First Law of Thermodynamics: Closed Systems - Lecture Notes

CHEN 205: Chemical Engineering Thermodynamics I
Topic 2: The First Law of Thermodynamics – Closed Systems
In this chapter, you will learn what the First Law is and how to apply it to systems for engineering-type calculations.
After this chapter, you should be able to (from the text):
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State and apply the first law of thermodynamics, making use of the appropriate sign conventions
Explain and employ the concepts of internal energy, enthalpy, state function, equilibrium, and reversible process
Explain the differences between state functions and path-dependent quantities such as heat and work
Relate changes in the internal energy and enthalpy of a substance to changes in temperature, with calculations
based on the appropriate heat capacity
Formulate a heat-capacity integral, decide whether to use 𝐶! or 𝐶" in it, and evaluate it with the heat capacity
expressed as a polynomial in temperature
Use a heat-capacity integral in an energy balance to determine the energy input required to achieve a given temperature change or to determine the temperature change that will result from a given energy input
In the 1840s, James P. Joule conducted experiments by performing shaft work (stirring) on a fluid in an insulated
container. He noted upon performance of shaft work, the temperature of the fluid increased in a predictable way. He
also noted that the temperature would go back down when he allowed the fluid to contact a cold body.
In particular, his shaft-work experiment related a change in potential energy to shaft work:
If the energy added from work can later be removed by heat transfer to
a cold body, where does that energy reside in between?
And how is that related to the temperature rise?
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CHEN 205: Chemical Engineering Thermodynamics I
To put it simply, energy cannot be created or destroyed, only change form. More precisely:
In the integrated form (the way the book presents it):
What is a system?
How does energy get in and out of a system?
What are the surroundings?
With this in mind, the first law turns into an energy balance (a “master equation” like the mass balance)
In a closed system, we use the integrated form of the energy balance (assume no reaction):
By convention, heat and work are positive when adding energy to the system (so they appear as “addition”)
Note: “W” and “Q” can refer to both the
“total” heat and work (in units of J) or the
“specific” or “molar” heat or work (J/kg,
J/mol). It depends on the context.
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CHEN 205: Chemical Engineering Thermodynamics I
Constant pressure, no shaft work case (enthalpy):
Constant volume, no shaft work case (internal energy):
Example 2.1 from text:
The Niagara River, separating the US from Canada, flows from Lake Erie to Lake Ontario. These lakes differ in
elevation by about 100 m. Most of this drop occurs over Niagara Falls and in the rapids just above and below the falls,
creating a natural opportunity for hydroelectric power generation. The Robert Moses hydroelectric power plant draws
water from the river well above the falls and discharges it well below them. It has a peak capacity of 2.3 × 106 kW at
a maximum water flow of 3.1 × 106 kg/s. In the following, take 1 kg of water as the system.
a)
What is the potential energy of the water flowing out of Lake Erie, relative to the surface of Lake Ontario?
b) At peak capacity, what fraction of this potential energy is converted to electrical energy in the Robert Moses
power plant?
c)
If the temperature of the water is unchanged in the overall process, how much heat flows to or from it?
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CHEN 205: Chemical Engineering Thermodynamics I
Example 2.2 from text:
A typical industrial-scale wind turbine has a peak efficiency of about 0.44 for a wind speed of 9 m/s. That is, it converts
about 44% of the kinetic energy of the wind approaching it into usable electrical energy. The total air flow impinging
on such a turbine with a rotor diameter of 43 m is about 15,000 kg/s for the given wind speed.
a)
How much electrical energy is produced when 1 kg of air passes through the turbine?
b) What is the power output of the turbine?
c)
If there is no heat transferred to the air, and if its temperature remains unchanged, what is its change in speed
upon passing through the turbine?
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CHEN 205: Chemical Engineering Thermodynamics I
Systems that are in equilibrium are not changing. Typical types of equilibria:
Because systems at equilibrium are static, this means there are no driving forces for change. Almost all of the situations
we study in this class are at equilibrium. Furthermore, most of the thermodynamic variables we will be dealing with
are state functions. In other words, they do not depend on the past history of the system, only the current state.
Exercise 2.1: Different amount of work, same state
Consider the piston/cylinder set ups shown below. In both cases, the starting point is 𝑃# , 𝑉# and the final point is 𝑃$ , 𝑉$ .
Which of these two cases has more work? Which has the larger change in internal energy?
Case B: slowly push down harder and harder
Case A: suddenly drop a weight on top
𝑃# , 𝑉#
𝑃# , 𝑉#
𝑃$ , 𝑉$
𝑃$ , 𝑉$
Exercise 2.2: What gives?
How can the two Cases above both have the same change in internal energy? They start at the same point, but have
different amounts of work put in.
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CHEN 205: Chemical Engineering Thermodynamics I
Case A: suddenly drop a weight on top
𝑃# , 𝑉#
𝑃"#.'" , 𝑉"#.'"
𝑃$ , 𝑉$
Case B: slowly push down harder and harder
Note: if heat flows out “fast enough,” the intermediate state won’t be seen. In practice, though,
heat flow out of a container is slow compared to
the sudden compression.
𝑃# , 𝑉#
𝑃$ , 𝑉$
So the thermodynamic states are the same at the start and finish, but the two path-dependent “energy transfers” between the two Cases are quite different. We will see that Case A is an irreversible process, and Case B is a reversible
process. Both processes allow for heat flow out, so they are not adiabatic, they are diathermic.
Example 2.4 from the text
When a system is taken from state a to state b in the accompanying figure along path acb, 100 J of heat flows into the
system and the system does 40 J of work.
a)
How much heat flows into the system along path aeb if the work done by the
system is 20 J?
b) The system returns from b to a along path bda. If the work done on the system is
30 J, does the system absorb or liberate heat? How much?
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CHEN 205: Chemical Engineering Thermodynamics I
The idea of a reversible process is important in thermodynamics. Reversible processes represent idealizations, and
they often lead to optimal work or efficiency, and they are usually easier to calculate. In the real world, true reversible
processes do not occur often. They must occur very slowly and by only infinitesimal changes…that way, the internal
equilibrium is not upset, and the system is always in a state of equilibrium.
In general, reversible processes:
Example: a piston with small grains of sand (or small
weights) being slowly added (removed) one at a time
Vs.: a piston with a weight suddenly dropped on it (or
removed from it) all at once
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CHEN 205: Chemical Engineering Thermodynamics I
Note that in a reversible process with a movable
boundary, the external pressure (surroundings) is always the same as the internal pressure (system).
Therefore:
Enthalpy
The closed system energy balance (with no shaft
work) can be written as:
If this is constant pressure:
This grouping arises so often in thermodynamics (and especially because constant pressure systems are common), that
H. Kamerlingh Onnes called it “enthalpy”.
E-bal for closed, constant volume systems:
E-bal for closed, constant pressure systems:
Δ𝑈 = 𝑄
Δ𝐻 = 𝑄
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CHEN 205: Chemical Engineering Thermodynamics I
Example 2.6 from the text:
Calculate Δ𝑈 and Δ𝐻 for 1 kg of water when it is vaporized at the constant temperature of 100oC and the constant
pressure of 101.33 kPa. The specific volumes of water (ℓ) and (v) at these conditions are 0.00104 and 1.673 m3/kg,
respectively. For this change, heat in the amount of 2256.9 kJ is added to the water.
Exercise:
Poll: which is valid?
a)
𝑈(𝑇, 𝑉)
b) 𝑈(𝑇, 𝑃)
c)
d) All are valid
e)
None are valid
𝑈(𝑃, 𝑉)
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CHEN 205: Chemical Engineering Thermodynamics I
Thus, according to the “total differential” seen in Calculus III:
Because both internal energy and enthalpy are strong functions of temperature, it makes sense to focus on their partials
with respect to temperature (ie, how much 𝑈, 𝐻 increase when the temperature increases):
Constant volume heat capacity:
Constant pressure heat capacity:
Furthermore, in most cases, the second term can be effectively ignored:
For large changes in volume or
pressure, you cannot ignore the second term for a real gas (Topic 06)
So if volume is (nearly) constant, or if internal energy is (nearly) independent of volume (IG, solids, liquids), then:
The heat that goes into raising the temperature of a substance is called “sensible heat” (because you can feel it)
Temperature Dependence of Heat Capacity
If we no longer assume that 𝐶" , 𝐶! are constant wrt T, then how do they vary wrt T?
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CHEN 205: Chemical Engineering Thermodynamics I
Heat Capacity in Ideal Gas State
The heat capacities of ideal gases can be determined from kinetic theories:
Example 4.2 from text
Calculate the heat required to raise the temperature of 1 mol of methane from 260 to 600°C in a steady-flow process
at a pressure sufficiently low that the ideal-gas state is a suitable approximation for methane.
You don’t have to do it in Excel, but there will come a day where you must do this in reverse: for a given 𝑄 and
starting temperature 𝑇# , find 𝑇$ . Then, Solver is your friend.
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CHEN 205: Chemical Engineering Thermodynamics I
Ideal gases form the foundation of a lot of our thermodynamic calculations. They have very convenient properties:
Let’s delve more deeply into the thermodynamic behavior of ideal gases.
Motivation: Consider a real gas (i.e., a non-ideal gas) undergoing a process from (𝑇# , 𝑃# ) to (𝑇$ , 𝑃$ ). How to calculate
property changes such as Δ𝑈, Δ𝑆, etc?
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CHEN 205: Chemical Engineering Thermodynamics I
Exercise 2.3: Reversible, adiabatic compression of an ideal gas
Consider an insulated, frictionless piston containing an ideal gas. The gas is then compressed reversibly by doubling
the pressure from 𝑃# to 𝑃$ bar. Which of the following is true?
𝑃𝑉 = 𝑅𝑇
A. Volume decreases, but temperature stays the same
B. Volume decreases, temperature increases
C. Volume decreases, temperature decreases
𝑃# , 𝑉# , 𝑇#
D. Volume stays the same, temperature increases
𝑃$ , 𝑉$ , 𝑇$
Process calculations for the IG state
Here we have all three 𝑃𝑉𝑇 variables, but only two are independent. Can use 𝑃𝑉 () = 𝑅𝑇 to eliminate one.
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Eliminate 𝑃 (if you don’t know 𝛥𝑃):
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Eliminate 𝑉 () (if you don’t know 𝛥𝑉):
Consider the four typical paths (which these equations could apply/be simplified: Isochoric, Isobaric, Isothermal, and
Adiabatic):
1.
Isochoric (const V) process:
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CHEN 205: Chemical Engineering Thermodynamics I
2.
Isobaric (const P) process:
3.
Isothermal (const T) process:
4.
Adiabatic processes (Q = 0):
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For an ideal gas, reversible
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For a closed system, with constant heat capacity:
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CHEN 205: Chemical Engineering Thermodynamics I
Note: 𝐶! = 𝑅 ⁄(𝛾 − 1), so you may see
that form substituted in. For example,
the text presents it this way.
Also (for below), 𝐶" = 𝑅 𝛾⁄(𝛾 − 1).
Example 3.3 from the text:
Air is compressed from an initial state of 1 bar and 298.15 K to a final state of 3 bar and 298.15 K by three different
mechanically reversible processes in a closed system:
a)
Heating at constant volume followed by cooling at constant pressure
b) Isothermal compression
c)
Adiabatic compression followed by cooling at constant volume
These processes are shown in the figure. We assume air to be in its ideal-gas state and the heat capacities are constant,
𝐶"() = 20.785 and 𝐶!() = 29.100 J/(mol K). Calculate the work required, heat transferred, and the changes in internal
energy and enthalpy of the air for each process.
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CHEN 205: Chemical Engineering Thermodynamics I
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CHEN 205: Chemical Engineering Thermodynamics I
Example 3.4 from the text:
A gas in its ideal-gas state undergoes the following sequence of mechanically reversible processes in a closed system:
a)
From an initial state of 70°C and 1 bar, it is compressed adiabatically to 150°C.
b) It is then cooled from 150 to 70°C at constant pressure.
c)
Finally, it expands isothermally to its original state.
Calculate 𝑊, 𝑄, 𝛥𝑈 () , and 𝛥𝐻() for each of the three processes and for
the entire cycle. Take 𝐶*() = 12.471 and 𝐶+() = 20.785 J/(mol K).
For the entire cycle, add each component up. Of course, Δ𝑈 () = Δ𝐻() = 0 for the cycle. (Why?)
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