the entire path.
Example 3.1: PV work in Expansion
A cylinder fitted with a piston contains 1 liters of gas. The
piston has a 1-in diameter and weighs 5 kg. How much
work is needed to expand the gas reversibly to twice its
volume against the pressure of the atmosphere (1.013
bar)?
Hint: the gas expands against the combined pressure of
the piston and the atmosphere.
The weight of the piston is Mg.
The area of the piston is πD2/4.
The pressure(Pꞌ) exerted by the piston is Mg/(πD2/4).
The total pressure(P) in the cylinder is Pꞌ + P0, which remains constant during the process.
Example 3.2: PV work using the SoaveRedlich-Kwong equation of state
Ethylene is compressed reversibly in a closed system. The
compression is conducted isothermally at 350 K, from initial
pressure 20 bar to final pressure 55 bar. Calculate the work
using the SRK equation of state.
Hint: estimate V points at the corresponding P points on the
PV diagram.
Internal Energy and Heat
• Molecules possess energy of various forms. It includes kinetic
energy due to the motion of the center of mass in space
(translational kinetic energy). In case of polyatomic molecules
possess rotational kinetic energy, vibrational energy, potential
energy (as a result of interaction between different molecules
as well as between atoms of the same molecule).
• Matter, regardless of chemical composition or phase, is
capable of storing energy internally. The combined storage
modes as internal energy (U) is a state function and for a pure
substance, it is a function of pressure and temperature.
• The molecular nature of matter gives rise to a different type of
energy transfer, heat.
Heat shares some important characteristics with work
1.
It is a transient form of energy that is observed during a change of state
(process); once the system is in equilibrium with its surroundings there is
no net heat transfer because both system and surroundings are at the
same temperature.
2.
As a transient form, heat is not a storable mode of energy: once it enters a
system, it is stored as internal energy. It is incorrect to say that a system
“contains heat,” or to speak of energy that “is converted into heat.”
3.
It has a direction, from the system to the surroundings, or vice versa.
4.
It is a path function whose value is determined by the entire path of the
process. This property of heat is not obvious at the moment but will
become so in the next section.
5.
Heat is positive if it is transferred to the system from the surroundings.
First Law for a Closed System
• All material systems possess internal energy. They may
possess various other forms of energy as well. Changes
of a state invariably require the exchange of energy
between the system and the surroundings.
• The exchanges of energy in the state transition must be
in the form of work (PV, shaft, or both) and/or heat: ∆EAB
= Q + W. Energy balance should be expressed as, ∆E (U
+ Ek + Ep) = Q + W. The contribution of Ek and Ep is
negligible. Therefore, ∆U = Q + W (operator ∆ is a
shorthand notation for a change of a property between
two states).
Example 3.4: Analysis of Joule’s Experiment
In this example we consider a variation of Joule’s experiment: a
thermally insulated vessel contains 10 kg of water. The liquid is
stirred by an impeller driven by a 1 kW motor. If the motor runs
for 1 min and all of the work it produces is transferred to the
liquid, analyze the experiment of the basis of the first law and
report the relevant amounts of heat, work, and internal energy.
Ans (hint):
∆U = W
Let’s consider elementary paths in different states
Constant-Volume Heating
• Let’s think of the process in which a closed system is
heated under constant volume.
• No PV work with the fixed volume of the system as well
as no shaft work.
• The first law gives, Q = ∆U (constant volume process).
• The amount of heat that is exchanged under constant
volume is equal to the change of internal energy,
provided that no shaft work is present.
Example 3.5: Constant-volume cooling?
A sealed metal cylinder contains steam at 1 bar, 500 C. How
much heat must be removed at constant volume in order to
produce saturated vapor?
Ans (hint):
∆U = W
Steam table
Constant-Pressure Heating
Example 3.6: Constant-pressure cooling of steam?
8.5 kg of steam at 600 C, 15 bar, are cooled at constant
pressure by removing 6200 kj of heat. Determine the final
temperature and the amount of PV work that is exchanged with
the surroundings.
Ans (hint):
∆U = W
Steam table
Constant-Temperature Process
An isothermal process is one
conducted
at
constant
temperature. Experimentally,
it is conducted by placing the
system inside a heat bath.
Its path is represented by an
isotherm.
Example 3.7: Isothermal compression of steam?
Steam is compressed isothermally at 350 C in a closed system.
Compression is conducted reversibly from initial pressure 20 bar
to final pressure 40 bar. Determine the amounts of work and
heat exchanged between the system and its surroundings.
Ans (hint):
∆UAB = Q + W
Steam table
Sensible Heat-Heat Capacities
• In the absence of phase transitions, the exchange of heat is
accompanied with a temperature change. Such heat is called
sensible because it can be sensed with a thermometer.
• The amount of heat needed to produce a given temperature
change varies from one substance to another and is quantified by
the heat capacity.
• The amount of heat depends on the path of the heating process:
constant-volume and constant-pressure path.
Constant-Volume Heat Capacity
Even though the definition of the heat capacity was
motivated by heat, the heat capacity is a partial derivative
of a state function (internal energy), and itself a state
function.
Constant-Pressure Heat Capacity
Both Cp and Cv has dimensions of energy per mass per
temperature. The term specific heat is sometimes used
when the heat capacity is reported on a per-mass basis.
Why are Cv and Cp different?
• Mathematical view: they are different due to different
partial derivatives
• Thermodynamic view: they are different as associated
with heating along different paths
• Physical argument view: In constant-volume heating the
amount Qv is used to increase temperature by ∆T. In
constant-pressure heating the amount Qp is used to
increase temperature by the same amount and to
expand the volume against the surroundings.
Qp > Qv -- Cp > Cv
Example 3.8: Heat capacities and heat?
According to the relation, C = Q/T, heat capacities are defined in
terms of heat. Does this make Cv and Cp path functions?
Ans (hint):
No. Why?
Practical Utility of the Heat Capacities
1. They allow to calculate heat in constant-volume and
constant-pressure processes. It is useful in energy
balance.
2. They allow to calculate changes in internal energy and
enthalpy. It allow to calculate these properties using
equations rather than tables or to obtain their values in
states that are not found in tables.
Effect of Pressure and
Temperature on Heat Capacity
• As state functions, Cv and Cp depend on pressure and
temperature.
• Heat capacity is a strong function of pressure in the
vapor phase but almost independent of pressure in the
liquid.
• Cp is fairly sensitive on temperature in both liquid and
gas phase.
Ideal-Gas Heat Capacity
• Ideal-gas heat capacities have been compiled for a large
number of pure components and the data are usually
presented in the form of empirical correlations that give
Cpig as a function of temperature.
Heat of Vaporization
Lever rule for enthalpy:
H = xLHL + xVHV
xL + xV = 1
H = HV - xL∆Hvap
= HL + xV∆Hvap
Pitzer correlation for ∆Hvap
Example 3.9: Energy balances with phase change:
using tables
Steam at 1.013 bar, 200 C is cooled under constant pressure by
removing an amount of heat equal to 1000 kJ/kg. Determine the
final state.
Ans (hint):
H=Q
Ideal-Gas State
In the ideal-gas state, intermolecular interactions are unimportant
because distances between molecules are large, beyond the range
such interactions. If the volume of the system is increased at constant
temperature, there should be no change in internal energy: since
temperature remains constant, the kinetic energy of the molecules
(including translational, rotational, vibrational, etc.) is the same before
and after the expansion; and since there is no interaction between
molecules, there no other mode of energy storage available to the
system. Therefore, internal energy remains constant as long as
temperature is not changed. In other words, the internal energy in the
ideal-gas state is a function of temperature only:
Uig = Uig (T), Hig = Hig(T)
Reversible adiabatic process
Example 3.10: Heat Transfer
A 10 kg piece of hot copper at 450 C is quenched in an open tub
that contains 10 kg of water at 20 C. Calculate the final
temperature assuming no heat losses to the environment. The
heat capacities of copper and water are Cpc = 0.38 kJ/kg K, Cpw
= 4.184 kJ/ kg K, and may be assumed independent of
temperature.
Ans (hint):
Q = 0 = ∆H tot
Energy Balances and Irreversible Processes
•
As a statement of energy conservation, the first law is
applicable to any process, whether it is reversible or not.
• If the process is irreversible and involves work, one must
be careful because dWrev = -PdV is not applicable.
• If the process does not involve work, the calculation is
done in the usual way.
• Implicit in all of these calculations is the assumption that
the system is internally uniform so that its state can be
described by a uniform pressure and temperature.
Example 3.11: Irreversible expansion against
vacuum
A closed insulated cylinder is divided into parts by a piston that is
held into place by latches. One compartment is evacuated, the
other contains steam at 7.5 bar, 300 C. The latches are removed
and the gas expands to fill the entire volume of the cylinder.
When the system reaches equilibrium, its pressure is 1 bar.
Determine the final temperature.
Ans (hint):
0 = ∆U12
Example 3.12: Irreversible expansion against
pressure
An insulated cylinder fitted with a weight less latched piston
contains steam at 5 bar, 300 C. The piston is unlatched and the
steam expands against ambient air at P =1 bar. The temperature
of steam in the final state is 142 C. Determine the energy
balance for this process.
Ans (hint):
W = ∆UAB
In this problem we were able to calculate the work from the
energy balance because ∆U and Q were both known.
Initial state
Final state
P(bar)
5
1
T(C)
300
142
U(kJ / kg)
2803.3
2570.6